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@@ -0,0 +1,8659 @@
+[
+    {
+        "id":  "1",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面各数，只读一个零的是( )。",
+        "options":  [
+                        "A. 105000",
+                        "B. 1056000",
+                        "C. 10010100"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据整数的读法：从高位到低位，一级一级地读;每级末尾的零不读，每级中有一个零或连续几个零时，只读一个零。由此逐项读出，判断是否符合\"只读一个零\"的要求.\n【详解】A。105000，读作十万五千，不读零，不符合题意;B。1056000，读作一百零五万六千，读出一个零，符合题意;C。10010100，读作一千零一万零一百，读出两个零，不符合题意。故答案为:B",
+        "knowledge_points":  [
+                                 "万以内数的读，写法"
+                             ],
+        "source":  "25-26四年级上·甘肃临夏·期末"
+    },
+    {
+        "id":  "2",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面各数，只读一个零的是( )。",
+        "options":  [
+                        "A. 56000230",
+                        "B. 2608000",
+                        "C. 5020809"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】根据整数的读法规则，从高位到低位一级一级地读，每一级末尾的0都不读出来，其他数位有一个0或连续几个0都只读一个\"零\"。需要逐一读出每个选项的数，判断读出的零的个数.\n【详解】A。56000230 读作：五千六百万零二百三十(读出一个零)。B。2608000 读作：二百六十万八千(没有读出零)。C。5020809 读作：五百零二万零八百零九(读出三个零)。故答案为:A",
+        "knowledge_points":  [
+                                 "亿以内数的读，写法"
+                             ],
+        "source":  "25-26四年级上·辽宁葫芦岛·期末"
+    },
+    {
+        "id":  "3",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.94,
+        "question":  "用量角器量角时，角的顶点应与( )重合。",
+        "options":  [
+                        "A. 量角器的边缘",
+                        "B. 量角器的中心",
+                        "C. 0 刻度线"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】量角的步骤：先把量角器的中心与角的顶点重合,0°刻度线与角的一条边重合。再看角的另一边所对的量角器上的刻度，就是这个角的度数.\n【详解】用量角器量角时，角的顶点应与量角器的中心重合。故答案为:B",
+        "knowledge_points":  [
+                                 "角的度量"
+                             ],
+        "source":  "25-26四年级上·黑龙江齐齐哈尔·期末"
+    },
+    {
+        "id":  "4",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "两个相乘的因数同时扩大到原来的10倍，积扩大( )倍。",
+        "options":  [
+                        "A. 10",
+                        "B. 20",
+                        "C. 100"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】根据积的变化规律，两个相乘的因数同时扩大到原来的10倍，积扩大到原来的100倍.\n【详解】两个相乘的因数同时扩大到原来的10倍，积扩大100倍。故答案为:C",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "5",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.94,
+        "question":  "620×50 的积的末尾有( )个0。",
+        "options":  [
+                        "A. 2",
+                        "B. 3",
+                        "C. 4"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】计算620×50时，可先算62×5，再在积末尾添上两个0.\n【详解】62×5=310，再添2个0得31000，积末尾共有3个0。故答案为:B",
+        "knowledge_points":  [
+                                 "三位数乘两位数，乘数末尾有0"
+                             ],
+        "source":  "24-25 四年级上·西藏林芝·期末"
+    },
+    {
+        "id":  "6",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.94,
+        "question":  "用一个放大10倍的放大镜看一个5°的角，看到的角的度数是( )。",
+        "options":  [
+                        "A. 5°",
+                        "B. 15°",
+                        "C. 50°"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】角的大小只与两边张开的大小有关，与边长无关。放大镜只改变图形大小，不改变角度.\n【详解】看到的角仍是5°。故答案为:A",
+        "knowledge_points":  [
+                                 "角的度量"
+                             ],
+        "source":  "25-26四年级上·安徽合肥·期中"
+    },
+    {
+        "id":  "7",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.94,
+        "question":  "比较1000000和999999的大小:( )。",
+        "options":  [
+                        "A. 1000000大",
+                        "B. 999999大",
+                        "C. 一样大"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】比较整数大小先比位数，位数多的数大.\n【详解】1000000是七位数,999999是六位数，所以1000000大。故答案为:A",
+        "knowledge_points":  [
+                                 "大数的比较"
+                             ],
+        "source":  "24-25 四年级上·重庆丰都·期末"
+    },
+    {
+        "id":  "8",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "《三国演义》(2019年第四版)字数四舍五入到千位约是717千字。根据这个数据，有四位同学发表了看法，下面说法正确的是( )。",
+        "options":  [
+                        "A. 佳佳说：\"《三国演义》的字数最多是717999字.\"",
+                        "B. 笑笑说：\"《三国演义》的字数最少是717499字.\"",
+                        "C. 思思说：\"如果四舍五入到万位,《三国演义》的字数约是71万.\"",
+                        "D. 妙妙说：\"如果省略十万后面的尾数,《三国演义》的字数约是七十万.\""
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】四舍五入到千位约为717千字，原数范围应在716500到717499之间.\n【详解】A把最大值写错;B把最小值写错;C四舍五入到万位约为72万;D省略十万位后尾数约为70万，正确。故答案为:D",
+        "knowledge_points":  [
+                                 "亿以内数的近似数"
+                             ],
+        "source":  "24-25 四年级上·重庆秀山·期末"
+    },
+    {
+        "id":  "9",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "《长城谣》歌词中\"四万万同胞心一样\"，下面关于\"四万万\"理解错误的是( )。",
+        "options":  [
+                        "A. \"四万万\"是4亿",
+                        "B. \"四万万\"是4个十万",
+                        "C. \"四万万\"是一个近似数",
+                        "D. 从0开始，\"千万，千万\"地数，要数40次"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】\"四万万\"即4亿=400000000;4个十万是400000，不相等.\n【详解】A正确,C正确,D正确，错误的是B。故答案为:B",
+        "knowledge_points":  [
+                                 "亿以上数的组成",
+                                 "整数的数级，数位和计数单位的认识"
+                             ],
+        "source":  "25-26 四年级上·河南南阳·期中"
+    },
+    {
+        "id":  "10",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面各数中，读出两个0的是( )。",
+        "options":  [
+                        "A. 404000",
+                        "B. 400404",
+                        "C. 400400",
+                        "D. 404004"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】按整数读法逐个读出，统计\"零\"的个数.\n【详解】A不读零;B读作\"四十万零四百零四\"，读两个零;C读一个零;D读一个零。故答案为:B",
+        "knowledge_points":  [
+                                 "亿以内数的读，写法"
+                             ],
+        "source":  "25-26 四年级上·新疆和田·期末"
+    },
+    {
+        "id":  "11",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面4个数中最接近1亿的数是( )。",
+        "options":  [
+                        "A. 1.1亿",
+                        "B. 0.99亿",
+                        "C. 1.01亿",
+                        "D. 9999万"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】比较各数与1亿的差，差最小者最接近.\n【详解】A差0.1亿;B差0.01亿;C差0.01亿;D=0.9999亿，差0.0001亿，最小。故答案为:D",
+        "knowledge_points":  [
+                                 "大数的比较"
+                             ],
+        "source":  "2025·湖南邵阳·小升初模拟"
+    },
+    {
+        "id":  "12",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "A/B=5。2，如果A,B都乘10，那么商( ),余数( )。",
+        "options":  [
+                        "A. 乘10;除以10",
+                        "B. 除以10;不变",
+                        "C. 不变;乘10"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】被除数和除数同时乘同一个数(0除外),商不变，余数也乘这个数.\n【详解】A,B都乘10，商不变，余数乘10。故答案为:C",
+        "knowledge_points":  [
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "25-26四年级上·甘肃定西·期末"
+    },
+    {
+        "id":  "13",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "甲数/乙数=8。5，如果把甲，乙两数都扩大到原来的10倍，那么商和余数各是( )。",
+        "options":  [
+                        "A. 8,5",
+                        "B. 80,5",
+                        "C. 8,50",
+                        "D. 80,50"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】被除数和除数同时乘同一个数，商不变，余数也乘这个数.\n【详解】商仍是8，余数变为50。故答案为:C",
+        "knowledge_points":  [
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "25-26 四年级上·辽宁鞍山·期末"
+    },
+    {
+        "id":  "14",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "A/B=5。2，如果A,B都扩大到原来的10倍，那么商( ),余数( )。",
+        "options":  [
+                        "A. 扩大到原来的10倍，不变",
+                        "B. 缩小到原来的1÷10，扩大到原来的10倍",
+                        "C. 扩大到原来的2倍，不变",
+                        "D. 不变，扩大到原来的10倍"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】依据商不变规律：被除数，除数同时扩大相同倍数，商不变;余数也扩大相同倍数.\n【详解】商不变，余数扩大到原来的10倍。故答案为:D",
+        "knowledge_points":  [
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "15",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "我们在学习条形统计图时，经历了怎样的学习过程。( )",
+        "options":  [
+                        "A. 收集数据--整理数据--分析数据--作出决策",
+                        "B. 整理数据--收集数据--分析数据--作出决策",
+                        "C. 收集数据--分析数据--整理数据--作出决策",
+                        "D. 作出决策--整理数据--收集数据--分析数据"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】学习统计的一般流程是先收集，再整理，再分析，最后决策.\n【详解】正确过程为\"收集数据→整理数据→分析数据→作出决策\"。故答案为:A",
+        "knowledge_points":  [
+                                 "数据的搜集与整理"
+                             ],
+        "source":  "25-26 四年级上·福建龙岩·月考"
+    },
+    {
+        "id":  "16",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "小明每分钟走64米，小刚走630米用9分钟，小红每分钟比小刚少走4米，三人中( )的速度最快。",
+        "options":  [
+                        "A. 小刚",
+                        "B. 小明",
+                        "C. 小红"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】先求小刚速度，再求小红速度，与小明比较.\n【详解】小刚:630÷9=70(米/分);小红:70-4=66(米/分);小明64(米/分)。70\u003e66\u003e64，最快是小刚。故答案为:A",
+        "knowledge_points":  [
+                                 "几百几十或几千几百除以一位数的除法",
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26 四年级上·河北唐山·期末"
+    },
+    {
+        "id":  "17",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "计算140÷35时，可以这样想:140÷35=(140×2)÷(35×2)=280÷70=4，这种算法的依据是( )。",
+        "options":  [
+                        "A. 除法与乘法的关系",
+                        "B. 积不变的规律",
+                        "C. 商不变的规律"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】被除数和除数同时乘同一个非0数，商不变.\n【详解】式中把被除数和除数都乘2，商保持不变，依据是商不变规律。故答案为:C",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)",
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "25-26 四年级上·河北唐山·期末"
+    },
+    {
+        "id":  "18",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面说法正确的是( )。",
+        "options":  [
+                        "A. 12时30分，钟面上时针与分针形成的角是平角。",
+                        "B. 把圆平均分成360份，每一份所对的角的大小是1°。",
+                        "C. 用一个3倍的放大镜去看一个50°的角，看到的角是150°。"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】平角是180°;钟面12:30时夹角不是180°;放大镜不改变角度大小;圆360等分每份1°.\n【详解】A错,B对,C错。故答案为:B",
+        "knowledge_points":  [
+                                 "角的大小比较",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "25-26 四年级上·河北唐山·期末"
+    },
+    {
+        "id":  "19",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面面积最大的是( )。",
+        "options":  [
+                        "A. 18平方千米",
+                        "B. 1800平方米",
+                        "C. 108公顷"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】统一换算后比较:1平方千米=100公顷=1000000平方米.\n【详解】18平方千米=18000000平方米;108公顷=1080000平方米;1800平方米最小。最大是18平方千米。故答案为:A",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·河北唐山·期末"
+    },
+    {
+        "id":  "20",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "大于999997且小于1000001的整数有( )个。",
+        "options":  [
+                        "A. 2",
+                        "B. 3",
+                        "C. 4",
+                        "D. 5"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】列出区间内整数即可.\n【详解】满足条件的有999998,999999,1000000，共3个。故答案为:B",
+        "knowledge_points":  [
+                                 "大数的比较"
+                             ],
+        "source":  "25-26 四年级上·天津滨海新·期末"
+    },
+    {
+        "id":  "21",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列各数中，一个零也不读出来的是( )。",
+        "options":  [
+                        "A. 1009000",
+                        "B. 10090000",
+                        "C. 10900000"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】按整数读法逐项读出，判断是否读\"零\".\n【详解】A读作一百万九千，不读零;B,C都读一个零。故答案为:A",
+        "knowledge_points":  [
+                                 "亿以内数的读，写法"
+                             ],
+        "source":  "25-26 四年级上·黑龙江绥化·期末"
+    },
+    {
+        "id":  "22",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "已知\"21÷5=4。1\"，那么算式\"2100÷500\"的正确结果是( )。",
+        "options":  [
+                        "A. 商4余1",
+                        "B. 商4余100",
+                        "C. 商400余100"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】有余数除法中，被除数和除数同时乘同一个数，商不变，余数也乘该数.\n【详解】21÷5=4。1，同乘100得2100÷500=4。100。故答案为:B",
+        "knowledge_points":  [
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "25-26四年级上·黑龙江齐齐哈尔·期末"
+    },
+    {
+        "id":  "23",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面各数中，要读出两个零的是( )。",
+        "options":  [
+                        "A. 38073000",
+                        "B. 200368509",
+                        "C. 50239600",
+                        "D. 84005600"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】先分级再读数，每级末尾0不读，中间连续0只读一个\"零\".\n【详解】A读一个零;B读作\"二亿零三十六万八千五百零九\"，读两个零;C读一个零;D不读零。故答案为:B",
+        "knowledge_points":  [
+                                 "亿以内数的读，写法",
+                                 "亿以上数的读，写法"
+                             ],
+        "source":  "24-25 四年级上·湖北武汉·期末"
+    },
+    {
+        "id":  "24",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面各数中只读一个零的数是( )。",
+        "options":  [
+                        "A. 8307000",
+                        "B. 8030700",
+                        "C. 8037000",
+                        "D. 8000307"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】按整数读法分别读出各数并统计\"零\"的个数.\n【详解】A不读零;B读两个零;C读一个零;D读两个零。故答案为:C",
+        "knowledge_points":  [
+                                 "亿以内数的读，写法"
+                             ],
+        "source":  "24-25四年级上·贵州遵义·期末"
+    },
+    {
+        "id":  "25",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "扎西利用步长测量操场长度，已知一步约50厘米，共走400步，操场大约长( )米。",
+        "options":  [
+                        "A. 20000",
+                        "B. 2000",
+                        "C. 200"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】步长×步数得总厘米数，再换算成米.\n【详解】50×400=20000(厘米)=200(米)。故答案为:C",
+        "knowledge_points":  [
+                                 "三位数与整十数的口算乘法",
+                                 "厘米和米之间的进率与换算",
+                                 "长度的估测"
+                             ],
+        "source":  "24-25四年级上·西藏林芝·期末"
+    },
+    {
+        "id":  "26",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面度数的角不能用一副三角尺拼成的是( )。",
+        "options":  [
+                        "A. 120°",
+                        "B. 105°",
+                        "C. 75°",
+                        "D. 35°"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】一副三角尺常用角有30°,45°,60°,90°,可通过相加得到部分角度.\n【详解】120°=90°+30°,105°=60°+45°,75°=45°+30°,35°无法由这些角相加得到。故答案为:D",
+        "knowledge_points":  [
+                                 "用三角尺画角"
+                             ],
+        "source":  "24-25四年级上·新疆巴音郭楞·期末"
+    },
+    {
+        "id":  "27",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "时间为3点整时，时针与分针所成的角是( )。",
+        "options":  [
+                        "A. 锐角",
+                        "B. 直角",
+                        "C. 钝角"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】钟面12等分，每大格30°,3点整两针相差3格.\n【详解】3×30°=90°,是直角。故答案为:B",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "25-26 四年级上·甘肃甘南·期末"
+    },
+    {
+        "id":  "28",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "235×16 的积是( )位数。",
+        "options":  [
+                        "A. 三",
+                        "B. 四",
+                        "C. 五"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】先算出乘积再判断位数.\n【详解】235×16=3760，是四位数。故答案为:B",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "29",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面算式中,( )的积最接近10000。",
+        "options":  [
+                        "A. 201×52",
+                        "B. 199×49",
+                        "C. 199×50",
+                        "D. 201×49"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】分别求积，并比较与10000的差，差最小者最接近.\n【详解】A=10452(差452);B=9751(差249);C=9950(差50);D=9849(差151)。差最小是C。故答案为:C",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法",
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "三位数乘两位数，三位数中间有0"
+                             ],
+        "source":  "25-26 四年级上·河南驻马店·期中"
+    },
+    {
+        "id":  "30",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一个因数比最大两位数多4，另一个因数是最小的两位数，它们的积是( )。",
+        "options":  [
+                        "A. 1030",
+                        "B. 1040",
+                        "C. 990"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】最大两位数99，最小两位数10，先求因数再相乘.\n【详解】(99+4)×10=103×10=1030。故答案为:A",
+        "knowledge_points":  [
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "三位数乘两位数，三位数中间有0",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25 四年级上·河北衡水·期末"
+    },
+    {
+        "id":  "31",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下列各数中，只读一个零的是( )。",
+        "options":  [
+                        "A. 208507000",
+                        "B. 2080500700",
+                        "C. 200805070",
+                        "D. 268005700"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】按亿级，万级，个级分级读数，统计读\"零\"的次数.\n【详解】A读作\"二亿零八百五十万七千\"，读一个零;B读两个零;C读两个零;D不读零。故答案为:A",
+        "knowledge_points":  [
+                                 "亿以上数的读，写法"
+                             ],
+        "source":  "24-25 四年级上·辽宁鞍山·期末"
+    },
+    {
+        "id":  "32",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面说法中，错误的是( )。",
+        "options":  [
+                        "A. 一条直线长6千米",
+                        "B. 大于0°且小于90°的角是锐角",
+                        "C. 钟面上6时整，分针和时针成平角"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】直线无限长，没有\"长度\"这一说;B与C符合定义.\n【详解】A错误;B正确;C正确(6时整两针成180°平角)。故答案为:A",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "25-26 四年级上·山西晋中·期中"
+    },
+    {
+        "id":  "33",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "明明家到公路有三条笔直小路，长度分别是360米,390米,310米。其中有一条与公路垂直，这条小路的长度是( )。",
+        "options":  [
+                        "A. 390米",
+                        "B. 360米",
+                        "C. 310米",
+                        "D. 无法确定"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】从直线外一点到直线的距离是垂线段长度，且垂线段最短.\n【详解】390\u003e360\u003e310，最短是310米，所以垂直那条是310米。故答案为:C",
+        "knowledge_points":  [
+                                 "点到直线的距离"
+                             ],
+        "source":  "24-25 四年级上·云南昭通·期末"
+    },
+    {
+        "id":  "34",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个七位数，它的最高位是( )位。它的计数单位是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "百万;百万",
+        "solution":  "【分析】一个七位数从右到左依次是个位，十位，百位，千位，万位，十万位，百万位，所以最高位是百万位;对应计数单位是百万.\n【详解】一个七位数，它的最高位是百万位，计数单位是百万。",
+        "knowledge_points":  [
+                                 "整数的数级，数位和计数单位的认识"
+                             ],
+        "source":  "25-26四年级上·甘肃陇南·期末"
+    },
+    {
+        "id":  "35",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "世界上最大的海洋是太平洋，总面积约是一亿八千一百三十四万四千平方千米。最大的沙漠是撒哈拉沙漠，总面积约9320000平方千米。(1)\"一亿八千一百三十四万四千\"写作( ),省略亿位后面的尾数约是( )亿。(2)\"9320000\"读作( ),改写成\"万\"为单位的数是( )万。",
+        "options":  [
+
+                    ],
+        "answer":  "181344000;2;九百三十二万;932",
+        "solution":  "【分析】按整数写法可得\"一亿八千一百三十四万四千\"写作181344000;省略亿位后面的尾数看千万位,8\u003e=5进一得2亿。9320000按分级读作九百三十二万;改写为\"万\"作单位是932万.\n【详解】\"一亿八千一百三十四万四千\"写作181344000，省略亿位后面的尾数约是2亿;\"9320000\"读作九百三十二万，改写成\"万\"为单位是932万。",
+        "knowledge_points":  [
+                                 "亿以内数的读，写法",
+                                 "亿以上数的读，写法",
+                                 "亿以内数的改写",
+                                 "亿以上数的近似数"
+                             ],
+        "source":  "25-26四年级上·甘肃临夏·期末"
+    },
+    {
+        "id":  "36",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "已知236÷8=29。4，那么2360÷80的商是( ),余数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "29;40",
+        "solution":  "【分析】被除数和除数同时乘相同的数(0除外),商不变，余数也乘相同的数.\n【详解】236÷8=29。4，两边同时乘10得2360÷80=29。40，所以商是29，余数是40。",
+        "knowledge_points":  [
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "24-25 四年级上·天津蓟州·期末"
+    },
+    {
+        "id":  "37",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "两个因数的积是125，其中一个因数乘2，另一个因数也乘2，积是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "500",
+        "solution":  "【分析】两个因数都乘同一个数，积会乘这个数的平方.\n【详解】125×2×2=500，所以积是500。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·湖南娄底·期中"
+    },
+    {
+        "id":  "38",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "112×15 的积是一个( )位数。",
+        "options":  [
+
+                    ],
+        "answer":  "四",
+        "solution":  "【分析】先计算乘积，再判断位数.\n【详解】112×15=1680,1680是四位数。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "24-25 四年级上·新疆吐鲁番·期末"
+    },
+    {
+        "id":  "39",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "9时整，钟面上时针和分针形成的较小角是( )角，再过10分钟，时针和分针形成的较小角是( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "直;钝",
+        "solution":  "【分析】钟面每大格30°。9时整两针相差3格，为90°;9:10时较小角在120°到150°之间.\n【详解】9时整是直角;再过10分钟形成钝角。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "整点时间与钟面指针的位置",
+                                 "一般时间与钟面指针的位置"
+                             ],
+        "source":  "25-26四年级上·甘肃平凉·期末"
+    },
+    {
+        "id":  "40",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "长江干流全长约6300000米，流域面积为178271500公顷。(1)6300000改写成用\"万\"作单位的数是( )。(2)178271500由( )个亿,( )个万和( )个一组成，省略万位后面的尾数约是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "630万;1;7827;1500;17827万",
+        "solution":  "【分析】6300000去掉末尾4个0得630万。178271500分级为1|7827|1500，即1个亿,7827个万,1500个一;省略万位后尾数看千位，千位是1，小于5，直接舍去.\n【详解】(1)630万;(2)1个亿,7827个万,1500个一，约17827万。",
+        "knowledge_points":  [
+                                 "亿以上数的组成",
+                                 "亿以内数的改写",
+                                 "亿以上数的近似数"
+                             ],
+        "source":  "25-26 四年级上·河北唐山·期末"
+    },
+    {
+        "id":  "41",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "由3个千万,7个十万,8个千和5个百组成的数是( ),它是一个( )位数，省略\"万\"位后面的尾数约是( )万。",
+        "options":  [
+
+                    ],
+        "answer":  "30708500;八;3071",
+        "solution":  "【分析】按数位组成写数：千万位3，十万位7，千位8，百位5，其余补0，得30708500;共8位。省略万位后尾数看千位,8\u003e=5进一.\n【详解】该数是30708500，是八位数，约3071万。",
+        "knowledge_points":  [
+                                 "亿以内数的组成",
+                                 "亿以内数的近似数"
+                             ],
+        "source":  "25-26四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "42",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "702是26的( )倍，从216里连续减去( )个27，结果是0。",
+        "options":  [
+
+                    ],
+        "answer":  "27;8",
+        "solution":  "【分析】求一个数是另一个数的几倍用除法;连续减去多少个27本质也是除法.\n【详解】702÷26=27,216÷27=8，所以填27和8。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "43",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "要使303×()2的积是四位数,()里最大填( );要使358÷3□的商是一位数,□里最大填( ),如果商是两位数,□里最小应填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "3;9;0",
+        "solution":  "【分析】303×32=9696为四位数,303×42=12726为五位数，所以最大填3.358÷3□若商一位数需3□\u003e35，等价看前两位与除数关系可得最大填3;若商两位数需3□\u003c=35且满足前两位够除，对应最小填4.\n【详解】依次可得三个空分别是3,3,4。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法",
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "25-26四年级上·吉林·期末"
+    },
+    {
+        "id":  "44",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据下列算式规律:9999×11=109989,9999×12=119988,9999×13=129987，可知9999×14=( ),积是159984的算式是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "139986;9999×16",
+        "solution":  "【分析】观察规律：第二因数每增1，积前两位随之变化，后两位为100减第二因数.\n【详解】9999×14=139986;积为159984时对应第二因数是16，即9999×16。",
+        "knowledge_points":  [
+                                 "算式的规律(整数)"
+                             ],
+        "source":  "25-26四年级上·福建龙岩·期中"
+    },
+    {
+        "id":  "45",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "23☐*55的积是( )位数;要使积的末尾至少有一个0，方框里可以填的数有( )。",
+        "options":  [
+
+                    ],
+        "answer":  "五;0、2、4、6、8",
+        "solution":  "【分析】23☐范围是230到239，乘55均为五位数。积末尾至少一个0，需23☐为偶数与5相乘末位为0.\n【详解】积是五位数;方框可填0,2,4,6,8。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "25-26 四年级上·福建龙岩·期中"
+    },
+    {
+        "id":  "46",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上\"\u003e\"\"\u003c\"或\"=\":36821( )12571,120750489( )102750489,480÷12( )480÷3÷4。",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;\u003e;=",
+        "solution":  "【分析】前两空按整数比较大小;第三空分别计算后比较.\n【详解】36821\u003e12571;120750489\u003e102750489;480÷12=40,480÷3÷4=160÷4=40，所以填\"=\"。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "24-25四年级上·河北保定·期末"
+    },
+    {
+        "id":  "47",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "2024年清明假期国内游客出游花费约53950000000元。这个数省略亿位后面的尾数约是( )元。",
+        "options":  [
+
+                    ],
+        "answer":  "540亿",
+        "solution":  "【分析】省略到亿位看千万位，千万位是5，按四舍五入进一.\n【详解】53950000000≈540亿。",
+        "knowledge_points":  [
+                                 "亿以上数的近似数"
+                             ],
+        "source":  "24-25 四年级上·广东中山·期末"
+    },
+    {
+        "id":  "48",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个平底锅每次最多烙2张饼，两面都烙，每面3分钟。烙好8张饼至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "24",
+        "solution":  "【分析】8张饼共16个\"面\"，每次可烙2个面，共需8次，每次3分钟.\n【详解】8×2=16(面),16÷2=8(次),8×3=24(分),至少24分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "25-26 四年级上·河北唐山·期末"
+    },
+    {
+        "id":  "49",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个四边形有一组对边平行，另一组对边不平行，它是( )形。",
+        "options":  [
+
+                    ],
+        "answer":  "梯形",
+        "solution":  "【分析】只有一组对边平行的四边形叫梯形.\n【详解】该图形是梯形。",
+        "knowledge_points":  [
+                                 "梯形的概念及特点"
+                             ],
+        "source":  "25-26 四年级上·河北唐山·期末"
+    },
+    {
+        "id":  "50",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个乘法算式的积是46，一个因数乘13，另一个因数不变，积是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "598",
+        "solution":  "【分析】一个因数不变，另一个因数乘几，积也乘几.\n【详解】46×13=598。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·河北唐山·期末"
+    },
+    {
+        "id":  "51",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "4公顷=( )平方米,180000平方米=( )公顷,32000000平方米=( )公顷=( )平方千米。",
+        "options":  [
+
+                    ],
+        "answer":  "40000;18;3200;32",
+        "solution":  "【分析】1公顷=10000平方米,1平方千米=100公顷=1000000平方米.\n【详解】4公顷=40000平方米;180000平方米=18公顷;32000000平方米=3200公顷=32平方千米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·河北唐山·期末"
+    },
+    {
+        "id":  "52",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "要使376÷□7的商是一位数,□里最小填( );要使□28÷62的商是两位数,□里最小填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "4;6",
+        "solution":  "【分析】三位数除两位数，判断商位数看被除数前两位与除数比较.\n【详解】376÷□7要商一位数需37\u003c□7，最小填4;□28÷62要商两位数需□2\u003e=62，最小填6。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "25-26四年级上·天津河西·期末"
+    },
+    {
+        "id":  "53",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "地球与太阳的距离约是149600000千米，横线上的数读作( )。这是一个( )位数,9在( )位上，表示( )。",
+        "options":  [
+
+                    ],
+        "answer":  "一亿四千九百六十万;九;百万;9个百万",
+        "solution":  "【分析】149600000按分级读作一亿四千九百六十万;共有9位;从右往左数,9在百万位，表示9个百万.\n【详解】填：一亿四千九百六十万，九，百万,9个百万。",
+        "knowledge_points":  [
+                                 "亿以上数的读，写法",
+                                 "整数的数级，数位和计数单位的认识"
+                             ],
+        "source":  "25-26四年级上·辽宁盘锦·期末"
+    },
+    {
+        "id":  "54",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "要统计\"数算洮北 智赢未来\"数学活动中各校参加人数情况，应绘制( )统计图。",
+        "options":  [
+
+                    ],
+        "answer":  "条形",
+        "solution":  "【分析】比较不同学校人数属于分类比较，条形统计图最直观.\n【详解】应绘制条形统计图。",
+        "knowledge_points":  [
+                                 "1格表示一个单位的单式条形统计图"
+                             ],
+        "source":  "25-26四年级上·吉林白城·期末"
+    },
+    {
+        "id":  "55",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "仔细观察规律:1×9=9,12×9=108,123×9=1107,1234×9=11106,12345×9=( ),( )×9=111111102。",
+        "options":  [
+
+                    ],
+        "answer":  "111105;12345678",
+        "solution":  "【分析】第一个因数由1开始连续递增，乘9后结果前面连续1的个数递增，倒数第二位为0，最后一位按规律递减.\n【详解】12345×9=111105;对应111111102的算式是12345678×9。",
+        "knowledge_points":  [
+                                 "算式的规律(整数)"
+                             ],
+        "source":  "25-26 四年级上·辽宁鞍山·期末"
+    },
+    {
+        "id":  "56",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "比较大小:7430760( )953670,340200000( )342000000,900000000( )9亿。",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;\u003c;=",
+        "solution":  "【分析】前两空按位数和高位比较;第三空先把900000000改写成\"亿\".\n【详解】7430760\u003e953670;340200000\u003c342000000;900000000=9亿。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "亿以上数的改写"
+                             ],
+        "source":  "25-26 四年级上·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "57",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "□81÷48，要使商是两位数,□里最小填( );要使商是一位数,□里最大填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "4;3",
+        "solution":  "【分析】看被除数前两位□8与48比较.\n【详解】商两位数需□8\u003e=48，最小填4;商一位数需□8\u003c48，最大填3。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "25-26四年级上·黑龙江绥化·期末"
+    },
+    {
+        "id":  "58",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "☐67÷82，如果商是一位数,☐里最大填( );如果商是两位数,☐里最小填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "7;8",
+        "solution":  "【分析】判断商位数看前两位☐6与82比较.\n【详解】商一位数需☐6\u003c82，最大填7;商两位数需☐6\u003e=82，最小填8。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "25-26四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "59",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个数由三十亿，五百万和六个一组成，这个数写作( ),读作( )。",
+        "options":  [
+
+                    ],
+        "answer":  "3005000006;三十亿零五百万零六",
+        "solution":  "【分析】按数位组成写数：三十亿是3000000000，五百万是5000000，再加6.\n【详解】写作3005000006，读作三十亿零五百万零六。",
+        "knowledge_points":  [
+                                 "亿以上数的读，写法",
+                                 "亿以上数的组成"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "60",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "136×24 的积的个位上是( ),125×40 的积的末尾有( )个0。",
+        "options":  [
+
+                    ],
+        "answer":  "4;3",
+        "solution":  "【分析】先算乘积再看指定数位或末尾0个数.\n【详解】136×24=3264，个位是4;125×40=5000，末尾有3个0。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法",
+                                 "三位数乘两位数，乘数末尾有0"
+                             ],
+        "source":  "25-26 四年级上·湖南岳阳·期中"
+    },
+    {
+        "id":  "61",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "最大的四位数是( ),最小的五位数是( ),这两个数相差( )。",
+        "options":  [
+
+                    ],
+        "answer":  "9999;10000;1",
+        "solution":  "【分析】最大四位数9999，最小五位数10000，相差用减法.\n【详解】10000-9999=1。",
+        "knowledge_points":  [
+                                 "万以内数的读，写法",
+                                 "万以内数的组成"
+                             ],
+        "source":  "24-25 四年级上·云南文山·期末"
+    },
+    {
+        "id":  "62",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "诗句\"万水千山只等闲\"中，包含的计数单位之间的进率是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "10",
+        "solution":  "【分析】诗句中有\"万\"和\"千\"，十进制下1万=10千.\n【详解】进率是10。",
+        "knowledge_points":  [
+                                 "整数的数级，数位和计数单位的认识",
+                                 "十进制计数法"
+                             ],
+        "source":  "25-26四年级上·河南南阳·期中"
+    },
+    {
+        "id":  "63",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "2025年\"五一\"期间，湖北省A级旅游景区共接待游客17813300人次。横线上的数读作( ),省略万位后面的尾数约是( )万。",
+        "options":  [
+
+                    ],
+        "answer":  "一千七百八十一万三千三百;1781",
+        "solution":  "【分析】17813300分级读作一千七百八十一万三千三百;省略万位后尾数看千位是3，小于5，直接舍去.\n【详解】读作一千七百八十一万三千三百，约1781万。",
+        "knowledge_points":  [
+                                 "亿以内数的读，写法",
+                                 "亿以内数的近似数"
+                             ],
+        "source":  "2025·湖北武汉·小升初真题"
+    },
+    {
+        "id":  "64",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "在同一平面内，不相交的两条直线叫做( ),或者说这两条直线( )平行。",
+        "options":  [
+
+                    ],
+        "answer":  "平行线;互相",
+        "solution":  "【分析】平行线定义：同一平面内不相交的两条直线.\n【详解】叫做平行线，也可以说两条直线互相平行。",
+        "knowledge_points":  [
+                                 "平行的特征及性质"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "65",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "钟面上10时整，时针和分针所夹的锐角是( )°;( )时整，时针和分针成直角。",
+        "options":  [
+
+                    ],
+        "answer":  "60;3或9",
+        "solution":  "【分析】钟面12等分，每格30°。10时整两针相差2格，角为60°;直角90°对应3格.\n【详解】锐角是60°;3时或9时整成直角。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "25-26 六年级·全国·随堂练习"
+    },
+    {
+        "id":  "66",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "有一个角的度数比直角大35°,这个角是( )角，度数是( )°。还有一个角的度数比直角小15°,这个角是( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "钝;125;锐",
+        "solution":  "【分析】直角是90°。比直角大35°是90°+35°=125°,属于钝角;比直角小15°是90°-15°=75°,属于锐角.\n【详解】依次填：钝,125，锐。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "直角，钝角，锐角的认识及特征"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "67",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个数由5个十万,6个万,7个千和8个十组成，这个数写作( ),读作( ),省略\"万\"位后面的尾数约是( )万。",
+        "options":  [
+
+                    ],
+        "answer":  "567080;五十六万七千零八十;57",
+        "solution":  "【分析】按组成写数：十万位5，万位6，千位7，十位8，其余补0，得567080。读数按万级，个级读。省略万位后尾数看千位7，进一.\n【详解】写作567080，读作五十六万七千零八十，约57万。",
+        "knowledge_points":  [
+                                 "亿以内数的读，写法",
+                                 "整数的数级，数位和计数单位的认识",
+                                 "亿以内数的近似数"
+                             ],
+        "source":  "25-26四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "68",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "∠1和一个65°角组成了平角，则∠1是( )度。",
+        "options":  [
+
+                    ],
+        "answer":  "115",
+        "solution":  "【分析】平角是180°,两角和为180°.\n【详解】180°-65°=115°,所以∠1是115度。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "25-26四年级上·吉林·期末"
+    },
+    {
+        "id":  "69",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "早晨6时整，钟面上时针与分针的夹角是( )°;晚上10时整，时针与分针形成的较小夹角是( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "180;锐",
+        "solution":  "【分析】钟面每大格30°。6时整两针相差6格，角是180°;10时整相差2格，角是60°,为锐角.\n【详解】依次填180和锐。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "25-26四年级上·甘肃陇南·期末"
+    },
+    {
+        "id":  "70",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "计算348×26时，\"3\"和\"2\"相乘表示( )×( )。",
+        "options":  [
+
+                    ],
+        "answer":  "300;20",
+        "solution":  "【分析】在348中，\"3\"在百位，表示300;在26中，\"2\"在十位，表示20.\n【详解】\"3\"和\"2\"相乘表示300×20。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "24-25 四年级上·云南文山·期末"
+    },
+    {
+        "id":  "71",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果□47÷52的商是两位数,□里的数不小于( )。",
+        "options":  [
+
+                    ],
+        "answer":  "5",
+        "solution":  "【分析】三位数除以两位数，若被除数前两位大于或等于除数，商是两位数.\n【详解】要使□47÷52的商是两位数，需□4\u003e=52，所以□最小是5，即不小于5。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25 四年级上·广东江门·期末"
+    },
+    {
+        "id":  "72",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "某省份常住人口约为126002500人。画线部分的数读作( ),省略亿位后面的尾数约是( )亿。",
+        "options":  [
+
+                    ],
+        "answer":  "一亿二千六百万二千五百;1",
+        "solution":  "【分析】126002500分级为1|2600|2500，读作一亿二千六百万二千五百;省略亿位后尾数看千万位是2，小于5，舍去.\n【详解】读作一亿二千六百万二千五百，约是1亿。",
+        "knowledge_points":  [
+                                 "亿以上数的读，写法",
+                                 "亿以上数的近似数"
+                             ],
+        "source":  "25-26四年级上·广东珠海·期中"
+    },
+    {
+        "id":  "73",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "两条平行线之间的距离是6厘米，在这两条平行线之间作一条垂线段，这条垂线段的长度是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】两条平行线间的距离处处相等，且等于它们之间垂线段的长度.\n【详解】垂线段长度是6厘米。",
+        "knowledge_points":  [
+                                 "垂直的特征",
+                                 "平行的特征及性质"
+                             ],
+        "source":  "25-26 四年级上·黑龙江绥化·期末"
+    },
+    {
+        "id":  "74",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "上午9时整，钟面上时针与分针形成的较小角是( )°;( )时整，钟面上时针与分针形成的角是平角。",
+        "options":  [
+
+                    ],
+        "answer":  "90;6",
+        "solution":  "【分析】钟面每大格30°。9时整时针与分针相差3格，角是90°;6时整两针相对，角是180°(平角).\n【详解】填90和6。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "平角，周角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25四年级上·安徽淮北·期末"
+    },
+    {
+        "id":  "75",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "钟面上的分针从数字2走到数字7，它走过的路程所对应的角度是150°。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】钟面一周360°,12个大格，每格30°。从2到7经过5格.\n【详解】5×30°=150°,说法正确。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "平角，周角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "25-26四年级上·天津南开·期末"
+    },
+    {
+        "id":  "76",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "把1235000000四舍五入到\"亿\"位约是13亿。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】四舍五入到亿位要看千万位,1235000000的千万位是3，小于5.\n【详解】应约为12亿，不是13亿，故错误。",
+        "knowledge_points":  [
+                                 "亿以上数的近似数"
+                             ],
+        "source":  "24-25 四年级上·重庆大足·期末"
+    },
+    {
+        "id":  "77",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "平角的度数是直角的度数的2倍。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】直角90°,平角180°.\n【详解】180÷90=2，所以说法正确。",
+        "knowledge_points":  [
+                                 "除数是整十数的笔算除法",
+                                 "求一个数是另一个数的几倍",
+                                 "平角，周角的认识及特征",
+                                 "角的分类及换算"
+                             ],
+        "source":  "25-26四年级上·河北唐山·期末"
+    },
+    {
+        "id":  "78",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "平行线之间的线段长度都相等。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平行线间\"距离处处相等\"指的是垂直线段长度相等，不是任意线段都相等.\n【详解】斜线段一般更长，所以该说法错误。",
+        "knowledge_points":  [
+                                 "平行的特征及性质"
+                             ],
+        "source":  "25-26 四年级上·辽宁鞍山·期末"
+    },
+    {
+        "id":  "79",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.94,
+        "question":  "万家乐超市12月份的营业额比11月份增加了60万元。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】题干给出\"增加了60万元\"，该表述本身成立.\n【详解】说法正确。",
+        "knowledge_points":  [
+                                 "亿以内数的改写"
+                             ],
+        "source":  "24-25 六年级上·河南信阳·期末"
+    },
+    {
+        "id":  "80",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "两个数的商是9，被除数和除数都扩大到原来的4倍，商就变成36。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】被除数和除数同时乘同一个非0数，商不变.\n【详解】商仍是9，不是36，故错误。",
+        "knowledge_points":  [
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "24-25 四年级上·云南文山·期末"
+    },
+    {
+        "id":  "81",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "计算350÷37时，可以把37看作40来试商，商会偏小。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】把除数估大，试得的商会偏小.\n【详解】37看作40属于估大，商偏小，故正确。",
+        "knowledge_points":  [
+                                 "多位数除以两位数的试商"
+                             ],
+        "source":  "24-25四年级上·新疆吐鲁番·期末"
+    },
+    {
+        "id":  "82",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "两个乘数的末尾都没有0，积的末尾也一定没有0。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】末尾没有0的两个数相乘，积末尾可能有0(如125×24=3000).\n【详解】\"一定没有0\"错误，故该说法错误。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "25-26 四年级上·河南南阳·期中"
+    },
+    {
+        "id":  "83",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "三位数乘两位数，积一定是五位数。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】三位数乘两位数，积可能是四位数，也可能是五位数.\n【详解】如100×10=1000(四位数),所以说法错误。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "25-26 四年级上·河南南阳·期中"
+    },
+    {
+        "id":  "84",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "两个乘数同时乘10，积也乘10。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】两个乘数都乘10，积应乘10×10=100.\n【详解】不是乘10而是乘100，故错误。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法",
+                                 "积的变化规律"
+                             ],
+        "source":  "25-26 四年级上·河南南阳·期中"
+    },
+    {
+        "id":  "85",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "某篮球场的面积是420平方米，宽15米，长是多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "28米",
+        "solution":  "【分析】长方形面积=长×宽，已知面积和宽，用面积/宽求长.\n【详解】420÷15=28(米)。答：长是28米。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "长方形的面积"
+                             ],
+        "source":  "25-26 四年级上·辽宁鞍山·期末"
+    },
+    {
+        "id":  "86",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "冰雪节期间,75公顷暖棚可产水稻900吨。照这样计算，若展区暖棚面积为60000平方米，可以产水稻多少吨。",
+        "options":  [
+
+                    ],
+        "answer":  "72吨",
+        "solution":  "【分析】先把60000平方米换算成公顷:60000平方米=6公顷。再求1公顷产量，最后乘6.\n【详解】900÷75=12(吨/公顷),12×6=72(吨)。答：可以产水稻72吨。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "单归一问题"
+                             ],
+        "source":  "25-26 四年级上·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "87",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "长峰小学共有24个班级，每班原来有120本藏书。学校又拿出360本平均分给各班，每班现在有多少本。",
+        "options":  [
+
+                    ],
+        "answer":  "135本",
+        "solution":  "【分析】先求每班分到多少本，再加原有120本.\n【详解】360÷24=15(本),120+15=135(本)。答：每班现在有135本。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "除加，除减混合运算"
+                             ],
+        "source":  "24-25 四年级上·云南昭通·期末"
+    },
+    {
+        "id":  "88",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "有一条宽6米的人行道，占地面积900平方米。道路宽度增加到12米，长不变。拓宽后面积是多少平方米。",
+        "options":  [
+
+                    ],
+        "answer":  "1800平方米",
+        "solution":  "【分析】长不变，宽由6变12，宽扩大2倍，面积也扩大2倍.\n【详解】12÷6=2,900×2=1800(平方米)。答：拓宽后面积是1800平方米。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "24-25 四年级上·山东济宁·期末"
+    },
+    {
+        "id":  "89",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "某工厂要生产3600面小红旗，已加工25天，平均每天加工108面，还剩多少面没有加工。",
+        "options":  [
+
+                    ],
+        "answer":  "900面",
+        "solution":  "【分析】先算已加工数量，再用总量减去已加工数量.\n【详解】25×108=2700(面),3600-2700=900(面)。答：还剩900面。",
+        "knowledge_points":  [
+                                 "三位数乘两位数，三位数中间有0",
+                                 "乘加，乘减混合运算"
+                             ],
+        "source":  "25-26 四年级上·山东济宁·期中"
+    },
+    {
+        "id":  "90",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "老师给同学买演出服，一件上衣138元，一条裤子99元，共36名同学，买演出服一共要花多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "8532元",
+        "solution":  "【分析】每人一套，先求每套价格，再乘人数.\n【详解】(138+99)×36=237×36=8532(元)。答：一共要花8532元。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "25-26 四年级上·辽宁葫芦岛·期末"
+    },
+    {
+        "id":  "91",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "花店购进140盆兰花，杜鹃花数量比兰花的15倍还多36盆。花店购进多少盆杜鹃花。",
+        "options":  [
+
+                    ],
+        "answer":  "2136盆",
+        "solution":  "【分析】先求兰花15倍，再加36.\n【详解】140×15+36=2100+36=2136(盆)。答：购进2136盆杜鹃花。",
+        "knowledge_points":  [
+                                 "乘加，乘减混合运算",
+                                 "求一个数的几倍是多少"
+                             ],
+        "source":  "24-25 四年级上·山东济宁·期末"
+    },
+    {
+        "id":  "92",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "某劳动实践基地一块菜地长35米，面积875平方米。扩建后长70米，宽不变，扩建后面积是多少平方米。",
+        "options":  [
+
+                    ],
+        "answer":  "1750平方米",
+        "solution":  "【分析】先用原面积/原长求宽，再用新长×宽求新面积.\n【详解】875÷35=25(米),70×25=1750(平方米)。答：扩建后面积是1750平方米。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "长方形的面积"
+                             ],
+        "source":  "24-25 四年级上·重庆开州·期末"
+    },
+    {
+        "id":  "93",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "公园里有银杏树306棵，桂花树棵数是银杏树的15倍。公园里有桂花树多少棵。",
+        "options":  [
+
+                    ],
+        "answer":  "4590棵",
+        "solution":  "【分析】\"15倍\"用乘法计算.\n【详解】306×15=4590(棵)。答：有桂花树4590棵。",
+        "knowledge_points":  [
+                                 "三位数乘两位数，三位数中间有0"
+                             ],
+        "source":  "24-25四年级上·新疆阿克苏·期末"
+    },
+    {
+        "id":  "94",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "甲，乙两地相距936千米，一辆汽车3小时行驶216千米。照这样的速度，从甲地到乙地需要多少小时。",
+        "options":  [
+
+                    ],
+        "answer":  "13小时",
+        "solution":  "【分析】先求速度，再用总路程除以速度.\n【详解】216÷3=72(千米/时),936÷72=13(小时)。答：需要13小时。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26四年级上·河北唐山·期末"
+    },
+    {
+        "id":  "95",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "冬冬要买一辆740元的自行车，已存390元，每周能存25元，至少还要存几周。",
+        "options":  [
+
+                    ],
+        "answer":  "14周",
+        "solution":  "【分析】先求还差多少钱，再除以每周存的钱数.\n【详解】740-390=350(元),350÷25=14(周)。答：至少还要存14周。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "25-26 四年级上·河北唐山·期末"
+    },
+    {
+        "id":  "96",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "14棵苹果树一年产苹果980千克，果园种了9棵这样的苹果树，一年可产苹果多少千克。",
+        "options":  [
+
+                    ],
+        "answer":  "630千克",
+        "solution":  "【分析】先求1棵苹果树年产量，再乘9.\n【详解】980÷14=70(千克),70×9=630(千克)。答：一年可产苹果630千克。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "单归一问题"
+                             ],
+        "source":  "25-26 四年级上·河北唐山·期末"
+    },
+    {
+        "id":  "97",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "某学校一块长方形菜园，若把长减少15米，面积减少525平方米;减少后菜园正好是正方形。原来面积是多少平方米。",
+        "options":  [
+
+                    ],
+        "answer":  "1750平方米",
+        "solution":  "【分析】面积减少=减少的长×原宽，所以原宽=525÷15=35米。减少后是正方形，说明减少后的长也为35米，所以原长=35+15=50米。原面积=50×35.\n【详解】525÷15=35(米),35+15=50(米),50×35=1750(平方米)。答：原来面积是1750平方米。",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "正方形的面积",
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·辽宁葫芦岛·期末"
+    },
+    {
+        "id":  "98",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "冰雕团队需雕刻395个组件，连续工作17天后还剩55个，平均每天雕刻多少个组件。",
+        "options":  [
+
+                    ],
+        "answer":  "20个",
+        "solution":  "【分析】先求17天已完成数量，再除以17求平均每天数量.\n【详解】395-55=340(个),340÷17=20(个)。答：平均每天雕刻20个。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "25-26 四年级上·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "99",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "万家乐超市运来苹果400箱，每箱35千克。卖出205箱后，还剩下多少千克。",
+        "options":  [
+
+                    ],
+        "answer":  "6825千克",
+        "solution":  "【分析】先求剩余箱数，再乘每箱质量.\n【详解】400-205=195(箱),195×35=6825(千克)。答：还剩6825千克。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法",
+                                 "三位数乘两位数的实际问题"
+                             ],
+        "source":  "25-26四年级上·河南驻马店·期中"
+    },
+    {
+        "id":  "100",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "胜利小学教学楼共有4层，每层有6个班级，共有课桌960张，平均每班有多少张课桌。",
+        "options":  [
+
+                    ],
+        "answer":  "40张",
+        "solution":  "【分析】先求总班级数，再用总课桌数除以班级数.\n【详解】6×4=24(个班),960÷24=40(张/班)。答：平均每班有40张课桌。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "24-25四年级上·山东济宁·期末"
+    },
+    {
+        "id":  "101",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "两条直线相交形成的 4 个角可能都是( )。",
+        "options":  [
+                        "A. 锐角",
+                        "B. 钝角",
+                        "C. 直角"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】两条直线相交形成的4个角和是360°。若都是锐角，则和小于360°;若都是钝角，则和大于360°;若都是直角，则4×90°=360°,符合条件.\n【详解】A 锐角不可能;B 钝角不可能;C 直角可能。故答案为:C.",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征"
+                             ],
+        "source":  "25-26四年级上·辽宁葫芦岛·期末"
+    },
+    {
+        "id":  "102",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "A/B=4。6，如果 A 和 B 都乘 10，那么结果是( )。",
+        "options":  [
+                        "A. 40。6",
+                        "B. 4。6",
+                        "C. 4。60",
+                        "D. 40。60"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】被除数和除数同时乘同一个数(0除外),商不变，余数也乘同一个数.\n【详解】原式余数6，乘10后余数变60，商仍是4，所以结果是4。60。故答案为:C.",
+        "knowledge_points":  [
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "25-26四年级上·吉林·期末"
+    },
+    {
+        "id":  "103",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "A/B=15。10，如果 A,B 都乘 10，余数是( )。",
+        "options":  [
+                        "A. 10",
+                        "B. 100",
+                        "C. 1"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】被除数和除数同时乘同一个数(0除外),商不变，余数也乘同一个数.\n【详解】10×10=100，所以余数由10变为100。故答案为:B.",
+        "knowledge_points":  [
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "24-25 四年级上·云南曲靖·期末"
+    },
+    {
+        "id":  "104",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面四个数中，读数时一个零也不读的数是( )。",
+        "options":  [
+                        "A. 34060000",
+                        "B. 1740000",
+                        "C. 13000009",
+                        "D. 120060000"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】按整数读法：每级末尾的0不读，每级中间有一个或连续几个0只读一个.\n【详解】A 读作三千四百零六万(读1个零);B 读作一百七十四万(不读零);C 读作一千三百万零九(读1个零);D 读作一亿二千零六万(读1个零)。故答案为:B.",
+        "knowledge_points":  [
+                                 "亿以内数的读，写法",
+                                 "亿以上数的读，写法"
+                             ],
+        "source":  "24-25四年级上·重庆璧山·期末"
+    },
+    {
+        "id":  "105",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "沿直线行驶的汽车车轮留下的两条印迹( )。",
+        "options":  [
+                        "A. 互相平行",
+                        "B. 互相垂直",
+                        "C. 相交",
+                        "D. 无法确定"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】同一平面内永不相交的两条直线叫平行线。汽车沿直线行驶时，两轮轨迹方向一致，间距不变.\n【详解】两条印迹互相平行。故答案为:A.",
+        "knowledge_points":  [
+                                 "平行的特征及性质"
+                             ],
+        "source":  "25-26 四年级上·新疆和田·期末"
+    },
+    {
+        "id":  "106",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "在同一平面内，与一条直线平行且相距 2 厘米的直线有( )条。",
+        "options":  [
+                        "A. 无数",
+                        "B. 3",
+                        "C. 2",
+                        "D. 1"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】与已知直线平行且距离固定为2厘米，只能在已知直线两侧各有一条.\n【详解】共有2条。故答案为:C.",
+        "knowledge_points":  [
+                                 "平行的特征及性质"
+                             ],
+        "source":  "25-26 四年级上·福建漳州·期中"
+    },
+    {
+        "id":  "107",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "从点 O 到直线 L 画了三条线段，分别长 8 厘米,6 厘米,10 厘米，如果其中有一条线段与直线垂直，那么这条与直线 L 垂直的线段长( )厘米。",
+        "options":  [
+                        "A. 8",
+                        "B. 6",
+                        "C. 10",
+                        "D. 无法确定"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】点到直线的垂线段最短.\n【详解】8,6,10中最短是6厘米，所以垂线段长6厘米。故答案为:B.",
+        "knowledge_points":  [
+                                 "点到直线的距离"
+                             ],
+        "source":  "24-25四年级上·四川凉山·期末"
+    },
+    {
+        "id":  "108",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "小华在用量角器量角时，误把外圈刻度看成了内圈刻度，量出的度数是130°,那么这个角实际应该是( )。",
+        "options":  [
+                        "A. 50°",
+                        "B. 40°",
+                        "C. 30°"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】量角器内外圈对应数和为180°.\n【详解】实际角度=180°-130°=50°。故答案为:A.",
+        "knowledge_points":  [
+                                 "角的度量"
+                             ],
+        "source":  "25-26四年级上·新疆阿克苏·期末"
+    },
+    {
+        "id":  "109",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "\"五一\"小长假期间，海口美兰机场运送旅客368500人次，这里的368500中的\"8\"表示( )。",
+        "options":  [
+                        "A. 8个千",
+                        "B. 8个百",
+                        "C. 8个万"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】368500中\"8\"在千位.\n【详解】表示8个千。故答案为:A.",
+        "knowledge_points":  [
+                                 "整数的数级，数位和计数单位的认识"
+                             ],
+        "source":  "25-26 四年级上·黑龙江鸡西·期末"
+    },
+    {
+        "id":  "110",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "小明用计算器计算126×8时，错误输成126×80，下面的做法中，可以弥补他的错误的是( )。",
+        "options":  [
+                        "A. 除以10",
+                        "B. 乘10",
+                        "C. 加10",
+                        "D. 减10"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】因数8误输成80，相当于乘数扩大10倍，积也扩大10倍.\n【详解】要还原正确结果，应把错误结果除以10。故答案为:A.",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "24-25四年级上·四川凉山·期末"
+    },
+    {
+        "id":  "111",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.94,
+        "question":  "小明看到墙上的钟面刚好是6时整，时针和分针组成的角是一个( )。",
+        "options":  [
+                        "A. 锐角",
+                        "B. 直角",
+                        "C. 平角",
+                        "D. 周角"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】6时整时，分针指12，时针指6，方向相反在同一直线上.\n【详解】形成180°,是平角。故答案为:C.",
+        "knowledge_points":  [
+                                 "平角，周角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "25-26四年级上·广西贵港·期中"
+    },
+    {
+        "id":  "112",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "653640中左边的\"6\"表示的数比右边的\"6\"表示的数多( )。",
+        "options":  [
+                        "A. 1000",
+                        "B. 599400",
+                        "C. 59940"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】左边6在十万位表示600000，右边6在百位表示600.\n【详解】600000-600=599400。故答案为:B.",
+        "knowledge_points":  [
+                                 "整数的数级，数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级上·西藏林芝·期末"
+    },
+    {
+        "id":  "113",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "每年的4月23日为\"世界读书日\"，为了培养同学们的读书习惯，四(1)班准备给班级图书角添置图书，你认为最需要收集同学们( )的信息。",
+        "options":  [
+                        "A. 阅读时间",
+                        "B. 阅读速度",
+                        "C. 阅读习惯",
+                        "D. 喜欢图书种类"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】图书角选书应优先匹配同学兴趣.\n【详解】最应收集\"喜欢图书种类\"信息。故答案为:D.",
+        "knowledge_points":  [
+                                 "数据的搜集与整理"
+                             ],
+        "source":  "24-25四年级上·贵州遵义·期末"
+    },
+    {
+        "id":  "114",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "把一个平角分成两个角，若其中一个是钝角，则另一个一定是( )。",
+        "options":  [
+                        "A. 钝角",
+                        "B. 直角",
+                        "C. 锐角",
+                        "D. 周角"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】平角是180°,钝角在(90°,180°)之间.\n【详解】另一个角=180°-钝角，结果在(0°,90°)之间，为锐角。故答案为:C.",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "25-26 四年级上·辽宁抚顺·期末"
+    },
+    {
+        "id":  "115",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.94,
+        "question":  "十亿位的左边是( )位。",
+        "options":  [
+                        "A. 亿",
+                        "B. 千万",
+                        "C. 百亿"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】数位顺序中,。亿位，十亿位，百亿位。.\n【详解】十亿位左边是百亿位。故答案为:C.",
+        "knowledge_points":  [
+                                 "整数的数级，数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级上·宁夏吴忠·期末"
+    },
+    {
+        "id":  "116",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "从小强家到河边修一条最短的路，这条路( )。",
+        "options":  [
+                        "A. 与小河平行",
+                        "B. 与小河垂直",
+                        "C. 与小河无关",
+                        "D. 可任意修"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】点到直线最短距离是垂线段.\n【详解】最短路应与小河垂直。故答案为:B.",
+        "knowledge_points":  [
+                                 "垂直的特征",
+                                 "点到直线的距离"
+                             ],
+        "source":  "24-25四年级上·云南玉溪·期末"
+    },
+    {
+        "id":  "117",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "两个因数的积是270，一个因数不变，另一个因数除以3，积是( )。",
+        "options":  [
+                        "A. 810",
+                        "B. 90",
+                        "C. 270"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】一个因数不变，另一个因数除以几，积也除以几.\n【详解】270÷3=90。故答案为:B.",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26四年级上·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "118",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "两条直线相交，如果其中一个角是90°,那么其他三个角是( )。",
+        "options":  [
+                        "A. 锐角",
+                        "B. 钝角",
+                        "C. 直角",
+                        "D. 可能是锐角也可能是钝角"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】两直线相交时邻角互补，对顶角相等。若一个角是90°,其邻角为180°-90°=90°,其对顶角也为90°.\n【详解】四个角都为直角。图示:![](images/8d5f1632a3994880441b1af59c80e49fb045397585e9f2381128ba35c94334a4.jpg)。故答案为:C.",
+        "knowledge_points":  [
+                                 "垂直的特征",
+                                 "直角，钝角，锐角的认识及特征"
+                             ],
+        "source":  "25-26 四年级上·广东东莞·月考"
+    },
+    {
+        "id":  "119",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "4700÷500 的结果正确的是( )。",
+        "options":  [
+                        "A. 9。2",
+                        "B. 9。20",
+                        "C. 9。200"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】用\"被除数=商×除数+余数\"验算.\n【详解】A:9×500+2=4502;B:9×500+20=4520;C:9×500+200=4700，正确。故答案为:C.",
+        "knowledge_points":  [
+                                 "有余数除法的认识",
+                                 "有余数的除法验算"
+                             ],
+        "source":  "25-26 四年级上·新疆喀什·期末"
+    },
+    {
+        "id":  "120",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面说法正确的是( )。",
+        "options":  [
+                        "A. 正方形相邻的两条边互相平行",
+                        "B. 平行四边形具有稳定性",
+                        "C. 长方形是特殊的平行四边形",
+                        "D. 梯形是特殊的平行四边形"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】正方形相邻边垂直;平行四边形不稳定;长方形两组对边平行;梯形只有一组对边平行.\n【详解】只有\"长方形是特殊的平行四边形\"正确。故答案为:C.",
+        "knowledge_points":  [
+                                 "长方形的概念及特点",
+                                 "正方形的概念及特点",
+                                 "平行四边形的不稳定性及应用",
+                                 "梯形的概念及特点"
+                             ],
+        "source":  "25-26四年级上·天津滨海新·期末"
+    },
+    {
+        "id":  "121",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "已知13÷2=6。1，那么1300÷200=( )。",
+        "options":  [
+                        "A. 6。1",
+                        "B. 6。10",
+                        "C. 6。100",
+                        "D. 600。100"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】被除数，除数同时乘100，商不变，余数也乘100.\n【详解】13→1300,2→200，余数1→100，所以1300÷200=6。100。故答案为:C.",
+        "knowledge_points":  [
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "25-26 四年级上·天津滨海新·期末"
+    },
+    {
+        "id":  "122",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "与240×15结果不同的是( )。",
+        "options":  [
+                        "A. 24×150",
+                        "B. 5×720",
+                        "C. 80×45",
+                        "D. 480×30"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】利用积不变规律判断.\n【详解】A,B,C都可等价变形为240×15;D=(240×2)×(15×2)=240×15×4，结果不同。故答案为:D.",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26四年级上·天津滨海新·期末"
+    },
+    {
+        "id":  "123",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "根据我国常用的分级法，数987610025应该分为( )。",
+        "options":  [
+                        "A. 三级",
+                        "B. 四级",
+                        "C. 五级"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】每四位一级：个级，万级，亿级.\n【详解】987610025分为9 | 8761 | 0025，共三级。故答案为:A.",
+        "knowledge_points":  [
+                                 "整数的数级，数位和计数单位的认识"
+                             ],
+        "source":  "24-25 四年级上·重庆丰都·期末"
+    },
+    {
+        "id":  "124",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "已知X×Y=350，若X乘10,Y除以5，则积变为( )。",
+        "options":  [
+                        "A. 17500",
+                        "B. 6500",
+                        "C. 1300",
+                        "D. 700"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】一个因数乘10，另一个因数除以5，积相当于乘10再除以5.\n【详解】350×10÷5=700。故答案为:D.",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·浙江杭州·期中"
+    },
+    {
+        "id":  "125",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "小明在本子上画了3条直线，分别是a,b,c,如果a∥b,b∥c,那么a和c的位置关系为( )。",
+        "options":  [
+                        "A. 互相平行",
+                        "B. 互相垂直",
+                        "C. 相交",
+                        "D. 无法确定"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】平行具有传递性：若a∥b且b∥c,则a∥c.\n【详解】a和c互相平行。故答案为:A.",
+        "knowledge_points":  [
+                                 "平行的特征及性质"
+                             ],
+        "source":  "24-25四年级上·重庆璧山·期末"
+    },
+    {
+        "id":  "126",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.94,
+        "question":  "一个30°的角，放在10倍的放大镜下观察，度数为( )。",
+        "options":  [
+                        "A. 3°",
+                        "B. 30°",
+                        "C. 300°"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】角的大小只与两边张开程度有关，与边长无关.\n【详解】放大镜只放大图形，不改变角度，所以仍是30°。故答案为:B.",
+        "knowledge_points":  [
+                                 "角的概念及表示方法"
+                             ],
+        "source":  "24-25 六年级上·河北石家庄·期末"
+    },
+    {
+        "id":  "127",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "一个锐角加一个锐角，得到的角( )。",
+        "options":  [
+                        "A. 不可能是锐角",
+                        "B. 不可能是钝角",
+                        "C. 可能是钝角，也可能是直角，还可能是锐角"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】两个锐角都小于90°,其和可小于90°,等于90°或大于90°且小于180°.\n【详解】和可能是锐角，直角或钝角。故答案为:C.",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "直角，钝角，锐角的认识及特征"
+                             ],
+        "source":  "25-26 四年级上·河南三门峡·期中"
+    },
+    {
+        "id":  "128",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "已知12×15=180，那么1200×15=( )。",
+        "options":  [
+                        "A. 1800",
+                        "B. 18000",
+                        "C. 180000"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】12变成1200是乘100，另一个因数不变，积也乘100.\n【详解】180×100=18000。故答案为:B.",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·安徽合肥·期中"
+    },
+    {
+        "id":  "129",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面说法正确的是( )。",
+        "options":  [
+                        "A. 大于90°的角一定是钝角",
+                        "B. 用一个放大3倍的放大镜看一个30°的角，这个角就成了90°",
+                        "C. 钟面上8时整，分针和时针组成的角是钝角"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】钝角是大于90°且小于180°;角度不会因放大镜改变;8时整夹角为120°.\n【详解】A错(周角等也大于90°);B错(角度不变);C对。故答案为:C.",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "角的大小比较",
+                                 "角的分类及换算",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "25-26 四年级上·江西宜春·期中"
+    },
+    {
+        "id":  "130",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "用一副三角尺不能画出( )的角。",
+        "options":  [
+                        "A. 80°",
+                        "B. 135°",
+                        "C. 15°",
+                        "D. 75°"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】一副三角尺常用角有30°,45°,60°,90°,通过和差能得到15°,75°,105°,120°,135°等15°的倍数角.\n【详解】80°不能由这些角的和差得到。故答案为:A.",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "用三角尺画角"
+                             ],
+        "source":  "24-25 四年级上·四川凉山·期末"
+    },
+    {
+        "id":  "131",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "3时30分时，钟面上时针和分针所成的角是( )。",
+        "options":  [
+                        "A. 钝角",
+                        "B. 直角",
+                        "C. 锐角",
+                        "D. 不能确定"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】每大格30°。3:30时分针在6，时针在3和4中间，二者夹角小于90°.\n【详解】所以是锐角。故答案为:C.",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "半点时间与钟面指针的位置"
+                             ],
+        "source":  "2025·广东揭阳·小升初模拟"
+    },
+    {
+        "id":  "132",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个数由3个亿,7个百万,9个千组成，这个数是( ),读作( ),省略万位后面的尾数约是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "307009000;三亿零七百万九千;30701万",
+        "solution":  "【分析】3个亿在亿位写3,7个百万在百万位写7,9个千在千位写9，其余位补0;按分级读数;省略万位后尾数看千位四舍五入.\n【详解】该数是307009000，读作三亿零七百万九千，约是30701万。",
+        "knowledge_points":  [
+                                 "亿以上数的读，写法",
+                                 "亿以上数的组成",
+                                 "整数的数级，数位和计数单位的认识",
+                                 "亿以上数的近似数"
+                             ],
+        "source":  "25-26 四年级上·甘肃临夏·期末"
+    },
+    {
+        "id":  "133",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "( )是用直条的高低或长短来表示数量的多少。",
+        "options":  [
+
+                    ],
+        "answer":  "条形统计图",
+        "solution":  "【分析】条形统计图通过直条长度(或高度)表达数量大小，条越长(高)表示数量越多.\n【详解】![](images/34061b0076c059d146ad6a5099a3d66070eff4a723d519c4157eca2e4de853e3.jpg) 所以应填\"条形统计图\"。",
+        "knowledge_points":  [
+                                 "1 格表示多个单位的单式条形统计图"
+                             ],
+        "source":  "25-26 四年级上·黑龙江齐齐哈尔·期末"
+    },
+    {
+        "id":  "134",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "580700000 读作( ),7 在( )位上，表示 7 个( )。把这个数改写成用\"万\"作单位的数是( ),省略亿后面的尾数约是()。",
+        "options":  [
+
+                    ],
+        "answer":  "五亿八千零七十万;十万;十万;58070万;6亿",
+        "solution":  "【分析】整数读法按数级读;7所在数位是十万位，表示7个十万。整万数改写成\"万\"为单位时去掉末尾4个0加\"万\";省略亿后尾数看千万位四舍五入.\n【详解】580700000读作五亿八千零七十万;7在十万位，表示7个十万;改写为58070万;省略亿后面的尾数约是6亿。",
+        "knowledge_points":  [
+                                 "亿以上数的读，写法",
+                                 "整数的数级，数位和计数单位的认识",
+                                 "亿以内数的改写",
+                                 "亿以上数的近似数"
+                             ],
+        "source":  "25-26 四年级上·黑龙江齐齐哈尔·期末"
+    },
+    {
+        "id":  "135",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "已知12×5=60，那么12×50=( ),12×10=( )。",
+        "options":  [
+
+                    ],
+        "answer":  "600;120",
+        "solution":  "【分析】一个因数不变，另一个因数乘几，积也乘几.\n【详解】12×50=12×(5×10)=60×10=600;12×10=12×(5×2)=60×2=120。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "136",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "立冬是二十四节气中的第19个节气，也是冬季的第一个节气,2025年11月7日12时03分将迎来立冬节气，这时钟面上时针和分针形成的较小角是( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "锐",
+        "solution":  "【分析】钟面每大格30°,每小格6°。12:03时分针在12后第3小格，时针也在12和1之间，二者夹角小于30°.\n【详解】夹角大于0°且小于90°,所以是锐角。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "一般时间与钟面指针的位置"
+                             ],
+        "source":  "25-26四年级上·新疆喀什·期末"
+    },
+    {
+        "id":  "137",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\"\u003e\"\"\u003c\"或\"=\":15亿( )240000000;99公顷( )10平方千米;95÷5( )9500÷500;450×30( )450×20。",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;\u003c;=;\u003e",
+        "solution":  "【分析】先统一单位再比较;商不变规律可比较95÷5与9500÷500;最后直接计算乘积比较.\n【详解】15亿=1500000000\u003e240000000;10平方千米=1000公顷，所以99公顷\u003c10平方千米;95÷5=9500÷500;450×30=13500\u003e9000=450×20。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "除数是两位数的笔算除法",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·新疆喀什·期末"
+    },
+    {
+        "id":  "138",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "20000平方米=()公顷,200平方千米=( )公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "2;20000",
+        "solution":  "【分析】10000平方米=1公顷,1平方千米=100公顷.\n【详解】20000÷10000=2(公顷);200×100=20000(公顷)。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·新疆喀什·期末"
+    },
+    {
+        "id":  "139",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "2025年春晚人形机器人扭秧歌震撼全场，观看春晚的人高达1738040000人次，据不完全统计，网络上关于机器人表演的相关话题达五亿八千七百四十万次.\"1738040000\"读作( ).\"五亿八千七百四十万\"写作( ),省略\"亿\"后面的尾数是( )亿。",
+        "options":  [
+
+                    ],
+        "answer":  "十七亿三千八百零四万;587400000;6",
+        "solution":  "【分析】按数级读写大数;省略亿后尾数看千万位四舍五入.\n【详解】1738040000读作十七亿三千八百零四万;五亿八千七百四十万写作587400000;587400000≈6亿。",
+        "knowledge_points":  [
+                                 "亿以上数的读，写法",
+                                 "亿以上数的近似数"
+                             ],
+        "source":  "25-26四年级上·新疆喀什·期末"
+    },
+    {
+        "id":  "140",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "钟面上3时整，时针和分针形成的角是( )角;6时整，时针和分针形成的角是( )角;( )时整，时针和分针形成的角是周角。",
+        "options":  [
+
+                    ],
+        "answer":  "直;平;12",
+        "solution":  "【分析】钟面相邻两数夹角30°。3时相差3格为90°,6时相差6格为180°,12时两针重合按一周记为360°.\n【详解】3时是直角,6时是平角,12时是周角。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征",
+                                 "角的分类及换算",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "25-26四年级上·吉林白城·期末"
+    },
+    {
+        "id":  "141",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "45°与∠2的角可以拼成一个直角，这个角的度数是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "45",
+        "solution":  "【分析】直角是90°.\n【详解】∠2=90°-45°=45°。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "直角，钝角，锐角的认识及特征"
+                             ],
+        "source":  "25-26 四年级上·新疆和田·期末"
+    },
+    {
+        "id":  "142",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在数位顺序表中，从右边数起，第( )位是亿位，第七位是( )位。与百万位相邻的两个数位分别是( )位和( )位。",
+        "options":  [
+
+                    ],
+        "answer":  "九;百万;十万;千万",
+        "solution":  "【分析】数位从右往左依次为个位，十位，百位，千位，万位，十万位，百万位，千万位，亿位。.\n【详解】亿位是第九位;第七位是百万位;与百万位相邻的是十万位和千万位。",
+        "knowledge_points":  [
+                                 "整数的数级，数位和计数单位的认识"
+                             ],
+        "source":  "25-26四年级上·浙江宁波·期中"
+    },
+    {
+        "id":  "143",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "一个十位数，它的最高位是( ),一个八位数，它的最高计数单位是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "十亿位;千万",
+        "solution":  "【分析】十位数最高位是十亿位;八位数最高位是千万位，其计数单位是千万.\n【详解】填\"十亿位;千万\"。",
+        "knowledge_points":  [
+                                 "整数的数级，数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级上·广西玉林·期末"
+    },
+    {
+        "id":  "144",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把一张正方形的纸沿着对角线对折后再对折，打开后两条折痕互相( )。",
+        "options":  [
+
+                    ],
+        "answer":  "垂直",
+        "solution":  "【分析】对折后两条折痕相交成直角.\n【详解】图示:![](images/0bc261d7f46eb8caedf558417f81cab84d12b2b381f53e4638d657e9e0670e99.jpg);所以两条折痕互相垂直。",
+        "knowledge_points":  [
+                                 "垂直的特征"
+                             ],
+        "source":  "25-26 四年级上·黑龙江齐齐哈尔·期末"
+    },
+    {
+        "id":  "145",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "1个周角=( )个平角=( )个直角。",
+        "options":  [
+
+                    ],
+        "answer":  "2;4",
+        "solution":  "【分析】周角360°,平角180°,直角90°.\n【详解】360°=2×180°=4×90°,所以1个周角=2个平角=4个直角。",
+        "knowledge_points":  [
+                                 "角的分类及换算"
+                             ],
+        "source":  "14-15 四年级上·全国·单元测试"
+    },
+    {
+        "id":  "146",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "电动伸缩门上有很多( ),这种图形具有( )的特性。",
+        "options":  [
+
+                    ],
+        "answer":  "平行四边形;易变形",
+        "solution":  "【分析】平行四边形具有不稳定性，容易变形.\n【详解】伸缩门上有很多平行四边形，这种图形具有易变形的特性。",
+        "knowledge_points":  [
+                                 "平行四边形的不稳定性及应用"
+                             ],
+        "source":  "24-25 四年级上·重庆开州·期末"
+    },
+    {
+        "id":  "147",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "○÷△=13。16，除数最小是( ),此时被除数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "17;237",
+        "solution":  "【分析】有余数除法中余数\u003c除数，所以除数最小为16+1=17;被除数=商×除数+余数.\n【详解】13×17+16=237。",
+        "knowledge_points":  [
+                                 "余数与除数的关系",
+                                 "有余数的除法验算"
+                             ],
+        "source":  "24-25四年级上·云南文山·期末"
+    },
+    {
+        "id":  "148",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据算式12345679×9=111111111，直接填空:12345679×18=( );( )×12345679=777777777。",
+        "options":  [
+
+                    ],
+        "answer":  "222222222;63",
+        "solution":  "【分析】18是9的2倍，所以积也乘2;777777777是111111111的7倍，对应另一个因数是9×7.\n【详解】12345679×18=222222222;63×12345679=777777777。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "24-25 四年级上·重庆大足·期末"
+    },
+    {
+        "id":  "149",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "直角的2倍是( )角，周角=( )个平角。",
+        "options":  [
+
+                    ],
+        "answer":  "平;2",
+        "solution":  "【分析】直角90°,平角180°,周角360°.\n【详解】90°*2=180°,即平角;360°=2×180°,即周角等于2个平角。",
+        "knowledge_points":  [
+                                 "角的分类及换算"
+                             ],
+        "source":  "25-26 四年级上·河南濮阳·期末"
+    },
+    {
+        "id":  "150",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "《九章算术》题意：若长方形田\"广\"30步，\"从\"16步，则田的面积为( )亩。",
+        "options":  [
+
+                    ],
+        "answer":  "2",
+        "solution":  "【分析】已知\"广15步，从16步\"为1亩，则1亩=15×16=240平方步。新田面积30×16=480平方步.\n【详解】480÷240=2(亩),所以面积为2亩。",
+        "knowledge_points":  [
+                                 "长方形的面积"
+                             ],
+        "source":  "24-25四年级上·重庆开州·期末"
+    },
+    {
+        "id":  "151",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "伸缩门应用了平行四边形( )的特点。",
+        "options":  [
+
+                    ],
+        "answer":  "不稳定性",
+        "solution":  "【分析】平行四边形受力时形状易改变.\n【详解】伸缩门正是利用平行四边形的不稳定性实现伸缩。",
+        "knowledge_points":  [
+                                 "平行四边形的不稳定性及应用"
+                             ],
+        "source":  "25-26 四年级上·新疆阿克苏·期末"
+    },
+    {
+        "id":  "152",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在笔直的公路上有三条小路通往小明家，长度分别是125米,207米,112米，其中有一条小路与公路垂直，那么这条小路的长度是( )米。",
+        "options":  [
+
+                    ],
+        "answer":  "112",
+        "solution":  "【分析】点到直线的垂线段最短.\n【详解】三条中最短是112米，所以垂直的小路长度是112米。",
+        "knowledge_points":  [
+                                 "点到直线的距离"
+                             ],
+        "source":  "25-26 四年级上·福建漳州·期中"
+    },
+    {
+        "id":  "153",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "6时整，分针和时针的夹角是( )度，也叫作( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "180;平角",
+        "solution":  "【分析】6时整时两针方向相反，形成平角。题图:![](images/8bb1c9bf220bc15e2f359e02280956a7ae40421381aeccab79d78228764b657f.jpg)\n【详解】夹角是180°,叫平角。",
+        "knowledge_points":  [
+                                 "平角，周角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·四川凉山·期末"
+    },
+    {
+        "id":  "154",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "小明和小红玩扑克牌比大小(点数大获胜),三局两胜。小明有5,7,9，小红有4,6,8。小明先出9，小红要想最终获胜应出( )。",
+        "options":  [
+
+                    ],
+        "answer":  "4",
+        "solution":  "【分析】小红任意牌都输给9，需\"弃一局保两局\".\n【详解】先出4对9(输),再用6对5,8对7(连赢两局),所以应先出4。",
+        "knowledge_points":  [
+                                 "田忌赛马问题"
+                             ],
+        "source":  "25-26 四年级上·天津滨海新·期末"
+    },
+    {
+        "id":  "155",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "两个数相除商是6，如果除数不变，被除数乘2，商就变成( );如果被除数不变，除数乘2，商就变成( )。",
+        "options":  [
+
+                    ],
+        "answer":  "12;3",
+        "solution":  "【分析】除数不变时，被除数乘几商也乘几;被除数不变时，除数乘几商就除以几.\n【详解】6×2=12;6÷2=3。",
+        "knowledge_points":  [
+                                 "商的变化规律及应用"
+                             ],
+        "source":  "25-26四年级上·天津滨海新·期末"
+    },
+    {
+        "id":  "156",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "2024年全国粮食总产量是706500000吨，横线上的数改写成用\"万\"作单位的数是( ),用\"四舍五入\"法省略亿位后面的尾数约是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "70650万;7亿",
+        "solution":  "【分析】改写\"万\"为单位去掉末4位并加\"万\";省略亿后尾数看千万位.\n【详解】706500000=70650万;千万位是0，故约是7亿。",
+        "knowledge_points":  [
+                                 "亿以上数的改写",
+                                 "亿以上数的近似数"
+                             ],
+        "source":  "25-26 四年级上·天津滨海新·期末"
+    },
+    {
+        "id":  "157",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "由6个千万,7个十万和8个百组成的数是( ),这个数读作( )。",
+        "options":  [
+
+                    ],
+        "answer":  "60700800;六千零七十万零八百",
+        "solution":  "【分析】按数位写数：千万位6，十万位7，百位8，其余补0.\n【详解】组成的数是60700800，读作六千零七十万零八百。",
+        "knowledge_points":  [
+                                 "亿以内数的读，写法",
+                                 "亿以内数的组成",
+                                 "整数的数级，数位和计数单位的认识"
+                             ],
+        "source":  "25-26四年级上·天津滨海新·期末"
+    },
+    {
+        "id":  "158",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "比较下面每组数中两个数的大小:600987( )5001987;111万( )111000。",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003e",
+        "solution":  "【分析】位数不同先比位数;111万先改写成1110000.\n【详解】600987是6位数,5001987是7位数，所以600987\u003c5001987;1110000\u003e111000，所以111万\u003e111000。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "亿以内数的改写"
+                             ],
+        "source":  "25-26四年级上·天津滨海新·期末"
+    },
+    {
+        "id":  "159",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "有2530千克石油，每50千克装一桶，这些油都装完需要( )个桶。",
+        "options":  [
+
+                    ],
+        "answer":  "51",
+        "solution":  "【分析】总量/每桶容量，若有余数还需再加1桶.\n【详解】2530÷50=50。30，所以需要50+1=51(桶)。",
+        "knowledge_points":  [
+                                 "有余数除法的实际应用"
+                             ],
+        "source":  "24-25 四年级上·江西宜春·期末"
+    },
+    {
+        "id":  "160",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在一幅条形统计图中，如果用3格代表6人，那么1格代表( )人;用5格可以代表( )人。",
+        "options":  [
+
+                    ],
+        "answer":  "2;10",
+        "solution":  "【分析】先求每格代表人数，再乘5格.\n【详解】6÷3=2(人);2×5=10(人)。",
+        "knowledge_points":  [
+                                 "1格表示多个单位的单式条形统计图"
+                             ],
+        "source":  "25-26四年级上·新疆阿克苏·期末"
+    },
+    {
+        "id":  "161",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据11×28=308，写出下面各算式的积:110×28=( ),110×2800=( ),11×14=( )。",
+        "options":  [
+
+                    ],
+        "answer":  "3080;308000;154",
+        "solution":  "【分析】根据因数变化与积变化规律推算.\n【详解】110×28=(11×10)×28=308×10=3080;110×2800=(11×10)×(28×100)=308×1000=308000;11×14=11×28÷2=154。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "24-25 四年级上·新疆巴音郭楞·期末"
+    },
+    {
+        "id":  "162",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "端点分别在平行的两条直线上，与平行线垂直的线段有( )条，这些线段的长度都相等。",
+        "options":  [
+
+                    ],
+        "answer":  "无数",
+        "solution":  "【分析】平行线间可从一条线上任一点向另一条线作垂线段，点有无数个;且平行线间距离处处相等.\n【详解】这样的垂直线段有无数条，且长度都相等。",
+        "knowledge_points":  [
+                                 "垂直的特征",
+                                 "平行的特征及性质"
+                             ],
+        "source":  "25-26 四年级上·山东菏泽·月考"
+    },
+    {
+        "id":  "163",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一块长方形菜地长40米，宽25米。( )块这样的菜地面积是1公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "10",
+        "solution":  "【分析】先算一块菜地面积，再与1公顷=10000平方米比较.\n【详解】40×25=1000(平方米);10000÷1000=10(块)。",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "20-21 四年级上·江西吉安·期末"
+    },
+    {
+        "id":  "164",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个平角是( )°,一个周角的度数等于( )个平角度数之和。",
+        "options":  [
+
+                    ],
+        "answer":  "180;2",
+        "solution":  "【分析】平角180°,周角360°.\n【详解】360°=180°+180°,所以一个平角是180°,一个周角等于2个平角。",
+        "knowledge_points":  [
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·广西玉林·期末"
+    },
+    {
+        "id":  "165",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "只有一个端点，并且可无限延伸的是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "射线",
+        "solution":  "【分析】线段有两个端点，直线无端点，射线有一个端点且向一端无限延伸.\n【详解】应填\"射线\"。",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25四年级上·重庆丰都·期末"
+    },
+    {
+        "id":  "166",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "乐乐通过拉伸平行四边形的框架，发现平行四边形具有( )性(填\"稳定\"或\"不稳定\"),在实际生活中( )也是运用了这一特性。",
+        "options":  [
+
+                    ],
+        "answer":  "不稳定;伸缩门",
+        "solution":  "【分析】平行四边形容易变形，具有不稳定性，常见应用是伸缩门.\n【详解】第一空填\"不稳定\"，第二空可填\"伸缩门\"。",
+        "knowledge_points":  [
+                                 "平行四边形的不稳定性及应用"
+                             ],
+        "source":  "25-26四年级上·广东东莞·月考"
+    },
+    {
+        "id":  "167",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "根据42×14=588，直接写出下面算式的结果:42×140=( );( )×140=588000。",
+        "options":  [
+
+                    ],
+        "answer":  "5880;4200",
+        "solution":  "【分析】42×140是在42×14基础上把14乘10，积也乘10;第二空由5880到588000乘了100，所以42也乘100.\n【详解】42×140=5880;4200×140=588000。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "24-25 四年级上·北京朝阳·期末"
+    },
+    {
+        "id":  "168",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在横线上填上合适的单位：教室的面积约是50( );我国陆地领土面积约是960万( );数学书的封面面积约是4( );一棵大树高约15( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方米;平方千米;平方分米;米",
+        "solution":  "【分析】按常见量级选单位：教室面积用平方米，国土面积用平方千米，书封面面积用平方分米，树高用米.\n【详解】依次填：平方米，平方千米，平方分米，米。",
+        "knowledge_points":  [
+                                 "长度单位的选择",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 四年级上·甘肃兰州·期末"
+    },
+    {
+        "id":  "169",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "两个因数的积是360，如果一个因数不变，另一个因数除以4，积是( );如果一个因数乘10，另一个因数除以10，积是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "90;360",
+        "solution":  "【分析】一个因数不变，另一个除以几，积也除以几;一个乘10另一个除以10，积不变.\n【详解】360÷4=90;第二种情况积仍为360。",
+        "knowledge_points":  [
+                                 "几百几十或几千几百除以一位数的除法",
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "170",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "已知A×B=260，如果A,B同时扩大到原来的3倍，那么积是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "2340",
+        "solution":  "【分析】两个因数都乘3，积要乘3×3=9.\n【详解】260×9=2340，所以积是2340。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·湖南岳阳·期中"
+    },
+    {
+        "id":  "171",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个40°的角，透过放大5倍的放大镜看，这个角是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "40°",
+        "solution":  "【分析】角的大小与边长无关，只与两边张开程度有关.\n【详解】放大镜只放大图形，不改变角度，所以仍是40°。",
+        "knowledge_points":  [
+                                 "角的度量"
+                             ],
+        "source":  "2025·四川绵阳·小升初真题"
+    },
+    {
+        "id":  "172",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在数位顺序表中，从右起第七位是( )位，这个数位右面一位是( )位，左面一位是( )位。",
+        "options":  [
+
+                    ],
+        "answer":  "百万;十万;千万",
+        "solution":  "【分析】从右往左依次是个位，十位，百位，千位，万位，十万位，百万位，千万位。.\n【详解】第七位是百万位，右面一位是十万位，左面一位是千万位。",
+        "knowledge_points":  [
+                                 "整数的数级，数位和计数单位的认识"
+                             ],
+        "source":  "24-25 四年级上·河北衡水·期末"
+    },
+    {
+        "id":  "173",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "9时整，钟面上时针和分针形成的角是( )角，它的度数是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "直;90",
+        "solution":  "【分析】钟面12大格，每大格30°;9时整两针相差3大格.\n【详解】30×3=90°,是直角。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "25-26 四年级上·河南信阳·期中"
+    },
+    {
+        "id":  "174",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "依依家到公路有三条笔直的小路，长度分别是270米,210米,150米，其中有一条小路与公路互相垂直，这条小路的长度是( )米。",
+        "options":  [
+
+                    ],
+        "answer":  "150",
+        "solution":  "【分析】点到直线的垂线段最短.\n【详解】三条路中最短的是150米，所以垂直的小路长150米。",
+        "knowledge_points":  [
+                                 "垂直的特征",
+                                 "点到直线的距离"
+                             ],
+        "source":  "25-26 四年级上·辽宁抚顺·期末"
+    },
+    {
+        "id":  "175",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "直线是不可以测量的，但线段和射线都有端点，是可以测量出长度的。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】线段可测量长度;射线虽有一个端点，但向一端无限延伸，不能测量长度.\n【详解】题干把射线也说成可测量，错误，所以判断为*。",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "25-26 四年级上·黑龙江齐齐哈尔·期末"
+    },
+    {
+        "id":  "176",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "喀什地区占地面积是16.2万平方千米。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】计量较大区域面积通常用平方千米作单位.\n【详解】\"16.2万平方千米\"作为地区面积单位与量级都合理，判断为√。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 四年级上·新疆喀什·期末"
+    },
+    {
+        "id":  "177",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "60060000中两个6表示的意义不相同。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】不同数位上的同一数字表示的计数单位不同.\n【详解】一个6在千万位表示6个千万，另一个6在万位表示6个万，意义不同，判断为√。",
+        "knowledge_points":  [
+                                 "整数的数级，数位和计数单位的认识"
+                             ],
+        "source":  "24-25 四年级上·新疆克孜勒苏·期末"
+    },
+    {
+        "id":  "178",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "把一条10厘米长的线段向两端各延长5厘米，得到一条直线。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】直线无限长;线段向两端有限延长后仍是线段.\n【详解】只延长5厘米仍有两个端点，不是直线，判断为*。",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "25-26四年级上·河南南阳·期中"
+    },
+    {
+        "id":  "179",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.94,
+        "question":  "延长角的两边，角的度数就变大了。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】角的大小由两边张开程度决定，与边长无关.\n【详解】延长边只改变长度，不改变夹角大小，判断为*。",
+        "knowledge_points":  [
+                                 "角的概念及表示方法"
+                             ],
+        "source":  "25-26四年级上·新疆喀什·期末"
+    },
+    {
+        "id":  "180",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "直线比射线长。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】直线向两端无限延伸，射线向一端无限延伸，二者都无法度量长度.\n【详解】无限长不能比较长短，题干错误，判断为*。",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "2015·湖北武汉·小升初真题"
+    },
+    {
+        "id":  "181",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "小明画了一条长5米的直线。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】直线没有端点，无限长，不能用\"长5米\"描述.\n【详解】应是\"5米长的线段\"，所以该说法错误，判断为*。",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "25-26 四年级上·河南三门峡·期中"
+    },
+    {
+        "id":  "182",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "在线段，射线和直线中，只有线段有两个端点，且有固定的长度。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】线段有两个端点且有限长;射线一个端点无限长;直线无端点无限长.\n【详解】题干说法正确，判断为√。",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "25-26 四年级上·辽宁葫芦岛·期末"
+    },
+    {
+        "id":  "183",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "在900900900中,3个9表示的数一样大。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】三个9在不同数位，表示的数值不同.\n【详解】分别表示9亿,9十万,9百，不相等，判断为*。",
+        "knowledge_points":  [
+                                 "整数的数级，数位和计数单位的认识"
+                             ],
+        "source":  "25-26 四年级上·黑龙江齐齐哈尔·期末"
+    },
+    {
+        "id":  "184",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "在算盘上表示出5370000，需要拨出的算珠颗数是7。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】5370000中：百万位5需1颗上珠;十万位3需3颗下珠;万位7需1上2下共3颗，其余位0不拨.\n【详解】总颗数1+3+3=7，判断为√。",
+        "knowledge_points":  [
+                                 "算盘的认识与使用",
+                                 "亿以内数的组成"
+                             ],
+        "source":  "25-26四年级上·天津滨海新·期末"
+    },
+    {
+        "id":  "185",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "个位，十位，百位，千位，万位。亿位都是计数单位。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】\"个位，十位。.\"是数位;\"个，十，百，千。.\"才是计数单位.\n【详解】题干把数位当成计数单位，错误，判断为*。",
+        "knowledge_points":  [
+                                 "整数的数级，数位和计数单位的认识"
+                             ],
+        "source":  "25-26 四年级上·天津滨海新·期末"
+    },
+    {
+        "id":  "186",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "直角一定比锐角大，所以大于锐角的角一定是直角。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】大于锐角的角还可能是钝角，平角等.\n【详解】\"大于锐角\"不一定是直角，题干错误，判断为*。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "角的大小比较"
+                             ],
+        "source":  "24-25 四年级上·云南文山·期末"
+    },
+    {
+        "id":  "187",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "亿级上的全部数位有亿，十亿，百亿，千亿。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】亿，十亿，百亿，千亿是计数单位，不是数位名称.\n【详解】数位应为亿位，十亿位，百亿位，千亿位，所以题干错误，判断为*。",
+        "knowledge_points":  [
+                                 "整数的数级，数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级上·广东江门·期末"
+    },
+    {
+        "id":  "188",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "不相交的两条直线叫做平行线。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平行线定义需\"在同一平面内且不相交\".\n【详解】题干缺少\"同一平面内\"条件，表述不完整，判断为*。",
+        "knowledge_points":  [
+                                 "平行的特征及性质"
+                             ],
+        "source":  "25-26四年级上·辽宁铁岭·期末"
+    },
+    {
+        "id":  "189",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "在乘法里，两个因数都除以10，积也除以10。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】两个因数都除以10，积应除以100.\n【详解】例如20×30=600,(20÷10)×(30÷10)=2×3=6,600÷100=6，所以题干错误，判断为*。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "190",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.94,
+        "question":  "在同一个平面内，只可以画一条直线与已知直线平行。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】在同一平面内，与一条已知直线平行的直线有无数条.\n【详解】题干说\"只可以画一条\"错误，判断为*。",
+        "knowledge_points":  [
+                                 "平行的特征及性质"
+                             ],
+        "source":  "24-25 四年级上·天津蓟州·期末"
+    },
+    {
+        "id":  "191",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "过直线外一点画已知直线的平行线，可以画无数条。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平行公理：过直线外一点，有且只有一条直线与已知直线平行.\n【详解】不是无数条，所以判断为*。",
+        "knowledge_points":  [
+                                 "平行的特征及性质"
+                             ],
+        "source":  "24-25 四年级上·河南郑州·期末"
+    },
+    {
+        "id":  "192",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.75,
+        "question":  "两条平行线之间可以画无数条垂直线段，这些垂直线段的长度都相等。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】平行线间距离处处相等，从任意点作垂线段都可得到该距离.\n【详解】可作无数条，且长度都相等，判断为√。",
+        "knowledge_points":  [
+                                 "平行的特征及性质;垂直与平行的特征及性质"
+                             ],
+        "source":  "24-25 四年级上·山东济南·期末"
+    },
+    {
+        "id":  "193",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.78,
+        "question":  "在同一平面内，两条直线不是平行就是垂直。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】同一平面内两条直线还可能斜交(相交但不垂直).\n【详解】题干遗漏斜交情况，判断为*。",
+        "knowledge_points":  [
+                                 "平行的特征及性质;垂直与平行的特征及性质"
+                             ],
+        "source":  "24-25 四年级上·江苏南京·期末"
+    },
+    {
+        "id":  "194",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.94,
+        "question":  "小举动成就大环境。如果每台空调在制冷情况下由26℃调至27℃,每年可节约用电22千瓦时，那么一个社区的356台空调都采取这一措施，每年可以节约用电多少千瓦时。",
+        "options":  [
+
+                    ],
+        "answer":  "7832千瓦时",
+        "solution":  "【分析】每台每年节约22千瓦时,356台就是356个22.\n【详解】356×22=7832(千瓦时)。答：每年可以节约用电7832千瓦时。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "25-26 四年级上·湖南岳阳·期中"
+    },
+    {
+        "id":  "195",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "商店计划从水果市场批发58箱苹果，每箱苹果128元。商店准备8000元钱够吗。",
+        "options":  [
+
+                    ],
+        "answer":  "够",
+        "solution":  "【分析】先算总价，再与8000比较.\n【详解】128×58=7424(元),7424\u003c8000。答：准备8000元够用。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "25-26 四年级上·广东珠海·期中"
+    },
+    {
+        "id":  "196",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "小明家和学校相距840米，早上他从家到学校走了14分钟。放学按原路返回，步行速度比上学时每分钟多走10米，小明放学回家需要多少分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "12分钟",
+        "solution":  "【分析】先求上学速度，再求放学速度，最后用时间=路程/速度.\n【详解】840÷14=60(米/分);60+10=70(米/分);840÷70=12(分)。答：放学回家需要12分钟。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26四年级上·天津滨海新·期末"
+    },
+    {
+        "id":  "197",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "动物园里的一匹成年斑马每天要吃150千克青草。饲养员准备了4吨青草，这些青草够这匹斑马吃25天吗。",
+        "options":  [
+
+                    ],
+        "answer":  "够",
+        "solution":  "【分析】先求25天需要的青草，再把4吨换算成千克比较.\n【详解】150×25=3750(千克);4吨=4000千克;3750\u003c4000。答：这些青草够这匹斑马吃25天。",
+        "knowledge_points":  [
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "吨，千克之间的换算与比较"
+                             ],
+        "source":  "25-26 四年级上·天津滨海新·期末"
+    },
+    {
+        "id":  "198",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "童心小学开展爱心助学活动，计划采购钢笔捐赠给帮扶学校。钢笔每盒有6支，每支售价16元。学校准备了960元采购经费，一共可以购买多少盒钢笔。",
+        "options":  [
+
+                    ],
+        "answer":  "10盒",
+        "solution":  "【分析】先求每盒价格，再用总经费除以每盒价格.\n【详解】6×16=96(元/盒);960÷96=10(盒)。答：一共可以购买10盒钢笔。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "25-26四年级上·天津滨海新·期末"
+    },
+    {
+        "id":  "199",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "王大伯的桃园占地3公顷，为方便采摘，按每6平方米种一棵的标准规划种植，这个桃园一共能种多少棵桃树。",
+        "options":  [
+
+                    ],
+        "answer":  "5000棵",
+        "solution":  "【分析】先把公顷换算成平方米，再按\"总面积/每棵占地\"计算棵数.\n【详解】3公顷=30000平方米;30000÷6=5000(棵)。答：这个桃园一共能种5000棵桃树。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·天津滨海新·期末"
+    },
+    {
+        "id":  "200",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "光明小学组织192名少先队员参加研学实践活动。为了方便管理和开展任务，老师们将同学们平均编成8个大组，每个大组又分为4个合作小组。平均每个合作小组有多少名队员。",
+        "options":  [
+
+                    ],
+        "answer":  "6名",
+        "solution":  "【分析】先求每个大组人数，再求每个合作小组人数.\n【详解】192÷8=24(名);24÷4=6(名)。答：平均每个合作小组有6名队员。",
+        "knowledge_points":  [
+                                 "多位数与一位数的连除"
+                             ],
+        "source":  "25-26 四年级上·辽宁盘锦·期末"
+    },
+    {
+        "id":  "201",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "如果把两个角拼成一个角，下面说法不正确的是( )。",
+        "options":  [
+                        "A. 锐角+锐角 = 直角",
+                        "B. 锐角+直角 = 钝角",
+                        "C. 直角+直角 = 平角",
+                        "D. 平角+平角 = 周角"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】根据角的定义：锐角小于90°,直角等于90°,钝角大于90°且小于180°,平角等于180°,周角等于360°。据此逐项分析即可解题.\n【详解】A。两个锐角相加可能小于90°(仍是锐角),等于90°(直角)或大于90°小于180°(钝角),因此\"锐角+锐角 = 直角\"不一定成立;B。锐角加直角的和大于90°且小于180°,一定是钝角;C。直角加直角等于180°,是平角;D。平角加平角等于360°,是周角。故答案为A。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征，平角，周角的认识及特征，角的分类及换算"
+                             ],
+        "source":  "25-26四年级上·福建龙岩·期中"
+    },
+    {
+        "id":  "202",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "关于直线，射线，线段的描述正确的是( )。",
+        "options":  [
+                        "A. 直线最长，线段最短。",
+                        "B. 直线，射线及线段的长度都无法测量。",
+                        "C. 直线没有端点，射线有一个端点，线段有两个端点。"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】直线没有端点，两边可无限延长，没有长度可言;射线有一个端点，另一端可无限延长，没有长度可言;线段有两个端点，端点间的距离就是线段长度.\n【详解】A项错误，直线没有长度可言;B项错误，线段可以测量长度;C项正确。故答案为C。",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25四年级上·天津津南·期末"
+    },
+    {
+        "id":  "203",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "小明用计算器计算6500×23时，不小心把6500输成650，但他仍然算出了正确结果，他输入的算式可能是( )。",
+        "options":  [
+                        "A. 650×230",
+                        "B. 650×23",
+                        "C. 650×2300"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】两个数相乘，一个乘数除以几(0除外),另一个乘数乘相同的数，积不变.\n【详解】A项中6500变为650(÷10),23变为230(*10),积不变;B项仅一个乘数/10，积改变;C项一个乘数/10另一个*100，积改变。故答案为A。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·河南三门峡·期中"
+    },
+    {
+        "id":  "204",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "用一个5倍的放大镜看一个20°的角，这个角是( )°。",
+        "options":  [
+                        "A. 100",
+                        "B. 20",
+                        "C. 50",
+                        "D. 80"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】角的大小与两边张开程度有关，放大镜只会放大图形大小，不改变角的度数.\n【详解】20°的角通过放大镜观察后仍是20°。故答案为B。",
+        "knowledge_points":  [
+                                 "角的初步认识及辨认，角的大小比较"
+                             ],
+        "source":  "2025·广东揭阳·小升初模拟"
+    },
+    {
+        "id":  "205",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面角的度数不能用一副三角板拼出来的是( )。",
+        "options":  [
+                        "A. 75°",
+                        "B. 115°",
+                        "C. 120°"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】一副三角板常见角有30°,45°,60°,90°,可通过加减得到若干角度.\n【详解】75°可由45°+30°得到,120°可由90°+30°得到,115°无法由这些角度加减得到。故答案为B。",
+        "knowledge_points":  [
+                                 "角度的计算"
+                             ],
+        "source":  "25-26四年级上·吉林白城·期末"
+    },
+    {
+        "id":  "206",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.94,
+        "question":  "把一条15米长的线段向两端各延长20米，得到的是一条( )。",
+        "options":  [
+                        "A. 线段",
+                        "B. 射线",
+                        "C. 直线"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】线段有两个端点且长度可测;射线和直线长度不可测.\n【详解】向两端各延长20米后仍有确定长度，仍是线段。故答案为A。",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "25-26 四年级上·黑龙江齐齐哈尔·期末"
+    },
+    {
+        "id":  "207",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面不能用一副三角尺画出的是( )的角。",
+        "options":  [
+                        "A. 75°",
+                        "B. 15°",
+                        "C. 135°",
+                        "D. 100°"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】一副三角尺常用角度为30°,45°,60°,90°,能画出的角通常是这些角的和或差.\n【详解】75°=45°+30°,15°=45°-30°,135°=90°+45°,100°不能由上述角度加减得到。故答案为D。",
+        "knowledge_points":  [
+                                 "角度的计算"
+                             ],
+        "source":  "25-26四年级上·浙江杭州·期中"
+    },
+    {
+        "id":  "208",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "用一副三角尺不能拼出的角是( )。",
+        "options":  [
+                        "A. 170°",
+                        "B. 75°",
+                        "C. 120°",
+                        "D. 105°"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】三角尺可用角有30°,45°,60°,90°,通过相加或相减可拼出部分角.\n【详解】75°=30°+45°,120°=90°+30°,105°=60°+45°,170°无法由这些角相加减得到。故答案为A。",
+        "knowledge_points":  [
+                                 "角度的计算，用三角尺画角"
+                             ],
+        "source":  "25-26 四年级上·辽宁盘锦·期末"
+    },
+    {
+        "id":  "209",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "将一个钝角分成两个角，这两个角( )。",
+        "options":  [
+                        "A. 一定都是锐角",
+                        "B. 一个是直角一个是锐角",
+                        "C. 不确定大小"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】钝角大于90°且小于180°,分成两个角后可能出现多种情况.\n【详解】两个角可能都锐，也可能一个锐一个钝，或一个锐一个直，无法唯一确定。故答案为C。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征，角的分类及换算"
+                             ],
+        "source":  "25-26四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "210",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "两个因数的乘积是48，如果一个因数不变，另一个因数乘3，则积会( )。",
+        "options":  [
+                        "A. 乘3",
+                        "B. 不变",
+                        "C. 除以3"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】在乘法中，一个因数不变，另一个因数扩大几倍，积也扩大几倍.\n【详解】另一个因数乘3，则积也乘3。故答案为A。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26四年级上·甘肃临夏·期末"
+    },
+    {
+        "id":  "211",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "小辰家到万达商城有四条路，长度分别是480米,520米,450米,380米，其中有一条路与万达商城是垂直的。这条路的长度是( )。",
+        "options":  [
+                        "A. 480米",
+                        "B. 520米",
+                        "C. 450米",
+                        "D. 380米"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】从直线外一点到直线的所有线段中，垂线段最短.\n【详解】四条路中最短的是380米，所以垂直那条路长380米。故答案为D。",
+        "knowledge_points":  [
+                                 "1000以内数的比大小，垂直的特征"
+                             ],
+        "source":  "25-26 四年级上·甘肃临夏·期末"
+    },
+    {
+        "id":  "212",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.94,
+        "question":  "3164000000这个数上的\"6\"表示( )。",
+        "options":  [
+                        "A. 6个百万",
+                        "B. 6个亿",
+                        "C. 6个千万",
+                        "D. 6个十"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】根据整数数位顺序表判断\"6\"所在数位及对应计数单位.\n【详解】3164000000中的\"6\"在千万位，表示6个千万。故答案为C。",
+        "knowledge_points":  [
+                                 "整数的数级，数位和计数单位的认识"
+                             ],
+        "source":  "24-25 四年级上·重庆璧山·期末"
+    },
+    {
+        "id":  "213",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "文文在一张纸上画了两条直线，这两条直线和同一条直线平行，这两条直线( )。",
+        "options":  [
+                        "A. 互相平行",
+                        "B. 互相垂直",
+                        "C. 相交"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】同一平面内，两条直线都平行于第三条直线，则这两条直线互相平行.\n【详解】题中两条直线都和同一条直线平行，所以它们互相平行。故答案为A。",
+        "knowledge_points":  [
+                                 "平行的特征及性质"
+                             ],
+        "source":  "25-26四年级上·新疆喀什·期末"
+    },
+    {
+        "id":  "214",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面哪个度数的角不能用一副三角尺画出来。( )",
+        "options":  [
+                        "A. 15°",
+                        "B. 25°",
+                        "C. 150°"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】一副三角尺可用角度有30°,45°,60°,90°,可通过加减组合.\n【详解】15°=45°-30°,150°=90°+60°,25°无法由上述角度加减得到。故答案为B。",
+        "knowledge_points":  [
+                                 "角度的计算"
+                             ],
+        "source":  "25-26 四年级上·吉林白城·期末"
+    },
+    {
+        "id":  "215",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.94,
+        "question":  "一个九位数，它的最高位上的计数单位是( )。",
+        "options":  [
+                        "A. 十亿",
+                        "B. 亿",
+                        "C. 千万",
+                        "D. 百万"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】九位数最高位是第九位，即亿位;亿位对应计数单位是亿.\n【详解】一个九位数最高位上的计数单位是亿。故答案为B。",
+        "knowledge_points":  [
+                                 "整数的数级，数位和计数单位的认识"
+                             ],
+        "source":  "24-25 四年级上·重庆大足·期末"
+    },
+    {
+        "id":  "216",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "小明通过测量发现,100张纸摞在一起的高度大约是1厘米，那么一万张同样的纸摞在一起大约是( )。",
+        "options":  [
+                        "A. 10厘米",
+                        "B. 10分米",
+                        "C. 10米",
+                        "D. 100米"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】10000÷100=100，说明有100个\"1厘米\"，总高100厘米，再换算单位.\n【详解】100厘米=10分米。故答案为B。",
+        "knowledge_points":  [
+                                 "毫米，厘米，分米，米之间的进率与换算"
+                             ],
+        "source":  "25-26四年级上·内蒙古鄂尔多斯·期中"
+    },
+    {
+        "id":  "217",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面说法正确的是( )。",
+        "options":  [
+                        "A. 角的两条边越长，角就越大",
+                        "B. 大于90°的角是钝角",
+                        "C. 把圆分成360等份，每一份所对的角的大小是1°",
+                        "D. 直线，射线和线段三者比较长度，直线最长"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】分别根据角大小，钝角定义，度数定义和线段直线射线特征判断.\n【详解】A错(角大小与边长无关);B错(还可能是平角或周角);C对;D错(直线和射线无法度量长度)。故答案为C。",
+        "knowledge_points":  [
+                                 "角的分类及换算，角的概念及表示方法，线段，直线，射线的认识及特征，角的度量"
+                             ],
+        "source":  "25-26 四年级上·广东珠海·期中"
+    },
+    {
+        "id":  "218",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "用一副三角尺能拼出最大角是( )。",
+        "options":  [
+                        "A. 180°",
+                        "B. 150°",
+                        "C. 120°"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】三角尺中有30°,45°,60°,90°,可拼出若干角度，比较最大值.\n【详解】可拼出180°(如90°+90°),且为最大。故答案为A。",
+        "knowledge_points":  [
+                                 "用三角尺画角"
+                             ],
+        "source":  "25-26 四年级上·黑龙江绥化·期末"
+    },
+    {
+        "id":  "219",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面各数中的\"8\"，表示8个百万的是( )。",
+        "options":  [
+                        "A. 98300",
+                        "B. 57586001",
+                        "C. 3890012",
+                        "D. 8359753"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】看\"8\"所在数位，所在数位对应计数单位即其表示意义.\n【详解】A为千位,B为万位,C为十万位,D为百万位。故答案为D。",
+        "knowledge_points":  [
+                                 "整数的数级，数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级上·甘肃临夏·期末"
+    },
+    {
+        "id":  "220",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.94,
+        "question":  "在76000800中的\"7\"表示7个( )。",
+        "options":  [
+                        "A. 千万",
+                        "B. 百万",
+                        "C. 十万"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】根据数位确定\"7\"所在位置.\n【详解】76000800中\"7\"在千万位，表示7个千万。故答案为A。",
+        "knowledge_points":  [
+                                 "整数的数级，数位和计数单位的认识"
+                             ],
+        "source":  "24-25 四年级上·新疆阿克苏·期末"
+    },
+    {
+        "id":  "221",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "已知195×22=4290，那么195×11=( )。",
+        "options":  [
+                        "A. 1950",
+                        "B. 3900",
+                        "C. 2145",
+                        "D. 2035"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】一个因数不变，另一个因数减半，积也减半.\n【详解】11=22÷2，所以195×11=4290÷2=2145。故答案为C。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "222",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "a×b=720，要使积变为7200，下面不正确的是( )。",
+        "options":  [
+                        "A. a乘10,b不变",
+                        "B. a乘5,b乘2",
+                        "C. a乘10,b除以10"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】要从720变到7200即乘10，判断各选项对积的总倍数影响.\n【详解】A使积*10,B使积×(5×2)=*10,C使积*10再/10，积不变仍720。故不正确的是C。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级下·全国·课前预习"
+    },
+    {
+        "id":  "223",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "我们通常采用的画角方法有：量角器画角和三角尺拼角。下面的角中，只能用量角器来画的是( )。",
+        "options":  [
+                        "A. 75°",
+                        "B. 100°",
+                        "C. 105°",
+                        "D. 135°"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】先看角度是否可由三角尺常见角30°,45°,60°,90°组合得到.\n【详解】75°=30°+45°,105°=60°+45°,135°=90°+45°均可拼出;100°不能拼出，只能用量角器画。故答案为B。",
+        "knowledge_points":  [
+                                 "用量角器画角，用三角尺画角"
+                             ],
+        "source":  "24-25四年级上·重庆璧山·期末"
+    },
+    {
+        "id":  "224",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "两个数相乘，一个因数乘10，要使积乘30，另一个因数( )。",
+        "options":  [
+                        "A. 乘30",
+                        "B. 乘3",
+                        "C. 乘10"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】积变化倍数=两个因数变化倍数之积.\n【详解】已知一个因数乘10，要使积乘30，则另一个因数应乘3(10×3=30)。故答案为B。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·湖南娄底·期中"
+    },
+    {
+        "id":  "225",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "在条形统计图中，如果用1格表示5人，那么表示30人需要画( )格。",
+        "options":  [
+                        "A. 6",
+                        "B. 7",
+                        "C. 8"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】每格表示5人，总人数30人，格数=30÷5.\n【详解】30÷5=6(格)。故答案为A。",
+        "knowledge_points":  [
+                                 "1格表示多个单位的单式条形统计图"
+                             ],
+        "source":  "24-25 四年级上·新疆吐鲁番·期末"
+    },
+    {
+        "id":  "226",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "两个锐角不可能拼成( )。",
+        "options":  [
+                        "A. 钝角",
+                        "B. 直角",
+                        "C. 周角"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】两个锐角都小于90°,和一定小于180°,不可能达到360°.\n【详解】两个锐角可拼成直角或钝角，但不可能拼成周角。故答案为C。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征，平角，周角的认识及特征"
+                             ],
+        "source":  "25-26四年级上·甘肃临夏·期末"
+    },
+    {
+        "id":  "227",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.94,
+        "question":  "分针指向12，时针指向4，分针与时针所形成的较小夹角是( )。",
+        "options":  [
+                        "A. 直角",
+                        "B. 钝角",
+                        "C. 平角",
+                        "D. 周角"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】钟面每大格30°,从12到4有4个大格，夹角是4×30°=120°.\n【详解】120°大于90°小于180°,是钝角。故答案为B。",
+        "knowledge_points":  [
+                                 "角度的计算，整点时间与钟面指针的位置"
+                             ],
+        "source":  "25-26四年级上·河南驻马店·期中"
+    },
+    {
+        "id":  "228",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "用一副三角尺不能画出下面( )的角。",
+        "options":  [
+                        "A. 75°",
+                        "B. 25°",
+                        "C. 135°"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】检查角度是否能由30°,45°,60°,90°通过加减得到.\n【详解】75°=30°+45°,135°=90°+45°可画;25°不能由这些角度组合得到。故答案为B。",
+        "knowledge_points":  [
+                                 "角度的计算，用三角尺画角"
+                             ],
+        "source":  "25-26四年级上·辽宁铁岭·期末"
+    },
+    {
+        "id":  "229",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面的叙述中，错误的是( )。",
+        "options":  [
+                        "A. 经过两点只能画一条直线",
+                        "B. 一条射线长10米",
+                        "C. 小红画了一条长8厘米的线段"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】射线有一个端点，可向一端无限延长，不能度量长度;线段可度量.\n【详解】A正确,C正确;B错误，因为射线不能说\"长10米\"。故答案为B。",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "25-26四年级上·河南三门峡·期中"
+    },
+    {
+        "id":  "230",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一副三角板不能拼出( )的角。",
+        "options":  [
+                        "A. 75°",
+                        "B. 120°",
+                        "C. 100°"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】三角板角度有30°,45°,60°,90°,可通过组合拼出部分角.\n【详解】75°=30°+45°,120°=90°+30°可拼出;100°不能拼出。故答案为C。",
+        "knowledge_points":  [
+                                 "用三角尺画角"
+                             ],
+        "source":  "24-25四年级上·云南曲靖·期末"
+    },
+    {
+        "id":  "231",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.94,
+        "question":  "一个平角等于( )个直角。",
+        "options":  [
+                        "A. 2",
+                        "B. 3",
+                        "C. 4"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】平角180°,直角90°,用180°/90°求个数.\n【详解】180°/90°=2。故答案为A。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征，平角，周角的认识及特征，角的分类及换算"
+                             ],
+        "source":  "25-26 四年级上·安徽合肥·期中"
+    },
+    {
+        "id":  "232",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "用竖式计算143×25时,143与十位上的2相乘的积是( )。",
+        "options":  [
+                        "A. 268",
+                        "B. 2680",
+                        "C. 286",
+                        "D. 2860"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】25十位上的2表示2个十，即20，应计算143×20.\n【详解】143×20=2860。故答案为D。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "25-26四年级上·天津南开·期末"
+    },
+    {
+        "id":  "233",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面各数中，要读出两个零的是( )。",
+        "options":  [
+                        "A. 78000450",
+                        "B. 805407000",
+                        "C. 870040500",
+                        "D. 805070400"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】按整数读法：每级末尾0不读，其他连续0只读一个零，逐项读数判断.\n【详解】A读一个零;B读一个零;C读作\"八亿七千零四万零五百\"，读两个零;D读三个零。故答案为C。",
+        "knowledge_points":  [
+                                 "亿以内数的读，写法，亿以上数的读，写法"
+                             ],
+        "source":  "25-26 四年级上·天津南开·期末"
+    },
+    {
+        "id":  "234",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "海海在绘制条形统计图时，用宽0.5cm,高3cm的直条表示化肥产量1200t,则用宽0.5cm,高4.5cm的直条表示的化肥产量是( )t。",
+        "options":  [
+
+                    ],
+        "answer":  "1800",
+        "solution":  "【分析】直条宽度相同，高度倍数就是产量倍数。先求4.5是3的几倍，再用1200乘这个倍数.\n【详解】4.5÷3=1.5,1200×1.5=1800(t)。所以填1800。",
+        "knowledge_points":  [
+                                 "1格表示多个单位的单式条形统计图"
+                             ],
+        "source":  "25-26 六年级·全国·随堂练习"
+    },
+    {
+        "id":  "235",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "钟面上3时30分，时针和分针所夹的较小角是( )°。时针和分针所夹的较小角是60°时，是( )时。",
+        "options":  [
+
+                    ],
+        "answer":  "75;2或10",
+        "solution":  "【分析】钟面12个大格，每大格30°。3时30分时分针指向6，时针在3和4中间，较小夹角为2.5个大格。题干分析图:![](images/25e6371f80483e1abe42743e7c491b71de695a78cbec08ace449c990aada4028.jpg).\n【详解】2.5×30°=75°。夹角60°对应2个大格，整点时可能是2时或10时。",
+        "knowledge_points":  [
+                                 "角度的计算，整点时间与钟面指针的位置，半点时间与钟面指针的位置"
+                             ],
+        "source":  "25-26四年级上·浙江宁波·期中"
+    },
+    {
+        "id":  "236",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "将一张圆形纸对折3次，得到( )°的角，两个这样的角拼成( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "45;直角",
+        "solution":  "【分析】圆周角是360°,每对折一次角度减半，连续对折3次即连续除以2三次.\n【详解】360°/2÷2÷2=45°;45°+45°=90°,是直角。",
+        "knowledge_points":  [
+                                 "角度的计算，直角，钝角，锐角的认识及特征"
+                             ],
+        "source":  "25-26四年级上·河南驻马店·期中"
+    },
+    {
+        "id":  "237",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "已知a×b=100，如果a乘4，则积是( );如果b除以2，则积是( );如果a乘4,b除以2，则积是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "400;50;200",
+        "solution":  "【分析】一个因数乘或除以几(0除外),积也相应乘或除以几;两个因数同时变化时，按倍数连乘处理.\n【详解】a乘4:100×4=400;b除以2:100÷2=50;同时变化:100×4÷2=200。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "238",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "A与B的积是36，如果A扩大到原来的2倍,B扩大到原来的8倍，积是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "576",
+        "solution":  "【分析】两个因数分别扩大2倍和8倍，积扩大2×8=16倍.\n【详解】36×16=576。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·山东济宁·期中"
+    },
+    {
+        "id":  "239",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "11时整，分针和时针构成的角中，最小的角是( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "锐角",
+        "solution":  "【分析】钟面每大格30°,11时整时针在11，分针在12，相差1大格.\n【详解】最小角为30°,30°是锐角。",
+        "knowledge_points":  [
+                                 "角度的计算，直角，钝角，锐角的认识及特征，整点时间与钟面指针的位置"
+                             ],
+        "source":  "25-26 四年级上·黑龙江绥化·期末"
+    },
+    {
+        "id":  "240",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "3时整时，钟面上时针与分针形成的角是( )°;3时30分时，时针与分针形成( )角;6时整时，钟面上时针与分针形成的角是( )°,是( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "90;锐角;180;平角",
+        "solution":  "【分析】钟面12等分，每份30°。3时是3个大格,6时是6个大格;3时30分时较小夹角小于90°.\n【详解】3时:3×30°=90°;3时30分是锐角;6时:6×30°=180°,是平角。",
+        "knowledge_points":  [
+                                 "角度的计算，直角，钝角，锐角的认识及特征，平角，周角的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·江西上饶·期末"
+    },
+    {
+        "id":  "241",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "画一个60°的角的步骤。(1)画出一条射线，使量角器的中心和射线的端点重合，并使( )和射线重合。(2)在量角器( )刻度线的地方点一个点。(3)从射线的端点出发，通过刚才标记的点，再画一条( )。",
+        "options":  [
+
+                    ],
+        "answer":  "0刻度线;60°;射线",
+        "solution":  "【分析】量角器画角三步：对中心，对零线，找刻度，连射线.\n【详解】依次填:0刻度线,60°,射线。",
+        "knowledge_points":  [
+                                 "用量角器画角"
+                             ],
+        "source":  "25-26四年级上·安徽合肥·期中"
+    },
+    {
+        "id":  "242",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在数位顺序表中，从个位起第( )位是万位，与千万位相邻的数位是( )和( )。",
+        "options":  [
+
+                    ],
+        "answer":  "五;百万位;亿位",
+        "solution":  "【分析】数位从右到左依次为个，十，百，千，万，十万，百万，千万，亿。.\n【详解】万位是第5位;千万位相邻的是百万位和亿位。",
+        "knowledge_points":  [
+                                 "整数的数级，数位和计数单位的认识"
+                             ],
+        "source":  "25-26 四年级上·广东珠海·期中"
+    },
+    {
+        "id":  "243",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "两个数的积是72，如果一个因数不变，另一个因数除以3，那么所得的结果是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "24",
+        "solution":  "【分析】一个因数不变，另一个因数除以3，积也除以3.\n【详解】72÷3=24。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·河南三门峡·期中"
+    },
+    {
+        "id":  "244",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "已知A×B=260，如果A扩大到原来的3倍,B不变，那么积是( );如果A,B同时扩大到原来的2倍，那么积是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "780;1040",
+        "solution":  "【分析】一个因数扩大3倍，积扩大3倍;两个因数都扩大2倍，积扩大4倍.\n【详解】260×3=780;260×4=1040。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26四年级上·新疆喀什·期末"
+    },
+    {
+        "id":  "245",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "用10倍的放大镜看一个17°的角，看到的角的度数是( )°,它比直角小( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "17;73",
+        "solution":  "【分析】放大镜不改变角度大小;直角是90°.\n【详解】看到仍是17°;90°-17°=73°。",
+        "knowledge_points":  [
+                                 "角度的计算，直角，钝角，锐角的认识及特征"
+                             ],
+        "source":  "25-26四年级上·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "246",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "∠1与25°的角组成一个直角,∠1=( ),它是( )角。∠2与75°的角组成一个平角,∠2=( ),它是( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "65°;锐;105°;钝",
+        "solution":  "【分析】直角90°,平角180°;用减法求未知角，再按角的分类判断.\n【详解】∠1=90°-25°=65°,是锐角;∠2=180°-75°=105°,是钝角。",
+        "knowledge_points":  [
+                                 "角度的计算，直角，钝角，锐角的认识及特征，平角，周角的认识及特征，角的分类及换算"
+                             ],
+        "source":  "25-26四年级上·湖北咸宁·期中"
+    },
+    {
+        "id":  "247",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "\"草长莺飞二月天，拂堤杨柳醉春烟。儿童散学归来早，忙趁东风放纸鸢.\"诗中的纸鸢也就是风筝。欢欢和乐乐参加放风筝比赛，她们的风筝线都是30米长，且都放完了。欢欢的风筝线与地面的夹角是30°,乐乐的风筝线与地面的夹角是75°,( )的风筝飞得高。",
+        "options":  [
+
+                    ],
+        "answer":  "乐乐",
+        "solution":  "【分析】当风筝线长度相等时，与地面夹角越大，风筝高度越高。题中示意图:![](images/aff8f5d4b5e4df66c424a80b24433b1a9c5a8cf69a256ef8ea651138b3b00fdd.jpg).\n【详解】30°\u003c75°,所以乐乐的风筝飞得更高。",
+        "knowledge_points":  [
+                                 "角的大小比较"
+                             ],
+        "source":  "25-26四年级上·河南南阳·期中"
+    },
+    {
+        "id":  "248",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "10个百万是( );( )个千万是一亿。",
+        "options":  [
+
+                    ],
+        "answer":  "一千万;10",
+        "solution":  "【分析】相邻计数单位进率是10.\n【详解】10个百万=一千万;10个千万=一亿。",
+        "knowledge_points":  [
+                                 "整数的数级，数位和计数单位的认识"
+                             ],
+        "source":  "25-26 四年级上·河北张家口·期中"
+    },
+    {
+        "id":  "249",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "两个数相乘的积是240，将其中一个因数扩大到原来的5倍，积是( ),若在这时将另一个因数除以10，积是( )。这个积与原来的积相比，变( )了。",
+        "options":  [
+
+                    ],
+        "answer":  "1200;120;小",
+        "solution":  "【分析】先把积乘5，再除以10，最后与原积比较大小.\n【详解】240×5=1200;1200÷10=120;120\u003c240，所以变小了。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "250",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把一根木料锯成3段需要6分钟，照这样计算，锯成7段需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "18",
+        "solution":  "【分析】锯成n段需锯n-1次。先求每锯1次用时，再求锯6次用时.\n【详解】3段需锯2次,6÷2=3(分钟/次);7段需锯6次,6×3=18(分钟)。",
+        "knowledge_points":  [
+                                 "间隔问题"
+                             ],
+        "source":  "24-25四年级上·河北保定·期末"
+    },
+    {
+        "id":  "251",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个直角是( )度，一个平角是( )度,1个周角=( )个平角=( )个直角。",
+        "options":  [
+
+                    ],
+        "answer":  "90;180;2;4",
+        "solution":  "【分析】直角90°,平角180°,周角360°.\n【详解】360°/180°=2,360°/90°=4，所以填90,180,2,4。",
+        "knowledge_points":  [
+                                 "几百几十或几千几百除以一位数的除法，直角，钝角，锐角的认识及特征，平角，周角的认识及特征，角的分类及换算"
+                             ],
+        "source":  "25-26 四年级上·内蒙古兴安盟·期末"
+    },
+    {
+        "id":  "252",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "甲，乙，丙三人同时到银行办理业务，甲需要12分钟，乙需要7分钟，丙需要5分钟。银行中午只有1人值班，要使三人等待时间和最少，应先给( )办理业务，再给( )办理业务，最后给( )办理业务。",
+        "options":  [
+
+                    ],
+        "answer":  "丙;乙;甲",
+        "solution":  "【分析】要使总等待时间最少，应按办理时长从短到长排队.\n【详解】5\u003c7\u003c12，所以顺序为丙→乙→甲。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "25-26四年级上·甘肃陇南·期末"
+    },
+    {
+        "id":  "253",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "钟表上，从下午3时到5时，时针旋转形成的角是( )°。( )时整，时针和分针的夹角是180°。",
+        "options":  [
+
+                    ],
+        "answer":  "60;6",
+        "solution":  "【分析】钟面每大格30°。3时到5时走2格;180°对应6格.\n【详解】2×30°=60°;180°/30°=6，所以是6时整。",
+        "knowledge_points":  [
+                                 "角度的计算，整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25四年级上·四川绵阳·期末"
+    },
+    {
+        "id":  "254",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "学校礼堂按照3盆月季花,4盆牡丹花和5盆兰花的顺序一共摆放了111盆花，最后一盆是( )花。",
+        "options":  [
+
+                    ],
+        "answer":  "月季",
+        "solution":  "【分析】一个周期共3+4+5=12盆，求111除以12的余数定位到周期内位置.\n【详解】111÷12=9。3，余3对应周期中的第3盆，仍在\"月季\"段内，所以最后一盆是月季花。",
+        "knowledge_points":  [
+                                 "有余数除法的实际应用，周期问题"
+                             ],
+        "source":  "25-26四年级上·天津蓟州·期末"
+    },
+    {
+        "id":  "255",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "比一比\"直角，周角，钝角，平角，锐角\"的大小。( )\u003e( )\u003e( )\u003e( )\u003e( )",
+        "options":  [
+
+                    ],
+        "answer":  "周角;平角;钝角;直角;锐角",
+        "solution":  "【分析】按角度范围比较：周角360°,平角180°,钝角在90°到180°之间，直角90°,锐角小于90°.\n【详解】周角\u003e平角\u003e钝角\u003e直角\u003e锐角。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征，角的大小比较，平角，周角的认识及特征"
+                             ],
+        "source":  "25-26 四年级上·甘肃张掖·期末"
+    },
+    {
+        "id":  "256",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一道除法算式的商是11，余数是28，除数最小是( ),这时被除数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "29;347",
+        "solution":  "【分析】余数必须小于除数，所以最小除数是29。被除数=除数×商+余数.\n【详解】最小除数:28+1=29;被除数:29×11+28=347。",
+        "knowledge_points":  [
+                                 "两位数与一位数的加法(不进位),余数与除数的关系，有余数的除法验算，乘加，乘减混合运算"
+                             ],
+        "source":  "25-26四年级上·辽宁盘锦·期末"
+    },
+    {
+        "id":  "257",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.7,
+        "question":  "一道除法算式的商是23，余数是16，除数最小是( ),这时被除数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "17;407",
+        "solution":  "【分析】有余数除法中，除数\u003e余数;被除数=除数×商+余数.\n【详解】最小除数16+1=17;被除数17×23+16=407。",
+        "knowledge_points":  [
+                                 "余数与除数的关系，有余数的除法验算"
+                             ],
+        "source":  "24-25 四年级上·湖北武汉·期末"
+    },
+    {
+        "id":  "258",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据15×17=255，填出下面算式的得数。15×170=( ),15×34=( ),150×17=( )。",
+        "options":  [
+
+                    ],
+        "answer":  "2550;510;2550",
+        "solution":  "【分析】利用积的变化规律进行倍数变换.\n【详解】15×170=15×(17×10)=255×10=2550;15×34=15×(17×2)=255×2=510;150×17=(15×10)×17=255×10=2550。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "24-25 四年级上·河北衡水·期末"
+    },
+    {
+        "id":  "259",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "两个因数的积是450，其中一个因数乘3，另一个因数不变，积是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "1350",
+        "solution":  "【分析】一个因数乘3，积也乘3.\n【详解】450×3=1350。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "24-25 四年级上·云南玉溪·期末"
+    },
+    {
+        "id":  "260",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "钟面上9时整，时针与分针的夹角是( )°,是( )角，此时钟面上的时针与分针互相( )。",
+        "options":  [
+
+                    ],
+        "answer":  "90;直角;垂直",
+        "solution":  "【分析】钟面每大格30°,9时整时分针在12，时针在9，相差3格.\n【详解】3×30°=90°,是直角;两针互相垂直。",
+        "knowledge_points":  [
+                                 "垂直的特征，直角，钝角，锐角的认识及特征，平角，周角的认识及特征，整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·江西宜春·期末"
+    },
+    {
+        "id":  "261",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把一张圆形纸片先对折一次，再对折一次，打开后可以得到( )个大小完全一样的角，每个角都是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "4;90",
+        "solution":  "【分析】圆先对折成2份，再对折成4份，圆周角360°平均分成4份.\n【详解】得到4个相同角，每个角是360°/4=90°。",
+        "knowledge_points":  [
+                                 "图形的折叠问题，直角，钝角，锐角的认识及特征"
+                             ],
+        "source":  "25-26 四年级上·广西贵港·期中"
+    },
+    {
+        "id":  "262",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个乘法算式的积是60，将其中一个因数乘12，另一个因数除以2，得到的积是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "360",
+        "solution":  "【分析】新积=原积*12÷2.\n【详解】60×12÷2=360。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26四年级上·北京丰台·期末"
+    },
+    {
+        "id":  "263",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "从个位起，第( )位是万位;一个十位数，它的最高位是( )位。",
+        "options":  [
+
+                    ],
+        "answer":  "五;十亿位",
+        "solution":  "【分析】数位顺序从右往左：个，十，百，千，万。第5位是万位;十位数最高位是第10位.\n【详解】第一空填五;第二空填十亿。",
+        "knowledge_points":  [
+                                 "整数的数级，数位和计数单位的认识"
+                             ],
+        "source":  "24-25 四年级上·重庆璧山·期末"
+    },
+    {
+        "id":  "264",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果∠1和∠2组成一个平角,∠1是∠2的3倍，那么∠2=( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "45",
+        "solution":  "【分析】平角180°,设∠2=x,则∠1=3x,3x+x=180°.\n【详解】4x=180°,x=45°,所以∠2=45°。",
+        "knowledge_points":  [
+                                 "角度的计算，平角，周角的认识及特征，角的分类及换算"
+                             ],
+        "source":  "25-26 四年级上·内蒙古鄂尔多斯·期中"
+    },
+    {
+        "id":  "265",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用放大100倍的放大镜看一个45°的角，看到的角的度数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "45°",
+        "solution":  "【分析】角的大小由两边张开程度决定，与边长无关.\n【详解】放大镜只改变视觉大小，不改变角度，仍是45°。",
+        "knowledge_points":  [
+                                 "角的度量"
+                             ],
+        "source":  "25-26 四年级上·吉林延边·期末"
+    },
+    {
+        "id":  "266",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "1个周角等于( )个直角;两个锐角度数和是130°,那么其中较大的角一定小于( )°,较小的角一定大于( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "4;90;40",
+        "solution":  "【分析】周角360°,直角90°;锐角都小于90°。和为130°时，若较大角接近90°,较小角就超过40°.\n【详解】360°/90°=4;较大角一定\u003c90°;较小角=130°-较大角\u003e130°-90°=40°。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征，平角，周角的认识及特征"
+                             ],
+        "source":  "25-26 四年级上·辽宁鞍山·期末"
+    },
+    {
+        "id":  "267",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个钟面上9时整，这时分针和时针成( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "直角",
+        "solution":  "【分析】9时整时，分针在12，时针在9，相差3个大格.\n【详解】3×30°=90°,是直角。",
+        "knowledge_points":  [
+                                 "角度的计算，直角，钝角，锐角的认识及特征"
+                             ],
+        "source":  "25-26四年级上·吉林·期末"
+    },
+    {
+        "id":  "268",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "比直角小20°的角是( ),是( )角。(填\"锐\"或\"钝\")",
+        "options":  [
+
+                    ],
+        "answer":  "70°;锐",
+        "solution":  "【分析】直角90°,比它小20°用减法.\n【详解】90°-20°=70°;70°\u003c90°,所以是锐角。",
+        "knowledge_points":  [
+                                 "角度的计算，直角，钝角，锐角的认识及特征，角的大小比较"
+                             ],
+        "source":  "25-26 四年级上·辽宁铁岭·期末"
+    },
+    {
+        "id":  "269",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在62°,180°,90°,35°,135°,360°,178°这些角中,( )是锐角,( )是直角,( )是钝角,( )是平角,( )是周角。",
+        "options":  [
+
+                    ],
+        "answer":  "62°,35°;90°;135°,178°;180°;360°",
+        "solution":  "【分析】锐角:0°到90°之间;直角:90°;钝角:90°到180°之间;平角:180°;周角:360°.\n【详解】锐角是62°,35°;直角是90°;钝角是135°,178°;平角是180°;周角是360°。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征，平角，周角的认识及特征，角的分类及换算"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "270",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把一条5厘米长的线段向两端各延伸10000米，得到的是一条( ),再接着把一端无限延长，就得到一条( )。",
+        "options":  [
+
+                    ],
+        "answer":  "线段;射线",
+        "solution":  "【分析】线段有两个端点且长度可测;射线有一个端点并向一端无限延伸.\n【详解】先向两端各延伸10000米后仍是有限长度，所以是线段;再把一端无限延长，就成为射线。",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "25-26四年级上·湖南常德·期中"
+    },
+    {
+        "id":  "271",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "如果∠1+直角+45°=平角，那么∠1=( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "45",
+        "solution":  "【分析】直角90°,平角180°,未知角=180°-90°-45°.\n【详解】∠1=45°。",
+        "knowledge_points":  [
+                                 "角度的计算，直角，钝角，锐角的认识及特征，平角，周角的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·河北衡水·期末"
+    },
+    {
+        "id":  "272",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把一张长方形纸对折两次，折痕互相( );对折三次，折痕互相( )。",
+        "options":  [
+
+                    ],
+        "answer":  "垂直或平行;垂直或平行",
+        "solution":  "【分析】折叠方向相同会得到平行折痕，不同方向会得到垂直折痕.\n【详解】两次或三次对折都可能出现\"垂直或平行\"。",
+        "knowledge_points":  [
+                                 "垂直的特征，平行的特征及性质"
+                             ],
+        "source":  "25-26 四年级上·新疆和田·期末"
+    },
+    {
+        "id":  "273",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上合适的面积单位。我们的校园占地3( ),校园里有两间机房，每间面积是96( )。信息技术课上，我坐在屏幕面积约1046( )的计算机前，通过网络尽情浏览\"南海明珠--海南岛\"约3.39万( )的迷人风光，心中充满了自豪与向往。",
+        "options":  [
+
+                    ],
+        "answer":  "公顷;平方米;平方厘米;平方千米",
+        "solution":  "【分析】校园占地常用公顷，机房面积常用平方米，屏幕面积常用平方厘米，省域面积常用平方千米.\n【详解】依次填：公顷，平方米，平方厘米，平方千米。",
+        "knowledge_points":  [
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 五年级上·海南省直辖县级单位·期末"
+    },
+    {
+        "id":  "274",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个角是91度，它是( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "钝角",
+        "solution":  "【分析】钝角大于90°且小于180°.\n【详解】91°满足范围，所以是钝角。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征，角的大小比较"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "275",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在同一平面内，两条直线a,b相交成( )角，就说这两条直线互相垂直，可以记作( )。",
+        "options":  [
+
+                    ],
+        "answer":  "直角;a⊥b",
+        "solution":  "【分析】两条直线相交成90°时互相垂直，记号用\"⊥\".\n【详解】应填\"直角\";记作\"a⊥b\"。",
+        "knowledge_points":  [
+                                 "垂直的特征，直角，钝角，锐角的认识及特征"
+                             ],
+        "source":  "25-26四年级上·甘肃临夏·期末"
+    },
+    {
+        "id":  "276",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在( )里填上适当的面积单位。(1)学校操场的面积约是7000( )。(2)天安门广场的占地面积约是44( )。(3)东莞市的陆地总面积是2460( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方米;公顷;平方千米",
+        "solution":  "【分析】操场面积常用平方米，广场占地常用公顷，城市面积常用平方千米.\n【详解】依次填：平方米，公顷，平方千米。",
+        "knowledge_points":  [
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26四年级上·广东东莞·月考"
+    },
+    {
+        "id":  "277",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "当钟表上显示时间为5时整，时针与分针的较小夹角是( )角;时针与分针的夹角为直角时，可能是( )时整。",
+        "options":  [
+
+                    ],
+        "answer":  "钝;3或9",
+        "solution":  "【分析】每大格30°。5时整夹角为5×30°=150°,是钝角;直角90°对应3个大格.\n【详解】5时整是钝角;直角整点可能是3时或9时。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征，平角，周角的认识及特征，角的分类及换算"
+                             ],
+        "source":  "25-26四年级上·黑龙江齐齐哈尔·期末"
+    },
+    {
+        "id":  "278",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个周角=( )个平角=( )个直角=( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "2;4;360",
+        "solution":  "【分析】周角360°,平角180°,直角90°.\n【详解】360°=2个平角=4个直角。",
+        "knowledge_points":  [
+                                 "平角，周角的认识及特征，角的分类及换算"
+                             ],
+        "source":  "25-26 四年级上·广东东莞·月考"
+    },
+    {
+        "id":  "279",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "按规律填空。21×9=189;321×9=2889;4321×9=38889;54321×9=( );( )×9=8888888889。",
+        "options":  [
+
+                    ],
+        "answer":  "488889;987654321",
+        "solution":  "【分析】乘数从高位到低位依次递减1，乘9后结果首位减1，中间连续8，末位9.\n【详解】54321×9=488889;对应九位递减数是987654321。",
+        "knowledge_points":  [
+                                 "两，三位数与一位数连续进位的乘法，算式的规律(整数)"
+                             ],
+        "source":  "24-25 四年级上·北京朝阳·期末"
+    },
+    {
+        "id":  "280",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "线段有( )个端点，直线( )端点，射线只有( )个端点。",
+        "options":  [
+
+                    ],
+        "answer":  "2;无;1",
+        "solution":  "【分析】线段两端固定，直线两端无限延伸，射线一端固定一端无限延伸.\n【详解】线段2个端点，直线无端点，射线1个端点。",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "25-26 四年级上·安徽合肥·期中"
+    },
+    {
+        "id":  "281",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "声音在空气中传播的速度是340米/秒，打雷后经过了19秒被A城居民听到，比B城居民听到雷声早7秒,B城与打雷地点的距离是( )米。",
+        "options":  [
+
+                    ],
+        "answer":  "8840",
+        "solution":  "【分析】B城听到时间=19+7=26秒，路程=速度×时间.\n【详解】340×26=8840(米)。",
+        "knowledge_points":  [
+                                 "三位数乘两位数，乘数末尾有0，基础行程问题"
+                             ],
+        "source":  "25-26四年级上·天津南开·期末"
+    },
+    {
+        "id":  "282",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果4□8÷48的商是一位数，那么□里最大填( );如果商是两位数,□里有( )种填法。",
+        "options":  [
+
+                    ],
+        "answer":  "7;2",
+        "solution":  "【分析】三位数除两位数，比较前两位4□与48：小于48则商一位数，大于等于48则商两位数.\n【详解】商一位数时4□\u003c48,□最大7;商两位数时4□可为48或49，共2种。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法，判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "25-26四年级上·天津南开·期末"
+    },
+    {
+        "id":  "283",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "文具店对精装笔记本开展促销活动，原价每本12元，现在\"买3本送1本\"。李老师用144元最多能买到( )本精装笔记本。",
+        "options":  [
+
+                    ],
+        "answer":  "16",
+        "solution":  "【分析】先算可付费买到的本数，再按每3本送1本计算赠送本数.\n【详解】144÷12=12(本),12÷3=4(本赠送),共12+4=16(本)。",
+        "knowledge_points":  [
+                                 "经济问题，除数是两位数的笔算除法"
+                             ],
+        "source":  "25-26四年级上·天津南开·期末"
+    },
+    {
+        "id":  "284",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "阳光农业生态观光园的场地呈长方形，长约500米，宽约240米，它的占地面积大约是( )公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "12",
+        "solution":  "【分析】长方形面积=长×宽，再把平方米换算为公顷(10000平方米=1公顷).\n【详解】500×240=120000(平方米)=12(公顷)。",
+        "knowledge_points":  [
+                                 "三位数乘两位数，乘数末尾有0，长方形的面积，公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·天津南开·期末"
+    },
+    {
+        "id":  "285",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\"\u003e\"\"\u003c\"或\"=\"。8亿( )88808800;360÷15( )360÷12;84×300( )420×60;1平方千米-50公顷( )5000平方米。",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;\u003c;=;\u003e",
+        "solution":  "【分析】分别用大数比较，同被除数商大小规律，积变化规律，面积单位换算判断.\n【详解】8亿=800000000\u003e88808800;15\u003e12所以360÷15\u003c360÷12;84×300=420×60;1平方千米-50公顷=50公顷=500000平方米\u003e5000平方米。",
+        "knowledge_points":  [
+                                 "大数的比较，除数是两位数的笔算除法，面积单位间的进率及换算，积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·天津南开·期末"
+    },
+    {
+        "id":  "286",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.94,
+        "question":  "用放大2倍的放大镜看45°的角仍是45°。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】角的大小由两边张开程度决定，与边长无关.\n【详解】放大镜只改变边长，不改变角度，所以仍是45°,说法正确。",
+        "knowledge_points":  [
+                                 "角的大小比较"
+                             ],
+        "source":  "25-26四年级上·湖北咸宁·期中"
+    },
+    {
+        "id":  "287",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "两数相乘，一个因数乘一个数，另一个因数除以相同的数，积不变。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】该规律成立需\"这个数不为0且可作除数\".\n【详解】题干未限定非0，若乘0则另一个因数需除以0(不成立),所以该说法错误。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·江西宜春·期中"
+    },
+    {
+        "id":  "288",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "两个锐角的和一定比直角大。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】两个锐角和可能小于，等于或大于90°.\n【详解】例如30°+15°=45°,小于直角，所以\"一定比直角大\"错误。",
+        "knowledge_points":  [
+                                 "100以内数的比大小，两位数与整十数的加法，直角，钝角，锐角的认识及特征"
+                             ],
+        "source":  "15-16 四年级上·全国·月考"
+    },
+    {
+        "id":  "289",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.94,
+        "question":  "线段和射线都没有直线长。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】线段有限长;射线和直线都可无限延伸.\n【详解】线段比直线短，但射线不比直线短，题干把两者并列为\"都没有直线长\"错误。",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "25-26四年级上·河南驻马店·期中"
+    },
+    {
+        "id":  "290",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.94,
+        "question":  "用一个能放大5倍的放大镜看一个25°的角，所看到的角是125°。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】放大镜不改变角度大小.\n【详解】25°经放大镜观察仍是25°,不是125°,说法错误。",
+        "knowledge_points":  [
+                                 "角的概念及表示方法"
+                             ],
+        "source":  "24-25 四年级上·广西玉林·期末"
+    },
+    {
+        "id":  "291",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "乘法算式中，一个因数扩大到原来的2倍，另一个因数扩大到原来的3倍，积扩大到原来的5倍。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】两个因数分别扩大a倍,b倍，积扩大a×b倍.\n【详解】应扩大2×3=6倍，不是5倍，错误。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "24-25四年级上·江西萍乡·期末"
+    },
+    {
+        "id":  "292",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "地球与月球间的平均距离约为384000千米。384000中的\"8\"在算盘上表示出来，需要拨4颗珠子。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】\"8\"在万位，算盘上8=1颗上珠(5)+3颗下珠(3).\n【详解】共需4颗珠子，题干正确。",
+        "knowledge_points":  [
+                                 "算盘的认识与使用，整数的数级，数位和计数单位的认识"
+                             ],
+        "source":  "25-26四年级上·陕西渭南·期末"
+    },
+    {
+        "id":  "293",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "在同一平面内两条直线不是相交就是垂直。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】同一平面内两直线关系是\"相交或平行\"，垂直属于相交的特殊情况.\n【详解】题干漏掉\"平行\"，错误。",
+        "knowledge_points":  [
+                                 "垂直的特征，平行的特征及性质"
+                             ],
+        "source":  "25-26 四年级上·甘肃陇南·期末"
+    },
+    {
+        "id":  "294",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.94,
+        "question":  "过一点可以画无数条已知直线的垂线。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】过定点作已知直线的垂线方向唯一.\n【详解】无论点在直线上还是直线外，都只能作一条垂线，所以错误。",
+        "knowledge_points":  [
+                                 "画垂线"
+                             ],
+        "source":  "24-25 四年级上·新疆克孜勒苏·期末"
+    },
+    {
+        "id":  "295",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "条形统计图可以通过直条的高低或长短表示数量的多少。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】条形统计图核心就是用条形长度(高度)反映数据大小.\n【详解】直条越长(高)数量越多，题干正确。",
+        "knowledge_points":  [
+                                 "1格表示一个单位的单式条形统计图"
+                             ],
+        "source":  "24-25 四年级上·重庆大足·期末"
+    },
+    {
+        "id":  "296",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "15度的角在放大10倍的放大镜下看变为150°的角了。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】放大镜只改变图形大小，不改变角度.\n【详解】15°仍是15°,不是150°,错误。",
+        "knowledge_points":  [
+                                 "角的度量，直角，钝角，锐角的认识及特征"
+                             ],
+        "source":  "25-26 四年级上·河北张家口·期中"
+    },
+    {
+        "id":  "297",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "用一个10倍的放大镜看8°的角，这个角变成了80°。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】角度由张开程度决定，与放大倍数无关.\n【详解】8°仍是8°,题干错误。",
+        "knowledge_points":  [
+                                 "角的度量，角的大小与边的关系"
+                             ],
+        "source":  "24-25 四年级上·广西河池·期末"
+    },
+    {
+        "id":  "298",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "体育馆每个看台有52排，每排有35个座位，一共有6个看台，这个体育馆一共有多少个座位。",
+        "options":  [
+
+                    ],
+        "answer":  "10920个",
+        "solution":  "【分析】先算一个看台座位数，再乘看台总数.\n【详解】52×35=1820(个),1820×6=10920(个)。答：一共有10920个座位。",
+        "knowledge_points":  [
+                                 "两，三位数与一位数连续进位的乘法，两位数乘两位数的进位乘法，用两步连乘解决实际问题"
+                             ],
+        "source":  "24-25 四年级上·河南商丘·期末"
+    },
+    {
+        "id":  "299",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "天安门广场上，观礼群众分成12个区域，每个区域有15排座位，每排可坐28名观礼群众，广场上一共可容纳多少名观礼群众。",
+        "options":  [
+
+                    ],
+        "answer":  "5040名",
+        "solution":  "【分析】每个区域人数=15×28，总人数=每区域人数*12.\n【详解】15×28=420(名),420×12=5040(名)。答：广场上一共可容纳5040名观礼群众。",
+        "knowledge_points":  [
+                                 "用两步连乘解决实际问题"
+                             ],
+        "source":  "25-26 四年级上·山东济宁·期中"
+    },
+    {
+        "id":  "300",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "校园科技节期间，学校采购了720套科技实验器材，平均分发给全校24个班级进行探究活动，每个班有3个实验小组。平均每个实验小组能分到多少套器材。",
+        "options":  [
+
+                    ],
+        "answer":  "10套",
+        "solution":  "【分析】先求每班分到多少套，再求每组分到多少套.\n【详解】720÷24=30(套/班),30÷3=10(套/组)。答：平均每个实验小组能分到10套器材。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "25-26 四年级上·天津南开·期末"
+    },
+    {
+        "id":  "301",
+        "unit":  "第一单元大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "在同一平面内画两条直线，如果它们都和这个平面内的第三条直线相交成直角，那么这两条直线( )。",
+        "options":  [
+                        "A. 互相垂直",
+                        "B. 互相平行",
+                        "C. 既不垂直也不平行"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据平行的判断可知：在同一平面内，两条直线都垂直于同一条直线，那么这两条直线互相平行.\n【详解】在同一平面内画两条直线，如果它们都和这个平面内的第三条直线相交成直角，那么这两条直线互相平行。故答案为:B",
+        "knowledge_points":  [
+                                 "垂直的特征",
+                                 "平行的特征及性质"
+                             ],
+        "source":  "25-26 四年级上·新疆和田·期末"
+    },
+    {
+        "id":  "302",
+        "unit":  "第一单元大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.94,
+        "question":  "过一点可以画( )直线。",
+        "options":  [
+                        "A. 0 条",
+                        "B. 1 条",
+                        "C. 2 条",
+                        "D. 无数条"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】直线没有端点，可以无限延伸。根据直线的这特点，过一个点向不同方向画直线，有无数个方向，所以过一点可以画无数条直线。据此解答.\n【详解】根据分析，过一点可以画无数条直线。故答案为:D",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·云南玉溪·期末"
+    },
+    {
+        "id":  "303",
+        "unit":  "第一单元大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "用计算器计算480×25 时，错误地输成了 480×250，下面可以弥补错误的是( )。",
+        "options":  [
+                        "A. 积乘 10",
+                        "B. 积除以 10",
+                        "C. 积除以 100"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】正确计算为480×25，错误输入为480×250，相当于把另一个因数扩大到原来的10倍。根据积的变化规律，错误积是正确积的10倍，因此将错误积除以10即可得到正确积.\n【详解】一个因数480不变，另一个因数25错误输入为250相当于扩大到原来的10倍，积也扩大到原来的10倍，错误积是正确积的10倍。因此，将错误积除以10即可得到正确积。故答案为:B",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26四年级上·黑龙江齐齐哈尔·期末"
+    },
+    {
+        "id":  "304",
+        "unit":  "第一单元大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "从12:30到13:00，分针旋转了( )。",
+        "options":  [
+                        "A. 90°",
+                        "B. 180°",
+                        "C. 60°"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】由题意得,12:30时，分针指向6;13:00时，分针指向12。由图可知从12:30到13:00，分针旋转了半圈。整圈是360°，半圈是180°。图示:![](images/86bda59a9f603ba4f7091abf67d0ebdb1c59c3e014ffd99c574613338c8bd8ac.jpg) \n【详解】由分析可得，从12:30到13:00，分针旋转了180°。故答案为:B",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "整点时间与钟面指针的位置",
+                                 "半点时间与钟面指针的位置"
+                             ],
+        "source":  "25-26四年级上·甘肃天水·期末"
+    },
+    {
+        "id":  "305",
+        "unit":  "第一单元大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "\"一旦出发，勇往直前.\"在数学中可以用这句话表示( )的特征。",
+        "options":  [
+                        "A. 线段",
+                        "B. 射线",
+                        "C. 直线",
+                        "D. 以上都可"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】\"一旦出发，勇往直前\"表示有一个端点，且向一个方向无限延伸。线段有两个端点，直线没有端点，射线有一个端点且无限长.\n【详解】由分析可知，这符合射线的特征，所以在数学中可以用这句话表示射线的特征。故答案为:B",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·江西萍乡·期末"
+    },
+    {
+        "id":  "306",
+        "unit":  "第一单元大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "校园里操场的占地面积约为 1( )。",
+        "options":  [
+                        "A. 平方米",
+                        "B. 公顷",
+                        "C. 米"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】操场属于较大面积，常用公顷计量;平方米偏小，\"米\"是长度单位.\n【详解】由分析可知：校园里操场的占地面积约为1公顷。故答案为:B",
+        "knowledge_points":  [
+                                 "面积单位的选择"
+                             ],
+        "source":  "24-25 四年级上·新疆吐鲁番·期末"
+    },
+    {
+        "id":  "307",
+        "unit":  "第一单元大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面信息中，不适合用条形统计图来表示的是( )。",
+        "options":  [
+                        "A. 莆田市本周不同景区游客量的情况",
+                        "B. 4名男生收集废矿泉水瓶情况",
+                        "C. 龙岩市区某天天气变化情况",
+                        "D. 四年级3个班获得流动红旗的数量"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】条形统计图适合比较不同类别数量，不适合体现随时间变化趋势.\n【详解】A,B,D都属于不同对象数量比较，适合条形统计图;C为天气随时间变化，更适合折线等趋势图。故答案为:C",
+        "knowledge_points":  [
+                                 "1格表示多个单位的单式条形统计图"
+                             ],
+        "source":  "25-26 四年级上·福建龙岩·月考"
+    },
+    {
+        "id":  "308",
+        "unit":  "第一单元大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "公路上有三条小路通往明明家，长度分别是125米,205米,117米，其中有一条小路是和公路垂直的，它的长度是( )米。",
+        "options":  [
+                        "A. 205",
+                        "B. 125",
+                        "C. 117",
+                        "D. 无法确定"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】从直线外一点到这条直线的线段中，垂直线段最短，所以与公路垂直的小路应是最短的一条.\n【详解】117米\u003c125米\u003c205米，所以垂直小路长度是117米。故答案为:C",
+        "knowledge_points":  [
+                                 "点到直线的距离"
+                             ],
+        "source":  "25-26 四年级上·天津蓟州·期末"
+    },
+    {
+        "id":  "309",
+        "unit":  "第一单元大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下午3时，钟面上的时针和分针( )。",
+        "options":  [
+                        "A. 互相平行",
+                        "B. 互相垂直",
+                        "C. 重合",
+                        "D. 无法确定"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】3时整分针指向12，时针指向3，相差3个大格，每格30°,夹角90°.\n【详解】360÷12=30(度),3×30=90(度),所以时针和分针互相垂直。故答案为:B",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "垂直的特征",
+                                 "平行的特征及性质",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "25-26四年级上·辽宁盘锦·期末"
+    },
+    {
+        "id":  "310",
+        "unit":  "第一单元大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "钟面上6时整，时针和分针成( )角。",
+        "options":  [
+                        "A. 平角",
+                        "B. 直角",
+                        "C. 钝角"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】6时整时，分针指向12，时针指向6，夹角180°,是平角.\n【详解】钟面上6时整，分针和时针成平角。故答案为:A",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "25-26 四年级上·全国·课后作业"
+    },
+    {
+        "id":  "311",
+        "unit":  "第一单元大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "早在两千多年前，我国劳动人民就会计算土地的面积。我国还用( )作土地面积单位，这样的一个单位约等于667平方米，现在这个单位已经不是我国的法定计量单位了。",
+        "options":  [
+                        "A. \"亩\"",
+                        "B. \"方\"",
+                        "C. \"丈\""
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】\"亩\"是传统土地面积单位，约667平方米;\"方\"常作体积单位，\"丈\"为长度单位.\n【详解】根据分析可知，该单位是\"亩\"。故答案为:A",
+        "knowledge_points":  [
+                                 "面积单位的选择"
+                             ],
+        "source":  "24-25 四年级上·云南文山·期末"
+    },
+    {
+        "id":  "312",
+        "unit":  "第一单元大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "小李，小张和小赵跳绳。现在只有一根跳绳，可以单人跳，也可以双人跳，每次跳1分钟。他们每人都要跳两次，至少需要( )分钟。",
+        "options":  [
+                        "A. 3",
+                        "B. 4",
+                        "C. 5",
+                        "D. 6"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】每次尽量双人跳可最省时：李张1次，李赵1次，张赵1次，共3次，每次1分钟.\n【详解】3×1=3(分钟),至少需要3分钟。故答案为:A",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "25-26 四年级上·天津河北·期末"
+    },
+    {
+        "id":  "313",
+        "unit":  "第一单元大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "妈妈开车送明明到学校，用时15分钟，平均车速900米/分。,爸爸从家到单位的路程是多少米。横线上应补充的条件是( )。",
+        "options":  [
+                        "A. 明明和妈妈出发时，爸爸已经步行了900米",
+                        "B. 爸爸每天比明明和妈妈早出发10分钟",
+                        "C. 爸爸步行到单位的时间比妈妈开车送明明的时间多3分钟",
+                        "D. 明明家到学校的路程是爸爸到单位路程的2倍"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】先由速度和时间算出明明家到学校路程，再需一个与爸爸路程的明确数量关系。只有D给出可计算关系.\n【详解】妈妈路程=900×15=13500米。若\"明明家到学校路程是爸爸到单位路程的2倍\"，则爸爸路程=13500÷2=6750米。故答案为:D",
+        "knowledge_points":  [
+                                 "提问题/填条件问题",
+                                 "三位数乘两位数(乘数末尾有0)",
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26 四年级上·天津河北·期末"
+    },
+    {
+        "id":  "314",
+        "unit":  "第一单元大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "1300÷200正确的结果是( )。",
+        "options":  [
+                        "A. 600。100",
+                        "B. 600。1",
+                        "C. 6。100",
+                        "D. 6。1"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】被除数和除数末尾都有0，可用商不变规律同时去掉两个0，先算13÷2=6。1，再还原余数.\n【详解】1300÷200=6。100。故答案为:C",
+        "knowledge_points":  [
+                                 "被除数和除数末尾都有0的除法",
+                                 "商不变规律及应用"
+                             ],
+        "source":  "25-26 四年级上·天津河北·期末"
+    },
+    {
+        "id":  "315",
+        "unit":  "第一单元大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面各数中，最接近100万的数是( )。",
+        "options":  [
+                        "A. 9999999",
+                        "B. 999990",
+                        "C. 1000001",
+                        "D. 999909"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】与1000000比较差值，差越小越接近.\n【详解】A差8999999,B差10,C差1,D差91;最小是1，对应1000001。故答案为:C",
+        "knowledge_points":  [
+                                 "亿以内数的读写法",
+                                 "大数比较"
+                             ],
+        "source":  "25-26四年级上·天津河北·期末"
+    },
+    {
+        "id":  "316",
+        "unit":  "第一单元大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面等式成立的是( )。",
+        "options":  [
+                        "A. 3000公顷=3平方千米",
+                        "B. 20平方分米=2平方米",
+                        "C. 6公顷=600平方米",
+                        "D. 2000000平方米=2平方千米"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】用面积单位进率换算:1平方千米=100公顷=1000000平方米,1平方米=100平方分米,1公顷=10000平方米.\n【详解】逐项换算后仅D成立。故答案为:D",
+        "knowledge_points":  [
+                                 "平方厘米/平方分米/平方米换算",
+                                 "公顷/平方千米换算"
+                             ],
+        "source":  "25-26 四年级上·天津河北·期末"
+    },
+    {
+        "id":  "317",
+        "unit":  "第一单元大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面四个数中，一个零也不读出来的是( )。",
+        "options":  [
+                        "A. 30000300",
+                        "B. 50030000",
+                        "C. 80005000"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】按分级读数规则判断是否需要读\"零\".\n【详解】A读作\"三千万零三百\";B读作\"五千零三万\";C读作\"八千万五千\"，不读零。故答案为:C",
+        "knowledge_points":  [
+                                 "亿以内数的读写法"
+                             ],
+        "source":  "25-26 四年级上·新疆省直辖县级单位·期末"
+    },
+    {
+        "id":  "318",
+        "unit":  "第一单元大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "李叔叔计划从伊犁市区前往那拉提草原游玩，途中他用2小时行了120千米，李叔叔可能是( )。",
+        "options":  [
+                        "A. 骑自行车",
+                        "B. 步行",
+                        "C. 乘汽车"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】先求速度:120÷2=60千米/时，再与常见交通方式速度比较.\n【详解】60千米/时更接近汽车速度。故答案为:C",
+        "knowledge_points":  [
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26 四年级上·新疆省直辖县级单位·期末"
+    },
+    {
+        "id":  "319",
+        "unit":  "第一单元大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "小华用步测法测量操场长度，共走120步，平均步长50厘米，操场长约( )。",
+        "options":  [
+                        "A. 60米",
+                        "B. 50米",
+                        "C. 600米",
+                        "D. 500米"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】总长度=步数×步长，再把厘米换算成米.\n【详解】120×50=6000厘米,6000厘米=60米。故答案为:A",
+        "knowledge_points":  [
+                                 "三位数乘两位数(乘数末尾有0)",
+                                 "毫米厘米分米米换算"
+                             ],
+        "source":  "25-26四年级上·天津蓟州·期末"
+    },
+    {
+        "id":  "320",
+        "unit":  "第一单元大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "计量天津市的面积时，用( )作单位比较合适。",
+        "options":  [
+                        "A. 平方分米",
+                        "B. 平方米",
+                        "C. 公顷",
+                        "D. 平方千米"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】城市行政区域面积较大，常用平方千米.\n【详解】天津市面积应用平方千米计量。故答案为:D",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 四年级上·天津蓟州·期末"
+    },
+    {
+        "id":  "321",
+        "unit":  "第一单元大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "400米跑道所围成的操场面积大约是1( )。",
+        "options":  [
+                        "A. 平方米",
+                        "B. 公顷",
+                        "C. 平方千米"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】操场面积一般远大于1平方米，远小于1平方千米，常用公顷.\n【详解】400米跑道围成操场面积约1公顷。故答案为:B",
+        "knowledge_points":  [
+                                 "面积单位的选择",
+                                 "公顷，平方千米的认识"
+                             ],
+        "source":  "25-26 五年级上·天津北辰·期末"
+    },
+    {
+        "id":  "322",
+        "unit":  "第一单元大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "小新在用计算器计算15×204时，发现数字键\"4\"失灵了。如果他仍用这个计算器计算，输入( )不能得到正确的结果。",
+        "options":  [
+                        "A. 30×102",
+                        "B. 15×210-6",
+                        "C. 15×205-15"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】先算原式15×204=3060，再分别比较各选项结果.\n【详解】A=3060,C=3060,B=15×210-6=3144!=3060，故不能得到正确结果的是B。故答案为:B",
+        "knowledge_points":  [
+                                 "三位数乘两位数(三位数中间有0)",
+                                 "乘加乘减混合运算",
+                                 "计算器初步认识与使用"
+                             ],
+        "source":  "25-26四年级上·新疆喀什·期末"
+    },
+    {
+        "id":  "323",
+        "unit":  "第一单元大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "太阳系的八大行星中，海王星到太阳的平均距离最远，约是4504000000千米。关于这个数，下面描述错误的是( )。",
+        "options":  [
+                        "A. 这个数的最高位是十亿位",
+                        "B. 这个数由45个亿和400个万组成",
+                        "C. 这个数省略亿位后面的尾数约是50亿"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】4504000000是十位数，最高位十亿位;省略亿位后面的尾数看千万位.\n【详解】4504000000≈45亿，不是50亿，因此错误描述是C。故答案为:C",
+        "knowledge_points":  [
+                                 "亿以上数的组成",
+                                 "整数数级数位和计数单位",
+                                 "亿以上数近似数"
+                             ],
+        "source":  "25-26 四年级上·新疆喀什·期末"
+    },
+    {
+        "id":  "324",
+        "unit":  "第一单元大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列说法正确的是( )。①万位，百万位，亿位，十亿位都是计数单位。②丽丽把5080000改写成了508万，改写后的数与原数相等。③伸缩衣架应用了平行四边形不稳定的特性。④一只平底锅最多能煎两条小黄鱼，用它煎1条小黄鱼需要2分钟(正，反面各1分钟),那么煎3条小黄鱼至少需要3分钟。",
+        "options":  [
+                        "A. ①②③",
+                        "B. ①③④",
+                        "C. ②③④"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】①把\"数位\"和\"计数单位\"混淆，错误;②5080000=508万，正确;③正确;④可用交替翻面方案3分钟完成，正确.\n【详解】正确的是②③④,故选C。故答案为:C",
+        "knowledge_points":  [
+                                 "亿以内数的改写",
+                                 "烙饼问题",
+                                 "平行四边形不稳定性及应用",
+                                 "整数数级数位和计数单位"
+                             ],
+        "source":  "25-26四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "325",
+        "unit":  "第一单元大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "由3个亿,5个十万和35个百组成的数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "300503500",
+        "solution":  "【分析】按整数写法从高位到低位写，空位补0.\n【详解】由3个亿,5个十万和35个百组成的数是300503500。",
+        "knowledge_points":  [
+                                 "亿以内数的组成"
+                             ],
+        "source":  "25-26 四年级上·天津南开·期末"
+    },
+    {
+        "id":  "326",
+        "unit":  "第一单元大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "天津上合峰会某论坛活动前五天，线上预约参与的人数达到了三十六万五千七百一十人次，横线上的数写作( ),用\"四舍五入\"法省略万位后面的尾数约是( )万。",
+        "options":  [
+
+                    ],
+        "answer":  "365710;37",
+        "solution":  "【分析】三十六万五千七百一十写作365710;省略万位后面的尾数看千位，千位是5，向万位进1.\n【详解】写作365710，约是37万。",
+        "knowledge_points":  [
+                                 "亿以内数的读写法",
+                                 "亿以内数近似数"
+                             ],
+        "source":  "25-26 四年级上·天津南开·期末"
+    },
+    {
+        "id":  "327",
+        "unit":  "第一单元大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在( )里填上合适的单位。一个鸡蛋的质量约50( );举世闻名的布达拉宫，耸立在西藏拉萨市红山之上，海拔3700多( ),占地总面积36万余( )。",
+        "options":  [
+
+                    ],
+        "answer":  "克/g;米/m;平方米/m²",
+        "solution":  "【分析】依据生活经验：鸡蛋质量用克;海拔高度用米;占地面积用平方米.\n【详解】依次填：克(g),米(m),平方米(m²)。",
+        "knowledge_points":  [
+                                 "质量单位的选择",
+                                 "长度单位的选择",
+                                 "面积单位的选择"
+                             ],
+        "source":  "2025·江西吉安·小升初真题"
+    },
+    {
+        "id":  "328",
+        "unit":  "第一单元大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "将一张圆形纸片对折3次，可以得到的角是( )°。8时整，分针和时针所成的较小角是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "45;120",
+        "solution":  "【分析】圆周角360°,对折3次即连续除以2三次;8时整两针相差4大格，每格30°.\n【详解】360÷2÷2÷2=45,30×4=120，故填45和120。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "平角周角认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "25-26 四年级上·江西赣州·期中"
+    },
+    {
+        "id":  "329",
+        "unit":  "第一单元大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个乘法算式的积是13，一个因数不变，另一个因数乘10，积是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "130",
+        "solution":  "【分析】一个因数不变，另一个因数乘10，积也乘10.\n【详解】13×10=130，所以积是130。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "24-25 四年级上·新疆吐鲁番·期末"
+    },
+    {
+        "id":  "330",
+        "unit":  "第一单元大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "9时整，钟面上的时针和分针形成( )角;( )时整，钟面上的时针和分针形成平角。",
+        "options":  [
+
+                    ],
+        "answer":  "直角;6",
+        "solution":  "【分析】钟面每大格30°。9时整两针相差3格是90°,为直角;平角180°对应相差6格，即6时整.\n【详解】9时整形成直角;6时整形成平角。",
+        "knowledge_points":  [
+                                 "直角钝角锐角认识及特征",
+                                 "平角周角认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "19-20 四年级上·湖南长沙·期中"
+    },
+    {
+        "id":  "331",
+        "unit":  "第一单元大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "钟面上6时整，时针和分针形成的角是( )角。( )时或( )时整，时针和分针形成的是直角。3时30分，时针和分针的夹角(较小角)是( )度。",
+        "options":  [
+
+                    ],
+        "answer":  "平角;3;9;75",
+        "solution":  "【分析】每大格30°。6时整相差6格=180°(平角);直角是90°对应相差3格，为3时或9时;3:30时两针较小角为75°.\n【详解】依次填：平,3,9,75。",
+        "knowledge_points":  [
+                                 "一般时间与钟面指针位置",
+                                 "角的分类及换算",
+                                 "角度的计算"
+                             ],
+        "source":  "25-26 四年级上·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "332",
+        "unit":  "第一单元大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "两个数的积是420，一个因数除以5，另一个因数乘10，则积是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "840",
+        "solution":  "【分析】积变化因子为*10÷5=*2.\n【详解】420÷5×10=840，所以积是840。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "333",
+        "unit":  "第一单元大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "ab∥表示直线a和b互相( );ab⊥表示直线a和b互相( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平行;垂直",
+        "solution":  "【分析】\"∥\"表示平行，\"⊥\"表示垂直.\n【详解】ab∥表示直线a和b互相平行;ab⊥表示直线a和b互相垂直。",
+        "knowledge_points":  [
+                                 "垂直的特征",
+                                 "平行的特征及性质"
+                             ],
+        "source":  "25-26 四年级上·辽宁铁岭·期末"
+    },
+    {
+        "id":  "334",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "从小明家到某条地铁线有三条小路可以走，它们的长度分别是880米,720米,950米，其中一条小路与这条地铁线垂直，这条小路的长度是( )米。",
+        "options":  [
+
+                    ],
+        "answer":  "720",
+        "solution":  "【分析】从直线外一点到这条直线的线段中，垂直线段最短，这条垂直线段的长度叫做点到直线的距离。比较三条小路长度，最短的就是与地铁线垂直的小路长度.\n【详解】950米\u003e880米\u003e720米，所以垂直小路长度是720米。",
+        "knowledge_points":  [
+                                 "1000以内数的比大小",
+                                 "点到直线的距离"
+                             ],
+        "source":  "25-26 四年级上·吉林松原·期末"
+    },
+    {
+        "id":  "335",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "9时整时，钟面上时针与分针成( );9时30分时，时针与分针成( )。(填\"锐角\"\"直角\"或\"钝角\")",
+        "options":  [
+
+                    ],
+        "answer":  "直角;钝角",
+        "solution":  "【分析】钟面12大格，每格30°。9时整时针与分针夹角为3×30°=90°,是直角。9时30分时，分针在6，时针在9和10中间，夹角是105°,是钝角.\n【详解】9时整成直角;9时30分成钝角。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "整点时间与钟面指针的位置",
+                                 "半点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·湖南长沙·期末"
+    },
+    {
+        "id":  "336",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "30°的角是( )角,30°的3倍是( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "锐;直",
+        "solution":  "【分析】锐角是大于0°小于90°的角，直角是90°。30°是锐角,30°*3=90°是直角.\n【详解】30°是锐角;30°的3倍是直角。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "角的分类及换算"
+                             ],
+        "source":  "24-25 四年级上·西藏林芝·期末"
+    },
+    {
+        "id":  "337",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "学校举行\"图书漂流\"捐书活动，小沙要绘制一幅各班捐书情况的条形统计图。如果用8厘米高的直条表示32本，那么6厘米高的直条表示( )本;如果要表示40本，那么直条的高度应是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "24;10",
+        "solution":  "【分析】8厘米对应32本，所以1厘米表示32÷8=4本。6厘米表示6×4=24本;40本对应高度40÷4=10厘米.\n【详解】6厘米高表示24本;表示40本需10厘米高。",
+        "knowledge_points":  [
+                                 "1格表示多个单位的单式条形统计图"
+                             ],
+        "source":  "25-26 四年级上·黑龙江齐齐哈尔·期末"
+    },
+    {
+        "id":  "338",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "填上合适的单位。暑假小明和爸爸妈妈去北京游玩，小明用手机搜索后，知道了北京市的面积是16410( )。在北京游玩的第一天，他们参观了占地面积约72( )的故宫博物院。晚上他们到宾馆休息，他们住的房间面积约是15( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方千米(km²);公顷(ha/hm²);平方米(m²)",
+        "solution":  "【分析】城市面积常用平方千米，较大景区占地常用公顷，房间面积常用平方米.\n【详解】北京市面积填平方千米;故宫博物院占地填公顷;宾馆房间面积填平方米。",
+        "knowledge_points":  [
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 四年级上·江西赣州·期中"
+    },
+    {
+        "id":  "339",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "3:30时，时针和分针的夹角是( )角;6:05时，时针与分针的夹角是( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "锐角;钝角",
+        "solution":  "【分析】3:30时，分针在6，时针在3和4之间，夹角为锐角(图示:![](images/f9b5d7bfbb86fc87b0f7ba1f1363f5402fa6a2a3ae04461bca1a238c8850cb78.jpg))。6:05时，分针在1，时针在6和7之间，夹角为钝角(图示:![](images/faa91b9fae80e8dd53d1aef17dacb9a35851883e9069d04c8796dd888b5684f9.jpg)).\n【详解】3:30是锐角;6:05是钝角。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "半点时间与钟面指针的位置",
+                                 "一般时间与钟面指针的位置"
+                             ],
+        "source":  "25-26四年级上·吉林延边·期末"
+    },
+    {
+        "id":  "340",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "从3时到6时，分针转了( )°,时针转了( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "1080;90",
+        "solution":  "【分析】分针1小时转360°,3小时转3圈:360°*3=1080°。时针1小时转30°,3小时转30°*3=90°.\n【详解】分针转1080°,时针转90°。",
+        "knowledge_points":  [
+                                 "旋转与旋转现象",
+                                 "角度的计算",
+                                 "平角，周角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "25-26 四年级上·新疆喀什·期末"
+    },
+    {
+        "id":  "341",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "小明在计算一道乘法算式时，把其中一个因数35看成了5，结果得到的积是80，正确的积是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "560",
+        "solution":  "【分析】35看成5，相当于这个因数缩小到原来的1÷7，积也缩小到原来的1÷7，所以正确积=80×7.\n【详解】35÷5=7,80×7=560，正确积是560。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·江西赣州·期中"
+    },
+    {
+        "id":  "342",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上合适的单位名称。8月1日，读四年级的甜甜小朋友跟着身高178( )的爸爸，来到面积为16400( )的北京游玩，他们先去故宫参观，一进故宫，就看到一棵高约9( )的梧桐树。妈妈从面积为50( )的文创店给甜甜买了一支长2( )的钢笔和一本封面面积约400( )的书送给她。",
+        "options":  [
+
+                    ],
+        "answer":  "厘米(cm);平方千米(km²);米(m);平方米(m²);分米(dm);平方厘米(cm²)",
+        "solution":  "【分析】身高常用厘米;城市面积常用平方千米;树高常用米;店铺面积常用平方米;钢笔长度可用分米;书封面面积可用平方厘米.\n【详解】依次填：厘米，平方千米，米，平方米，分米，平方厘米。",
+        "knowledge_points":  [
+                                 "长度单位的选择",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 四年级上·河南郑州·期中"
+    },
+    {
+        "id":  "343",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个乘法算式的积是40，一个因数乘12，另一个因数不变，积是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "480",
+        "solution":  "【分析】一个因数不变，另一个因数乘12，积也乘12.\n【详解】40×12=480，积是480。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·辽宁盘锦·期末"
+    },
+    {
+        "id":  "344",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上适当的单位。(1)北京天坛公园的占地面积约为273( );(2)杭州西湖湖面的面积约为638( );(3)兰兰家阳台的面积约是15( );(4)香港特别行政区的陆地面积约1106( )。",
+        "options":  [
+
+                    ],
+        "answer":  "公顷;公顷;平方米;平方千米",
+        "solution":  "【分析】公园，湖面这类中等土地面积常用公顷;阳台面积用平方米;特别行政区陆地面积较大用平方千米.\n【详解】依次填：公顷，公顷，平方米，平方千米。",
+        "knowledge_points":  [
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 四年级上·湖南常德·期中"
+    },
+    {
+        "id":  "345",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "已知A×B=800，如果B乘50,A不变，那么积是( );如果A缩小到原数的一半,B不变，那么积是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "40000;400",
+        "solution":  "【分析】一个因数不变，另一个因数乘(或除以)几，积也乘(或除以)几.\n【详解】800×50=40000;800÷2=400。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·吉林·期末"
+    },
+    {
+        "id":  "346",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "伸缩门的制作利用了平行四边形的( )的特性。",
+        "options":  [
+
+                    ],
+        "answer":  "容易变形",
+        "solution":  "【分析】平行四边形在受力时形状易改变，具有不稳定性(容易变形).\n【详解】伸缩门利用了平行四边形容易变形的特性。",
+        "knowledge_points":  [
+                                 "平行四边形的不稳定性及应用"
+                             ],
+        "source":  "25-26 四年级上·吉林白城·期末"
+    },
+    {
+        "id":  "347",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果一个因数不变，另一个因数除以3，积是150，那么原来的积是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "450",
+        "solution":  "【分析】另一个因数除以3，积也除以3。已知新积150，则原积应乘3.\n【详解】150×3=450，原来的积是450。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26四年级上·江西赣州·期中"
+    },
+    {
+        "id":  "348",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "汽车车灯发出的光线可以看作一条( )。(填\"直线\"\"射线\"或\"线段\")理由:( )。",
+        "options":  [
+
+                    ],
+        "answer":  "射线;汽车车灯是一个端点，光向一个方向无限延伸",
+        "solution":  "【分析】线段有2个端点且有限长;直线无端点且向两端无限延伸;射线有1个端点且向一端无限延伸.\n【详解】车灯发光可看作以车灯为端点向前方无限延伸，所以是射线。理由：汽车车灯是一个端点，光向一个方向无限延伸。",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "25-26 四年级上·甘肃平凉·期末"
+    },
+    {
+        "id":  "349",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "两数相乘，积是60，其中一个因数乘5，另一个因数不变，则积是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "300",
+        "solution":  "【分析】一个因数不变，另一个因数乘5，积也乘5.\n【详解】60×5=300。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·天津河北·期末"
+    },
+    {
+        "id":  "350",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "活动晾衣架，伸缩门是利用了平行四边形( )的特点设计的。",
+        "options":  [
+
+                    ],
+        "answer":  "不稳定性",
+        "solution":  "【分析】平行四边形受力容易变形，具有不稳定性.\n【详解】活动晾衣架和伸缩门都利用了平行四边形不稳定性的特点。",
+        "knowledge_points":  [
+                                 "平行四边形的不稳定性及应用"
+                             ],
+        "source":  "25-26 四年级上·吉林·期末"
+    },
+    {
+        "id":  "351",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "两个数相乘的积是180，如果一个因数乘3，另一个因数不变，积是( );如果两个因数都除以2，积是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "540;45",
+        "solution":  "【分析】一个因数乘3，积乘3;两个因数都除以2，积要除以2×2=4.\n【详解】180×3=540;180÷4=45。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26四年级上·吉林白城·期末"
+    },
+    {
+        "id":  "352",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "每套演出服102元，买20套需要( )元，这里是已知单价和数量，求( )。",
+        "options":  [
+
+                    ],
+        "answer":  "2040;总价",
+        "solution":  "【分析】总价=单价×数量.\n【详解】102×20=2040(元),所求是总价。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "25-26 四年级上·新疆阿克苏·期末"
+    },
+    {
+        "id":  "353",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "计算647÷39时，可以把39看作( )来试商。",
+        "options":  [
+
+                    ],
+        "answer":  "40",
+        "solution":  "【分析】两位数除法试商时，常把除数看成最接近的整十数.\n【详解】39可看作40来试商。",
+        "knowledge_points":  [
+                                 "多位数除以两位数的试商"
+                             ],
+        "source":  "25-26 四年级上·新疆阿克苏·期末"
+    },
+    {
+        "id":  "354",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在一张梯形纸上剪一刀，分成两个图形。如果其中一个是平行四边形，那么另一个图形可能是( )形，也可能是( )形。",
+        "options":  [
+
+                    ],
+        "answer":  "三角形;梯形",
+        "solution":  "【分析】梯形剪一刀后，若其中一部分是平行四边形，另一部分常见可能是三角形或梯形.\n【详解】另一个图形可能是三角形，也可能是梯形。",
+        "knowledge_points":  [
+                                 "平行四边形的概念及特点",
+                                 "梯形的概念及特点",
+                                 "平面图形的分割"
+                             ],
+        "source":  "25-26 四年级上·新疆阿克苏·期末"
+    },
+    {
+        "id":  "355",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上适当的面积单位或数。故宫博物院的占地面积约72( )。乌鲁木齐市的面积约13800( )。5平方千米=( )公顷。1周角=( )直角。",
+        "options":  [
+
+                    ],
+        "answer":  "公顷;平方千米;500;4",
+        "solution":  "【分析】故宫面积常用公顷，城市面积常用平方千米。1平方千米=100公顷，所以5平方千米=500公顷。1周角360°,1直角90°,所以1周角=4直角.\n【详解】依次填：公顷，平方千米,500,4。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征",
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26四年级上·新疆阿克苏·期末"
+    },
+    {
+        "id":  "356",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "由四个百万，六个十万和五个千组成的数是( ),这个数读作( )。",
+        "options":  [
+
+                    ],
+        "answer":  "4605000;四百六十万五千",
+        "solution":  "【分析】四个百万是4000000，六个十万是600000，五个千是5000，相加得4605000。按数位读作\"四百六十万五千\".\n【详解】组成的数是4605000，读作四百六十万五千。",
+        "knowledge_points":  [
+                                 "亿以内数的读，写法"
+                             ],
+        "source":  "25-26 四年级上·新疆阿克苏·期末"
+    },
+    {
+        "id":  "357",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一辆货车以每小时65千米的速度从甲城开往乙城。行驶3小时后，货车在过中点20千米处休息。此时，货车距离乙城还有( )千米。",
+        "options":  [
+
+                    ],
+        "answer":  "155",
+        "solution":  "【分析】已行路程65×3=195千米。又知\"过中点20千米\"，则半程是195-20=175千米。离乙城还剩半程减20千米，即175-20=155千米.\n【详解】货车距离乙城还有155千米。",
+        "knowledge_points":  [
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26 四年级上·天津河北·期末"
+    },
+    {
+        "id":  "358",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\"\u003e\"\"\u003c\"或\"=\"。65780( )67890;10020( )9978;950000( )95万;25×6( )200;480÷40( )10;350÷32( )9。",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003e;=;\u003c;\u003e;\u003e",
+        "solution":  "【分析】按整数大小比较法比较前3组;后3组先算结果再比较.\n【详解】65780\u003c67890;10020\u003e9978;950000=95万;25×6=150\u003c200;480÷40=12\u003e10;350÷32约10余30，故\u003e9。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "两，三位数与一位数连续进位的乘法",
+                                 "除数是两位数的笔算除法",
+                                 "亿以内数的改写"
+                             ],
+        "source":  "25-26 四年级上·天津河北·期末"
+    },
+    {
+        "id":  "359",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "足球的单价70元/个,370元能买( )个足球，还剩( )元。",
+        "options":  [
+
+                    ],
+        "answer":  "5;20",
+        "solution":  "【分析】用总钱数除以单价，商是个数，余数是剩余钱数.\n【详解】370÷70=5(个)。20(元),能买5个，还剩20元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "有余数除法的实际应用",
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "25-26 四年级上·天津河北·期末"
+    },
+    {
+        "id":  "360",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "□52÷28，商是一位数且没有余数,□里面填( ),商是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "2;9",
+        "solution":  "【分析】商是一位数，前两位□5要小于28,□可试1或2.152÷28有余数，不合题意;252÷28=9，且无余数.\n【详解】□填2，商是9。",
+        "knowledge_points":  [
+                                 "三位数除以两位数的有余数除法",
+                                 "除数是两位数的笔算除法",
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "25-26 四年级上·天津河北·期末"
+    },
+    {
+        "id":  "361",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "已知A×B=600，如果B不变,A×4，则积是( );如果A×4,B/2，则积是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "2400;1200",
+        "solution":  "【分析】第一空:A乘4，积乘4。第二空:A乘4且B除以2，积相当于乘4再除以2，即乘2.\n【详解】600×4=2400;600×4÷2=1200。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "362",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "两个因数的积是120，如果一个因数不变，另一个因数除以3，积是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "40",
+        "solution":  "【分析】一个因数不变，另一个因数除以3，积也除以3.\n【详解】120÷3=40。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "24-25 四年级上·新疆克孜勒苏·期末"
+    },
+    {
+        "id":  "363",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "2700006是由( )个百万,( )个十万和( )个一组成的。",
+        "options":  [
+
+                    ],
+        "answer":  "2;7;6",
+        "solution":  "【分析】按数位看,2700006中百万位是2，十万位是7，个位是6.\n【详解】由2个百万,7个十万和6个一组成。",
+        "knowledge_points":  [
+                                 "亿以内数的组成"
+                             ],
+        "source":  "25-26四年级上·新疆省直辖县级单位·期末"
+    },
+    {
+        "id":  "364",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "国庆来临，伊宁市举办\"喜迎国庆\"主题文化展，期间累计接待来自全国各地的游客达20850000人次，横线上的数读作( ),改写成用\"万\"作单位的数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "二千零八十五万;2085万",
+        "solution":  "【分析】按整数读法,20850000读作二千零八十五万。改写成\"万\"作单位，去掉末尾4个0，得2085万.\n【详解】读作二千零八十五万，改写是2085万。",
+        "knowledge_points":  [
+                                 "亿以内数的读，写法",
+                                 "亿以内数的改写"
+                             ],
+        "source":  "25-26 四年级上·新疆省直辖县级单位·期末"
+    },
+    {
+        "id":  "365",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在( )里填上\"\u003e\"\"\u003c\"或\"=\"。46900000( )469万;240×30( )120×60;240分( )3时;420÷21( )60÷3;750÷31( )750÷29;5平方千米( )498公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;=;\u003e;=;\u003c;\u003e",
+        "solution":  "【分析】469万=4690000;积的变化规律可判第2组;3时=180分;420÷21与60÷3都等于20;同被除数时除数越大商越小;1平方千米=100公顷.\n【详解】46900000\u003e469万;240×30=120×60;240分\u003e3时;420÷21=60÷3;750÷31\u003c750÷29;5平方千米=500公顷\u003e498公顷。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "公顷，平方千米的进率与换算",
+                                 "积的变化规律(整数乘法)",
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "25-26四年级上·天津蓟州·期末"
+    },
+    {
+        "id":  "366",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈煎一块牛排需要6分钟(正，反两面各需要3分钟),平底锅上每次最多能煎2块牛排，冬冬一家三口每人一块，妈妈煎牛排至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "9",
+        "solution":  "【分析】三块牛排编号①②③:第1个3分钟煎①②正面;第2个3分钟煎①反面和③正面;第3个3分钟煎②③反面.\n【详解】共3个阶段，每阶段3分钟，最少9分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "25-26四年级上·天津蓟州·期末"
+    },
+    {
+        "id":  "367",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "□28×26，如果积是四位数，那么□里最大填( );如果积是五位数，那么□里最小填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "3;4",
+        "solution":  "【分析】依次试:128×26=3328,228×26=5928,328×26=8528(四位数);428×26=11128(五位数).\n【详解】四位数时□最大是3;五位数时□最小是4。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "25-26 四年级上·天津蓟州·期末"
+    },
+    {
+        "id":  "368",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "新安镇培育了一片长方形\"和谐林\"，长是400米，宽是200米，这片\"和谐林\"的面积是( )公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "8",
+        "solution":  "【分析】长方形面积=长×宽，先算平方米，再换算成公顷(10000平方米=1公顷).\n【详解】400×200=80000(平方米),80000平方米=8公顷。",
+        "knowledge_points":  [
+                                 "三位数与整十数的口算乘法",
+                                 "长方形的面积",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·天津蓟州·期末"
+    },
+    {
+        "id":  "369",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "四年级有238名学生去研学，每辆大巴车最多能坐45名学生，学校至少要租( )辆大巴车。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】总人数除以每车人数，若有余数需要再加1辆.\n【详解】238÷45=5。13，余13人还需1辆，所以至少租6辆。",
+        "knowledge_points":  [
+                                 "有余数除法的实际应用",
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "25-26 四年级上·天津蓟州·期末"
+    },
+    {
+        "id":  "370",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "猎豹奔跑的速度是每小时120千米，它的奔跑速度可以写成( )。照这样计算，它30分钟跑了( )千米。",
+        "options":  [
+
+                    ],
+        "answer":  "120千米/时;60",
+        "solution":  "【分析】速度可写成\"千米/时\"。30分钟是1小时的一半，路程也取一半.\n【详解】120÷2=60，所以30分钟跑60千米。",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "时，分，秒之间的换算与比较"
+                             ],
+        "source":  "25-26 四年级上·天津蓟州·期末"
+    },
+    {
+        "id":  "371",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个平行四边形的周长是36厘米，其中一条边的长度是10厘米，那么和它相邻的一条边的长度是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "8",
+        "solution":  "【分析】平行四边形周长=(相邻两边和)×2，所以相邻两边和=36÷2=18(厘米),再减去已知边10厘米即可.\n【详解】36÷2-10=18-10=8(厘米)。",
+        "knowledge_points":  [
+                                 "平行四边形的周长"
+                             ],
+        "source":  "25-26 四年级上·天津蓟州·期末"
+    },
+    {
+        "id":  "372",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个数除以54商是6，且有余数，余数最大是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "53",
+        "solution":  "【分析】有余数除法中，余数一定小于除数，最大余数=除数-1.\n【详解】54-1=53。",
+        "knowledge_points":  [
+                                 "余数与除数的关系"
+                             ],
+        "source":  "24-25 四年级上·西藏林芝·期末"
+    },
+    {
+        "id":  "373",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "平行四边形的一条边长12厘米，比它的邻边短3厘米，这个平行四边形的周长是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "54",
+        "solution":  "【分析】邻边长度=12+3=15(厘米),平行四边形周长=(12+15)×2.\n【详解】(12+15)×2=27×2=54(厘米)。",
+        "knowledge_points":  [
+                                 "平行四边形的概念及特点",
+                                 "平行四边形的周长"
+                             ],
+        "source":  "25-26四年级上·新疆喀什·期末"
+    },
+    {
+        "id":  "374",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "328000000这个数中,2在( )位上，表示( )。",
+        "options":  [
+
+                    ],
+        "answer":  "千万;2个千万",
+        "solution":  "【分析】328000000从高位起依次是亿位，千万位，百万位。数字2在第二位即千万位.\n【详解】2在千万位上，表示2个千万。",
+        "knowledge_points":  [
+                                 "亿以上数的组成",
+                                 "整数的数级，数位和计数单位的认识"
+                             ],
+        "source":  "25-26 四年级上·新疆喀什·期末"
+    },
+    {
+        "id":  "375",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\"\u003e\"\"\u003c\"或\"=\"。8010184000( )898108400;60公顷( )6平方千米;48×500( )480×50;702÷27( )705÷15。",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;\u003c;=;\u003c",
+        "solution":  "【分析】位数多的数更大;1平方千米=100公顷;分别计算乘除法结果比较.\n【详解】8010184000是十位数,898108400是九位数，所以\u003e;6平方千米=600公顷，所以60公顷\u003c6平方千米;48×500=24000,480×50=24000，所以=;702÷27=26,705÷15=47，所以\u003c。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "除数是两位数的笔算除法",
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "大数的比较"
+                             ],
+        "source":  "25-26四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "376",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "同一平面内，两条直线分别垂直于第三条直线，那么两条直线一定互相平行。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】同一平面内，垂直于同一条直线的两条直线互相平行.\n【详解】题干说法正确。图示:![](images/214f649ec0af733e078e11a267e4a10e.jpg)",
+        "knowledge_points":  [
+                                 "垂直的特征",
+                                 "平行的特征及性质"
+                             ],
+        "source":  "25-26 四年级上·河南南阳·期中"
+    },
+    {
+        "id":  "377",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "钟面上9时30分，时针和分针所组成的角是钝角。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】9时30分时，分针指向6，时针在9和10之间，夹角为105°,大于90°小于180°.\n【详解】105°是钝角，题干说法正确。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "平角，周角的认识及特征",
+                                 "角的分类及换算",
+                                 "半点时间与钟面指针的位置"
+                             ],
+        "source":  "25-26 四年级上·辽宁鞍山·期末"
+    },
+    {
+        "id":  "378",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.94,
+        "question":  "两条平行线长20厘米。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平行线是两条直线，直线没有端点，可无限延伸，无法度量长度.\n【详解】题干把平行线当成有固定长度，错误。",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征",
+                                 "平行的特征及性质"
+                             ],
+        "source":  "25-26 四年级上·甘肃张掖·期末"
+    },
+    {
+        "id":  "379",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "同一平面内的两条直线，不平行就一定相交。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】在同一平面内，两条直线只有平行和相交两种位置关系.\n【详解】不平行就一定相交，题干正确。图示:![](images/518f073d7cf76638f2eaf10c8c59c220.jpg)",
+        "knowledge_points":  [
+                                 "平行的特征及性质",
+                                 "垂直的特征"
+                             ],
+        "source":  "24-25 四年级上·山东济宁·期末"
+    },
+    {
+        "id":  "380",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "所有的钝角都大于90°,那么大于90°的角叫钝角。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】钝角是大于90°且小于180°的角。大于90°还可能是平角，周角等.\n【详解】\"大于90°\"不一定是钝角，题干错误。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "25-26 四年级上·湖南常德·期中"
+    },
+    {
+        "id":  "381",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "两条线段在同一个平面内不平行，就一定相交。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】线段长度有限，即使方向不同(不平行)也可能彼此错开，没有交点.\n【详解】不平行的两条线段不一定相交，题干错误。",
+        "knowledge_points":  [
+                                 "平行的特征及性质"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "382",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "条形统计图能更清楚更直观地表示数据的大小。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】条形统计图用条形长度表示数量，便于比较大小.\n【详解】题干说法正确。",
+        "knowledge_points":  [
+                                 "1格表示一个单位的单式条形统计图"
+                             ],
+        "source":  "25-26 四年级上·辽宁抚顺·期末"
+    },
+    {
+        "id":  "383",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.94,
+        "question":  "一个因数不变，另一个因数乘3，积也乘3。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】根据积的变化规律，一个因数不变，另一个因数乘几，积也乘几.\n【详解】题干符合规律，正确。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·湖南常德·期中"
+    },
+    {
+        "id":  "384",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "周末，红红帮妈妈做家务，擦桌子2分钟，扫地5分钟，烧水8分钟，倒垃圾3分钟，她完成这四件事至少需要10分钟。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】烧水8分钟可与扫地5分钟，倒垃圾3分钟并行完成;擦桌子2分钟需另做.\n【详解】最少时间=8+2=10(分钟),题干正确。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "25-26四年级上·新疆阿克苏·期末"
+    },
+    {
+        "id":  "385",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "把50600000改写成用\"万\"作单位的数是506万。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】改写成\"万\"作单位要去掉末尾4个0.\n【详解】50600000=5060万，不是506万，题干错误。",
+        "knowledge_points":  [
+                                 "亿以内数的改写"
+                             ],
+        "source":  "25-26 四年级上·新疆阿克苏·期末"
+    },
+    {
+        "id":  "386",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.94,
+        "question":  "520□1354000≈520亿，方框中最小能填1。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】近似到亿位看千万位(□)。若□\u003c5，约为520亿;若□\u003e=5，约为521亿.\n【详解】要约成520亿,□可取0,1,2,3,4，最小应是0，不是1，题干错误。",
+        "knowledge_points":  [
+                                 "亿以上数的近似数"
+                             ],
+        "source":  "25-26 四年级上·新疆省直辖县级单位·期末"
+    },
+    {
+        "id":  "387",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "每两个计数单位之间的进率都是10。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】相邻计数单位之间进率是10;不相邻的不是10，如个到百进率是100.\n【详解】题干说\"每两个\"都为10不正确。",
+        "knowledge_points":  [
+                                 "十进制计数法"
+                             ],
+        "source":  "25-26四年级上·新疆省直辖县级单位·期末"
+    },
+    {
+        "id":  "388",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.94,
+        "question":  "聪聪量出一条直线长50米。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】直线无限长，没有端点，不能量出固定长度.\n【详解】题干错误。",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "25-26四年级上·辽宁盘锦·期末"
+    },
+    {
+        "id":  "389",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "把线段向两端无限延伸，就得到一条直线，直线没有端点。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】线段有两个端点，向两端无限延伸得到直线，直线无端点.\n【详解】题干说法正确。",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·河南新乡·期末"
+    },
+    {
+        "id":  "390",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.78,
+        "question":  "小华把一条长3厘米线段的一端延长6厘米，就得到了一条长9厘米的射线。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】射线向一端无限延伸，不能说\"长9厘米\".\n【详解】把3厘米线段一端延长6厘米得到的是9厘米线段，不是射线，题干错误。",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "391",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.8,
+        "question":  "平角不是一条直线。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】平角是一个角(有顶点和两条边),度数180°;直线不是角.\n【详解】平角不是一条直线，题干正确。",
+        "knowledge_points":  [
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·辽宁鞍山·期末"
+    },
+    {
+        "id":  "392",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.75,
+        "question":  "射线有一个端点，直线没有端点，所以直线比射线长。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】直线和射线都可无限延伸，长度都无法度量，不能比较长短.\n【详解】\"直线比射线长\"错误。",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·山东济宁·期末"
+    },
+    {
+        "id":  "393",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "图书管理员计划整理950本图书，已经整理了230本。如果每小时能整理15本，整理完所有图书还需要几小时。",
+        "options":  [
+
+                    ],
+        "answer":  "48小时",
+        "solution":  "【分析】先求剩余本数，再除以每小时整理本数.\n【详解】(950-230)÷15=720÷15=48(小时)。答：整理完所有图书还需要48小时。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "25-26 四年级上·天津河北·期末"
+    },
+    {
+        "id":  "394",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "一件毛衣95元，一条裤子80元。买这样的12套服装，需要多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "2100元",
+        "solution":  "【分析】每套价格=95+80，再乘12;也可分别算毛衣总价和裤子总价后相加.\n【详解】95×12+80×12=1140+960=2100(元)。答：需要2100元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "乘加，乘减混合运算"
+                             ],
+        "source":  "25-26 四年级上·天津河北·期末"
+    },
+    {
+        "id":  "395",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "四年级同学准备了一些彩带制作花束。如果每把花束用彩带18米，准备的这些彩带正好做25把花束。如果每把花束用彩带15米，这些彩带可以做多少把花束。",
+        "options":  [
+
+                    ],
+        "answer":  "30把",
+        "solution":  "【分析】先求彩带总长度，再除以每把15米.\n【详解】18×25÷15=450÷15=30(把)。答：这些彩带可以做30把花束。",
+        "knowledge_points":  [
+                                 "归总问题"
+                             ],
+        "source":  "25-26四年级上·天津河北·期末"
+    },
+    {
+        "id":  "396",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "学校体操队共200人，如果28人组成一个表演方阵，可以组成几个这样的方阵。还剩几人。",
+        "options":  [
+
+                    ],
+        "answer":  "7个方阵，还剩4人",
+        "solution":  "【分析】总人数除以每个方阵人数，商是方阵数，余数是剩余人数.\n【详解】200÷28=7(个)。4(人)。答：可以组成7个方阵，还剩4人。",
+        "knowledge_points":  [
+                                 "有余数除法的实际应用",
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "25-26 四年级上·天津河北·期末"
+    },
+    {
+        "id":  "397",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.94,
+        "question":  "伊宁市家乡好超市做促销活动,3件衣服共153元，每件衣服多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "51元",
+        "solution":  "【分析】单价=总价/数量.\n【详解】153÷3=51(元)。答：每件衣服51元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "三位数除以一位数，首位不够除"
+                             ],
+        "source":  "25-26 四年级上·新疆省直辖县级单位·期末"
+    },
+    {
+        "id":  "398",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "超市促销活动时，张阿姨购买了3箱同样的牛奶，李阿姨也买了7箱这样的牛奶，李阿姨比张阿姨多支付256元。每箱牛奶多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "64元",
+        "solution":  "【分析】李阿姨比张阿姨多买7-3=4箱，这4箱对应多付256元，所以单价=256÷4.\n【详解】256÷(7-3)=256÷4=64(元)。答：每箱牛奶64元。",
+        "knowledge_points":  [
+                                 "经济问题"
+                             ],
+        "source":  "25-26 五年级上·甘肃兰州·期末"
+    },
+    {
+        "id":  "399",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "小区里有一个平行四边形的广告牌，相邻两边的长度是140厘米和60厘米，现在要在广告牌的四周钉一圈铝条，至少需要多少厘米的铝条。如果每米铝条的成本是65元，这块广告牌围一圈铝条需要多少钱。(不考虑接合处损耗)",
+        "options":  [
+
+                    ],
+        "answer":  "400厘米;260元",
+        "solution":  "【分析】先求平行四边形周长，再换算成米，最后乘每米单价.\n【详解】(140+60)×2=200×2=400(厘米)。400厘米=4米。4×65=260(元)。答：至少需要400厘米铝条，费用是260元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "平行四边形的概念及特点",
+                                 "两，三位数与一位数的不进位乘法",
+                                 "两，三位数与一位数连续进位的乘法"
+                             ],
+        "source":  "25-26 四年级上·新疆喀什·期末"
+    },
+    {
+        "id":  "400",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "动画片是通过连续播放一系列的画面，给视觉造成连续变化的图画。如果1秒播放24帧画面，那么一段时长6分钟的动画片需要播放多少帧画面。",
+        "options":  [
+
+                    ],
+        "answer":  "8640帧",
+        "solution":  "【分析】先把6分钟换算成秒，再乘每秒帧数.\n【详解】6分钟=360秒,24×360=8640(帧)。答：需要播放8640帧画面。",
+        "knowledge_points":  [
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "时，分，秒之间的换算与比较"
+                             ],
+        "source":  "25-26 四年级上·新疆喀什·期末"
+    },
+    {
+        "id":  "401",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面选项中，错误的是( )。",
+        "options":  [
+                        "A. 过一点能画无数条直线。",
+                        "B. 当两条直线相交成直角时，这两条直线就互相垂直。",
+                        "C. 9 时 15 分，钟面上分针和时针所夹的角是一个平角。",
+                        "D. 两个锐角的和一定比平角小。"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】根据所学的直线，垂直，钟表角度和角的性质，逐个分析选项并作答.\n【详解】A。直线可以向任意方向无限延伸，过一点可画无数条直线，该选项说法正确。B。当两条直线相交成直角时，这两条直线就互相垂直，此说法正确。C。9时15分时针与分针所成角小于180°,不是平角，该项错误。D。两个锐角和小于180°,一定比平角小。故答案为:C.",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征",
+                                 "垂直的特征",
+                                 "平角，周角的认识及特征",
+                                 "角的分类及换算"
+                             ],
+        "source":  "25-26 四年级上·天津河西·期末"
+    },
+    {
+        "id":  "402",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一种蚂蚁,100 只的质量大约是 4 克,1 亿只这样的蚂蚁质量大约是()。",
+        "options":  [
+                        "A. 1 吨",
+                        "B. 4 千克",
+                        "C. 40 千克",
+                        "D. 4 吨"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】先求1亿里有多少个100，再乘每100只的质量4克，最后把克换算为吨.\n【详解】100000000÷100=1000000(组),1000000×4=4000000(克)=4000千克=4吨。故答案为:D.",
+        "knowledge_points":  [
+                                 "亿以上数的组成",
+                                 "一亿有多大",
+                                 "克，千克之间的换算与比较",
+                                 "吨，千克之间的换算与比较"
+                             ],
+        "source":  "25-26 四年级上·天津河西·期末"
+    },
+    {
+        "id":  "403",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面各数中，读两个\"零\"的是( )。",
+        "options":  [
+                        "A. 9230705",
+                        "B. 9230075",
+                        "C. 9023750",
+                        "D. 9200375"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】按整数读法分级读数，判断各数读\"零\"的次数.\n【详解】A读作\"九百二十三万零七百零五\"，读两个\"零\";B,C,D都只读一个\"零\"。故答案为:A.",
+        "knowledge_points":  [
+                                 "亿以内数的读，写法"
+                             ],
+        "source":  "25-26 四年级上·天津河西·期末"
+    },
+    {
+        "id":  "404",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面各数中，一个零也不读出来的数是( )。",
+        "options":  [
+                        "A. 20030000",
+                        "B. 20300000",
+                        "C. 23000000"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】按\"分级读数，每级末尾0不读，其他位置连续0只读一个0\"的规则判断.\n【详解】A读作\"二千零三万\",B读作\"二千零三十万\"，都读出一个零;C读作\"二千三百万\"，一个零也不读。故答案为:C.",
+        "knowledge_points":  [
+                                 "亿以内数的读，写法"
+                             ],
+        "source":  "25-26 四年级上·甘肃兰州·期末"
+    },
+    {
+        "id":  "405",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一盒牙膏的原价是 24元，现在商场搞促销活动，每盒优惠 2 元，现在买 3 盒牙膏需要多少钱。下面列式正确的有( )个.\n①24-2×3 ②(24-2)×3 ③24×3-2 ④24×3-2×3",
+        "options":  [
+                        "A. 0",
+                        "B. 1",
+                        "C. 2"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】可先求现价再乘数量，也可先求原总价再减总优惠.\n【详解】正确列式是②(24-2)×3和④24×3-2×3，共2个。故答案为:C.",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "乘加，乘减混合运算",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "25-26 四年级上·甘肃平凉·期末"
+    },
+    {
+        "id":  "406",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "两个数的商是 20，要使商变为 2，下面的说法正确的是( )。",
+        "options":  [
+                        "A. 被除数不变，除数除以 10",
+                        "B. 被除数不变，除数扩大到原来的10倍",
+                        "C. 除数不变，被除数扩大到原来的10倍"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】利用商的变化规律：被除数不变，除数扩大几倍，商缩小到原来的几分之一.\n【详解】要把20变成2，应使除数扩大10倍。故答案为:B.",
+        "knowledge_points":  [
+                                 "商的变化规律及应用"
+                             ],
+        "source":  "25-26 四年级上·甘肃陇南·期末"
+    },
+    {
+        "id":  "407",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "32 个 128 的和大约是( )。",
+        "options":  [
+                        "A. 3600",
+                        "B. 3900",
+                        "C. 4800"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】估算32×128，可把32看作30,128看作130.\n【详解】30×130=3900，所以大约是3900。故答案为:B.",
+        "knowledge_points":  [
+                                 "三位数乘两位数的估算"
+                             ],
+        "source":  "25-26 四年级上·甘肃陇南·期末"
+    },
+    {
+        "id":  "408",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "在数字 7 与 8 之间添( )个 0，就组成了七亿零八。",
+        "options":  [
+                        "A. 7",
+                        "B. 6",
+                        "C. 5"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】\"七亿零八\"写作700000008.\n【详解】7与8之间共有7个0。故答案为:A.",
+        "knowledge_points":  [
+                                 "亿以上数的读，写法",
+                                 "亿以上数的组成"
+                             ],
+        "source":  "25-26 四年级上·甘肃陇南·期末"
+    },
+    {
+        "id":  "409",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "星期天爸爸要做以下家务：擦玻璃20分钟，收拾厨房15分钟，用全自动洗衣机洗衣服30分钟，晾衣服10分钟。爸爸做完这些事最少需要( )分钟。",
+        "options":  [
+                        "A. 75",
+                        "B. 45",
+                        "C. 55"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】洗衣机可与其他家务并行，晾衣服需在洗完后进行.\n【详解】擦玻璃+收拾厨房共35分钟，可覆盖洗衣30分钟;再晾衣10分钟，共45分钟。故答案为:B.",
+        "knowledge_points":  [
+                                 "两位数与整十数的加法",
+                                 "沏茶问题"
+                             ],
+        "source":  "25-26 四年级上·甘肃张掖·期末"
+    },
+    {
+        "id":  "410",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "亚洲的面积约是4458 万( )。",
+        "options":  [
+                        "A. 平方米",
+                        "B. 公顷",
+                        "C. 平方千米"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】计量很大区域面积常用平方千米.\n【详解】亚洲面积应以平方千米计。故答案为:C.",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识"
+                             ],
+        "source":  "25-26 四年级上·甘肃甘南·期末"
+    },
+    {
+        "id":  "411",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "一个长方形的面积是120平方米，若它的长扩大到10倍，宽不变，面积是()平方米。",
+        "options":  [
+                        "A. 1200",
+                        "B. 2400",
+                        "C. 12000"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】面积=长×宽，一个因数扩大10倍，积也扩大10倍.\n【详解】120×10=1200(平方米)。故答案为:A.",
+        "knowledge_points":  [
+                                 "三位数与整十数的口算乘法",
+                                 "长方形的面积",
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·甘肃甘南·期末"
+    },
+    {
+        "id":  "412",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "a / b = 45，当 b 除以5 时，要使商不变,a 应( )。",
+        "options":  [
+                        "A. 乘 5",
+                        "B. 乘 10",
+                        "C. 除以 5"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】被除数和除数同时乘或除以同一个非0数，商不变.\n【详解】b除以5,a也应除以5。故答案为:C.",
+        "knowledge_points":  [
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "25-26 四年级上·甘肃甘南·期末"
+    },
+    {
+        "id":  "413",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "北京奥林匹克森林公园是亚洲最大的城市绿化景观，位于北京中轴延长线的最北端，占地约 680( )。",
+        "options":  [
+                        "A. 米",
+                        "B. 千米",
+                        "C. 平方米",
+                        "D. 公顷"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】题目要求面积单位，且规模较大.\n【详解】\"680公顷\"符合实际，米和千米是长度单位,680平方米过小。故答案为:D.",
+        "knowledge_points":  [
+                                 "面积单位的选择",
+                                 "公顷，平方千米的认识"
+                             ],
+        "source":  "25-26四年级上·北京西城·期末"
+    },
+    {
+        "id":  "414",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "读下面的数，只用读出一个零的是( )。",
+        "options":  [
+                        "A. 606600",
+                        "B. 600606",
+                        "C. 6000660",
+                        "D. 6060660"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】按整数读法逐个判断读\"零\"的次数.\n【详解】A不读零;B读两个零;C读作\"六百万零六百六十\"，读一个零;D读两个零。故答案为:C.",
+        "knowledge_points":  [
+                                 "亿以内数的读，写法"
+                             ],
+        "source":  "2025·广东揭阳·小升初模拟"
+    },
+    {
+        "id":  "415",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "码头上有三艘轮船等待卸货，甲船要10小时卸空，乙船要12小时卸空，丙船要9小时卸空。因人手不足，只能一艘一艘地卸，要使三艘船的等待时间总和最少，应该按怎样的顺序卸货。( )。",
+        "options":  [
+                        "A. 丙，甲，乙",
+                        "B. 乙，甲，丙",
+                        "C. 乙，丙，甲"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】等待总时间最少时，按处理时间从短到长安排.\n【详解】9\u003c10\u003c12，应先丙，再甲，最后乙。故答案为:A.",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "25-26四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "416",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列说法中，正确的是( )。",
+        "options":  [
+                        "A. 梯形只有两条高。",
+                        "B. 一个15°的角放在10倍的放大镜下看就变成150°。",
+                        "C. 画一个60°的角，既可以用量角器，也可以用三角尺。"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】梯形有无数条高;放大镜不改变角度大小;60°可由量角器或三角尺作出.\n【详解】A,B错误,C正确。故答案为:C.",
+        "knowledge_points":  [
+                                 "用三角尺画角",
+                                 "用量角器画角",
+                                 "角的初步认识及辨认",
+                                 "梯形的高及画法"
+                             ],
+        "source":  "25-26四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "417",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "要使 206*□2 的积是五位数,□里最小填( )。",
+        "options":  [
+                        "A. 4",
+                        "B. 5",
+                        "C. 6"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】分别试算42,52,62与206相乘，找出最小满足五位数的情况.\n【详解】206×42=8652(四位);206×52=10712(五位),故最小是5。故答案为:B.",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "418",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "145×24，因数十位上的2去乘145，得( )。",
+        "options":  [
+                        "A. 290",
+                        "B. 2900",
+                        "C. 29000"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】24十位上的2表示2个十，即20.\n【详解】145×20=2900。故答案为:B.",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "25-26 四年级上·甘肃临夏·期末"
+    },
+    {
+        "id":  "419",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一个数四舍五入到万位后约等于 8 万，这个数最大是( )。",
+        "options":  [
+                        "A. 84999",
+                        "B. 85000",
+                        "C. 79999"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】四舍五入到万位得8万的范围是[79500, 84999].\n【详解】最大是84999。故答案为:A.",
+        "knowledge_points":  [
+                                 "亿以内数的近似数"
+                             ],
+        "source":  "25-26 四年级上·甘肃临夏·期末"
+    },
+    {
+        "id":  "420",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "把一个平角分成两个角，其中一个是钝角，另一个角是( )。",
+        "options":  [
+                        "A. 锐角",
+                        "B. 直角",
+                        "C. 钝角"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】平角180°,钝角大于90°小于180°.\n【详解】另一个角=180°-钝角，小于90°,是锐角。故答案为:A.",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "421",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "AI 工厂生产了 386个智能芯片，每 30 个装一盒，可以装满( )盒。",
+        "options":  [
+                        "A. 12",
+                        "B. 13",
+                        "C. 14"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】求\"装满\"盒数，取整数商.\n【详解】386÷30=12。26，可装满12盒。故答案为:A.",
+        "knowledge_points":  [
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "422",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面各数中，最接近 AI 芯片算力\"10 亿次/秒\"的是()。",
+        "options":  [
+                        "A. 99999999",
+                        "B. 999999999",
+                        "C. 1001000000"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】比较各数与1000000000的差.\n【详解】A差900000001,B差1,C差1000000，最小是B。故答案为:B.",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "亿以上数的改写"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "423",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "( )公顷=260000 平方千米;8 公顷 40 平方米=( )平方米;54000000 平方米=( )公顷=( )平方千米。",
+        "options":  [
+
+                    ],
+        "answer":  "26000000;80040;5400;54",
+        "solution":  "【分析】用面积单位进率:1平方千米=100公顷,1公顷=10000平方米,1平方千米=1000000平方米.\n【详解】260000×100=26000000(公顷);8×10000+40=80040(平方米);54000000÷10000=5400(公顷),54000000÷1000000=54(平方千米)。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "424",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个整数四舍五入到万位约是 50 万，这个数最小是( );一个整数四舍五入到亿位约是 6 亿，这个数最大是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "495000;649999999",
+        "solution":  "【分析】按四舍五入边界值求最小与最大.\n【详解】到万位约50万，最小是495000;到亿位约6亿，最大是649999999。",
+        "knowledge_points":  [
+                                 "亿以内数的近似数",
+                                 "亿以上数的近似数"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "425",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在下面的括号里填上合适的单位名称。小明家卧室的面积是 15( );动物园的占地面积约是 3( );中国原始森林总面积约为 16 万( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方米;公顷;平方千米",
+        "solution":  "【分析】按面积大小选择合适单位.\n【详解】卧室用平方米，动物园占地常用公顷，国家级大面积用平方千米。",
+        "knowledge_points":  [
+                                 "平方厘米，平方分米，平方米的认识",
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "426",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "2025年\"十一\"旅游数据：全国国内出游888000000人次，国内出游总花费八千零九十亿零六百万元.\n(1) 888000000读作( ),省略亿位后面的尾数约是( ).\n(2) 八千零九十亿零六百万写作( ),其中\"9\"表示( )。",
+        "options":  [
+
+                    ],
+        "answer":  "八亿八千八百万;9亿 809006000000;9个十亿",
+        "solution":  "【分析】考查亿以上数读写法，数位意义与近似数.\n【详解】888000000读作八亿八千八百万，约9亿;八千零九十亿零六百万写作809006000000，\"9\"在十亿位，表示9个十亿。",
+        "knowledge_points":  [
+                                 "亿以上数的读，写法",
+                                 "整数的数级，数位和计数单位的认识",
+                                 "亿以上数的近似数"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "427",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "光伏累计装机量十一亿一千七百二十三万千瓦，横线上的数写作( )。用\"四舍五入\"法省略亿后面的尾数约是( )亿。",
+        "options":  [
+
+                    ],
+        "answer":  "1117230000;11",
+        "solution":  "【分析】先写出数，再看千万位四舍五入到亿位.\n【详解】十一亿一千七百二十三万写作1117230000;省略亿后尾数约11亿。",
+        "knowledge_points":  [
+                                 "亿以上数的读，写法",
+                                 "亿以上数的近似数"
+                             ],
+        "source":  "25-26四年级上·天津河西·期末"
+    },
+    {
+        "id":  "428",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算786÷12时，把12看作( )试商，商是( )位数，余数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "10;两;6",
+        "solution":  "【分析】试商时常把12看作10，再计算精确结果.\n【详解】786÷12=65。6，因此看作10试商，商是两位数，余数6。",
+        "knowledge_points":  [
+                                 "三位数除以两位数的有余数除法",
+                                 "多位数除以两位数的试商"
+                             ],
+        "source":  "25-26 四年级上·新疆和田·期末"
+    },
+    {
+        "id":  "429",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个正方形花园的面积是1公顷，现在边长增加100米，增加后的花园面积是( )公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "4",
+        "solution":  "【分析】1公顷=10000平方米，对应正方形边长100米;边长增到200米后求新面积并换算.\n【详解】200×200=40000平方米=4公顷。",
+        "knowledge_points":  [
+                                 "正方形的面积",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·新疆和田·期末"
+    },
+    {
+        "id":  "430",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个十位数，最高位上是6，百万位和百位上是5，其它数位上都是最小的自然数，这个数写作( ),省略亿位后面的尾数是( )亿。",
+        "options":  [
+
+                    ],
+        "answer":  "6005000500;60",
+        "solution":  "【分析】按数位写数;省略亿位后尾数看千万位四舍五入.\n【详解】该数为6005000500，约60亿。",
+        "knowledge_points":  [
+                                 "亿以上数的读，写法",
+                                 "亿以上数的近似数"
+                             ],
+        "source":  "25-26 四年级上·新疆和田·期末"
+    },
+    {
+        "id":  "431",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在横线上填\"\u003e\"\"\u003c\"或\"=\".\n560×30( )56×300;7平方千米( )7000公顷;480÷32( )480÷4÷8;300300( )303000。",
+        "options":  [
+
+                    ],
+        "answer":  "=;\u003c;=;\u003c",
+        "solution":  "【分析】分别用积不变规律，单位换算，除法性质和整数大小比较判断.\n【详解】560×30=56×300;7平方千米=700公顷\u003c7000公顷;480÷32=480÷4÷8;300300\u003c303000。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "除数是两位数的笔算除法",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26四年级上·甘肃兰州·期末"
+    },
+    {
+        "id":  "432",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "由9个亿,6个千万,4个万,8个百和3个一组成的数写作( ),读作( ),省略亿位后面的尾数约是( )亿。",
+        "options":  [
+
+                    ],
+        "answer":  "960040803;九亿六千零四万零八百零三;10",
+        "solution":  "【分析】按\"先亿级，再万级，后个级\"写数并读数，再按亿位近似.\n【详解】写作960040803，读作九亿六千零四万零八百零三;省略亿位后尾数约10亿。",
+        "knowledge_points":  [
+                                 "亿以上数的读，写法",
+                                 "亿以上数的组成",
+                                 "整数的数级，数位和计数单位的认识",
+                                 "亿以上数的近似数"
+                             ],
+        "source":  "25-26四年级上·甘肃兰州·期末"
+    },
+    {
+        "id":  "433",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个数由8个千万,6个十万和5个百组成，这个数写作( ),读作( )。",
+        "options":  [
+
+                    ],
+        "answer":  "80600500;八千零六十万零五百",
+        "solution":  "【分析】根据计数单位确定各数位，其余数位补0，再按分级读数.\n【详解】写作80600500，读作八千零六十万零五百。",
+        "knowledge_points":  [
+                                 "亿以内数的读，写法"
+                             ],
+        "source":  "25-26 四年级上·新疆和田·期末"
+    },
+    {
+        "id":  "434",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个平底锅，每次最多只能煎2个鸡蛋，两面都要煎，每面2分钟。煎3个鸡蛋至少要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】采用交替煎法可减少总时间.\n【详解】先煎第1,2个正面2分钟;再煎第1个反面和第3个正面2分钟;最后煎第2个反面和第3个反面2分钟，共6分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "2014四年级·全国·课后作业"
+    },
+    {
+        "id":  "435",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "4时整，时针与分针的较小夹角是( )角;10时整，时针与分针的较小夹角是( )角。(选填\u0027锐\u0027\u0027钝\u0027或\u0027直\u0027)",
+        "options":  [
+
+                    ],
+        "answer":  "钝角;锐角",
+        "solution":  "【分析】钟面每大格是30°.\n【详解】4时夹角为4×30°=120°,是钝角;10时较小夹角为2×30°=60°,是锐角。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "25-26四年级上·甘肃陇南·期末"
+    },
+    {
+        "id":  "436",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "李明走一步的距离是58厘米，他从家到学校一共走了650步，他家到学校大约有( )米。",
+        "options":  [
+
+                    ],
+        "answer":  "390",
+        "solution":  "【分析】题目求\u0027大约\u0027,可估算58≈60.\n【详解】60×650=39000(厘米)=390(米),所以约390米。",
+        "knowledge_points":  [
+                                 "三位数乘两位数的估算",
+                                 "厘米和米之间的进率与换算"
+                             ],
+        "source":  "25-26四年级上·甘肃陇南·期末"
+    },
+    {
+        "id":  "437",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上合适的数.\n5000000平方米=( )平方千米=( )公顷;( )÷40=6。38;740÷( )=9。20。",
+        "options":  [
+
+                    ],
+        "answer":  "5;500;278;80",
+        "solution":  "【分析】面积单位换算:1平方千米=1000000平方米,1公顷=10000平方米;有余数除法：被除数=除数×商+余数.\n【详解】5000000平方米=5平方千米=500公顷;6×40+38=278;(740-20)÷9=80。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "25-26四年级上·甘肃张掖·期末"
+    },
+    {
+        "id":  "438",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在算盘上，用一颗上珠和两颗下珠表示最小的七位数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "1000006",
+        "solution":  "【分析】上珠表示5，下珠表示1;同时要满足是最小七位数.\n【详解】百万位放1个下珠得1000000;个位放1上珠和1下珠得6，合起来是1000006。",
+        "knowledge_points":  [
+                                 "算盘的认识与使用"
+                             ],
+        "source":  "25-26 四年级上·甘肃张掖·期末"
+    },
+    {
+        "id":  "439",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上合适的单位。国家游泳中心(\u0027水立方\u0027)的占地面积约6( )。课桌面的面积约40( )。教室地面的面积约56( )。",
+        "options":  [
+
+                    ],
+        "answer":  "公顷;平方分米;平方米",
+        "solution":  "【分析】根据生活实际选择合适面积单位.\n【详解】\u0027水立方\u0027占地较大用公顷;课桌面较小用平方分米;教室地面用平方米。",
+        "knowledge_points":  [
+                                 "面积单位的选择",
+                                 "公顷，平方千米的认识",
+                                 "平方厘米，平方分米，平方米的认识"
+                             ],
+        "source":  "25-26四年级上·甘肃张掖·期末"
+    },
+    {
+        "id":  "440",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "截止2025年2月12日深中通道总计车流量约19360400车次，读作:( )车次。",
+        "options":  [
+
+                    ],
+        "answer":  "一千九百三十六万零四百",
+        "solution":  "【分析】按万级和个级读数.\n【详解】19360400读作\u0027一千九百三十六万零四百\u0027。",
+        "knowledge_points":  [
+                                 "亿以内数的读，写法"
+                             ],
+        "source":  "25-26 四年级上·甘肃张掖·期末"
+    },
+    {
+        "id":  "441",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "聪聪帮妈妈做家务，整理房间用7分钟，烧开水用9分钟，沏茶用3分钟，扫地用2分钟，他做完这些事至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "12",
+        "solution":  "【分析】能同时做的事情同时进行.\n【详解】烧开水9分钟时可完成整理房间7分钟和扫地2分钟;之后沏茶3分钟，总计9+3=12分钟。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25 四年级上·湖南长沙·期末"
+    },
+    {
+        "id":  "442",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "14平方米=( )平方分米;8平方千米=( )公顷;40000平方米=( )公顷;5平方米=( )平方厘米;12公顷=( )平方米;800平方厘米=( )平方分米。",
+        "options":  [
+
+                    ],
+        "answer":  "1400;800;4;50000;120000;8",
+        "solution":  "【分析】用面积单位进率:1平方米=100平方分米,1平方分米=100平方厘米,1平方千米=100公顷,1公顷=10000平方米.\n【详解】依次计算得到1400,800,4,50000,120000,8。",
+        "knowledge_points":  [
+                                 "平方厘米，平方分米，平方米之间的进率与换算",
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·甘肃甘南·期末"
+    },
+    {
+        "id":  "443",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "括号里最大能填几。\n( )×16\u003c330;( )×82\u003c720;( )×48\u003c248;53×( )\u003c302。",
+        "options":  [
+
+                    ],
+        "answer":  "20;8;5;5",
+        "solution":  "【分析】用除法求上限，取不超过条件的最大整数.\n【详解】330÷16=20。10,720÷82=8。64,248÷48=5。8,302÷53=5。37，所以填20,8,5,5。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "25-26 四年级上·甘肃甘南·期末"
+    },
+    {
+        "id":  "444",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "有249朵花，按5朵红花,9朵黄花,13朵绿花的顺序排列，最后一朵花是( )色的。这249朵花中，有红花( )朵，黄花( )朵，绿花( )朵。",
+        "options":  [
+
+                    ],
+        "answer":  "黄;50;82;117",
+        "solution":  "【分析】一个循环有5+9+13=27朵.\n【详解】249÷27=9。6，余6落在新一轮黄花区间，所以最后一朵是黄花。红花=9×5+5=50;黄花=9×9+1=82;绿花=9×13=117。",
+        "knowledge_points":  [
+                                 "三位数除以两位数的有余数除法",
+                                 "乘加，乘减混合运算",
+                                 "周期问题"
+                             ],
+        "source":  "25-26四年级上·甘肃甘南·期末"
+    },
+    {
+        "id":  "445",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "个级的数位包括( ),( ),( ),( )。",
+        "options":  [
+
+                    ],
+        "answer":  "个位;十位;百位;千位",
+        "solution":  "【分析】个级共有四个数位.\n【详解】依次为个位，十位，百位，千位。",
+        "knowledge_points":  [
+                                 "整数的数级，数位和计数单位的认识"
+                             ],
+        "source":  "25-26 四年级上·甘肃甘南·期末"
+    },
+    {
+        "id":  "446",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "2024年广东省GDP约是十四万一千六百三十三亿元，写作( ),省略万亿位后面的尾数约是( )万亿。",
+        "options":  [
+
+                    ],
+        "answer":  "14163300000000;14",
+        "solution":  "【分析】先按数位写数，再按万亿位取近似数.\n【详解】写作14163300000000;千亿位是1，小于5，省略后约14万亿。",
+        "knowledge_points":  [
+                                 "亿以上数的近似数",
+                                 "亿以上数的改写",
+                                 "亿以上数的读，写法"
+                             ],
+        "source":  "2025·广东揭阳·小升初模拟"
+    },
+    {
+        "id":  "447",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "5公顷45平方米=( )平方米;60平方千米=( )公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "50045;6000",
+        "solution":  "【分析】1公顷=10000平方米,1平方千米=100公顷.\n【详解】5公顷45平方米=5×10000+45=50045平方米;60平方千米=60×100=6000公顷。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "448",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "王老师早上起床后要做很多事情：洗脸刷牙5分钟，听广播20分钟，吃早饭10分钟，收碗抹桌子2分钟，王老师做完这些事情至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "20",
+        "solution":  "【分析】将可并行的短任务放到听广播期间完成.\n【详解】5+10+2=17分钟，小于20分钟，因此总时长由听广播决定，为20分钟。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "449",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "李奶奶到超市买汇源果汁，一箱要15元，买4箱送1箱，李奶奶一次买4箱，每箱果汁可以便宜( )元。",
+        "options":  [
+
+                    ],
+        "answer":  "3",
+        "solution":  "【分析】买4送1表示按4箱的钱拿到5箱.\n【详解】总价15×4=60元，平均每箱60÷5=12元，比原价便宜15-12=3元。",
+        "knowledge_points":  [
+                                 "经济问题"
+                             ],
+        "source":  "25-26四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "450",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "李叔叔用一根50厘米的铁丝围成了一个平行四边形框架，其中一条边长是18厘米，与它相邻的另一条边长( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "7",
+        "solution":  "【分析】平行四边形周长=2×(邻边和).\n【详解】邻边和=50÷2=25，另一边=25-18=7厘米。",
+        "knowledge_points":  [
+                                 "平行四边形的周长",
+                                 "除加，除减混合运算"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "451",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "钟面上5时整时针和分针组成的较小的角是( )°;3:30，时针和分针组成较小的角是( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "150;锐角",
+        "solution":  "【分析】每大格30°.\n【详解】5时:5×30°=150°。3:30时较小夹角为75°,属于锐角。",
+        "knowledge_points":  [
+                                 "半点时间与钟面指针的位置",
+                                 "整点时间与钟面指针的位置",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "角度的计算"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "452",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在下面括号里填上合适的数.\n4600平方分米=( )平方米;8000公顷=( )平方千米;26平方千米=( )公顷=( )平方米。",
+        "options":  [
+
+                    ],
+        "answer":  "46;80;2600;26000000",
+        "solution":  "【分析】利用面积单位进率换算.\n【详解】4600÷100=46;8000÷100=80;26×100=2600;2600×10000=26000000。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "平方厘米，平方分米，平方米之间的进率与换算",
+                                 "三位数乘两位数，乘数末尾有0"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "453",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "亚洲最大的岛屿加里曼丹岛面积是七十四万三千三百平方千米，横线上的数字写作( ),省略\u0027万\u0027位后面的尾数约是( )万。",
+        "options":  [
+
+                    ],
+        "answer":  "743300;74",
+        "solution":  "【分析】先写数，再按万位四舍五入.\n【详解】写作743300;省略\u0027万\u0027位后面的尾数约74万。",
+        "knowledge_points":  [
+                                 "亿以内数的近似数",
+                                 "亿以内数的读，写法"
+                             ],
+        "source":  "25-26四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "454",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在下面括号里填上\u0027\u003e\u0027\u0027\u003c\u0027或\u0027=\u0027.\n899999( )89999;67450000( )6745万;70000001( )70万;87000000( )870万。",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;=;\u003e;\u003e",
+        "solution":  "【分析】比较整数大小先看位数，再看高位;带\u0027万\u0027的先还原.\n【详解】899999\u003e89999;67450000=6745万;70000001\u003e70万;87000000\u003e870万。",
+        "knowledge_points":  [
+                                 "亿以内数的改写",
+                                 "大数的比较"
+                             ],
+        "source":  "25-26四年级上·甘肃临夏·期末"
+    },
+    {
+        "id":  "455",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在一道没有余数的除法算式中，当被除数加上一个( )数时，商就增加1。",
+        "options":  [
+
+                    ],
+        "answer":  "除数",
+        "solution":  "【分析】被除数=除数×商.\n【详解】商增加1，表示被除数要增加1个除数，所以填\u0027除数\u0027。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "乘，除法的意义和各部分间的关系"
+                             ],
+        "source":  "25-26 四年级上·甘肃临夏·期末"
+    },
+    {
+        "id":  "456",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "5公顷=( )平方米;7平方千米=( )公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "50000;700",
+        "solution":  "【分析】1公顷=10000平方米,1平方千米=100公顷.\n【详解】5×10000=50000;7×100=700。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·甘肃临夏·期末"
+    },
+    {
+        "id":  "457",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "两个数相除的商是23，把被除数和除数都扩大到原来的7倍，商是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "23",
+        "solution":  "【分析】商不变性质：被除数和除数同时乘同一个非0数，商不变.\n【详解】所以商仍是23。",
+        "knowledge_points":  [
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "25-26 四年级上·甘肃临夏·期末"
+    },
+    {
+        "id":  "458",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "速度×时间=( );工作总量/( )=工作时间。",
+        "options":  [
+
+                    ],
+        "answer":  "路程;工作效率",
+        "solution":  "【分析】套用行程和工程基本数量关系.\n【详解】速度×时间=路程;工作总量/工作效率=工作时间。",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "有具体量的工程问题"
+                             ],
+        "source":  "25-26 四年级上·甘肃临夏·期末"
+    },
+    {
+        "id":  "459",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "刘老师买了36本同样的图书，付出600元，找回60元，每本图书单价是( )元。",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】单价=总价/数量，先求实付总价.\n【详解】600-60=540(元),540÷36=15(元)。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "25-26四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "460",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "从甲地到乙地的距离是840千米，汽车每小时行70千米,( )小时到达。如果要提前2小时到达，每小时要行( )千米。",
+        "options":  [
+
+                    ],
+        "answer":  "12;84",
+        "solution":  "【分析】先求原时间，再按提前后的时间求新速度.\n【详解】840÷70=12(小时);提前2小时后用时10小时,840÷10=84(千米/时)。",
+        "knowledge_points":  [
+                                 "除数是整十数的口算除法",
+                                 "除数是整十数的笔算除法",
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "461",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "农场的种植面积是3平方千米，合( )公顷;其中智能灌溉区占46公顷，合( )平方米。",
+        "options":  [
+
+                    ],
+        "answer":  "300;460000",
+        "solution":  "【分析】1平方千米=100公顷,1公顷=10000平方米.\n【详解】3平方千米=300公顷;46公顷=460000平方米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "462",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个九位数，最高位是8，百万位是6，十位是3，其余各位都是0，这个数是( ),读作( ),省略亿位后面的尾数约是( )亿。",
+        "options":  [
+
+                    ],
+        "answer":  "806000030;八亿零六百万零三十;8",
+        "solution":  "【分析】按数位写数并读数，近似到亿位看千万位.\n【详解】写作806000030，读作八亿零六百万零三十，约是8亿。",
+        "knowledge_points":  [
+                                 "亿以上数的读，写法",
+                                 "亿以上数的组成",
+                                 "整数的数级，数位和计数单位的认识",
+                                 "亿以上数的近似数"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "463",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个数由80个万,4个千,95个一组成，这个数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "804095",
+        "solution":  "【分析】分别换算后相加.\n【详解】80个万=800000,4个千=4000,95个一=95，总和804095。",
+        "knowledge_points":  [
+                                 "亿以内数的组成",
+                                 "整数的数级，数位和计数单位的认识"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "464",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "排序题",
+        "difficulty":  0.85,
+        "question":  "按从大到小的顺序排列各数:76890,24508,136780,( )\u003e( )\u003e( )。",
+        "options":  [
+
+                    ],
+        "answer":  "136780;76890;24508",
+        "solution":  "【分析】位数多的数更大;同位数比较高位.\n【详解】136780是六位数最大;76890与24508比较,7万位大于2万位，所以136780\u003e76890\u003e24508。",
+        "knowledge_points":  [
+                                 "大数的比较"
+                             ],
+        "source":  "25-26 四年级上·甘肃临夏·期末"
+    },
+    {
+        "id":  "465",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "排序题",
+        "difficulty":  0.85,
+        "question":  "把809000,890000,800090,80009这四个数按照从大到小的顺序排列为( )\u003e( )\u003e( )\u003e( )。",
+        "options":  [
+
+                    ],
+        "answer":  "890000;809000;800090;80009",
+        "solution":  "【分析】先比位数，再比高位.\n【详解】六位数中890000\u003e809000\u003e800090，五位数80009最小，所以890000\u003e809000\u003e800090\u003e80009。",
+        "knowledge_points":  [
+                                 "大数的比较"
+                             ],
+        "source":  "25-26四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "466",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.94,
+        "question":  "除法算式中，被除数和除数同时除以5，商不变。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】被除数和除数同时除以同一个非0数，商不变.\n【详解】该说法正确。",
+        "knowledge_points":  [
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "25-26四年级上·新疆和田·期末"
+    },
+    {
+        "id":  "467",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "一个办公室的面积约为100平方米。100个这样的办公室约为1公顷。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】1公顷=10000平方米.\n【详解】100×100=10000(平方米)=1公顷，所以说法正确。",
+        "knowledge_points":  [
+                                 "三位数与整十数的口算乘法",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·新疆和田·期末"
+    },
+    {
+        "id":  "468",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "100块1平方米的正方形绿地的总面积是1公顷。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】先求总面积再与1公顷比较.\n【详解】100块1平方米共100平方米，而1公顷=10000平方米，不相等，所以错误。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 五年级上·河北张家口·期末"
+    },
+    {
+        "id":  "469",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "把一个长方形框架拉成一个平行四边形，周长变小，面积不变。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】框架拉动时边长不变，所以周长不变;高通常变小，面积变小.\n【详解】\u0027周长变小，面积不变\u0027不成立，故错误。",
+        "knowledge_points":  [
+                                 "平行四边形的概念及特点"
+                             ],
+        "source":  "25-26 五年级上·广东汕头·期末"
+    },
+    {
+        "id":  "470",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "计算315÷39时，把39看作40试商，初商偏小，商应调大。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】把39看作40，除数变大，试得的商往往偏小.\n【详解】315÷39试商7时,39×7=273，余42，余数大于除数，说明商7偏小，应调大为8，所以说法正确。",
+        "knowledge_points":  [
+                                 "多位数除以两位数的试商"
+                             ],
+        "source":  "25-26 四年级上·甘肃陇南·期末"
+    },
+    {
+        "id":  "471",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "早在两千多年前，我国劳动人民用\u0027亩\u0027作土地面积单位，一亩等于667平方米。现在\u0027亩\u0027这个单位已经不是我国的法定计量单位了。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】\u0027亩\u0027是传统面积单位，不属于我国现行法定计量单位.\n【详解】题干表述正确，故判断为√。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识"
+                             ],
+        "source":  "25-26 四年级上·甘肃张掖·期末"
+    },
+    {
+        "id":  "472",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.94,
+        "question":  "在数字4和6之间填上5个0。这个数读作四十六万。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】在4和6之间填5个0得到4000006.\n【详解】4000006读作\u0027四百万零六\u0027,不是\u0027四十六万\u0027,所以错误。",
+        "knowledge_points":  [
+                                 "亿以内数的读，写法"
+                             ],
+        "source":  "25-26 四年级上·甘肃张掖·期末"
+    },
+    {
+        "id":  "473",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "与100万相邻的两个数是10万和1000万。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】\u0027相邻\u0027指前后相差1.\n【详解】100万=1000000，与它相邻的是999999和1000001，不是10万和1000万，故错误。",
+        "knowledge_points":  [
+                                 "亿以内数的改写"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "474",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "从4:00到5:00，时针所转的角是180度。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】钟面一周360°,12小时转一周，每小时转30°.\n【详解】从4:00到5:00是1小时，时针只转30°,不是180°,所以错误。",
+        "knowledge_points":  [
+                                 "角的度量",
+                                 "角度的计算"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "475",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "360÷24和360÷4÷6的结果相同。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】可分别计算比较.\n【详解】360÷24=15;360÷4÷6=90÷6=15，结果相同，故正确。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "多位数与一位数的连除"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "476",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "计算题",
+        "difficulty":  0.94,
+        "question":  "672是12的多少倍。",
+        "options":  [
+
+                    ],
+        "answer":  "56",
+        "solution":  "【分析】求一个数是另一个数的几倍，用除法.\n【详解】672÷12=56，所以672是12的56倍。",
+        "knowledge_points":  [
+                                 "求一个数是另一个数的几倍",
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "25-26 四年级上·甘肃临夏·期末"
+    },
+    {
+        "id":  "477",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "被除数是5960，除数是48，商和余数各是多少。",
+        "options":  [
+
+                    ],
+        "answer":  "商124，余数8",
+        "solution":  "【分析】直接进行有余数除法计算.\n【详解】5960÷48=124。8，所以商是124，余数是8。",
+        "knowledge_points":  [
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "25-26四年级上·甘肃临夏·期末"
+    },
+    {
+        "id":  "478",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "妙想正在读一本590页的故事书，不小心合上了，她记得刚读完的连续两页页码之和是99.\n(1) 妙想刚读完的两页页码分别是( )页和( )页.\n(2) 如果之后妙想每天读20页，剩下的几天能读完。",
+        "options":  [
+
+                    ],
+        "answer":  "49页和50页;27天",
+        "solution":  "【分析】连续两页相差1，且和为99，可用和差求数;再算剩余页数除以每天阅读页数.\n【详解】(1)较小页=(99-1)÷2=49，较大页=50.\n(2)剩余页数=590-50=540(页),540÷20=27(天)。",
+        "knowledge_points":  [
+                                 "和差倍问题",
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "25-26四年级上·天津河西·期末"
+    },
+    {
+        "id":  "479",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "奇思一家从天津出发自驾游，行驶13时后到达距天津884千米的吉林省辽源市，奇思一家开车行驶的平均速度是多少。",
+        "options":  [
+
+                    ],
+        "answer":  "68千米/时",
+        "solution":  "【分析】速度=路程/时间.\n【详解】884÷13=68(千米/时),所以平均速度是68千米/时。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26 四年级上·天津河西·期末"
+    },
+    {
+        "id":  "480",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "张师傅每天能加工134个零件，照这样计算，他本月(23个工作日)能完成3000个零件的加工任务吗。",
+        "options":  [
+
+                    ],
+        "answer":  "能",
+        "solution":  "【分析】先算23天共加工多少个，再与3000比较.\n【详解】134×23=3082(个),3082\u003e3000，所以能完成。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "25-26 四年级上·天津河西·期末"
+    },
+    {
+        "id":  "481",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一辆洒水车，每分钟行250米，洒水的宽度8米，洒水车行驶5小时能洒多少公顷的路面。",
+        "options":  [
+
+                    ],
+        "answer":  "60公顷",
+        "solution":  "【分析】先求每分钟洒水面积，再求5小时总面积，最后换算成公顷.\n【详解】每分钟面积=250×8=2000(平方米).\n5小时=300分钟，总面积=2000×300=600000(平方米).\n600000÷10000=60(公顷)。",
+        "knowledge_points":  [
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "长方形的面积",
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·新疆和田·期末"
+    },
+    {
+        "id":  "482",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "和田市到一牧场亚门景区约220千米，早晨8:00乐乐全家坐长途客车从和田出发去亚门景区游玩，中午12:00到达，长途客车平均每小时行多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "55千米",
+        "solution":  "【分析】先求行驶时间，再用速度=路程/时间.\n【详解】12:00-8:00=4(小时),220÷4=55(千米/时)。",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "时，分，秒有关的计算"
+                             ],
+        "source":  "25-26 四年级上·新疆和田·期末"
+    },
+    {
+        "id":  "483",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "公园有一个长24米，面积360平方米的长方形花坛。为了种植更多品种植物，花坛的宽不变，将长增加到72米。扩建后花坛的面积是多少平方米。",
+        "options":  [
+
+                    ],
+        "answer":  "1080平方米",
+        "solution":  "【分析】先由原面积和原长求宽，再求新面积.\n【详解】宽=360÷24=15(米).\n新面积=72×15=1080(平方米)。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "带有小括号的混合运算",
+                                 "长方形的面积"
+                             ],
+        "source":  "25-26 四年级上·甘肃兰州·期末"
+    },
+    {
+        "id":  "484",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "甲，乙两城之间的高速公路长480千米。早上8:30，马阿姨开车从甲城去往乙城，平均每小时行驶90千米,12:30到达高速公路服务区。从高速公路服务区出发，马阿姨还要行驶多少千米才能到达乙城。",
+        "options":  [
+
+                    ],
+        "answer":  "120千米",
+        "solution":  "【分析】先求已行驶时间与路程，再用总路程减去已行路程.\n【详解】12:30-8:30=4(小时),已行路程=90×4=360(千米).\n剩余路程=480-360=120(千米)。",
+        "knowledge_points":  [
+                                 "整十，整百，整千数与一位数的乘法",
+                                 "基础行程问题",
+                                 "24时计时法时间的计算"
+                             ],
+        "source":  "25-26 四年级上·新疆和田·期末"
+    },
+    {
+        "id":  "485",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.94,
+        "question":  "一辆汽车每小时行70千米,4小时行多少千米。李东骑自行车每分钟行225米,10分钟行多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "280千米;2250米",
+        "solution":  "【分析】路程=速度×时间.\n【详解】汽车路程:70×4=280(千米).\n自行车路程:225×10=2250(米)。",
+        "knowledge_points":  [
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26 四年级上·新疆和田·期末"
+    },
+    {
+        "id":  "486",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "12本《成语故事》需要288元，照这样计算，买36本《成语故事》一共需要多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "864元",
+        "solution":  "【分析】先求单价，再求36本总价.\n【详解】单价=288÷12=24(元),总价=24×36=864(元)。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "两位数乘两位数的进位乘法",
+                                 "除数是整十数的笔算除法"
+                             ],
+        "source":  "25-26四年级上·甘肃陇南·期末"
+    },
+    {
+        "id":  "487",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "农场今年新开辟了一块长方形葡萄园，它的长是800米，是宽的16倍，葡萄园的面积是多少公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "4公顷",
+        "solution":  "【分析】先由倍数关系求宽，再求面积并换算公顷.\n【详解】宽=800÷16=50(米).\n面积=800×50=40000(平方米).\n40000÷10000=4(公顷)。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "长方形的面积",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26四年级上·甘肃陇南·期末"
+    },
+    {
+        "id":  "488",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "甲乙两地相距200千米，汽车行驶全程要5小时，步行全程要40小时。小强从甲地出发，先乘汽车4小时，然后改步行，他还需要几小时到达乙地。",
+        "options":  [
+
+                    ],
+        "answer":  "8小时",
+        "solution":  "【分析】先求汽车速度和步行速度，再求剩余路程所需步行时间.\n【详解】汽车速度=200÷5=40(千米/时);步行速度=200÷40=5(千米/时).\n先乘车4小时行40×4=160(千米),剩余200-160=40(千米).\n步行时间=40÷5=8(小时)。",
+        "knowledge_points":  [
+                                 "除数是整十数的口算除法",
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26 四年级上·甘肃张掖·期末"
+    },
+    {
+        "id":  "489",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "李叔叔承包的杏园面积为2公顷，如果每4平方米可种一株杏树，那么这个杏园一共可种杏树多少株。",
+        "options":  [
+
+                    ],
+        "answer":  "5000株",
+        "solution":  "【分析】先把公顷换算成平方米，再用总面积除以每株占地面积.\n【详解】2公顷=20000平方米,20000÷4=5000(株)。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "公顷，平方千米的进率与换算",
+                                 "公顷，平方千米的实际问题"
+                             ],
+        "source":  "25-26 四年级上·甘肃张掖·期末"
+    },
+    {
+        "id":  "490",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.7,
+        "question":  "王叔叔有一个长方形果园，长800米，是宽的4倍，这个果园的面积是多少平方米。合多少公顷。如果每公顷种果树2500棵，一共可以种多少棵果树。",
+        "options":  [
+
+                    ],
+        "answer":  "160000平方米;16公顷;40000棵",
+        "solution":  "【分析】先求宽，再求面积;平方米换算为公顷后再乘每公顷棵数.\n【详解】宽=800÷4=200(米).\n面积=800×200=160000(平方米).\n160000平方米=16公顷.\n可种果树=16×2500=40000(棵)。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "公顷与平方米的进率与换算",
+                                 "长方形面积的计算",
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "24-25 四年级上·山东济南·期末"
+    },
+    {
+        "id":  "491",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "商场上午卖出台灯29个，下午卖出同样的台灯40个，比上午多收入638元，每个台灯的售价是多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "58元",
+        "solution":  "【分析】多收入对应多卖出的数量.\n【详解】多卖数量=40-29=11(个),单价=638÷11=58(元)。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "25-26四年级上·甘肃甘南·期末"
+    },
+    {
+        "id":  "492",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "学校图书馆买了25本《科技天地》,每本24元;买了36本《学画画》,每本18元。买这些书一共花了多少钱。如果这两本书算一套,270元钱能买几套。还剩多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "1248元;6套;18元",
+        "solution":  "【分析】总价分别相加;一套按24+18计算，再做有余数除法.\n【详解】总价=24×25+18×36=600+648=1248(元).\n一套42元,270÷42=6。18，所以能买6套，剩18元。",
+        "knowledge_points":  [
+                                 "两位数乘两位数的进位乘法",
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "25-26四年级上·甘肃甘南·期末"
+    },
+    {
+        "id":  "493",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "学校准备发作业本，发给24个班，每班106本，还要留70本作为备用。学校应买多少本作业本。",
+        "options":  [
+
+                    ],
+        "answer":  "2614本",
+        "solution":  "【分析】先求24个班需要的本数，再加备用本数.\n【详解】24个班共需106×24=2544(本),再加70本,2544+70=2614(本)。",
+        "knowledge_points":  [
+                                 "三位数乘两位数，三位数中间有0"
+                             ],
+        "source":  "25-26 四年级上·甘肃甘南·期末"
+    },
+    {
+        "id":  "494",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "学校组织四年级同学参加团建活动，相关信息:①参加团建的学生有176人，每人订一件T恤;②每人活动预算420元;③张老师带了8000元钱;④每件T恤38元;⑤每辆大客车最多可乘36人.\n(1) 求张老师带的钱够不够买所有人的服装，需要用到的信息是( )(写序号).\n(2) 如果够，剩多少元。如果不够，还差多少元。\n(3) 团建第一项先进行跑步拉练,4分钟跑了436米，照这样的速度，继续跑28分钟，这次拉练一共跑了多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "①③④;够，剩1312元;3488米",
+        "solution":  "【分析】(1)判断买服装是否够钱只需人数，单价和带的钱;(2)算总价后比较;(3)先求每分钟路程，再求28分钟路程并加上已跑436米.\n【详解】(1)用①③④.\n(2)服装总价=38×176=6688(元),8000-6688=1312(元),够买，剩1312元.\n(3)速度=436÷4=109(米/分);继续28分钟跑109×28=3052(米);总路程=3052+436=3488(米)。",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "三位数与两位数的乘法",
+                                 "提问题与填条件问题"
+                             ],
+        "source":  "25-26四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "495",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "王叔叔将购买的茶礼发快递。快递车从仓库出发，每小时行65千米，行驶了14小时到达目的地。原路返回时提前了1小时到达，返回时平均每小时行多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "70千米",
+        "solution":  "【分析】先求去程路程，再除以回程时间.\n【详解】去程路程=65×14=910(千米).\n回程时间=14-1=13(小时).\n回程速度=910÷13=70(千米/时)。",
+        "knowledge_points":  [
+                                 "两位数乘两位数的进位乘法",
+                                 "除数是两位数的笔算除法",
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "496",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "四年级学生共102人去茶园参观研学，每人的车票和门票共48元。带5000元买票够吗。",
+        "options":  [
+
+                    ],
+        "answer":  "够",
+        "solution":  "【分析】算出总票款与5000比较.\n【详解】总票款=102×48=4896(元),4896\u003c5000，所以够。",
+        "knowledge_points":  [
+                                 "三位数乘两位数，三位数中间有0"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "497",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一个长方形果园，长400米，宽250米，这个果园的面积是多少公顷。如果每公顷施肥300千克，一共需要施肥多少千克。",
+        "options":  [
+
+                    ],
+        "answer":  "10公顷;3000千克",
+        "solution":  "【分析】先求面积(平方米),再换算公顷，最后乘每公顷施肥量.\n【详解】面积=400×250=100000(平方米)=10(公顷).\n施肥总量=10×300=3000(千克)。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "长方形的面积",
+                                 "三位数乘两位数，乘数末尾有0"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "498",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "小明步行从家出发到学校，他每分钟行75米，行了8分钟，刚好到全程的中点，他家到学校的路程是多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "1200米",
+        "solution":  "【分析】先求8分钟走的路程，这是全程一半.\n【详解】75×8=600(米)是半程，全程=600×2=1200(米)。",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "两，三位数与一位数连续进位的乘法",
+                                 "整十，整百，整千数与一位数的乘法"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "499",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一批零件，如果每箱装20个，可以装840箱，如果改成每25个装一箱，可以装多少箱。",
+        "options":  [
+
+                    ],
+        "answer":  "672箱",
+        "solution":  "【分析】先求零件总数，再按新装法求箱数.\n【详解】总数=20×840=16800(个).\n新箱数=16800÷25=672(箱)。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "归总问题"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "500",
+        "unit":  "第一单元 大数的认识",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一条路是长35米，宽12米的长方形，用面积是60平方分米的方砖铺路，需要多少块方砖。",
+        "options":  [
+
+                    ],
+        "answer":  "700块",
+        "solution":  "【分析】先求路面总面积并统一到平方分米，再除以每块方砖面积.\n【详解】路面面积=35×12=420(平方米)=42000(平方分米).\n方砖块数=42000÷60=700(块)。",
+        "knowledge_points":  [
+                                 "平方厘米，平方分米，平方米之间的进率与换算",
+                                 "长方形的面积",
+                                 "除数是整十数的笔算除法",
+                                 "两位数乘两位数的进位乘法"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    }
+]
diff --git "a/\345\233\233\345\271\264\347\272\247\351\242\230\351\233\206/\345\233\233\345\271\264\347\272\247\344\270\212/\345\233\233\345\271\264\347\272\247\344\270\212\347\254\254\344\270\203\347\253\240\350\212\202\346\235\241\345\275\242\347\273\237\350\256\241\345\233\276.json" "b/\345\233\233\345\271\264\347\272\247\351\242\230\351\233\206/\345\233\233\345\271\264\347\272\247\344\270\212/\345\233\233\345\271\264\347\272\247\344\270\212\347\254\254\344\270\203\347\253\240\350\212\202\346\235\241\345\275\242\347\273\237\350\256\241\345\233\276.json"
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index 0000000000000000000000000000000000000000..f22b13d13764dfc00cfdac4d21e55063f2638bce
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+++ "b/\345\233\233\345\271\264\347\272\247\351\242\230\351\233\206/\345\233\233\345\271\264\347\272\247\344\270\212/\345\233\233\345\271\264\347\272\247\344\270\212\347\254\254\344\270\203\347\253\240\350\212\202\346\235\241\345\275\242\347\273\237\350\256\241\345\233\276.json"
@@ -0,0 +1,1448 @@
+[
+  {
+    "id": "1",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.85,
+    "question": "小东在绘制一幅条形统计图时，如果用 6 个方格表示 24 人，那么 3 个方格表示()人。",
+    "options": [],
+    "answer": "12",
+    "solution": "【分析】用6个方格表示24人，那么用24除以6即可求出每个方格表示的人数，最后乘 3 即可求解。\n【详解】 24÷6×3 = 4×3 = 12 (人) 3个方格表示12人。",
+    "knowledge_points": [
+      "1格表示多个单位的单式条形统计图"
+    ],
+    "source": "24-25四年级上·河北保定·期末"
+  },
+  {
+    "id": "2",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.85,
+    "question": "在一幅表示某仓库羽绒服数量的条形统计图中，如果纵轴上 4 格代表中长款羽绒服有 2000 件，那么短款羽绒服有 3500 件，在纵轴上应该用( )格表示。",
+    "options": [],
+    "answer": "7",
+    "solution": "【分析】4 格代表 2000 件，用 2000÷4 计算出 1 格代表多少，结果是 500 件，用 3500÷500 即可计算出 3500 应该用几格表示，计算时可以利用商不变的规律进行计算。\n【详解】 2000÷4 = 500 (件) 3500÷500 = (3500÷100)÷(500÷100) = 35÷5 = 7 (格) 在一幅表示某仓库羽绒服数量的条形统计图中，如果纵轴上4格代表中长款羽绒服有2000件，那么短款羽绒服有3500件，在纵轴上应该用7格表示。",
+    "knowledge_points": [
+      "整十，整百，整千数除以一位数的除法",
+      "1格表示多个单位的单式条形统计图",
+      "商不变的规律及应用"
+    ],
+    "source": "24-25 四年级上·河南许昌·期末"
+  },
+  {
+    "id": "3",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "在一个条形图中，用3 厘米直条表示 60 人，用( )厘米直条表示 200 人。",
+    "options": [],
+    "answer": "10",
+    "solution": "【分析】根据题意，用 60÷3 = 20 人，求出 1 厘米表示的人数，要求表示200人需要多少厘米的直条，就是求200里面有多少个20，有多少个20就是需要多少厘米的直条，用200÷20即可求出用多少厘米直条表示 200 人。\n【详解】 60÷3 = 20 (人) 200÷20 = 10 (厘米) 在一个条形图中，用3厘米直条表示60人，用10厘米直条表示 200 人。",
+    "knowledge_points": [
+      "1格表示多个单位的单式条形统计图",
+      "除数是整十数的口算除法"
+    ],
+    "source": "24-25 四年级上·河南南阳·期末"
+  },
+  {
+    "id": "4",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.85,
+    "question": "( )统计图能直观地反应出数量的多少。",
+    "options": [],
+    "answer": "条形",
+    "solution": "【分析】条形统计图是一种常用的统计图表，其通过长短不一的直条表示不同组别的数据，能够直观地展示数量的多少和比较各组之间的数量差异；据此填空即可。\n【详解】例:\n\n![](images/c87dbac2b6cc4c54e66cfb63dd200621c632d14475565371735569b6129c7b74.jpg) \n四(1)班数学期中测试成绩统计图通过上图可直观看出四(1)班期中成绩的优秀，良好，及格，不及格的人数。因此，条形统计图能够直观地反映出数量的多少。",
+    "knowledge_points": [
+      "1格表示一个单位的单式条形统计图",
+      "1格表示多个单位的单式条形统计图"
+    ],
+    "source": "24-25四年级上·黑龙江齐齐哈尔·期末"
+  },
+  {
+    "id": "5",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "如果条形统计图用 2 格表示 30 人，那么平均每格表示( )人，照这样计算，要表示 120 人需要画( )格，画 6 格表示( )人。",
+    "options": [],
+    "answer": "15;8;90",
+    "solution": "【分析】条形统计图用 2 格表示 30 人，用 30 人除以 2 算出平均每格表示多少人，要表示120人需要画几格，用120除以1格表示的人数，再用1格表示的人数乘6即可求出6格表示的人数。\n【详解】 30÷2 = 15 (人) 120÷15 = 8 (格) 15×6 = 90 (人) 如果条形统计图用2格表示30人，那么平均每格表示15人，照这样计算，要表示 120 人需要画 8 格，画 6 格表示 90 人。",
+    "knowledge_points": [
+      "1 格表示多个单位的单式条形统计图"
+    ],
+    "source": "24-25 四年级上·吉林·期末"
+  },
+  {
+    "id": "6",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "条形统计图中用 1 厘米高的直条表示 80 人，那么 5 厘米高的直条表示 人，在这个统计图上有一个直条上标有1200人，那这个直条高度是 厘米。",
+    "options": [],
+    "answer": "400;15",
+    "solution": "【分析】根据条形统计图的特点可知，5厘米高的直条表示5个80人；用一个直条上标有的总人数除以1厘米高的直条表示的人数，即可计算出这个直条的高度。\n【详解】 80×5 = 400 (人) 1200÷80 = 15 (厘米) 5厘米高的直条表示400人，在这个统计图上有一个直条上标有1200人，那这个直条高度是15厘米。",
+    "knowledge_points": [
+      "1 格表示多个单位的单式条形统计图"
+    ],
+    "source": "23-24 四年级上·河南南阳·期末"
+  },
+  {
+    "id": "7",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.85,
+    "question": "王芳家今年前3 个月的电费是 225 元，平均每月电费( )元，照这样计算，一年的电费是( )元。",
+    "options": [],
+    "answer": "75;900",
+    "solution": "【分析】前 3 个月的电费是 225 元，用总钱数除以 3，求出平均每个月的电费，然后再乘12即可求出王芳家一年的电费是多少钱。\n【详解】 225÷3 = 75 (元) 75×12 = 900 (元) 王芳家今年前 3 个月的电费是 225 元，平均每月电费 75 元，照这样计算，一年的电费是 900 元。",
+    "knowledge_points": [
+      "平均数的意义及求法"
+    ],
+    "source": "23-24四年级上·内蒙古呼伦贝尔·期末"
+  },
+  {
+    "id": "8",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "一幅反映苹果销量的统计图，用1厘米长的直条表示200千克，表示800千克的直条应该画( )厘米长;一个直条长3.5厘米，表示销售苹果( )千克。",
+    "options": [],
+    "answer": "4;700",
+    "solution": "【分析】因为1厘米长的直条表示200千克，求表示800千克的直条应该画几厘米长，即求800里面含有几个200，用除法解答即可；1厘米长的直条表示200千克，则3厘米长的直条表示3个200 千克是 600 千克，又因 1 厘米长直条表示 200 千克，则0.5 厘米长的直条表示 100 千克，然后把 3 厘米长的直条表示的千克数加上0.5厘米长的直条表示的千克数，即可解答。\n【详解】 800÷200 = 4 (厘米) 200×3 + 100 = 600 + 100 = 700 (千克) 一幅反映苹果销量的统计图，用1厘米长的直条表示200千克，表示800千克的直条应该画4厘米长；一个直条长3.5厘米，表示销售苹果 700 千克。",
+    "knowledge_points": [
+      "1格表示多个单位的单式条形统计图"
+    ],
+    "source": "24-25四年级上·河南洛阳·期中"
+  },
+  {
+    "id": "9",
+    "unit": "第七章节 条形统计图",
+    "question_type": "判断题",
+    "difficulty": 0.65,
+    "question": "条形统计图是用长短不同的直条表示各种数量的多少。( )说明理由或改正:",
+    "options": [],
+    "answer": "√ 单式条形统计图用于比较一个物体的数量，用直条的长短表示数量的多少，能直观地，形象地反映数量的多少，便于比较",
+    "solution": "【分析】单式条形统计图用于比较一个物体的数量，用直条的长短表示数量的多少，能直观地，形象地反映数量的多少，据此解答。\n【详解】根据解析可知，条形统计图是用长短不同的直条表示各种数量的多少，原题表达正确。",
+    "knowledge_points": [
+      "1格表示一个单位的单式条形统计图",
+      "1格表示多个单位的单式条形统计图"
+    ],
+    "source": "23-24 四年级上·全国·期末"
+  },
+  {
+    "id": "10",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "在统计人数时，用长2厘米长方形直条表示特长学生人数60人，那少先队员共 120 人，应画长( )厘米的长方形直条;长 3 厘米长方形直条表示的人数为( )人。",
+    "options": [],
+    "answer": "4;90",
+    "solution": "【分析】根据题意可知，用长2厘米长方形直条表示特长学生的人数，60÷2 = 30，即1厘米长方形直条表示30人。总人数÷1厘米长方形直条表示的人数 = 应画长方形直条的长度，1厘米长方形直条表示的人数×3 = 3 厘米长方形直条表示的人数，依此解答。\n【详解】 60÷2 = 30 (人) 120÷30 = 4 (厘米) 30×3 = 90 (人) 少先队员共 120 人，应画长 4 厘米的长方形直条；长 3 厘米长方形直条表示的人数为 90 人。",
+    "knowledge_points": [
+      "1 格表示多个单位的单式条形统计图"
+    ],
+    "source": "23-24四年级下·湖南益阳·期末"
+  },
+  {
+    "id": "11",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "一个牧场有马 200 匹，羊 300 只，牛 80 头，绘制条形统计图时，表示牛的直条高4 厘米，表示马的直条高应是( )厘米，表示羊的直条高应是( )厘米。",
+    "options": [],
+    "answer": "10;15",
+    "solution": "【分析】一个牧场有马200匹，羊300只，牛80头，绘制条形统计图时，表示牛的直条高4厘米，说明1厘米高的直条表示数量 80÷4 = 20，用马和羊的数量除以1厘米所代表的数量，分别求出马和羊的直条高度。\n【详解】1 厘米高代表数量: 80÷4 = 20 马的直条高: 200÷20 = 10 (厘米) 羊的直条高: 300÷20 = 15 (厘米)",
+    "knowledge_points": [
+      "1格表示多个单位的单式条形统计图"
+    ],
+    "source": "2020·广西柳州·小升初模拟"
+  },
+  {
+    "id": "12",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.85,
+    "question": "条形统计图是用一个单位长度表示一定的( ),根据数量的多少画成长短不同的( )。",
+    "options": [],
+    "answer": "数量;直条",
+    "solution": "【分析】根据条形统计图的意义及特点，条形统计图用于比较一个物体的数量，用直条的长短表示数量的多少，能直观地，形象地反映数量的多少，据此填空即可。\n【详解】条形统计图是用一个单位长度表示一定的数量，根据数量的多少画成长短不同的直条。",
+    "knowledge_points": [
+      "1格表示多个单位的单式条形统计图"
+    ],
+    "source": "23-24 四年级上·山东菏泽·期末"
+  },
+  {
+    "id": "13",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "条形统计图用 2 格表示 10 人，平均每格表示( )人，照这样计算，要表示60 人需要画( )格。",
+    "options": [],
+    "answer": "5;12",
+    "solution": "【分析】2 格表示 10 人，求每格表示多少人，可用除法列式计算:10÷2(人)；用 60 人除以每格表示的人数即可。\n【详解】 10÷2 = 5 (人) 60÷5 = 12 (格) 所以平均每格表示 5 人，照这样计算，要表示 60 人需要画12 格。",
+    "knowledge_points": [
+      "归一问题",
+      "整十，整百，整千数除以一位数的除法",
+      "1 格表示多个单位的单式条形统计图"
+    ],
+    "source": "23-24 四年级上·山东菏泽·期末"
+  },
+  {
+    "id": "14",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "塘塘在绘制一幅条形统计图时，如果用 6 厘米高的直条表示 24 人，那么 3 厘米高的直条表示( )人;如果要表示72人，那么直条的高度应该是( )厘米。",
+    "options": [],
+    "answer": "12;18",
+    "solution": "【分析】用6厘米高的直条表示24人，1厘米表示 24÷6 = 4 人，则3厘米高的直条表示: 3×4 = 12 人；求72人里面有几个4人就用几厘米长的直条表示72人，据此解答即可。\n【详解】 24÷6 = 4 (人) 3×4 = 12 (人) 72÷4 = 18 (厘米) 则 3 厘米高的直条表示 12 人，要表示 72 人，直条的高度应是18厘米。",
+    "knowledge_points": [
+      "1格表示多个单位的单式条形统计图"
+    ],
+    "source": "23-24 四年级上·浙江·期末"
+  },
+  {
+    "id": "15",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "条形统计图可直观地表示出( )。",
+    "options": [],
+    "answer": "数量的多少",
+    "solution": "【分析】条形统计图能很容易看出数量的多少，条状越高，数量越多。\n【详解】根据统计图的特点可知：条形统计图可直观地表示出数量的多少。",
+    "knowledge_points": [
+      "1格表示多个单位的单式条形统计图",
+      "1格表示一个单位的单式条形统计图"
+    ],
+    "source": "23-24 四年级上·山东济宁·期末"
+  },
+  {
+    "id": "16",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "条形统计图可以直观地看出数据的( ),直条越高所表示的数据( ),直条越低所表示的数据( )。",
+    "options": [],
+    "answer": "大小;越大;越小",
+    "solution": "【分析】条形统计图可以直观地看出6月份每种天气的天数，天数越多直条越高，天数越少直条越低。\n【详解】条形统计图可以直观地看出数据的大小，直条越高所表示的数据越大，直条越低所表示的数据越小。如图：条形统计图可以直观地看出6月份每种天气的天数，天数越多直条越高，天数越少直条越低。\n\n![](images/f545efae88916b235a329cffeb1cead5a7b330f4061faa1512994f9e5d5b0645.jpg)",
+    "knowledge_points": [
+      "1 格表示一个单位的单式条形统计图",
+      "1 格表示多个单位的单式条形统计图"
+    ],
+    "source": "23-24 四年级上·河南周口·期末"
+  },
+  {
+    "id": "17",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.85,
+    "question": "条形统计图用2格表示 50 人，平均每格表示( )人，照这样计算，要表示150 人需要画( )格。",
+    "options": [],
+    "answer": "25;6",
+    "solution": "【分析】条形统计图上每格表示的人数相等，2格表示50人，要求平均每格表示的人数，用50除以2即可；用要表示的总人数除以每格表示的人数就是需要画的格数。\n【详解】 50÷2 = 25 (人) 150÷25 = 6 (格) 条形统计图用2格表示50人，平均每格表示25人，照这样计算，要表示150人需要画6格。",
+    "knowledge_points": [
+      "1格表示多个单位的单式条形统计图"
+    ],
+    "source": "22-23四年级上·江苏宿迁·期末"
+  },
+  {
+    "id": "18",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.85,
+    "question": "根据排列方式的不同，条形统计图可分为( )和( )。",
+    "options": [],
+    "answer": "纵向条形统计图;横向条形统计图",
+    "solution": "【分析】本题考查条形统计图的直观特点，需要理解直条长度与数据大小之间的对应关系。\n【详解】根据排列方式的不同，条形统计图可分为纵向条形统计图和横向条形统计图。\n\n![](images/26b0c3b97cb188a4e0ebc62fdc75f110a0b846771123ca79356b3abd27aceb00.jpg) \n纵向条形统计图\n\n![](images/ca7abcc87e1f06e13b9de551cf473c9cf9499c56e260b0f765f2dbba908aacc7.jpg) \n横向条形统计图",
+    "knowledge_points": [
+      "1 格表示多个单位的单式条形统计图"
+    ],
+    "source": "22-23 四年级上·河南周口·期末"
+  },
+  {
+    "id": "19",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.85,
+    "question": "在一幅条形统计图中，如果用 1 格表示 4kg,那么 12kg 用( )格表示。",
+    "options": [],
+    "answer": "3",
+    "solution": "【分析】根据条形统计图的特点可知，用总重量除以1格表示的重量，即可计算出表示这些重量的格数，依此计算并填空。\n【详解】 12÷4 = 3 (格)，即 12kg 用 3 格表示。",
+    "knowledge_points": [
+      "1 格表示多个单位的单式条形统计图"
+    ],
+    "source": "22-23 四年级上·四川宜宾·期末"
+  },
+  {
+    "id": "20",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "绘制条形统计图时要先画出两条( )的射线。",
+    "options": [],
+    "answer": "互相垂直",
+    "solution": "【分析】本题考查条形统计图的绘制步骤，需要先建立坐标系框架。\n【详解】绘制条形统计图时要先画出两条互相垂直的射线，再确定横轴和纵轴所表示的项目。例如，绘制四(1)班同学\"最喜欢的贵州特色美食\"统计图如下:\n\n![](images/ca36b4d0e5fa8b22397b5ea83a9c170c69ec38fba9aa3f7315fbdb123072aee9.jpg)",
+    "knowledge_points": [
+      "1 格表示多个单位的单式条形统计图"
+    ],
+    "source": "22-23 四年级上·海南省直辖县级单位·期末"
+  },
+  {
+    "id": "21",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.85,
+    "question": "在绘制条形统计图时，注意写清( ),确定( )轴和( )轴所表示的对象，确定( )轴 1 格所表示的大小及单位。",
+    "options": [],
+    "answer": "单位;横;纵;纵",
+    "solution": "【分析】本题考查绘制条形统计图的注意事项，需要掌握规范的制图步骤。\n【详解】例：某校四年级(2)班同学对本班学生双休日活动调查结果统计如下表:\n\n<table><tr><td>类别</td><td>平均每个学生所用时间/分</td></tr><tr><td>写作业</td><td>60</td></tr><tr><td>看电视</td><td>180</td></tr><tr><td>阅读</td><td>50</td></tr><tr><td>参加兴趣班</td><td>90</td></tr><tr><td>锻炼身体</td><td>40</td></tr></table>\n\n![](images/e9525aa2c5da2a3f68251f048eafba00c8716b3e26064fd97760edbec13ff20f.jpg) \n四年级(2)班学生双休日活动时间统计图。故在绘制条形统计图时，注意写清单位，确定横轴和纵轴所表示的对象，确定纵轴 1 格所表示的大小及单位。",
+    "knowledge_points": [
+      "1格表示多个单位的单式条形统计图",
+      "1格表示一个单位的单式条形统计图"
+    ],
+    "source": "22-23四年级下·辽宁·课前预习"
+  },
+  {
+    "id": "22",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "在一幅条形统计图中，用 2 厘米长的直条表示 8 人，用( )厘米长的直条表示 48 人，用 5 厘米长的直条表示( )人。",
+    "options": [],
+    "answer": "12;20",
+    "solution": "【分析】2 厘米长的直条表示 8 人，用 8÷2 = 4 人，求出 1 厘米表示 4 人；48 里面有几个 4 就是几厘米长的直条；1 厘米表示 4 人，5 厘米表示 5×4 = 20 人。\n【详解】 8÷2 = 4 (人)；48÷4 = 12 (厘米)；5×4 = 20(人) 在一幅条形统计图中，用2厘米长的直条表示8人，用(12)厘米长的直条表示 48 人，用 5 厘米长的直条表示 (20)人。",
+    "knowledge_points": [
+      "归一问题",
+      "1 格表示多个单位的单式条形统计图"
+    ],
+    "source": "21-22 四年级上·云南保山·期末"
+  },
+  {
+    "id": "23",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "在条形统计图中，若 3 厘米高的条形表示 9 人，则 5 厘米高的条形表示( )人。",
+    "options": [],
+    "answer": "15",
+    "solution": "【分析】根据除法的意义，先求出1厘米表示多少人；再用乘法求出 5 厘米高的条形表示多少人即可。\n【详解】 9÷3×5 = 3×5 = 15 (人) 在条形统计图中，若3厘米高的条形表示9人，则5厘米高的条形表示15人。",
+    "knowledge_points": [
+      "1格表示多个单位的单式条形统计图"
+    ],
+    "source": "22-23四年级上·湖南常德·期末"
+  },
+  {
+    "id": "24",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "在一幅条形统计图中，如果用3格代表18人，那么6格代表( )人;用()格代表30人。",
+    "options": [],
+    "answer": "36;5",
+    "solution": "【分析】3 格代表 18 人，那么可以用除法求出 1 格表示的人数，再用乘法计算出 6 格表示的人数；用 30 除以 1 格表示的人数，得出格数；据此解答。\n【详解】根据分析:18÷3 = 6 (人)，6×6 = 36 (人)，所以 6 格代表 36 人；30÷6 = 5 (格)，用5格代表30人。",
+    "knowledge_points": [
+      "统计图表的综合应用",
+      "1格表示多个单位的单式条形统计图"
+    ],
+    "source": "22-23四年级上·四川广元·期末"
+  },
+  {
+    "id": "25",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.94,
+    "question": "统计表和( )都可以用来表示数量的多少，从( )中能更清楚地看各项数量的多少。",
+    "options": [],
+    "answer": "条形统计图;条形统计图",
+    "solution": "【分析】统计表和条形统计图都可以用来表示数据，条形统计图是在统计图表的基础上加工而得，条形统计图更直观，可以更清楚地看出各项数量的多少。\n【详解】统计表和条形统计图都可以用来表示数量的多少，从条形统计图中能更清楚地看各项数量的多少。",
+    "knowledge_points": [
+      "1 格表示一个单位的单式条形统计图"
+    ],
+    "source": "20-21四年级上·海南省直辖县级单位·期末"
+  },
+  {
+    "id": "26",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "在一幅条形统计图中，如果用3厘米的条形高度表示18人，那么6厘米的条形高度可以表示( )人;用( )厘米的条形高度表示30人。",
+    "options": [],
+    "answer": "36;5",
+    "solution": "【分析】根据题意，3厘米的条形高度表示18人，用18除以3计算出1厘米的条形高度表示多少人，再进行解答即可。\n【详解】 18÷3 = 6 (人) 6×6 = 36 (人) 30÷6 = 5 (厘米) 在一幅条形统计图中，如果用 3 厘米的条形高度表示 18 人，那么6厘米的条形高度可以表示(36)人；用(5)厘米的条形高度表示 30 人。",
+    "knowledge_points": [
+      "1 格表示多个单位的单式条形统计图"
+    ],
+    "source": "21-22 四年级下·陕西安康·期末"
+  },
+  {
+    "id": "27",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "统计图在生活中有着广泛的应用，条形统计图和统计表相比较，它的特点是()。",
+    "options": [],
+    "answer": "能很容易看出数量的多少",
+    "solution": "【分析】本题考查条形统计图与统计表的对比，需要理解两种统计方式各自的特点和优势。\n【详解】统计表的特点是使统计资料条理化，简明清晰，便于检查数据的完整性和准确性，以及对比分析；统计图在生活中有着广泛的应用，条形统计图和统计表相比较，它的特点是能很容易看出数量的多少。",
+    "knowledge_points": [
+      "单式统计表的特点及填补",
+      "1格表示多个单位的单式条形统计图"
+    ],
+    "source": "21-22 四年级上·山东济宁·期末"
+  },
+  {
+    "id": "28",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "一幅反映某些地区面积多少的条形统计图，如果用 1 格的直条表示 50 公顷，那么表示 1 平方千米的直条占( )格。",
+    "options": [],
+    "answer": "2",
+    "solution": "【分析】先把平方千米换算成以公顷为单位，1平方千米=100公顷，再用 100÷50，即可求出需要几格的直条。\n【详解】1平方千米 = 100 公顷 100÷50 = 2 (格) 即表示1平方千米的直条占2格。",
+    "knowledge_points": [
+      "1格表示多个单位的单式条形统计图",
+      "公顷，平方千米的进率与换算"
+    ],
+    "source": "21-22四年级上·贵州黔西·期末"
+  },
+  {
+    "id": "29",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "五年级四个班开展\"学雷锋，做好事\"活动，为了清楚地表示三月份各班做好事件数的多少，应制作( )统计图。",
+    "options": [],
+    "answer": "条形",
+    "solution": "【分析】条形统计图反映数据的多少，据此解答即可。\n【详解】五年级四个班开展\"学雷锋，做好事\"活动，为了清楚地表示三月份各班做好事件数的多少，应制作条形统计图。",
+    "knowledge_points": [
+      "统计图的选择",
+      "1格表示多个单位的单式条形统计图"
+    ],
+    "source": "19-20 五年级下·广西贺州·期末"
+  },
+  {
+    "id": "30",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.85,
+    "question": "在一幅条形统计图上，纵轴 1 厘米表示 20 千克，表示 100 千克的直条应画()厘米长。",
+    "options": [],
+    "answer": "5",
+    "solution": "【分析】根据\"纵轴 1 厘米表示 20 千克\"，100 千克里面有 5 个 20 千克，所以表示 100 千克的直条画 5 个 1 厘米；据此解答。\n【详解】 100÷20 = 5 5×1 = 5 (厘米) 则表示100千克的直条应画5厘米。",
+    "knowledge_points": [
+      "1 格表示多个单位的单式条形统计图"
+    ],
+    "source": "21-22四年级上·山东菏泽·期末"
+  },
+  {
+    "id": "31",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.94,
+    "question": "能清楚地表示数量多少，除了统计表外，还有( )统计图。",
+    "options": [],
+    "answer": "条形",
+    "solution": "【分析】本题考查能清楚表示数量多少的统计工具，需要区分不同统计图表的功能特点。\n【详解】条形统计图特点是用一个单位长度表示一定的数量，用直条的长短表示数量的多少，作用是从图中能清楚地看出各种数量的多少，便于相互比较；条形统计图比统计表更加直观，形象；据此可知，能清楚地表示数量多少，除了统计表外，还有条形统计图。",
+    "knowledge_points": [
+      "单式统计表的特点及填补",
+      "1格表示一个单位的单式条形统计图",
+      "1格表示多个单位的单式条形统计图"
+    ],
+    "source": "21-22 四年级上·河北承德·期末"
+  },
+  {
+    "id": "32",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.94,
+    "question": "与统计表相比，条形统计图表示数据更加( )。",
+    "options": [],
+    "answer": "直观;具体",
+    "solution": "【分析】条形统计图的特点：能清楚地表示出数量的多少；与统计表相比，条形统计图表示数据更加直观和具体。\n【详解】与统计表相比，条形统计图表示数据更加直观、具体。\n(本题答案不唯一)",
+    "knowledge_points": [
+      "1格表示一个单位的单式条形统计图",
+      "1格表示多个单位的单式条形统计图"
+    ],
+    "source": "21-22 四年级上·四川德阳·期末"
+  },
+  {
+    "id": "33",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.85,
+    "question": "为了清楚地看出各年级人数应采用( )统计图。",
+    "options": [],
+    "answer": "条形",
+    "solution": "【分析】条形统计图能清楚地看出各个数量的多少；折线统计图不仅能看出数量的多少，还能看出数量的变化情况；扇形统计图能清楚地看出部分与整体之间的关系；据此作答。\n【详解】为了清楚地看出各年级人数应采用条形统计图。",
+    "knowledge_points": [
+      "统计图的选择"
+    ],
+    "source": "21-22 六年级上·广东汕尾·期末"
+  },
+  {
+    "id": "34",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.85,
+    "question": "条形统计图可以通过直条的( )清楚地看出数量的多少。",
+    "options": [],
+    "answer": "长短",
+    "solution": "【分析】条形统计图是用一个单位长度表示一定的数量，根据数量的多少画成长短不同的直条，然后把这些按照一定的顺序排列起来，据此解答。\n【详解】条形统计图可以通过直条的(长短)清楚地看出数量的多少。",
+    "knowledge_points": [
+      "1 格表示多个单位的单式条形统计图"
+    ],
+    "source": "21-22 四年级上·四川凉山·期末"
+  },
+  {
+    "id": "35",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "甲乙两数的平均数是 30，甲数比乙数多 8，甲数是( )。",
+    "options": [],
+    "answer": "34",
+    "solution": "【分析】甲乙两数的平均数是30，则甲乙两数的和是 30×2 = 60。甲数比乙数多 8，则甲数是(60+8)÷2。\n【详解】(60 + 8)÷2 = 68÷2 = 34，则甲数是 34。本题考查平均数的求法以及和差问题，平均数×份数 = 总数量，据此求出两个数的和。和差问题中，大数=(和+差)÷2，小数 = (和-差)÷2。",
+    "knowledge_points": [
+      "和差倍问题",
+      "平均数的意义及求法"
+    ],
+    "source": "20-21 四年级上·河北承德·期末"
+  },
+  {
+    "id": "36",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.85,
+    "question": "冬冬做跳绳练习，第一次跳 71 下，第二次跳 86 下，他要想三次平均成绩达到83 下，第三次至少要跳( )下。",
+    "options": [],
+    "answer": "92",
+    "solution": "【分析】平均成绩乘 3 等于三次的成绩和，减去前两次跳的下数即等于第三次跳的下数。\n【详解】83×3 - 71 - 86 = 249 - 71 - 86 = 92 (下)。本题主要考查学生对平均数知识的掌握和灵活运用。",
+    "knowledge_points": [
+      "平均数的意义及求法"
+    ],
+    "source": "21-22 四年级上·江苏泰州·期末"
+  },
+  {
+    "id": "37",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.85,
+    "question": "在条形统计图中，如果一格代表 5 人，那么两格代表( )人;如果要表示出 75 人，需要( )格。",
+    "options": [],
+    "answer": "10;15",
+    "solution": "【分析】每格代表的人数乘直条的格数等于直条代表的人数，人数除以每格代表的人数等于直条的格数。\n【详解】 5×2 = 10 (人) 75÷5 = 15 (格)",
+    "knowledge_points": [
+      "1 格表示多个单位的单式条形统计图"
+    ],
+    "source": "21-22 四年级上·黑龙江佳木斯·期末"
+  },
+  {
+    "id": "38",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "妈妈想要做鱼，程序如下：杀鱼，洗鱼5分钟;烧鱼10分钟;淘米2分钟。做米饭 15 分钟妈妈最少用( )分钟做好饭菜。",
+    "options": [],
+    "answer": "17",
+    "solution": "【分析】可以先淘米需要 2 分钟，然后做米饭需要 15 分钟，在做米饭的同时，可以进行杀鱼、洗鱼5分钟，烧鱼10分钟，这样安排不相互冲突，所用的时间最少，据此即可解答。\n【详解】先淘米，再做饭，做饭的15分钟时间同时杀鱼、洗鱼和烧鱼，这样可以节约 10 + 5 = 15 分钟时间，所以这样安排最少需要的时间为: 2 + 15 = 17 (分钟)。",
+    "knowledge_points": [
+      "优化问题"
+    ],
+    "source": "20-21四年级上·湖南怀化·单元测试"
+  },
+  {
+    "id": "39",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "17名工人坐船渡河，渡口处只有一条能载5人的小船(无船工)。他们全部渡河，至少需要这条小船渡河( )次。",
+    "options": [],
+    "answer": "7",
+    "solution": "【分析】根据题意可知，无船工，那么需要自己找人撑船，来回使用小船2次，只渡河4人，最后一次可以渡5人，据此解答。\n【详解】(17 - 5)÷4×2 + 1 = 12÷4×2 + 1 = 3×2 + 1 = 6 + 1 = 7 (次)。答：至少需要这条小船渡河 7 次。",
+    "knowledge_points": [
+      "优化问题"
+    ],
+    "source": "20-21四年级上·湖南怀化·单元测试"
+  },
+  {
+    "id": "40",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "要想知道全国各省一天新冠肺炎病人出院人数多少，应选用( )统计图。",
+    "options": [],
+    "answer": "条形",
+    "solution": "【分析】条形统计图用直条的长短表示数量的多少，从图中直观地看出数量的多少，据此解答。\n【详解】根据分析，要想知道全国各省一天新冠肺炎病人出院人数多少，应选用条形统计图。",
+    "knowledge_points": [
+      "统计图的选择"
+    ],
+    "source": "20-21 四年级上·湖南怀化·单元测试"
+  },
+  {
+    "id": "41",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "16,18,21,37,43 这五个数的平均数是( )。",
+    "options": [],
+    "answer": "27",
+    "solution": "【分析】先将这5个数相加求和，再用和除以5，求出这五个数的平均数。\n【详解】(16+18+21+37+43)÷5 = 135÷5 = 27，则这五个数的平均数是 27。",
+    "knowledge_points": [
+      "平均数的意义及求法"
+    ],
+    "source": "21-22 四年级上·山西大同·期中"
+  },
+  {
+    "id": "42",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.85,
+    "question": "条形图中，用 2 厘米长的直条表示 10 人，用( )厘米的直条表示 30 人。",
+    "options": [],
+    "answer": "6",
+    "solution": "【分析】根据题意，2厘米长的直条表示10人，那么用30除以10看能分出几个10人，再根据一个10人是2厘米，然后再乘2厘米即可。\n【详解】 30÷10 = 3 3×2 = 6 (厘米) 条形图中，用2厘米长的直条表示10人，用(6)厘米的直条表示 30 人。",
+    "knowledge_points": [
+      "1格表示多个单位的单式条形统计图"
+    ],
+    "source": "18-19 四年级上·四川绵阳·期末"
+  },
+  {
+    "id": "43",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.85,
+    "question": "条形统计图中，用直条的( )表示数量的多少。",
+    "options": [],
+    "answer": "长短",
+    "solution": "【分析】条形统计图能很容易看出数量的多少，条形统计图的特点是用直条的长短表示数量的多少，直条越长表示数量越多，直条越短表示数量越少，若直条长度相等，数量就相等。\n【详解】条形统计图中，用直条的长短表示数量的多少。",
+    "knowledge_points": [
+      "1格表示一个单位的单式条形统计图",
+      "1格表示多个单位的单式条形统计图"
+    ],
+    "source": "21-22 四年级上·全国·期末"
+  },
+  {
+    "id": "44",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "小芳，小亮和小敏三人排队等老师批改作业，小芳用 2 分钟，这时小亮和小敏各等了( )分钟。全部批改完成后，他们三人一共等了( )分钟。",
+    "options": [],
+    "answer": "2;12",
+    "solution": "【分析】小芳用 2 分钟，小芳在等的同时，小亮和小敏也在等，每人等了 2 分钟；小芳批改完成之后，再批改下一个人的，这时剩余的2人都要等2分钟，下一人批改完成之后，批改最后一人的作业，最后一人再等2分钟。\n【详解】 2×3 = 6 (分钟) 2×3 + 2×2 + 2 = 6 + 4 + 2 = 12 (分钟) 小芳，小亮和小敏三人排队等老师批改作业，小芳用2分钟，这时小亮和小敏各等了(2)分钟。全部批改完成后，他们三人一共等了12分钟。",
+    "knowledge_points": [
+      "优化问题"
+    ],
+    "source": "20-21 四年级上·湖南怀化·单元测试"
+  },
+  {
+    "id": "45",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "6 个病人排队看病，杨会排在第一位，医生替他看病需要5分钟，他后面的病人一共等待了( )分钟。",
+    "options": [],
+    "answer": "75",
+    "solution": "【分析】根据题意，第二个病人等了5分钟，第三个病人等了10分钟，第四个病人等了15分钟，第五个病人等了20分钟，第六个病人等了25分钟，再把他们5人等待的时间相加即可。\n【详解】5×(1 + 2 + 3 + 4 + 5) = 5×15 = 75 (分钟)，所以他后面的病人一共等待了 75 分钟。本题属于合理安排时间的问题，6个病人排队看病，杨会排在第一位，后面还有5个病人，因此他们的等待时间总和就是5个人的等待时间总和；分别计算出杨会后面5位病人各自等待的时间，是解答此题的关键。",
+    "knowledge_points": [
+      "优化问题",
+      "乘加，乘减混合运算"
+    ],
+    "source": "20-21 四年级上·湖南怀化·单元测试"
+  },
+  {
+    "id": "46",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "煮熟一个鸡蛋要 9 分钟，一只锅子一次可以煮 20 个鸡蛋，那么煮熟这 20个鸡蛋至少需要( )分钟。",
+    "options": [],
+    "answer": "9",
+    "solution": "【分析】一只锅子一次可以煮 20 个鸡蛋，那么就只需要煮一次，据此解答。\n【详解】煮熟一个鸡蛋要9分钟，一次同时煮熟这20个鸡蛋是一样的时间，至少需要 9 分钟。",
+    "knowledge_points": [
+      "优化问题"
+    ],
+    "source": "20-21 四年级上·湖南怀化·单元测试"
+  },
+  {
+    "id": "47",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "唐阿姨要用平锅烙 5 张饼，平锅每次只能烙两张饼，两面都要烙，每面 3分钟，最少要( )分钟才能烙好。10 张饼至少要( )分钟可以烙好。",
+    "options": [],
+    "answer": "15;30",
+    "solution": "【分析】要使烙饼的总时间最少，那么每一次锅内就应有两张饼，然后计算出时间总和即可；第一次：第 1 张饼的正面，第 2 张饼的正面，用时 3 分钟；第二次：第 1 张饼的反面，第 3 张饼的正面，用时 3 分钟；第三次：第 3 张饼的反面，第 2 张饼的反面，用时 3 分钟；第四次：第 4 张饼的正面，第 5 张饼的正面，用时 3 分钟；第五次：第 5 张饼的反面，第 4 张饼的反面，用时 3 分钟；因此当烙两张及以上的饼，并且每两张饼一起烙时，烙饼所需要的最短时间 = 烙一面需要的时间×张数。\n【详解】 3×5 = 15 (分钟) 3×10 = 30 (分钟)",
+    "knowledge_points": [
+      "优化问题"
+    ],
+    "source": "20-21 四年级上·湖南怀化·单元测试"
+  },
+  {
+    "id": "48",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.85,
+    "question": "条形统计图是用一个单位长度表示一定的( ),根据数量的多少画成长短不同的( ),然后把这些直条按照一定的顺序排列起来。",
+    "options": [],
+    "answer": "数量;直条",
+    "solution": "【分析】本题考查条形统计图的基本定义和构成要素，需要掌握条形统计图的制作原理。\n【详解】条形统计图是用一个单位长度表示一定的数量，根据数量的多少画成长短不同的直条，然后把这些直条按照一定的顺序排列起来。",
+    "knowledge_points": [
+      "复式条形统计图",
+      "1 格表示一个单位的单式条形统计图",
+      "1 格表示多个单位的单式条形统计图"
+    ],
+    "source": "20-21 四年级上·湖南怀化·单元测试"
+  },
+  {
+    "id": "49",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.94,
+    "question": "条形统计图可以清楚地反映出( )的情况。",
+    "options": [],
+    "answer": "数量",
+    "solution": "【分析】本题考查条形统计图的核心功能，需要区分三种常见统计图(条形图，折线图，扇形图)的不同特点。\n【详解】条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系。",
+    "knowledge_points": [
+      "1格表示一个单位的单式条形统计图",
+      "1格表示多个单位的单式条形统计图"
+    ],
+    "source": "2021 四年级上·全国·专题练习"
+  },
+  {
+    "id": "50",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "李阿姨用平底锅煎小黄鱼，一只平底锅每次只能煎2条小黄鱼。煎1条小黄鱼需要4 分钟(正，反面各 2 分钟)。李阿姨煎 5 条小黄鱼至少需要 分钟。",
+    "options": [],
+    "answer": "10",
+    "solution": "【分析】此类问题中，尽量使每次都有 2 条小黄鱼在煎，由此进行合理安排即可解决问题。\n【详解】1. 先煎第一条小黄鱼和第二条小黄鱼的正面 2 分钟；2. 再煎第一条小黄鱼和第二条小黄鱼的反面 2 分钟；3. 再煎第三条小黄鱼和第四条小黄鱼正面 2 分钟；4. 再煎第三条小黄鱼的反面和第五条小黄鱼的正面 2 分钟；5. 再煎第四条小黄鱼和第五条小黄鱼的反面 2 分钟；至少需要: 2×5 = 10 (分钟)。",
+    "knowledge_points": [
+      "优化问题"
+    ],
+    "source": "18-19四年级上·全国·期末"
+  },
+  {
+    "id": "51",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "一只平底锅一次只能煎 2 个鸡蛋，用它煎 1 个鸡蛋需要 2 分钟(正，反面各 1分钟),煎 3 个鸡蛋至少需要( )分钟。",
+    "options": [],
+    "answer": "3",
+    "solution": "【分析】要使煎 3 个鸡蛋所用时间最少，就要交替煎鸡蛋，这样用时最少。\n【详解】3 个鸡蛋分别为 1 号、2 号、3 号。第1分钟：1 号正面、2 号正面；第2分钟：1 号反面、3 号正面；第3分钟：2 号反面、3 号反面。3×1 = 3 (分)，煎3个鸡蛋至少需要3分钟。",
+    "knowledge_points": [
+      "优化问题"
+    ],
+    "source": "2022·黑龙江七台河·小升初真题"
+  },
+  {
+    "id": "52",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.85,
+    "question": "用统计图表示数量之间的关系，比较( ),使人( )。",
+    "options": [],
+    "answer": "清晰;一目了然",
+    "solution": "【分析】根据统计图的特点：条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；由此根据情况选择即可。\n【详解】用统计图表示数量之间的关系比较形象，常用统计图有条形统计图、折线统计图和扇形统计图。所以用统计图表示数量之间的关系，比较清晰，使人一目了然。",
+    "knowledge_points": [
+      "复式条形统计图",
+      "1格表示一个单位的单式条形统计图"
+    ],
+    "source": "21-22 四年级上·全国·单元测试"
+  },
+  {
+    "id": "53",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.85,
+    "question": "要绘制小丽 1 ~ 12 岁身高变化情况统计图，用 统计图比较合适;要绘制新华书店一周来每天的销售情况统计图，用 统计图比较合适。",
+    "options": [],
+    "answer": "折线;条形",
+    "solution": "【分析】条形统计图很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；由此解答即可。\n【详解】要绘制小丽 1 ~ 12 岁身高变化情况统计图，用折线统计图比较合适；要绘制新华书店一周每天的销售情况，用条形统计图比较合适。",
+    "knowledge_points": [
+      "统计图的选择"
+    ],
+    "source": "20-21 五年级下·全国·单元测试"
+  },
+  {
+    "id": "54",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.85,
+    "question": "反映参加大渡口区智力运动会各校男女选手人数情况，应绘制( )统计图。",
+    "options": [],
+    "answer": "条形",
+    "solution": "【分析】条形统计图可以很容易地反映各个数量的多少。据此填空即可。\n【详解】反映参加大渡口区智力运动会各校男女选手人数情况，应绘制条形统计图。",
+    "knowledge_points": [
+      "1 格表示一个单位的单式条形统计图",
+      "1 格表示多个单位的单式条形统计图"
+    ],
+    "source": "2021 六年级下·全国·专题练习"
+  },
+  {
+    "id": "55",
+    "unit": "第七章节 条形统计图",
+    "question_type": "判断题",
+    "difficulty": 0.64,
+    "question": "两数相除，余数是 26，商是32，则除数最小只能是27。",
+    "options": [],
+    "answer": "正确",
+    "solution": "【分析】解答此题应明确：余数总比除数小，即除数最小为：余数 + 1。\n【详解】根据在有余数的除法中，余数总比除数小，即除数最小为：余数 + 1，由此解答即可。\n\n解：除数最小为: 26 + 1 = 27，故答案为正确。",
+    "knowledge_points": [
+      "在有余数的除法中，余数 < 除数(余数总比除数小)"
+    ],
+    "source": "2014 三年级·全国·课后作业"
+  },
+  {
+    "id": "56",
+    "unit": "第七章节 条形统计图",
+    "question_type": "判断题",
+    "difficulty": 0.64,
+    "question": "36×9÷36×9 = 1 。",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】本题考查乘除混合运算的顺序。同级运算应按照从左到右的顺序计算。\n【详解】36×9÷36×9 = 324÷36×9 = 9×9 = 81，不等于 1。故答案为 ×。",
+    "knowledge_points": [
+      "乘除是同级运算"
+    ],
+    "source": "2014 四年级·全国·课后作业"
+  },
+  {
+    "id": "57",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.85,
+    "question": "条形统计图中直条的高低反映了数据的( ),它与统计表比较更( )一些。",
+    "options": [],
+    "answer": "多少;直观",
+    "solution": "【分析】本题考查条形统计图的直观特点，需要对比条形统计图与统计表的优缺点。\n【详解】条形统计图中直条的高低反映了数据的多少，它与统计表比较更直观一些。",
+    "knowledge_points": [
+      "1格表示一个单位的单式条形统计图"
+    ],
+    "source": "20-21四年级上·四川·单元测试"
+  },
+  {
+    "id": "58",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.85,
+    "question": "条形统计图是用()表示( ),可以直观的反映出数量的多少。",
+    "options": [],
+    "answer": "直条的长短;数量的多少",
+    "solution": "【分析】条形统计图一般简称\"条形图\"，也叫\"直条图\"，使用直条的长短表示数量的多少，从条形统计图中可直观地看出各个数量的多少。\n【详解】条形统计图是用直条的长短表示数量的多少，可以直观地反映出数量的多少。",
+    "knowledge_points": [
+      "1格表示一个单位的单式条形统计图",
+      "1格表示多个单位的单式条形统计图"
+    ],
+    "source": "18-19 四年级下·云南临沧·期末"
+  },
+  {
+    "id": "59",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "要统计光明小学各年级学生的人数情况，可选择( )统计图。",
+    "options": [],
+    "answer": "条形",
+    "solution": "【分析】条形统计图：从图中直观地看出数量的多少，便于比较；折线统计图：不仅能看清数量的多少，还能反映数量的增减变化情况；扇形统计图：清楚地看出各部分数量与总数量之间的关系。据此解答即可。\n【详解】由分析可知：选择条形统计图比较合适。\n【点睛】本题主要考查统计图的选择，一般来说，如果几个数量是并列的，只要求表示数量的多少时，选条形统计图。如果表示一个量或几个量增减变化和发展变化趋势，则选折线统计图。如果要求表示各部分数量与总数量之间的关系，则选扇形统计图。",
+    "knowledge_points": [
+      "统计图的选择"
+    ],
+    "source": "19-20 六年级下·全国·单元测试"
+  },
+  {
+    "id": "60",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.85,
+    "question": "统计数据除了可以制成 表以外，还可以制成 图。",
+    "options": [],
+    "answer": "统计;统计",
+    "solution": "【分析】本题考查统计数据的呈现方式，了解常见的统计工具。\n【详解】统计数据收集整理后，除了可以制成统计表以外，还可以制成统计图。统计表用表格形式呈现数据，数据精确，便于查找具体数；统计图用图形形式呈现数据，包括条形统计图、折线统计图、扇形统计图等，直观形象，便于比较分析。两种呈现方式各有优势，可根据实际需要选择使用。",
+    "knowledge_points": [
+      "数据的搜集与整理"
+    ],
+    "source": "19-20 三年级下·江苏·单元测试"
+  },
+  {
+    "id": "61",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "条形统计图用 2 格表示 10 人，平均每格表示( )人，照这样计算，要表示 70 人需要画( )格。",
+    "options": [],
+    "answer": "5;14",
+    "solution": "【分析】条形统计图上每格都是平均分的，2格表示10人，则平均每格表示的人数用除法计算即可。要表示的总人数÷每格表示的人数就是需要画的格数。\n【详解】 10÷2 = 5 (人) 70÷5 = 14 (格) 即平均每格表示5人，照这样计算，要表示70人需要画14格。",
+    "knowledge_points": [
+      "两位数除以一位数，首位除不尽",
+      "1 格表示一个单位的单式条形统计图"
+    ],
+    "source": "19-20 四年级上·山东·期末"
+  },
+  {
+    "id": "62",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "条形统计图可以清楚地看出( ),因为条形统计图直条的长短表示( )。",
+    "options": [],
+    "answer": "数量的多少;数量的多少",
+    "solution": "【分析】本题考查条形统计图的核心特点和表示方法，理解直条长度与数量的对应关系。\n【详解】条形统计图可以清楚地看出数量的多少，因为条形统计图直条的长短表示数量的多少。直条越长，表示数量越大；直条越短，表示数量越小。这种直观对比是条形统计图最大的优势。",
+    "knowledge_points": [
+      "1 格表示多个单位的单式条形统计图"
+    ],
+    "source": "19-20 四年级上·全国·单元测试"
+  },
+  {
+    "id": "63",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "当数据的种类不多，但数据又比较大时，用( )统计图表示比较方便。",
+    "options": [],
+    "answer": "横向",
+    "solution": "【分析】本题考查横向条形统计图的适用场景，需要根据数据特点选择合适的统计图类型。\n【详解】当数据的种类不多，但数据又比较大时，用横向统计图表示比较方便。原因：纵向条形统计图若数据太大，直条会画得很长，超出纸张范围；横向条形统计图将直条水平放置，可以充分利用纸张的宽度，更适合展示较大数值的数据。",
+    "knowledge_points": [
+      "复式条形统计图",
+      "1格表示多个单位的单式条形统计图"
+    ],
+    "source": "19-20 四年级上·全国·单元测试"
+  },
+  {
+    "id": "64",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "条形统计图根据直条的方向可分为( )统计图和( )统计图。",
+    "options": [],
+    "answer": "纵向;横向",
+    "solution": "【分析】本题考查条形统计图的分类标准，根据直条的排列方向进行划分。\n【详解】条形统计图根据直条的方向可分为两种：纵向统计图，直条竖直排列，类别在横轴，数量在纵轴；横向统计图，直条水平排列，类别在纵轴，数量在横轴。两种形式只是方向不同，都能直观表示数量的多少。",
+    "knowledge_points": [
+      "1格表示多个单位的单式条形统计图",
+      "1格表示一个单位的单式条形统计图",
+      "复式条形统计图"
+    ],
+    "source": "19-20 四年级上·全国·单元测试"
+  },
+  {
+    "id": "65",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "在一幅条形统计图上，纵轴上 1 格表示 5 人，表示 20 人的直条应画 格。",
+    "options": [],
+    "answer": "4",
+    "solution": "【分析】因为1格表示5人，所以20里面有几个5就需要画几格。\n【详解】 20÷5 = 4 (格)，所以表示20人的直条应画4格。故答案为:4。",
+    "knowledge_points": [
+      "简单条形统计图"
+    ],
+    "source": "19-20 四年级上·全国·单元测试"
+  },
+  {
+    "id": "66",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.85,
+    "question": "条形统计图能够清楚地表示( )的多少，比统计表更形象，直观。",
+    "options": [],
+    "answer": "数量",
+    "solution": "【分析】本题考查条形统计图与统计表的对比优势，理解条形统计图的直观性特点。\n【详解】条形统计图能够清楚地表示数量的多少。与统计表相比，条形统计图更形象、直观。统计表用数字呈现，需要仔细阅读比较；条形统计图用直条长短呈现，一眼就能看出数量大小和差异。",
+    "knowledge_points": [
+      "1 格表示多个单位的单式条形统计图",
+      "1 格表示一个单位的单式条形统计图",
+      "复式条形统计图"
+    ],
+    "source": "19-20 四年级上·全国·期末"
+  },
+  {
+    "id": "67",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "条形统计图是用一个( )表示( ),根据( )的多少画成( ),然后把这些( )按照( )排列起来。",
+    "options": [],
+    "answer": "单位长度;一定的数量;数量;长短不同的直条;直条;一定的顺序",
+    "solution": "【分析】本题考查条形统计图的完整定义和制作步骤，需要掌握条形统计图的构成要素。\n【详解】条形统计图是用一个单位长度表示一定的数量，根据数量的多少画成长短不同的直条，然后把这些直条按照一定的顺序排列起来。通过直条的长短对比，直观展示各类别数量的差异。",
+    "knowledge_points": [
+      "条形统计图基本概念"
+    ],
+    "source": "19-20四年级上·全国·单元测试"
+  },
+  {
+    "id": "68",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "在一幅条形统计图中，用 2 厘米长的条形表示 80 人，那么 120 人要用( )厘米的条形表示。",
+    "options": [],
+    "answer": "3",
+    "solution": "【分析】本题考查条形统计图中长度与数量的比例关系，需要先求出单位长度代表的人数。\n【详解】第一步：求1厘米表示多少人，80人÷2厘米 = 40人/厘米。第二步：求120人需要多少厘米，120人÷40人/厘米 = 3厘米。或列比例式:2:80 = x:120，解得x = 3。",
+    "knowledge_points": [
+      "1格表示多个单位的单式条形统计图"
+    ],
+    "source": "19-20 四年级上·全国·单元测试"
+  },
+  {
+    "id": "69",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "统计表和( )都可以用来表示数据，从( )能更清楚地看出各种数据的多少",
+    "options": [],
+    "answer": "条形图;条形图",
+    "solution": "【分析】本题考查数据的表示方式，对比统计表和条形统计图的特点。\n【详解】统计表和条形图都可以用来表示数据，从条形图能更清楚地看出各种数据的多少。统计表用数字呈现，数据精确但不够直观；条形图用直条呈现，直观形象，便于比较数量大小。",
+    "knowledge_points": [
+      "1格表示多个单位的单式条形统计图",
+      "1格表示个单位的单式条形统计图",
+      "复式统计表的特点及填补",
+      "单式统计表的特点及填补"
+    ],
+    "source": "19-20 四年级上·全国·单元测试"
+  },
+  {
+    "id": "70",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.85,
+    "question": "在制统计表或统计图时应先进行数据的( )和 ( )。",
+    "options": [],
+    "answer": "收集;整理",
+    "solution": "【分析】本题考查制作统计表或统计图的前期准备工作，了解统计的基本流程。\n【详解】在制统计表或统计图时应先进行数据的收集和整理。收集：通过调查、测量、查阅资料等方式获取原始数据；整理：对收集到的数据进行分类、汇总、计算，使其条理化。只有经过收集和整理，才能制作出准确、清晰的统计表或统计图。",
+    "knowledge_points": [
+      "数据的搜集与整理"
+    ],
+    "source": "19-20 四年级上·全国·单元测试"
+  },
+  {
+    "id": "71",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.85,
+    "question": "条形统计图和折线统计图都有( )式和( )式两种。",
+    "options": [],
+    "answer": "单;复",
+    "solution": "【分析】本题考查条形统计图和折线统计图的分类，根据数据组数进行划分。\n【详解】条形统计图和折线统计图都有单式和复式两种。单式：只表示一组数据；复式：同时表示多组数据，便于对比分析。例如：单式条形图只画一种颜色的直条，复式条形图用不同颜色或花纹的直条表示不同组的数据。",
+    "knowledge_points": [
+      "简单条形统计图"
+    ],
+    "source": "19-20 六年级上·全国·单元测试"
+  },
+  {
+    "id": "72",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.85,
+    "question": "在条形统计图中，如果每格表示 20 万人，则 60 万人要画( )格,2 格表示()人，半格表示( )人",
+    "options": [],
+    "answer": "3;40 万;10 万",
+    "solution": "【分析】本题考查条形统计图中单位量与格数的换算，涉及整数格和半格的理解。\n【详解】每格表示20万人，60万人需要:60÷20 = 3格；2格表示:2×20 = 40万人；半格表示:20÷2 = 10万人。",
+    "knowledge_points": [
+      "1 格表示一个单位的单式条形统计图"
+    ],
+    "source": "19-20四年级上·全国·单元测试"
+  },
+  {
+    "id": "73",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.85,
+    "question": "条形统计图有竖式和( )式两种。",
+    "options": [],
+    "answer": "横",
+    "solution": "【分析】本题考查条形统计图按方向分类的两种形式。\n【详解】条形统计图有竖式和横式两种。竖式即纵向条形统计图，直条竖直向上；横式即横向条形统计图，直条水平向右。两种形式可根据实际需要选择使用。",
+    "knowledge_points": [
+      "1 格表示一个单位的单式条形统计图"
+    ],
+    "source": "19-20 四年级上·全国·单元测试"
+  },
+  {
+    "id": "74",
+    "unit": "第七章节 条形统计图",
+    "question_type": "选择题",
+    "difficulty": 0.65,
+    "question": "( )和( )都可以用来表示数据。",
+    "options": [
+      "A. 统计表",
+      "B. 条形图",
+      "C. 画\"正\"字"
+    ],
+    "answer": "A;B",
+    "solution": "【分析】本题考查数据的表示方式，区分不同统计工具的特点。\n【详解】统计表和条形图都可以用来表示数据。\"画'正'字\"是数据收集时的一种计数方法，不是最终的数据呈现方式。因此答案选A(统计表)和B(条形图)。",
+    "knowledge_points": [
+      "单式统计表的特点及填补",
+      "1 格表示一个单位的单式条形统计图"
+    ],
+    "source": "19-20 四年级上·全国·单元测试"
+  },
+  {
+    "id": "75",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "一个格子表示多少个单位与所画条形统计图表示的数据有关，数据大，每格所表示的单位，数据小，每格所表示的单位 。",
+    "options": [],
+    "answer": "大;小",
+    "solution": "【分析】本题考查绘制条形统计图时单位长度的选择原则，需要根据数据大小灵活确定。\n【详解】一个格子表示多少个单位与所画条形统计图表示的数据有关。数据大，每格所表示的单位大；数据小，每格所表示的单位小。目的是使统计图大小适中，既清晰美观又便于观察。",
+    "knowledge_points": [
+      "1 格表示多个单位的单式条形统计图"
+    ],
+    "source": "19-20四年级上·全国·单元测试"
+  },
+  {
+    "id": "76",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "条形统计图是用一个( )表示( )。",
+    "options": [],
+    "answer": "直条的长短;数量的多少",
+    "solution": "【分析】条形统计图用直条的长短表示数量的多少，能直观地、形象地反映数量的多少，便于比较。\n【详解】条形统计图是用直条的长短表示数量的多少。",
+    "knowledge_points": [
+      "1格表示多个单位的单式条形统计图"
+    ],
+    "source": "19-20 六年级上·全国·单元测试"
+  },
+  {
+    "id": "77",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "当绘制条形统计图时，一格代表的数据可根据( )确定一个合适的数。",
+    "options": [],
+    "answer": "收集的数据的大小",
+    "solution": "【分析】本题考查绘制条形统计图时如何确定一格代表的数据，需要根据数据特点灵活选择。\n【详解】当绘制条形统计图时，一格代表的数据可根据收集的数据的大小确定一个合适的数。数据较小时，一格可代表1、2、5等较小单位。目的是使画出的直条长度适中，既不太长也不太短，便于观察和比较。",
+    "knowledge_points": [
+      "简单条形统计图"
+    ],
+    "source": "19-20 四年级上·全国·单元测试"
+  },
+  {
+    "id": "78",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "如果要统计上海世博会每天观众的变化情况，那么该选用( )统计图进行统计",
+    "options": [],
+    "answer": "折线",
+    "solution": "【分析】折线统计图是用一个单位长度表示一定的数量，根据数量的多少描出各点，然后把各点用线段顺次连接起来。折线统计图不但可以表示出数量的多少，而且能够清楚地表示出数量增减变化的情况。\n【详解】要统计上海世博会每天观众的变化情况，重点突出的是数据的变化情况，所以应绘制折线统计图。故答案为：折线。",
+    "knowledge_points": [
+      "统计图的选择"
+    ],
+    "source": "19-20四年级上·全国·单元测试"
+  },
+  {
+    "id": "79",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "某幼儿园大班有 100名小朋友，中班有80名小朋友，小班有60名小朋友，在作业本上画条形统计图表示这些数据时，每格表示( )名小朋友比较合适。",
+    "options": [],
+    "answer": "10或20",
+    "solution": "【分析】本题考查条形统计图中单位长度的选择。需要根据数据特点和作业本大小，选择合适的每格代表数量，使统计图既清晰又美观。\n【详解】已知数据：大班100人，中班80人，小班60人，三个数据都是整十数，公因数有10、20等。若每格表示10人:100÷10 = 10格，80÷10 = 8格，60÷10 = 6格，画图适中；若每格表示20人:100÷20 = 5格，80÷20 = 4格，60÷20 = 3格，更简洁。因此每格表示10人或20人比较合适。",
+    "knowledge_points": [
+      "1格表示多个单位的单式条形统计图"
+    ],
+    "source": "19-20 四年级上·全国·单元测试"
+  },
+  {
+    "id": "80",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "条形统计图的特点是能清楚，直观地表示出( )的多少",
+    "options": [],
+    "answer": "各种数据",
+    "solution": "【分析】本题考查条形统计图的基本特点，需要理解条形统计图与其他统计图相比的优势。\n【详解】条形统计图是用直条的长短来表示数量多少的统计图，其最大特点是能清楚、直观地表示出各种数据的多少，便于直观比较不同类别之间数量的差异。",
+    "knowledge_points": [
+      "1 格表示多个单位的单式条形统计图"
+    ],
+    "source": "19-20 四年级上·全国·单元测试"
+  },
+  {
+    "id": "81",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "在一幅条形统计图中，如果用 4 厘米高的直条表示200本，那么5厘米高的直条表示( )本;如果表示 500 本，那么要画( )厘米高的直条。",
+    "options": [],
+    "answer": "250;10",
+    "solution": "【分析】本题考查条形统计图中直条高度与数量的比例关系，需要先求出单位长度代表的数量，再进行换算。\n【详解】第一步：求1厘米表示多少本，200本÷4厘米 = 50本/厘米。第二步：求5厘米表示多少本，5厘米×50本/厘米 = 250本。第三步：求500本需要多少厘米，500本÷50本/厘米 = 10厘米。",
+    "knowledge_points": [
+      "简单条形统计图"
+    ],
+    "source": "19-20 四年级上·全国·单元测试"
+  },
+  {
+    "id": "82",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "条形统计图是用一个直条表示一定数量，它较容易看出( )",
+    "options": [],
+    "answer": "数量的多少",
+    "solution": "【分析】本题考查条形统计图的核心优势，需要理解直条长度与数量的对应关系。\n【详解】条形统计图中，直条越长表示数量越大，直条越短表示数量越小。通过直条的长短对比，可以较容易看出数量的多少，这是条形统计图相比统计表最直观的优势。",
+    "knowledge_points": [
+      "条形统计图"
+    ],
+    "source": "19-20六年级上·全国·单元测试"
+  },
+  {
+    "id": "83",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.85,
+    "question": "我们学过的条形统计图可以是( )向的，可以是( )向的。",
+    "options": [],
+    "answer": "横;纵",
+    "solution": "【分析】本题考查条形统计图的分类，按直条方向可分为两种形式。\n【详解】我们学过的条形统计图可以是横向的，可以是纵向的。纵向条形统计图：直条竖直排列，类别在横轴，数量在纵轴，是最常用的形式；横向条形统计图：直条水平排列，类别在纵轴，数量在横轴，适合类别名称较长或数据较大的情况。两种形式只是方向不同，都能直观表示数量的多少，便于比较。",
+    "knowledge_points": [
+      "1格表示一个单位的单式条形统计图",
+      "复式条形统计图"
+    ],
+    "source": "18-19四年级上·广东江门·单元测试"
+  },
+  {
+    "id": "84",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "条形统计图根据形式的不同分为( )式和( )式。",
+    "options": [],
+    "answer": "横;纵",
+    "solution": "【分析】纵向条形统计图：直条竖直排列，类别在横轴，数量在纵轴，最常用；横向条形统计图：直条水平排列，类别在纵轴，数量在横轴，适合类别名称较长的情况。两种形式只是方向不同，都能直观表示数量多少。\n【详解】条形统计图从形式上分为横式条形统计图和纵式条形统计图，据此解答。",
+    "knowledge_points": [
+      "简单条形统计图"
+    ],
+    "source": "2019 六年级下·全国·专题练习"
+  },
+  {
+    "id": "85",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.85,
+    "question": "统计表用( )呈现数据，条形统计图用( )呈现数据。",
+    "options": [],
+    "answer": "表格;直条",
+    "solution": "【分析】本题考查统计表与条形统计图呈现数据方式的区别，需要理解两种统计工具的特点。\n【详解】统计表用表格呈现数据，条形统计图用直条呈现数据。下面是A市2021年8月的天气情况图。\n\n<table><tr><td>日</td><td>一</td><td>二</td><td>三</td><td>四</td><td>五</td><td>六</td></tr><tr><td>1</td><td>2</td><td>3</td><td>4</td><td>5</td><td>6</td><td>7</td></tr><tr><td>8</td><td>9</td><td>10</td><td>11</td><td>12</td><td>13</td><td>14</td></tr><tr><td>15</td><td>16</td><td>17</td><td>18</td><td>19</td><td>20</td><td>21</td></tr><tr><td>22</td><td>23</td><td>24</td><td>25</td><td>26</td><td>27</td><td>28</td></tr><tr><td>29</td><td>30</td><td>31</td><td></td><td></td><td></td><td></td></tr></table>\n\n可以用统计表来表示:\n\n<table><tr><td>天气情况</td><td>晴</td><td>阴</td><td>多云</td><td>阵雨</td><td>雷阵雨</td></tr><tr><td>天数</td><td>9</td><td>6</td><td>9</td><td>5</td><td>2</td></tr></table>\n\n也可以用条形统计图来表示；用直条直观表示数据，便于比较分析。\n\n![](images/808db55352642ae8b7a506ba2005ce5c63d11826e45f0340b165eeab2929103f.jpg)",
+    "knowledge_points": [
+      "1格表示一个单位的单式条形统计图",
+      "单式统计表的特点及填补"
+    ],
+    "source": "16-17 四年级上·贵州毕节·月考"
+  },
+  {
+    "id": "86",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "在一个条形统计图中，如果用 1cm 长的直条表示 30 人，那么应该用( )cm 长的直条表示120人。",
+    "options": [],
+    "answer": "4",
+    "solution": "【分析】本题考查条形统计图中数量与直条长度的正比例关系，用比例法或归一法求解。\n【详解】方法一：归一法。1厘米表示30人，120人里面有几个30人:120÷30 = 4，所以需要4厘米。方法二：比例法。设需要x厘米，1:30 = x:120，30x = 120，x = 4。",
+    "knowledge_points": [
+      "1 格表示多个单位的单式条形统计图"
+    ],
+    "source": "2019六年级下·全国·专题练习"
+  },
+  {
+    "id": "87",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "条形统计图可以分为条形统计图 条形统计图。",
+    "options": [],
+    "answer": "单式;复式",
+    "solution": "【分析】根据常用的条形统计图的种类进行解答即可。\n【详解】条形统计图分为：单式条形统计图和复式条形统计图，前者只表示1个项目的数据，后者可以同时表示多个项目的数据。",
+    "knowledge_points": [
+      "复式条形统计图"
+    ],
+    "source": "16-17 五年级·全国·课后作业"
+  },
+  {
+    "id": "88",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "小明在统计本校人数时，用长3厘米，宽0.5厘米长方形直条表示三年级学生人数60人，那么六年级共120人，应画长 厘米，宽 厘米的长方形直条。",
+    "options": [],
+    "answer": "6;0.5",
+    "solution": "【分析】长3厘米的直条表示60人，则1厘米的直条表示60÷3 = 20 (人)。用六年级总人数除以 1 厘米的直条表示的人数，求出需要画直条长度。而直条的宽度不变，是0.5厘米。\n【详解】第一步：求1厘米直条表示多少人，60人÷3厘米 = 20人/厘米。第二步：求六年级120人需要多长的直条，120人÷20人/厘米 = 6厘米。第三步：确定直条宽度，直条宽度应保持一致，仍为0.5厘米。因此，六年级应画长6厘米，宽0.5厘米的长方形直条。",
+    "knowledge_points": [
+      "1格表示多个单位的单式条形统计图"
+    ],
+    "source": "18-19 四年级上·全国·课后作业"
+  },
+  {
+    "id": "89",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "在一幅表示某学校学生人数的条形统计图中，纵轴\"5格\"表示一年级有250人，那么五年级有 300 人，在纵轴上应该用 格表示。",
+    "options": [],
+    "answer": "6",
+    "solution": "【分析】用 250 人除以 5 求出 1 格表示的人数，然后用 300人除以一格表示的人数即可求出格数。\n【详解】 250÷5 = 50 (人) 300÷50 = 6 (格)",
+    "knowledge_points": [
+      "1 格表示多个单位的单式条形统计图"
+    ],
+    "source": "2014·全国·小升初模拟"
+  },
+  {
+    "id": "90",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "( )统计图能让我们清楚地看出数量的多少",
+    "options": [],
+    "answer": "条形",
+    "solution": "【分析】本题考查不同类型统计图的特点对比，需要明确条形统计图、折线统计图、扇形统计图各自的优势。\n【详解】条形统计图：用直条长短表示数量，能清楚地看出数量的多少，便于比较；折线统计图：用点线表示数量变化，能反映数量的增减变化趋势；扇形统计图：用扇形面积表示比例，能显示各部分与整体的关系。题目强调\"清楚地看出数量的多少\"，这是条形统计图的核心特点。",
+    "knowledge_points": [
+      "简单条形统计图"
+    ],
+    "source": "18-19 四年级下·辽宁·单元测试"
+  },
+  {
+    "id": "91",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.85,
+    "question": "从条形统计图可以清晰的看出不同数量的( )。",
+    "options": [],
+    "answer": "多少",
+    "solution": "【分析】本题考查条形统计图的核心特点，需要理解条形统计图与其他统计图相比最突出的优势。\n【详解】条形统计图的特点是能很清晰地看出不同数量的多少，据此解答。例如：下图是希望小学新学期购进球类统计图。从图中能看出每种球类的数量。\n\n![](images/8337e95442567137ff539579a0e4a6a17347d3cd87a424364d9a599233b2d190.jpg)",
+    "knowledge_points": [
+      "1格表示一个单位的单式条形统计图"
+    ],
+    "source": "15-16 四年级上·全国·课后作业"
+  },
+  {
+    "id": "92",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "( )统计图是用长直条表示数量的，从图中很容易看出( )。",
+    "options": [],
+    "answer": "条形;数量的多少",
+    "solution": "【分析】本题考查条形统计图的基本定义和主要功能，需要准确描述其构成特点。\n【详解】条形统计图是用长直条表示数量的统计图。每个直条代表一个类别，直条的长度(或高度)代表该类别的数量大小。由于直条长短对比鲜明，从图中很容易看出数量的多少，这是条形统计图最显著的优势，便于进行数量间的比较分析。",
+    "knowledge_points": [
+      "简单条形统计图"
+    ],
+    "source": "2019 六年级下·全国·专题练习"
+  },
+  {
+    "id": "93",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.85,
+    "question": "在条形统计图中，如果每格表示 20 万人，则 80 万人要画( )格，半格表示()人。",
+    "options": [],
+    "answer": "4;10 万",
+    "solution": "【分析】本题考查条形统计图中单位量与格数的换算关系，涉及整数格和半格的理解。\n【详解】每格表示20万人，80万人需要:80÷20 = 4格；半格表示:20÷2 = 10万人。",
+    "knowledge_points": [
+      "数据的搜集与整理",
+      "简单条形统计图"
+    ],
+    "source": "18-19四年级上·全国·单元测试"
+  },
+  {
+    "id": "94",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "甲，乙的平均数是 85，甲，乙，丙的平均数是 79，那么丙数是( )。",
+    "options": [],
+    "answer": "67",
+    "solution": "【分析】本题考查平均数的概念及计算方法，利用\"总数=平均数×个数\"的关系求解。\n【详解】甲、乙的总数:85×2 = 170；甲、乙、丙的总数:79×3 = 237；丙数:237 - 170 = 67。",
+    "knowledge_points": [
+      "平均数，中位数，众数"
+    ],
+    "source": "17-18 四年级上·河南新乡·阶段练习"
+  },
+  {
+    "id": "95",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.85,
+    "question": "条形统计图用2格表示30人，平均每格表示( )人，照这样计算，要表示120 人需要画( )格。",
+    "options": [],
+    "answer": "15;8",
+    "solution": "【分析】本题考查条形统计图中单位量的计算及应用，先求每格代表的人数，再计算所需格数。\n【详解】平均每格表示:30÷2 = 15人；表示120人需要:120÷15 = 8格。",
+    "knowledge_points": [
+      "统计图"
+    ],
+    "source": "17-18四年级上·河南新乡·阶段练习"
+  },
+  {
+    "id": "96",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "条形统计图是用一个( )表示( ),根据( )画成( ),然后把这些( )按一定的顺序排列起来。",
+    "options": [],
+    "answer": "单位长度;一定的数量;数量的多少;长短不同的长条;直条",
+    "solution": "【分析】本题考查条形统计图的完整定义和构成要素，需要掌握条形统计图的制作方法。\n【详解】条形统计图是用一个单位长度表示一定的数量，根据数量的多少画成长短不同的长条，然后把这些直条按一定的顺序排列起来。",
+    "knowledge_points": [
+      "统计图"
+    ],
+    "source": "18-19 四年级上·全国·课后作业"
+  },
+  {
+    "id": "97",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.85,
+    "question": "在一幅条形统计图上，纵轴上 1 厘米表示 20 万元，表示 120 万元的直条应画( )厘米长，一个直条长 4 厘米，它表示( )万元。",
+    "options": [],
+    "answer": "6;80",
+    "solution": "【分析】本题考查条形统计图中单位长度与实际数量的换算关系，涉及正比例计算。\n【详解】1厘米表示20万元，表示120万元需要:120÷20 = 6厘米；4厘米表示:4×20 = 80万元。",
+    "knowledge_points": [
+      "统计"
+    ],
+    "source": "18-19 四年级上·全国·期末"
+  },
+  {
+    "id": "98",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "在画条形统计图时，用2 厘米长的直条表示 500 万元，要表示 2000 万元产值，直条应画 厘米。",
+    "options": [],
+    "answer": "8",
+    "solution": "【分析】本题考查条形统计图中单位长度的计算，先求出1厘米代表的数量，再计算所需长度。\n【详解】根据\"用2厘米长的直条表示500万元\"，可以求出用 1 厘米长的直条表示: 500÷2 万元，再用2000 万元除以 500÷2 的商即可。解:2000÷(500÷2) = 2000÷250 = 8 (厘米)，故答案为 8。关键是求出用1厘米长的直条表示的钱数。",
+    "knowledge_points": [
+      "1格表示多个单位的单式条形统计图",
+      "被除数和除数末尾都有0的除法"
+    ],
+    "source": "2014 六年级·全国·课后作业"
+  },
+  {
+    "id": "99",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.65,
+    "question": "条形统计图用2格表示20 人，平均每格表示( )人，照这样计算，要表示 120 人需要画( )格。",
+    "options": [],
+    "answer": "10;12",
+    "solution": "【分析】因为用 2 格表示 20 人，平均每格表示 10 人，则可求出数据120人的条形高。\n【详解】用2格表示20人，则每个格代表 20÷2 = 10 (人)，则可求出表示数据120人的条形高为 120÷10 = 12 (格)。",
+    "knowledge_points": [
+      "1格表示多个单位的单式条形统计图"
+    ],
+    "source": "16-17四年级上·全国·期中"
+  },
+  {
+    "id": "100",
+    "unit": "第七章节 条形统计图",
+    "question_type": "填空题",
+    "difficulty": 0.85,
+    "question": "条形统计图能清楚地反映( )",
+    "options": [],
+    "answer": "各种数量的多少",
+    "solution": "【分析】本题考查条形统计图的核心功能，需要理解条形统计图最主要的用途。\n【详解】条形统计图可以清楚地看出各种数量的多少。",
+    "knowledge_points": [
+      "1格表示多个单位的单式条形统计图",
+      "1格表示一个单位的单式条形统计图"
+    ],
+    "source": "15-16四年级上·山东青岛·课后作业"
+  }
+]
diff --git "a/\345\233\233\345\271\264\347\272\247\351\242\230\351\233\206/\345\233\233\345\271\264\347\272\247\344\270\212/\345\233\233\345\271\264\347\272\247\344\270\212\347\254\254\344\270\211\347\253\240\350\247\222\347\232\204\345\272\246\346\225\260\346\230\257.json" "b/\345\233\233\345\271\264\347\272\247\351\242\230\351\233\206/\345\233\233\345\271\264\347\272\247\344\270\212/\345\233\233\345\271\264\347\272\247\344\270\212\347\254\254\344\270\211\347\253\240\350\247\222\347\232\204\345\272\246\346\225\260\346\230\257.json"
new file mode 100644
index 0000000000000000000000000000000000000000..81b7bb0b4fca8493499ae9716d868cc638880932
--- /dev/null
+++ "b/\345\233\233\345\271\264\347\272\247\351\242\230\351\233\206/\345\233\233\345\271\264\347\272\247\344\270\212/\345\233\233\345\271\264\347\272\247\344\270\212\347\254\254\344\270\211\347\253\240\350\247\222\347\232\204\345\272\246\346\225\260\346\230\257.json"
@@ -0,0 +1,9081 @@
+[
+    {
+        "id":  "1",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.94,
+        "question":  "过一点可以画( )条直线，过两点可以画( )条直线。",
+        "options":  [
+                        "A. 1;1",
+                        "B. 1;无数",
+                        "C. 无数;1",
+                        "D. 无数;无数"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】根据点和直线的关系可知：过一点可以画无数条直线，同时通过两点只能画1条直线。\n【详解】由分析可得：过一点可以画无数条直线，过两点可以画1条直线。故答案为:C",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·江西吉安·期中"
+    },
+    {
+        "id":  "2",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "经过两点可以画()条线段，可以画( )条直线。",
+        "options":  [
+                        "A. 1;1",
+                        "B. 2;无数",
+                        "C. 无数;1"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】根据题意，过两点只能画一条线段。过点画直线，可以向任何方向画，所以能画无数条。过两点画直线，两个点已经确定了直线的方向，所以只能画一条。以此选择即可。\n【详解】根据分析可知：经过两点可以画1条线段，可以画1条直线。故答案为:A",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·青海海东·期中"
+    },
+    {
+        "id":  "3",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面是课堂上小军作的笔记，其中有一条被小军写错了，请问是哪一条( )。",
+        "options":  [
+                        "A. 线段有两个端点，可以量出长度",
+                        "B. 射线有一个端点，汽车灯射出的光可以看作射线",
+                        "C. 经过两点能画出无数条直线",
+                        "D. 直线和射线都是无限长"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】线段：有两个端点，可以测量长度，不能向两端无限延伸；射线：有一个端点，可以向一端无限延伸，不能测量长度；直线：没有端点，可以向两端无限延伸，不能测量长度；经过一点可以画无数条直线，但是经过两点只能画出一条直线；据此解答。\n【详解】A. 线段有两个端点，可以量出长度，说法正确，不符合题意；B. 射线有一个端点，汽车灯射出的光可以看作射线，说法正确，不符合题意；C. 经过两点只能画出一条直线，选项中说法错误，符合题意；D. 直线和射线都是无限长，说法正确，不符合题意。故答案为:C",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·贵州六盘水·期中"
+    },
+    {
+        "id":  "4",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "关于线段，射线和直线的描述，错误的是( )。",
+        "options":  [
+                        "A. 直线比射线长",
+                        "B. 过一点能画无数条直线",
+                        "C. 线段有两个端点，射线有一个端点，直线没有端点"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】根据线段，射线和直线的特点，线段有两个端点，不能向两端无限延长，长度有限，可以测量长度；射线只有一个端点，可以向一端无限延长，长度是无限的，无法测量长度；直线没有端点，可以向两端无限延长，长度是无限的，无法测量长度；过两点只能画一条直线，因为这两点固定了直线的方向；过一点可以画无数条直线，因为这一点没有固定直线的方向。据此解答。\n【详解】A. 直线和射线的长度都是无限的，无法测量长度，原说法错误；B. 过一点能画无数条直线，原说法正确；C. 线段有两个端点，射线有一个端点，直线没有端点，原说法正确。故答案为:A",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25四年级上·广东潮州·期中"
+    },
+    {
+        "id":  "5",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "钟面上秒针旋转一周得到的是( )。",
+        "options":  [
+                        "A. 直角",
+                        "B. 钝角",
+                        "C. 平角",
+                        "D. 周角"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】根据题意，秒针旋转一周形成的角度是 360°，根据角的分类，周角为 360°，因此秒针旋转一周得到的是周角。以此选择即可。\n【详解】根据分析可知：钟面上秒针旋转一周得到的是周角。故答案为:D",
+        "knowledge_points":  [
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·广西百色·期中"
+    },
+    {
+        "id":  "6",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "一个圆形纸片对折 3次后展开，任意两条折痕所夹的角不可能是( )。",
+        "options":  [
+                        "A. 45°",
+                        "B. 60°",
+                        "C. 90°",
+                        "D. 135°"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】将圆形纸片对折 1 次，圆心角 360°变为 180°；对折 2 次，变为 90°；对折 3 次，变为 45°。因此，对折 3 次后展开，圆心角被分成 8 份，每份为 45°。那么折痕形成的角度均为 45° 的整数倍。\n【详解】 360°÷8 = 45°。A. 45°÷45° = 1，可能是 45°；B. 60°÷45° = 1……15°，不可能是 60°；C. 90°÷45° = 2，可能是 90°；D. 135°÷45° = 3，可能是 135°。故答案为:B",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "图形的折叠问题",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·江西赣州·期中"
+    },
+    {
+        "id":  "7",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "把一个平角分成两个角，这两个角可能是( )。",
+        "options":  [
+                        "A. 一个锐角和一个直角",
+                        "B. 两个锐角",
+                        "C. 一个锐角和一个钝角",
+                        "D. 两个钝角"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】1平角 = 180°，锐角小于 90°，钝角大于 90° 而小于 180°，1直角 = 90°。A. 直角 90° 加一个小于 90° 的角一定小于 180°，不可能为平角；B. 两个锐角相加和也小于 180°；C. 如 30° 与 150° 相加为 180°，可能；D. 两个钝角相加一定大于 180°。\n【详解】A、B、D都不可能，只有一个锐角和一个钝角可能将平角分成两部分。故答案为:C",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·江西赣州·期中"
+    },
+    {
+        "id":  "8",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "把两个锐角拼成一个较大的角，下面说法正确的是( )。",
+        "options":  [
+                        "A. 一定是锐角",
+                        "B. 一定是直角",
+                        "C. 一定是钝角",
+                        "D. 无法确定"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】若两个锐角都是 10°，和为 20°，仍是锐角；若都是 45°，和为 90°，是直角；若都是 60°，和为 120°，是钝角，所以结果不唯一。\n【详解】两个锐角拼成的角可能是锐角、直角或钝角，因此无法确定。故答案为:D",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "角度的计算"
+                             ],
+        "source":  "24-25 四年级上·福建龙岩·期中"
+    },
+    {
+        "id":  "9",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "一个钝角不能分成()。",
+        "options":  [
+                        "A. 2 个锐角",
+                        "B. 1个钝角和1个锐角",
+                        "C. 2个钝角"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】钝角大于 90° 小于 180°，锐角小于 90°，直角等于 90°。两个钝角相加一定大于 180°，不可能由一个钝角拆分得到。\n【详解】A、B都可能，只有C不可能。故答案为:C",
+        "knowledge_points":  [
+                                 "角的分类及换算",
+                                 "直角，钝角，锐角的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·广东广州·期中"
+    },
+    {
+        "id":  "10",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "用一副三角尺可以画出( )的角。",
+        "options":  [
+                        "A. 140°",
+                        "B. 105°",
+                        "C. 95°",
+                        "D. 40°"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】一副三角板的角有 30°、45°、60°、90°。可组合出 15°、75°、105°、135°、120°、150° 等。\n【详解】A:无法拼成 140°；B:45°+60°=105°，可以画；C:无法拼成 95°；D:无法拼成 40°。故答案为:B",
+        "knowledge_points":  [
+                                 "用三角尺画角"
+                             ],
+        "source":  "24-25 四年级上·福建漳州·期中"
+    },
+    {
+        "id":  "11",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "用一副三角尺不能拼出( )。",
+        "options":  [
+                        "A. 15°",
+                        "B. 30°",
+                        "C. 40°",
+                        "D. 105°"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】一副三角尺的角为 30°、45°、60°、90°。A:45°-30°=15°；B:有 30° 本身；D:60°+45°=105°；C:无法由上述角的加减得到 40°。\n【详解】只能选C。故答案为:C",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "用三角尺画角"
+                             ],
+        "source":  "24-25四年级上·浙江·期中"
+    },
+    {
+        "id":  "12",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面说法正确的有 ( )个.\n\n①用一个 8 倍的放大镜看 20°角，这个角就变成了 160° \n②一个数四舍五入到亿位是10亿，这个数最大是999999999 \n③最小的自然数是1，自然数的个数是无限的 \n④十进制中，相邻的两个计数单位间的进率是 10",
+        "options":  [
+                        "A. 1",
+                        "B. 2",
+                        "C. 3",
+                        "D. 4"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】①角大小只与张开程度有关，与放大无关，错；②四舍五入到亿位是10亿，最大应为1049999999，错；③最小自然数是0，不是1，错；④相邻计数单位进率为10，对。\n【详解】只有④正确，共1个。故答案为:A",
+        "knowledge_points":  [
+                                 "角的度量",
+                                 "十进制计数法",
+                                 "自然数的认识",
+                                 "亿以上数的近似数"
+                             ],
+        "source":  "25-26四年级上·福建厦门·月考"
+    },
+    {
+        "id":  "13",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "105°,135°,160°,150°中，一副三角板能画出的角有( )个。",
+        "options":  [
+                        "A. 1",
+                        "B. 2",
+                        "C. 3",
+                        "D. 4"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】三角板角为 30°、45°、60°、90°。105°:60°+45°；135°:90°+45°；150°:90°+60°，160°无法拼出。\n【详解】能画出105°、135°、150°共3个。故答案为:C",
+        "knowledge_points":  [
+                                 "用三角尺画角"
+                             ],
+        "source":  "24-25 四年级上·河南·期中"
+    },
+    {
+        "id":  "14",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面的说法错误的是( )。",
+        "options":  [
+                        "A. 平角的两条边组成一条直线",
+                        "B. 大于 90 度的角都是钝角",
+                        "C. 一条线段长 10 厘米"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】平角两边组成一条直线，正确；钝角是 90° 到 180° 之间的角，大于90度但未到180度才是钝角，\"大于90度的角都是钝角\"忽略了平角和周角，错误；线段可度量长度，10厘米合理。\n【详解】错误的是B。故答案为:B",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·湖南长沙·期中"
+    },
+    {
+        "id":  "15",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "钟面上 3 时整，分针和时针成( )度。",
+        "options":  [
+                        "A. 90",
+                        "B. 100",
+                        "C. 120"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】整点时分针指向12。3时整，时针指向3，12与3之间相隔3大格，每格 30°，共 3×30° = 90°。\n【详解】分针与时针成90度。故答案为:A",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "平角，周角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·湖南长沙·期中"
+    },
+    {
+        "id":  "16",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "用一副三角尺不能拼出的角是( )。",
+        "options":  [
+                        "A. 120°",
+                        "B. 65°",
+                        "C. 15°",
+                        "D. 150°"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】三角尺角度为 30°、45°、60°、90°。120°=30°+90°；15°=45°-30°；150°=90°+60°；65°无法由这些角加减得到。\n【详解】不能拼出的是65°。故答案为:B",
+        "knowledge_points":  [
+                                 "用三角尺画角"
+                             ],
+        "source":  "24-25 四年级上·福建漳州·期中"
+    },
+    {
+        "id":  "17",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面各角度中，不能用一副三角尺拼出来的是( )。",
+        "options":  [
+                        "A. 75°",
+                        "B. 70°",
+                        "C. 135°"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】一副三角板有角 30°、45°、60°、90°。75°=30°+45°；135°=90°+45°；70°无法由这些角度的加减得到。\n【详解】不能拼出的角是70°。故答案为:B",
+        "knowledge_points":  [
+                                 "用三角尺画角"
+                             ],
+        "source":  "24-25 四年级上·江西赣州·期中"
+    },
+    {
+        "id":  "18",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "把一个平角分成两个角，其中一个是钝角，那么另一个角一定是( )。",
+        "options":  [
+                        "A. 锐角",
+                        "B. 直角",
+                        "C. 钝角",
+                        "D. 周角"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】平角是 180°，若一个角是钝角(大于 90° 小于 180°)，则另一个角必小于 90°，为锐角。\n【详解】例如 100° 和 80°，后者为锐角。故答案为:A",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "2022 四年级上·全国·专题练习"
+    },
+    {
+        "id":  "19",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面的角中,( )不能用一副三角尺画出来。",
+        "options":  [
+                        "A. 15°",
+                        "B. 135°",
+                        "C. 120°",
+                        "D. 145°"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】三角尺角度 30°、45°、60°、90°。15°=45°-30°；135°=90°+45°；120°=30°+90°；145°无法由这些角的加减得到。\n【详解】不能画出的是145°。故答案为:D",
+        "knowledge_points":  [
+                                 "用三角尺画角"
+                             ],
+        "source":  "24-25 四年级上·湖南长沙·期中"
+    },
+    {
+        "id":  "20",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "过两点能画( )条直线。",
+        "options":  [
+                        "A. 1",
+                        "B. 2",
+                        "C. 无数"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】过两点有且仅有一条直线。\n【详解】因此只能画1条直线。故答案为:A",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·广东汕头·期中"
+    },
+    {
+        "id":  "21",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.94,
+        "question":  "用一个100 倍的放大镜看 18°的角，此时该角的度数是( )。",
+        "options":  [
+                        "A. 100°",
+                        "B. 180°",
+                        "C. 118°",
+                        "D. 18°"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】放大镜改变长度，不改变两边夹角大小，所以角的度数不变。\n【详解】看到的仍是 18°。故答案为:D",
+        "knowledge_points":  [
+                                 "角的度量"
+                             ],
+        "source":  "24-25四年级上·海南省直辖县级单位·期中"
+    },
+    {
+        "id":  "22",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "经过平面上的两个点可以画( )条直线。",
+        "options":  [
+                        "A. 1",
+                        "B. 2",
+                        "C. 无数",
+                        "D. 无法确定"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】直线性质：过两点有且仅有一条直线。\n【详解】所以只能画1条。故答案为:A",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25四年级上·海南省直辖县级单位·期中"
+    },
+    {
+        "id":  "23",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面不能用一副三角尺拼出的角是( )。",
+        "options":  [
+                        "A. 150°",
+                        "B. 135°",
+                        "C. 115°",
+                        "D. 75°"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】三角尺角为 30°、45°、60°、90°。150°=90°+60°；135°=90°+45°；75°=30°+45°；115°无法由这些角组合。\n【详解】所以不能拼出115°。故答案为:C",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "用三角尺画角"
+                             ],
+        "source":  "24-25 四年级上·湖南岳阳·期中"
+    },
+    {
+        "id":  "24",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "将一张圆形纸片对折三次后，得到的角是( )。",
+        "options":  [
+                        "A. 45°",
+                        "B. 90°",
+                        "C. 60°"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】周角 360°，对折3次等分为8份，每份 360°÷8 = 45°。\n【详解】得到的角是45°。故答案为:A",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "图形的折叠问题",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25四年级上·湖南邵阳·期中"
+    },
+    {
+        "id":  "25",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.94,
+        "question":  "在一个能放大10倍的放大镜下看 10°的角，看到的角的度数是( )°。",
+        "options":  [
+                        "A. 20",
+                        "B. 10",
+                        "C. 100"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】角大小由两边张开程度决定，与边长无关，放大不改变角度。\n【详解】仍为10°。故答案为:B",
+        "knowledge_points":  [
+                                 "角的度量",
+                                 "角的概念及表示方法"
+                             ],
+        "source":  "24-25 四年级上·内蒙古通辽·期中"
+    },
+    {
+        "id":  "26",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "用一副三角尺不可以画出的角的度数是( )。",
+        "options":  [
+                        "A. 75°",
+                        "B. 105°",
+                        "C. 115°",
+                        "D. 135°"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】三角尺角 30°、45°、60°、90°。75°=30°+45°；105°=60°+45°；135°=90°+45°；115°无法组合。\n【详解】不能画出的是115°。故答案为:C",
+        "knowledge_points":  [
+                                 "用三角尺画角"
+                             ],
+        "source":  "24-25 四年级上·内蒙古通辽·期中"
+    },
+    {
+        "id":  "27",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.94,
+        "question":  "通过 5 倍放大镜看到了一个 20° 的角，这个角的实际度数是( )。",
+        "options":  [
+                        "A. 20°",
+                        "B. 4°",
+                        "C. 100°"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】放大镜只放大长度，不改变角的大小。\n【详解】实际度数仍为20°。故答案为:A",
+        "knowledge_points":  [
+                                 "角的概念及表示方法"
+                             ],
+        "source":  "25-26 四年级上·山东德州·月考"
+    },
+    {
+        "id":  "28",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.94,
+        "question":  "小学生平时用的三角尺中没有( )度的角。",
+        "options":  [
+                        "A. 60",
+                        "B. 45",
+                        "C. 75"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】三角尺自带角为 30°、45°、60°、90°，75°是拼合得到，不是自带角。\n【详解】因此三角尺\"本身\"没有75°。故答案为:C",
+        "knowledge_points":  [
+                                 "用三角尺画角"
+                             ],
+        "source":  "25-26四年级上·山东德州·月考"
+    },
+    {
+        "id":  "29",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "用100倍的放大镜看 30° 的角，这时该角为( )。",
+        "options":  [
+                        "A. 30°",
+                        "B. 100°",
+                        "C. 130°",
+                        "D. 300°"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】角的大小与长度无关，放大镜只改变边长，不改变夹角。\n【详解】该角仍为30°。故答案为:A",
+        "knowledge_points":  [
+                                 "角的概念及表示方法"
+                             ],
+        "source":  "24-25四年级上·辽宁鞍山·期中"
+    },
+    {
+        "id":  "30",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列说法错误的是( )。",
+        "options":  [
+                        "A. 计数单位和数位的意义不同",
+                        "B. 最小的七位数比最大的六位数大1",
+                        "C. 角的两边越长，角的度数越大"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】A正确：数位和计数单位概念不同；B正确:1000000与999999差1；C错误：角大小只与张开程度有关，与边长无关。\n【详解】错误的是\"角的两边越长，角的度数越大\"。故答案为:C",
+        "knowledge_points":  [
+                                 "角的度量",
+                                 "整数的数级，数位和计数单位的认识",
+                                 "大数的比较"
+                             ],
+        "source":  "25-26 四年级上·甘肃天水·阶段练习"
+    },
+    {
+        "id":  "31",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面的说法中，正确的是( )。",
+        "options":  [
+                        "A. 计算器上的这个键 MC 是储存键",
+                        "B. 锐角+锐角=平角",
+                        "C. 测量一间会议室的面积，要用\"公顷\"作单位",
+                        "D. 人脑大约有 140 亿个细胞，横线上的数是近似数"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】MC是清除存储键；两个锐角和小于180°，不一定平角；会议室面积用平方米，不用公顷；\"大约\"表明显示的是近似数。\n【详解】只有D正确。故答案为:D",
+        "knowledge_points":  [
+                                 "亿以上数的近似数",
+                                 "面积单位的选择",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "计算器的初步认识与使用"
+                             ],
+        "source":  "24-25 四年级上·安徽阜阳·期中"
+    },
+    {
+        "id":  "32",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "时针和分针组成的较小角是 120° 的是( )。",
+        "options":  [
+                        "A. 4时整",
+                        "B. 5 时整",
+                        "C. 10 时整",
+                        "D. 6 时整"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】整点时分针指向12。4时整，时针指向4，间隔4大格:4×30° = 120°。\n【详解】4时整时针分针夹角为120°。故答案为:A",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·福建龙岩·期中"
+    },
+    {
+        "id":  "33",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "小明在观看奥运会时用一个10倍的放大镜看运动场上一个 30° 的角，看到的角是( )。",
+        "options":  [
+                        "A. 3°",
+                        "B. 30°",
+                        "C. 300°",
+                        "D. 无法确定"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】放大镜改变的是视觉大小，不改变两边的方向，故夹角不变。\n【详解】看到的角仍为30°。故答案为:B",
+        "knowledge_points":  [
+                                 "角的概念及表示方法"
+                             ],
+        "source":  "24-25 四年级上·贵州黔南·期中"
+    },
+    {
+        "id":  "34",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "射线的特点可以用下面的哪个词语来形容。( )",
+        "options":  [
+                        "有始无终",
+                        "无始无终",
+                        "有始有终"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】射线是指由线段的一端无限延长所形成的直的线。射线有一个端点，这个端点就是\"始\"，然后从这个端点向一个方向无限延伸，没有\"终\"点，所以可以用\"有始无终\"来形容。\"无始无终\"，这是直线的特点，直线没有端点，向两端无限延伸。\"有始有终\"，这是线段的特点，线段有两个端点，不能延伸。\n【详解】根据分析可知，射线的特点可以用\"有始无终\"来形容。故答案为:A",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·广东珠海·期中"
+    },
+    {
+        "id":  "35",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "将一张圆形纸片对折3次后展开，任意两条折痕所夹的角不可能是( )。",
+        "options":  [
+                        "45°",
+                        "60°",
+                        "90°",
+                        "135°"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】将圆形纸片对折1次，就是把圆形纸片平均分成2份，圆心角360°变为360°÷2=180°；对折2次，就是把圆形纸片平均分成4份，圆心角360°变为360°÷4=90°；对折3次，就是把圆形纸片平均分成8份，圆心角360°变为360°÷8=45°，即每份为45°，那么折痕形成的角度均为45°的整数倍，据此解答即可。\n【详解】A. 45°÷45°=1，可能是45°；B. 60°÷45°=1……15°，不可能是60°；C. 90°÷45°=2，可能是90°；D. 135°÷45°=3，可能是135°。将一张圆形纸片对折3次后展开，任意两条折痕所夹的角不可能是60°。故答案为:B",
+        "knowledge_points":  [
+                                 "图形的折叠问题",
+                                 "角的分类及换算"
+                             ],
+        "source":  "24-25四年级上·浙江·期中"
+    },
+    {
+        "id":  "36",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "把一条 6 厘米长的线段向两端各延长 100 厘米，得到一条( )。",
+        "options":  [
+                        "直线",
+                        "线段",
+                        "射线",
+                        "不确定"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】线段有两个端点，长度有限；直线没有端点，长度无限；射线有一个端点，长度无限。将线段向两端延长后，仍有两个端点，长度变为6+100+100=206(厘米)，仍为线段。\n【详解】原线段长6厘米，向两端各延长100厘米后，总长度为6+100+100=206(厘米)。延长后的图形有两个端点，且长度有限，因此仍为线段。故答案为:B",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·浙江·期中"
+    },
+    {
+        "id":  "37",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面说法正确的是( )。",
+        "options":  [
+                        "直线和射线都能无限延伸",
+                        "过两点能画两条直线",
+                        "射线比直线短"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】射线有一个端点，可以向一个方向无限延伸，不可测量长度；直线无端点，可以向两个方向无限延伸，不可测量长度；过两点有且只有一条直线。\n【详解】A. 直线没有端点，可以两端无限延伸，射线有一个端点，可以向一端无限延伸，所以直线和射线都能无限延伸，故该选项正确；B. 根据直线的性质，过两点有且只有一条直线，也就是过两点只能画一条直线，不是两条，故该选项错误；C. 直线和射线都是无限长的，无法比较长短，所以不能说射线比直线短，故该选项错误。故答案为:A",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·河北石家庄·期中"
+    },
+    {
+        "id":  "38",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面三种角,( )的度数最大。",
+        "options":  [
+                        "平角",
+                        "周角",
+                        "直角"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】已知平角等于180°，周角等于360°，直角等于90°，据此解答。\n【详解】360°>180°>90°，所以，周角>平角>直角。所以，三种角中，周角的度数最大。故答案为:B",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "角的大小比较",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25四年级上·河北石家庄·期中"
+    },
+    {
+        "id":  "39",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "小明画了一条5厘米长的( )。",
+        "options":  [
+                        "直线",
+                        "射线",
+                        "线段"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】直线没有端点，射线只有一个端点，二者都不能量得其长度，只有线段有两个端点，可以测量长度。\n【详解】因直线没有端点，射线只有一个端点，二者都不能量得其长度，而线段有两个端点，能量得长度，所以应为小明画了一条5厘米长的线段。故答案选:C",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25四年级上·湖南永州·期中"
+    },
+    {
+        "id":  "40",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面( )不能看作是射线。",
+        "options":  [
+                        "射灯",
+                        "角的边",
+                        "手电筒发的光",
+                        "木棍"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】根据题意，射线有一个起点(端点)，但没有终点，向一个方向无限延伸。以此逐项分析即可。\n【详解】根据分析可知:A. 射灯发出的光线从一个点(灯源)开始，向一个方向传播，可以抽象为射线；B. 在几何中，角是由两条射线共享一个公共端点(顶点)形成的，因此角的边本身就是射线；C. 与射灯类似，手电筒的光从一个点(光源)开始，向一个方向延伸，可以看作是射线；D. 木棍是一个实物，有固定的长度和两个端点(起点和终点)，不能无限延伸，因此它更像一条线段，而不是射线。木棍不能看作是射线。故答案为:D",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·山西忻州·期中"
+    },
+    {
+        "id":  "41",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "春晚是全国人民的文化盛宴，每年春晚从晚上8时开始，此时钟面上时针和分针形成的较小角是( )°角，它是一个( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "120;钝",
+        "solution":  "【分析】根据题意，在钟面上，8时整时，时针指向8，分针指向12。钟面一圈为360°，每大格(即相邻数字之间)为30°。从12到8较小的角有4大格，所以时针与分针的夹角为4×30°=120°，120°大于90°且小于180°，因此是钝角。\n【详解】根据分析可知:4×30°=120°。春晚是全国人民的文化盛宴，每年春晚从晚上8时开始，此时钟面上时针和分针形成的较小角是120°角，它是一个钝角。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "25-26四年级上·河北唐山·期末"
+    },
+    {
+        "id":  "42",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "小军在用量角器量两个角的度数时，犯了两个错误.\n(1)量第一个角时，角的一条边没有与 0 刻度线重合，而是与10°的刻度线重合了，这时角被他量成了90°,实际这个角的度数是 度.\n(2)量第二个角时，将本应该读的内圈读成了外圈的度数，这个角被他量成了120°,实际这个角的度数是 度。",
+        "options":  [
+
+                    ],
+        "answer":  "80;60",
+        "solution":  "【分析】如果角的起始边不是与0刻度线重合，角的度数为两条边所对的刻度之差。这个角的实际度数等于90°-10°=80°；同一刻度线上，内圈的读数加外圈的读数等于180°，由于看错了内外圈，一个锐角被读成了120°的钝角，这个锐角的度数等于180°-120°=60°；据此即可解答。\n【详解】(1)90°-10°=80°。量第一个角时，角的一条边没有与0刻度线重合，而是与10°的刻度线重合了，这时角被他量成了90°，实际这个角的度数是80度。(2)180°-120°=60°。量第二个角时，将本应该读的内圈读成了外圈的度数，这个角被他量成了120°，实际这个角的度数是60度。",
+        "knowledge_points":  [
+                                 "角的度量"
+                             ],
+        "source":  "24-25 四年级上·贵州六盘水·期中"
+    },
+    {
+        "id":  "43",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小军做完作业之后，看到墙上钟面的时刻刚好是下午4时整，这时时针和分针形成的角是( )角，它的度数是( )°;他做完作业出去踢球，回来看到时针和分针形成的角是平角，这时是( )时整。",
+        "options":  [
+
+                    ],
+        "answer":  "钝;120;6",
+        "solution":  "【分析】钟面上12个数字，以表芯为旋转点，表针转一圈是360°，被12个数字平均分成12份，每一份也就是两数之间夹角是30°。下午4时整时针指4，分针指12，中间有4大格，也就是30°×4=120°，大于90°小于180°的角叫钝角；时针和分针形成的角是平角，说明在一条直线上，此时是6时整。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·贵州六盘水·期中"
+    },
+    {
+        "id":  "44",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "6时整，钟面上分针与时针所成的角是( )°,叫做( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "180;平",
+        "solution":  "【分析】钟面是一个周角，总度数为360°，被12个大格平均分成12份。每个大格对应的角度为360°÷12=30°。6时整时，时针指向6，分针指向12，两者之间间隔6个大格。夹角度数为30°×6=180°，180°的角定义为平角。6时整，时针指向6，分针指向12，此时时针和分针成180°角，也就是平角。\n【详解】由分析可知:6时整，钟面上分针与时针所成的角是180°，叫做平角。",
+        "knowledge_points":  [
+                                 "平角，周角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25四年级上·广东潮州·期中"
+    },
+    {
+        "id":  "45",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "从 5:00 到 5:25，分针走了( )°,3 时整，时针和分针形成( )°的角,( )时整，时针和分针形成周角。",
+        "options":  [
+
+                    ],
+        "answer":  "150;90;12",
+        "solution":  "【分析】已知钟面上有12个大格，把钟面看作一个周角，是360°，因12×30°=360°，所以指针走过1个大格是30°。5:00分针指着12，5:25分针指着5，分针一共走过了5个大格，走过的度数是30°×5=150°；3时整，分针指着12，时针指着3，中间一共有3个大格，共的度数是30°×3=90°，所以时针和分针形成90°的角，即直角；已知周角的两条边重合在一起，形成一条射线；当时间为整时的时候，钟面上分针应指着12，要使时针和分针形成周角，则时针也应指着12，此时的时间为12时整。据此解答。\n【详解】根据分析可知:\n12×30°=360°\n30°×5=150°\n30°×3=90°。\n所以，从5:00到5:25，分针走了150°；3时整，时针和分针形成90°的角；12时整，时针和分针形成周角。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征",
+                                 "角的分类及换算",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·江西九江·期中"
+    },
+    {
+        "id":  "46",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "钟面上9时整，时针和分针所组成的是( )角，是( )度。当钟面上时针和分针成平角时，正好是( )时。",
+        "options":  [
+
+                    ],
+        "answer":  "直;90;6",
+        "solution":  "【分析】钟面上有12大格，每一大格对应的夹角是30度，9时整，时针和分针所成的夹角对应的大格数为3，所以时针和分针的夹角等于30×3=90(度)，是直角；6时整，分针指向12，时针指向6，时针和分针在一条直线上，时针和分针所成的夹角是平角。\n【详解】30×3=90(度)。钟面上9时整，时针和分针所组成的是直角，是90度。当钟面上时针和分针成平角时，正好是6时。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25四年级上·浙江杭州·期中"
+    },
+    {
+        "id":  "47",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "2时整，时针和分针成( )角,6 时整，时针和分针成( )角;9时整，时针和分针成( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "锐;平;直",
+        "solution":  "【分析】钟面1大格是30°，先计算出角度，再根据锐角小于90°，钝角大于90°而小于180°，直角等于90°，平角的两条边在同一条直线上，周角的两边重合，据此来确定这是一个什么角。2时整，时针指向2，分针指向12，时针与分针相隔2大格，30°乘2即可求出度数；6时整，时针指向6，分针指向12，时针与分针在一条直线上，此时所成的角是平角；9时整，时针指向9，分针指向12，时针与分针相隔3大格，30°乘3即可求出度数。\n【详解】30°×2=60°，30°×3=90°。2时整，时针和分针成锐角，6时整，时针和分针成平角；9时整，时针和分针成直角。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25四年级上·广东汕尾·期中"
+    },
+    {
+        "id":  "48",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "( )时整时针和分针形成一个平角;从2:00到2:20分针转动了( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "6;120",
+        "solution":  "【分析】钟面上有12个大格，每1个大格之间的度数是30°；平角是180°的角，180°÷30°=6，转了6个大格，也就是时针从12转到6，即6时整；根据经过时间=结束时间-开始时间，可知从2:00到2:20，共经过了20分钟，分针走一大格是5分钟，则2:00到2:20，分针共走20÷5=4大格，也就是分针转动了4×30°，据此解答即可。\n【详解】180°÷30°=6；2时20分-2时=20分；20÷5×30°=4×30°=120°。所以6时整时针和分针形成一个平角；从2:00到2:20分针转动了120°。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "平角，周角的认识及特征",
+                                 "整点时间与钟面指针的位置",
+                                 "时，分，秒有关的计算"
+                             ],
+        "source":  "24-25四年级上·江西赣州·期中"
+    },
+    {
+        "id":  "49",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "6 时整，钟面上时针和分针所形成的夹角是( )°;从早上6时到中午12时，时针相当于走了 2 个( )角的度数。",
+        "options":  [
+
+                    ],
+        "answer":  "180;直",
+        "solution":  "【分析】钟面一周360°，平均分12个大格，每个大格角度为360°÷12。6时整，时针指向6，分针指向12，两者相隔6个大格，用6乘每个大格的角度，即可求得时针和分针所形成的夹角。从早上6时到中午12时，先计算时针走了几小时，也就是经过了几个大格，用经过的大格数量乘每个大格的角度，即可求得走了多少角度，据此解答。\n【详解】360÷12=30°；6×30=180°；12-6=6(个)；6×30=180°。一个直角=90°，所以180°相当于走了2个直角的度数。因此，6时整，钟面上时针和分针所形成的夹角是180°；从早上6时到中午12时，时针相当于走了2个直角的度数。",
+        "knowledge_points":  [
+                                 "角的度量",
+                                 "角的分类及换算",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·安徽阜阳·期中"
+    },
+    {
+        "id":  "50",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "一个角比直角小30°,这个角是( )度。",
+        "options":  [
+
+                    ],
+        "answer":  "60",
+        "solution":  "【分析】直角是90°，题目中的角比直角小30°，因此用90°减去30°即可求出这个角的度数。\n【详解】已知直角为90°，这个角比直角小30°，则这个角的度数为:90°-30°=60°。所以这个角是60度。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "角度的计算"
+                             ],
+        "source":  "24-25四年级上·贵州·期中"
+    },
+    {
+        "id":  "51",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "108°的角和( )°的角可以拼成一个平角，和( )°的角可以拼成一个周角。",
+        "options":  [
+
+                    ],
+        "answer":  "72;252",
+        "solution":  "【分析】平角是180°，周角是360°。已知一个角是108°，求另一个角时，分别用平角和周角的度数减去108°即可。\n【详解】180°-108°=72°，108°的角和72°的角可以拼成一个平角；360°-108°=252°，108°的角和252°的角可以拼成一个周角。",
+        "knowledge_points":  [
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·安徽·期中"
+    },
+    {
+        "id":  "52",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "钟面上6时整，时针和分针形成的角是( )角。时针在钟面上走12小时所转过的角是( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "平;周",
+        "solution":  "【分析】根据题意，因为钟表上的刻度是把一个圆平均分成了12等份，每一份是30度，钟面上6时整，时针指向数字6，分针指向数字12。从12到6有6个大格，用大格数6乘30即可求出度数；时针在钟面上走12小时所转过的角，依据钟面的特点可知，1圈为360度；根据角的度数判断角的种类，锐角大于0度小于90度，直角等于90度，钝角大于90度小于180度，平角等于180度，周角等于360度；据此解答。\n【详解】根据分析可知:6×30=180(度)。钟面上6时整，时针和分针形成的角是平角。时针在钟面上走12小时所转过的角是周角。",
+        "knowledge_points":  [
+                                 "平角，周角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25四年级上·浙江·期中"
+    },
+    {
+        "id":  "53",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在30°,100°,90°,1°,89°,91°和300°七个角中,( )的角是锐角,( )的角是钝角,( )的角是直角。",
+        "options":  [
+
+                    ],
+        "answer":  "30°,1°,89°;100°,91°;90°",
+        "solution":  "【分析】大于0°小于90°的角叫做锐角，直角的度数等于90°，大于90°小于180°的角叫做钝角。由题意得，30°、1°和89°的角大于0°且小于90°，所以它们都是锐角；90°的角是直角；91°和100°的角大于90°且小于180°，所以它们都是钝角。\n【详解】在30°、100°、90°、1°、89°、91°和300°七个角中，30°、1°、89°的角是锐角，100°、91°的角是钝角，90°的角是直角。",
+        "knowledge_points":  [
+                                 "角的分类及换算"
+                             ],
+        "source":  "25-26 四年级上·安徽宿州·月考"
+    },
+    {
+        "id":  "54",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "21时整，钟面上时针和分针的夹角是( )角;18时整，时针和分针的夹角是( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "直;平",
+        "solution":  "【分析】24时计时法与普通计时法：加上时间限制词(如凌晨、早晨、上午、下午、晚上等)，数字大于12的需要减掉12。由题意得，21时整，21-12=9，所以21时=晚上9时。晚上9时整，分针指向数字12，时针指向数字9。由图可知，时针和分针的夹角是直角。18时整，18-12=6，所以18时=晚上6时。晚上6时整，分针指向数字12，时针指向数字6，由图可知，时针和分针的夹角是平角。\n【详解】21时整，钟面上时针和分针的夹角是直角；18时整，时针和分针的夹角是平角。",
+        "knowledge_points":  [
+                                 "24时计时法与普通计时法的互化",
+                                 "平角，周角的认识及特征",
+                                 "角的分类及换算",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "25-26四年级上·安徽宿州·月考"
+    },
+    {
+        "id":  "55",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "比周角小200°的角是( )°,比150°大( )°的角是平角。",
+        "options":  [
+
+                    ],
+        "answer":  "160;30",
+        "solution":  "【分析】周角是360°，平角是180°，据此计算。\n【详解】360°-200°=160°；180°-150°=30°。所以比周角小200°的角是160°；比150°大30°的角是平角。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "25-26 四年级上·福建厦门·月考"
+    },
+    {
+        "id":  "56",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个角是由( )个顶点和( )条边组成的。三角板上的直角是( )度。",
+        "options":  [
+
+                    ],
+        "answer":  "1或一;2或两;90",
+        "solution":  "【分析】角的概念：从一点引出两条射线所形成的图形叫做角。由图可知，一个角是由1个顶点和2条边组成的。直角的度数为90度，所以三角板上的直角是90度。\n【详解】一个角是由1个顶点和2条边组成的。三角板上的直角是90度。",
+        "knowledge_points":  [
+                                 "角的概念及表示方法",
+                                 "角的分类及换算"
+                             ],
+        "source":  "25-26 四年级上·福建厦门·月考"
+    },
+    {
+        "id":  "57",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "钟面上经过15分钟，分针旋转的角度是( )度;钟面上经过半小时，分针旋转的角度是( )度。",
+        "options":  [
+
+                    ],
+        "answer":  "90;180",
+        "solution":  "【分析】根据题意，钟面一圈为360度，被平均分成12个大格，每个大格对应30度。因此可通过时间计算分针转动的角度。以此答题即可。\n【详解】360÷12=30(度)；15÷5=3(格)；30×3=90(度)；30÷5=6(格)；30×6=180(度)。钟面上经过15分钟，分针旋转的角度是90度；钟面上经过半小时，分针旋转的角度是180度。",
+        "knowledge_points":  [
+                                 "角的度量",
+                                 "一般时间与钟面指针的位置"
+                             ],
+        "source":  "25-26四年级上·福建厦门·月考"
+    },
+    {
+        "id":  "58",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "当钟表上显示时间为10点半时，时针与分针的较小夹角是( )角。当钟表上显示时间为( )时整，时针与分针的夹角是平角。15:00 时，时针与分针的指针相互( )。(填\"平行\"或\"垂直\")",
+        "options":  [
+
+                    ],
+        "answer":  "钝;6;垂直",
+        "solution":  "【分析】角的大小与开口的大小有关系，角的开口越小，角就越小，开口越大，角就越大。10点半时，时针指向10和11之间，分针指向6，时针与分针的较小夹角比直角大，是钝角。一条射线绕它的端点旋转，当始边和终边在同一条直线上，方向相反时，所构成的角叫平角。当6时整时，时针指向6，分针指向12，时针和分针在一条直线上时，时针与分针的夹角是平角。15:00时，是下午3时，时针指向3，分针指向12时，时针和分针所形成的较小夹角是直角，时针与分针的指针相互垂直。\n【详解】当钟表上显示时间为10点半时，时针与分针的较小夹角是(钝)角。当钟表上显示时间为(6)时整，时针与分针的夹角是平角。15:00时，时针与分针的指针相互(垂直)。",
+        "knowledge_points":  [
+                                 "24时计时法与普通计时法的互化",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征",
+                                 "半点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·河南郑州·期中"
+    },
+    {
+        "id":  "59",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "9时整，时针和分针组成的较小角是( )角,2时整，时针和分针组成的较小角是( )角。(填\"锐\"\"钝\"\"平\"\"直\"\"周\")",
+        "options":  [
+
+                    ],
+        "answer":  "直;锐",
+        "solution":  "【分析】因为钟面上12个数字，以表芯为顶点，表针转一圈是360°，被12个数字平均分成12份，每一份也就是两数之间夹角是30°。9时整，时针和分针之间有3个大格，角度为3×30=90°，是直角。2时整，时针和分针组成的较小角比直角小，是锐角。\n【详解】9时整，时针和分针组成的较小角是(直)角，2时整，时针和分针组成的较小角是(锐)角。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征"
+                             ],
+        "source":  "24-25四年级上·江西宜春·期中"
+    },
+    {
+        "id":  "60",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "乐乐用量角器量角，误把外圈刻度当成内圈刻度，读得的度数为65°,正确的度数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "115",
+        "solution":  "【分析】量角器同一刻度线内圈和外圈的两个度数和一定是180°，用180°减去误读的度数，即可求出正确的度数。\n【详解】根据分析，正确的度数加上错误的读数等于180°:180°-65°=115°，所以正确的度数是115°。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "角的度量"
+                             ],
+        "source":  "24-25四年级上·福建漳州·期中"
+    },
+    {
+        "id":  "61",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "钟面上 6 时整，时针和分针成( )角;4 时整，时针和分针的夹角是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "平;120",
+        "solution":  "【分析】6时整，分针指向数字12，时针指着数字6。由图可知，6时整，时针和分针成平角。4时整，分针指向数字12，时针指着数字4。由图可知，时针和分针之间有4个大格。钟面上每个大格对应的度数是30°，直接用30°乘4即可算出它们的夹角是多少度。\n【详解】30°×4=120°。钟面上6时整，时针和分针成平角；4时整，时针和分针的夹角是120°。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "平角，周角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·江西赣州·期中"
+    },
+    {
+        "id":  "62",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "∠1+25°是一个直角，则∠1是( )°,它是( )角;∠2+45°是一个平角，则∠2是( )°,它是( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "65;锐;135;钝",
+        "solution":  "【分析】根据直角的度数是90°，平角的度数是180°，分别用90°减去25°得到∠1的度数，用180°减去45°得到∠2的度数，再根据角的分类标准判断角的类型。锐角是小于90°的角；钝角是大于90°，小于180°的角；直角是等于90°的角；平角是等于180°的角；周角是等于360°的角；据此解答即可。\n【详解】直角是90°，所以∠1=90°-25°=65°。65°小于90°，因此∠1是锐角。平角是180°，所以∠2=180°-45°=135°。135°大于90°且小于180°，因此∠2是钝角。∠1+25°是一个直角，则∠1是(65)°，它是(锐)角；∠2+45°是一个平角，则∠2是(135)°，它是(钝)角。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·江西赣州·期中"
+    },
+    {
+        "id":  "63",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把一个圆形纸片对折3次，得到的角是( )°,它是一个( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "45;锐",
+        "solution":  "【分析】根据题意，圆形纸片对折一次，将圆心角360°平均分成2份，每份是180°；对折两次，将180°再平均分成2份，每份是90°；对折三次，将90°再平均分成2份，每份是45°。45°小于90°，属于锐角。以此答题即可。\n【详解】根据分析可知：对折1次:360°÷2=180°；对折2次:180°÷2=90°；对折3次:90°÷2=45°。把一个圆形纸片对折3次，得到的角是45°，它是一个锐角。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·湖南长沙·期中"
+    },
+    {
+        "id":  "64",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "90°的角是( )角，一个平角是( )度,360°的角是( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "直;180;周",
+        "solution":  "【分析】锐角的度数小于90°；大于90°而小于180°的角叫做钝角；90°的角是直角；180°的角是平角；360°的角是周角。\n【详解】根据分析:90°的角是直角，一个平角是180度，360°的角是周角。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·广东汕头·期中"
+    },
+    {
+        "id":  "65",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "把一张圆形的纸对折两次后，得到的 4 个角的和是( )度。",
+        "options":  [
+
+                    ],
+        "answer":  "360",
+        "solution":  "【分析】圆形纸片的周角为360度，每次对折都会将当前的角度平均分成两等份。第一次对折后得到两个180度的角，第二次对折后将每个180度的角再平均分成两个90度的角，共得到4个90度的角。无论对折多少次，所有分出的角的度数之和始终等于周角360度。\n【详解】根据分析，把一张圆形的纸对折两次后，得到的4个角的和是360度。",
+        "knowledge_points":  [
+                                 "图形的折叠问题",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·湖南衡阳·期中"
+    },
+    {
+        "id":  "66",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "3时整，时针和分针成( )角,6 时整，时针和分针成( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "直;平",
+        "solution":  "【分析】钟面一周为360°，钟表上有12个数字，分12大格，每相邻两个数字之间的夹角为30°，每个大格30°。3时整，时针指向3，分针指向12，时针走了3个大格，即3×30°=90°，90°的角是直角；6时整，时针走了6个大格，即6×30°=180°，180°的角是平角，据此解答即可。\n【详解】3时整，时针和分针成直角，6时整，时针和分针成平角。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25四年级上·辽宁鞍山·期中"
+    },
+    {
+        "id":  "67",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "火把节是彝族的传统盛大节日之一。2024年7月29日在凉山彝族火把节西昌主会场，群众跳起彝族达体舞。开始时间是晚上8:00，这时钟面上的时针和分针所形成的较小的夹角是( )°;晚上11:00结束，这时时针和分针所形成的较小的夹角是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "120;30",
+        "solution":  "【分析】钟面一周为360°，共分12大格，每大格为360°÷12=30°。计算整点时，时针和分针的夹角，只需看时针和分针之间间隔几个大格，用大格数乘30°即可得到夹角的度数。\n【详解】8:00时，时针在8，分针在12，相差4大格，夹角4×30°=120°(较小角)。11:00时，时针在11，分针在12，相差1大格，夹角1×30°=30°(较小角)。开始时间是晚上8:00，这时钟面上的时针和分针所形成的较小的夹角是120°；晚上11:00结束，这时时针和分针所形成的较小的夹角是30°。",
+        "knowledge_points":  [
+                                 "角的度量",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25四年级上·陕西西安·期中"
+    },
+    {
+        "id":  "68",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "( )角>平角>( )角>直角>锐角。",
+        "options":  [
+
+                    ],
+        "answer":  "周;钝",
+        "solution":  "【分析】根据角的分类，周角为360°，平角为180°，钝角大于90°且小于180°，直角为90°，锐角小于90°，据此解答。\n【详解】根据分析可知，按从大到小排列应为：周角>平角>钝角>直角>锐角。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "角的大小比较",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·海南省直辖县级单位·期中"
+    },
+    {
+        "id":  "69",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "( )角>平角>( )角>直角>锐角。",
+        "options":  [
+
+                    ],
+        "answer":  "周;钝",
+        "solution":  "【分析】锐角：小于90°的角；钝角：大于90°小于180°的角；直角：等于90°的角；平角：等于180°的角；周角：等于360°的角。\n【详解】从大到小的顺序为：周角>平角>钝角>直角>锐角。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "角的大小比较",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·海南省直辖县级单位·期中"
+    },
+    {
+        "id":  "70",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "北京时间 2024 年 8月12日，巴黎奥运会闭幕式主题为《记录》的闭幕式表演于北京时间 3 时开始，这时时针和分针所形成的角是( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "直",
+        "solution":  "【分析】钟面一圈为360°，被平均分成12个大格，那么一个大格的角度是360°÷12=30°；3时整，时针指向3，分针指向12，时针和分针之间间隔了3个大格；90°的角是直角。\n【详解】30°×3=90°。3时时针和分针所形成的角是直角。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·贵州黔南·期中"
+    },
+    {
+        "id":  "71",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "4 时整，时针和分针组成的角是( )°,这个角是一个( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "120;钝",
+        "solution":  "【分析】锐角是小于90度的角；钝角是大于90度，小于180度的角；直角是等于90度的角。因为钟面上有12个数字，表针转一圈是360度，被12个数字平均分成12份(12个大格)，每一份(每1个大格)也就是两数之间夹角是30度。4时整，时针指向数字4，分针指向数字12，时针和分针之间有4个大格，时针和分针所成的夹角是4×30度=120度，这个角是一个钝角。据此解答。\n【详解】4时整，时针和分针组成的角是120°，这个角是钝角。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·广东珠海·期中"
+    },
+    {
+        "id":  "72",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "∠1 与 25°的角组成一个直角,∠1=( )°;∠2 与 75°的角组成一个平角,∠2=( )°;1 个周角=( )个平角=( )个直角=( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "65;105;2;4;360",
+        "solution":  "【分析】根据直角，平角，周角的定义及角度计算求解。直角为90°，平角为180°，周角为360°，利用减法计算未知角，并确定周角与平角、直角的倍数关系。\n【详解】∠1=90°-25°=65°；∠2=180°-75°=105°；360°÷180°=2；360°÷90°=4。∠1 与25°的角组成一个直角，∠1=65°；∠2 与75°的角组成一个平角，∠2=105°；1 个周角=2 个平角=4 个直角=360°。",
+        "knowledge_points":  [
+                                 "平角，周角的认识及特征",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "角度的计算"
+                             ],
+        "source":  "24-25 四年级上·江西赣州·期中"
+    },
+    {
+        "id":  "73",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "北京时间 2024 年 4 月 25 日 20 时 59 分，搭载神舟十八号载人飞船的长征二号 F 遥十八运载火箭在酒泉卫星发射中心点火发射，约在 21 时 9 分，神舟十八号载人飞船与火箭成功分离。从 20 时 59 分到 21 时 9 分，经过了( )分钟，分针旋转了( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "10;60",
+        "solution":  "【分析】经过时间=结束时刻-开始时刻；钟面上有12个大格，每大格有5个小格，总共60个小格，每个小格360°÷60=6°。经过几分钟就是几乘6°，据此计算。\n【详解】21 时9 分-20 时59 分=10 分；360°÷60=6°；6°×10=60°。从20 时59 分到21 时9 分，经过了10 分钟，分针旋转了60°。",
+        "knowledge_points":  [
+                                 "角的度量",
+                                 "时，分，秒有关的计算"
+                             ],
+        "source":  "24-25 四年级上·江西赣州·期中"
+    },
+    {
+        "id":  "74",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在花样滑冰中，有一个动作叫做\"阿克塞尔\"。这个动作要求选手起跳后在空中旋转四周半，即旋转( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "1620",
+        "solution":  "【分析】根据题意，选手在空中旋转四周半，也就是四圈半，可分别算出4圈和半圈的度数，再相加算出总度数。\n【详解】四圈:360°×4=1440°；半圈:360°÷2=180°；1440°+180°=1620°。故旋转了1620°。",
+        "knowledge_points":  [
+                                 "两，三位数与一位数连续进位的乘法",
+                                 "两位数除以一位数，首位除不尽",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25四年级上·浙江·期中"
+    },
+    {
+        "id":  "75",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "1周角=( )平角=( )直角;一个锐角大于( )度，而小于( )度;度量角的大小可以使用( )。",
+        "options":  [
+
+                    ],
+        "answer":  "2;4;0;90;量角器",
+        "solution":  "【分析】小于90°的角叫锐角，大于90°而小于180°的角叫钝角，0°<锐角<90°，直角=90°，90°<钝角<180°，平角=180°，周角=360°，1 个平角=2 个直角，1 个周角=2 个平角=4 个直角。量角器可以度量角的大小。\n【详解】1周角=2 平角=4 直角；一个锐角大于0度，而小于90度；度量角的大小可以使用量角器。",
+        "knowledge_points":  [
+                                 "角的度量",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征",
+                                 "角的分类及换算"
+                             ],
+        "source":  "24-25四年级上·河北沧州·期中"
+    },
+    {
+        "id":  "76",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "3时整，时针和分针所组成的角是( )角。6时整，时针和分针所组成的角是( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "直;平",
+        "solution":  "【分析】钟面上有12个大格，每个大格对应的角度是30度。3时整时，时针和分针之间有3个大格，利用3×30°=90°，这是一个直角；6时整时，时针和分针之间有6个大格，就用6×30°=180°，这是一个平角。\n【详解】3时整，3×30°=90°，这是一个直角；6时整，6×30°=180°，这是一个平角。3时整，时针和分针所组成的角是直角；6时整，时针和分针所组成的角是平角。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25四年级上·湖南永州·期中"
+    },
+    {
+        "id":  "77",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "125°比平角小( )°;比43°大( )°是直角。",
+        "options":  [
+
+                    ],
+        "answer":  "55;47",
+        "solution":  "【分析】平角是180°，用180°减125°可算出125°比平角小多少；直角是90°，用90°减43°可算出比43°大多少度的是直角。\n【详解】180°-125°=55°；90°-43°=47°。所以125°比平角小55°；比43°大47°的是直角。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·湖北孝感·期中"
+    },
+    {
+        "id":  "78",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "3 时整，时针和分针构成的角是 度，这个角是一个 角,6 时整，时针和分针构成的角是 度，这个角是一个 角。",
+        "options":  [
+
+                    ],
+        "answer":  "90;直;180;平",
+        "solution":  "【分析】钟面上，时针与分针的夹角可根据整时时刻判断。3时整，分针指向12，时针指向3，中间有3大格，每格30°，所以是90°，为直角。6时整，分针指向12，时针指向6，中间有6大格，所以是180°，为平角。\n【详解】3时整:30°×3=90°，是直角。6时整:30°×6=180°，是平角。3 时整，时针和分针构成的角是90 度，这个角是一个直角；6 时整，时针和分针构成的角是180 度，这个角是一个平角。",
+        "knowledge_points":  [
+                                 "整点时间与钟面指针的位置",
+                                 "角的分类及换算",
+                                 "角的度量",
+                                 "角度的计算"
+                             ],
+        "source":  "18-19 四年级上·河北·期中"
+    },
+    {
+        "id":  "79",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "用一副三角尺可以画出 150°的角，也可以画出 15°的角。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】一副三角尺六个角，分别是30°、60°、90°和45°、45°、90°；选择其中的两个角度相加或相减，看能否拼出150°和15°的角；据此解答。\n【详解】60°+90°=150°，可以画出150°的角；60°-45°=15°，可以画出15°的角。所以，用一副三角尺可以画出150°的角，也可以画出15°的角。原题说法正确。答案为:√",
+        "knowledge_points":  [
+                                 "角度的计算"
+                             ],
+        "source":  "24-25 四年级上·天津南开·期末"
+    },
+    {
+        "id":  "80",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.94,
+        "question":  "角的两条边无限延长，角的大小不变。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】角的大小由两边叉开的大小决定，与边的长度无关。无限延长角的边只会增加边的长度，不会改变两边叉开的大小，因此角的大小不变。\n【详解】角的大小取决于两边叉开的程度，与边的长度无关。角的两条边无限延长，角的大小保持不变。故答案为:√",
+        "knowledge_points":  [
+                                 "角的初步认识及辨认",
+                                 "角的大小比较"
+                             ],
+        "source":  "19-20四年级上·全国·期中"
+    },
+    {
+        "id":  "81",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "平面内，过一点可以画无数条直线，过两点只能画一条直线。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】在平面内，若有一个点，以该点为固定点，直线可以向任意方向延伸，因此能画出无数条直线。若有两点，两点确定一条且仅有一条直线，因此只能画一条直线。\n【详解】由分析得，平面内，过一点可以画无数条直线，过两点只能画一条直线。原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·广东汕头·期中"
+    },
+    {
+        "id":  "82",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "9 时 30 分，钟面上的时针和分针形成直角。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】9时30分时，时针指向9和10的正中间，分针指向6。钟面一周为360°，共分12个大格，每个大格为360°÷12=30°，此时时针和分针之间有3个半大格子，3个大格是90°，多半格，所以是钝角，据此判断。\n【详解】9时30分，钟面上的时针和分针形成钝角，原题说法错误。故答案为: ×。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "一般时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·福建·期中"
+    },
+    {
+        "id":  "83",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "用一副三角板可以拼出一个 150°的角。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】一副三角板的角度有30°、45°、60°、90°。要拼出150°的角，可以用两个角相加，或者用一个角减去另一个角。\n【详解】取90°角和60°角拼在一起:90°+60°=150°，所以可以拼出150°的角。故答案为:√",
+        "knowledge_points":  [
+                                 "用三角尺画角",
+                                 "角度的计算"
+                             ],
+        "source":  "24-25四年级上·陕西·期中"
+    },
+    {
+        "id":  "84",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.94,
+        "question":  "用一个10倍的放大镜观察 15°角，这个角还是15°。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】角的大小由角两边张开的大小决定，与边的长短无关。放大镜只会放大边的长度，不会改变角两边张开的大小，因此角度不变。\n【详解】用10倍的放大镜观察15°的角，虽然角的两边被放大了，但角两边张开的大小没有改变，因此这个角仍然是15°，原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "角的概念及表示方法",
+                                 "角的初步认识及辨认"
+                             ],
+        "source":  "24-25四年级上·贵州黔西·期中"
+    },
+    {
+        "id":  "85",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "平角比直角大。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】根据角的分类，直角是90度，平角是180度。比较两者的度数即可判断大小关系。\n【详解】直角是90 度，平角是180 度。因为180 度大于90 度，所以平角比直角大，说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "角的大小比较",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·贵州黔西·期中"
+    },
+    {
+        "id":  "86",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "探照灯射出的光线可以看作是一条线段。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】线段有两个端点，长度有限；射线有一个端点，可向一端无限延伸；直线没有端点，可向两端无限延伸。探照灯的光线从光源出发，向一个方向无限延伸，符合射线的特征。\n【详解】探照灯射出的光线可以看作是一条射线。题干说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·广东汕头·期中"
+    },
+    {
+        "id":  "87",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "平角就是一条直线，周角就是一条射线。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】角是由一个顶点和两条边(射线)组成的图形。平角的两边在一条直线上，周角的两边重合，但都需有顶点。直线和射线本身没有顶点，不能直接称为角。\n【详解】平角是由一个顶点引出的两条方向相反的射线组成的角，度数为180°，虽然两边在一条直线上，但必须有顶点。周角是两边重合的角，度数为360°，同样需要顶点。直线和射线只是平角或周角的边，不能等同于角本身。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "2015·全国·小升初模拟"
+    },
+    {
+        "id":  "88",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "用 10 倍的放大镜看一个 5°的角，看到的角是50°。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】用放大镜观察角时，放大的是角的边的长度，而角的大小由两边张开的角度决定，与边的长度无关。因此，无论放大多少倍，角的度数不变。\n【详解】用10倍的放大镜看一个5°的角，放大后的角的两边长度被放大，但两边张开的角度未改变，因此角的度数仍为5°。题目中认为看到的角是50°。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "角的概念及表示方法"
+                             ],
+        "source":  "24-25 四年级上·湖南邵阳·期中"
+    },
+    {
+        "id":  "89",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "9 时 30 分，钟面上时针和分针所成的角是直角。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】根据题意，钟面1大格是30°。9时30分，分针指向6，时针位于9和10中间。计算两针之间的较小夹角：有3大格半，3个大格是90°，多半格，即3×30°+30°÷2=90°+15°=105°，大于90°，属于钝角，以此判断即可。\n【详解】3×30°+30°÷2=90°+15°=105°。9时30分，两针之间较小的夹角为105°，105°>90°，是钝角。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "一般时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·福建漳州·期中"
+    },
+    {
+        "id":  "90",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "一条射线长 5 米。( ) 理由:",
+        "options":  [
+
+                    ],
+        "answer":  "× 射线有一个端点，可以向另一端无限延伸，无法测量长度",
+        "solution":  "【分析】射线的定义：射线是由线段的一端无限延长所形成的直的线，射线仅有一个端点，另一端可以无限延伸。\n【详解】题目中\"一条射线长5米\"的说法是错误的。理由：射线有一个端点，可以向另一端无限延伸，无法测量其长度。故答案为: ×",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·海南省直辖县级单位·期中"
+    },
+    {
+        "id":  "91",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "只用三角尺也能画出 15°角。( ) 理由:",
+        "options":  [
+
+                    ],
+        "answer":  "√ 15°可以由三角尺的45°角和30°角相减得到",
+        "solution":  "【分析】我们知道三角尺有两种，一种是等腰直角三角尺，它的两个锐角都是45°；另一种三角尺的锐角分别是30°和60°。15°可以通过这两种三角尺的45°角和30°角相减得到。\n【详解】45°-30°=15°。用三角尺的45°角和30°角组合：先画一个45°的角，再在内部用30°角的一边对齐，剩余部分即为15°的角。因此，只用三角尺能画出15°角。故答案为:√",
+        "knowledge_points":  [
+                                 "用三角尺画角"
+                             ],
+        "source":  "24-25四年级上·江西赣州·期中"
+    },
+    {
+        "id":  "92",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.94,
+        "question":  "用10倍的放大镜看一个 5°的角，这个角就变成了50°。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】角的大小由角两边张开的大小决定，与边的长短无关。放大镜只会放大边的长度，不会改变角两边张开的程度，因此角的度数不变。\n【详解】用放大镜观察角时，角两边的长度被放大，但角两边张开的程度未改变，因此角的度数仍为5°。原题说法错误；故答案为: ×",
+        "knowledge_points":  [
+                                 "角的度量",
+                                 "角的初步认识及辨认",
+                                 "角的大小比较"
+                             ],
+        "source":  "24-25 四年级上·江西赣州·期中"
+    },
+    {
+        "id":  "93",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "小彤画了一条长 14厘米的射线。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】根据射线的定义，射线只有一个端点，另一端可以无限延伸，因此无法测量其长度。题目中提到的\"14厘米的射线\"与射线的特性矛盾。\n【详解】射线是有一个端点，另一端无限延伸的图形，其长度无法度量。题目中说\"画了一条长14厘米的射线\"，这与射线的定义不符，原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·广东广州·期中"
+    },
+    {
+        "id":  "94",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "可以用一副三角尺画出 100°,75°和 135°的角。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】一副三角尺，一个三角尺的角有30°、60°、90°，等腰直角三角尺的角有45°、45°、90°，用它们进行拼组，看是否能得出这几个角度即可。\n【详解】45°+90°=135°，用一副三角尺可以画出135°的角；30°+45°=75°，可以画出75°的角；无论怎么拼组，一副三角尺拼不出100°的角。可以用一副三角尺画出75°和135°的角，画不出100°的角。故答案为: ×",
+        "knowledge_points":  [
+                                 "用三角尺画角"
+                             ],
+        "source":  "24-25 四年级上·河北石家庄·期中"
+    },
+    {
+        "id":  "95",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.94,
+        "question":  "小王老师在黑板上画了一条 8 厘米长的射线。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】射线有一个端点，另一端可以无限延伸，无法测量长度。题目中描述画了8厘米长的射线，与射线的定义矛盾。\n【详解】根据射线的定义，射线只有一个端点，另一端无限延长，无法测量其长度。因此，画一条8厘米长的射线是不可能的。故答案为: ×",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25四年级上·广东汕头·期中"
+    },
+    {
+        "id":  "96",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "射线比直线短，线段最短。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】根据线段，直线，射线的特征，直线和射线都是无限长的，无法比较长短，而线段是有限长的。因此，\"射线比直线短\"的说法错误，\"线段最短\"的说法正确。\n【详解】直线没有端点，可以向两端无限延伸，无法测量长度；射线有一个端点，只能向一端无限延伸，同样无法测量长度。因此，射线和直线都是无限长的，无法比较长短。线段有两个端点，长度有限，是三者中最短的。题目中\"射线比直线短\"错误，\"线段最短\"正确，但整体命题错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "两点间线段最短与两点间的距离",
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·湖北黄冈·期中"
+    },
+    {
+        "id":  "97",
+        "unit":  "第三章角的度数是",
+        "question_type":  "作图题",
+        "difficulty":  0.65,
+        "question":  "从健康角度考虑，座椅靠背与座椅座面的夹角是 115°时最舒适。请你画出一个 115°角。",
+        "options":  [
+
+                    ],
+        "answer":  "见详解",
+        "solution":  "【分析】画115°角的步骤：画一条射线：先画一条射线，作为角的一条边，确定角的顶点为 O。量角器对齐：将量角器的中心与顶点O重合，同时让量角器的0°刻度线和这条射线重合。标记角度：在量角器的115°刻度的位置，用铅笔点一个点。画出另一条边：从顶点O出发，通过刚才标记的点画一条射线，这样就得到了一个115°的角。\n【详解】如下图所示画出115°角:\n![](images/fbd4b7ebf11e0abb1a71b04f7a0313fbfaef02fe09610f5673f1fea21f65b383.jpg)",
+        "knowledge_points":  [
+                                 "用量角器画角"
+                             ],
+        "source":  "25-26四年级上·河北唐山·期末"
+    },
+    {
+        "id":  "98",
+        "unit":  "第三章角的度数是",
+        "question_type":  "作图题",
+        "difficulty":  0.85,
+        "question":  "请画出一个 80° 的角，并标出角的度数。",
+        "options":  [
+
+                    ],
+        "answer":  "见详解",
+        "solution":  "【分析】用量角器画一个80°的角，画角时，先从一点画一条射线，使量角器的中心和射线的端点重合，零刻度线和射线重合，在量角器80°的地方点一个点，然后以量角器的中心为端点，通过刚刚画的点，再画一条射线，这两条射线所夹的角就是我们所要画的角。\n【详解】如下图所示画出80°角并标出度数:\n![](images/e87cac97b521819ee931620557dc56488bc78e43445372bae8324aad0b65b5c8.jpg)",
+        "knowledge_points":  [
+                                 "用量角器画角"
+                             ],
+        "source":  "24-25四年级上·安徽阜阳·期中"
+    },
+    {
+        "id":  "99",
+        "unit":  "第三章角的度数是",
+        "question_type":  "作图题",
+        "difficulty":  0.65,
+        "question":  "以点 O 为顶点，画出一个比平角小75°的角。",
+        "options":  [
+
+                    ],
+        "answer":  "见详解",
+        "solution":  "【分析】平角等于180°，用180°-75°=105°。先从O点画条射线，使量角器的中心和射线的端点重合，零刻度线和射线重合，在量角器105°的地方点一个点，然后以量角器的中心为端点，通过刚刚画的点，再画一条射线，这两条射线所夹的角就是我们所要画的角。\n【详解】180°-75°=105°。如下图所示画出105°角:\n![](images/b5ff1d790741ff84d661a068c7aa8c9be77b4525ff40d73e4d5f27ee33bedf13.jpg)",
+        "knowledge_points":  [
+                                 "平角，周角的认识及特征",
+                                 "角的分类及换算",
+                                 "用量角器画角"
+                             ],
+        "source":  "24-25 四年级上·浙江·期中"
+    },
+    {
+        "id":  "100",
+        "unit":  "第三章角的度数是",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "小丽用两个角拼成了一个 275°的角，其中一个角是125°,另一个角是多少度。",
+        "options":  [
+
+                    ],
+        "answer":  "150°",
+        "solution":  "【分析】已知用两个角拼成了一个275°的角，其中一个角是125°，则用275°减去125°，即可求得另一个角是多少度。\n【详解】275°-125°=150°。答：另一个角是150°。",
+        "knowledge_points":  [
+                                 "角度的计算"
+                             ],
+        "source":  "24-25 四年级上·海南省直辖县级单位·期中"
+    },
+    {
+        "id":  "101",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "《新闻联播》在晚上7时播出，此时时针和分针所成的较小的角是().\n\n![](images/e23f5958c7e556e3e38d7739f2a6031b4cf2232ae80117c33169bee02efd0eaf.jpg)",
+        "options":  [
+                        "直角",
+                        "锐角",
+                        "钝角"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】钟面上每个大格是30°，晚上7时整，时针和分针之间的较小角有5个大格，因此用30°乘5即可。钝角>直角>锐角，据此解题。\n【详解】30°×5=150°。《新闻联播》在晚上7时播出，此时时针和分针所成的较小的角是钝角。故答案为:C",
+        "knowledge_points":  [
+                                 "角度的计算，直角，钝角，锐角的认识及特征，整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25四年级上·湖北省直辖县级单位·期中"
+    },
+    {
+        "id":  "102",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "关于\"角\"，下面说法中正确的是( )。",
+        "options":  [
+                        "小于90°的角是锐角，大于90°的角就是钝角",
+                        "两个锐角的和一定比直角大",
+                        "把一个平角分成两个角，不可能分成两个钝角"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】锐角是大于0°小于90°的角；直角=90°；钝角是大于90°小于180°的角；平角=180°。\n【详解】A. 锐角应是大于0°小于90°，钝角是大于90°小于180°，说法错误；B. 两个锐角相加不一定超过90°，如15°+15°=30°，说法错误；C. 因为两个钝角的和一定大于180°，所以一个平角不可能分成两个钝角，说法正确。故答案为:C",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征，平角，周角的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·河北邢台·期中"
+    },
+    {
+        "id":  "103",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.94,
+        "question":  "黑板上直角的大小()三角尺上的直角的大小。",
+        "options":  [
+                        "大于",
+                        "小于",
+                        "等于"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】无论画在什么地方，只要是直角，度数都是90°。\n【详解】黑板上直角的大小等于三角尺上直角的大小，都是90°。故答案为:C",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征，角的大小比较"
+                             ],
+        "source":  "24-25 四年级上·河北邢台·期中"
+    },
+    {
+        "id":  "104",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "用两把不同的三角尺可以拼出( )的角。",
+        "options":  [
+                        "60°和90°",
+                        "150°和105°",
+                        "75°和125°"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】两把不同的三角尺的角度为：一块是90°、45°、45°，另一块是90°、60°、30°。用它们可以拼出15°、30°、45°、60°、75°、105°、120°、135°、150°等角。\n【详解】A. 可以得到60°=90°-30°，但不能用一副三角尺拼出新的90°，不符合题意；B. 90°+60°=150°，45°+60°=105°，两角都能拼出，符合题意；C. 75°=45°+30°可以拼出，但125°无法由这些角度的和或差得到，不符合题意。故答案为:B",
+        "knowledge_points":  [
+                                 "用三角尺画角"
+                             ],
+        "source":  "24-25 四年级上·山东临沂·期中"
+    },
+    {
+        "id":  "105",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面说法中，错误的是( )。",
+        "options":  [
+                        "一条直线长6千米",
+                        "小于90°的角是锐角",
+                        "钟面上6时整，分针和时针成平角",
+                        "向阳小学的占地面积约是3公顷"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】只有线段才有确切长度，直线和射线的长度都无法度量；小于90°的角是锐角；钟面上6时整分针指12，时针指6，两针成平角；学校占地用\"公顷\"作单位比较合理。\n【详解】A. 直线没有端点，长度无限，不能说\"一条直线长6千米\"，错误；B. 小于90°的角是锐角，正确；C. 6时整时针和分针成180°的角，是平角，正确；D. 学校占地面积一般以\"公顷\"计，3公顷合理，正确。故答案为:A",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征，直角，钝角，锐角的认识及特征，平角，周角的认识及特征，公顷，平方千米的认识"
+                             ],
+        "source":  "23-24 四年级上·山东菏泽·期中"
+    },
+    {
+        "id":  "106",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面的角中，能用一副三角尺画出的是()。",
+        "options":  [
+                        "10°",
+                        "15°",
+                        "25°"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】一副三角尺包含30°、60°、45°、90°四种角。通过加减可拼出15°(60°-45°)、75°(30°+45°)、105°、120°、135°、150°等角。\n【详解】A. 10°不能由30°、45°、60°、90°的和或差得到；B. 60°-45°=15°，可以拼出15°角；C. 25°也不能由这些角组合得到。故答案为:B",
+        "knowledge_points":  [
+                                 "用三角尺画角"
+                             ],
+        "source":  "25-26 四年级上·甘肃天水·月考"
+    },
+    {
+        "id":  "107",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面的几种说法中，哪个是正确的。( )",
+        "options":  [
+                        "没有最大的自然数",
+                        "93540900 中的两个\"9\"表示的意义相同",
+                        "平角就是一条直线"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】自然数从0开始，向右无限延伸，没有最大的自然数；同一个数中不同数位上的相同数字意义不同；平角是由两条成一直线的射线组成的角。\n【详解】A. 自然数的个数无限，没有最大的自然数，正确；B. 93540900中左边的\"9\"在千万位，右边的\"9\"在百位，意义不同，错误；C. 平角是两条射线组成的角，不是单独的一条直线，错误。故答案为:A",
+        "knowledge_points":  [
+                                 "整数的数级，数位和计数单位的认识，自然数的认识，平角，周角的认识及特征"
+                             ],
+        "source":  "23-24 四年级上·山东济宁·期中"
+    },
+    {
+        "id":  "108",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列说法中，正确的有( )个.\n\n① 手电筒发出的光可以近似地看作一条射线.\n② 边长是400米的正方形，面积是16公顷.\n③ 两个因数末尾共有几个0，积的末尾至少有几个0.\n④ 角的两边无限延长，角的大小不变。",
+        "options":  [
+                        "2",
+                        "3",
+                        "4"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】① 射线有一个端点，向一端无限延伸，手电筒光线符合这一特征；② 正方形面积=边长×边长=400×400=160000平方米=16公顷；③ 因数末尾共有几个0，积的末尾至少有几个0；④ 角的大小只与两边张开的程度有关，与边长无关，延长两边角度不变。\n【详解】四条说法均正确，所以正确的有4个。故答案为:C",
+        "knowledge_points":  [
+                                 "两位数乘整十数的口算乘法，线段，直线，射线的认识及特征，角的大小比较，公顷，平方千米的进率与换算"
+                             ],
+        "source":  "23-24 四年级上·山东临沂·期中"
+    },
+    {
+        "id":  "109",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "时针旋转一圈(12小时)与分针形成( )次直角。",
+        "options":  [
+                        "1",
+                        "22",
+                        "12"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】12小时内，时针转1圈，分针转12圈，分针比时针多转11圈。分针每比时针多转1圈，两针就形成2次直角。\n【详解】11×2=22，所以时针旋转一圈(12小时)与分针形成22次直角。故答案为:B",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征，时，分的认识"
+                             ],
+        "source":  "23-24四年级上·山东临沂·期中"
+    },
+    {
+        "id":  "110",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "用一副三角尺不能拼出( )的角。",
+        "options":  [
+                        "75°",
+                        "20°",
+                        "135°"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】一副三角板的角度有30°、60°、45°、90°。可以通过加减得到:75°=45°+30°，135°=90°+45°等。\n【详解】A. 75°=45°+30°，可以拼出；B. 20°不能由30°、45°、60°、90°的加减得到；C. 135°=90°+45°，可以拼出。所以不能拼出的角是20°。故答案为:B",
+        "knowledge_points":  [
+                                 "用三角尺画角"
+                             ],
+        "source":  "23-24 四年级上·陕西渭南·期中"
+    },
+    {
+        "id":  "111",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.94,
+        "question":  "用 4 个同样大小的角正好拼成一个周角，这个角的度数是( )。",
+        "options":  [
+                        "120°",
+                        "90°",
+                        "30°"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】周角的度数是360°，4个一样大的角拼成周角，每个角的度数为360°÷4。\n【详解】360°÷4=90°，所以这个角的度数是90°。故答案为:B",
+        "knowledge_points":  [
+                                 "平角，周角的认识及特征，角的分类及换算"
+                             ],
+        "source":  "24-25 四年级上·陕西延安·期中"
+    },
+    {
+        "id":  "112",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "笑笑要用三角尺画角，下面不能用三角尺画出的是( )。",
+        "options":  [
+                        "75°",
+                        "120°",
+                        "160°"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】一副三角尺的角度为90°、45°、45°和90°、30°、60°。\n【详解】A. 45°+30°=75°，可以画出75°；B. 30°+90°=120°，可以画出120°；C. 160°不能由30°、45°、60°、90°的加减得到，因此不能画出。故答案为:C",
+        "knowledge_points":  [
+                                 "用三角尺画角"
+                             ],
+        "source":  "24-25 四年级上·吉林白城·期中"
+    },
+    {
+        "id":  "113",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列说法不正确的是( ).\n\n① 万位，十万位，百万位，亿位，十亿位都是计数单位.\n② 要使 45785≈45□万,□里只能填 1,2,3,4 这 4 个数.\n③ 每相邻两个面积单位间的进率都是 100.\n④ 用一副三角尺不可以拼出145°的角。",
+        "options":  [
+                        "①②③",
+                        "①③④",
+                        "②③④",
+                        "①③"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】① 万位，十万位，百万位，亿位，十亿位是数位，不是计数单位；② 四舍五入到\"万位\"，千位小于5时都舍去，所以□可以填0、1、2、3、4；③ 面积单位之间的进率不都是100，如1公顷=10000平方米；④ 由三角尺角度组合无法得到145°。\n【详解】① 说法错误；② 说只能填1、2、3、4，忽略0，错误；③ 相邻面积单位进率不一定是100，错误；④ 三角尺可拼出的角中没有145°，说法正确。故不正确的是①②③，对应选项A。",
+        "knowledge_points":  [
+                                 "整数的数级，数位和计数单位的认识，用三角尺画角，公顷，平方千米的进率与换算，亿以内数的近似数"
+                             ],
+        "source":  "24-25 四年级上·福建莆田·期中"
+    },
+    {
+        "id":  "114",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "把一个30°的角放在放大镜下观察，看到的角是( )。",
+        "options":  [
+                        "20°",
+                        "30°",
+                        "50°"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】放大镜只改变图形的大小(边的长度)，不会改变角两边张开的程度，所以角度不变。\n【详解】把一个30°的角放在放大镜下观察，看到的角仍是30°。故答案为:B",
+        "knowledge_points":  [
+                                 "角的度量，角的大小比较"
+                             ],
+        "source":  "24-25 四年级上·陕西商洛·期中"
+    },
+    {
+        "id":  "115",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "2024 年霜降时间为 10 月 23 日 06:14:32，此时钟面上时针与分针所成的较小角是( )。",
+        "options":  [
+                        "锐角",
+                        "直角",
+                        "钝角",
+                        "平角"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】按角的大小分类:0°<锐角<90°，直角=90°，90°<钝角<180°，平角=180°。从图示时间位置看，此时两针夹角大于90°小于180°。\n【详解】根据钟面示意图，此时钟面上时针与分针所成的较小角是钝角。故答案为:C",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征，平角，周角的认识及特征，一般时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·福建厦门·期中"
+    },
+    {
+        "id":  "116",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "用 4 个同样大小的角正好拼成一个周角，这个角的度数是( )。",
+        "options":  [
+                        "120°",
+                        "90°",
+                        "30°"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】周角为360°，被4个相同的角平均分。\n【详解】360°÷4=90°，所以每个角是90°。故答案为:B",
+        "knowledge_points":  [
+                                 "角度的计算，平角，周角的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·陕西西安·期中"
+    },
+    {
+        "id":  "117",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "2020 年 11 月 24 日4时30分，长征五号遥五运载火箭点火升空。此时钟面上时针和分针组成的角是( )。",
+        "options":  [
+                        "锐角",
+                        "直角",
+                        "钝角",
+                        "平角"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】钟面被12个数字平均分成12大格，每格30°。4时30分时，分针指向6，时针在4和5之间，较小夹角小于90°，为锐角。\n【详解】从图示可知，此时钟面上时针和分针组成的角是锐角。故答案为:A",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征，一般时间与钟面指针的位置"
+                             ],
+        "source":  "25-26 四年级上·广东·期中"
+    },
+    {
+        "id":  "118",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "钟面上( )时整，时针与分针成60°的角。",
+        "options":  [
+                        "1",
+                        "2",
+                        "3"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】相邻两个数字间夹角是30°。整点时，分针指12，时针指对应数字。1时整夹角30°，2时整夹角60°，3时整夹角90°。\n【详解】2时整，2个大格，角度2×30°=60°。故答案为:B",
+        "knowledge_points":  [
+                                 "角的度量，整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25四年级上·新疆吐鲁番·期中"
+    },
+    {
+        "id":  "119",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "如果∠1 是一个平角,∠2 是一个钝角，那么∠1-∠2 是一个( )。",
+        "options":  [
+                        "锐角",
+                        "直角",
+                        "钝角",
+                        "无法确定"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】平角=180°，钝角在90°到180°之间。用一个180°的角减去一个大于90°小于180°的角，结果在0°到90°之间，为锐角。\n【详解】A. 锐角是大于0°小于90°的角，符合；B、C不符合；D可以确定类型。故答案为:A",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征，平角，周角的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·陕西安康·期中"
+    },
+    {
+        "id":  "120",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "度量一个角，角的一条边对着量角器上\"180\"的刻度，另一条边对着\"120\"的刻度，这个角是( )度。",
+        "options":  [
+                        "60",
+                        "90",
+                        "120",
+                        "180"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】量角器上两个刻度对应的角度之差即为该角的度数。\n【详解】180-120=60(度)，所以这个角的度数是60度。故答案为:A",
+        "knowledge_points":  [
+                                 "角的度量"
+                             ],
+        "source":  "24-25 四年级上·浙江嘉兴·期中"
+    },
+    {
+        "id":  "121",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.94,
+        "question":  "长方形中有4个()。",
+        "options":  [
+                        "锐角",
+                        "直角",
+                        "钝角"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】长方形的四个角都等于90°。\n【详解】长方形中有4个直角。故答案为:B",
+        "knowledge_points":  [
+                                 "长方形的概念及特点，直角，钝角，锐角的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·重庆黔江·期中"
+    },
+    {
+        "id":  "122",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.94,
+        "question":  "用一个放大10倍的放大镜看一个10°的角，看到的角的度数是( )。",
+        "options":  [
+                        "100°",
+                        "10°",
+                        "20°",
+                        "1°"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】角的大小只与两边张开的程度有关，与边的长度无关。放大镜只放大边长，不改变角度。\n【详解】用放大10倍的放大镜看10°的角，角的度数仍然是10°。故答案为:B",
+        "knowledge_points":  [
+                                 "角的度量，角的概念及表示方法"
+                             ],
+        "source":  "24-25四年级上·福建莆田·期中"
+    },
+    {
+        "id":  "123",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "用一副三角板可以拼出下面( )的角。",
+        "options":  [
+                        "55°",
+                        "75°",
+                        "106°"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】一副三角板的基本角度为30°、45°、60°、90°，通过加减可得到75°、105°、120°、135°、150°、15°等角。\n【详解】75°=30°+45°可以拼出；55°和106°都不能由30°、45°、60°、90°的加减得到。故答案为:B",
+        "knowledge_points":  [
+                                 "用三角尺画角"
+                             ],
+        "source":  "24-25 四年级上·新疆喀什·期中"
+    },
+    {
+        "id":  "124",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "用一副三角板不可画出的角是( )。",
+        "options":  [
+                        "15°",
+                        "50°",
+                        "75°",
+                        "105°"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】一副三角尺有30°、60°、90°和45°的角。\n【详解】15°=45°-30°可以画出；75°=30°+45°可以画出；105°=45°+60°可以画出；50°不能由30°、45°、60°、90°的加减得到，因此不可画出。故答案为:B",
+        "knowledge_points":  [
+                                 "用三角尺画角"
+                             ],
+        "source":  "24-25 四年级上·广东东莞·期中"
+    },
+    {
+        "id":  "125",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面语句正确的是( )。",
+        "options":  [
+                        "两个锐角之和一定是钝角",
+                        "小红在纸上画了一条10cm的直线",
+                        "用放大3倍的放大镜看30°的角是90°",
+                        "一个平角等于两个直角"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】A:两个锐角之和可能仍小于90°；B:直线无限长，不能有具体长度；C:放大镜不改变角度；D:直角90°，平角180°。\n【详解】A. 如10°+20°=30°<90°，不一定是钝角，错误；B. 应为\"10cm的线段\"，错误；C. 放大镜不会改变角度，错误；D. 90°+90°=180°，一个平角等于两个直角，正确。故答案为:D",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征，直角，钝角，锐角的认识及特征，平角，周角的认识及特征，角的分类及换算"
+                             ],
+        "source":  "24-25 六年级下·河北邢台·期末"
+    },
+    {
+        "id":  "126",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面说法正确的是( )。",
+        "options":  [
+                        "线段没有端点",
+                        "射线只有一个端点",
+                        "直线有两个端点"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】线段有两个端点，直线没有端点，射线只有一个端点。\n【详解】A. 线段有两个端点，错误；B. 射线只有一个端点，正确；C. 直线没有端点，错误。故答案为:B",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "2025 四年级上·海南海口·专题练习"
+    },
+    {
+        "id":  "127",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "在一条射线上截取8厘米长的线段，可以截取( )条。",
+        "options":  [
+                        "1",
+                        "10",
+                        "无数"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】射线一端为端点，另一端无限延伸，可在上面不断截取长度为8厘米的线段。\n【详解】射线无限长，因此在一条射线上可以截取无数条8厘米长的线段。故答案为:C",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "25-26四年级上·全国·课后作业"
+    },
+    {
+        "id":  "128",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "海海在黑板上和作业本上各画了一个80°的角。这两个角相比,( )。",
+        "options":  [
+                        "黑板上的角大",
+                        "作业本上的角大",
+                        "一样大"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】角的大小只与两边张开的程度有关，与画在什么地方，边长多长无关。\n【详解】两个角都是80°，大小相等。故答案为:C",
+        "knowledge_points":  [
+                                 "角的大小比较"
+                             ],
+        "source":  "25-26 四年级上·全国·课后作业"
+    },
+    {
+        "id":  "129",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "从5时15分到5时45分这段时间里，钟表的分针旋转( )。",
+        "options":  [
+                        "30°",
+                        "120°",
+                        "180°",
+                        "360°"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】分针一圈60分钟，对应360°，每分钟转6°。从5:15到5:45经过30分钟。\n【详解】每分钟转6°，30×6°=180°，所以分针旋转了180°。故答案为:C",
+        "knowledge_points":  [
+                                 "角的度量，一般时间与钟面指针的位置"
+                             ],
+        "source":  "24-25五年级下·天津津南·期末"
+    },
+    {
+        "id":  "130",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "时针从2时走到5时,( )时针旋转了( )。",
+        "options":  [
+                        "顺,90°",
+                        "顺,120°",
+                        "逆,90°",
+                        "逆,120°"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】时针转动方向为顺时针。一圈360°分成12大格，每大格30°。从2到5走了3大格。\n【详解】3×30°=90°，方向为顺时针，所以是\"顺,90°\"。故答案为:A",
+        "knowledge_points":  [
+                                 "旋转与旋转现象，角的度量，整点时间与钟面指针的位置"
+                             ],
+        "source":  "23-24 五年级下·安徽黄山·期末"
+    },
+    {
+        "id":  "131",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一副三角尺，最大的角是( )°,最小的角是( )°,一个最大的角和一个最小的角可以拼成一个( )角。(填\"锐\"\"直\"或\"钝\")",
+        "options":  [
+
+                    ],
+        "answer":  "90;30;钝",
+        "solution":  "【分析】一副三角板中的各个角的度数分别是30°、60°、45°、90°。最大的是90°，最小的是30°。90°+30°=120°，大于90°小于180°，为钝角。\n【详解】一副三角尺，最大的角是90°，最小的角是30°，一个最大的角和一个最小的角可以拼成一个钝角。",
+        "knowledge_points":  [
+                                 "角度的计算，直角，钝角，锐角的认识及特征"
+                             ],
+        "source":  "24-25四年级上·河北邢台·期中"
+    },
+    {
+        "id":  "132",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "如果∠1+162°=平角，那么∠1=( )°,是个( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "18;锐",
+        "solution":  "【分析】平角=180°，所以∠1=180°-162°。18°小于90°，是锐角。\n【详解】180°-162°=18°。如果∠1+162°=平角，那么∠1=18°，是个锐角。",
+        "knowledge_points":  [
+                                 "角度的计算，直角，钝角，锐角的认识及特征，平角，周角的认识及特征"
+                             ],
+        "source":  "24-25四年级上·河北邢台·期中"
+    },
+    {
+        "id":  "133",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把一条40 厘米长的( )向两端无限延长，就得到一条( ),向一端无限延长，就得到一条( )。",
+        "options":  [
+
+                    ],
+        "answer":  "线段;直线;射线",
+        "solution":  "【分析】线段有两个端点，长度有限；直线没有端点，向两端无限延长；射线有一个端点，向一端无限延长。\n【详解】把一条40厘米长的线段向两端无限延长，就得到一条直线；向一端无限延长，就得到一条射线。",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·湖北省直辖县级单位·期中"
+    },
+    {
+        "id":  "134",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "上午 9 时半，钟面上时针与分针所成的较小角是( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "钝",
+        "solution":  "【分析】钟面平均分为12大格，每大格30°。9时半时，时针在9和10之间，分针在6，较小角度大于3×30°且小于4×30°，即大于90°小于120°，所以是钝角。\n【详解】3×30°=90°；4×30°=120°，因此上午9时半时针与分针的较小夹角在90°与120°之间，是钝角。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "半点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·浙江嘉兴·期中"
+    },
+    {
+        "id":  "135",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "下午 2:30 开始考试,60 分钟结束，这段时间钟面上的分针转动( )度，时针转动了( )度。",
+        "options":  [
+
+                    ],
+        "answer":  "360;30",
+        "solution":  "【分析】分针60分钟走一圈，为360°；钟面有12大格，每大格30°，时针60分钟走一大格，即30°。\n【详解】分针：一圈为360°，60分钟正好一圈，转动360°；时针:360°÷12=30°，60分钟走一大格，转动30°。所以分针转动360°，时针转动30°。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "平角，周角的认识及特征",
+                                 "时，分的认识"
+                             ],
+        "source":  "24-25 四年级上·内蒙古通辽·期中"
+    },
+    {
+        "id":  "136",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "过一点可以画( )条直线，不在同一条直线上的三个点，过其中任意两点一共可以画( )条直线。",
+        "options":  [
+
+                    ],
+        "answer":  "无数;3",
+        "solution":  "【分析】平面内过一点可以向任意方向引直线，因此可以画无数条直线；不在同一直线上的三个点，两点确定一条直线，共有3种点对组合，因此可以画3条直线。\n【详解】过一点可以画无数条直线；不在同一条直线上的三个点，任意两点连成一条直线，一共可以画3条直线。",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·内蒙古通辽·期中"
+    },
+    {
+        "id":  "137",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个周角的度数是( )°;一个直角的一半是( )°;两个直角可以拼成( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "360;45;平",
+        "solution":  "【分析】周角=360°；直角=90°，一半为45°；两个直角和为180°，即平角。\n【详解】一个周角的度数是360°；一个直角的一半是45°；两个直角可以拼成平角。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·内蒙古通辽·期中"
+    },
+    {
+        "id":  "138",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个直角三角形中有一个锐角是35°,另一个锐角是( )°;一个钝角是105°,它比平角小( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "55;75",
+        "solution":  "【分析】直角三角形中两个锐角和为90°，所以另一锐角为90°-35°；平角=180°，平角减去105°即可得到它比平角小多少度。\n【详解】90°-35°=55°；180°-105°=75°。所以另一个锐角是55°，该钝角比平角小75°。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "25-26 四年级上·甘肃天水·月考"
+    },
+    {
+        "id":  "139",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\">\"\"<\"或\"=\".\n\n87620000( )870620000 1 公顷( )10000 平方米 钝角( )直角 1 平方千米( )100000 平方米 100000000( )1 千万",
+        "options":  [
+
+                    ],
+        "answer":  "<;=;>;>;>",
+        "solution":  "【分析】(1)比较多位数大小，先看位数，再从高位依次比较；(2)1公顷=10000平方米；(3)钝角>90°，直角=90°；(4)1平方千米=100公顷=1000000平方米>100000平方米；(5)100000000=1亿>一千万。\n【详解】87620000<870620000；1 公顷=10000 平方米；钝角>直角；1 平方千米>100000 平方米；100000000>1 千万。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "角的分类及换算",
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "23-24四年级上·山东临沂·期中"
+    },
+    {
+        "id":  "140",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "上午8时整，钟面上时针和分针组成的较小角是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "120",
+        "solution":  "【分析】钟面一圈360°，分成12大格，每大格30°。8时整时，时针指8，分针指12，相差4大格。\n【详解】30°×4=120°，所以较小角是120°。",
+        "knowledge_points":  [
+                                 "平角，周角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25四年级上·山东临沂·期中"
+    },
+    {
+        "id":  "141",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "37°角比直角小( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "53",
+        "solution":  "【分析】直角=90°，用90°减去37°。\n【详解】90°-37°=53°，所以37°角比直角小53°。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·山东菏泽·期中"
+    },
+    {
+        "id":  "142",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "工夫茶是广东潮汕地区汉族人民传统的饮茶风俗。每 3 克茶叶需要浸泡五分钟，这段时间分针转动了( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "30",
+        "solution":  "【分析】分针60分钟走一圈360°，每分钟6°。5分钟转动的角度为5×6°。\n【详解】360°÷60=6°；6°×5=30°。所以这段时间分针转动了30°。",
+        "knowledge_points":  [
+                                 "角度的计算"
+                             ],
+        "source":  "23-24 四年级上·山东菏泽·期中"
+    },
+    {
+        "id":  "143",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "晒青是茶叶制作过程中的一个重要环节，太阳射出来的光线，在数学中就可以看作是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "射线",
+        "solution":  "【分析】光线从光源出发，向一个方向延伸，有一个端点，向一端无限延长，符合射线的特征。\n【详解】太阳射出来的光线，在数学中可以看作是射线。",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "23-24 四年级上·山东菏泽·期中"
+    },
+    {
+        "id":  "144",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "3时30分时，钟面上时针和分针组成的较小的角是( );6时整，钟面上时针和分针组成的角是( );9时整，钟面上时针和分针组成的较小的角是( );5时整，钟面上时针和分针组成的较小的角是( )。(填\"锐角\"\"直角\"\"钝角\"或\"平角\")",
+        "options":  [
+
+                    ],
+        "answer":  "锐角;平角;直角;钝角",
+        "solution":  "【分析】3:30 分针在6，时针在3和4之间，较小角小于90°，为锐角；6:00 时针在6，分针在12，成180°平角；9:00 时针在9，分针在12，成90°直角；5:00 时针在5，分针在12，形成大于90°小于180°的钝角。\n【详解】3时30分时为锐角；6时整为平角；9时整为直角；5时整为钝角。",
+        "knowledge_points":  [
+                                 "平角，周角的认识及特征",
+                                 "角的分类及换算",
+                                 "整点时间与钟面指针的位置",
+                                 "半点时间与钟面指针的位置"
+                             ],
+        "source":  "23-24 四年级上·山东临沂·期中"
+    },
+    {
+        "id":  "145",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "2024年初，小区物业购进了一批太阳能路灯，路灯发出的光线可以看作是一条条的( )。",
+        "options":  [
+
+                    ],
+        "answer":  "射线",
+        "solution":  "【分析】路灯光线从灯头出发，向一个方向传播，有一个端点，向一端无限延伸，可看作射线。\n【详解】路灯发出的光线可以看作是一条条的射线。",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "23-24四年级上·山东菏泽·期中"
+    },
+    {
+        "id":  "146",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "钟面上 5 时整，分针和时针形成的最小角的度数是( ),是( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "150°;钝",
+        "solution":  "【分析】钟面12大格，每大格30°。5时整分针在12，时针在5，相隔5大格，夹角5×30°。大于90°小于180°是钝角。\n【详解】30°×5=150°，90°<150°<180°，所以是钝角。5时整分针和时针形成的最小角度为150°，是钝角。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "23-24 四年级上·陕西渭南·期中"
+    },
+    {
+        "id":  "147",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "1 时整，时针和分针组成的角是( )度，这个角是一个( )角;5 时整，时针和分针组成的角是( )度，这个角是一个( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "30;锐;150;钝",
+        "solution":  "【分析】每大格30°。1时整时，两针相差1大格，角度30°，小于90°为锐角；5时整相差5大格，角度150°，大于90°小于180°为钝角。\n【详解】1时整时针和分针的夹角是30°，是锐角；5时整夹角150°，是钝角。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "23-24 四年级上·陕西安康·期中"
+    },
+    {
+        "id":  "148",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "把一个平角分成两个角，其中一个是钝角，另外一个一定是( )角;长方形四个角的和正好是一个( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "锐;周",
+        "solution":  "【分析】平角=180°，若其中一个角是大于90°小于180°的钝角，则另一个角小于90°为锐角；长方形4个直角和为4×90°=360°，为周角。\n【详解】平角分成两个角，其中一个为钝角，另一个一定是锐角；长方形四个角的和是360°，是一个周角。",
+        "knowledge_points":  [
+                                 "长方形的概念及特点",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征",
+                                 "角的分类及换算"
+                             ],
+        "source":  "24-25四年级上·陕西西安·期中"
+    },
+    {
+        "id":  "149",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "5时整，时针与分针的最小夹角是( )°;时针与分针的夹角是一个平角，是( )时整;3:30 时针和分针成( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "150;6;锐",
+        "solution":  "【分析】5时整两针相差5大格，5×30°=150°；平角180°需相差6大格，对应6时整；3:30时分针在6，时针在3和4之间，较小角小于90°，为锐角。\n【详解】5时整最小夹角150°；平角对应6时整；3:30时，时针和分针成锐角。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25四年级上·四川绵阳·期中"
+    },
+    {
+        "id":  "150",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "5 平方千米=( )平方米 30000 平方米=( )公顷 6000 公顷=( )平方千米 1 周角=( )平角=( )直角",
+        "options":  [
+
+                    ],
+        "answer":  "5000000;3;60;2;4",
+        "solution":  "【分析】1平方千米=1000000平方米；1公顷=10000平方米；1平方千米=100公顷；1周角=360°，1平角=180°，1直角=90°。\n【详解】5平方千米=5×1000000=5000000 平方米；30000 平方米=3 公顷；6000 公顷=6000÷100=60 平方千米；1 周角=2 平角=4 直角。",
+        "knowledge_points":  [
+                                 "角的分类及换算",
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "24-25 四年级上·陕西安康·期中"
+    },
+    {
+        "id":  "151",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "3 时整，时针和分针所形成的较小角是( )°,是( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "90;直",
+        "solution":  "【分析】3时整分针在12，时针在3，相差3大格，3×30°=90°，为直角。\n【详解】3时整两针夹角为90°，是直角。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·陕西渭南·期中"
+    },
+    {
+        "id":  "152",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "把半圆沿中心平均分成 180 份，每一份所对应的角的度数是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "1",
+        "solution":  "【分析】半圆对应180°，平均分成180份，每份1°。\n【详解】180°÷180=1°，所以每份是1°。",
+        "knowledge_points":  [
+                                 "角的度量",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·浙江嘉兴·期中"
+    },
+    {
+        "id":  "153",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上\">\"\"<\"或\"=\".\n\n1 个平角( )2 个直角 25×40( )750×4 9999999( )1 亿 3640000( )32500000",
+        "options":  [
+
+                    ],
+        "answer":  "=;<;<;<",
+        "solution":  "【分析】平角180°，2个直角也是180°；25×40=1000，750×4=3000；9999999<100000000；3640000<32500000。\n【详解】1 个平角=2 个直角；25×40<750×4；9999999<1 亿；3640000<32500000。",
+        "knowledge_points":  [
+                                 "亿以上数的改写",
+                                 "平角，周角的认识及特征",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "大数的比较"
+                             ],
+        "source":  "24-25 四年级上·河北唐山·期中"
+    },
+    {
+        "id":  "154",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在时钟上时针与分针成90°的时刻是( )和( );时针和分针形成平角的时刻是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "3 时;9 时;6 时",
+        "solution":  "【分析】钟面每大格30°，成90°时，两针相差3大格。分针在12时，时针在3或9；平角180°时，相差6大格，对应6时整。\n【详解】成直角的时刻是3时和9时；成平角的时刻是6时。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25四年级上·河南信阳·期中"
+    },
+    {
+        "id":  "155",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "钟面上 3 时整，时针和分针成一个( )。6 时整，时针和分针组成一个( )。(选填\"锐角\"\"直角\"\"钝角\"或\"平角\")",
+        "options":  [
+
+                    ],
+        "answer":  "直角;平角",
+        "solution":  "【分析】3时整两针夹角90°，是直角；6时整夹角180°，是平角。\n【详解】3时整成直角，6时整成平角。",
+        "knowledge_points":  [
+                                 "平角，周角的认识及特征",
+                                 "角的分类及换算",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·陕西安康·期中"
+    },
+    {
+        "id":  "156",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "( )时整，时针和分针成平角;2时整，分针和时针成( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "6;锐",
+        "solution":  "【分析】平角180°对应6时整；2时整两针夹角为2×30°=60°，小于90°，是锐角。\n【详解】6时整时针和分针成平角；2时整分针和时针成锐角。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25四年级上·陕西商洛·期中"
+    },
+    {
+        "id":  "157",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "河马是世界上嘴巴最大的陆生哺乳动物之一，成年河马把嘴巴张开到最大时角度可接近180°!这是一个( )角，这个角等于( )个直角。",
+        "options":  [
+
+                    ],
+        "answer":  "平;2或两",
+        "solution":  "【分析】180°是平角；180°=2×90°，相当于2个直角。\n【详解】该角是平角，等于2个直角。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征",
+                                 "角的分类及换算"
+                             ],
+        "source":  "24-25四年级上·广西河池·期中"
+    },
+    {
+        "id":  "158",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "大雁结群南飞过冬，飞行过程中一会排成\"人\"字，一会排成\"一\"字，\"人\"字形大雁群近似为110°的角，这个\"人\"字形角是一个( )角，它和一个( )°的( )角能拼成一个\"一\"字形角。",
+        "options":  [
+
+                    ],
+        "answer":  "钝;70;锐",
+        "solution":  "【分析】110°大于90°小于180°，是钝角；平角180°，180°-110°=70°，70°<90°为锐角，两者拼成平角\"一\"字形。\n【详解】\"人\"字形角为钝角；180°-110°=70°，70°为锐角，所以它和一个70°的锐角能拼成\"一\"字形角(平角)。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "25-26 四年级上·广东·期中"
+    },
+    {
+        "id":  "159",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "周角里有( )个直角，直角度数的 2 倍是( )度。",
+        "options":  [
+
+                    ],
+        "answer":  "4;180",
+        "solution":  "【分析】周角360°，直角90°，360°÷90°=4；直角度数的2倍为2×90°。\n【详解】周角里有4个直角；直角的2倍是180°。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·江西南昌·期中"
+    },
+    {
+        "id":  "160",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "用直尺( )量出直线的长度;用一副三角尺( )画出75°的角。(填\"能\"或\"不能\")",
+        "options":  [
+
+                    ],
+        "answer":  "不能;能",
+        "solution":  "【分析】直线没有端点，无限长，不能量长度；一副三角尺包含30°和45°角，30°+45°=75°，可拼出75°角。\n【详解】用直尺不能量出直线的长度；用一副三角尺能画出75°的角。",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征",
+                                 "用三角尺画角"
+                             ],
+        "source":  "24-25 四年级上·重庆黔江·期中"
+    },
+    {
+        "id":  "161",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "钟面上的分针从12起，转动了30分，转过的角是( )角，度数是( )度;当钟面上的指针指向3时整时，时针和分针所夹的较小角是( )度，是( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "平;180;90;直",
+        "solution":  "【分析】30分钟等于分针走半圈，为180°平角；3时整两针相差3大格，3×30°=90°，为直角。\n【详解】分针从12起转30分钟，转过平角，度数180°；3时整时，两针夹角为90°，是直角。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25四年级上·重庆黔江·期中"
+    },
+    {
+        "id":  "162",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "青菜播种需要湿润的土壤环境。李伯伯早上7:30在田间播种青菜，此时时针与分针所成的较小的夹角是( )角，播种完的时间是 8:30，那么钟面上的分针旋转了( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "锐;360",
+        "solution":  "【分析】7:30时针在7和8之间，分针在6，经计算较小夹角为45°，小于90°，是锐角；7:30到8:30为60分钟，分针转一圈360°。\n【详解】7:30时两针的较小夹角为45°，是锐角；7:30到8:30分针旋转了360°。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "一般时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·山西长治·期中"
+    },
+    {
+        "id":  "163",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.4,
+        "question":  "钟面上 3:00，时针和分针所成的角是( )角,12:30 时针和分针所成的角是( )角,( )时，时针和分针所成的角是平角。",
+        "options":  [
+
+                    ],
+        "answer":  "直;钝;6",
+        "solution":  "【分析】3:00时两针相差3大格，90°直角；12:30时分针在6，时针在12和1之间，夹角大于90°小于180°，为钝角；平角180°对应6时整。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征",
+                                 "一般时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·贵州遵义·期中"
+    },
+    {
+        "id":  "164",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "1 个周角=( )°;1 个平角=( )°;1 个直角=( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "360;180;90",
+        "solution":  "【分析】按角的分类：周角=360°，平角=180°，直角=90°。\n【详解】1个周角=360°；1个平角=180°；1个直角=90°。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·新疆喀什·期中"
+    },
+    {
+        "id":  "165",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "1 周角=( )直角,2 平角=( )周角。",
+        "options":  [
+
+                    ],
+        "answer":  "4;1",
+        "solution":  "【分析】周角360°，直角90°，360°÷90°=4；平角180°，2个平角和为360°，等于1个周角。\n【详解】1 周角=4 个直角，2 平角=1 个周角。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征",
+                                 "角的分类及换算"
+                             ],
+        "source":  "24-25 四年级上·新疆吐鲁番·期中"
+    },
+    {
+        "id":  "166",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "用一个 10 倍的放大镜观察一个 12 度的角，看到的角是( )度。",
+        "options":  [
+
+                    ],
+        "answer":  "12",
+        "solution":  "【分析】放大镜只改变图形大小，不改变角两边张开的程度，角度不变。\n【详解】用10倍放大镜观察一个12度的角，看到的角仍然是12度。",
+        "knowledge_points":  [
+                                 "角的度量"
+                             ],
+        "source":  "24-25 四年级下·湖南永州·期末"
+    },
+    {
+        "id":  "167",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "钟面上3时整，时针和分针构成的最小的角是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "90",
+        "solution":  "【分析】3时整两针相差3大格，每大格30°。\n【详解】30°×3=90°，所以最小角是90°。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "25-26四年级上·全国·单元测试"
+    },
+    {
+        "id":  "168",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "1 平角=( )直角 1 周角=( )度 2 公顷=( )平方米 7 平方千米=( )公顷",
+        "options":  [
+
+                    ],
+        "answer":  "2;360;20000;700",
+        "solution":  "【分析】平角=180°，直角=90°，180°=2×90°；周角=360°；1公顷=10000平方米；1平方千米=100公顷。\n【详解】1 平角=2 直角；1 周角=360 度；2 公顷=2×10000=20000 平方米；7 平方千米=7×100=700 公顷。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "2025 四年级上·海南海口·专题练习"
+    },
+    {
+        "id":  "169",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "1 周角=( )直角 1 公顷=( )平方米 3 平方千米=( )公顷",
+        "options":  [
+
+                    ],
+        "answer":  "4;10000;300",
+        "solution":  "【分析】1周角=360°，1直角=90°，360°÷90°=4；1公顷=10000平方米；1平方千米=100公顷。\n【详解】1 周角=4 直角；1 公顷=10000 平方米；3 平方千米=3×100=300 公顷。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "2025 四年级上·海南海口·专题练习"
+    },
+    {
+        "id":  "170",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "把圆平均分成360份，其中1份的大小就是( )度，记作( );60份的大小是( )度，记作( )。",
+        "options":  [
+
+                    ],
+        "answer":  "1;1°;60;60°",
+        "solution":  "【分析】把一个圆平均分成360份，每一份就是1度，记作1°；60份就是60个1度，因此是60°，记作60°。\n【详解】如图所示:\n![](images/9cdf0271f43f7129e063cfacc6d97bd6f2a135484d626d12d3f6fb1b61e07a99.jpg)\n把圆平均分成360份，其中1份的大小就是1度，记作1°；60份的大小是60度，记作60°。",
+        "knowledge_points":  [
+                                 "角的度量"
+                             ],
+        "source":  "25-26四年级上·全国·单元测试"
+    },
+    {
+        "id":  "171",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "从 6:00 到 12:00，钟面上时针转动了( )°;从 4:00 到 4:20，钟面上分针转动了( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "180;120",
+        "solution":  "【分析】钟面一圈360°，12个大格，每格30°。6:00到12:00，时针从6走到12，共6大格；4:00到4:20，经过20分钟，分针每5分钟走1大格，共4大格。\n【详解】6:00到12:00:6×30°=180°，时针转动180°；4:00到4:20:20÷5=4(格)，4×30°=120°，分针转动120°。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级下·河南新乡·期末"
+    },
+    {
+        "id":  "172",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "乐乐在制作研学行程表时，将出发时间8:30在钟表模型上画出，分针与时针的夹角是( )度。",
+        "options":  [
+
+                    ],
+        "answer":  "75",
+        "solution":  "【分析】钟面一圈360°分12大格，每格30°。8:30时，分针在6，时针在8和9之间。可以看作从6到8是2大格，再加上时针多走的半格(15°).\n【详解】360÷12=30(度)。(8-6)×30+15=2×30+15=60+15=75(度)。所以分针与时针的夹角是75度。",
+        "knowledge_points":  [
+                                 "角的度量",
+                                 "一般时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 六年级下·山东济宁·期末"
+    },
+    {
+        "id":  "173",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "比直角小的角是( ),比直角大且比平角小的角是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "锐角;钝角",
+        "solution":  "【分析】按角的大小分类:0°<锐角<90°，直角=90°，90°<钝角<180°，平角=180°。\n【详解】比直角小的角是锐角，比直角大且比平角小的角是钝角。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25 四年级下·云南红河·期末"
+    },
+    {
+        "id":  "174",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "从 3:55 到 4:15，经过了( )分钟，分针转动了( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "20;120",
+        "solution":  "【分析】4:15-3:55=20分钟。分针5分钟走1大格，每大格30°，20分钟走4大格。\n【详解】4:15-3:55=20(分钟)；20÷5=4(格)；30°×4=120°。从3:55到4:15，经过20分钟，分针转动了120°。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "时，分，秒有关的计算"
+                             ],
+        "source":  "24-25 三年级下·山东济宁·期末"
+    },
+    {
+        "id":  "175",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把下面画一个75°的角的步骤补充完整.\n(1)先画一条( ),使量角器的( )和射线的( )重合,( )和射线重合.\n(2)在量角器( )°刻度线的地方点一个点.\n(3)以画出的射线的端点为端点，通过刚画的点，再画一条( )。",
+        "options":  [
+
+                    ],
+        "answer":  "射线;中心;端点;0°刻度线;75;射线",
+        "solution":  "【分析】用量角器画角的一般步骤：画射线→对齐中心和0°刻度线→在指定度数处点点→连成另一条射线。\n【详解】(1)先画一条射线，使量角器的中心和射线的端点重合，0°刻度线和射线重合。\n(2)在量角器75°刻度线的地方点一个点。\n(3)以画出的射线的端点为端点，通过刚画的点，再画一条射线。",
+        "knowledge_points":  [
+                                 "用量角器画角"
+                             ],
+        "source":  "25-26四年级上·全国·课后作业"
+    },
+    {
+        "id":  "176",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "钟面上分针从 6 走到 12 形成的角是( )°,分针从 9 走到 12 形成的角是( )°,分针从 2 走到 4 形成的角是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "180;90;60",
+        "solution":  "【分析】每大格30°。从6到12走6大格；9到12走3大格；2到4走2大格。\n【详解】(12-6)×30°=6×30°=180°；(12-9)×30°=3×30°=90°；(4-2)×30°=2×30°=60°。",
+        "knowledge_points":  [
+                                 "角度的计算"
+                             ],
+        "source":  "24-25 四年级上·安徽淮南·期末"
+    },
+    {
+        "id":  "177",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "两个角正好拼成一个平角，如果一个角是锐角，那么另一个角一定是钝角。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】平角=180°。其中一个角是锐角，小于90°，平角减去锐角得到的另一个角大于90°小于180°，必为钝角。\n【详解】设锐角α，0°<α<90°，另一个角为180°-α，则90°<180°-α<180°，是钝角。原命题正确，故答案为:√。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "25-26 四年级上·江西萍乡·期中"
+    },
+    {
+        "id":  "178",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "红领巾上有一个直角，两个锐角。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】红领巾的形状实际上有一个钝角和两个锐角，而不是一个直角和两个锐角。\n【详解】红领巾上是一个钝角和两个锐角，所以\"一个直角，两个锐角\"的说法错误。故答案为: ×。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征"
+                             ],
+        "source":  "24-25四年级上·内蒙古通辽·期中"
+    },
+    {
+        "id":  "179",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "量角时，角的一边与量角器内圈零刻度线重合，读角的度数时，对照量角器外圈或内圈的刻度读数都可以。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】当角的一边对齐内圈0刻度线时，必须读内圈刻度；若读外圈刻度，会得到错误的度数。\n【详解】例如某角一边与内圈0°重合，另一边指向内圈50°，外圈130°，正确度数应读内圈50°，若读外圈130°则错误。因此不能任意读内外圈刻度。原命题错误，故答案为: ×。",
+        "knowledge_points":  [
+                                 "角的度量"
+                             ],
+        "source":  "23-24 四年级上·山东济宁·期中"
+    },
+    {
+        "id":  "180",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "黑板上 40 度的角与练习本上 40 度的角一样大。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】角的大小只由两边张开的程度决定，与画在什么地方，边长多少无关。\n【详解】黑板上和练习本上的角度都为40°，两边张开的程度相同，角大小相等。故答案为:√。",
+        "knowledge_points":  [
+                                 "角的大小比较"
+                             ],
+        "source":  "24-25 四年级上·陕西渭南·期中"
+    },
+    {
+        "id":  "181",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "∠1+∠2=平角，如果∠1 是钝角，那么∠2 是锐角。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】平角=180°，钝角大于90°小于180°。若∠1为钝角，则180°-∠1小于90°大于0°，为锐角。\n【详解】∠1+∠2=180°，且90°<∠1<180°，所以0°<180°-∠1<90°，即∠2为锐角。命题正确，故答案为:√。",
+        "knowledge_points":  [
+                                 "角的分类及换算",
+                                 "平角，周角的认识及特征",
+                                 "直角，钝角，锐角的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·重庆黔江·期中"
+    },
+    {
+        "id":  "182",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.94,
+        "question":  "角的两条边越长，量角度时，角的度数就越大。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】角的大小只与两边张开的程度有关，与边的长度无关，延长边不会改变角度。\n【详解】即使两条边无限延长，只要张开的程度不变，角的度数不变。因此\"边越长，角度越大\"的说法错误。故答案为: ×。",
+        "knowledge_points":  [
+                                 "角的大小比较",
+                                 "角的度量"
+                             ],
+        "source":  "15-16 四年级上·安徽六安·月考"
+    },
+    {
+        "id":  "183",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "射线 AB 长 5 厘米。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】射线只有一个端点，向另一端无限延长，长度无法度量，有长度的是线段。\n【详解】\"长5厘米\"应描述的是线段AB，而非射线AB。原命题错误，故答案为: ×。",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·陕西安康·期中"
+    },
+    {
+        "id":  "184",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "平角比钝角大，但比周角小。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】钝角在90°到180°之间，平角=180°，周角=360°，所以周角>平角>钝角。\n【详解】平角180°大于任何钝角(在90°到180°之间)，小于周角360°，命题正确。故答案为:√。",
+        "knowledge_points":  [
+                                 "角的大小比较",
+                                 "角的分类及换算"
+                             ],
+        "source":  "24-25 四年级上·吉林白城·期中"
+    },
+    {
+        "id":  "185",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "一条射线只有一个端点，一条线段没有端点。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】射线有一个端点，直线没有端点，线段有两个端点。\n【详解】\"射线只有一个端点\"正确，但\"线段没有端点\"错误，线段有两个端点，所以整体命题错误。故答案为: ×。",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·河北唐山·期中"
+    },
+    {
+        "id":  "186",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "角的两条边延长，角的大小不变，但用放大镜看角，角的度数变大了。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】无论延长边还是用放大镜观察，都不会改变两边张开的程度，角度不变。\n【详解】放大镜只改变图形大小，不改变角两边张开的大小，角的度数不会变大。命题错误，故答案为: ×。",
+        "knowledge_points":  [
+                                 "角的初步认识及辨认"
+                             ],
+        "source":  "24-25 四年级上·新疆喀什·期中"
+    },
+    {
+        "id":  "187",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "小丽画了一条 6 分米长的直线。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】直线没有端点，长度无限；有具体长度的应是线段。\n【详解】小丽画的是6分米长的线段，而不是直线。原命题错误，故答案为: ×。",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "25-26 四年级上·湖南永州·月考"
+    },
+    {
+        "id":  "188",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "聪聪用一副三角尺拼出了很多的角，他不可能拼出 95°的角。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】一副三角尺的角有30°、45°、60°、90°，通过加减可得到15°、75°、105°、120°、135°、150°等，但得不到95°。\n【详解】所有由30°、45°、60°、90°相加或相减得到的角度中不包括95°，因此不可能拼出95°角。命题正确，故答案为:√。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "用三角尺画角"
+                             ],
+        "source":  "24-25 四年级下·四川乐山·期末"
+    },
+    {
+        "id":  "189",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "105°的角是钝角。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】钝角是大于90°小于180°的角，105°符合这一范围。\n【详解】因90°<105°<180°，所以105°是钝角。命题正确，故答案为:√。",
+        "knowledge_points":  [
+                                 "角的度量",
+                                 "直角，钝角，锐角的认识及特征"
+                             ],
+        "source":  "24-25 六年级下·贵州安顺·期末"
+    },
+    {
+        "id":  "190",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "用一个放大 5 倍的放大镜看10°的角，此时角度是50°。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】放大镜只改变边长，不改变角两边张开的程度，角度不变。\n【详解】用5倍放大镜看10°的角，角的度数仍为10°。题目认为是50°，说法错误，故答案为: ×。",
+        "knowledge_points":  [
+                                 "角的度量",
+                                 "角的概念及表示方法"
+                             ],
+        "source":  "24-25 四年级下·江西吉安·期末"
+    },
+    {
+        "id":  "191",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "一条射线长 20 m。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】射线只有一个端点，另一端无限延伸，长度不可度量。\n【详解】\"长20m\"应描述线段而非射线。原命题错误，故答案为: ×。",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25 六年级下·重庆渝北·期末"
+    },
+    {
+        "id":  "192",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "三个角组成一个平角，这三个角一定都是锐角。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平角180°可以由钝角+锐角+锐角，或直角+锐角+锐角等多种情况组成，不一定全是锐角。给出反例即可。\n【详解】例如120°+20°+40°=180°(钝角+锐角+锐角)，或90°+60°+30°=180°(直角+锐角+锐角)，说明三个角组成的平角不一定都是锐角。命题错误，故答案为: ×。",
+        "knowledge_points":  [
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "25-26 四年级上·全国·随堂练习"
+    },
+    {
+        "id":  "193",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "一条射线就是一个周角，一条直线就是一个平角。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】角须由一个顶点和两条射线组成；周角是两边重合的角，平角是两边成一条直线的角。射线和直线本身不是角。\n【详解】周角和平角都必须有顶点和两条边(射线)，而单独的一条射线或一条直线没有\"两边\"这一结构，不能等同于角。命题错误，故答案为: ×。",
+        "knowledge_points":  [
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "25-26 四年级上·全国·随堂练习"
+    },
+    {
+        "id":  "194",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "12:30，时针和分针形成的角是180°。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】12:30时分针在6，对应180°；时针已从12走向1，走了30分钟，对应15°，所以两针间夹角为180°-15°=165°，不是180°。\n【详解】分针180°，时针偏离12为15°，两针夹角165°，非180°，命题错误，故答案为: ×。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "角的度量",
+                                 "半点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 六年级下·湖北随州·期末"
+    },
+    {
+        "id":  "195",
+        "unit":  "第三章角的度数是",
+        "question_type":  "改错题",
+        "difficulty":  0.85,
+        "question":  "用 10 倍的放大镜看一个50°的角，角的度数变成500°。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】角的大小由两边张开的大小决定，与边长无关。放大镜只会放大边的长度，不会改变角两边张开的程度，因此角的度数不变。\n【详解】\"角的度数变成500°\"说法错误。改正理由是：放大镜只会放大边的长度，不会改变角两边张开的程度，因此角的度数仍为50°。",
+        "knowledge_points":  [
+                                 "角的概念及表示方法",
+                                 "角的大小比较"
+                             ],
+        "source":  "24-25 四年级上·广西河池·期中"
+    },
+    {
+        "id":  "196",
+        "unit":  "第三章角的度数是",
+        "question_type":  "作图题",
+        "difficulty":  0.65,
+        "question":  "画两个角，使这两个角相差60°。",
+        "options":  [
+
+                    ],
+        "answer":  "见详解",
+        "solution":  "【分析】要画两个相差60°的角，可以先任选一个角的度数，如30°，则另一个角为30°+60°=90°，或取任意一对差为60°的角度。画法：用量角器分别画出这两个角即可。\n【详解】示例：先画一个30°角，再画一个90°角，它们相差60°。如图所示:\n![](images/9fe3c35684acd74ee2675f74dd430f026ad965364e6b0e0b3bc43c0ddd95bf8e.jpg)\n![](images/56c73960d76126addd298eb0632fb30ecab916c5383c6da7de0f5b29ff67ebbe.jpg)\n(答案不唯一)",
+        "knowledge_points":  [
+                                 "用量角器画角"
+                             ],
+        "source":  "24-25 四年级上·河北邢台·期中"
+    },
+    {
+        "id":  "197",
+        "unit":  "第三章角的度数是",
+        "question_type":  "作图题",
+        "difficulty":  0.65,
+        "question":  "选择合适的方法画出度数为60°,115°和160°的角。",
+        "options":  [
+
+                    ],
+        "answer":  "见详解",
+        "solution":  "【分析】可以用量角器画，也可以用三角尺拼角。用量角器画角的一般步骤：画一条射线→对齐中心和0°刻度→在指定刻度处点点→连成另一条射线。\n【详解】示例画法:\n- 画60°角：用量角器在60°刻度处点点，再连成射线；- 画115°角：同理，在115°刻度处点点，连成射线；- 画160°角：在160°刻度处点点，连成射线。\n如图:\n![](images/89f867c2b99fb049c72a3ca4937c4a9a482e8728b9c0b92a38186faee1266015.jpg)\n(画法不唯一)",
+        "knowledge_points":  [
+                                 "用量角器画角"
+                             ],
+        "source":  "24-25 四年级上·四川绵阳·期中"
+    },
+    {
+        "id":  "198",
+        "unit":  "第三章角的度数是",
+        "question_type":  "作图题",
+        "difficulty":  0.85,
+        "question":  "分别画出60°,135°,75°的角。",
+        "options":  [
+
+                    ],
+        "answer":  "见详解",
+        "solution":  "【分析】用量角器画角：先画射线，使量角器中心与端点重合，0°刻度线与射线重合，在对应的角度刻度处点点，再连成射线。\n【详解】步骤同理，分别在60°、135°、75°刻度处点点，再画射线，即得到三个角。示意图:\n![](images/c5bbb0f7081bc9122e7b3992b53de16f84a89a72a52a4f84d924c36e8521c04a.jpg)",
+        "knowledge_points":  [
+                                 "用量角器画角"
+                             ],
+        "source":  "24-25 四年级上·新疆克孜勒苏·期中"
+    },
+    {
+        "id":  "199",
+        "unit":  "第三章角的度数是",
+        "question_type":  "作图题",
+        "difficulty":  0.65,
+        "question":  "画线段，射线各一条。",
+        "options":  [
+
+                    ],
+        "answer":  "见详解",
+        "solution":  "【分析】线段有两个端点，有限长；射线有一个端点，向一端无限延长。按定义画出即可。\n【详解】画图示例：一条有两个端点的线段和一条从某点出发向一端延伸的射线，如图:\n![](images/1f5eb06c73fb5d3d516211cf258137fc2e9c7d9c0e8a6d1a8a78a8d8422e98a7.jpg)\n(答案不唯一)",
+        "knowledge_points":  [
+                                 "用直尺画线段",
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·新疆喀什·期中"
+    },
+    {
+        "id":  "200",
+        "unit":  "第三章角的度数是",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "体育课上，体育老师喊口令\"向左转\"\"向右转\"\"向后转\"，完成这些口令的动作时，我们分别在原地转了多少度。(画草图说明)",
+        "options":  [
+
+                    ],
+        "answer":  "向左转90°;向右转90°;向后转180°",
+        "solution":  "【分析】可以画出人面向上(如正上方)时的示意图，再按口令依次旋转。向左转相当于左转90°，向右转相当于右转90°，向后转相当于转到反方向，为180°。\n【详解】如图所示:\n![](images/d8a49fb4502326e893a35f9c683e99bc43e0aaccaa9c6a33db80eb5adcd1087e.jpg)\n![](images/e548d8537c3ebd00878dccc4630ff66460db4d6aa73a9e816f1a72b473e78af0.jpg)\n![](images/ca75ffa8e116c6e5ccca31e2d320d05940670754b4d0b882ff8350de1eab3a39.jpg)\n答：向左转转了90°，向右转转了90°，向后转转了180°。",
+        "knowledge_points":  [
+                                 "旋转与旋转现象",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25四年级上·海南省直辖县级单位·期中"
+    },
+    {
+        "id":  "201",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "将一张圆形纸片对折3次后展开，任意两条折痕所夹的角不可能是( )。",
+        "options":  [
+                        "45°",
+                        "60°",
+                        "90°"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】将圆形纸片对折1次，圆心角360°变为180°；对折2次，变为90°；对折3次，变为45°。因此，对折3次后展开，圆心角被分成8份，每份为45°，折痕形成的角度均为45°的整数倍。\n【详解】360°÷8=45°。\nA. 45°÷45°=1，可能是45°；B. 60°÷45°=1……15°，不能是60°；C. 90°÷45°=2，可能是90°。故不可能是60°，答案为:B",
+        "knowledge_points":  [
+                                 "图形的折叠问题",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "25-26 四年级上·全国·课后作业"
+    },
+    {
+        "id":  "202",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "用3倍的放大镜看60°的角，看到的角的度数是( )。",
+        "options":  [
+                        "20°",
+                        "60°",
+                        "180°"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】角的大小与两边的长短无关，只与边叉开的大小有关。放大镜只会改变边的长度，不改变两边叉开的大小。\n【详解】用3倍的放大镜看60°的角，只是角的两边长度增大了，角的大小没变，因此看到的角的度数还是60°。故答案为:B",
+        "knowledge_points":  [
+                                 "角的度量",
+                                 "角的初步认识及辨认"
+                             ],
+        "source":  "23-24 四年级下·云南文山·期末"
+    },
+    {
+        "id":  "203",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "钟面上17:55，时针和分针组成的较大角接近于( )。",
+        "options":  [
+                        "锐角",
+                        "平角",
+                        "钝角"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】钟面圆周360°，12个数字，将圆分成12份，每份30°。17点即5点多，时针在5和6之间，角度在150°和180°之间；55分时分针在11上，角度为11×30°=330°。分针与时针之间的较大角在150°和180°之间，与330°的差接近180°。\n【详解】时针大致在150°~180°之间，分针在330°，二者差在150°~180°之间，更接近平角(180°)。故答案为:B",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征",
+                                 "一般时间与钟面指针的位置"
+                             ],
+        "source":  "2024·河北保定·小升初真题"
+    },
+    {
+        "id":  "204",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "用一副三角板不可以拼成( )的角。",
+        "options":  [
+                        "135°",
+                        "80°",
+                        "15°",
+                        "75°"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】一副三角板的角度有30°、60°、45°、90°。可通过它们的加减拼角。\n【详解】A. 90°+45°=135°，可以拼成135°；B. 无论怎样组合加减，均不能得到80°；C. 45°-30°=15°，可以拼成15°；D. 30°+45°=75°，可以拼成75°。所以不能拼成80°的角，答案为:B",
+        "knowledge_points":  [
+                                 "用三角尺画角"
+                             ],
+        "source":  "2024·江西吉安·小升初真题"
+    },
+    {
+        "id":  "205",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一条( )长3.5厘米。",
+        "options":  [
+                        "直线",
+                        "射线",
+                        "线段"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】直线没有端点，长度无限；射线有一个端点，向一端无限延长，长度也无限；线段有两个端点，长度有限，可以量出。\n【详解】\"长3.5厘米\"说明有确定长度，只能是线段。故答案为:C",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "2025六年级下·吉林·专题练习"
+    },
+    {
+        "id":  "206",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "2024年8月6日，巴黎奥运会女子10米跳台决赛中，全红婵夺得冠军。备战中每天苦练207C(向后翻腾三周半抱膝),\"向后翻腾三周半\"，即是转( )度。",
+        "options":  [
+                        "800",
+                        "1080",
+                        "1260"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】一周为360°，三周为3×360°，半周为360°÷2=180°，三周半是两者之和。\n【详解】3×360°=1080°；360°÷2=180°；1080°+180°=1260°。故答案为:C",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25四年级上·河北保定·期末"
+    },
+    {
+        "id":  "207",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "队列练习时，原地向左转2次，共转了( )度。",
+        "options":  [
+                        "90",
+                        "180",
+                        "270"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】向左转一次为90°，两次即2×90°。\n【详解】2×90°=180°，共转了180度。故答案为:B",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25四年级上·贵州六盘水·期末"
+    },
+    {
+        "id":  "208",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面四个时间中，钟面上时针和分针形成的较小角是钝角的是( )。",
+        "options":  [
+                        "3:30",
+                        "8:30",
+                        "9:30",
+                        "10:00"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】钟面有12大格，每格30°。\n3:30：时针在3和4之间，分针在6，较小角小于3格(90°)，为锐角；8:30：同理为锐角；9:30：时针在9和10之间，分针在6，较小角大于3格(90°)且小于4格(120°)，为钝角；10:00：两针相差2格，60°，为锐角。\n【详解】只有9:30时较小角是钝角。故答案为:C",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "整点时间与钟面指针的位置",
+                                 "一般时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·浙江嘉兴·期末"
+    },
+    {
+        "id":  "209",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "6 时整，钟面上的时针和分针所成的角是( )角。",
+        "options":  [
+                        "周",
+                        "平",
+                        "直"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】6时整分针指12，时针指6，两针成一条直线，夹角180°，为平角。\n【详解】6时整，时针和分针所成的角是平角。故答案为:B",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·贵州安顺·期末"
+    },
+    {
+        "id":  "210",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "用一副三角板不能画出( )的角。",
+        "options":  [
+                        "15°",
+                        "85°",
+                        "75°",
+                        "135°"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】一副三角板角度有30°、60°、45°、90°，可拼得15°、75°、105°、135°、120°、150°等。\n【详解】A. 45°-30°=15°，可画；B. 85°无法由30°、45°、60°、90°加减得到；C. 30°+45°=75°，可画；D. 90°+45°=135°，可画。故不能画出的是85°，答案为:B",
+        "knowledge_points":  [
+                                 "用三角尺画角"
+                             ],
+        "source":  "24-25 四年级上·贵州黔西·期末"
+    },
+    {
+        "id":  "211",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列时刻中，钟面上时针和分针所成的较小角比90°小的是( )。",
+        "options":  [
+                        "2 时 30 分",
+                        "3 时",
+                        "3 时 30 分",
+                        "9 时 30 分"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】每大格30°。\n2:30:3格半，3.5×30°=105°>90°；3:00:3×30°=90°；3:30:2格半，2.5×30°=75°<90°；9:30:3.5格，105°>90°。\n【详解】只有3时30分时较小角为75°，小于90°，为锐角。故答案为:C",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "整点时间与钟面指针的位置",
+                                 "一般时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·浙江台州·期末"
+    },
+    {
+        "id":  "212",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列关于角的说法正确的是( )。",
+        "options":  [
+                        "将圆分为360份，每份所对应的角是1°",
+                        "一条直线就是平角",
+                        "大于或等于90°的角叫作钝角",
+                        "角的大小与边的长短无关"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】圆应\"平均分成\"360份，每份对应1°；平角是两条射线成一直线的角，不是一条直线本身；钝角是大于90°且小于180°的角；角的大小只与两边张开的大小有关，与边长无关。\n【详解】只有\"角的大小与边的长短无关\"说法正确。故答案为:D",
+        "knowledge_points":  [
+                                 "角的度量",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "角的大小比较",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·河南许昌·期末"
+    },
+    {
+        "id":  "213",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "2024 年 10 月 30 日 4 时27分，搭载神舟十九号载人飞船的长征二号F遥十九运载火箭在酒泉卫星发射中心点火发射。此时钟面上时针和分针所成的角是( )。",
+        "options":  [
+                        "钝角",
+                        "锐角",
+                        "直角",
+                        "平角"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】4:27大致可看作4点多，分针在5与6之间。分针与时针的较小夹角约为1个大格左右，即约30°，小于90°，为锐角。\n【详解】根据钟面位置可知，此时两针夹角小于90°，为锐角。故答案为:B",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征",
+                                 "一般时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·河南许昌·期末"
+    },
+    {
+        "id":  "214",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "将一张圆形纸片对折3次后展开，不能得到的角是( )。",
+        "options":  [
+                        "180°",
+                        "90°",
+                        "60°",
+                        "45°"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】对折1次:360°÷2=180°；对折2次:180°÷2=90°；对折3次:90°÷2=45°，折痕形成的角度都是45°的整数倍，如45°、90°、180°等。\n【详解】45°、90°、180°都是45°的倍数，而60°不是45°的倍数，不能得到。故答案为:C",
+        "knowledge_points":  [
+                                 "图形的折叠问题",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·湖南常德·期末"
+    },
+    {
+        "id":  "215",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "过两点可以画( )条直线，过一点可以画( )条射线。",
+        "options":  [
+                        "1;无数",
+                        "2;1",
+                        "2;无数"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】两点确定一条直线，所以过两点只能画一条直线；过一点可以向任意方向引射线，因此可以画无数条射线。\n【详解】过两点可以画1条直线，过一点可以画无数条射线。故答案为:A",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·安徽安庆·期末"
+    },
+    {
+        "id":  "216",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "钟面上11:00，时针和分针形成的角是( );过了20分钟，这时时针和分针形成的角是( )。",
+        "options":  [
+                        "锐角;钝角",
+                        "直角;锐角",
+                        "钝角;直角"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】11:00时，时针在11，分针在12，相差1大格，30°，为锐角；加20分钟到11:20，分针到4，时针已向12方向移动一些，大约形成大于120°小于150°的角，为钝角。\n【详解】11:00时形成30°锐角；11:20时，两针夹角在120°~150°之间，为钝角。故答案为:A",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "整点时间与钟面指针的位置",
+                                 "时，分，秒时间的推算"
+                             ],
+        "source":  "24-25 四年级上·河南郑州·期末"
+    },
+    {
+        "id":  "217",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "\"向后翻腾两周半\"这个动作是指运动员在跳水过程中要转( )。",
+        "options":  [
+                        "900°",
+                        "540°",
+                        "720°"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】一周360°，两周为2×360°=720°；半周为180°，两周半为720°+180°。\n【详解】2×360°=720°；720°+180°=900°。故答案为:A",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·河南郑州·期末"
+    },
+    {
+        "id":  "218",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "在黑板上画一个50°的角和在纸上画一个50°的角相比,( )。",
+        "options":  [
+                        "一样大",
+                        "黑板上的角大",
+                        "纸上的角大"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】角的大小与两边长短无关，只与开口大小有关；度数相同，大小相同。\n【详解】两处画的角度数都为50°，两边叉开程度相同，角一样大。故答案为:A",
+        "knowledge_points":  [
+                                 "角的大小比较"
+                             ],
+        "source":  "24-25 四年级上·河南洛阳·期末"
+    },
+    {
+        "id":  "219",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "用一副三角板可以画出( )的角。",
+        "options":  [
+                        "35°",
+                        "120°",
+                        "130°"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】一副三角板的角有30°、60°、90°、45°。120°=90°+30°可以拼出；35°和130°都不能由这些角加减得到。\n【详解】A. 35°不能拼出；B. 30°+90°=120°可拼出；C. 130°不能拼出。故答案为:B",
+        "knowledge_points":  [
+                                 "用三角尺画角",
+                                 "角度的计算"
+                             ],
+        "source":  "24-25 四年级上·安徽芜湖·期末"
+    },
+    {
+        "id":  "220",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "用一张圆形纸折成一个45°的角，至少需要对折( )次。",
+        "options":  [
+                        "5",
+                        "4",
+                        "3",
+                        "2"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】圆周360°。对折1次:360°÷2=180°；再对折:180°÷2=90°；再对折:90°÷2=45°，共3次。\n【详解】对折1次得180°，对折2次得90°，对折3次得45°，至少对折3次。故答案为:C",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "角的分类及换算"
+                             ],
+        "source":  "24-25 四年级下·河南南阳·开学考试"
+    },
+    {
+        "id":  "221",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面说法错误的是( )。",
+        "options":  [
+                        "周长是400米的正方形草地，它的面积是1公顷",
+                        "在算盘上表示54，要在十位上拨4颗下珠，个位上拨1颗上珠",
+                        "用一个5倍的放大镜看12°的角，这个角的大小仍然是12°",
+                        "一副三角尺有两个，用它们可以拼出150°的角"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】A:周长400米，边长400÷4=100米，面积100×100=10000平方米=1公顷，正确；B:十位应拨1颗上珠(表示50)，个位拨4颗下珠(表示4)，题干说反，错误；C:放大镜不改变角度，正确；D:90°+60°=150°，可以拼出150°，正确。\n【详解】错误的是算盘表示54的说法。故答案为:B",
+        "knowledge_points":  [
+                                 "算盘的认识与使用",
+                                 "角的初步认识及辨认",
+                                 "用三角尺画角",
+                                 "正方形的面积"
+                             ],
+        "source":  "24-25 四年级上·广东佛山·期末"
+    },
+    {
+        "id":  "222",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "林放星期天上午出门时看了看钟表，正好是9:00，当他回到家时，发现时针已经转动了60°,而分针的位置不变，林放到家时是( )。",
+        "options":  [
+                        "9:10",
+                        "10:00",
+                        "11:00",
+                        "10:20"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】每大格30°，时针转动60°相当于转了2个大格，即2小时。9:00+2小时=11:00。\n【详解】60°÷30°=2(格)，9+2=11，所以是11:00。故答案为:C",
+        "knowledge_points":  [
+                                 "平角，周角的认识及特征",
+                                 "时，分，秒时间的推算"
+                             ],
+        "source":  "24-25 四年级上·河南信阳·期末"
+    },
+    {
+        "id":  "223",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面的度数不能用三角尺拼出来的是( )。",
+        "options":  [
+                        "135°",
+                        "70°",
+                        "105°",
+                        "15°"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】三角板角有30°、60°、45°、90°，可拼出15°、75°、105°、120°、135°等，为15°的整数倍。70°不是15°的整数倍。\n【详解】135°=90°+45°，可拼出；105°=60°+45°，可拼出；15°=60°-45°，可拼出；70°不能由这些角加减得到。故答案为:B",
+        "knowledge_points":  [
+                                 "用三角尺画角"
+                             ],
+        "source":  "24-25 四年级上·河南信阳·期末"
+    },
+    {
+        "id":  "224",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "钟表上,6时30分，时针和分针构成一个( )。",
+        "options":  [
+                        "锐角",
+                        "钝角",
+                        "周角"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】每大格30°，每小格6°。6:30时，分针在6(180°)，时针在6与7之间，偏离6约15°，两针较小夹角约15°，小于90°，为锐角。\n【详解】6:30时针和分针夹角为15°，是锐角。故答案为:A",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征",
+                                 "一般时间与钟面指针的位置"
+                             ],
+        "source":  "24-25四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "225",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "度量一个角时，角的一条边对着量角器上\"180°\"的刻度线，另一条边对着\"70°\"的刻度线，这个角是( )。",
+        "options":  [
+                        "70°",
+                        "180°",
+                        "110°"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】起始边不在0°刻度线，而在180°刻度线，角度为两刻度之差:180°-70°。\n【详解】180°-70°=110°，这个角是110°。故答案为:C",
+        "knowledge_points":  [
+                                 "角的度量"
+                             ],
+        "source":  "24-25 四年级上·湖北恩施·期末"
+    },
+    {
+        "id":  "226",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面说法错误的是( )。",
+        "options":  [
+                        "将圆平均分成360份，人们把其中的1份所对的角作为度量角的单位",
+                        "过一点可以画无数条直线",
+                        "算盘是我国的传统计算工具，算盘上的每一颗珠子都表示1",
+                        "最小的自然数是0，自然数的个数是无限的"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】A:圆\"平均分成\"360份，每份为1°，正确；B:过一点可以画无数条直线，正确；C:算盘上梁上珠表示5，梁下珠表示1，不是每颗都表示1，错误；D:最小自然数是0，自然数个数无限，正确。\n【详解】错误的是算盘珠子含义的说法。故答案为:C",
+        "knowledge_points":  [
+                                 "算盘的认识与使用",
+                                 "角的度量",
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·湖北黄冈·期末"
+    },
+    {
+        "id":  "227",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "把一个直角分成两个锐角，如果一个锐角是另一个锐角的2倍，那么较小的锐角是( )。",
+        "options":  [
+                        "20°",
+                        "60°",
+                        "30°",
+                        "45°"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】设较小锐角为x，较大锐角为2x，直角为90°，有x+2x=90°。\n【详解】3x=90°，x=30°，较小的锐角是30°。故答案为:C",
+        "knowledge_points":  [
+                                 "倍的认识",
+                                 "直角，钝角，锐角的认识及特征"
+                             ],
+        "source":  "23-24 四年级上·湖北随州·期末"
+    },
+    {
+        "id":  "228",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "用一副三角板不能拼成( )的角。",
+        "options":  [
+                        "180°",
+                        "105°",
+                        "85°"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】一副三角板角度:30°、45°、60°、90°，所有可拼角为15°的整数倍。180°、105°都是15°的整数倍，85°不是。\n【详解】180°=90°+90°；105°=60°+45°；85°不能由30°、45°、60°、90°的加减得到。故答案为:C",
+        "knowledge_points":  [
+                                 "用三角尺画角"
+                             ],
+        "source":  "2024·山西长治·小升初真题"
+    },
+    {
+        "id":  "229",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "关于直线，射线，线段，下面说法错误的是( )。",
+        "options":  [
+                        "过平面内两点，可以画2条线段",
+                        "过平面内一点，可以画无数条射线",
+                        "线段有两个端点，可以测量长度",
+                        "直线没有端点，可以无限延伸"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】两点间只能画1条线段；从一点可以向不同方向画无数射线；线段有两个端点且长度有限；直线无端点，无限延伸。\n【详解】A 说\"可以画2条线段\"错误，其余说法正确。故答案为:A",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·福建龙岩·期末"
+    },
+    {
+        "id":  "230",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列说法正确的是( )。",
+        "options":  [
+                        "锐角+锐角=直角",
+                        "钝角+锐角=平角",
+                        "直角+锐角=钝角"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】锐角<90°，直角=90°，钝角在90°~180°。两锐角相加结果不确定；钝角+锐角的和可能<、=或>180°；直角+锐角(>0°且<90°)的和必大于90°且小于180°，必为钝角。\n【详解】只有\"直角+锐角=钝角\"必然成立，故答案为:C",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·四川德阳·期末"
+    },
+    {
+        "id":  "231",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "钟面上的时间为10时，此时时针和分针的夹角是( )度。",
+        "options":  [
+
+                    ],
+        "answer":  "60",
+        "solution":  "【分析】每大格30°。10时整分针指12，时针指10，相差2大格。\n【详解】30°×2=60°，所以夹角是60°。",
+        "knowledge_points":  [
+                                 "整点时间与钟面指针的位置",
+                                 "角的度量"
+                             ],
+        "source":  "24-25 四年级上·新疆乌鲁木齐·期中"
+    },
+    {
+        "id":  "232",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一条射线，绕它的端点旋转一周所形成的角是( ),它是( )度，相当于( )个直角;一个角的两条边恰好在同一条直线上，那这个角是( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "周角;360;4;平",
+        "solution":  "【分析】一条射线绕端点旋转一周形成周角，周角=360°，直角=90°，360°=4×90°；两边在同一条直线上为平角，平角=180°。\n【详解】一条射线绕端点旋转一周形成的角是周角，它是360度，相当于4个直角；两条边恰好在同一条直线上的是平角。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·新疆乌鲁木齐·期末"
+    },
+    {
+        "id":  "233",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "1 平角=( )直角 1 周角=( )平角=( )直角",
+        "options":  [
+
+                    ],
+        "answer":  "2;2;4",
+        "solution":  "【分析】平角=180°，直角=90°，180°÷90°=2；周角=360°，360°÷180°=2，360°÷90°=4。\n【详解】1 平角=2 直角；1 周角=2 平角=4 直角。",
+        "knowledge_points":  [
+                                 "两，三位数与一位数的一次进位乘法",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "23-24 四年级上·广东肇庆·期中"
+    },
+    {
+        "id":  "234",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "高铁是我们重要的交通工具，往返于梅州西站和广州南站的高铁，除起点站和终点站外，中间还要停靠 4 个站，那么要准备( )种车票。",
+        "options":  [
+
+                    ],
+        "answer":  "30",
+        "solution":  "【分析】根据题意画线段图：先求出线段的条数，再计算车票的种数。\n\n![](images/0aa83c3c0ee3a266f2003b9d6819c3d4f7bdf392c4126409125b732bb13af1cd.jpg)\n\n【详解】根据线段的定义：可知图中共有线段有AC、AD、AE、AF、AB、CD、CE、CF、CB、DE、DF、DB、EF、EB、FB，一共 15 条。15×2=30(种) 因为每一段路程可以有往返两种车票，故要准备30种车票。",
+        "knowledge_points":  [
+                                 "组合问题",
+                                 "数图形(线段，直线，射线)"
+                             ],
+        "source":  "2024·广东梅州·小升初真题"
+    },
+    {
+        "id":  "235",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "7时15分，钟面上分针与时针所组成的角是( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "钝",
+        "solution":  "【分析】7时15分时，时针刚超过7，分针指着3，所以时针和分针之间超过4个大格不到5个大格，30°×4=120°，所以时针和分针所组成的角是钝角。\n【详解】(7-3)×30°=4×30°=120°，7时15分，钟面上分针与时针所组成的角大于120°，是钝角。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "一般时间与钟面指针的位置"
+                             ],
+        "source":  "2024·河南开封·小升初真题"
+    },
+    {
+        "id":  "236",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "阳阳，锦锦，华华分别用同一副三角板拼出了三个不同的钝角，他们拼出的钝角度数分别是( )°,( )°,( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "120;150;135",
+        "solution":  "【分析】根据题意，明确大于90°小于180°的角是钝角；一副三角板上的角的度数有:90°、60°、30°、45°；通过组合不同角度的三角板，计算拼出的钝角即可。\n【详解】根据分析可知: 90°+30°=120°；90°+60°=150°；90°+45°=135°；60°+45°=105°。阳阳，锦锦，华华分别用同一副三角板拼出了三个不同的钝角，他们拼出的钝角度数分别是120°、150°、135°。(答案不唯一)",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "用三角尺画角"
+                             ],
+        "source":  "24-25四年级上·河北保定·期末"
+    },
+    {
+        "id":  "237",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "立春是二十四节气中的第一个节气,2025 年 2 月 3 日 22 时 10 分将迎来立春节气，这时钟面上时针和分针形成的较小角是( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "钝",
+        "solution":  "【分析】钟面一圈为360°，被分成12个大格，每个大格之间是5个小格，每个大格的角度为360°÷12=30°，每个小格的度数是30°÷5=6°。22时10分，时针在10和11之间，分针指向2。计算时针和分针之间的大格数，再乘以每个大格的度数，就能得出夹角的度数，进而判断角的类型。锐角在0°-90°之间，直角是90°，钝角在90°-180°之间。\n【详解】结合分析可知，22时10分，时针刚刚过完10，应该快走完第一个小格了，此时距离11时还有四个小格多一点，分针指向2，则分针和时针之间接近四个大格，高于3个大格，所以这个角的度数介于3×30°和4×30°之间，也就是90°-120°之间，所以这个角是钝角。",
+        "knowledge_points":  [
+                                 "角的初步认识及辨认",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "一般时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·河北保定·期末"
+    },
+    {
+        "id":  "238",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "730000 平方米=( )公顷 3000 平方厘米=( )平方分米\n\n8平方千米=( )公顷 72000000 平方米=( )平方千米,600 公顷=( )平方千米 1 周角=( )平角=( )直角",
+        "options":  [
+
+                    ],
+        "answer":  "73;30;800;72;6;2;4",
+        "solution":  "【分析】平方米，平方分米，平方厘米相邻单位间的进率是100，平方米与公顷间的进率是10000，公顷与平方千米间的进率是100，平方米与平方千米间的进率是1000000；由高级单位化低级单位乘进率，反之除以进率。直角是90°，平角是180°，周角是360°。\n【详解】因为730000平方米里面含有73个10000平方米，所以730000平方米=73公顷；因为3000平方厘米里面含有30个100平方厘米，所以3000平方厘米=30平方分米；因为8×100=800(公顷)，所以8平方千米=800公顷；因为72000000平方米里面含有72个1000000平方米，所以72000000平方米=72平方千米；因为600公顷里面含有6个100公顷，所以600公顷=6平方千米；因为180°×2=360°，90°×4=360°，所以1周角=2平角=4直角。",
+        "knowledge_points":  [
+                                 "平角，周角的认识及特征",
+                                 "平方厘米，平方分米，平方米之间的进率与换算",
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "24-25 四年级上·河北保定·期末"
+    },
+    {
+        "id":  "239",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "钟面上 3:00，时针和分针所成的角是( )角,2:30 时针和分针所成的角是( )角,( )时，时针和分针所成的角是平角。",
+        "options":  [
+
+                    ],
+        "answer":  "直;钝;6",
+        "solution":  "【分析】钟面上12个数字把钟面平均分成12个大格，每个大格是30°。3:00时，12和3之间相差3个大格，夹角为3×30°=90°；2:30时，分针指向6，时针在2和3之间，夹角大于3×30°小于4×30°，是钝角；平角是180°，6×30°=180°，整时分针指向12，则时针指向6。\n【详解】根据分析:3×30°=90°，所以3:00为直角；4×30°=120°，所以2:30的较小角大于90°小于180°，是钝角；6×30°=180°，所以6时，时针和分针所成的角是平角。综上：直角，钝角，6时。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征",
+                                 "整点时间与钟面指针的位置",
+                                 "半点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·贵州黔西·期末"
+    },
+    {
+        "id":  "240",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在 80°,100°,135°这几个角中,( )可以直接用一副三角尺拼成。",
+        "options":  [
+
+                    ],
+        "answer":  "135°",
+        "solution":  "【分析】一副三角板的角有30°、45°、60°、90°，把它们进行组合加减，找出能得到的角度。\n【详解】根据分析可知:60°-45°=15°，60°+45°=105°，60°+90°=150°，90°+45°=135°，90°+30°=120°，30°+45°=75°，90°+90°=180°。所以80°、100°、135°中，135°角可以直接用一副三角尺拼成。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "用三角尺画角"
+                             ],
+        "source":  "24-25 四年级上·河南许昌·期末"
+    },
+    {
+        "id":  "241",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "李林观察钟面时发现,6时整时针与分针的夹角是( )度，是一个( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "180;平",
+        "solution":  "【分析】6时整分针指向12，时针指向6，两个指针在同一条直线上，夹角是一个平角，即180°。\n【详解】李林观察钟面时发现，6时整时针与分针的夹角是180度，是一个平角。",
+        "knowledge_points":  [
+                                 "平角，周角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25四年级上·贵州黔东南·期末"
+    },
+    {
+        "id":  "242",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "钟面上( )时整，时针和分针成 180°,是一个( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "6;平",
+        "solution":  "【分析】时针和分针成180°时，两针在一条直线上。整时分针指向12，则时针应指向6，所以是6时整；180°是平角。\n【详解】由分析可知：钟面上6时整，时针和分针成180°，是一个平角。",
+        "knowledge_points":  [
+                                 "平角，周角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·贵州铜仁·期末"
+    },
+    {
+        "id":  "243",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "奇奇用一个破损的量角器测量一个角，角的一条边和外圈的10°刻度线重合，读数时误读了内圈的刻度，读出的度数是110°,这个角的实际度数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "60°",
+        "solution":  "【分析】量角器内圈和外圈刻度互补，同一位置内外圈刻度之和是180°。先用180°减去误读的内圈刻度得到正确外圈读数，再减去起始对齐的外圈10°。\n【详解】180°-110°-10°=70°-10°=60°，因此，这个角的实际度数是60°。",
+        "knowledge_points":  [
+                                 "角的度量"
+                             ],
+        "source":  "24-25四年级上·河南郑州·期末"
+    },
+    {
+        "id":  "244",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "钟面上 4 时整，时针与分针所夹的角是( )度，是( )角;6 时整，时针和分针所成的角是( )度，是( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "120;钝;180;平",
+        "solution":  "【分析】钟表刻度把一个圆平均分成12等份，每一份是30°。4时整两针相差4个大格，夹角是4×30°；6时整相差6个大格，夹角是6×30°；再按角的分类填空。\n【详解】根据分析可知: 4×30°=120°，6×30°=180°。钟面上4时整夹角是120度，是钝角；6时整夹角是180度，是平角。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·河北张家口·期末"
+    },
+    {
+        "id":  "245",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把一个平角分成两个角，那么这两个角可能是( )角和( )角，也可能是()角和( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "锐角;钝角;直角;直角",
+        "solution":  "【分析】平角是180°。分成两个角：一种情况是两个直角(90°+90°=180°)；另一种情况是一个锐角(小于90°)加一个钝角(大于90°小于180°)。\n【详解】根据分析可知：把一个平角分成两个角，那么这两个角可能是锐角和钝角，也可能是直角和直角。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·浙江杭州·期末"
+    },
+    {
+        "id":  "246",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈看了一下手表时间是10:15，但是家里挂在墙上的钟慢了半小时，为了调整时间，妈妈需把分针顺时针旋转( )。",
+        "options":  [
+
+                    ],
+        "answer":  "180°或180度",
+        "solution":  "【分析】分针60分钟转一圈360°，每分钟转6°。钟慢半小时，需要顺时针拨30分钟，即转动30×6°。也可看作半小时分针从12走到6，走6个大格，每格30°。\n【详解】分针每分钟转6°，30分钟旋转角度为:6×30=180°；或6个大格×30°=180°。所以为了调整时间，妈妈需把分针顺时针旋转180°。",
+        "knowledge_points":  [
+                                 "旋转与旋转现象",
+                                 "时，分，秒时间的推算"
+                             ],
+        "source":  "24-25四年级上·浙江绍兴·期末"
+    },
+    {
+        "id":  "247",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "周日上午 9:30，聪聪开始做作业，此时，钟面上的时针和分针形成的较小角是( )角;上午 10:00，聪聪完成作业，从 9:30-10:00 分针在钟面上转过的角度是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "钝;180",
+        "solution":  "【分析】钟面每大格30°。9:30时针在9和10中间，分针指在6，夹角约为3个半大格:3×30°=90°，半个大格15°，共105°，为钝角；9:30到10:00分针走6个大格:6×30°=180°。\n【详解】3×30°=90°，30°÷2=15°，90°+15°=105°；6×30°=180°。所以较小角是钝角；分针转过180°。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征",
+                                 "半点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·浙江绍兴·期末"
+    },
+    {
+        "id":  "248",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "3 时 30 分，钟面上时针和分针形成( )角;5 时整，时针和分针又形成( )角。(填\"锐\"\"直\"或\"钝\")",
+        "options":  [
+
+                    ],
+        "answer":  "锐;钝",
+        "solution":  "【分析】锐角在0°-90°之间，直角是90°，钝角在90°-180°之间。钟面每大格30°。3:30时针在3和4中间，分针在6，夹角对应大格数小于3，即小于90°，是锐角；5:00夹角为5×30°=150°，是钝角。\n【详解】5×30°=150°。所以3时30分为锐角；5时整为钝角。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "整点时间与钟面指针的位置",
+                                 "一般时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·浙江绍兴·期末"
+    },
+    {
+        "id":  "249",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "跳水是一项优美的水上运动，跳水姿势有直体，屈体，抱膝和翻腾兼转体四种类型。在一次跳水比赛中，某运动员向前翻腾 3 周半屈体，这名运动员身体向前翻转了( )度。",
+        "options":  [
+
+                    ],
+        "answer":  "1260",
+        "solution":  "【分析】周角是360°，半周是180°。向前翻腾3周半=3个周角+1个平角。\n【详解】360°×3+180°=1080°+180°=1260°。所以该运动员身体向前翻转了1260度。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·河南漯河·期末"
+    },
+    {
+        "id":  "250",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\">\"\"<\"或\"=\".\n\n9716000( )13615000 871986( )875123\n\n2 平角( )1 周角 20000 公顷( )2 平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "<;<;=;>",
+        "solution":  "【分析】比较整数大小：先比位数，再从高位到低位逐位比较。2平角=2×180°=360°，1周角=360°，所以相等。1平方千米=100公顷，2平方千米=200公顷，20000公顷>200公顷。\n【详解】9716000<13615000；871986<875123；2平角=1周角；20000公顷>2平方千米。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "平角，周角的认识及特征",
+                                 "角的分类及换算",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·河南新乡·期末"
+    },
+    {
+        "id":  "251",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "2 时钟面上的时针和分针成( )角,9 时成( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "锐;直",
+        "solution":  "【分析】钟面12大格，每格30°。2时夹角为2×30°=60°，为锐角；9时夹角为3×30°=90°，为直角。\n【详解】360°÷12=30°；30°×2=60°；30°×3=90°。所以2时为锐角，9时为直角。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·湖南永州·期末"
+    },
+    {
+        "id":  "252",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在钟面上,1 时整时针和分针成( )角,5 时整时针和分针成( )角,9 时整时针和分针成( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "锐;钝;直",
+        "solution":  "【分析】钟面每大格30°。1时夹角1×30°=30°为锐角；5时夹角5×30°=150°为钝角；9时夹角3×30°=90°为直角。\n【详解】360°÷12=30°。1时：1×30°=30°，锐角；5时：5×30°=150°，钝角；9时：3×30°=90°，直角。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·安徽安庆·期末"
+    },
+    {
+        "id":  "253",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "将一个 120°角分成两个角，其中一个是锐角，另一个角可能是什么角。把可能的结果都写下来:( )。",
+        "options":  [
+
+                    ],
+        "answer":  "锐角、直角、钝角",
+        "solution":  "【分析】120°分成两个角，其中一个是锐角(0°<x<90°)。另一个角=120°-x：当x<30°时，另一个角>90°为钝角；当x=30°时，另一个角=90°为直角；当30°<x<90°时，另一个角<90°为锐角。\n【详解】由分析可知：另一个角可能是锐角、直角、钝角。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·河南新乡·期末"
+    },
+    {
+        "id":  "254",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "下午3:00开始考试，此时钟面上时针和分针所组成的较小的角是( )角(填\"锐角\"\"直角\"\"钝角\"或\"平角\")。60分钟后考试结束，这段时间时针转动了( )°，分针转动了( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "直角;30;360",
+        "solution":  "【分析】钟面12大格，每格30°。3:00时针指3，分针指12，相差3大格，夹角3×30°=90°为直角。60分钟=1小时，时针走1大格转30°，分针转一圈360°。\n【详解】360°÷12=30°；3×30°=90°，为直角；60分钟后时针转30°，分针转360°。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25四年级上·河南新乡·期末"
+    },
+    {
+        "id":  "255",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "6时整时针与分针的夹角是( )°;4 时整时针与分针的夹角是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "180;120",
+        "solution":  "【分析】钟面每大格30°。6时相差6大格，夹角6×30°=180°;4时相差4大格，夹角4×30°=120°.\n【详解】6×30°=180°;4×30°=120°。所以6时为180°,4时为120°。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "角的度量",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25四年级上·湖南岳阳·期末"
+    },
+    {
+        "id":  "256",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用两个角拼成了一个钝角。这两个角可能是( )角和( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "锐;锐或直或钝",
+        "solution":  "【分析】锐角：0°<x<90°；直角：90°；钝角：90°<x<180°。要拼成钝角(90°-180°之间)，其中一个是锐角时，另一个角可为锐角、直角或钝角(但两角和需在90°-180°之间)。\n【详解】两个角拼成钝角的情况举例：锐角+锐角；锐角+直角；锐角+钝角。\n\n![](images/a39c4c3b5535db99b1a0fde47a92d5458ee21b36e14fd7c5ef9e45fd46c4176e.jpg)\n\n![](images/c8aad0593a622590ea9638bfffc05c0ceacccb870cfb5d4d624fbcdde2ca58aa.jpg)\n\n![](images/aa606e7446f018faf0f8f33e7603f40f1fa8fb945d0261f17c34120e233a7318.jpg)",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "用不同的图形拼接"
+                             ],
+        "source":  "24-25四年级上·河南郑州·期末"
+    },
+    {
+        "id":  "257",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "12:15，钟面上的时针和分针组成的较小的角是( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "锐",
+        "solution":  "【分析】小于90°是锐角。钟面每大格30°。12:15分针指向3，时针走过12还不到12和1的正中间，时针和分针之间的大格数不到3个，因此夹角小于30°×3=90°。\n【详解】30°×3=90°，小于90°的是锐角。所以12:15时较小角是锐角。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征",
+                                 "一般时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·山东济南·期末"
+    },
+    {
+        "id":  "258",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "6时整，钟面上时针与分针所成的角是( )角。3时整，钟面上时针与分针所成的较小的角是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "平;90",
+        "solution":  "【分析】钟面每大格30°，几时整，时针与分针之间就有几个大格。6时夹角6×30°=180°是平角；3时夹角3×30°=90°。\n【详解】30°×6=180°；30°×3=90°。所以6时整是平角；3时整较小角是90°。",
+        "knowledge_points":  [
+                                 "平角，周角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25四年级上·广东汕头·期末"
+    },
+    {
+        "id":  "259",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "角的两条边都是( )线，线段有( )个端点，经过两点能画( )条直线。",
+        "options":  [
+
+                    ],
+        "answer":  "射;2或两;1或一",
+        "solution":  "【分析】由一点引出的两条射线所围成的图形叫角;线段有两个端点;两点确定一条直线.\n【详解】角的两条边都是射线，线段有两个端点，经过两点能画一条直线。",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征",
+                                 "角的初步认识及辨认"
+                             ],
+        "source":  "24-25 四年级上·广东汕头·期末"
+    },
+    {
+        "id":  "260",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小林在量一个角时，本应该读内圈刻度，他读成了外圈刻度是 35°,正确的读数应该是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "145°",
+        "solution":  "【分析】量角器内圈刻度和外圈刻度之和为180°。读成外圈35°,正确应读内圈:180°-35°.\n【详解】180°-35°=145°。所以正确读数是145°。",
+        "knowledge_points":  [
+                                 "角的度量"
+                             ],
+        "source":  "23-24四年级下·河南南阳·开学考试"
+    },
+    {
+        "id":  "261",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填\">\"\"<\"或\"=\"。\n\n98889( )100000 700010( )700100\n\n813204( )81 万 4 直角( )2 平角",
+        "options":  [
+
+                    ],
+        "answer":  "<;<;>;=",
+        "solution":  "【分析】比较整数大小先比位数再逐位比较。81万=810000。直角是90°，平角是180°，4直角=4×90°=360°，2平角=2×180°=360°。\n【详解】98889<100000；700010<700100；81万=810000，813204>810000，所以813204>81万；4直角=2平角。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征",
+                                 "亿以内数的改写"
+                             ],
+        "source":  "24-25四年级上·广东珠海·期末"
+    },
+    {
+        "id":  "262",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "2 时整，时针与分针组成的角是( )°;6 时整，时针与分针组成的角是( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "60;平",
+        "solution":  "【分析】钟面每大格30°。2时相差2大格，夹角2×30°=60°;6时相差6大格，夹角6×30°=180°为平角.\n【详解】2×30°=60°;6×30°=180°。所以2时是60°;6时是平角。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·河南信阳·期末"
+    },
+    {
+        "id":  "263",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "线段有( )个端点;度量角的单位是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "2或两;度",
+        "solution":  "【分析】线段是直线的一部分，有两个端点，有限长。角的度量单位是度，用\"°\"表示.\n【详解】线段有2个端点;度量角的单位是度。",
+        "knowledge_points":  [
+                                 "角的度量",
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·河南信阳·期末"
+    },
+    {
+        "id":  "264",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个周角减去一个平角，再减去一个锐角，剩下的是一个( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "钝",
+        "solution":  "【分析】周角=360°，平角=180°，锐角<90°。360°-180°=180°，再减去一个锐角后结果介于90°和180°之间，为钝角。\n【详解】周角减平角差是平角，再减去锐角后差大于90°小于180°，所以剩下的是钝角。",
+        "knowledge_points":  [
+                                 "角的分类及换算"
+                             ],
+        "source":  "24-25四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "265",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "钟面从 2 时走到 3时，分针转了( )度，时针转了( )度。",
+        "options":  [
+
+                    ],
+        "answer":  "360;30",
+        "solution":  "【分析】从2时到3时是1小时。分针1小时转一圈360°,时针走1个大格即30°.\n【详解】钟面从2时走到3时，分针转360度，时针转30度。",
+        "knowledge_points":  [
+                                 "角的度量",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "266",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "3时整，钟面上的时针和分针所形成的角是( )度,( )时整，时针和分针所构成的角是平角。",
+        "options":  [
+
+                    ],
+        "answer":  "90;6",
+        "solution":  "【分析】钟面每大格30°。3时整两针相差3大格，夹角3×30°=90°。平角是180°，两针相差6大格时为180°，整时分针指向12，则时针指向6。\n【详解】30°×3=90°，所以3时整夹角是90度；6时整两针相差6大格，夹角是180°为平角。",
+        "knowledge_points":  [
+                                 "平角，周角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25四年级上·河南信阳·期末"
+    },
+    {
+        "id":  "267",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "冬至如大年，人间小团圆。2024 年 12 月 21 日 17 时 21 分迎来冬至节气，这时钟面上时针和分针形成的较小角是( )角(填\"钝\"，\"锐\"或\"直\");再过 60 分钟，分针转动( )°,时针相应转动( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "锐;360;30",
+        "solution":  "【分析】17时21分时针在5和6之间，分针在4和5之间偏向5，较小角小于90°为锐角。再过60分钟=1小时，分针转一周360°,时针走1大格转30°.\n【详解】如图:\n\n![](images/c3084a393baf788cfb935e83b49d39b22f6c47fc54fdc9d9c1d4581c805beaf1.jpg)\n\n这时钟面上时针和分针形成的较小角是锐角;再过60分钟，分针转动360°,时针相应转动30°。",
+        "knowledge_points":  [
+                                 "角的度量",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "般时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "268",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "小华星期天的时间是这样安排的：早上 8:00 起床洗漱,8:30 吃早餐,9:00 阅读。这三个时刻钟面上的时针与分针所形成的夹角依次是( )角,( )角,( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "钝;锐;直",
+        "solution":  "【分析】钟面每大格30°。8:00两针相差4大格，夹角4×30°=120°为钝角；8:30相差2个大格加半个大格，夹角2×30°+15°=75°为锐角；9:00相差3大格，夹角3×30°=90°为直角。\n【详解】4×30°=120°；2×30°+30°÷2=60°+15°=75°；3×30°=90°。所以依次是钝角、锐角、直角。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "整点时间与钟面指针的位置",
+                                 "一般时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "269",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在 3 时整的时候，钟面上的分针和时针形成的较小角是( )角;7 时整的时候，钟面上的分针和时针所形成的较小角是( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "直;钝",
+        "solution":  "【分析】钟面每大格30°。3时整两针相差3大格，夹角3×30°=90°为直角;7时整较小夹角相差5大格，夹角5×30°=150°为钝角.\n【详解】3×30°=90°;5×30°=150°。所以3时整为直角;7时整为钝角。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·河南信阳·期末"
+    },
+    {
+        "id":  "270",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "钟面上，从 5 时到 6时，时针旋转所形成的角是( )°，分针旋转所形成的角是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "30;360",
+        "solution":  "【分析】从5时到6时，时针走了1个大格。钟面上有12个大格，一共是360度，用除法可求出1个大格的度数。从5时到6时，分针走了1圈，1圈是360度，是一个周角。\n【详解】360°÷12=30°，所以1个大格是30度。所以钟面上，从5时到6时，时针旋转所形成的角是30°，分针旋转所形成的角是360°。",
+        "knowledge_points":  [
+                                 "平角，周角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "271",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "2 时整，时针和分针所形成的较小角是( )°;6 时整，时针和分针形成的夹角是( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "60;平",
+        "solution":  "【分析】钟面一周是360°，被12个数字平均分成12份，相邻两数之间夹角是30°。2时整时两针相差2个大格；6时整相差6个大格，且平角是180°。\n【详解】2时整：30°×2=60°；6时整：30°×6=180°，是平角。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "平角，周角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·湖北荆州·期末"
+    },
+    {
+        "id":  "272",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个角是55°的2倍，它是( )°，是( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "110;钝",
+        "solution":  "【分析】用55°×2求出角的度数；大于90°小于180°的是钝角。\n【详解】55°×2=110°，且90°<110°<180°，所以它是110°，是钝角。",
+        "knowledge_points":  [
+                                 "求一个数的几倍是多少",
+                                 "直角，钝角，锐角的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·河南三门峡·期末"
+    },
+    {
+        "id":  "273",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "将一个半圆平均分成 180 份，这样的 135 份是( )度，它再减少( )度，就是直角。",
+        "options":  [
+
+                    ],
+        "answer":  "135;45",
+        "solution":  "【分析】半圆是平角180°，平均分成180份，每份是1°。直角是90°。\n【详解】135份=135°；135°-90°=45°。所以135份是135度，再减少45度就是直角。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·湖北荆州·期末"
+    },
+    {
+        "id":  "274",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "4时整，时针与分针的夹角是( )度;12 时 30 分，时针和分针的夹角是( )度。",
+        "options":  [
+
+                    ],
+        "answer":  "120;165",
+        "solution":  "【分析】钟面一圈360°，平均分成12大格，每格30°。4时整两针相差4大格：4×30°。12:30时分针指向6(180°)，时针从12走了半小时，相当于走了15°，因此夹角为180°-15°=165°。\n【详解】30°×4=120°；12:30夹角：180°-15°=165°。所以答案是120度和165度。",
+        "knowledge_points":  [
+                                 "角的度量",
+                                 "整点时间与钟面指针的位置",
+                                 "半点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25四年级上·湖北黄冈·期末"
+    },
+    {
+        "id":  "275",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "钟面上，时针和分针在下午 3 时形成的角是( )°,是( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "90;直",
+        "solution":  "【分析】钟面每大格30°，下午3时整两针相差3大格：3×30°=90°。等于90°的是直角。\n【详解】3×30°=90°，所以形成的角是90°，是直角。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25四年级上·湖北恩施·期末"
+    },
+    {
+        "id":  "276",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "6 公顷=( )平方米 ( )个直角 = 1 个周角",
+        "options":  [
+
+                    ],
+        "answer":  "60000;4",
+        "solution":  "【分析】1公顷=10000平方米；周角是360°，直角是90°。\n【详解】6公顷=6×10000=60000平方米；360°里面有4个90°，所以4个直角=1个周角。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "24-25 四年级上·湖北十堰·期末"
+    },
+    {
+        "id":  "277",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "用一个放大 10 倍的放大镜看一个 30 度角，这个角的度数是( )度。",
+        "options":  [
+
+                    ],
+        "answer":  "30",
+        "solution":  "【分析】角的大小与边的长短无关，只与两条边叉开的大小有关;放大镜只会放大边，不改变角的大小.\n【详解】用放大10倍的放大镜看一个30°角，这个角的度数仍然是30°。",
+        "knowledge_points":  [
+                                 "角的概念及表示方法"
+                             ],
+        "source":  "23-24 六年级下·内蒙古通辽·期中"
+    },
+    {
+        "id":  "278",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把一个平角分成2个角，如果一个角是锐角，另一个角是( )角;如果一个角是直角，另一个角是( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "钝;直",
+        "solution":  "【分析】平角是180°。锐角小于90°，所以180°减去一个锐角后必大于90°，为钝角；直角是90°，180°-90°=90°，另一个也是直角。\n【详解】把一个平角分成2个角：一个是锐角时，另一个是钝角；一个是直角时，另一个也是直角。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25四年级上·福建福州·期末"
+    },
+    {
+        "id":  "279",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "周末，明明和爸爸，妈妈去宁化县天鹅洞游玩,9:00从家门口坐车出发,9:25 到达天鹅洞，这段时间钟面上的分针走了( )°,它是一个( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "150;钝",
+        "solution":  "【分析】9:00到9:25经过25分钟。分针每分钟转6°，所以转过25×6°。90°<角度<180°为钝角。\n【详解】25分钟×6°=150°，且90°<150°<180°，所以分针走了150°，它是一个钝角。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "时，分，秒有关的计算"
+                             ],
+        "source":  "24-25四年级上·福建三明·期末"
+    },
+    {
+        "id":  "280",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "任意两个锐角度数的和小于一个平角的度数。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】锐角小于90°,平角是180°。两个锐角的度数都小于90°,它们的和一定小于180°.\n【详解】任意两个锐角度数的和小于一个平角的度数。这句话正确。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "角的大小比较",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·安徽淮南·期末"
+    },
+    {
+        "id":  "281",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "用一个 10 倍的放大镜来看一个 15°的角，所看到的角是 150°。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】角的大小与边长无关;放大镜放大的是角的边，不改变角的度数.\n【详解】用10倍放大镜看15°角，角的度数仍是15°,原题说法错误。",
+        "knowledge_points":  [
+                                 "角的度量"
+                             ],
+        "source":  "2024·新疆吐鲁番·小升初真题"
+    },
+    {
+        "id":  "282",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "一条直线长 50米，一条线段长 200 米，线段比直线长 150 米。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】直线没有端点，可以向两端无限延伸，没有确定长度，不能说\"直线长50米\"，也无法与线段比较长短.\n【详解】直线是无限长的，没有具体长度，因此原题结论错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·新疆乌鲁木齐·期末"
+    },
+    {
+        "id":  "283",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.94,
+        "question":  "射线只能向一端延长，而直线可以向两端延长，所以射线可以测量长度而直线不能测量。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】直线和射线都可以无限延长，都是无限长的，无法测量长度.\n【详解】直线能向两个方向无限延长，射线向一个方向无限延长，但二者都无限长，都不能测量长度，原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "2024·河北保定·小升初真题"
+    },
+    {
+        "id":  "284",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "用2倍的放大镜看 30° 的角，看到的角是 30°。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】角的大小只与两条边开口大小有关，与边长无关;放大镜不改变开口大小.\n【详解】用2倍放大镜看30°角，角度不变仍是30°。原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "角的度量",
+                                 "角的概念及表示方法"
+                             ],
+        "source":  "2024·河北邯郸·小升初真题"
+    },
+    {
+        "id":  "285",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "钟面上 3:00 时，时针和分针的夹角是直角。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】3:00时分针指12，时针指3，两针相差3大格，每大格30°,夹角为90°,是直角.\n【详解】3:00 时针和分针的位置如下:\n\n![](images/6d2711d20b920080a3cce474c9682526528e20c4efbc82ec06567c5372230b27.jpg)\n\n故答案为:√",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·河北保定·期末"
+    },
+    {
+        "id":  "286",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "透过放大镜看35°的角，这个角会变大。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】角的大小与射线长度无关，只与夹角大小有关;放大镜不能改变角度.\n【详解】用放大镜看一个角，角的度数不会改变。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "角的度量",
+                                 "角的概念及表示方法"
+                             ],
+        "source":  "24-25 四年级上·河南漯河·期末"
+    },
+    {
+        "id":  "287",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "线段是可以量出它的长度的。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】线段有两个端点，长度有限，可以测量.\n【详解】线段是可以测量出长度的。原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·浙江绍兴·期末"
+    },
+    {
+        "id":  "288",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "把正方形的四个角撕下来拼在一起可以拼成周角。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】正方形每个角是直角90°，4个直角拼在一起是360°，即周角。\n【详解】90°×4=360°，所以可以拼成周角。原说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "正方形的概念及特点",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25四年级上·河南新乡·期末"
+    },
+    {
+        "id":  "289",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "把线段向两端无限延伸，就得到一条直线，直线没有端点。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】线段有两个端点;向两端无限延伸得到直线，直线没有端点且无限长.\n【详解】把线段向两端无限延伸，就得到一条直线，直线没有端点。原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·河南新乡·期末"
+    },
+    {
+        "id":  "290",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "平角的两条边在一条直线上，周角的两条边重合成射线。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】平角的两边在同一直线上且方向相反;周角是一条射线绕端点旋转一周，两边重合.\n【详解】平角两条边在一条直线上，周角两条边重合成射线。题中说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·河北石家庄·期末"
+    },
+    {
+        "id":  "291",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "用放大 10 倍的放大镜看一个 9° 的角，看到的是一个直角。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】放大镜只放大角的边，不改变角的度数;角的大小与边长无关.\n【详解】用放大10倍的放大镜看9°角，看到的仍是9°,不是直角。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "角的度量",
+                                 "角的大小比较"
+                             ],
+        "source":  "24-25 四年级上·广东汕头·期末"
+    },
+    {
+        "id":  "292",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "\"舒适的椅子\"椅子面和靠背之间的夹角，最适合的度数是 100°。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】100°大于90°小于180°,是钝角。结合生活常识，靠背略向后倾斜更舒适.\n【详解】100°为钝角，椅子面和靠背之间夹角取100°较舒适，原说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "角的度量",
+                                 "直角，钝角，锐角的认识及特征"
+                             ],
+        "source":  "23-24 四年级下·河南南阳·开学考试"
+    },
+    {
+        "id":  "293",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.94,
+        "question":  "用能放大 2 倍的放大镜观察一个 20° 的角，这个角就变成 40°。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】角的大小与边的长短无关，只与两条边叉开的大小有关;放大镜不改变角的度数.\n【详解】用能放大2倍的放大镜观察20°的角，角的度数不变仍是20°,原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "角的概念表示方法"
+                             ],
+        "source":  "2024·广东汕头·小升初真题"
+    },
+    {
+        "id":  "294",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "一个钝角减去一个锐角，可能会得到一个直角。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】钝角在90°-180°之间，锐角小于90°。选择合适的角度相减可能得到90°.\n【详解】例如150°-60°=90°,能得到直角，所以原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·河南信阳·期末"
+    },
+    {
+        "id":  "295",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "小华把一条长 3 厘米线段的一端延长 6 厘米，就得到了一条长 9 厘米的射线。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】把线段向一端无限延伸得到射线，射线长度无限，不能说\"长9厘米\"。把一端延长6厘米仍然是一条线段.\n【详解】延长后得到的是长9厘米的线段，不是射线。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "296",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "把一个 40° 的角放在一个 3 倍的放大镜下，看到的是 120°的角。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】放大镜只改变角两边的长度，不改变角的大小;角的大小与边长无关.\n【详解】40°角在放大镜下仍是40°,不是120°。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "角的初步认识及辨认"
+                             ],
+        "source":  "24-25 四年级上·河南信阳·期末"
+    },
+    {
+        "id":  "297",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "一条射线绕它的端点，旋转一周，形成的角叫做平角。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平角是180°;射线绕端点旋转一周形成的是周角360°.\n【详解】射线旋转一周形成周角，不是平角。原说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25四年级上·河南信阳·期末"
+    },
+    {
+        "id":  "298",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "角的大小和两边的长短无关，跟两边叉开的大小有关。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】角的大小由两条边张开的程度决定，与边的长短无关.\n【详解】两条边开叉越大角越大，开叉越小角越小。原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "角的大小比较"
+                             ],
+        "source":  "15-16 四年级·全国·期末"
+    },
+    {
+        "id":  "299",
+        "unit":  "第三章角的度数是",
+        "question_type":  "作图题",
+        "difficulty":  0.65,
+        "question":  "比平角少25°的角是( )°,画出这个角。",
+        "options":  [
+
+                    ],
+        "answer":  "155;图见详解",
+        "solution":  "【分析】平角是180°,比平角少25°用180°-25°求出角的度数，再用量角器画出该角.\n【详解】180°-25°=155°.\n\n![](images/9ab012d773489804f1b6a607210e7ac0e2194b35ff910357d4e2c3525cc0cdcc.jpg)",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "用量角器画角"
+                             ],
+        "source":  "24-25四年级上·安徽淮南·期末"
+    },
+    {
+        "id":  "300",
+        "unit":  "第三章角的度数是",
+        "question_type":  "作图题",
+        "difficulty":  0.65,
+        "question":  "用量角器画一个135°的角，再把它分成一个直角和一个锐角。",
+        "options":  [
+
+                    ],
+        "answer":  "见详解",
+        "solution":  "【分析】用量角器画135°:中心对准端点,0刻度线与一边重合，在135°处点点并连成另一边。再用三角板的直角边辅助，在135°角内作出一个90°角，则剩余部分为锐角.\n【详解】作图如下:\n\n![](images/058e1eba1c38d344f5695e72745866cac8f381d28aa4b15c5a46f0b294166327.jpg)",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "用量角器画角"
+                             ],
+        "source":  "24-25 四年级上·湖南岳阳·期末"
+    },
+    {
+        "id":  "301",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面各角中，不能用一副三角尺画出来的是( )。",
+        "options":  [
+                        "A. 15°",
+                        "B. 105°",
+                        "C. 125°"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】一副三角尺中各个角的度数是30°、60°、90°和45°。可以用45°-30°得到15°，用60°+45°得到105°。检查所有这些角的和与差，都不能得到125°。\n【详解】A：15°=45°-30°，可以画出；B：105°=60°+45°，可以画出；C：125°不能由30°、45°、60°、90°通过加减得到，不能画出。故答案为:C.",
+        "knowledge_points":  [
+                                 "用三角尺画角"
+                             ],
+        "source":  "24-25 四年级上·湖北孝感·期末"
+    },
+    {
+        "id":  "302",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "2024年巴黎奥运会中国体育代表团取得辉煌成绩，在跳水比赛中，某运动员做了一个\"向后翻转两周半\"的屈体动作，这名运动员一共转了( )。",
+        "options":  [
+                        "A. 360°",
+                        "B. 450°",
+                        "C. 720°",
+                        "D. 900°"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】两周半=2个周角+半个周角。一个周角是360°，2个周角是360°×2，半个周角是360°÷2。\n【详解】360°×2+360°÷2=720°+180°=900°。所以这名运动员一共转了900°。故答案为:D.",
+        "knowledge_points":  [
+                                 "因数中间或末尾是0的乘法",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25四年级上·福建龙岩·期末"
+    },
+    {
+        "id":  "303",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "用一副三角尺可以画出( )的角。",
+        "options":  [
+                        "A. 110°",
+                        "B. 70°",
+                        "C. 75°",
+                        "D. 125°"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】一副三角尺由两个三角尺组成，一个是30°、60°、90°，另一个是45°、45°、90°。用这些角度相加或相减来判断能否得到题目中的角。\n【详解】A：110°不能由这些角度拼出；B：70°也不能由这些角度拼出；C：75°=45°+30°，可以画出；D：125°不能拼出。故答案为:C.",
+        "knowledge_points":  [
+                                 "用三角尺画角"
+                             ],
+        "source":  "24-25四年级上·湖南永州·期末"
+    },
+    {
+        "id":  "304",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "用一副三角尺不能画出的角是( )。",
+        "options":  [
+                        "A. 15°",
+                        "B. 150°",
+                        "C. 75°",
+                        "D. 155°"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】一副三角尺提供30°、60°、90°和45°的角，通过加减可以拼出许多角度。\n【详解】15°=45°-30°，可以画出；150°=60°+90°，可以画出；75°=30°+45°，可以画出；155°无法由30°、45°、60°、90°通过加减得到，不能画出。故答案为:D.",
+        "knowledge_points":  [
+                                 "用三角尺画角"
+                             ],
+        "source":  "24-25四年级上·湖南常德·期末"
+    },
+    {
+        "id":  "305",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "从11时到12时，分针走过的角是( )。",
+        "options":  [
+                        "A. 直角",
+                        "B. 平角",
+                        "C. 锐角",
+                        "D. 周角"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】钟面有12个大格，每大格30°。从11时到12时，分针走一整圈，走过12个大格。\n【详解】30°×12=360°，分针走过的角是360°，即周角。故答案为:D.",
+        "knowledge_points":  [
+                                 "角的度量",
+                                 "直角，钝角，锐角的认识及特征 ",
+                                 "平角，周角的认识及特征",
+                                 "时，分的认识"
+                             ],
+        "source":  "24-25四年级上·湖南永州·期末"
+    },
+    {
+        "id":  "306",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "用一副三角尺不能画出下面( )的角。",
+        "options":  [
+                        "A. 75°",
+                        "B. 23°",
+                        "C. 150°"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】一副三角尺中角的度数为90°、45°、45°和90°、30°、60°。可以由它们拼出135°、75°、105°、120°、150°、15°、180°等角，但不能得到23°。\n【详解】75°=30°+45°，可以画出；150°=60°+90°，可以画出；23°无法由30°、45°、60°、90°拼出。故答案为:B.",
+        "knowledge_points":  [
+                                 "用三角尺画角"
+                             ],
+        "source":  "24-25 四年级上·重庆渝北·期末"
+    },
+    {
+        "id":  "307",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "钟面上从 6:00 到 6:25，分针转动形成的是( )。",
+        "options":  [
+                        "A. 锐角",
+                        "B. 直角",
+                        "C. 钝角"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】从6:00到6:25，分针从12转到5，走了5个大格。每个大格30°.\n【详解】5×30°=150°,150°大于90°小于180°,是钝角。故答案为:C.",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "一般时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·重庆渝北·期末"
+    },
+    {
+        "id":  "308",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "把 5 厘米长的线段向两端各延长 10 米，得到的是一条( )。",
+        "options":  [
+                        "A. 直线",
+                        "B. 线段",
+                        "C. 射线",
+                        "D. 角"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】直线没有端点，可以向两端无限延伸;射线只有一个端点，只能向一端无限延伸;线段有两个端点，长度有限，可以量出长度.\n【详解】把一条5厘米长的线段向两端各延长10米，仍然只有两个端点，是一条线段。故答案为:B.",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·重庆南岸·期末"
+    },
+    {
+        "id":  "309",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "钟面上分针与时针正好相互垂直的时刻是( )。",
+        "options":  [
+                        "A. 6:15",
+                        "B. 12:00",
+                        "C. 18:00",
+                        "D. 9:00"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】钟面上相邻两个数字间夹角为30°。6:15时角度大于90°,12:00时两针重合夹角为0°,18:00时夹角为180°。9:00时分针指向12，时针指向9，相差3个大格.\n【详解】3×30°=90°,是直角，两针垂直。故答案为:D.",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征",
+                                 "整点时间与钟面指针的位置",
+                                 "一般时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·重庆巴南·期末"
+    },
+    {
+        "id":  "310",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "华华在用量角器量角时，误把外圈刻度看成了内圈刻度，量出的度数是130°,那么这个角实际应该是( )。",
+        "options":  [
+                        "A. 50°",
+                        "B. 40°",
+                        "C. 30°",
+                        "D. 10°"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】量角器同一条刻度线上的内外圈刻度之和是180°。误读外圈130°,正确读数应为180°-130°.\n【详解】180°-130°=50°,所以这个角的实际度数是50°。故答案为:A.",
+        "knowledge_points":  [
+                                 "角的度量"
+                             ],
+        "source":  "24-25 四年级上·湖南娄底·期末"
+    },
+    {
+        "id":  "311",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "钟面上9时30分，时针与分针形成的较小的角是( )。",
+        "options":  [
+                        "A. 锐角",
+                        "B. 钝角",
+                        "C. 直角",
+                        "D. 平角"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】钟面有12个大格，每大格30°。9:30时，时针在9和10的中间，分针指向6，两针之间有3个半大格。\n【详解】30°×3+30°÷2=90°+15°=105°。105°大于90°小于180°，是钝角。故答案为:B.",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征",
+                                 "半点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25四年级上·湖南怀化·期末"
+    },
+    {
+        "id":  "312",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "学完\"角的度量\"单元后，请你判断下面四个同学说法，正确的是( )。",
+        "options":  [
+                        "A. 张红：\"因为射线只能向一端无限延伸，所以射线比直线短.\"",
+                        "B. 李明：\"我画了一条 15 厘米的射线.\"",
+                        "C. 王欢：\"平角的两边在一条直线上，没有顶点.\"",
+                        "D. 龙强：\"用能放大10倍的放大镜看一个50°的角，它还是50°.\""
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】直线没有端点，可以向两端无限延伸;射线有一个端点，只能向一端无限延伸，两者都无限长，不能比较长短;线段有两个端点，长度有限;角的大小与两边张开的大小有关，与边长无关.\n【详解】A:直线和射线都无限长，不能比较谁长谁短，错误;B:射线无限长，不能说\"15厘米的射线\"，错误;C:平角有顶点，只是两边在同一直线上，错误;D:放大镜只改变边长，不改变角度，用10倍放大镜看50°的角仍是50°,正确。故答案为:D.",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征",
+                                 "角的大小比较",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25四年级上·湖南湘西·期末"
+    },
+    {
+        "id":  "313",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "把一条长 5 厘米的线段向两端各延长 10 米，得到的是一条( )。",
+        "options":  [
+                        "A. 直线",
+                        "B. 射线",
+                        "C. 线段"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】直线无端点，可向两端无限延伸;射线有一个端点，只能向一端无限延伸;线段有两个端点，长度有限.\n【详解】延长后仍然有两个端点，得到的是一条更长的线段。故答案为:C.",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "314",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "用一副三角尺不可能拼出来的角是( )。",
+        "options":  [
+                        "A. 25°",
+                        "B. 75°",
+                        "C. 120°",
+                        "D. 150°"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】一副三角尺的角度有30°、60°、90°和45°。可以用30°+45°=75°，30°+90°=120°，60°+90°=150°。\n【详解】没有两个角的和或差等于25°，因此25°不可能拼出，其余角都可以拼出。故答案为:A.",
+        "knowledge_points":  [
+                                 "用三角尺画角"
+                             ],
+        "source":  "24-25 四年级上·山东济宁·期末"
+    },
+    {
+        "id":  "315",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "把一个钝角分成两个角，其中一个是锐角，另一个角( )。",
+        "options":  [
+                        "A. 一定是锐角",
+                        "B. 一定是钝角",
+                        "C. 一定是直角",
+                        "D. 可能还是锐角"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】钝角大于90°小于180°,锐角大于0°小于90°。把一个钝角分成两个角，其中一个是锐角，另一个角可以是锐角，直角或钝角，情况不唯一.\n【详解】如图示例：锐角+锐角，锐角+直角，锐角+钝角都可能得到原来的钝角，所以另一个角不确定是哪一类，可能还是锐角。故答案为:D.",
+        "knowledge_points":  [
+                                 "角的分类及换算"
+                             ],
+        "source":  "24-25 四年级上·天津河北·期末"
+    },
+    {
+        "id":  "316",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "夜幕下的天津津湾广场灯火辉煌，可以把光源发出的光线看成( )。",
+        "options":  [
+                        "A. 直线",
+                        "B. 射线",
+                        "C. 线段",
+                        "D. 曲线"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】线段有两个端点，长度有限;射线有一个端点，向一端无限延伸;直线没有端点，向两端无限延伸.\n【详解】光从灯光处向外发散，有一个光源端点，向一个方向延伸，符合射线的特点。故答案为:B.",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·天津滨海新·期末"
+    },
+    {
+        "id":  "317",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "钟面上 4 时整，时针和分针形成的角(较小的)是( )°。",
+        "options":  [
+                        "A. 90",
+                        "B. 60",
+                        "C. 120"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】4时整，分针指12，时针指4。钟面有12个大格，每大格30°，两针之间有4个大格。\n【详解】30°×4=120°，所以较小的角是120°。故答案为:C.",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "318",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "用一副三角尺，不能拼出的角是( )。",
+        "options":  [
+                        "A. 180°",
+                        "B. 80°",
+                        "C. 75°",
+                        "D. 105°"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】一副三角尺有30°,60°,90°和45°四种角度。180°=90°+90°,75°=30°+45°,105°=60°+45°,都能拼成.\n【详解】80°不能由30°,45°,60°,90°通过加减得到，用一副三角尺不能拼成80°的角。故答案为:B.",
+        "knowledge_points":  [
+                                 "用三角尺画角"
+                             ],
+        "source":  "24-25 四年级上·广东东莞·期末"
+    },
+    {
+        "id":  "319",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "过同一个平面上的两个点，能画直线的条数是( )条。",
+        "options":  [
+                        "A. 0",
+                        "B. 1",
+                        "C. 2",
+                        "D. 无数"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】几何中\"过两点只能画一条直线\"，叫作两点确定一条直线.\n【详解】过同一个平面上的两个点，只能画一条直线。故答案为:B.",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25四年级上·广东东莞·期末"
+    },
+    {
+        "id":  "320",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "小明早上在校时间是 9 时至 12 时 30 分，钟面上时针旋转形成的角是( )。",
+        "options":  [
+                        "A. 锐角",
+                        "B. 直角",
+                        "C. 钝角",
+                        "D. 平角"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】钟面上相邻两个数字间的夹角是30°。从9时到12时，时针转过3个大格；从12时到12时30分，再转半个大格。\n【详解】9时到12时：3×30°=90°；再加上半个大格15°，共90°+15°=105°，105°是钝角。故答案为:C.",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "角的分类及换算",
+                                 "整点时间与钟面指针的位置",
+                                 "半点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·浙江温州·期末"
+    },
+    {
+        "id":  "321",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面角的度数，不能用一副三角尺拼成的是( )。",
+        "options":  [
+                        "A. 75°",
+                        "B. 105°",
+                        "C. 160°",
+                        "D. 150°"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】一副三角尺的角度为30°,45°,60°,90°。可以用这些角做加减运算.\n【详解】75°=30°+45°;105°=60°+45°;150°=90°+60°都可拼成;160°不能由30°,45°,60°,90°通过加减得到。故答案为:C.",
+        "knowledge_points":  [
+                                 "用三角尺画角"
+                             ],
+        "source":  "24-25 四年级上·浙江杭州·期末"
+    },
+    {
+        "id":  "322",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面三个角中，不能用一副三角尺画出的是( )的角。",
+        "options":  [
+                        "A. 135°",
+                        "B. 65°",
+                        "C. 15°"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】一副三角尺由30°,60°,90°和45°,90°,45°组成.\n【详解】135°=45°+90°,可以画出;15°=60°-45°,可以画出;65°不能由30°,45°,60°,90°通过加减得到。故答案为:B.",
+        "knowledge_points":  [
+                                 "用三角尺画角"
+                             ],
+        "source":  "24-25 四年级上·浙江杭州·期末"
+    },
+    {
+        "id":  "323",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "9:30时，钟面上分针和时针组成的较小角是( )。",
+        "options":  [
+                        "A. 钝角",
+                        "B. 直角",
+                        "C. 锐角",
+                        "D. 平角"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】9:30时，分针指6，时针在9和10中间，两针之间有3个半大格，每大格30°。\n【详解】30°×3+30°÷2=90°+15°=105°，105°是钝角。故答案为:A.",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征",
+                                 "一般时间与钟面指针的位置"
+                             ],
+        "source":  "24-25四年级上·湖北黄冈·期末"
+    },
+    {
+        "id":  "324",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "用一副三角尺不能拼成的角是( )。",
+        "options":  [
+                        "A. 120°",
+                        "B. 130°",
+                        "C. 75°",
+                        "D. 135°"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】一副三角尺有90°、45°、45°和90°、30°、60°。120°可以用90°+30°，75°可以用30°+45°，135°可以用90°+45°。\n【详解】130°不能由30°、45°、60°、90°通过加减得到，因此不能拼成130°的角。故答案为:B.",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "用三角尺画角"
+                             ],
+        "source":  "24-25 四年级上·湖北襄阳·期末"
+    },
+    {
+        "id":  "325",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "\"有始有终\"指做事能够坚持到底，在数学上符合\"有始有终\"的是( )。",
+        "options":  [
+                        "A. 直线",
+                        "B. 射线",
+                        "C. 线段"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】直线没有端点，有始无终;射线有一个端点，有始无终;线段有两个端点，既有开始又有结束.\n【详解】线段有两个端点，符合\"有始有终\"的含义。故答案为:C.",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25四年级上·江西吉安·期末"
+    },
+    {
+        "id":  "326",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "钟面上 6:00，时针和分针形成( )角；6:45 时针和分针形成( )角。正确答案是( )。",
+        "options":  [
+                        "A. 直，平",
+                        "B. 平，直",
+                        "C. 平，锐"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】6:00时，分针指12，时针指6，两针之间有6个大格，6×30°=180°，是平角；6:45时，时针在6和7之间，分针指9，两针之间不足3个大格，夹角小于90°，是锐角。\n【详解】所以6:00是平角，6:45是锐角，对应\"平，锐\"。故答案为:C.",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征",
+                                 "整点时间与钟面指针的位置",
+                                 "一般时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·广东广州·期末"
+    },
+    {
+        "id":  "327",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "小明同学画了一条4厘米长的( )。",
+        "options":  [
+                        "A. 直线",
+                        "B. 射线",
+                        "C. 线段",
+                        "D. 角"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】直线和射线都是无限长，不能用长度来表示;线段有两个端点，可以用长度描述;角的大小与边长无关，不能用\"厘米\"表示.\n【详解】\"4厘米长\"只能描述线段的长度，所以是画了一条4厘米长的线段。故答案为:C.",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征",
+                                 "角的概念及表示方法"
+                             ],
+        "source":  "24-25四年级上·山东菏泽·期末"
+    },
+    {
+        "id":  "328",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面说法中，正确的是( )。",
+        "options":  [
+                        "A. 过一点可以画无数条射线",
+                        "B. 直线的长度是射线的2倍",
+                        "C. 角的两边越长，这个角越大",
+                        "D. 不能画出50米的线段"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】过一点可以向任意方向画射线，条数没有限制;直线和射线都无限长，不能比较长度;角的大小与两边张开的大小有关，与边长无关;线段有两个端点，可以任意设定长度.\n【详解】A项正确，其余说法都与定义不符。故答案为:A.",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征",
+                                 "角的大小比较"
+                             ],
+        "source":  "24-25 四年级上·广东广州·期末"
+    },
+    {
+        "id":  "329",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "张叔叔每天早上 6时开始晨跑，此时钟面上的时针和分针形成的夹角是( )。",
+        "options":  [
+                        "A. 锐角",
+                        "B. 直角",
+                        "C. 钝角",
+                        "D. 平角"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】6时整，分针指12，时针指6，两针之间有6个大格，每大格30°.\n【详解】6×30°=180°,是平角。故答案为:D.",
+        "knowledge_points":  [
+                                 "整十，整百，整千数与一位数的乘法",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·广东广州·期末"
+    },
+    {
+        "id":  "330",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "角的两条边是两条( )。",
+        "options":  [
+                        "A. 射线",
+                        "B. 直线",
+                        "C. 线段"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】从一点引出两条射线所组成的图形叫做角，这一点是顶点，两条射线是角的两边.\n【详解】所以角的两条边是射线。故答案为:A.",
+        "knowledge_points":  [
+                                 "角的概念及表示方法"
+                             ],
+        "source":  "24-25 四年级上·河北沧州·期末"
+    },
+    {
+        "id":  "331",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "钟面上 9:30，时针与分针所成的角是( )。",
+        "options":  [
+                        "A. 锐角",
+                        "B. 钝角",
+                        "C. 直角",
+                        "D. 平角"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】9:30时，分针指6，时针在9和10中间，两针所成的较小角大于90°小于180°.\n【详解】如图所示，该角大于直角小于平角，是钝角。故答案为:B.",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "半点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·辽宁抚顺·期末"
+    },
+    {
+        "id":  "332",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "用一副三角尺画不出下面的哪个角。( )",
+        "options":  [
+                        "A. 75°",
+                        "B. 80°",
+                        "C. 120°",
+                        "D. 165°"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】一副三角尺中的各个角的度数是30°、45°、60°、90°。它们可以组合出75°、120°、165°等角度。\n【详解】75°=30°+45°可以画出；120°=30°+90°可以画出；165°=30°+90°+45°可以画出；80°不能由30°、45°、60°、90°通过加减得到。故答案为:B.",
+        "knowledge_points":  [
+                                 "用三角尺画角"
+                             ],
+        "source":  "24-25 四年级上·辽宁鞍山·期末"
+    },
+    {
+        "id":  "333",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "把一个平角分成两个角，其中一个角是锐角，另一个角( )。",
+        "options":  [
+                        "A. 可能是锐角，也可能是钝角",
+                        "B. 可能是直角",
+                        "C. 一定是钝角",
+                        "D. 不确定"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】平角是180°，锐角小于90°。把一个平角分成两个角，其中一个是锐角，另一个角=180°-锐角度数，一定大于90°小于180°。\n【详解】180°减去一个小于90°的角，结果必然是大于90°小于180°的角，即钝角，所以另一个角一定是钝角。故答案为:C.",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "23-24 四年级上·山东临沂·期末"
+    },
+    {
+        "id":  "334",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "将一张圆形纸片对折三次后，折成的角是( )。",
+        "options":  [
+                        "A. 135°",
+                        "B. 90°",
+                        "C. 45°"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】周角为360°。对折1次得到的角为360°÷2=180°；对折2次得到的角为180°÷2=90°；对折3次得到的角为90°÷2=45°。\n【详解】360°÷2=180°；180°÷2=90°；90°÷2=45°。所以将一张圆形纸片对折3次以后所形成的角是45°。故答案为:C.",
+        "knowledge_points":  [
+                                 "商末尾有 0 的除法",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "17-18 四年级上·四川德阳·期末"
+    },
+    {
+        "id":  "335",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "\"有始有终\"的意思是有开头也有结尾，指做事能坚持到底。在数学上可以用这个成语表示我们本学期所学的( )的特征。",
+        "options":  [
+                        "A. 线段",
+                        "B. 直线",
+                        "C. 射线",
+                        "D. 角"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】线段有两个端点，可以测量长度;直线没有端点，可以向两端无限延伸;射线只有一个端点，可以向一端无限延伸.\"有开头也有结尾\"对应\"有两个端点\".\n【详解】线段有两个端点，既有开始又有结束，符合\"有始有终\"的特征。故答案为:A.",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·福建福州·期中"
+    },
+    {
+        "id":  "336",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "用一个 30 倍的放大镜看一个30°的角，这个角是( )。",
+        "options":  [
+                        "A. 10°",
+                        "B. 30°",
+                        "C. 90°"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】角的大小与两边叉开的大小有关，与边的长短无关。放大镜只会改变角两边的长短，不会改变角的大小.\n【详解】用一个30倍的放大镜看一个30°的角，这个角仍然是30°。故答案为:B.",
+        "knowledge_points":  [
+                                 "角的大小比较"
+                             ],
+        "source":  "24-25 四年级上·河北邯郸·期末"
+    },
+    {
+        "id":  "337",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "用一副三角尺，不能拼出( )的角。",
+        "options":  [
+                        "A. 15°",
+                        "B. 20°",
+                        "C. 150°"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】一副三角尺中角的度数为90°,45°,45°和90°,30°,60°。两个角可以相加或相减来拼角.\n【详解】15°=45°-30°,可以拼出;150°=90°+60°,可以拼出;20°不能由30°,45°,60°,90°通过加减得到，因此不能拼出。故答案为:B.",
+        "knowledge_points":  [
+                                 "用三角尺画角"
+                             ],
+        "source":  "24-25 四年级上·河北邯郸·期末"
+    },
+    {
+        "id":  "338",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "成语\"有始有终\"的意思是有开头也有结尾，指做事能坚持到底。在数学上可以用这个成语表示我们本学期所学的( )的特征。",
+        "options":  [
+                        "A. 线段",
+                        "B. 直线",
+                        "C. 射线"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】线段有两个端点，可以测量长度;直线没有端点;射线只有一个端点.\"有开头也有结尾\"表示有两个端点.\n【详解】线段既有起点又有终点，符合\"有始有终\"的含义。故答案为:A.",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·黑龙江齐齐哈尔·期末"
+    },
+    {
+        "id":  "339",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面能量出长度的是( )。",
+        "options":  [
+                        "A. 直线",
+                        "B. 射线",
+                        "C. 线段"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】直线没有端点，可以向两端无限延伸，长度不能量出;射线只有一个端点，也可以无限延伸，长度不能量出;线段有两个端点，长度有限，可以测量.\n【详解】线段有两个端点，而两个端点间的距离就是线段的长度，能量出长度的是线段。故答案为:C.",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "23-24 四年级上·内蒙古赤峰·期末"
+    },
+    {
+        "id":  "340",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列时刻中，时针和分针成30°角的是( )。",
+        "options":  [
+                        "A. 11 时",
+                        "B. 9 时",
+                        "C. 3 时"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】周角为360°，钟面被12个数字平均分成12个大格，每个大格是360°÷12=30°。11时整分针指12，时针指11，两针相隔1个大格；9时或3时整两针相隔3个大格，为90°。\n【详解】11时：1×30°=30°，符合题意；9时：3×30°=90°，不符合；3时：3×30°=90°，不符合。故答案为:A.",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "平角，周角的认识及特征",
+                                 "整点时间与钟面指针的位置",
+                                 "半点时间与钟面指针的位置"
+                             ],
+        "source":  "23-24四年级上·内蒙古鄂尔多斯·期末"
+    },
+    {
+        "id":  "341",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "射线有( )个端点,( )有 2 个端点。",
+        "options":  [
+
+                    ],
+        "answer":  "1;线段",
+        "solution":  "【分析】直线没有端点，可以向两边无限延伸;射线只有一个端点，可以向一边无限延伸;线段有2个端点，有确定的长度.\n【详解】射线有1个端点，线段有2个端点。",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25 六年级下·海南海口·单元测试"
+    },
+    {
+        "id":  "342",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "从 8:00 到 8:15，分针转动( );从 4:00 到 9:00，时针转动( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "90°或90度;150°或150度",
+        "solution":  "【分析】钟面被分成12个大格，每个大格是30°。从8:00到8:15，分针走3大格；从4:00到9:00，时针从4走到9，共5大格。\n【详解】30°×3=90°；30°×5=150°。所以分针转动90°，时针转动150°。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "整点时间与钟面指针的位置",
+                                 "一般时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·重庆大渡口·期末"
+    },
+    {
+        "id":  "343",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "3时整，时针和分针所成的角是( )角;7时整，时针和分针所成的角是( )角;11时整，时针和分针所成的角是( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "直;钝;锐",
+        "solution":  "【分析】钟面12大格，每大格30°。3时整夹角3×30°=90°，是直角；7时整夹角5×30°=150°，是钝角；11时整夹角1×30°=30°，是锐角。\n【详解】3时整为直角；7时整为钝角；11时整为锐角。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25六年级下·海南海口·单元测试"
+    },
+    {
+        "id":  "344",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "过两点可以画( )条直线，过一点可以画( )条直线。从一点可引出( )条射线。",
+        "options":  [
+
+                    ],
+        "answer":  "一;无数;无数",
+        "solution":  "【分析】两点确定一条直线;过一点可以向不同方向画无数条直线;从一点可向不同方向引出无数条射线.\n【详解】过两点可以画一条直线;过一点可以画无数条直线;从一点可引出无数条射线。",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25六年级下·海南海口·单元测试"
+    },
+    {
+        "id":  "345",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "上午 9 时，钟面上的时针和分针成( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "直",
+        "solution":  "【分析】9时整，分针指12，时针指9，两针之间有3个大格，每大格30°。\n【详解】30°×3=90°，是直角。所以上午9时，时针和分针成直角。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·湖南永州·期末"
+    },
+    {
+        "id":  "346",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "3时30分，钟面上的时针和分针成( )角。(填\"锐\"\"直\"或\"钝\")",
+        "options":  [
+
+                    ],
+        "answer":  "锐",
+        "solution":  "【分析】3:30时，时针在3和4之间，分针指向6，较小角小于90°,为锐角.\n【详解】锐角是大于0°且小于90°的角。3时30分时针和分针组成的较小角是锐角。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "半点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25四年级上·湖南益阳·期末"
+    },
+    {
+        "id":  "347",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "李明用量角器量一个角时，角的顶点与量角器的中心重合，角的一边与量角器内圈的30°刻度线重合，另一条边与内圈85°刻度线重合，这个角是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "55",
+        "solution":  "【分析】角的一边在30°刻度线，另一边在85°刻度线，角的度数等于两读数之差.\n【详解】85°-30°=55°。所以这个角是55°。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "角的度量"
+                             ],
+        "source":  "24-25四年级上·重庆渝中·期末"
+    },
+    {
+        "id":  "348",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在钟面上,9 时整分针和时针所组成较小的夹角是( )。时针从9时转动到11时，时针转动了( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "直角或90°或90度;60",
+        "solution":  "【分析】钟面每大格30°。9时整时针和分针之间有3大格；从9时到11时，时针走2大格。\n【详解】30°×3=90°，是直角；30°×2=60°，所以时针从9时到11时转动了60°。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "角的分类及换算",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·重庆永川·期末"
+    },
+    {
+        "id":  "349",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "8 平方千米=( )公顷 1 周角=( )直角\n\n11000000 平方米=( )公顷=( )平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "800;4;1100;11",
+        "solution":  "【分析】1平方千米=100公顷。周角360°，直角90°。1公顷=10000平方米。\n【详解】8平方千米=8×100=800公顷；360°÷90°=4，所以1周角=4直角；11000000平方米÷10000=1100公顷；1100÷100=11平方千米。因此答案为:800;4;1100;11。",
+        "knowledge_points":  [
+                                 "角的分类及换算",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·山东临沂·期末"
+    },
+    {
+        "id":  "350",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "1 平角=( )周角=( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "1/2或半;180",
+        "solution":  "【分析】平角是180°，周角是360°。\n【详解】1平角=1/2周角=180°。",
+        "knowledge_points":  [
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·湖南长沙·期末"
+    },
+    {
+        "id":  "351",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "平角比220°小( )°，平角的一半是( )角。过一点可以画( )条直线，经过两点可以画( )条直线。线段有( )个端点。",
+        "options":  [
+
+                    ],
+        "answer":  "40;直;无数;一或1;两或2",
+        "solution":  "【分析】平角是180°，直角是90°。比较平角与220°，再根据直线与线段的定义判断直线条数和端点个数。\n【详解】220°-180°=40°，所以平角比220°小40°；180°÷2=90°，平角的一半是直角；过一点可以画无数条直线，经过两点可以画一条直线；线段有两个端点。",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征",
+                                 "平角，周角的认识及特征",
+                                 "角的分类及换算"
+                             ],
+        "source":  "24-25 四年级上·湖南株洲·期末"
+    },
+    {
+        "id":  "352",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "钟面上( )时整，时针和分针成平角;( )时整或( )时整，钟面上的时针和分针所夹的较小角是30°;钟面上9时整，时针和分针所夹的较小角是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "6;11;1;90",
+        "solution":  "【分析】钟面12大格，每大格30°。6时整两针相差6大格，为180°平角;11时整或1时整，两针相差1大格，为30°;9时整两针相差3大格，为90°.\n【详解】6时整成平角;11时整或1时整较小角为30°;9时整较小角为90°。",
+        "knowledge_points":  [
+                                 "角的度量",
+                                 "角的分类及换算",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·湖南张家界·期末"
+    },
+    {
+        "id":  "353",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "钟面上 3 时 30 分，时针和分针所夹的较小角是( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "锐",
+        "solution":  "【分析】3:30时，分针指6，时针在3和4之间，两针相差2个半大格。每大格30°，半格15°。\n【详解】30°÷2=15°；2×30°=60°；60°+15°=75°。75°大于0°小于90°，是锐角。所以较小角是锐角。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征",
+                                 "半点时间与钟面指针的位置"
+                             ],
+        "source":  "18-19 四年级上·山东德州·期中"
+    },
+    {
+        "id":  "354",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "钟面上，当时针从\"9\"走到\"10\"的时候，时针转动了( )°,分针转动形成的角是( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "30;周",
+        "solution":  "【分析】钟面有12大格，一圈360°，每大格30°。时针从9到10走了1大格；这一小时内分针转一整圈，是360°周角。\n【详解】360°÷12=30°，所以时针转动30°；分针转动形成360°，是周角。",
+        "knowledge_points":  [
+                                 "平角，周角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "355",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一副三角尺能拼成的最大的角是( )°,这是一个( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "180;平",
+        "solution":  "【分析】一副三角尺由两个三角尺组成，各有一个直角90°。要拼出最大的角，就把两个直角拼在一起.\n【详解】90°+90°=180°,是平角。因此一副三角尺能拼成的最大的角是180°,是平角。",
+        "knowledge_points":  [
+                                 "平角，周角的认识及特征",
+                                 "用三角尺画角"
+                             ],
+        "source":  "24-25 四年级上·湖南长沙·期末"
+    },
+    {
+        "id":  "356",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "钟面上,3时整，时针与分针形成的最小角是( )角，再过2个小时，时针与分针形成的最小角是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "直;150",
+        "solution":  "【分析】3时整时针指3，分针指12，两针相差3大格为90°直角。再过2小时是5时整，此时两针相差5大格.\n【详解】3×30°=90°,为直角;5×30°=150°,所以再过2个小时，最小角是150°。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "角的分类及换算",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25四年级上·湖南张家界·期末"
+    },
+    {
+        "id":  "357",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "∠1 加上一个60°的角是一个平角,∠1 是( )°;把一张圆形纸片对折( )次可以得到45°的角。",
+        "options":  [
+
+                    ],
+        "answer":  "120;3",
+        "solution":  "【分析】平角是180°,∠1+60°=180°,所以∠1=180°-60°。圆形纸片对折1次得到180°,再对折得到90°,再对折得到45°.\n【详解】180°-60°=120°,所以∠1是120°;对折3次可以得到45°的角。",
+        "knowledge_points":  [
+                                 "整十，整百，整千数除以一位数的除法",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25四年级上·山东济宁·期末"
+    },
+    {
+        "id":  "358",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "钟面2时整，时针和分针组成的角是( )角;( )时整，时针和分针组成的角是平角。",
+        "options":  [
+
+                    ],
+        "answer":  "锐;6",
+        "solution":  "【分析】2时整两针相差2大格，2×30°=60°，小于90°是锐角；6时整两针相差6大格，6×30°=180°，是平角。\n【详解】钟面2时整，时针和分针组成的角是锐角；6时整，时针和分针组成的角是平角。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征",
+                                 "整点时间与钟面指针的位置",
+                                 "一般时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·山东济宁·期末"
+    },
+    {
+        "id":  "359",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据角的度数大小，将角分为不同种类。下午3:00的时候，时针和分针的夹角是( )角，从下午3:00到4:00，分针转动形成的角是( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "直;周",
+        "solution":  "【分析】下午3:00时，两针相差3大格，3×30°=90°，为直角；3:00到4:00分针转一圈360°，为周角。\n【详解】下午3:00的时候，时针和分针的夹角是直角；从3:00到4:00，分针转一圈形成周角。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·天津滨海新·期末"
+    },
+    {
+        "id":  "360",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "比平角小40°的角是( )°,它比直角大( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "140;50",
+        "solution":  "【分析】平角是180°,直角是90°。比平角小40°的角度先用180°-40°,再和直角比较.\n【详解】180°-40°=140°;140°-90°=50°。所以这个角的度数是140°,它比直角大50°。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "角的分类及换算"
+                             ],
+        "source":  "24-25 四年级上·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "361",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把一条5厘米的线段向两端无限延长，得到一条( );把线段的一端无限延长，得到一条( )。",
+        "options":  [
+
+                    ],
+        "answer":  "直线;射线",
+        "solution":  "【分析】直线两端都没有端点，可以向两端无限延长;射线只有一个端点，可以向一端无限延长.\n【详解】把一条5厘米的线段向两端无限延长，得到一条直线;把线段的一端无限延长，得到一条射线。",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·天津北辰·期末"
+    },
+    {
+        "id":  "362",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一条射线绕它的端点旋转一周，形成的角是( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "周",
+        "solution":  "【分析】一条射线绕着端点旋转一周回到起始位置，形成的角叫周角，度数为360°.\n【详解】一条射线绕它的端点旋转一周，形成的角是周角。",
+        "knowledge_points":  [
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25四年级上·广东广州·期末"
+    },
+    {
+        "id":  "363",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "2时整，时针与分针成的角是( )度,4 时 30 分，时针与分针成的角是( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "60;锐",
+        "solution":  "【分析】2时整两针相差2大格,2×30°=60°;4:30时，分针指6，时针在4和5之间，两针之间小于3大格，夹角小于90°,是锐角.\n【详解】2时整，时针与分针成的角是60°;4时30分，时针与分针成的角是锐角。",
+        "knowledge_points":  [
+                                 "两位数与一位数的乘法口算",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25四年级上·浙江杭州·期末"
+    },
+    {
+        "id":  "364",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "有四个角的度数分别是89°,108°,150°,192°,其中锐角有( )个，钝角有( )个。",
+        "options":  [
+
+                    ],
+        "answer":  "1;2",
+        "solution":  "【分析】锐角：小于90°;直角：等于90°;钝角：大于90°小于180°;大于等于180°的是平角或反身角.\n【详解】89°是锐角，共1个;108°和150°是钝角，共2个;192°大于180°,不是锐角或钝角。故锐角1个，钝角2个。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·浙江杭州·期末"
+    },
+    {
+        "id":  "365",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "两个锐角可以拼成一个什么角。请将所有可能的情况举例说明。",
+        "options":  [
+
+                    ],
+        "answer":  "锐角、直角或钝角",
+        "solution":  "【分析】锐角小于90°。两个锐角的和可能小于90°，等于90°或大于90°但小于180°。\n【详解】例如：42°+42°=84°，是锐角；45°+45°=90°，是直角；45°+60°=105°，是钝角。所以两个锐角可能拼成一个锐角、一个直角或一个钝角(举例不唯一)。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "直角，钝角，锐角的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·浙江·期末"
+    },
+    {
+        "id":  "366",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "上午9:30，时针和分针组成的角是( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "钝",
+        "solution":  "【分析】9:30时分针指6，时针在9和10中间，两针之间有3个半大格：3×30°+30°÷2=90°+15°=105°，大于90°小于180°为钝角。\n【详解】105°是钝角，所以上午9:30，时针和分针组成的角是钝角。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "一般时间与钟面指针的位置"
+                             ],
+        "source":  "24-25四年级上·湖北武汉·期末"
+    },
+    {
+        "id":  "367",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "钟表上3:00到4:00，分针转了( )°,时针转了( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "360;30",
+        "solution":  "【分析】从3:00到4:00是1小时，分针转一圈是360°,时针走1大格是30°.\n【详解】钟表上3:00到4:00，分针转了360°,时针转了30°。",
+        "knowledge_points":  [
+                                 "角的度量",
+                                 "平角，周角的认识及特征",
+                                 "时，分的认识",
+                                 "时，分，秒有关的计算"
+                             ],
+        "source":  "24-25四年级上·浙江宁波·期末"
+    },
+    {
+        "id":  "368",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在下面的括号里填上\">\"\"<\"或\"=\"。\n\n804000( )800400 3 亿( )300000 万\n\n450000 平方米( )405 公顷 2 平角( )1 周角",
+        "options":  [
+
+                    ],
+        "answer":  ">;<;<;=",
+        "solution":  "【分析】(1)大数的比较：比较两个大数时，首先比较它们的位数。位数越多的数越大，当两个数位数相同的时候，就从最高位开始比较，如果最高位上的数字不同，数字大的那个数就大，若最高位上的数字相同，就接着比较下位数字的大小，依次类推，直到比较出大小为止。(2)1亿=10000万，所以3亿=30000万，再与300000万进行比较即可。(3)1公顷=10000平方米，450000平方米=45公顷，再与405公顷进行比较即可。(4)周角为360°，平角=180°，2平角=360°，所以2平角=1周角。\n【详解】804000>800400；3亿=30000万，30000万<300000万，所以3亿<300000万；450000平方米=45公顷，45公顷<405公顷，所以450000平方米<405公顷；2平角=1周角。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "一亿有多大",
+                                 "平角",
+                                 "周角的认识及特征",
+                                 "公顷",
+                                 "平方千米的进率与换算"
+                             ],
+        "source":  "24-25四年级上·湖北武汉·期末"
+    },
+    {
+        "id":  "369",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "钟面上 9 时整，时针和分针形成的较小的角是( )角;3 时 30 分时针和分针所成的较小的角是( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "直;锐",
+        "solution":  "【分析】9时整，两针相差3大格,3×30°=90°,是直角;3:30时，两针相差2个半大格，较小角为75°,是锐角.\n【详解】9时整时针和分针形成的较小角是直角;3时30分时针和分针所成的较小角是锐角。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·湖北襄阳·期末"
+    },
+    {
+        "id":  "370",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈看了一下钟表上的时间是8:30，此时钟表上的时针和分针的夹角(较小角)的度数是( )°,再添上( )°就能成为一个平角。",
+        "options":  [
+
+                    ],
+        "answer":  "75;105",
+        "solution":  "【分析】8:30时分针指向6，时针指在8和9的中间，钟面上一大格是30°,2个大格又多半格为30×2+30÷2=60+15=75°。平角是180°,用180°减75°可得再添上的度数.\n【详解】30×2+30÷2=75°;180°-75°=105°。所以较小角是75°,再添上105°就能成为一个平角。",
+        "knowledge_points":  [
+                                 "角的度量",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征",
+                                 "一般时间与钟面指针的位置"
+                             ],
+        "source":  "24-25四年级上·江西吉安·期末"
+    },
+    {
+        "id":  "371",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一副三角尺，一个最大的角和一个最小的角相差( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "60",
+        "solution":  "【分析】一副三角尺中最大角是90°,最小角是30°,用最大角减最小角即可.\n【详解】90°-30°=60°。所以相差60°。",
+        "knowledge_points":  [
+                                 "角度的计算"
+                             ],
+        "source":  "24-25四年级上·浙江宁波·期末"
+    },
+    {
+        "id":  "372",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "钟面上 3 时整，时针和分针所成的角是( )度,6时整时针和分针所成的角是( )度。",
+        "options":  [
+
+                    ],
+        "answer":  "90;180",
+        "solution":  "【分析】钟面每大格30°。3时整两针相差3大格，6时整相差6大格。\n【详解】30°×3=90°；30°×6=180°。所以分别是90度和180度。",
+        "knowledge_points":  [
+                                 "角的度量",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "23-24 四年级上·山东济宁·期末"
+    },
+    {
+        "id":  "373",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果一个角与55°的角合成一个平角，则这个角是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "125",
+        "solution":  "【分析】平角是180°，用180°减去55°得到所求角的度数。\n【详解】180°-55°=125°。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25四年级上·湖北孝感·期末"
+    },
+    {
+        "id":  "374",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "钟面上,9时整，分针与时针所成的角是( )°;6时整，分针与时针所成的角是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "90;180",
+        "solution":  "【分析】钟面每大格30°。9时整相差3大格;6时整相差6大格.\n【详解】3×30°=90°;6×30°=180°。所以分别是90°和180°。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25四年级上·湖南岳阳·期末"
+    },
+    {
+        "id":  "375",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "1 周角=( )平角=( )直角=( )。",
+        "options":  [
+
+                    ],
+        "answer":  "2;4;360°或360度",
+        "solution":  "【分析】周角是360°，平角是180°，直角是90°。360°=2×180°=4×90°。\n【详解】1周角=2平角=4直角=360°。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "23-24 四年级上·广东河源·期末"
+    },
+    {
+        "id":  "376",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "小飞用量角器量角时，错把内圈刻度看成外圈刻度，读出的度数是70°,这个角实际的度数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "110°",
+        "solution":  "【分析】量角器同一刻度线内外圈读数之和为180°。错读为70°,正确应为180°-70°.\n【详解】180°-70°=110°。",
+        "knowledge_points":  [
+                                 "角的度量"
+                             ],
+        "source":  "24-25四年级上·河北邯郸·期末"
+    },
+    {
+        "id":  "377",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在( )里填\">\"\"<\"或\"=\"。\n\n896370000( )863970000 直角( )钝角",
+        "options":  [
+
+                    ],
+        "answer":  ">;<",
+        "solution":  "【分析】比较整数大小：位数相同从最高位比起。直角为90°，钝角大于90°小于180°。\n【详解】896370000>863970000；直角<钝角。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "直角，钝角，锐角的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "378",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "《新闻联播》每晚7:00开播，一般7:30结束，结束时，钟面上的时针和分针所形成的较小的角是( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "锐",
+        "solution":  "【分析】钟面每大格30°。7:30时分针指6，时针在7和8之间，两针间隔1个半大格，夹角为30°+15°=45°，小于90°为锐角。\n【详解】30°+30°÷2=45°，45°<90°，所以较小角是锐角。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "一般时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "379",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个直角、35°的角和∠1合成一个平角，则∠1=( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "55",
+        "solution":  "【分析】∠1=平角-直角-35°，平角180°，直角90°。\n【详解】180°-90°-35°=55°。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25四年级上·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "380",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "40000 平方米=( )公顷\n\n5 平方千米=( )公顷\n\n1 周角=( )°=( )平角=( )直角",
+        "options":  [
+
+                    ],
+        "answer":  "4;500;360;2;4",
+        "solution":  "【分析】1公顷=10000平方米;1平方千米=100公顷;周角360°,平角180°,直角90°.\n【详解】40000平方米=4公顷;5平方千米=500公顷;1周角=360°=2平角=4直角。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "平角，周角的认识及特征",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "商末尾有0的除法"
+                             ],
+        "source":  "24-25 四年级上·新疆喀什·期末"
+    },
+    {
+        "id":  "381",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "春节前小明准备剪一朵美丽的花，他先将圆形纸片对折三次，再裁剪，对折后形成的角的度数是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "45",
+        "solution":  "【分析】圆心角360°对折三次相当于平均分成8份，每份角度为360°÷8。\n【详解】360°÷8=45°。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "图形的折叠问题",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·内蒙古呼伦贝尔·期末"
+    },
+    {
+        "id":  "382",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "已知一个角是130°,它比平角小( )度，比直角大( )度。",
+        "options":  [
+
+                    ],
+        "answer":  "50;40",
+        "solution":  "【分析】平角180°，直角90°。分别用180°-130°与130°-90°计算。\n【详解】180°-130°=50°；130°-90°=40°。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "23-24 四年级上·内蒙古赤峰·期末"
+    },
+    {
+        "id":  "383",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "队列练习时，原地向右转，转过一个( )°的角;原地向后转，转过一个( )°的角。",
+        "options":  [
+
+                    ],
+        "answer":  "90;180",
+        "solution":  "【分析】原地向右转形成直角90°;原地向后转形成平角180°.\n【详解】如图:\n\n![](images/54ff066344f6197728d90d40b04cf392b84648253816e7ac093fe4b095abcd4d.jpg)\n\n![](images/8c932a39b27eae9a2a90f68f42bcacb10d1273d2b9088713db356698787d3f36.jpg)\n\n所以向右转是90°,向后转是180°。",
+        "knowledge_points":  [
+                                 "平角，周角的认识及特征",
+                                 "角的分类及换算"
+                             ],
+        "source":  "24-25 四年级上·内蒙古鄂尔多斯·期末"
+    },
+    {
+        "id":  "384",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "张明测量出一条长度是 10cm 的直线。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】直线没有端点，可以向两端无限延伸，长度无法测量.\n【详解】直线无限长，不可能测量出直线的长度。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·重庆渝北·期末"
+    },
+    {
+        "id":  "385",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "把 5 厘米长的线段的一端延长 5000 米，还是一条线段。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】线段有两个端点，延长一端一定长度后仍有两个端点，仍是线段.\n【详解】把线段的一端延长5000米后仍然是线段。原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·山东济宁·期末"
+    },
+    {
+        "id":  "386",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "3 时 30 分，时针和分针所成的较小角是 90°。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】3:30时分针指6，时针在3和4之间，较小角不是直角.\n【详解】如图:\n\n![](images/697a328af6970dae5c016b814fb27b7f6cb50f2f16c31365310696d76d650ce9.jpg)\n\n由图可知较小角不是90°。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "角的分类及换算",
+                                 "半点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25 四年级上·山东济宁·期末"
+    },
+    {
+        "id":  "387",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "线段比射线短，射线比直线短。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】射线和直线都可以无限延伸，长度无法测量，不能与线段或彼此比较长短.\n【详解】线段可测量长度，但射线，直线长度无法测量，原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "16-17 四年级·全国·课后作业"
+    },
+    {
+        "id":  "388",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "用一副三角尺可以画出150°,135°,100°和75°的角。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】一副三角尺可拼出150°、135°、75°，但不能画出100°。\n【详解】60°+90°=150°；45°+90°=135°；30°+45°=75°；100°不能由30°、45°、60°、90°拼出。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "用三角尺画角"
+                             ],
+        "source":  "24-25 四年级上·湖北黄冈·期末"
+    },
+    {
+        "id":  "389",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "一条长5厘米的线段向两端各延长 1 厘米，就得到一条直线。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】直线是向两端无限延长的，延长有限长度仍是线段.\n【详解】向两端各延长1厘米后仍然是线段，不是直线。原说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25四年级上·湖北鄂州·期末"
+    },
+    {
+        "id":  "390",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "直线没有端点，射线有一个端点，线段有两个端点。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】直线两端无限延伸无端点;射线有一个端点;线段有两个端点.\n【详解】该说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·广东广州·期末"
+    },
+    {
+        "id":  "391",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "量角器既可以度量角的度数，又可以画指定度数的角。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】量角器用于度量角度，也可按刻度画出指定度数的角.\n【详解】该说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "角的度量",
+                                 "用量角器画角"
+                             ],
+        "source":  "24-25 四年级上·河北·期末"
+    },
+    {
+        "id":  "392",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "一个40°的角用 5 倍的放大镜看仍然是40°。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】角的大小与边长无关，只与两边张开的大小有关，放大镜不改变角度.\n【详解】40°角在5倍放大镜下仍是40°。原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "角的大小比较"
+                             ],
+        "source":  "24-25 四年级上·河北沧州·期末"
+    },
+    {
+        "id":  "393",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "把圆平均分成 360份，其中 1 份所对的角的大小是 1°。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】周角是360°,平均分成360份，每份对应1°.\n【详解】该说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "角的度量"
+                             ],
+        "source":  "24-25 四年级上·辽宁葫芦岛·期末"
+    },
+    {
+        "id":  "394",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "因为钝角都大于90°,所以大于90°的角都是钝角。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】钝角是大于90°且小于180°的角;平角是180°,也大于90°但不是钝角.\n【详解】大于90°的角不一定是钝角，还可能是平角等。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平角，周角的认识及特征",
+                                 "角的分类及换算"
+                             ],
+        "source":  "17-18 四年级上·全国·期中"
+    },
+    {
+        "id":  "395",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "平角不是一条直线。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】平角有顶点和两条边(在一条直线上方向相反);直线只是一条线，没有顶点.\n【详解】平角不是一条直线。原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25四年级上·辽宁鞍山·期末"
+    },
+    {
+        "id":  "396",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "小于180°的角一定是钝角。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】小于180°的角可能是锐角，直角或钝角.\n【详解】原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征",
+                                 "角的分类及换算"
+                             ],
+        "source":  "24-25 四年级上·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "397",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "直线，射线和线段的长度作比较，直线最长。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】直线和射线长度无限，无法测量，不能与线段比较长短.\n【详解】直线，射线和线段的长度无法比较。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·黑龙江鸡西·期末"
+    },
+    {
+        "id":  "398",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "老师在黑板上画了一个40°的角，同学们在纸上画了一个40°的角，这两个角的大小无法比较。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】角的大小只与度数有关，与画在什么地方，边长无关.\n【详解】40°=40°,两个角大小相等。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "角的大小比较"
+                             ],
+        "source":  "24-25 四年级上·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "399",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "钱老师在黑板上画了一条长 20 厘米的射线。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】射线只有一个端点，可以向一端无限延伸，长度无法测量.\n【详解】射线不能用\"长20厘米\"来描述。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·内蒙古呼伦贝尔·期末"
+    },
+    {
+        "id":  "400",
+        "unit":  "第三章角的度数是",
+        "question_type":  "作图题",
+        "difficulty":  0.65,
+        "question":  "用你喜欢的方法画 45°和 135°的角。",
+        "options":  [
+
+                    ],
+        "answer":  "见详解",
+        "solution":  "【分析】用量角器画角：先画一条射线，使量角器中心与射线端点重合,0°刻度线与射线重合，在45°刻度处点一点并连成另一条射线即可画出45°;同理画出135°.\n【详解】作图如下:\n\n![](images/2d737b6f37fa44788c6b2c73d994e2bbe4251b33a9e0b6c0f163e10be00fe2bd.jpg)",
+        "knowledge_points":  [
+                                 "用量角器画角"
+                             ],
+        "source":  "24-25四年级上·内蒙古鄂尔多斯·期末"
+    },
+    {
+        "id":  "401",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面关于直线，射线和线段说法正确的是( )。",
+        "options":  [
+                        "A. 直线比射线长",
+                        "B. 过一点只能画出一条直线",
+                        "C. 两点间的距离是两点间线段的长度",
+                        "D. 射线可以向两个方向无限延伸"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】线段有两个端点，有限长，可以度量，两点间的距离是两点间线段的长度；射线有一个端点，射线是无限长的，可以向一个方向无限延伸；直线无端点，直线是无限长的，可以向两个方向无限延伸。过一点可以画无数条直线。\n【详解】A。直线比射线长说法错误，两者均不可测量长度；B。过一点只能画出一条直线说法错误，过一点可以画无数条直线；C。两点间的距离是两点间线段的长度这句话正确；D。射线可以向两个方向无限延伸说法错误，射线可以向一个方向无限延伸。故答案为:C",
+        "knowledge_points":  [
+                                 "线段",
+                                 "直线",
+                                 "射线的认识及特征"
+                             ],
+        "source":  "23-24四年级上·内蒙古包头·期末"
+    },
+    {
+        "id":  "402",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "体育课上，体育老师让我们向后转时，右脚所转的角应该是一个( )角。",
+        "options":  [
+                        "A. 锐角",
+                        "B. 直角",
+                        "C. 平角",
+                        "D. 周角"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】体育课上我们如果向左转或向右转时，就是由原来的方向向左或右旋转90°，即转过的角是直角；如果向后转时，就是由原来的方向向后旋转180°，即转过的角是平角。\n【详解】根据分析：体育课上，体育老师让我们向后转时，右脚所转的角应该是一个平角。故答案为:C",
+        "knowledge_points":  [
+                                 "平角",
+                                 "周角的认识及特征"
+                             ],
+        "source":  "23-24四年级上·内蒙古赤峰·期末"
+    },
+    {
+        "id":  "403",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "王老师在操场上画了一条直直的长约 100 米的( )。",
+        "options":  [
+                        "A. 直线",
+                        "B. 射线",
+                        "C. 线段"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】直线是直的，没有端点，无限长，可以向两端无限延长。如图：射线是直的，有 1 个端点，无限长，可以向一端无限延长。如图：线段是直的，有 2 个端点，有限长。如图:\n【详解】王老师在操场上画了一条直直的长约 100 米的线段故答案为:C",
+        "knowledge_points":  [
+                                 "线段",
+                                 "直线",
+                                 "射线的认识及特征"
+                             ],
+        "source":  "23-24四年级上·内蒙古鄂尔多斯·期末"
+    },
+    {
+        "id":  "404",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "用一个能放大 100倍的放大镜看一个 65°的角，这个角是( )。",
+        "options":  [
+                        "A. 65°",
+                        "B. 650°",
+                        "C. 6500°"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】角的大小与两边的长短无关，只有边叉开的大小有关，叉开得越大，角越大，叉开得越小，角越小。放大镜只会改变角两边的长短，不会改变角的大小，依此选择即可。\n【详解】根据分析可得：一个100倍的放大镜看一个65°的角，这个角还是65°。故答案为:A",
+        "knowledge_points":  [
+                                 "角的初步认识及辨认"
+                             ],
+        "source":  "23-24四年级上·内蒙古赤峰·期末"
+    },
+    {
+        "id":  "405",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "经过一点可以画( )条直线。",
+        "options":  [
+                        "A. 2",
+                        "B. 1",
+                        "C. 无数"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】线段，直线，射线的认识及特征\n【详解】根据题意作图如下:\n\n![](images/4e4e9b85308fae488c370bc666b21a81288fd1d7c9b4cafbd4a92b6d891b13e8.jpg)\n\n由图可知，经过一点可以画无数条直线。故答案为:C",
+        "knowledge_points":  [
+                                 "线段",
+                                 "直线",
+                                 "射线的认识及特征"
+                             ],
+        "source":  "14-15一年级·全国·课后作业"
+    },
+    {
+        "id":  "406",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "同一平面内 A、B、C三点不在同一条直线上，过每两点画一条直线，可以画( )条。",
+        "options":  [
+                        "A. 一",
+                        "B. 两",
+                        "C. 三"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】直线两端是可以无限延长的，两点确定一条直线，据此解答即可.\n【详解】如图:\n\n![](images/ccf6b78afaa198393ac07c61e60610a4e33cf6d22f8e6b96d927b95dab65c577.jpg)\n\n所以同一平面内A,B,C三点不在同一条直线上，过每两点画一条直线，可以画三条。故答案为:C",
+        "knowledge_points":  [
+                                 "线段",
+                                 "直线",
+                                 "射线的认识及特征"
+                             ],
+        "source":  "23-24四年级上·吉林松原·期末"
+    },
+    {
+        "id":  "407",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "两个角的和是一个平角，其中一个角是15°，另一个角是( )。",
+        "options":  [
+                        "A. 75°",
+                        "B. 165°",
+                        "C. 345°"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】两个角的和是一个平角，用平角180°减去一个角的度数，就是另一个角的度数。\n【详解】180°-15°=165°。两个角的和是一个平角，其中一个角是15°，另一个角是165°。故答案为:B",
+        "knowledge_points":  [
+                                 "平角",
+                                 "周角的认识及特征"
+                             ],
+        "source":  "23-24四年级上·吉林松原·期末"
+    },
+    {
+        "id":  "408",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "钟面时间是 9 时 30 分时，时针与分针成( ).\n\n![](images/d738995df5afc42e290505d9897885d95e1f0a9a0cc5260858cdeed60bf11622.jpg)",
+        "options":  [
+                        "A. 锐角",
+                        "B. 直角",
+                        "C. 钝角"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】由题意得,9时30分时，时针在9和10的正中间，分针指着数字6(如下图)\n\n![](images/d738995df5afc42e290505d9897885d95e1f0a9a0cc5260858cdeed60bf11622.jpg)\n\n由图可知，时针与分针成钝角.\n【详解】由分析得，钟面时间是9时30分时，时针与分针成钝角。故答案为:C",
+        "knowledge_points":  [
+                                 "角的分类及换算",
+                                 "半点时间与钟面指针的位置"
+                             ],
+        "source":  "23-24四年级上·吉林白城·期末"
+    },
+    {
+        "id":  "409",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面按角的大小关系排列正确的是( )。",
+        "options":  [
+                        "A. 直角<锐角<钝角<平角<周角",
+                        "B. 周角>平角>直角>钝角>锐角",
+                        "C. 锐角<直角<钝角<平角<周角"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】锐角是大于0°小于90°的角，直角是等于90°的角，钝角是大于90°小于180°的角，平角是等于180°的角，周角是等于360°的角；据此解答。\n【详解】根据分析：按角的大小关系排列正确的是锐角<直角<钝角<平角<周角。故答案为:C",
+        "knowledge_points":  [
+                                 "直角",
+                                 "钝角",
+                                 "锐角的认识及特征",
+                                 "平角",
+                                 "周角的认识及特征",
+                                 "角的分类及换算"
+                             ],
+        "source":  "24-25四年级上·吉林松原·期末"
+    },
+    {
+        "id":  "410",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "用一个放大镜去看一个90°的角，所看到的角的度数( )。",
+        "options":  [
+                        "A. 变大了",
+                        "B. 变小了",
+                        "C. 不变"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】用放大镜看角，只改变角两边的长度，没有改变角两边叉开的大小，则角的度数不变。\n【详解】用一个放大镜去看一个90°的角，所看到的角的度数不变，故答案为:C",
+        "knowledge_points":  [
+                                 "角的初步认识及辨认",
+                                 "角的大小比较"
+                             ],
+        "source":  "23-24四年级上·吉林·期末"
+    },
+    {
+        "id":  "411",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "若一个角比平角小89°，则这个角是( )角。",
+        "options":  [
+                        "A. 钝角",
+                        "B. 直角",
+                        "C. 锐角",
+                        "D. 平角"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】平角是180°，用180°减去89°，即可求出比平角小89°的角的度数；等于90°的角是直角，小于90°的角是锐角，大于90°且小于180°的角是钝角，据此解答。\n【详解】180°-89°=91°。91°大于90°且小于180°，所以这个角是钝角。故答案为:A",
+        "knowledge_points":  [
+                                 "直角",
+                                 "钝角",
+                                 "锐角的认识及特征",
+                                 "平角",
+                                 "周角的认识及特征"
+                             ],
+        "source":  "24-25四年级上·吉林四平·期末"
+    },
+    {
+        "id":  "412",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面说法不正确的是( )。",
+        "options":  [
+                        "A. 直线没有端点，射线只有一个端点",
+                        "B. 线段可以量出长度",
+                        "C. 直线和射线都可以无限延伸",
+                        "D. 角的两条边是线段"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】线段有两个端点且有一定的长度;射线有一个端点，它可以向一个方向无限延伸;直线没有端点，它可以向两端无限延伸;从一点引出两条射线所组成的图形叫作角。据此解答.\n【详解】A。直线没有端点，射线只有一个端点。该选项说法正确。 B。线段有两个端点且有一定的长度，它的长度可以测量。该选项说法正确。 C。直线和射线都可以无限延伸。该选项说法正确。 D。角的两条边是射线。该选项说法错误。故答案为:D",
+        "knowledge_points":  [
+                                 "线段",
+                                 "直线",
+                                 "射线的认识及特征",
+                                 "角的概念及表示方法"
+                             ],
+        "source":  "23-24四年级上·北京石景山·期末"
+    },
+    {
+        "id":  "413",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "用一副三角尺不能拼出来的角是( )。",
+        "options":  [
+                        "A. 10°",
+                        "B. 75°",
+                        "C. 120°",
+                        "D. 135°"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】一副三角尺，其中一个三角尺的角有30°、60°、90°，另一个等腰直角三角尺的角有45°、45°、90°，用它们进行拼组，看是否能得出这几个角度即可。\n【详解】A。无论怎么拼组，都拼不出10°的角；B。30°+45°=75°，用一副三角尺可以拼成75°的角；C。30°+90°=120°，用一副三角尺可以拼成120°的角；D。45°+90°=135°，用一副三角尺可以拼成135°的角。因此用一副三角尺不能拼出来的角是10°。故答案为:A",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "角的度量"
+                             ],
+        "source":  "23-24四年级上·贵州遵义·期末"
+    },
+    {
+        "id":  "414",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "两个锐角拼在一起，不可能是( )。",
+        "options":  [
+                        "A. 锐角",
+                        "B. 直角",
+                        "C. 钝角",
+                        "D. 无法判断"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】根据锐角、直角和钝角的定义，大于0°小于90°的角，叫做锐角；等于90°的角，叫做直角；大于90°小于180°的角，叫做钝角。两个锐角拼在一起，例如：当两个锐角都是10°时，10°+10°=20°，还是一个锐角；两个锐角都是45°时，45°+45°=90°，是一个直角；两个锐角都是50°时，50°+50°=100°，是一个钝角；据此解答。\n【详解】根据分析可知：两个锐角拼在一起，可能是锐角、直角或钝角。所以不可能的结果无法判断。故答案为:D",
+        "knowledge_points":  [
+                                 "直角",
+                                 "钝角",
+                                 "锐角的认识及特征",
+                                 "角的分类及换算"
+                             ],
+        "source":  "24-25四年级上·湖南郴州·期末"
+    },
+    {
+        "id":  "415",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "李老师在黑板上画了一条 5 厘米长的( )。",
+        "options":  [
+                        "A. 线段",
+                        "B. 直线",
+                        "C. 射线"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】射线有一个端点，可以向一端无限延伸，长度不可测量;线段有2个端点，长度可以测量;直线没有端点，可以向两端无限延伸，长度不可测量;据此解答.\n【详解】根据对直线，射线和线段知识的认识可知：李老师在黑板上画了一条5厘米长的线段。故答案为:A",
+        "knowledge_points":  [
+                                 "线段",
+                                 "直线",
+                                 "射线的认识及特征"
+                             ],
+        "source":  "23-24四年级上·河北石家庄·期末"
+    },
+    {
+        "id":  "416",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "角的大小与( )有关。",
+        "options":  [
+                        "A. 边的长短",
+                        "B. 角两边张开的大小",
+                        "C. 角的位置"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】从一点引出两条射线所形成的图形叫做角。角的大小与两边的长短无关，只与两条边张开的大小有关。张开得越大，角越大，张开得越小，角越小。据此解答.\n【详解】A。角的大小与两条边的长短无关。 B。角的大小与两边张开的大小有关。 C。角的大小与角的位置无关。故答案为:B",
+        "knowledge_points":  [
+                                 "角的大小比较"
+                             ],
+        "source":  "23-24四年级上·四川绵阳·期末"
+    },
+    {
+        "id":  "417",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "若∠1+∠2=90°，∠2+∠3=180°，那么∠3一定是( )。",
+        "options":  [
+                        "A. 锐角",
+                        "B. 钝角",
+                        "C. 直角",
+                        "D. 平角"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据题意可知，∠1+∠2=90°，可得出0°<∠2<90°，又由于∠2+∠3=180°，所以90°<∠3<180°；大于90°且小于180°的角是钝角；据此解答。\n【详解】根据分析可知，若∠1+∠2=90°，∠2+∠3=180°，那么∠3大于90°且小于180°，则∠3一定是钝角。故答案为:B",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "直角",
+                                 "钝角",
+                                 "锐角的认识及特征"
+                             ],
+        "source":  "24-25四年级上·福建福州·期中"
+    },
+    {
+        "id":  "418",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面能用一副三角尺拼出的一组角是( )。",
+        "options":  [
+                        "A. 15°,55°,60°",
+                        "B. 75°,105°,160°",
+                        "C. 120°,135°,150°"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】一副三角尺上的度数有30°、45°、60°、90°，可以拼出的角有15°(45°-30°或60°-45°)、75°(30°+45°)、105°(45°+60°)、120°(30°+90°)、135°(45°+90°)、150°(60°+90°)、180°(90°+90°)；据此解答。\n【详解】A。用一副三角尺不能拼出55°的角；B。用一副三角尺不能拼出160°的角；C。30°+90°=120°，45°+90°=135°，60°+90°=150°，则用一副三角尺能拼出120°、135°、150°的角。故答案为:C",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "用三角尺画角"
+                             ],
+        "source":  "24-25四年级上·福建福州·期中"
+    },
+    {
+        "id":  "419",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "从一点出发可以画( )条射线，经过两个点可以画( )条直线。",
+        "options":  [
+                        "A. 1;无数",
+                        "B. 2;3",
+                        "C. 3;无数",
+                        "D. 无数;1"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】射线有一个端点，无限长，从一点出发可以作无数条射线;两点确定一条直线，经过两个点可以画一条直线，并且只能画一条直线;据此解答.\n【详解】根据分析可知，从一点出发可以画无数条射线，经过两个点可以画 1 条直线。故答案为:D",
+        "knowledge_points":  [
+                                 "线段",
+                                 "直线",
+                                 "射线的认识及特征"
+                             ],
+        "source":  "24-25四年级上·福建福州·期中"
+    },
+    {
+        "id":  "420",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一张圆形纸对折3次后展开，不可以得到角的度数是( )。",
+        "options":  [
+                        "A. 60°",
+                        "B. 90°",
+                        "C. 135°"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】把一张圆形纸看作一个周角，度数是360°，对折一次可得到360°÷2=180°的角，对折二次可得到180°÷2=90°的角，对折三次可得到90°÷2=45°的角；展开后，即可得到45°的1倍、2倍、3倍等度数的角。据此解答。\n【详解】360°÷2=180°；180°÷2=90°；90°÷2=45°。45°×1=45°，45°×2=90°，45°×3=135°。所以，一张圆形纸对折3次后展开，不可以得到角的度数是60°。故答案为:A",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "平角",
+                                 "周角的认识及特征"
+                             ],
+        "source":  "23-24四年级上·四川绵阳·期末"
+    },
+    {
+        "id":  "421",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "两个角正好能拼成一个平角，如果其中一个角是锐角，则另外一个角是( )。",
+        "options":  [
+                        "A. 锐角",
+                        "B. 直角",
+                        "C. 钝角",
+                        "D. 不能确定"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】大于0°小于90°的角是锐角，等于90°的角是直角，大于90°小于180°的角是钝角，平角等于180°，两个角组成一个平角，其中一个角是锐角，则另一个角应大于90°，是一个钝角，据此解答即可。\n【详解】两个角正好能拼成一个平角，如果其中一个角是锐角，则另外一个角是钝角。故答案为:C",
+        "knowledge_points":  [
+                                 "直角",
+                                 "钝角",
+                                 "锐角的认识及特征",
+                                 "平角",
+                                 "周角的认识及特征"
+                             ],
+        "source":  "23-24四年级上·重庆璧山·期末"
+    },
+    {
+        "id":  "422",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "把一条 5 厘米长的线段向两端各延长 10 厘米，得到一条( )。",
+        "options":  [
+                        "A. 直线",
+                        "B. 线段",
+                        "C. 射线",
+                        "D. 不确定"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】直线，射线，线段都是直的，线段有两个端点，长度有限;射线有一个端点，长度无限;直线没有端点，长度无限;线段向两端延长一定的长度，还是线段.\n【详解】根据分析可知：将一条线段的两端各延长10厘米，得到一条线段。故答案为:B",
+        "knowledge_points":  [
+                                 "线段",
+                                 "直线",
+                                 "射线的认识及特征"
+                             ],
+        "source":  "23-24四年级上·四川乐山·期末"
+    },
+    {
+        "id":  "423",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "用一副三角尺不能拼出的角是( )。",
+        "options":  [
+                        "A. 105°",
+                        "B. 150°",
+                        "C. 125°",
+                        "D. 75°"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】一副三角尺上的度数有30°、45°、60°、90°，可以拼出的角有15°(45°-30°或60°-45°)、75°(30°+45°)、105°(45°+60°)、120°(30°+90°)、135°(45°+90°)、150°(60°+90°)、180°(90°+90°)；据此解答。\n【详解】A。45°+60°=105°，则用一副三角尺能拼出105°的角；B。60°+90°=150°，则用一副三角尺能拼出150°的角；C。用一副三角尺不能拼出125°的角；D。30°+45°=75°，则用一副三角尺能拼出75°的角。故答案为:C",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "用三角尺画角"
+                             ],
+        "source":  "23-24三年级上·四川南充·期末"
+    },
+    {
+        "id":  "424",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面各角中，不能用一副三角尺画出来的是( )。",
+        "options":  [
+                        "A. 125°",
+                        "B. 105°",
+                        "C. 75°"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】一副三角尺上的角有45°、45°、90°、60°、30°、90°，用这些角相加减所能得到的相应度数的角，就能用一副三角尺画出来，否则就不能；据此解答。\n【详解】A。125°不能用三角尺上的角相加减得到，所以125°的角不能用一副三角尺画出来；B。105°=60°+45°，所以105°的角能用一副三角尺画出来；C。75°=45°+30°，所以75°的角能用一副三角尺画出来；所以，不能用一副三角尺画出来的是125°。故答案为:A",
+        "knowledge_points":  [
+                                 "用三角尺画角"
+                             ],
+        "source":  "23-24四年级上·河南信阳·期末"
+    },
+    {
+        "id":  "425",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "把一个平角分成两个角，这两个角可能是( )。",
+        "options":  [
+                        "A. 锐角，锐角",
+                        "B. 锐角，直角",
+                        "C. 锐角，钝角",
+                        "D. 锐角，平角"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】平角等于180°，直角等于90°，把平角分成两个角，这两个角可能是2个直角或者1个锐角和1个钝角。\n【详解】根据分析：A。把一个平角分成两个角，不可能分成两个锐角；B。把一个平角分成两个角，如果其中一个是直角，那另一个一定也是直角；C。把一个平角分成两个角，这两个角可能是一个锐角一个钝角；D。把一个平角分成两个角，不可能再分成其中一个是平角。故答案为:C",
+        "knowledge_points":  [
+                                 "直角",
+                                 "钝角",
+                                 "锐角的认识及特征",
+                                 "平角",
+                                 "周角的认识及特征",
+                                 "角的分类及换算"
+                             ],
+        "source":  "23-24四年级上·湖南娄底·期末"
+    },
+    {
+        "id":  "426",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "用一副三角尺不可以画出的是( )度的角。",
+        "options":  [
+                        "A. 75",
+                        "B. 105",
+                        "C. 115",
+                        "D. 135"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】一副三角板中的各个角的度数分别是30°、60°、45°、90°，把它们进行组合，相加或者相减后能得出的角都可以用一副三角板拼出。\n【详解】A。75°=30°+45°，它可以用一副三角板拼出；B。105°=60°+45°，它可以用一副三角板拼出；C。115°不可以用一副三角板拼出；D。135°=90°+45°，它可以用一副三角板拼出。故答案为:C",
+        "knowledge_points":  [
+                                 "用三角尺画角"
+                             ],
+        "source":  "23-24四年级上·湖南娄底·期末"
+    },
+    {
+        "id":  "427",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "用一副三角板不可以直接拼成( )。",
+        "options":  [
+                        "A. 150°",
+                        "B. 135°",
+                        "C. 120°",
+                        "D. 100°"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】一副三角尺上的度数有30°、45°、60°、90°，可以拼出的角有15°(45°-30°或60°-45°)、75°(30°+45°)、105°(45°+60°)、120°(30°+90°)、135°(45°+90°)、150°(60°+90°)、180°(90°+90°)；据此解答。\n【详解】A。60°+90°=150°，则能用一副三角尺拼出150°的角；B。45°+90°=135°，则能用一副三角尺拼出135°的角；C。30°+90°=120°，则能用一副三角尺拼出120°的角；D。不能用一副三角尺拼出100°的角。故答案为:D",
+        "knowledge_points":  [
+                                 "用三角尺画角"
+                             ],
+        "source":  "23-24四年级上·重庆綦江·期末"
+    },
+    {
+        "id":  "428",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "用一副三角板可以拼出( )的角。",
+        "options":  [
+                        "A. 105°",
+                        "B. 85°",
+                        "C. 140°"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】一副三角板有90°、45°、45°、90°、60°、30°几个角，分别从每个三角板中任取一个角，进行加或减，拼出不同的角。可得到的角有60°-45°=15°，60°+45°=105°，60°+90°=150°，90°+45°=135°，90°+30°=120°，30°+45°=75°。然后再作选择即可。\n【详解】A。45°+60°=105°，用一副三角板可以拼出105°的角，符合题意；B。用一副三角板不可以拼出85°的角，不符合题意；C。用一副三角板不可以拼出140°的角，不符合题意。故答案为:A",
+        "knowledge_points":  [
+                                 "用三角尺画角"
+                             ],
+        "source":  "19-20四年级上·山东临沂·期末"
+    },
+    {
+        "id":  "429",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "我们可以用尺子画一条 5 厘米长的( )。",
+        "options":  [
+                        "A. 直线",
+                        "B. 射线",
+                        "C. 线段"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】直线没有端点，是可以无限延伸的，长度不能被度量;射线只有一个端点，可以向一端无限延伸，长度也不能被度量;线段有两个端点，长度可以被度量，据此解答即可.\n【详解】根据分析可得，直线和射线都是无限长，所以我们可以用尺子画一条5厘米长的线段。故答案为:C",
+        "knowledge_points":  [
+                                 "线段",
+                                 "直线",
+                                 "射线的认识及特征"
+                             ],
+        "source":  "23-24四年级上·四川广元·期末"
+    },
+    {
+        "id":  "430",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面各个角中，不能用一副三角板画出来的是( )。",
+        "options":  [
+                        "A. 15°",
+                        "B. 80°",
+                        "C. 135°",
+                        "D. 150°"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】一副三角板有两个三角尺，一个三角尺的三个角的度数分别为90°、45°、45°，另一个三角尺的三个角的度数分别为90°、30°、60°，要用它们来画角，就是要把它们进行组合，据此解答即可。\n【详解】A。60°-45°=15°，可以画出15°的角。B。无论怎么组合也画不出80°的角。C。90°+45°=135°，可以画出135°的角。D。90°+60°=150°，可以画出150°的角。不能用一副三角板画出来的角是80°。故答案为:B",
+        "knowledge_points":  [
+                                 "用三角尺画角"
+                             ],
+        "source":  "22-23四年级上·四川乐山·期末"
+    },
+    {
+        "id":  "431",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "把一个平角分成两个角，这时所成的角可能是( )。\n\n① 一个钝角和一个锐角 ② 两个锐角 ③ 两个钝角 ④ 两个直角",
+        "options":  [
+                        "A. ①②",
+                        "B. ②③",
+                        "C. ①④"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】1平角=180°，锐角大于0°而小于90°，钝角大于90°而小于180°。①一个钝角与一个锐角的和有可能等于180°，如120°与60°的和是180°。②两个锐角的和一定小于180°。③两个钝角的和一定大于180°。④1直角=90°，2个90°的和就是180°。\n【详解】①120°+60°=180°，所以一个钝角和一个锐角是有可能的。②89°+89°=178°，两个锐角的和不可能是180°。③91°+91°=182°，两个钝角的和不可能是180°。④90°+90°=180°，两个直角的和是180°。把一个平角分成两个角，这时所成的角可能是一个钝角和一个锐角，或两个直角。因此，①④说法正确。故答案为:C",
+        "knowledge_points":  [
+                                 "直角",
+                                 "钝角",
+                                 "锐角的认识及特征",
+                                 "平角",
+                                 "周角的认识及特征"
+                             ],
+        "source":  "23-24四年级上·湖北咸宁·期末"
+    },
+    {
+        "id":  "432",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "小新用量角器量角时，错把内圈刻度看成了外圈刻度，读出的度数是65°，那么这个角实际是( )°。",
+        "options":  [
+                        "A. 115°",
+                        "B. 125°",
+                        "C. 35°",
+                        "D. 65°"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】根据量角器的构造，外圈刻度与内圈刻度的和是180°，误把外圈刻度当成了内圈刻度，读得度数是65°，正确的度数是180°-65°，由此即可解答。\n【详解】180°-65°=115°，这个角实际应该是115°。故答案为:A",
+        "knowledge_points":  [
+                                 "角的度量"
+                             ],
+        "source":  "24-25四年级上·浙江·期中"
+    },
+    {
+        "id":  "433",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面说法错误的有( )句。\n\n① 直线是无限长的。\n② 小于90°的角叫做锐角，大于90°的角是钝角。\n③ 个级包括的数位有个，十，百，千。\n④ 角的大小与边的长短无关，与两条边叉开的大小有关。",
+        "options":  [
+                        "A. 1",
+                        "B. 2",
+                        "C. 3",
+                        "D. 4"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】直线没有端点，可以向两端无限延长。大于0°小于90°的是锐角，大于90°小于180°的是钝角。角是由一个顶点和两条射线组成的，角的两边可以无限延长。数位顺序表从右起，每4个数位为一级，分别是个级、万级和亿级。每个数字所占的位置叫数位，个级的数位包括个位、十位、百位、千位。\n【详解】据分析，①正确。②错误，应该是大于0°小于90°的是锐角，大于90°小于180°的是钝角。③错误，个级应该包括个位、十位、百位、千位。④正确。一共有两句话错误。故答案为:B",
+        "knowledge_points":  [
+                                 "整数的数级",
+                                 "数位和计数单位的认识",
+                                 "线段",
+                                 "直线",
+                                 "射线的认识及特征",
+                                 "角的初步认识及辨认",
+                                 "直角",
+                                 "钝角",
+                                 "锐角的认识及特征"
+                             ],
+        "source":  "24-25四年级上·浙江·期中"
+    },
+    {
+        "id":  "434",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面的角能用一副三角尺拼出的是( )。",
+        "options":  [
+                        "A. 65°",
+                        "B. 75°",
+                        "C. 85°",
+                        "D. 95°"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】一副三角尺由两个三角尺组成，一个三角尺的三个角分别是90°、45°、45°，另一个三角尺的三个角分别是90°、60°、30°。由题意得，将两个三角尺的角度相加或相减看能否得到选项中的角度即可。\n【详解】A。不能用一副三角尺拼出65°的角；B。30°+45°=75°，则能用一副三角尺拼出75°的角；C。不能用一副三角尺拼出85°的角；D。不能用一副三角尺拼出95°的角。故答案为:B",
+        "knowledge_points":  [
+                                 "用三角尺画角"
+                             ],
+        "source":  "24-25四年级上·浙江·期中"
+    },
+    {
+        "id":  "435",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.4,
+        "question":  "如果1路公交车有10个汽车站，单程需要准备( )种不同的车票。",
+        "options":  [
+                        "A. 45",
+                        "B. 55",
+                        "C. 65",
+                        "D. 20"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】根据题意可知，一共10个汽车站，从第一站到其它各站有9种单程车票，同理从第二个站到其他站有8种单程车票，第三站到其他站有7种单程车票，从第四站到其他站有6种单程车票。依次类推，由此求解。\n【详解】9+8+7+6+5+4+3+2+1=(9+8+7)+(6+5+4)+(3+2+1)=24+15+6=45(种)。则如果1路公交车有10个汽车站，单程需要准备45种不同的车票。故答案为:A",
+        "knowledge_points":  [
+                                 "数图形(线段",
+                                 "直线",
+                                 "射线)"
+                             ],
+        "source":  "23-24四年级上·河南商丘·期末"
+    },
+    {
+        "id":  "436",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "时针走 3 小时转过的角度是( )。",
+        "options":  [
+                        "A. 30°",
+                        "B. 60°",
+                        "C. 90°",
+                        "D. 180°"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】钟面上时针走一圈是360°，钟面上有12个大格，所以每个大格是30°。时针1小时走1大格，3小时走3大格。据此可以知道走了多少度。\n【详解】走一圈是360°，钟面上有12个大格，所以每个大格是30°。30°×3=90°。所以，时针走3小时转过的角是90°。故答案为:C",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "平角",
+                                 "周角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "23-24四年级上·河南商丘·期末"
+    },
+    {
+        "id":  "437",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "用一副三角尺无法画出( )的角。",
+        "options":  [
+                        "A. 15°",
+                        "B. 145°",
+                        "C. 75°",
+                        "D. 105°"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】一副三角尺中的各个角的度数分别是30°,60°,45°,90°。将它们进行组合，可得到的角有60°-45°=15°,60°+45°=105°,60°+90°=150°,90°+45°=135°,90°+30°=120°,30°+45°=75°.\n【详解】A。60°-45°=15°, 15°的角能用一副三角尺拼出B。145°的角不能用一副三角尺拼出。C。30°+45°=75°, 75°的角能用一副三角尺拼出。D。60°+45°=105°, 105°的角能用一副三角尺拼出.\n\n用一副三角尺无法画出145°的角。故答案为:B",
+        "knowledge_points":  [
+                                 "用三角尺画角"
+                             ],
+        "source":  "23-24四年级上·重庆秀山·期末"
+    },
+    {
+        "id":  "438",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "将一张圆形纸片对折，对折，再对折，展开后折痕之间最小的夹角是( )。",
+        "options":  [
+                        "A. 45°",
+                        "B. 90°",
+                        "C. 120°"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】每对折一次就用原来角的度数除以2得到新形成角的度数。可拿纸实际操作一下。\n【详解】360°÷2÷2÷2=180°÷2÷2=90°÷2=45°。将一张圆形纸片对折，对折，再对折，展开后折痕之间最小的夹角是45°。故答案为:A",
+        "knowledge_points":  [
+                                 "图形的折叠问题",
+                                 "平角",
+                                 "周角的认识及特征"
+                             ],
+        "source":  "23-24四年级上·重庆万州·期末"
+    },
+    {
+        "id":  "439",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面说法正确的是( )。",
+        "options":  [
+                        "A. 一条直线长30米",
+                        "B. 线段有两个端点",
+                        "C. 过一点只能画一条直线"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】线段的特征：有两个端点，不能向两端延伸，可以量出长度;直线的特征：没有端点，可以向两端无限延伸，不能测量出长度;经过一点可以画无数条直线，经过两点只能画一条直线;据此解答即可.\n【详解】A。一条直线长 30 米。直线不能测量出长度，原题说法错误;B。线段有两个端点。说法正确;C。过一点只能画一条直线。经过一点可以画无数条直线，原题说法错误。说法正确的是线段有两个端点。故答案为:B",
+        "knowledge_points":  [
+                                 "线段",
+                                 "直线",
+                                 "射线的认识及特征"
+                             ],
+        "source":  "23-24四年级上·重庆万州·期末"
+    },
+    {
+        "id":  "440",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "用一副三角尺不能直接拼出的角的度数是( )。",
+        "options":  [
+                        "A. 105°",
+                        "B. 115°",
+                        "C. 120°"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】一副三角尺由两个三角尺组成，一个三角尺的三个角分别是90°、45°、45°，另一个三角尺的三个角分别是90°、60°、30°。由题意得，将两个三角尺的角度相加或相减看能否得到选项中的角度即可。\n【详解】A。60°+45°=105°，即105°的角可以用一副三角尺拼出。B。115°的角无法通过一副三角尺拼出。C。90°+30°=120°，即120°的角可以用一副三角尺拼出。故答案为:B",
+        "knowledge_points":  [
+                                 "用三角尺画角"
+                             ],
+        "source":  "23-24四年级上·重庆万州·期末"
+    },
+    {
+        "id":  "441",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面度数的角中，不能用一副三角板直接拼出的是( )。",
+        "options":  [
+                        "A. 135°",
+                        "B. 120°",
+                        "C. 85°"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】一副三角尺分为直角三角尺和等腰直角三角尺。直角三角尺的角的度数分别是30°、60°、90°，等腰直角三角尺的度数是45°、90°、45°，用它们进行拼组，然后再与各选项比较，即可解答。\n【详解】A。135°，90°+45°=135°，可以拼出；B。120°，90°+30°=120°，可以拼出；C。85°不能用一副三角板直接拼出。不能用一副三角板直接拼出的角是85°。故答案为:C",
+        "knowledge_points":  [
+                                 "用三角尺画角"
+                             ],
+        "source":  "24-25四年级上·广西贵港·期中"
+    },
+    {
+        "id":  "442",
+        "unit":  "第三章角的度数是",
+        "question_type":  "选择题",
+        "difficulty":  0.94,
+        "question":  "用直尺把两点连接起来，就得到一条( )。",
+        "options":  [
+                        "A. 线段",
+                        "B. 直线",
+                        "C. 射线"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】直线上任意两点之间的一段叫做线段。连接两点的线段的长度叫做两点间的距离。线段有两个端点，任意两点间的一段都可以看作一条线段.\n【详解】用直尺把纸上的两个点连接起来，就得到一条线段。故答案为:A",
+        "knowledge_points":  [
+                                 "线段",
+                                 "直线",
+                                 "射线的认识及特征"
+                             ],
+        "source":  "24-25四年级上·广西贵港·期中"
+    },
+    {
+        "id":  "443",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "钟面上3时整，时针和分针形成( )角，钟面上 6 时整，时针和分针形成( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "直;平",
+        "solution":  "【分析】钟面上12个数字，以表芯为旋转点，表针转一圈是360°，被12个数字平均分成12份，每一大格也就是两数之间夹角是30°。3时整，时针指向3，分针指向12，时针和分针之间有3个大格。6时整，分针指向12，时针指向6，时针和分针之间有6个大格。然后根据直角和平角的含义：等于90°的角叫直角；等于180°的角叫平角；进行解答即可。\n【详解】30°×3=90°，90°是直角。30°×6=180°，180°是平角。钟面上3时整，时针和分针形成直角，钟面上6时整，时针和分针形成平角。",
+        "knowledge_points":  [
+                                 "直角",
+                                 "钝角",
+                                 "锐角的认识及特征",
+                                 "平角",
+                                 "周角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25四年级上·吉林·期末"
+    },
+    {
+        "id":  "444",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "周角是平角的( )倍，是( )角的 4 倍。",
+        "options":  [
+
+                    ],
+        "answer":  "2;直",
+        "solution":  "【分析】根据周角、平角、直角的定义可知，周角=360°，平角=180°，直角=90°；一个数里面有几个另一个数，则这个数是另一个数的几倍。\n【详解】360°里面有2个180°，所以周角是平角的2倍；360°÷4=90°，所以周角是直角的4倍。",
+        "knowledge_points":  [
+                                 "倍的认识",
+                                 "直角",
+                                 "钝角",
+                                 "锐角的认识及特征",
+                                 "平角",
+                                 "周角的认识及特征"
+                             ],
+        "source":  "23-24四年级上·吉林松原·期末"
+    },
+    {
+        "id":  "445",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个角是85°，它是( )角，一个平角等于( )个直角，一个周角等于( )个平角，一个周角等于( )个直角。",
+        "options":  [
+
+                    ],
+        "answer":  "锐;2;2;4",
+        "solution":  "【分析】大于0°小于90°的角叫作锐角；大于90°小于180°的角叫作钝角。直角的度数等于90°，平角的度数等于180°，周角的度数等于360°。据此解答。\n【详解】0°<85°<90°，故85°的角是锐角。90°×2=180°，即一个平角等于2个直角。180°×2=360°，即一个周角等于2个平角。90°×4=360°，即一个周角等于4个直角。一个角是85°，它是锐角，一个平角等于2个直角，一个周角等于2个平角，一个周角等于4个直角。",
+        "knowledge_points":  [
+                                 "平角",
+                                 "周角的认识及特征",
+                                 "角的分类及换算"
+                             ],
+        "source":  "23-24四年级上·吉林白城·期末"
+    },
+    {
+        "id":  "446",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "( )和( )都可以看作是直线的一部分。",
+        "options":  [
+
+                    ],
+        "answer":  "线段;射线",
+        "solution":  "【分析】结合所学知识，直线是线段向两端无限延伸的，没有端点;射线是由线段的一端无限延长所形成的直的线，有一个端点;线段是指直线上两点间的有限部分，有两个端点。据此解答即可.\n【详解】结合分析可知，线段和射线都可以看作是直线的一部分。",
+        "knowledge_points":  [
+                                 "线段",
+                                 "直线",
+                                 "射线的认识及特征"
+                             ],
+        "source":  "23-24四年级上·吉林松原·期末"
+    },
+    {
+        "id":  "447",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "明明在用量角器的外圈刻度准确度量一个角时，角的一边与180°刻度线重合，角的另一边指向50°刻度线，这个角是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "130",
+        "solution":  "【分析】角的度量方法：用量角器量角时，先把量角器的中心与角的顶点重合，0°刻度线与角的一条边重合，角的另一条边所对的量角器上的刻度，就是这个角的度数；如果角的起始边不是与0°刻度线重合，角的度数为两条边所对的刻度之差。\n【详解】180°-50°=130°。明明在用量角器的外圈刻度准确度量一个角时，角的一边与180°刻度线重合，角的另一边指向50°刻度线，这个角是130°。",
+        "knowledge_points":  [
+                                 "角的度量"
+                             ],
+        "source":  "23-24四年级上·河北衡水·期末"
+    },
+    {
+        "id":  "448",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "平角的度数是直角度数的( )倍,( )角的度数是平角度数的2倍。",
+        "options":  [
+
+                    ],
+        "answer":  "2;周",
+        "solution":  "【分析】直角等于90°，平角等于180°，用180°÷90°即可求出平角的度数是直角度数的几倍；周角等于360°，用180°×2，即可求出哪种角的度数是平角度数的2倍。\n【详解】180°÷90°=2；180°×2=360°，360°的角是周角。平角的度数是直角度数的2倍，周角的度数是平角度数的2倍。",
+        "knowledge_points":  [
+                                 "求一个数是另一个数的几倍",
+                                 "求一个数的几倍是多少",
+                                 "直角",
+                                 "钝角",
+                                 "锐角的认识及特征",
+                                 "平角",
+                                 "周角的认识及特征"
+                             ],
+        "source":  "23-24四年级上·北京丰台·期末"
+    },
+    {
+        "id":  "449",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "钟面上 6 时整，时针和分针所成的角是( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "平",
+        "solution":  "【分析】钟面上有12大格，每一大格所对应的夹角等于30°，6时整，时针指向6，分针指向12，6到12有6大格，时针和分针所形成的角等于30°×6=180°，再根据大于0°小于90°的角是锐角，90°的角是直角，大于90°小于180°的角是钝角，180°是平角，360°的角是周角判断后填空即可。\n【详解】30°×6=180°，180°的角是平角。钟面上6时整，时针和分针所成的角是平角。",
+        "knowledge_points":  [
+                                 "直角",
+                                 "钝角",
+                                 "锐角的认识及特征",
+                                 "平角",
+                                 "周角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25四年级上·吉林白城·期末"
+    },
+    {
+        "id":  "450",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\">\"或\"<\"。\n\n30 公顷( )30 平方千米 1 周角( )160°\n\n7 公顷( )6000 平方米 7650420( )76005420",
+        "options":  [
+
+                    ],
+        "answer":  "<;>;>;<",
+        "solution":  "【分析】整数大小的比较方法：位数不同，位数多的数就大；位数相同，左起第一位上的数大的那个数就大，如果左起第一位上的数相同，就比较左起第二位上的数，直到比较出大小为止。单位不统一时，先化成相同单位，再比较大小。1平方千米=100公顷，1公顷=10000平方米，1周角=360°。\n【详解】30平方千米=3000公顷，30公顷<3000公顷，30公顷<30平方千米；1周角=360°，360°>160°，1周角>160°；7公顷=70000平方米，70000平方米>6000平方米，7公顷>6000平方米；7650420<76005420。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "平角",
+                                 "周角的认识及特征",
+                                 "公顷",
+                                 "平方千米的进率与换算"
+                             ],
+        "source":  "24-25四年级上·陕西商洛·期末"
+    },
+    {
+        "id":  "451",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "钟面上,2 时整，时针和分针形成的角是( )°;6 时整，时针和分针形成的角是( )角。形成 90 ° 角的是( )时整或( )时整。",
+        "options":  [
+
+                    ],
+        "answer":  "60;平;3;9",
+        "solution":  "【分析】时钟上12个数字把钟面平均分成12个大格，每个大格是 30 circ 。钟面上2时整，时针和分针之间有2个大格，则时针和分针的夹角是 2×30 circ 。6时整，时针和分针之间有6 个大格，则时针和分针的夹角是 6×30 circ ,即 180 circ ,是一个平角。 3×30 circ = 90 circ ,整时，分针指向 12，要使时针和分针的夹角是 90 ° ,时针和分针之间应该有3个大格，那么时针应指向 3 或 9 .\n【详解】 2×30 circ = 60 circ\n\n6×30 circ = 180 circ \n 3×30 circ = 90 circ\n\n钟面上,2 时整，时针和分针形成的角是 60°;6 时整，时针和分针形成的角是平角。形成 90 circ 角的是3时整或9时整。",
+        "knowledge_points":  [
+                                 "平角",
+                                 "周角的认识及特征",
+                                 "角的分类及换算",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "23-24四年级上·浙江宁波·期末"
+    },
+    {
+        "id":  "452",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明在用量角器测量一个钝角的度数时，角的一条边与0刻度线重合，读数时误把内圈刻度读成外圈刻度，读出的度数是 30 circ ,则正确的度数是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "150",
+        "solution":  "【分析】根据量角器的构造即可求解，注意外圈刻度与内圈刻度的和是 180 circ ,误把内圈刻度读成外圈刻度，读出的度数是 30 circ ,因此用 180 circ 减去 30 circ ,就是正确的度数;据此解答即可.\n【详解】 180 circ - 30 circ = 150 circ则正确的度数是 150 circ mathfrak c 。",
+        "knowledge_points":  [
+                                 "角的度量"
+                             ],
+        "source":  "23-24四年级上·浙江宁波·期末"
+    },
+    {
+        "id":  "453",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "根据( )原理，人们制作了度量角的工具( )。",
+        "options":  [
+
+                    ],
+        "answer":  "将圆平均分成;360 份的 量角器",
+        "solution":  "【分析】人们把圆平均分成360份，将其中1份所对的角作为度量角的单位，它的大小是1度，记作 1 circ ,根据这一原理，人们制作了度量角的工具量角器。据此解答.\n【详解】根据将圆平均分成360份的原理，人们制作了度量角的工具。",
+        "knowledge_points":  [
+                                 "角的度量"
+                             ],
+        "source":  "24-25四年级上·河北沧州·期中"
+    },
+    {
+        "id":  "454",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "方方测量一个钝角时，角的顶点对准量角器的中心，一条边与量角器的内圈0 circ 刻度线对齐，看到另一条边对应的是 60 circ 刻度线，就写下了 60 circ ,这个角实际是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "120",
+        "solution":  "【分析】锐角小于 90 circ ,钝角大于 90 circ 而小于 180 circ 。 。量角器上同一刻度线对应的两个度数和是 180 circ ,所以用 180 circ 减 60 circ 即可求出这个钝角的度数.\n【详解】 180 circ - 60 circ = 120 circ方方测量一个钝角时，角的顶点对准量角器的中心，一条边与量角器的内圈 0 circ 刻度线对齐，看到另一条边对应的是60 circ 刻度线，就写下了 60 circ ,这个角实际是 120 circ 。",
+        "knowledge_points":  [
+                                 "角的度量"
+                             ],
+        "source":  "24-25四年级上·河北邯郸·期中"
+    },
+    {
+        "id":  "455",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "钟面上,3 时整时，时针和分针成( )角,( )时整时，时针和分针成平角;4 时整时，时针和分针成( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "直;6;钝",
+        "solution":  "【分析】钟面上一共有 12 个大格，每两个大格之间的夹角是 30°。3 时整时，时针指向 3，分针指向 12，钟面上的时针和分针之间有3个大格，夹角是 3×30 circ = 90 circ ,等于 90 ° 的角叫做直角;6时整时，时针指向6，分针指向12，钟面上的时针和分针之间有6个大格，夹角是 6×30 circ = 180 circ ,等于180°的角叫做平角;4时整时，时针指向4，分针指向12，时针和分针之间有4个大格，夹角是 4×30 circ = 120 circ ,大于90°小于 180 circ 的角叫做钝角。据此填空即可.\n【详解】由分析可知:3 时整时，夹角是 3×30 circ = 90 circ;6时整时，夹角是 6×30 circ = 180 circ;4时整时，夹角是 4×30 circ = 120 circ;因此，钟面上,3时整时，时针和分针成直角,6时整时，时针和分针成平角;4 时整时，时针和分针成钝角。",
+        "knowledge_points":  [
+                                 "直角",
+                                 "钝角",
+                                 "锐角的认识及特征",
+                                 "平角",
+                                 "周角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25四年级上·浙江温州·期中"
+    },
+    {
+        "id":  "456",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "下午4时，时针和分针成 角，是 度。",
+        "options":  [
+
+                    ],
+        "answer":  "钝;120",
+        "solution":  "【分析】本题主要考查学生对角的度量和分类知识的掌握。钟面上有 12 大格，每一大格对应的夹角是 30 度;4 时整，时针指向4，分针指向12,4到12有4大格，时针和分针的角是 30×4 = 120 (度),根据大于 90 度小于 180 度的角是钝角，以此判断即可.\n【详解】根据分析可知:30 circ×4 = 120 (度)180\u003e120\u003e90下午4时，时针和分针成钝角，是120度。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "直角",
+                                 "钝角",
+                                 "锐角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "23-24四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "457",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个直角和一个锐角可以拼成一个( )角;从一个平角里剪去一个钝角，可以得到一个( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "钝;锐",
+        "solution":  "【分析】直角是 90 circ c 。大于 0 circ 小于 90 circ 的是锐角。平角是180 circ ,大于 90 circ 小于 180 circ 是钝角。据此解答.\n【详解】因为锐角大于 0 circ 小于 90 circ ,直角是 90 circ 。它们相加大于 90 circ 小于 180 circ ,所以一个直角和一个锐角可以拼成一个钝角。 平角是 180 circ ,钝角大于 90 circ 小于 180 circ 。它们相减大于 0 circ 小于90 circ ,所以从一个平角里剪去一个钝角，可以得到一个锐角。",
+        "knowledge_points":  [
+                                 "直角",
+                                 "钝角",
+                                 "锐角的认识及特征",
+                                 "平角",
+                                 "周角的认识及特征"
+                             ],
+        "source":  "24-25四年级上·福建福州·期中"
+    },
+    {
+        "id":  "458",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "从12时整到1时整，分针旋转了( )°,是一个( )角;时针旋转了( )°是一个( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "360;周;30;锐",
+        "solution":  "【分析】钟面上有12大格，每一大格对应的夹角是 30 circ ,旋转一圈就是 360 circ 。从题目可知,12时整到1时整，分针旋转了 1 周，时针旋转了一大格。根据小于 90 circ 的角是锐角，等于 90 circ 的角是直角，大于90°的角是锐角，等于180°的角是平角，等于 360°的角是周角。据此解答.\n【详解】从12时整到1时整，分针旋转了一圈，一圈等于360 circ ,周角等于 360 circ 。同时时针转动一大格，一大格是30 circ ,即时针旋转了 30 circ 是一个锐角。",
+        "knowledge_points":  [
+                                 "直角",
+                                 "钝角",
+                                 "锐角的认识及特征",
+                                 "平角",
+                                 "周角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25四年级上·福建福州·期中"
+    },
+    {
+        "id":  "459",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "∠1与∠2组成了一个直角，且∠1=40°,那么∠2=( )°;∠2+∠3=180°,则∠3=( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "50;130",
+        "solution":  "【分析】根据题意可知,∠1+∠2=90°,则∠2=90°-∠1;∠3=180°-∠2;代入数据计算即可.\n【详解】\n90° - 40° = 50° \n\n180° - 50° = 130°\n\n\n则∠1与∠2组成了一个直角，且∠1=40°,那么∠2=50°;∠2+∠3=180°,则∠3=130°。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "直角",
+                                 "钝角",
+                                 "锐角的认识及特征"
+                             ],
+        "source":  "23-24四年级上·河北衡水·期末"
+    },
+    {
+        "id":  "460",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "中央电视台《新闻联播》每天19:00开始,19:30结束。开始时，钟面上时针和分针所形成的较小角是( )°,结束时，钟面上时针和分针所形成的较小角是( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "150;锐",
+        "solution":  "【分析】钟表一圈为 360 circ ,被平均分成了12个大格，那么每一个大格的角度是 30 circ 。因为19:00时，时针指向7，分针指向 12，它们之间间隔了 12 - i 7 = 5 (个)大格，又因为每一个大格是 30 circ ,所以时针和分针所夹的较小的角的度数用乘法计算。又因为19:30时，时针指向7和8之间，分针指向6，它们之间间隔了一个半大格，又因为每一个大格是 30 circ ,所以时针和分针所夹的较小的角的度数 30 circ + 30 circ 2，根据小于90°的角是锐角，等于 90 circ 的角是直角，大于 90 circ 而小于180°的角是钝角.\n【详解】根据分析可知:\n12 - i 7 = 5 (个)\n\n\n\n30 circ×5 = 150 circ\n\n\n19:00钟面上时针和分针所形成的较小角是 150 circ circ\n\n30 circ + 30 circ 2 \n = 30 circ + 15 circ \n = 45 circ\n\n45°,90°,19:30钟面上时针和分针所形成的较小角是锐角。中央电视台《新闻联播》每天19:00开始,19:30结束。开始时，钟面上时针和分针所形成的较小角是 150 circ ,结束时，钟面上时针和分针所形成的较小角是锐角。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "直角",
+                                 "钝角",
+                                 "锐角的认识及特征",
+                                 "整点时间与钟面指针的位置",
+                                 "一般时间与钟面指针的位置"
+                             ],
+        "source":  "23-24四年级上·四川绵阳·期末"
+    },
+    {
+        "id":  "461",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "3 点半时，时针和分针所形成的角是( )角,( )时整，时针和分针的夹角是180°。",
+        "options":  [
+
+                    ],
+        "answer":  "锐;6",
+        "solution":  "【分析】钟面上12个数字，以表芯为旋转点，表针转一圈是 360 circ ,被12个数字平均分成12份，每一份也就是一大格之间的夹角是 30 circ 。当时针和分针之间较小的夹角有3大格时, 30 circ×3 = 90 circ ,是个直角;当时针和分针之间较小的夹角小于3大格时，是个锐角;3时半时，分针指向6，时针指在3与4中间，它们之间较小的夹角小于3大格，此时时针和分针所形成的角是锐角;要使时针和分针的夹角是180 circ , 30×6 = 180 circ ,则时针和分针之间有6个大格。整时，分针指向 12，则时针应指向 6 ‰ .\n【详解】3点半时，时针和分针所形成的角是锐角,6时整，时针和分针的夹角是 180 circ circ 。",
+        "knowledge_points":  [
+                                 "直角",
+                                 "钝角",
+                                 "锐角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "23-24四年级上·河南焦作·期末"
+    },
+    {
+        "id":  "462",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "钟面上,( )时整，时针和分针组成的较小的角是直角;4时整，时针和分针组成的较小的角的度数是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "3 或 9;120",
+        "solution":  "【分析】根据直角的含义：等于 90 circ 的角叫做直角，并结合实际，时钟上 12 个数字把钟面平均分成 12 个大格，每个大格的度数是 30 circ ,整时，分针指向12，当时针指向3或9时，时针与分针之间有3个大格，时针和分针所形成的角是30 circ×3 = 90 circ ,是直角;4时整，时针指向4，分针指向12，时针与分针之间有4个大格，时针和分针所形成的角是 30 circ×4 = 120 circ;据此解答即可.\n【详解】 30 circ×3 = 90 circ\n30 circ×4 = 120 circ\n钟面上,3 或 9 时整，时针和分针组成的较小的角是直角;4时整，时针和分针组成的较小的角的度数是 120 circ 。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "角的分类及换算",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "23-24四年级上·四川乐山·期末"
+    },
+    {
+        "id":  "463",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "钟面上,3时整，时针与分针成( )角;从3时到3时30分，分针转动形成的角是( )角;5 时整，时针与分针成( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "直;平;钝",
+        "solution":  "【分析】钟面一周为 360 circ ,共分12大格，每格为 30 circ ,3时整，分针与时针相差3个大格，所以钟面上时针与分针形成的夹角是: 30 circ×3 = 90 circ ,是直角;从3时到3时30分，分针转动6大格，形成的角是 30 circ×6 = 180 circ ,是平角;5时整，分针与时针相差5个大格，所以钟面上时针与分针形成的夹角是: 30 circ×5 = 150 circ ,是钝角。据此解答即可.\n【详解】钟面上,3时整，时针与分针成直角;从3时到3时30分，分针转动形成的角是平角;5时整，时针与分针成钝角。",
+        "knowledge_points":  [
+                                 "角的度量",
+                                 "直角",
+                                 "钝角",
+                                 "锐角的认识及特征",
+                                 "平角",
+                                 "周角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "23-24四年级上·重庆綦江·期末"
+    },
+    {
+        "id":  "464",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小红用量角器量角时，误把内圈刻度看成了外圈刻度，量出的角是 135 circ ,这个角实际是 °。",
+        "options":  [
+
+                    ],
+        "answer":  "45",
+        "solution":  "【分析】在量角器上的内圈刻度和外圈刻度数字相加的和是 180 circ ,所以，用 180 circ 减看错的外圈刻度 135 circ ,即得到内圈刻度，也就是这个角实际的度数。据此解答.\n【详解】 180 circ - 135 circ = 45 circ所以，这个角实际是 45 circ circ",
+        "knowledge_points":  [
+                                 "用量角器画角"
+                             ],
+        "source":  "23-24三年级上·四川南充·期末"
+    },
+    {
+        "id":  "465",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在( )里填上合适的数.\n\n200 公顷=( )平方千米 ( )个直角 = 1 个周角\n\n3 平方千米=( )公顷=( )平方米",
+        "options":  [
+
+                    ],
+        "answer":  "2;4;300;3000000",
+        "solution":  "【分析】公顷与平方千米间的进率是100，平方千米与平方米间的进率是1000000，由高级单位化低级单位乘进率，反之除以进率;直角 = 90 circ ,周角 = 360 circ;据此进行解答即可.\n【详解】因为 200 公顷里面含有 2 个 100 公顷，所以 200 公顷 = 2 平方千米;因为 4×90 circ = 360 circ ,所以 4 个直角 = 1 个周角;因为 3×100 = 300 (公顷), 3×1000000 = 3000000 (平方米),所以3 平方千米 = 300 公顷 = 3000000 平方米。",
+        "knowledge_points":  [
+                                 "角的分类及换算",
+                                 "公顷",
+                                 "平方千米的进率与换算"
+                             ],
+        "source":  "23-24四年级上·四川广元·期末"
+    },
+    {
+        "id":  "466",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把一张圆形纸片对折3次，所得到的角是( )度，这是一个( )角。把一个平角分成两个角，其中一个是直角，另一个角是()角。",
+        "options":  [
+
+                    ],
+        "answer":  "45;锐;直",
+        "solution":  "【分析】把一张圆形纸片对折1次，这张圆形纸片被平均分成2份，得到的角是平角。把这张圆形纸片再对折1次，平角被平均分成2份，得到的角是直角。把这张圆形纸片再对折1次，直角被平均分成2份，得到的角的度数是 45 circ;大于 0 circ 小于 90 circ 的角是锐角。把平角分成两个角，其中一个是直角，据此作图如下：由图可知，另一个角也是直角.\n【详解】 360 circ 2 = 180 circ\n180 circ 2 = 90 circ\n\n\n\n90 circ 2 = 45 circ\n把一张圆形纸片对折3次，所得到的角是45度，这是一个锐角。把一个平角分成两个角，其中一个是直角，另一个角是直角。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "平角",
+                                 "周角的认识及特征",
+                                 "角的分类及换算"
+                             ],
+        "source":  "23-24四年级上·重庆綦江·期末"
+    },
+    {
+        "id":  "467",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "钟面上是 3 时整时，时针与分针所成的较小的角是( )角，半小时后，时针与分针所成的较小的角是( )角。(填\"锐\"\"钝\"\"直\"\"平\"或\"周\")",
+        "options":  [
+
+                    ],
+        "answer":  "直;锐",
+        "solution":  "【分析】因为钟表上的刻度是把一个圆平均分成了12等份，每一份是 30 circ ,钟面上3时整时，时针指向3，分针指向12，时针和分针之间相差3个大格数，它们之间的夹角度数是 3×30 circ = 90 circ;半小时 = 30 分，也就是半小时后是3时30分,3时30分，时针指在3与4之间，分针指向6，时针与分针之间间隔2大格还有半个大格，钟面1大格是 30 circ , 30 circ 乘2，再加 15 circ 即可求出夹角度数;大于 0 circ 小于 90 ° 的角叫做锐角;等于 90 circ 的角叫做直角;大于 90 circ 小于 180 circ 的角叫做钝角;等于180°的角叫做平角;等于 360 circ 的角叫做周角;据此根据角的分类判断即可.\n【详解】 3×30 circ = 90 circ\n\n\n2×30 circ + 30 circ 2 \n = 60 circ + 15 circ \n = 75 circ\n\n\n钟面上是3时整时，时针与分针所成的较小的角是直角，半小时后，时针与分针所成的较小的角是锐角。",
+        "knowledge_points":  [
+                                 "角的分类及换算",
+                                 "整点时间与钟面指针的位置",
+                                 "半点时间与钟面指针的位置"
+                             ],
+        "source":  "23-24四年级上·四川广元·期末"
+    },
+    {
+        "id":  "468",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "4时整，时针与分针所形成的较小的夹角是( )°,1时整，时针与分针所形成的较小的夹角是( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "120;锐",
+        "solution":  "【分析】钟表上的 12 个数字把钟表平均分成了 12 个大格，一个大格是 30 circ ,如图所示,4 时整，时针和分针形成的最小的夹角是钝角，从 12 到 4 是四个大格，即用 30 circ×4 即可计算出夹角;1时整时，分针指向12，时针指向 1，如图时针与分针所形成的较小的夹角是锐角.\n\n![](images/8bd0d48da70cb43358abd9797303a9698c5a38af1b02edb4df24b7bae9627d07.jpg)\n\n![](images/d106fa23071ee550a7ecd29a8d5f83f24c12d89dccc5946a892167470a538fb3.jpg)\n\n【详解】 30 circ×4 = 120 circ4时整，时针与分针所形成的较小的夹角是 120 circ ,1时整，时针与分针所形成的较小的夹角是锐角。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "直角",
+                                 "钝角",
+                                 "锐角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "22-23四年级上·陕西延安·期末"
+    },
+    {
+        "id":  "469",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "3 时整时，钟面上时针与分针成( );9 时 30 分时，时针与分针成( )。(填\"锐角\" \"直角\"或\"钝角\")",
+        "options":  [
+
+                    ],
+        "answer":  "直角;钝角",
+        "solution":  "【分析】钟表一圈为 360 circ ,被平均分成了12个大格，那么每一个大格的角度是 36012 = 30 \\ circ 。 。3时整，时针指向3，分针指向12，中间有3个大格，所以它们所成的角度是\n30 circ×3 = 90 circ\n当9时30分时，分针指向6，时针在9和10正中间，时针和分针之间有3个半大格,3个格的角度是 30 circ×3 = 90 circ ,半个格的角度是 30 circ 2 = 15 circ ,所以此时时针和分针的夹角是90 circ + 15 circ = 105 circ 。锐角是小于 90 circ 的角，直角是等于 90 ° 的角，钝角是大于 90 circ 而小于 180 circ 的角。据此解答.\n【详解】3 时整时，钟面上时针与分针成 30 circ×3 = 90 circ ,是直角.\n\n30 circ×3 = 90 circ \n 30 circ 2 = 15 circ\n\n9 时 30 分时，时针与分针成 90 circ + 15 circ = 105 circ , 0 circ \u003c 105 circ \u003c 180 circ ,所以 105°是钝角.\n3时整时，钟面上时针与分针成直角;9时30分时，时针与分针成钝角。",
+        "knowledge_points":  [
+                                 "直角",
+                                 "钝角",
+                                 "锐角的认识及特征",
+                                 "角的分类及换算",
+                                 "整点时间与钟面指针的位置",
+                                 "半点时间与钟面指针的位置"
+                             ],
+        "source":  "23-24四年级上·西藏林芝·期末"
+    },
+    {
+        "id":  "470",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "《新闻联播》在每天的19:00播出,19:30结束。开始时，钟面上时针和分针所夹的较小的角是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "150",
+        "solution":  "【分析】钟表一圈为360°,被平均分成了12个大格，那么每一个大格的角度是 360°/12=30°。因为19:00时，时针指向7，分针指向12，它们之间间隔了 12-7=5(个)大格，又因为每一个大格是30°,所以时针和分针所夹的较小的角的度数是 30°*5=150°.\n【详解】\n360°/12=30° \n\n12-7=5 (个) \n\n30°*5=150°\n\n\n《新闻联播》在每天的19:00播出,19:30结束。开始时，钟面上时针和分针所夹的较小的角是150°。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "23-24四年级上·江西赣州·期末"
+    },
+    {
+        "id":  "471",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "平角等于( )度，周角等于( )度。",
+        "options":  [
+
+                    ],
+        "answer":  "180;360",
+        "solution":  "【分析】一条射线绕它的端点旋转半周，形成的角叫作平角;1 平角 = 180 circ;条射线绕它的端点旋转一周，形成的角叫作周角;1 周角= 360 circ;据此即可解答.\n【详解】平角等于 180 度，周角等于 360 度。",
+        "knowledge_points":  [
+                                 "平角",
+                                 "周角的认识及特征"
+                             ],
+        "source":  "23-24四年级上·河南商丘·期末"
+    },
+    {
+        "id":  "472",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "6时整，时针和分针的夹角是一个( )角;从 6 时到 7 时分针走了( )度。",
+        "options":  [
+
+                    ],
+        "answer":  "平;360",
+        "solution":  "【分析】6时整，分针指向12，时针指向6，两者在同一条直线上，方向相反时，所构成的角叫平角。从6时到7时，过了 1 小时，分针走了一圈，即 360 度.\n【详解】6 时整，时针和分针的夹角是一个(平)角;从 6时到 7 时分针走了(360)度。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "平角",
+                                 "周角的认识及特征"
+                             ],
+        "source":  "23-24四年级上·重庆万州·期末"
+    },
+    {
+        "id":  "473",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "将一张圆形纸片对折 1 次，可以得到的角是( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "平",
+        "solution":  "【分析】周角等于 360 circ ,平角等于 180 circ ,一张圆形纸片的圆心角是一个 360 circ 的周角，对折一次，就是把周角平均分成了 3602 = 180 circ 的平角，据此解答即可.\n【详解】 3602 = 180 circ将一张圆形纸片对折 1 次，可以得到的角是平角。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "图形的折叠问题",
+                                 "平角",
+                                 "周角的认识及特征"
+                             ],
+        "source":  "24-25四年级上·广西贵港·期中"
+    },
+    {
+        "id":  "474",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在钟面上，从 10 点到 12 点，时针转动的度数是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "60",
+        "solution":  "【分析】把钟面看作一个周角是 360 circ ,钟面上一共有 12 个大格，每个大格是 30 circ;10点时针指着10,12点时针指着12，从10点到12点，时针则从10转到12，转过了12-10= 2 (个)大格，再用每个大格的度数 30 circ 乘2，即得到时针转动的度数。据此解答.\n【详解】 12 - 10 = 2 (个)n30 circ×2 = 60 circn所以，在钟面上，从 10 点到 12 点，时针转动的度数是60°。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "角的度量",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "24-25四年级上·广西贵港·期中"
+    },
+    {
+        "id":  "475",
+        "unit":  "第三章角的度数是",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "线段是直的，有()个端点;将一条线段向两端无限延伸，就得到一条( )线;把一条线段向一个方向无限延伸，就形成了一条( )线。",
+        "options":  [
+
+                    ],
+        "answer":  "2;直;射",
+        "solution":  "【分析】根据线段的特点可知，线段是直的，有两个端点，长度是可测量的。射线只有一个端点，可以向没有端点的端无限延长，直线没有端点，可以向两端无限延长。据此解答.\n【详解】如下图所示，线段是直的，有2个端点。将一条线段向两端无限延伸，就得到一条射线;把一条线段向一个方向无限延伸，就形成了一条直线。线段：射线：直线:",
+        "knowledge_points":  [
+                                 "线段",
+                                 "直线",
+                                 "射线的认识及特征"
+                             ],
+        "source":  "24-25四年级上·广西贵港·期中"
+    },
+    {
+        "id":  "476",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "直角，平角和周角中，周角最大，不存在比周角还大的角。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】锐角：大于 0 circ 小于 90 circ 的角;钝角：大于 90 circ 。 小于180°的角;直角：等于 90 circ 的角;平角：等于 180 circ 的角;周角：等于360°的角;我们学过的角周角是最大的角，据此解答即可.\n【详解】由由分析知：直角\u003c平角\u003c周角，所以直角，平角和周角中，周角最大，不存在比周角还大的角。原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "直角",
+                                 "钝角",
+                                 "锐角的认识及特征",
+                                 "平角",
+                                 "周角的认识及特征"
+                             ],
+        "source":  "24-25四年级上·内蒙古呼伦贝尔·期末"
+    },
+    {
+        "id":  "477",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "用一副三角板不能画出 15°的角。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】一副三角尺由两个三角尺组成，一个三角尺的三个角分别是 90 circ , 45 circ , 45 circ ,另一个三角尺的三个角分别是90 circ , 60 circ , 30 circ 。由题意得，将两个三角尺的角度相加或相减看能否得到 15 circ 即可.\n【详解】 45 circ - 30 circ = 15 circ ,即用一副三角板可以画出15°的角。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "用三角尺画角"
+                             ],
+        "source":  "23-24四年级上·吉林白城·期末"
+    },
+    {
+        "id":  "478",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "把半圆平均分成180份，每一份所对的角的大小是 1°。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】一个半圆的度数就是平角，一个平角的度数是180 circ ,因此把半圆平均分成180份，每一份所对的角的大小就是1°。据此进行判断即可.\n【详解】把半圆平均分成180份，每一份所对的角的大小是1°。该说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "平角",
+                                 "周角的认识及特征"
+                             ],
+        "source":  "23-24四年级上·吉林松原·期末"
+    },
+    {
+        "id":  "479",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "用 10 倍的放大镜看一个 20 circ 的角，结果是 200 circ 。 。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】因为角的大小与角的两边的长短没有关系，只与两边叉开的大小有关;用放大镜看一个角，只能把两边的粗细和长度放大，两边叉开的角度不变，所以看到的角大小不变，仍然是 20 circ mathfrak c 。据此判断.\n【详解】由分析可得：用 10 倍的放大镜看一个 20 circ 的角，结果是 20 circ 。 。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "角的概念及表示方法"
+                             ],
+        "source":  "24-25四年级上·黑龙江齐齐哈尔·期末"
+    },
+    {
+        "id":  "480",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "小红画了一条 30 厘米长的射线。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】根据直线，射线和线段的含义：线段有2个端点，有限长，可以度量;射线有一个端点，无限长，不可度量;直线无端点，无限长，不可度量;进而解答即可.\n【详解】据分析可知可以说小红画了一条30厘米长的线段，原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "线段",
+                                 "直线",
+                                 "射线的认识及特征"
+                             ],
+        "source":  "23-24四年级上·浙江宁波·期末"
+    },
+    {
+        "id":  "481",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "两个锐角的度数和一定比 90 circ 大。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】根据锐角，直角和钝角的定义：大于 0 circ 小于 90 circ 的角叫做锐角;等于 90 circ 的角叫做直角;大于 90 circ 。 小于 180 circ 的角叫做钝角;进行举例判断即可.\n【详解】假设两个锐角分别是 30 circ 和 20 circ , 30 circ + 20 circ = 50 circ;50 circ \u003c 90 circ ,这两个锐角的度数和比 90 circ 小;假设两个锐角分别是 30 circ 和 60 circ , 30 circ + 60 circ = 90 circ;90 circ = 90 circ ,这两个锐角的度数和等于 90 circ;再假设这两个锐角分别是 50 circ 和 60 circ , 50 circ + 60 circ = 110 circ;110 circ \u003e 90 circ ,这两个锐角的度数和比 90 circ 大;所以两个锐角的度数和可能比 90 circ 小，也可能等于 90 circ ,还可能大于 90 °;故原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "直角",
+                                 "钝角",
+                                 "锐角的认识及特征",
+                                 "角的大小比较"
+                             ],
+        "source":  "22-23四年级上·山东日照·期末"
+    },
+    {
+        "id":  "482",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "钟面上 4 时整，分针和时针组成的较小的夹角是钝角。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】因为钟面上12个数字，以表芯为旋转点，表针转圈是 360 circ ,被12个数字平均分成12份，每一份也就是两数之间夹角是 30 circ;当钟面上4时整，时针指向4，分针指向12，时针与分针之间较小角有4个大格，夹角的度数是4×30 circ ,据此解答即可.\n【详解】钟面上 4 时整，时针指向 4，分针指向 12，时针与分针之间较小角有4个大格夹角的度数是 4×30 circ = 120 circ 90 circ \u003c 120 circ \u003c 180 circ 所以时针和分针组成的较小的角是钝角。故答案为:√",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "直角",
+                                 "钝角",
+                                 "锐角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "23-24四年级上·河北保定·期末"
+    },
+    {
+        "id":  "483",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "在线段，射线和直线中，直线最长，线段最短。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】本题主要考查学生对线段，射线，直线的概念及特征的掌握和灵活运用。直线上任意两点之间的一段叫做线段;把线段的一端无限延长，得到一条射线;把线段的两端无限延长，得到一条直线;线段的长度是有限的，直线和射线的长度都是无限的。据此即可解答.\n【详解】根据分析可知，线段的长度可以测量，直线和射线的长度都是无限的，不能测量;所以判断错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "线段",
+                                 "直线",
+                                 "射线的认识及特征"
+                             ],
+        "source":  "23-24四年级上·江西赣州·期末"
+    },
+    {
+        "id":  "484",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "小尼画了一条长 1米的直线。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】把线段的两端无限延长，得到一条直线，直线不可度量，据此解答.\n【详解】根据分析可知：因为直线不可度量，所以小尼画了一条长1米的直线说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "线段",
+                                 "直线",
+                                 "射线的认识及特征"
+                             ],
+        "source":  "23-24四年级上·陕西安康·期末"
+    },
+    {
+        "id":  "485",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "两个锐角可以拼成一个锐角，也可以拼成一个直角，还可以拼成一个钝角。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】锐角是指大于 0 circ 而小于 90 circ 的角。直角是等于 90 circ 的角。钝角是大于 90 ° 而小于 180 circ 的角。两个锐角拼成的角，度数应小于 180 circ ,可以是锐角，可以是直角，也可以是钝角。据此举例判断.\n【详解】例如 30 circ + 40 circ = 70 circ ,拼成一个锐角。30 circ + 60 circ = 90 circ ,拼成一个直角。60 circ + 70 circ = 130 circ ,拼成一个钝角。两个锐角可以拼成一个锐角，也可以拼成一个直角，还可以拼成一个钝角。说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "直角",
+                                 "钝角",
+                                 "锐角的认识及特征"
+                             ],
+        "source":  "24-25四年级上·吉林白山·期末"
+    },
+    {
+        "id":  "486",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "用一个锐角和一个钝角一定能拼成平角。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】小于 90 ° 的角是锐角，大于 90 circ 且小于 180 circ 的角是钝角,1平角是 180 circ ,一个锐角和一个钝角拼成的角应大于90 circ 且小于 270 circ ,可能是钝角，也可能是平角，据此解答.\n【详解】假设一个锐角是 20 circ ,一个钝角是 100 circ , 20 circ + 100 circ = 120 circ ,此时拼成的是一个钝角，所以用一个锐角和一个钝角不一定能拼成平角，原说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "直角",
+                                 "钝角",
+                                 "锐角的认识及特征",
+                                 "平角",
+                                 "周角的认识及特征"
+                             ],
+        "source":  "24-25四年级上·陕西宝鸡·期末"
+    },
+    {
+        "id":  "487",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "直角 + 锐角 = 钝 角。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】首先明确锐角是小于 90 circ 的角;直角是等于 90 ° 的角;钝角是大于 90 circ 小于 180 circ 的角;90 circ 加上小于 90 circ 的角应该是大于 90 circ 小于 180 circ 的角，所以锐角+直角 = 钝角。以此答题即可.\n【详解】根据分析可知：直角+锐角 = 钝角。说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "直角",
+                                 "钝角",
+                                 "锐角的认识及特征"
+                             ],
+        "source":  "24-25四年级上·河北秦皇岛·期末"
+    },
+    {
+        "id":  "488",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "线段和射线都可以看成直线上的一部分，所以射线比直线短。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】线段有两个端点，可以测量出长度;射线只有一个端点，不可以测量出长度;直线没有端点，是无限长的;因此射线和直线无法比较长短，依此即可判断.\n【详解】线段和射线都可以看成直线上的一部分，由于射线和直线都可以无限延长，因此不能比较射线和直线的长短。故答案为: ×",
+        "knowledge_points":  [
+                                 "线段",
+                                 "直线",
+                                 "射线的认识及特征"
+                             ],
+        "source":  "23-24四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "489",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "时针从钟面的数字\"12\"走到\"1\"，分针走了 360°。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】钟面有 12 个大格，每个大格有 5 个小格，一共有60个小格;时针每走1大格就是1小时，分针每走1小格就是 1 分，分针走一圈是 60 个小格就是 60 分,1 时 = 60 分;时针从钟面的数字\"12\"走到\"1\"，走过了1大格，就是1小时，即60分，所以分针走了一圈，可以看作走了一个周角，即 360 circ 。据此判断.\n【详解】根据分析可知：时针从钟面的数字\"12\"走到\"1\"，分针走了 360 circ mathfrak c故答案为:√",
+        "knowledge_points":  [
+                                 "平角",
+                                 "周角的认识及特征",
+                                 "角的分类及换算",
+                                 "时",
+                                 "分的认识"
+                             ],
+        "source":  "23-24三年级上·四川南充·期末"
+    },
+    {
+        "id":  "490",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "小明画了一组 1 米长的平行线。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】同一平面内不相交的两条直线互相平行，其中条直线是另一条直线的平行线。直线没有端点，可以向两端无限延长，它的长度不能被度量，据此解答即可.\n【详解】平行线是直线，直线是无限长的，可以说成小明画了一组平行线，所以原说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "线段",
+                                 "直线",
+                                 "射线的认识及特征"
+                             ],
+        "source":  "23-24四年级上·河北唐山·期末"
+    },
+    {
+        "id":  "491",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "3 时半时，钟面上时针与分针所成的角是直角。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】钟面上一圈为 360 circ ,被平均分成了12个大格，那么每一个大格的角度是 360 circ 12 = 30 circ 。3时半时，时针在3和4的正中间，分针指向6。此时时针和分针之间共有2个加上半个大格。2个大格的角度是30 circ×2 = 60 circ ,半个大格的角度是 30 circ 2 = 15 circ ,它们所成角的度数是 60 circ + 15 circ 。算出角度再根据\"锐角是小于 90 circ 的角，直角是等于 90 circ 的角，钝角是大于 90 ° 而小于 180 circ 的角.\"判断钟面上时针与分针所成的角是什么角.\n【详解】 360 circ 12 = 30 circn30 circ×2 = 60 circnnn30 circ 2 = 15 circnn60 circ + 15 circ = 75 circn因为 75 circ \u003c 90 circ ,所以 75 circ 是锐角。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "直角",
+                                 "钝角",
+                                 "锐角的认识及特征",
+                                 "半点时间与钟面指针的位置"
+                             ],
+        "source":  "19-20四年级上·山东·期末"
+    },
+    {
+        "id":  "492",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "因为射线有 1 个端点，所以射线比直线短，比线段长。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】直线没有端点，两边可无限延长;射线有一端有端点，另一端可无限延长;据此解答.\n【详解】因为直线和射线都无限长，所以无法比较长短。原题干说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "线段",
+                                 "直线",
+                                 "射线的认识及特征"
+                             ],
+        "source":  "23-24四年级上·湖南娄底·期末"
+    },
+    {
+        "id":  "493",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "王红画的一条长10cm 的线段比李佳画的一条长 3cm 的射线长。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】根据直线，射线和线段的含义：线段有2个端点，有限长，可以度量;射线有一个端点，无限长;直线无端点，无限长;进而解答即可.\n【详解】根据分析可知，射线不能度量长度，而原题中说李佳画的一条长 3cm 的射线，说法错误.\n故答案为: ×",
+        "knowledge_points":  [
+                                 "线段",
+                                 "直线",
+                                 "射线的认识及特征"
+                             ],
+        "source":  "23-24四年级上·重庆綦江·期末"
+    },
+    {
+        "id":  "494",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "经过一点可以画无数条直线，经过两点只能画一条直线。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】根据直线的特征：经过两个点可以画一条直线，并且只能画一条直线，如下图所示，过两点只能画一条直线，过一点可以画无数条直线，据此判断.\n\n![](images/7ef8fac9d0479a5da482d1e39faca0dd771f6aa2394b6dbaa8d3dbf56cacfcbf.jpg)\n\n【详解】由分析可知，经过一点可以画无数条直线，经过两点只能画一条直线;原题干说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "线段",
+                                 "直线",
+                                 "射线的认识及特征"
+                             ],
+        "source":  "15-16四年级·河南郑州·期中"
+    },
+    {
+        "id":  "495",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "透过放大镜看15度的角，这个角变大。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】角的大小与两边的长短无关，只有边叉开的大小有关，叉开得越大，角越大，叉开得越小，角越小。角的大小只和两条边张开的大小有关，与边的长短无关。据此判断.\n【详解】透过放大镜看 15 度的角，角的边变长，角度不变还是15度。原说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "角的度量",
+                                 "角的大小比较"
+                             ],
+        "source":  "23-24四年级上·湖北咸宁·期末"
+    },
+    {
+        "id":  "496",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "1 周角 = 360 circ 。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】周角的两条边重合,1个周角的度数是 360 circ ,据此解答.\n【详解】1周角 = 360 circ ,这句话是正确的。故答案为:√",
+        "knowledge_points":  [
+                                 "平角",
+                                 "周角的认识及特征"
+                             ],
+        "source":  "23-24四年级上·青海果洛·期末"
+    },
+    {
+        "id":  "497",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "张鹏用一副三角尺能画出一个 105°的角。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】一副三角尺上的度数有 30 circ , 45 circ , 60 circ , 90 circ ,可以拼出 15 circ ( · 45 circ - 30 circ 或 60 circ - 45 circ ) ),75°(30°+45°),105°(45°+60°),120°(30°+90°),135°(45°+ 90 circ , ),150°(60°+90°), 180 circ (90°+90°);据此判断.\n【详解】根据分析可知, 45 circ + 60 circ = 105 circ ,则张鹏用一副三角尺能画出一个 105 circ 的角。所以原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "用三角尺画角"
+                             ],
+        "source":  "23-24四年级上·江西赣州·期末"
+    },
+    {
+        "id":  "498",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "12 时半，钟面上时针和分针形成的角是平角。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】在钟面上，一共有 12 个大格，时针与分针所夹的每一个大格是 30 circ ,12时半，时针指向12和1之间，分针指向6，钟面上时针和分针形成的角相差5个大格和半大格，列式计算，再根据平角是 180 circ ,判断即可.\n【详解】由分析可得:12 时半，时针指向 12 和 1 之间，分针指向6，钟面上时针和分针形成的角相差5个大格半。n30 circ×5 = 150 circnn30 circ 2 = 15 circnnn150 circ + 15 circ = 165 circnnn165 circ \u003c 180 circn12时半，钟面上时针和分针形成的角不是平角。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平角",
+                                 "周角的认识及特征",
+                                 "一般时间与钟面指针的位置"
+                             ],
+        "source":  "23-24四年级上·江西赣州·期末"
+    },
+    {
+        "id":  "499",
+        "unit":  "第三章角的度数是",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "一条直线长 560 米。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】直线没有端点，向两端无限延长，长度是无法度量的.\n【详解】一条直线长 560 米在概念上错误，直线长度无法衡量。故答案为: ×",
+        "knowledge_points":  [
+                                 "线段",
+                                 "直线",
+                                 "射线的认识及特征"
+                             ],
+        "source":  "24-25四年级上·广西贵港·期中"
+    },
+    {
+        "id":  "500",
+        "unit":  "第三章角的度数是",
+        "question_type":  "作图题",
+        "difficulty":  0.65,
+        "question":  "选择合适的工具，分别画一个 40 circ 的角和一个 120°的角。",
+        "options":  [
+
+                    ],
+        "answer":  "见详解",
+        "solution":  "【分析】用量角器画角，先画一条射线，使量角器的中心和射线的端点重合,0刻度线和射线重合，分别在量角器40°, 120 circ 的地方点一个点，以画出的射线的端点为端点，通过刚画的点，再分别画一条射线，最后标出度数即可.\n【详解】如下图:\n\n![](images/43866f89f10c5c67fd2ebe426f27c9ac35bca2d81691cb4966c06aed6e99b534.jpg)",
+        "knowledge_points":  [
+                                 "用量角器画角"
+                             ],
+        "source":  "23-24四年级上·河南商丘·期末"
+    }
+]
diff --git "a/\345\233\233\345\271\264\347\272\247\351\242\230\351\233\206/\345\233\233\345\271\264\347\272\247\344\270\212/\345\233\233\345\271\264\347\272\247\344\270\212\347\254\254\344\272\214\345\215\225\345\205\203\345\271\263\346\226\271\345\215\203\347\261\263.json" "b/\345\233\233\345\271\264\347\272\247\351\242\230\351\233\206/\345\233\233\345\271\264\347\272\247\344\270\212/\345\233\233\345\271\264\347\272\247\344\270\212\347\254\254\344\272\214\345\215\225\345\205\203\345\271\263\346\226\271\345\215\203\347\261\263.json"
new file mode 100644
index 0000000000000000000000000000000000000000..70e14b8cb99d816acfbfb3f50d5e863a12c79532
--- /dev/null
+++ "b/\345\233\233\345\271\264\347\272\247\351\242\230\351\233\206/\345\233\233\345\271\264\347\272\247\344\270\212/\345\233\233\345\271\264\347\272\247\344\270\212\347\254\254\344\272\214\345\215\225\345\205\203\345\271\263\346\226\271\345\215\203\347\261\263.json"
@@ -0,0 +1,8883 @@
+[
+    {
+        "id":  "1",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一个正方形花坛的边长是 10 米，下面说法正确的是( )。",
+        "options":  [
+                        "A. 10 个这样的花坛的面积是 1 公顷",
+                        "B. 100 个这样的花坛的面积是 1 公顷",
+                        "C. 100 个这样的花坛的面积是 1 平方千米",
+                        "D. 1000 个这样的花坛的面积是 1 平方千米"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】首先根据正方形面积公式：面积 = 边长×边长，计算单个正方形花坛的面积，再明确面积单位换算:1公顷=10000平方米,1平方千米=1000000平方米。分别计算每一个选项，并进行判断即可解答.\n【详解】 10×10 = 100 (平方米)A。10 个花坛面积为 10×100 = 1000 平方米，不等于 1 公顷，错误。 B。100 个花坛面积为 100×100 = 10000 平方米，等于 1 公顷，正确。 C。100 个花坛面积为 1 公顷，远小于 1 平方千米，错误。 D。1000 个花坛面积为 1000×100 = 100000 平方米，不等于1 平方千米，错误。故答案为:B",
+        "knowledge_points":  [
+                                 "正方形的面积",
+                                 "公顷",
+                                 "平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "2",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "阳光小学校园的占地面积约为5公顷,( )个这样的校园占地面积约为1平方千米。",
+        "options":  [
+                        "A. 20",
+                        "B. 5",
+                        "C. 10",
+                        "D. 100"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】1平方千米 = 100 公顷，因此100公顷里面有几个5公顷，则几个这样的校园面积大约是1平方千米，依此换算并计算即可.\n【详解】1平方千米 = 100 公顷 100÷5 = 20 (个)20 个这样的校园面积大约是 1 平方千米。故答案为:A",
+        "knowledge_points":  [
+                                 "公顷",
+                                 "平方千米的进率与换算"
+                             ],
+        "source":  "25-26四年级上·浙江杭州·期中"
+    },
+    {
+        "id":  "3",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "长江发源于\"世界屋脊\"-- 青藏高原的唐古拉山脉各拉丹冬峰西南侧，流域面积达180000000公顷，长江流域面积合( )万平方千米。",
+        "options":  [
+                        "A. 18",
+                        "B. 180",
+                        "C. 1800",
+                        "D. 18000"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据100公顷 = 1 平方千米,180000000 公顷里面有 1800000 个 1 平方千米，所以 180000000 公顷=1800000平方千米。再根据把一个整数改写成用\"万\"作单位的数，把个级里4个0去掉同时在后面写上\"万\"字，据此即可求出合多少万平方千米。据此解答.\n【详解】根据分析得出:\n\n180000000 公顷=1800000 平方千米 1800000 平方千米 = 180 万平方千米 所以长江发源于\"世界屋脊\" 青藏高原的唐古拉山脉 各拉丹冬峰西南侧，流域面积达180000000公顷，长江流域 面积合180万平方千米。故答案为:B",
+        "knowledge_points":  [
+                                 "亿以内数的改写",
+                                 "公顷",
+                                 "平方千米的进率与换算"
+                             ],
+        "source":  "25-26四年级上·湖南岳阳·期中"
+    },
+    {
+        "id":  "4",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "天安门广场是当今世界上最大的城市广场，也是中国的文化象征之一，还是中国举办重大庆典的重要场所。其占地面积为 44 公顷，合( )平方米。",
+        "options":  [
+                        "A. 440000",
+                        "B. 44000",
+                        "C. 4400"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】根据题意,1 公顷=10000 平方米,44 公顷就是 44个 10000 平方米，据此转换即可.\n【详解】44 公顷就是 44 个 10000 平方米，是 440000 平方米。所以占地面积为 44 公顷，合 440000 平方米。故答案为:A",
+        "knowledge_points":  [
+                                 "公顷",
+                                 "平方千米的进率与换算"
+                             ],
+        "source":  "25-26四年级上·河南三门峡·期中"
+    },
+    {
+        "id":  "5",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面选项中，两个面积单位之间的进率不是 100 的是( )。",
+        "options":  [
+                        "A. 平方千米和公顷",
+                        "B. 公顷和平方米",
+                        "C. 平方米和平方分米",
+                        "D. 平方分米和平方厘米"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据 1 平方千米 = 100 公顷,1 公顷 = 10000 平方米,1平方米 = 100 平方分米,1平方分米 = 100 平方厘米，据此判断即可.\n【详解】A。平方千米和公顷之间的进率是 100;B。公顷和平方米之间的进率是10000;C。平方米和平方分米之间的进率是 100;D。平方分米和平方厘米之间的进率是 100;所以选项中，两个面积单位之间的进率不是100的是公顷和平方米。故答案为:B",
+        "knowledge_points":  [
+                                 "平方厘米，平方分米，平方米之间的进率与换算",
+                                 "公顷",
+                                 "平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·福建福州·期中"
+    },
+    {
+        "id":  "6",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "测量淮北市的面积用( )作单位比较合适。",
+        "options":  [
+                        "A. 平方米",
+                        "B. 公顷",
+                        "C. 平方千米",
+                        "D. 平方分米"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】根据生活经验和实际数据相结合即可解答，这里选择合适的面积单位。面积的常用单位有公顷，平方千米，平方米，平方分米等,1公顷比一个标准足球场稍大一点,1平方千米大约是小型湖泊的湖面面积,1平方米大约是1张方桌面的大小,1平方分米大约是1个成人手掌面的大小，国土面积的表示通常用平方千米作单位比较合适;据此解答.\n【详解】由分析可知，测量淮北市的面积用平方千米作单位比较合适。故答案为:C",
+        "knowledge_points":  [
+                                 "面积单位的选择",
+                                 "公顷",
+                                 "平方千米的认识"
+                             ],
+        "source":  "24-25 四年级上·安徽淮北·期末"
+    },
+    {
+        "id":  "7",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "北京奥林匹克公园的占地面积约是 11590000 平方米，合( )公顷。",
+        "options":  [
+                        "A. 1159000",
+                        "B. 115900",
+                        "C. 1159"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】1 公顷 = 10000 平方米,11590000 平方米是 1159 个10000平方米，也就是1159个1公顷.\n【详解】11590000 平方米是 1159 个 1 公顷，是 1159 公顷。那么北京奥林匹克公园的占地面积约是11590000平方米，合 1159 公顷。故答案为:C",
+        "knowledge_points":  [
+                                 "公顷",
+                                 "平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级·全国·假期作业"
+    },
+    {
+        "id":  "8",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "地处于北京奥林匹克公园中心区南部的国家体育场\"鸟巢\"，为 2008 年北京奥运会的主体育场，占地面积约 20( )。",
+        "options":  [
+                        "A. 平方千米",
+                        "B. 公顷",
+                        "C. 平方米",
+                        "D. 平方分米"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】平方千米常用在大面积的地方，比如国土面积，城市面积，海洋面积等;根据生活经验和实际数据相结合可知，计量2008年北京奥运会的主体育场的占地面积，作为大型体育场用公顷比较合适.\n【详解】国家体育场\"鸟巢\"，地处于北京奥林匹克公园中心区南部，为2008年北京奥运会的主体育场，占地面积约 20 公顷。故答案为:B",
+        "knowledge_points":  [
+                                 "公顷",
+                                 "平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "24-25 四年级上·重庆秀山·期末"
+    },
+    {
+        "id":  "9",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "一块长方形果园长500 米，宽 40 米，面积是( )公顷。",
+        "options":  [
+                        "A. 20000",
+                        "B. 20000",
+                        "C. 20",
+                        "D. 2"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】长方形的面积 = 长×宽，依此计算出长方形果园的面积，最后再根据\"10000 平方米 = 1 公顷\"将单位化成公顷即可.\n【详解】 500×40 = 20000 (平方米)20000 平方米 = 2 公顷这个长方形的面积是 2 公顷。故答案为:D",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "公顷",
+                                 "平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·广东珠海·期中"
+    },
+    {
+        "id":  "10",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "边长 50 米的正方形广场占地面积是( )平方米,( )个这样的广场占地面积是 1 公顷;计量一个国家的领土面积，一般用( )作单位比较合适。",
+        "options":  [
+
+                    ],
+        "answer":  "2500;4 平方千米",
+        "solution":  "【分析】第一个空需要计算边长50米的正方形面积，使用正方形面积公式(边长×边长),求出它的占地面积是2500平方米;第二个空需要将1公顷换算为10000平方米，再看10000平方米里面有几个2500平方米，得到个数;第三个空:1平方千米大约是小型湖泊的湖面面积，故一个国家的领土面积用平方千米作单位比较合适.\n【详解】 50×50 = 2500 (平方米)1公顷 = 10000 平方米 2500×4 = 10000 (平方米)边长50米的正方形广场占地面积是2500平方米,4个这样的广场占地面积是1公顷;计量一个国家的领土面积，一般用平方千米作单位比较合适。",
+        "knowledge_points":  [
+                                 "正方形的面积",
+                                 "公顷",
+                                 "平方千米的认识",
+                                 "平方千米的进率与换算",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 四年级上·河北唐山·期末"
+    },
+    {
+        "id":  "11",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个学校的面积大约是2公顷,( )个这样的学校的面积大约是1平方千米。",
+        "options":  [
+
+                    ],
+        "answer":  "50",
+        "solution":  "【分析】根据1平方千米 = 100 公顷，只需看100公顷里面有多少个2公顷，就可得到几个这样的学校的面积约是1平方千米.\n【详解】1 平方千米 = 100 公顷 2×50 = 100 (公顷)一个学校的面积大约是2公顷,50个这样的学校的面积大约是 1 平方千米。",
+        "knowledge_points":  [
+                                 "公顷",
+                                 "平方千米的进率与换算"
+                             ],
+        "source":  "25-26四年级上·黑龙江齐齐哈尔·期末"
+    },
+    {
+        "id":  "12",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在( )里填上\"\u003e\"、\"\u003c\"或\"=\"。`r`n`r`n480000000( )5亿`r`n`r`n30500000( )3500000`r`n`r`n90公顷( )900000平方米",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003e;=",
+        "solution":  "【分析】比较整数的大小，先看整数的位数，位数多的那个数就大;如果位数相同，就看最高位，最高位上的数大，那个数就大;如果最高位上的数相同，就看次高位，次高位上的数大，那个数就大，依此类推。单位不同的要先统一单位再比较。`r`n【详解】5亿=500000000，480000000和500000000都是九位数，480000000的最高位是4，500000000的最高位是5，所以480000000\u003c5亿;30500000是八位数，3500000是七位数，所以30500000\u003e3500000;1公顷=10000平方米，90公顷=900000平方米，所以90公顷=900000平方米。",
+        "knowledge_points":  [
+                                 "亿以上数的改写",
+                                 "公顷",
+                                 "平方千米的进率与换算",
+                                 "大数的比较"
+                             ],
+        "source":  "25-26 四年级上·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "13",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上合适的单位名称.\n\n(1)埃及最大的金字塔的占地面积约是 5( )。 \n(2)大连的长兴岛面积约是 252( )。",
+        "options":  [
+
+                    ],
+        "answer":  "公顷;平方千米",
+        "solution":  "【分析】(1)公顷适合计量稍大的土地面积，边长是 100米的正方形的面积是1公顷，所以计量埃及最大的金字塔的占地面积用\"公顷\"作单位比较合适;(2)平方千米适合计量特别大的土地面积，边长是1千米的正方形的面积是1平方千米，所以计量大连的长兴岛的面积用\"平方千米\"作单位比较合适.\n【详解】(1)埃及最大的金字塔的占地面积约是 5 公顷。(2)大连的长兴岛面积约是 252 平方千米。",
+        "knowledge_points":  [
+                                 "公顷",
+                                 "平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 四年级上·黑龙江绥化·期末"
+    },
+    {
+        "id":  "14",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "42000000 平方米=( )公顷 = ( )平方千米。",
+        "options":  [
+
+                    ],
+        "answer":  "4200;42",
+        "solution":  "【分析】根据 1 平方千米 = 100 公顷,1 公顷 = 10000 平方米，据此换算单位即可.\n【详解】42000000 平方米里面有 4200 个 10000 平方米，所以 42000000 平方米 = 4200 公顷;4200 公顷里面有 42 个 100 公顷，所以 4200 公顷 = 42 平方千米;所以 42000000 平方米 = 4200 公顷 = 42 平方千米。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "公顷",
+                                 "平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "15",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "填合适的单位或数。教室的占地面积约是 60( ),我国领土面积约是 960 万(),数学书的封面面积约是 4( ).\n\n5 平方千米=( )公顷,800 公顷=( )平方千米。",
+        "options":  [
+
+                    ],
+        "answer":  "平方米;平方千米;平方分米;500;8",
+        "solution":  "【分析】根据题意，根据生活经验和实际数据相结合，即可解答。常用的面积单位有公顷，平方千米，平方米，平方分米等，计量一般大小的面积用平方米作单位，计量教室的占地面积用平方米作单位比较合适;计量比较大的土地的面积，常用平方千米作单位，计量我国领土面积用平方千米作单位比较合适;边长1分米的正方形的面积是1平方分米，计量数学书的封面的面积用平方分米作单位比较合适;1平方千米 = 100 公顷，据此进行单位换算即可，即:5平方千米 = 500 公顷,800公顷 = 8 平方千米.\n【详解】根据分析可知：教室的占地面积约是 60 平方米;我国领土面积约是 960 万平方千米;数学书的封面面积约是4平方分米。5平方千米 = 500 公顷800 公顷 = 8 平方千米",
+        "knowledge_points":  [
+                                 "平方厘米，平方分米，平方米的认识",
+                                 "公顷",
+                                 "平方千米的认识",
+                                 "平方千米的进率与换算",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 四年级上·吉林白城·期末"
+    },
+    {
+        "id":  "16",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "吉林市是中国唯一一个省，市同名的城市，幅员总面积约 27120( );被誉为\"滑雪胜地，粉雪天堂\"的北大湖滑雪场地处世界黄金滑雪带，雪道总面积约 239( )。截至2024年末，户籍总人口三百九十一万九千人，写作( ),精确到万位约是()人。",
+        "options":  [
+
+                    ],
+        "answer":  "平方千米;公顷;3919000;392万",
+        "solution":  "【分析】计量较大的土地面积，常用公顷和平方千米做单位。边长是100米的正方形的面积为1公顷，边长是1千米的正方形的面积是1平方千米。计量大型场馆或公园的面积，常用公顷做单位。由题意得，北大湖滑雪场地的占地面积较大，选择公顷作单位比较合适。计量城市或行政区的面积，常用平方千米作单位.\n【详解】三百九十一万九千写作:3919000.3919000 的千位上是 9，需要进一，所以 3919000≈392 万。吉林市是中国唯一一个省，市同名的城市，幅员总面积约27120 平方千米;被誉为\"滑雪胜地，粉雪天堂\"的北大湖滑雪场地处世界黄金滑雪带，雪道总面积约 239 公顷。截至2024 年末，户籍总人口三百九十一万九千人，写作3919000，精确到万位约是 392 万。",
+        "knowledge_points":  [
+                                 "亿以内数的读，写法",
+                                 "面积单位的选择",
+                                 "亿以内数的近似数"
+                             ],
+        "source":  "25-26四年级上·吉林·期末"
+    },
+    {
+        "id":  "17",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "2 平方千米=( )公顷=( )平方米。",
+        "options":  [
+
+                    ],
+        "answer":  "200;2000000",
+        "solution":  "【分析】1 平方千米 = 100 公顷,1 公顷 = 10000 平方米，高级单位换算低级单位用乘法，据此解答.\n【详解】 2×100 = 200;2平方千米 = 200 公顷;200×10000 = 2000000;200 公顷 = 2000000 平方米所以,2 平方千米=200 公顷=2000000 平方米。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "公顷",
+                                 "平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "18",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上适当的单位.\n\n我国领土面积为 960 万( )。鸟巢的占地面积约是20( ).\n\n某小学校园的面积约为 1( )。一间教室的面积约为63( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方千米;公顷;公顷;平方米",
+        "solution":  "【分析】根据生活经验以及对面积单位和数据大小的认识，计量比较大的土地的面积，常用平方千米作单位，计量我国领土面积用\"平方千米\"作单位;计量土地的面积，可以用公顷为单位，即：计量鸟巢的占地面积用\"公顷\"作单位，计量某小学校园的面积用\"公顷\"作单位;计量一般大小的面积用平方米作单位，计量一间教室的面积用\"平方米\"作单位.\n【详解】由分析可知，我国领土面积为960万平方千米。鸟巢的占地面积约是20公顷。某小学校园的面积约为1公顷。一间教室的面积约为63平方米。",
+        "knowledge_points":  [
+                                 "平方厘米，平方分米，平方米的认识",
+                                 "公顷",
+                                 "平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26四年级上·吉林松原·期末"
+    },
+    {
+        "id":  "19",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "重庆市高新区是1991 年 3 月经国务院批准设立的全国首批 27 个国家高新区之一，总面积约为 316 平方千米，约合( )公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "31600",
+        "solution":  "【分析】1 平方千米 = 100 公顷;高级单位换算低级单位，乘进率，据此解答.\n【详解】 316×100 = 31600 (公顷)重庆市高新区是1991年3月经国务院批准设立的全国首批27个国家高新区之一，总面积约为316平方千米，约合31600 公顷。",
+        "knowledge_points":  [
+                                 "公顷",
+                                 "平方千米的进率与换算"
+                             ],
+        "source":  "24-25五年级上·重庆沙坪坝·期末"
+    },
+    {
+        "id":  "20",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "下面是奇奇的旅行笔记，请你在括号里填上合适的面积单位.\n\n(1)\"故人西辞黄鹤楼，烟花三月下扬州\"中的\"黄鹤楼\"建筑面积约 3219( );诗词中的\"扬州\"指的是江苏省扬州，总面积约 6591( ).\n\n(2)\"会当凌绝顶，一览众山小\"形容的是泰山，泰山总面积约 25000( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方米;平方千米;公顷",
+        "solution":  "【分析】根据生活经验，对面积单位和数据大小的认识可知,(1)1平方米大约有一大块正方形地砖那么大，所以计量黄鹤楼的面积用\"平方米\"作单位比较合适;平方千米常用在大面积的地方，比如国土面积，城市面积，海洋面积等，所以计量江苏省扬州市用\"平方千米\"作单位比较合适;(2)公顷是测量土地面积的常用单位，比如中等面积的土地，如公园或景区，所以计量泰山总面积用\"公顷\"作单位比较合适.\n【详解】(1)由分析知，\"黄鹤楼\"建筑面积约 3219 平方米;江苏省扬州市总面积约 6591 平方千米。 (2)由分析知，泰山总面积约 25000 公顷。",
+        "knowledge_points":  [
+                                 "平方厘米，平方分米，平方米的认识",
+                                 "公顷",
+                                 "平方千米的认识"
+                             ],
+        "source":  "25-26四年级上·广东广州·期末"
+    },
+    {
+        "id":  "21",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上合适的单位。杭州是浙江省省会和经济，文化，科教中心，长江三角洲中心城市，行政辖区总面积为 16850( ),辖区内浙江大学是一所历史悠久，声誉卓著的高等学府，拥有 7 个校区，总占地面积约 700( ),其中学校的一个图书馆面积约有 700()。",
+        "options":  [
+
+                    ],
+        "answer":  "平方千米;公顷;平方米",
+        "solution":  "【分析】根据生活经验，对面积单位和数据大小的认识可知，边长是1千米的正方形的面积是1平方千米,1平方千米的面积比2个天安门广场的面积还要大一些，所以计量杭州行政辖区总面积用\"平方千米\"作单位比较合适;边长是100米的正方形的面积是1公顷,400米跑道围起来的部分的面积大约是1公顷，所以计量浙江大学占地面积用\"公顷\"作单位比较合适;边长是1米的正方形的面积是1平方米。双人课桌面的面积大约是1平方米，所以计量一个图书馆的面积用\"平方米\"作单位比较合适.\n【详解】由分析得出，杭州是浙江省省会和经济，文化，科教中心，长江三角洲中心城市，行政辖区总面积为16850平方千米，辖区内浙江大学是一所历史悠久，声誉卓著的高等学府，拥有7个校区，总占地面积约700公顷，其中学校的一个图书馆面积约有700平方米。",
+        "knowledge_points":  [
+                                 "平方厘米，平方分米，平方米的认识",
+                                 "公顷",
+                                 "平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 四年级上·浙江杭州·期中"
+    },
+    {
+        "id":  "22",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\"\u003e\"\"\u003c\"或\"=\".\n\n5 公顷( )500 平方米 \n\n200 公顷( )1 平方千米\n\n900 平方米( )8 平方千米 \n\n10 平方千米-300 公顷( )700 公顷",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;\u003e;\u003c;=",
+        "solution":  "【分析】1公顷 = 10000 平方米，那么5公顷就是5个10000平方米。据此转换再和500平方米作比较。1平方千米 = 100 公顷，再和200公顷比较即可。1平方千米 = 100 公顷 = 1000000 平方米，那么8平方千米就是8个1000000平方米，据此转换再和900平方米作比较。1 平方千米 = 100 公顷，那么 10 平方千米就是 10 个 100 公顷，也是1000公顷。然后再与300公顷相减算出结果和700 公顷比较即可。 \n【详解】5 公顷就是 5 个 10000 平方米，是 50000 平方米,50000 平方米 \u003e 500 平方米。所以 5 公顷 \u003e 500 平方米。 1 平方千米 = 100 公顷,200 公顷\u003e100 公顷，所以 200 公顷{ \u003e } 1 平方千米。 8 平方千米就是 8 个 1000000 平方米，是 8000000 平方米,900平方米\u003c8000000平方米。所以900平方米\u003c8平方千米。 10平方千米就是10个100公顷，也是1000公顷。1000-300 = 700 (公顷),所以10平方千米-300公顷 = 700 公顷。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "公顷",
+                                 "平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·江西吉安·期中"
+    },
+    {
+        "id":  "23",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上\"\u003e\"\"\u003c\"或\" = \".\n\n30170000000( )3017000000 \n\n421010000( )420999000\n\n1000000平方米( )1平方千米 \n\n50公顷( )5平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;\u003e;=;\u003c",
+        "solution":  "【分析】多位数的大小比较，数位不同时，位数多的就大。数位相同时，从最高位比起，如果最高位上的数字相同，就比较下一位，依次类推，直到比较出大小。据此比较30170000000 和 3017000000,421010000 和 420999000 的大小。1 平方千米 = 100 公顷 = 1000000 平方米，据此比较 1000000平方米和1平方千米的大小。1平方千米 = 100 公顷,5平方千米就是5个100公顷，据此转换再和50公顷比较.\n【详解】根据多位数的大小比较:30170000000\u003e3017000000,421010000\u003e420999000。根据分析,1 平方千米 = 1000000 平方米。5 平方千米就是 5 个 100 公顷，是 500 公顷,50 公顷\u003c500公顷。那么 50 公顷 \u003c 5 平方千米。30170000000\u003e3017000000;421010000\u003e420999000;1000000 平方米 = 1 平方千米;50 公顷\u003c5 平方千米。",
+        "knowledge_points":  [
+                                 "公顷",
+                                 "平方千米的进率与换算",
+                                 "大数的比较"
+                             ],
+        "source":  "25-26四年级上·河南三门峡·期中"
+    },
+    {
+        "id":  "24",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个正方形训练场的边长是 200 米,( )个这样的训练场的面积是 1 平方千米。",
+        "options":  [
+
+                    ],
+        "answer":  "25",
+        "solution":  "【分析】正方形的面积 = 边长×边长，先计算出这个正方形的面积是多少平方米，再根据10000平方米 = 1 公顷,1平方千米 = 100 公顷，换算单位，然后用100除以这个正方形训练场的面积，即可算出多少个这样的训练场面积是1平方千米。据此解答.\n【详解】 200×200 = 40000 (平方米)40000 平方米 = 4 公顷1 平方千米 = 100 公顷100÷4 = 25 (个)一个正方形训练场的边长是200米,25个这样的训练场的面积是 1 平方千米。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "公顷",
+                                 "平方千米的进率与换算",
+                                 "正方形的面积"
+                             ],
+        "source":  "25-26 四年级上·河南三门峡·期中"
+    },
+    {
+        "id":  "25",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "5公顷+( )平方米=64000平方米 \n5 平方千米+( )公顷=7平方千米 \n3 平方千米 + 4 公顷=( )公顷 \n4 平方米+( )平方厘米=6平方米",
+        "options":  [
+
+                    ],
+        "answer":  "14000;200;304;20000",
+        "solution":  "【分析】1公顷 = 10000 平方米,1平方千米 = 100 公顷,1平方米 = 10000 平方厘米，结合题目数据进行单位变换，再计算。 \n【详解】5公顷 +=( )平方米=64000平方米,5公顷 = 50000 平方米,64000平方米-50000平方米 = 14000 平方米，故填:14000;5平方千米 +( )公顷 = 7 平方千米,5平方千米+2平方千米 = 7 平方千米,2 平方千米 = 200 公顷，故填:200;3 平方千米 +4 公顷( )公顷,3 平方千米 = 300 公顷,300 公顷+4 公顷 = 304 公顷，故填:304;4平方米 + ( )平方厘米 = 6 平方米,4平方米+2平方米= 6 平方米,2 平方米 = 20000 平方厘米，故填:20000;5 公顷+14000 平方米 = 64000 平方米 5平方千米 +200 公顷 = 7 平方千米3 平方千米 +4 公顷=304 公顷 4 平方米 + 20000平方厘米 = 6 平方米",
+        "knowledge_points":  [
+                                 "公顷",
+                                 "平方千米的进率与换算",
+                                 "平方厘米，平方分米，平方米之间的进率与换算"
+                             ],
+        "source":  "25-26四年级上·湖北咸宁·期中"
+    },
+    {
+        "id":  "26",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "7000 公顷=( )平方千米 \n\n60 平方千米=( )平方米\n\n64000000 平方米=( )公顷=( )平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "70;60000000;6400;64",
+        "solution":  "【分析】(1)100 公顷 = 1 平方千米，可以想 7000 公顷里面有多少个 100 公顷，那么就对应多少平方千米。(2)1平方千米 = 100 公顷,1公顷 = 10000 平方米，所以1平方千米 = 100 公顷 = 1000000 平方米。可以想60平方千米里面有多少个1平方千米，那么就对应着多少个1000000平方米。(3)10000 平方米 = 1 公顷，可以想64000000平方米里面有多少个 10000 平方米，那么就对应着多少公顷，然后再根据 100 公顷 = 1 平方千米将得数转化为多少平方千米即可.\n【详解】100公顷 = 1 平方千米,7000公顷中有70个1平方千米，所以 7000 公顷 = 70 平方千米。1 平方千米 = 1000000 平方米,60 平方千米中有 60 个1000000 平方米，所以 60 平方千米 = 60000000 平方米。10000 平方米 = 1 公顷，所以 64000000 平方米 = 6400 公顷。100 公顷 = 1 平方千米，所以 6400 公顷 = 64 平方千米，即64000000 平方米 = 6400 公顷 = 64 平方千米。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "公顷",
+                                 "平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·湖南常德·期中"
+    },
+    {
+        "id":  "27",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "天坛是世界上最大的祭祀性建筑群，占地面积约 3( )。颐和园是中国四大古典园林之一，占地面积约 290( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方千米;公顷",
+        "solution":  "【分析】根据生活经验，对面积单位和数据的大小的认识可知：1平方千米大约是140个标准足球场的大小，所以天坛的占地面积约3平方千米比较合适；1公顷是边长为100米的正方形面积的大小，所以颐和园的占地面积约290公顷比较合适。`r`n【详解】天坛是世界上最大的祭祀性建筑群，占地面积约3平方千米。颐和园是中国四大古典园林之一，占地面积约290公顷。",
+        "knowledge_points":  [
+                                 "面积单位的选择",
+                                 "公顷",
+                                 "平方千米的认识"
+                             ],
+        "source":  "24-25四年级上·河北衡水·期末"
+    },
+    {
+        "id":  "28",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果 1 平方米能摆放3盆花，按照这样摆放,1公顷能摆放 盆花;1 平方千米能摆放 盆花。",
+        "options":  [
+
+                    ],
+        "answer":  "30000;3000000",
+        "solution":  "【分析】本题考查面积单位换算的应用。已知1平方米能摆放3盆花，需要计算1公顷和1平方千米能摆放的盆数。1公顷等于 10000 平方米，因此盆数为 10000 乘 3;1 平方千米等于1000000平方米，因此盆数为1000000乘3。计算时直接相乘即可.\n【详解】1公顷 = 10000 平方米10000×3 = 30000 (盆)1平方千米 = 1000000 平方米 1000000×3 = 3000000 (盆)因此,1公顷能摆放30000盆花;1平方千米能摆放3000000 盆花。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "公顷",
+                                 "平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·重庆大足·期末"
+    },
+    {
+        "id":  "29",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "中国国家图书馆总馆约占地7公顷，建筑面积140000平方米。据2020年10月该图书馆官网信息显示，馆藏文献 39086300 册.\n\n(1)7 公顷= 平方米。 \n(2)140000 读作;39086300 读作 。 \n(3)在 39086300 这个数中 9 在 位上，它表示 。",
+        "options":  [
+
+                    ],
+        "answer":  "70000;十四万;三千九百零八万六千三百;百万;9个百万",
+        "solution":  "【分析】(1)1 公顷 = 10000 平方米，根据面积单位的进率，即可换算解答。(2)读亿以内的数时，先读万级，再读个级，万级的数要按照个级的数的读法来读，再在末尾加上一个\"万\"，每级末尾不管有几个 0，都不读，其余数位有 1 个 0 或连续几个0，都只读1个。(3)从右边起，第一位是个位，计数单位是一(个),第二位是十位，计数单位是十，第三位是百位，计数单位是百，第四位是千位，计数单位是千，第五位是万位，计数单位是万，第六位是十万位，计数单位是十万，第七位是百万位，计数单位是百万，第八位是千万位，计数单位是千万;哪个数位上有数字几，就表示有几个这样的计数单位。据此解答.\n【详解】(1)1 公顷 = 10000 平方米,7 公顷 = 70000 平方米。(2)140000 读作：十四万;39086300 读作：三千九百零八万六千三百。 (3)39086300 这个数中,9 在百万位上，表示 9 个百万。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "整数的数级，数位和计数单位的认识",
+                                 "亿以内数的组成",
+                                 "亿以内数的读，写法"
+                             ],
+        "source":  "24-25四年级上·重庆大足·期末"
+    },
+    {
+        "id":  "30",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "边长为 200 米的正方形花圃的面积是( )公顷,( )个这样的花圃的面积和是1平方千米。",
+        "options":  [
+
+                    ],
+        "answer":  "4;25",
+        "solution":  "【分析】根据正方形面积 = 边长×边长进行计算，然后再根据1公顷 = 10000 平方米进行单位换算;根据1平方千米=100公顷，用100公顷除以正方形花圃的面积即可求得多少个正方形花圃的面积和是1平方千米.\n【详解】 200×200 = 40000 (平方米)40000 平方米 = 4 公顷1 平方千米 = 100 公顷100÷4 = 25 (个)所以，边长为200米的正方形花圃的面积是4公顷,25个这样的花圃的面积和是 1 平方千米。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "公顷",
+                                 "平方千米的进率与换算",
+                                 "正方形的面积"
+                             ],
+        "source":  "24-25 四年级上·辽宁鞍山·期末"
+    },
+    {
+        "id":  "31",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在( )里填上合适的面积单位.\n\n(1)武汉市经开(汉南区)面积约 489.6( )。 \n(2)400 米标准跑道的操场面积约 1( ),操场内足球场面积约 7000( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方千米;公顷;平方米",
+        "solution":  "【分析】(1)根据生活经验，对面积单位，长度单位和数据大小的认识可知，边长是1千米的正方形的面积是1平方千米,1平方千米的面积比2个天安门广场的面积还要大一些，所以计量武汉市经开(汉南区)的面积用\"平方千米\"作单位比较合适。(2)边长是100米的正方形的面积是1公顷，所以400米跑道围起来的部分的面积大约是1公顷;边长是1米的正方形的面积是1平方米，双人课桌面的面积大约是1平方米，所以计量操场内足球场面积用\"平方米\"作单位比较合适.\n【详解】(1)武汉市经开(汉南区)面积约 489.6 平方千米。(2)400米标准跑道的操场面积约1公顷，操场内足球场面积约7000平方米。",
+        "knowledge_points":  [
+                                 "面积单位的选择",
+                                 "公顷",
+                                 "平方千米的认识",
+                                 "平方厘米，平方分米，平方米的认识"
+                             ],
+        "source":  "24-25 四年级上·湖北武汉·期末"
+    },
+    {
+        "id":  "32",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "340000 平方米=( )公顷 20 平方千米 = ( )公顷",
+        "options":  [
+
+                    ],
+        "answer":  "34;2000",
+        "solution":  "【分析】根据1公顷 = 10000 平方米，因此平方米换算成公顷要除以 10000;1 平方千米 = 100 公顷，因此平方千米换算成公顷要乘 100，即可解答.\n【详解】1 公顷 = 10000 平方米，所以 340000÷10000 = 34，即340000 平方米 = 34 公顷。1平方千米 = 100 公顷，所以 20×100 = 2000，即20平方千米=2000 公顷。",
+        "knowledge_points":  [
+                                 "公顷",
+                                 "平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·湖北武汉·期末"
+    },
+    {
+        "id":  "33",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个长方形果园的面积是 2 公顷，这个果园的长是 400 米，宽是( )米。( )个这样的果园占地面积是1平方千米。",
+        "options":  [
+
+                    ],
+        "answer":  "50;50",
+        "solution":  "【分析】先根据1公顷 = 10000 平方米，把长方形果园的面积换算成平方米，再根据长方形的面积 = 长×宽，用长方形果园的面积除以长得出这个果园的宽;再根据1平方千米 = 100公顷，用100公顷除以一个长方形果园的面积2公顷得出几个这样的果园占地面积是1平方千米。据此解答.\n【详解】2 公顷 = 20000 平方米 20000÷400 = 50 (米)1平方千米 = 100 公顷100÷2 = 50 (个)\n所以，一个长方形果园的面积是 2 公顷，这个果园的长是400 米，宽是 50 米。50 个这样的果园占地面积是 1 平方千米。",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "公顷",
+                                 "平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·四川绵阳·期末"
+    },
+    {
+        "id":  "34",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "请在( )里填上合适的长度单位或面积单位。路桥区位于浙江省台州市的沿海中部，是中国黄金海岸中段的一个重要区域。路桥区总面积为 274( ),建城区面积为 29.25( )。其中，十里长街是路桥的文脉之根，总长度在 3( )以上;路桥区博物馆是集收藏，展览，研究，考古，教育，文化交流于一体的综合性博物馆，建筑面积近 10000 平方米，合 1( ),陈列展厅总面积约4000( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方千米;平方千米;千米;公顷;平方米",
+        "solution":  "【分析】1平方千米 = 100 公顷,1公顷 = 10000 平方米。边长是100米的正方形的面积是10000平方米，即1公顷。公顷适合计量较大的土地面积，例如：广场面积，校园面积，湖泊面积;边长为 1000 米的正方形的面积是 1 平方千米，即100公顷。平方千米适合计量特别大的土地面积，例如：国土面积，城市面积;边长为1米的正方形的面积是1平方米，平方米适合计量房屋占地面积，例如：教室等。标准跑道两圈半的长度是 1 千米。据此解答.\n【详解】描述一个区的总面积，通常使用较大的面积单位，平方千米是用于衡量较大区域面积的合适单位，所以路桥区总面积为 274 平方千米。建城区面积相对区总面积较小，但仍较大，用平方千米合适，因此建城区面积为 29.25 平方千米。描述街道的长度，常用千米作为单位，十里长街总长度在 3 千米以上。因为 1 公顷 = 10000 平方米，所以建筑面积近 10000 平方米，合 1 公顷。陈列展厅面积相对较小，用平方米合适，即陈列展厅总面积约 4000 平方米。综上，路桥区位于浙江省台州市的沿海中部，是中国黄金海岸中段的一个重要区域。路桥区总面积为 274 平方千米，建城区面积为 29.25 平方千米。其中，十里长街是路桥的文脉之根，总长度在 3 千米以上;路桥区博物馆是集收藏，展览，研究，考古，教育，文化交流于一体的综合性博物馆，建筑面积近 10000 平方米，合 1 公顷，陈列展厅总面积约4000 平方米。",
+        "knowledge_points":  [
+                                 "长度单位的选择",
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位的选择"
+                             ],
+        "source":  "24-25四年级上·浙江台州·期末"
+    },
+    {
+        "id":  "35",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "20 公顷=( )平方米\n\n700 公顷 = ( )平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "200000;7",
+        "solution":  "【分析】根据 1 公顷 = 10000 平方米,100 公顷 = 1 平方千米进行单位换算.\n【详解】20 公顷里面有 20 个 1 公顷，所以 20 公顷 = 200000 平方米700 公顷里面有 7 个 100 公顷，所以 700 公顷 = 7 平方千米",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·安徽芜湖·期末"
+    },
+    {
+        "id":  "36",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "7 公顷=( )平方米\n\n5800 公顷 = ( )平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "70000;58",
+        "solution":  "【分析】根据题意，明确单位间的进率,1公顷 = 10000 平方米,1平方千米 = 100 公顷，高级单位换算成低级单位，乘进率;低级单位换算成高级单位，除以进率。以此答题即可.\n【详解】根据分析可知:\n\n1公顷 = 10000 平方米,7公顷 = 70000 平方米1 平方千米 = 100 公顷,5800 公顷 = 58 平方千米",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·云南昭通·期末"
+    },
+    {
+        "id":  "37",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "某个岛屿的面积约 43平方千米，合( )公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "4300",
+        "solution":  "【分析】根据1平方千米 = 100 公顷,43平方千米就是43个 100 公顷。依此解答.\n【详解】43平方千米就是43个100公顷，是4300公顷，所以面积约 43 平方千米，合 4300 公顷。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级·全国·假期作业"
+    },
+    {
+        "id":  "38",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "我国国土总面积约为 960 万( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方千米",
+        "solution":  "【分析】计量城市面积，国土面积等，常用平方千米作单位。我国国土总面积约 960 万平方千米。据此解答.\n【详解】由分析可知，我国国土总面积约为 960 万平方千米。",
+        "knowledge_points":  [
+                                 "面积单位的选择",
+                                 "公顷，平方千米的认识"
+                             ],
+        "source":  "24-25 四年级上·云南曲靖·期末"
+    },
+    {
+        "id":  "39",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "23000000 平方米=()公顷=( )平方千米\n\n10 平方千米-10 公顷=( )公顷",
+        "options":  [
+
+                    ],
+        "answer":  "2300;23;990",
+        "solution":  "【分析】1平方千米 = 100 公顷=1000000平方米，小单位换大单位除以进率，大单位换小单位乘进率，根据面积单位间的进率，即可解答.\n【详解】1000000 平方米 = 100 公顷 = 1 平方千米,23000000 平方米里面有 23 个 1000000 平方米，所以\n\n23000000 平方米 = 2300 公顷 = 23 平方千米;10 平方千米 = 1000 公顷,1000 公顷-10 公顷 = 990 公顷，所以 10 平方千米-10 公顷 = 990 公顷。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·云南曲靖·期末"
+    },
+    {
+        "id":  "40",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "74 平方千米=()公顷 60 公顷=( )平方米",
+        "options":  [
+
+                    ],
+        "answer":  "7400;600000",
+        "solution":  "【分析】根据1平方千米=100公顷，1公顷=10000平方米，大单位到小单位乘进率，小单位到大单位除以进率，计算即可。`r`n【详解】因为1平方千米=100公顷，所以74平方千米=74×100公顷=7400公顷。因为1公顷=10000平方米，所以60公顷=60×10000平方米=600000平方米。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·贵州遵义·期末"
+    },
+    {
+        "id":  "41",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "请在下面的括号里填上合适的面积单位.\n\n一间教室的占地面积约是 63( ).\n\n安阳市龙安公园的占地面积约是 16( ).\n\n安阳市的土地面积约是 7413( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方米;公顷;平方千米",
+        "solution":  "【分析】(1)1 平方米大约是边长是 1 米的正方形面积，结合数据\"63\"可知，所以计量一间教室的占地面积用\"平方米\"作单位比较合适;(2)一般小学的操场面积大约是 1 公顷，结合数据\"16\"可知，所以计量安阳市龙安公园的占地面积用 \"公顷\"作单位比较合适;(3)1平方千米是边长为1千米的正方形土地的面积，结合数据\"7413\"可知，所以计量安阳市的土地面积用\"平方千米\"作单位比较合适。 \n【详解】一间教室的占地面积约是 63 平方米。安阳市龙安公园的占地面积约是 16 公顷。安阳市的土地面积约是 7413 平方千米。",
+        "knowledge_points":  [
+                                 "面积单位的选择",
+                                 "公顷，平方千米的认识",
+                                 "平方厘米，平方分米，平方米的认识"
+                             ],
+        "source":  "24-25 四年级上·河南安阳·期末"
+    },
+    {
+        "id":  "42",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号内填入正确的答案.\n\n9 公顷=( )平方米 8000000 平方米=( )公顷=()平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "90000;800;8",
+        "solution":  "【分析】把9公顷换算成平方米，因为1公顷 = 10000 平方米,9 公顷里面有 9 个 10000 平方米，即 90000 平方米;把8000000平方米换算成公顷，因为1公顷 = 10000 平方米,8000000 平方米里面有 800 个 10000 平方米，即 800 公顷;把8000000平方米换算成平方千米，因为1平方千米 = 1000000 平方米,8000000 平方米里面有 8 个 1000000 平方米，即 8 平方千米.\n【详解】9 公顷 = 90000 平方米;8000000 平方米 = 800 公顷 = 8 平方千米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "公顷，平方千米的认识"
+                             ],
+        "source":  "24-25 四年级上·广西玉林·期末"
+    },
+    {
+        "id":  "43",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上\"\u003e\"、\"\u003c\"或\"=\"。`r`n`r`n7万( )700002`r`n`r`n34620089( )34260098`r`n`r`n9公顷( )1平方千米`r`n`r`n140×3( )14×30",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003e;\u003c;=",
+        "solution":  "【分析】比较7万和700002，根据整数大小的比较方法进行解答即可：位数不同，位数多的数就大;位数相同，左起第一位上的数大的那个数就大，如果左起第一位上的数相同，就比较左起第二位上的数，直到比较出大小为止。比较34620089和34260098，根据整数大小的比较方法进行解答即可：位数不同，位数多的数就大;位数相同，左起第一位上的数大的那个数就大，如果左起第一位上的数相同，就比较左起第二位上的数，直到比较出大小为止。比较9公顷和1平方千米，根据1平方千米 = 100 公顷，把1平方千米换算成公顷再比较;比较 140×3 和 14×30，根据整数乘法中因数末尾有0的乘法的计算方法进行解答即可。两个因数相乘，可先把0前面的数相乘，再看因数中一共有几个 0，再在积的末尾添上几个0.\n【详解】7万=70000因为70000\u003c700002，所以7万\u003c700002;34620089\u003e34260098;1平方千米=100公顷因为9公顷\u003c100公顷，所以9公顷\u003c1平方千米;140×3=420 :14×30=420因为420=420，所以140×3=14×30。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "两位数乘整十数的口算乘法",
+                                 "整十，整百，整千数与一位数的乘法",
+                                 "大数的比较"
+                             ],
+        "source":  "24-25四年级上·西藏林芝·期末"
+    },
+    {
+        "id":  "44",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "3 公顷=( )平方米 \n\n4000 公顷 = ( )平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "30000;40",
+        "solution":  "【分析】已知1公顷 = 10000 平方米，所以把3公顷换算成平方米时，用 3 乘 10000;又因为 1 平方千米 = 100 公顷，把 4000 公顷换算成平方千米时，用 4000 除以 100 .\n【详解】1公顷 = 10000 平方米, 3×10000 = 30000 (平方米),所以 3 公顷 = 30000 平方米;1平方千米 = 100 公顷, 4000÷100 = 40 (平方千米),所以4000 公顷 = 40 平方千米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "24-25 四年级上·山东济宁·期末"
+    },
+    {
+        "id":  "45",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "160000 平方米=()公顷 \n\n7500 公顷=( )平方千米\n\n2平方千米-75公顷=( )公顷",
+        "options":  [
+
+                    ],
+        "answer":  "16;75;125",
+        "solution":  "【分析】此题考查的面积单位的换算，根据 1 公顷 = 10000平方米,1平方千米 = 100 公顷，小单位换大单位除以进率，大单位换小单位乘进率，据此进行解答.\n【详解】根据分析,160000 平方米换成公顷，需要除以 10000，那么 160000 平方米 = 16 公顷7500 公顷换成平方千米需要除以 100，那么 7500 公顷 = 75 平方千米;2 平方千米 = 200 公顷,2 平方千米-75 公顷 = 200 公顷-75 公顷 = 125 公顷",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "24-25 四年级上·四川凉山·期末"
+    },
+    {
+        "id":  "46",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "6 平方千米=( )公顷 400 公顷 = ( )平方千米\n\n14000000 平方米=( )公顷=( )平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "600;4;1400;14",
+        "solution":  "【分析】1 平方千米 = 100 公顷 = 1000000 平方米,1 公顷=10000 平方米，据此解答.\n【详解】1平方千米 = 100 公顷,6平方千米里面包含6个1平方千米，故 6 平方千米 = 600 公顷;1平方千米 = 100 公顷,400公顷里面包含4个100公顷，故400 公顷 = 4 平方千米;1公顷 = 10000 平方米,14000000 平方米里面包含 1400 个10000 平方米，那么 14000000 平方米=1400 公顷,1 平方千米 = 1000000 平方米,14000000平方米里面包含14个1000000 平方米，那么 14000000 平方米 = 14 平方千米，故14000000 平方米 = 1400 公顷 = 14 平方千米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "24-25 四年级上·河南商丘·期末"
+    },
+    {
+        "id":  "47",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "5 公顷=( )平方米 \n\n800 公顷=( )平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "50000;8",
+        "solution":  "【分析】根据\"1公顷 = 10000 平方米,1平方千米 = 100 公顷\"，进行单位换算即可.\n【详解】由分析可知,1 公顷 = 10000 平方米,5 公顷是 5 个 1 公顷，所以 5 公顷=50000 平方米;1平方千米 = 100 公顷,800公顷是8个100公顷，所以800公顷 = 8 平方千米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "24-25 四年级上·北京朝阳·期末"
+    },
+    {
+        "id":  "48",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "湖南省博物馆的占地面积约 5( ),( )个湖南省博物馆的占地面积约为 1 平方千米。",
+        "options":  [
+
+                    ],
+        "answer":  "公顷;20",
+        "solution":  "【分析】根据题意，根据生活常识，博物馆的占地面积较大，常用公顷为单位;1平方千米 = 100 公顷，用100除以5，就是多少个湖南省博物馆的占地面积约为1平方千米。以此答题即可.\n【详解】根据分析可知:1平方千米 = 100 公顷 100÷5 = 20 (个)湖南省博物馆的占地面积约5公顷,20个湖南省博物馆的占地面积约为1平方千米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位的选择"
+                             ],
+        "source":  "24-25 四年级上·浙江杭州·期末"
+    },
+    {
+        "id":  "49",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一块盐田长 600 米，宽 100 米,( )块这样的盐田占地 6 平方千米。",
+        "options":  [
+
+                    ],
+        "answer":  "100",
+        "solution":  "【分析】要求多少块盐田占地6平方千米，需要先计算一块盐田的面积，长方形的面积 = 长×宽，由于单位不同，需将6 平方千米转换为平方米,1 平方千米=1000000 平方米，再用总面积除以一块盐田的面积，即可得到块数.\n【详解】一块盐田的面积: 600×100 = 60000 (平方米)6平方千米有6个1平方千米，也就是6个1000000平方米，是 6000000 平方米。需要块数: 6000000÷60000 = 100 (块)因此，一块盐田长 600 米，宽 100 米,100 块这样的盐田占地 6 平方千米。",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·江西赣州·期末"
+    },
+    {
+        "id":  "50",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "2023 年 12 月 20 日澳门回归祖国 24 周年。澳门由澳门半岛，氹仔岛和路环岛组成，陆地面积约为 33.3( )。澳门是我国首个提供 3 ~ 18 岁学生免费教育的地区。澳门的金莲花广场面积约有 2()。",
+        "options":  [
+
+                    ],
+        "answer":  "平方千米;公顷",
+        "solution":  "【分析】1平方千米是边长为1千米的正方形土地的面积，结合数据“33.3”，所以计量澳门的陆地面积用“平方千米”作单位比较合适；一般小学的操场大约是10000平方米，因为1公顷 = 10000平方米，一般小学的操场面积大约是1公顷，结合数据“2”，所以计量澳门金莲花广场面积用“公顷”作单位比较合适。`r`n【详解】2023年12月20日是澳门回归祖国24周年。澳门由澳门半岛、氹仔岛和路环岛组成，陆地面积约为33.3平方千米。澳门是我国首个提供3~18岁学生免费教育的地区。澳门的金莲花广场面积约有2公顷。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "24-25 四年级上·江西上饶·期末"
+    },
+    {
+        "id":  "51",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "(1)600000 公顷 = ( )平方千米;(2)如果 1 平方米能站 12 人,1 公顷大约能站( )人。",
+        "options":  [
+
+                    ],
+        "answer":  "6000;120000",
+        "solution":  "【分析】(1)1平方千米 = 100 公顷，据此进行单位换算;(2)1 公顷 = 10000 平方米，在 12 后面加 4 个 0 可求出大约能站多少人.\n【详解】(1)1平方千米 = 100 公顷,600000 公顷 = 6000平方千米;(2)1 公顷 = 10000 平方米，所以 1 公顷大约能站 120000人。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "24-25 四年级上·重庆璧山·期末"
+    },
+    {
+        "id":  "52",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "(1)故宫博物院的占地面积约 72( )。 \n(2)一间教室的面积约 63( )。",
+        "options":  [
+
+                    ],
+        "answer":  "公顷;平方米",
+        "solution":  "【分析】边长是100米的正方形的面积是1公顷，所以故宫博物院的占地面积约72公顷，比较合理;1平方米是边长为 1 米的正方形的面积的大小，一间教室的面积约 63 平方米比较合理.\n【详解】(1)故宫博物院的占地面积约 72 公顷。(2)一间教室的面积约 63 平方米。",
+        "knowledge_points":  [
+                                 "平方厘米，平方分米，平方米的认识",
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "24-25 四年级上·重庆璧山·期末"
+    },
+    {
+        "id":  "53",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "(1)教室的面积约 56( )。 \n(2)开州区汉丰湖是因长江三峡工程建设而形成的人工湖，是三峡库区最大的内陆湖，是国家 AAAA 级旅游景区，水域面积 15( ).\n\n(3)一本字典的厚度约 40( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方米;平方千米;毫米",
+        "solution":  "【分析】根据生活经验以及对面积单位，长度单位和数据大小的认识，结合实际情况，选择合适的计量单位填空即可。(1)面积单位的选择:1张书桌的占地面积为1平方米，由此即可选择;(2)面积单位的选择：描述国家土地，省，市的面积一般用平方千米来描述，由此即可选择;(3)长度单位的选择:1张身份证的厚度为1毫米，由此即可选择.\n【详解】(1)教室面积通常以平方米为单位，即教室的面积约 56 平方米;(2)较大水域面积常用平方千米为单位，即开州区汉丰湖是因长江三峡工程建设而形成的人工湖，是三峡库区最大的内陆湖，是国家 AAAA 级旅游景区，水域面积 15 平方千米;(3)较小厚度常用毫米为单位，即一本字典的厚度约40毫米。",
+        "knowledge_points":  [
+                                 "毫米和分米的认识",
+                                 "长度单位的选择",
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "24-25 四年级上·重庆开州·期末"
+    },
+    {
+        "id":  "54",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "( )平方米 = 5 公顷 3000000 平方米=( )平方千米=( )公顷",
+        "options":  [
+
+                    ],
+        "answer":  "50000;3;300",
+        "solution":  "【分析】1公顷 = 10000 平方米,5乘进率10000换算单位。1平方千米 = 1000000 平方米,3000000 除以进率 1000000即可;1平方千米 = 100 公顷,3乘进率100即可.\n【详解】5×10000=50000(平方米),故50000平方米=5公顷。3000000÷1000000=3(平方千米),3×100=300(公顷),故3000000平方米=3平方千米=300公顷。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·重庆开州·期末"
+    },
+    {
+        "id":  "55",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "(1)一个标准的篮球场的面积约是 420( )。 \n(2)我国陆地的领土面积约是 960 万( )。 \n(3)天安门广场位于北京市中心，占地面积约是 44( ),可容纳 100 万人举行盛大集会。",
+        "options":  [
+
+                    ],
+        "answer":  "平方米;平方千米;公顷",
+        "solution":  "【分析】常用的面积单位有平方千米，公顷，平方米，平方分米，平方厘米等。计量国土或城市的面积通常用平方千米作单位，计量广场，学校等场所的面积常用公顷作单位。据此解答.\n【详解】(1)一个标准的篮球场的面积约是 420 平方米;(2)我国陆地的领土面积约是 960 万平方千米;(3)天安门广场位于北京市中心，占地面积约是 44 公顷，可容纳 100 万人举行盛大集会。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 四年级上·安徽合肥·期中"
+    },
+    {
+        "id":  "56",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上“\u003e”“\u003c”或“=”。`r`n`r`n400公顷( )40平方千米`r`n`r`n8平方千米( )80公顷`r`n`r`n7000000平方米( )7平方千米`r`n`r`n500平方千米( )5公顷",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003e;=;\u003e",
+        "solution":  "【分析】1平方千米 = 100公顷，1000000平方米 = 1平方千米，大单位转化成小单位要乘进率，小单位转化成大单位要除以进率，统一单位后进行比较。`r`n【详解】400公顷 = 4平方千米，4平方千米\u003c40平方千米，所以400公顷\u003c40平方千米；8平方千米 = 800公顷，800公顷\u003e80公顷，所以8平方千米\u003e80公顷；7000000平方米 = 7平方千米，所以7000000平方米=7平方千米；500平方千米 = 50000公顷，50000公顷\u003e5公顷，所以500平方千米\u003e5公顷。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·安徽合肥·期中"
+    },
+    {
+        "id":  "57",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "如果1平方米土地能栽种10株草莓，那么1公顷土地能栽种( )株草莓,1平方千米土地能栽种( )株草莓。",
+        "options":  [
+
+                    ],
+        "answer":  "100000;10000000",
+        "solution":  "【分析】根据 1 公顷 = 10000 平方米，用 10000 乘 10 算出 1公顷土地能栽种多少株草莓;根据1平方千米 = 1000000 平方米，用 1000000 乘 10 算出 1平方千米土地能栽种多少株草莓;【详解】 10×10000 = 100000 (株);10×1000000=10000000(株);所以如果1平方米土地能栽种10株草莓，那么1公顷土地能栽种100000株草莓,1平方千米土地能栽种10000000株草莓。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26四年级上·安徽合肥·期中"
+    },
+    {
+        "id":  "58",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上合适的面积单位。一间教室的面积约 60( )。 颐和园的占地面积约 300()。",
+        "options":  [
+
+                    ],
+        "answer":  "平方米;公顷",
+        "solution":  "【分析】(1)1 平方米大约是边长是 1 米的正方形面积。结合数据\"60\"可知，所以计量一间教室的面积用\"平方米\"作单位比较合适;(2)一般小学的操场大约是10000平方米，因为1公顷=10000平方米，一般小学的操场面积大约是1公顷，而颐和园的占地面积相对于操场大得多，再结合数据\"300\"，所以计量颐和园的占地面积用\"公顷\"作单位比较合适.\n【详解】一间教室的面积约60平方米。颐和园的占地面积约300公顷。",
+        "knowledge_points":  [
+                                 "平方厘米，平方分米，平方米的认识",
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "24-25四年级上·宁夏吴忠·期末"
+    },
+    {
+        "id":  "59",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "北京\"水立方\"占地约 6( ),约为( )平方米。",
+        "options":  [
+
+                    ],
+        "answer":  "公顷;60000",
+        "solution":  "【分析】测量土地的面积可以用“公顷”作单位。边长为100米的正方形的面积是1公顷，那么计量“水立方”的面积用“公顷”比较合适。1公顷 = 10000平方米，据此转换即可。`r`n【详解】北京“水立方”占地约6公顷，6公顷 = 60000平方米。综上可知：北京“水立方”占地约6公顷，约为60000平方米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位的选择"
+                             ],
+        "source":  "24-25 四年级上·重庆丰都·期末"
+    },
+    {
+        "id":  "60",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "在括号里填上合适的单位.\n\n动物园的占地面积约是 3( ).\n\n张老师家的住房面积约是 120( ).\n\n香港特别行政区的陆地面积约是 1106( )。",
+        "options":  [
+
+                    ],
+        "answer":  "公顷;平方米;平方千米",
+        "solution":  "【分析】根据面积单位相关知识和数据的大小的认知结合生活经验进行分析。边长是100米的正方形面积是1公顷，所以计量动物园的占地面积用公顷作单位比较合适；边长是1米的正方形的面积是1平方米，所以计量住房面积用平方米作单位比较合适；边长是1000米的正方形的面积是1平方千米，所以计量香港特别行政区的陆地面积用平方千米作单位比较合适。据此解答。`r`n【详解】动物园的占地面积约是3公顷。张老师家的住房面积约是120平方米。香港特别行政区的陆地面积约是1106平方千米。",
+        "knowledge_points":  [
+                                 "平方厘米，平方分米，平方米的认识",
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "24-25 四年级上·重庆秀山·期末"
+    },
+    {
+        "id":  "61",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "250000000 平方米=( )公顷 = ( )平方千米。",
+        "options":  [
+
+                    ],
+        "answer":  "25000;250",
+        "solution":  "【分析】根据\"1 平方千米 = 100 公顷 = 1000000 平方米\"，先把 250000000 平方米换算成公顷,250000000 平方米里面有 25000 个 10000 平方米，即 25000 公顷，再把 25000 公顷换算成平方千米,25000公顷里面有250个100公顷，即250平方千米.\n【详解】由分析可知,250000000 平方米=25000 公顷 = 250 平方千米",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "24-25 四年级上·重庆秀山·期末"
+    },
+    {
+        "id":  "62",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "在横线上填上合适的面积单位.\n\n中国的国土面积是 960 万;小明家的住房面积约是 130;一块地板砖的面积约 1;香港特别行政区的陆地面积约 1106 。",
+        "options":  [
+
+                    ],
+        "answer":  "平方千米;平方米;平方分米;平方千米",
+        "solution":  "【分析】根据生活经验，对面积单位和数据大小的认识可知：1平方千米大约是140个标准足球场的大小，所以计量中国国土面积以及香港特别行政区的陆地面积用“平方千米”作单位比较合适；1平方米是边长1米的正方形面积的大小，所以计量小明家的住房面积用“平方米”作单位比较合适；1平方分米大约是一个手掌面的大小，所以计量一块地板砖的面积用“平方分米”作单位比较合适。据此解答。`r`n【详解】(1)中国的国土面积是960万平方千米。(2)小明家的住房面积约是130平方米。(3)一块地板砖的面积约1平方分米。(4)香港特别行政区的陆地面积约1106平方千米。",
+        "knowledge_points":  [
+                                 "平方厘米，平方分米，平方米的认识",
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26四年级上·河南南阳·期中"
+    },
+    {
+        "id":  "63",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "12 平方千米=( )公顷 \n\n4 平方千米 = ( )平方米",
+        "options":  [
+
+                    ],
+        "answer":  "1200;4000000",
+        "solution":  "【分析】根据1平方千米 = 100公顷，1平方千米 = 1000000平方米，高级单位换低级单位乘进率，低级单位换高级单位除以进率，据此解答。`r`n【详解】因为12×100 = 1200，所以12平方千米 = 1200公顷；因为4×1000000 = 4000000，所以4平方千米 = 4000000平方米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·山东济宁·期中"
+    },
+    {
+        "id":  "64",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "天安门广场东西宽500 米，南北长 880 米，广场占地面积约( )万平方米。阅兵仪式上出场的特号国旗，面积为 165( )。",
+        "options":  [
+
+                    ],
+        "answer":  "44;平方米",
+        "solution":  "【分析】天安门广场东西宽500米，南北长880米，说明天安门广场是一个长方形，长方形面积 = 长×宽，算出后再根据1万平方米 = 10000平方米进行换算；1平方米是边长1米的正方形面积的大小，结合数据165可知，计量阅兵仪式上出场的特号国旗用“平方米”作单位比较合适。`r`n【详解】880×500 = 440000（平方米），440000平方米 = 44万平方米，所以天安门广场占地面积约44万平方米。阅兵仪式上出场的特号国旗，面积为165平方米。",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "面积单位的选择",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·山东济宁·期中"
+    },
+    {
+        "id":  "65",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "6 平方千米=( )公顷 \n\n920 公顷=( )平方米\n\n8 公顷 80 平方米=( )平方米 \n\n3 平方千米 12 公顷=( )公顷",
+        "options":  [
+
+                    ],
+        "answer":  "600;9200000;80080;312",
+        "solution":  "【分析】1平方千米 = 100 公顷,6乘进率100，即可把6平方千米换算成几公顷。1 公顷 = 10000 平方米,920 乘进率 10000，即可把 920 公顷换算成几平方米。1 公顷 = 10000 平方米,8 乘进率 10000，可以把 8 公顷换算成 80000 平方米，再加上 80 平方米即可。 1平方千米 = 100 公顷,3乘进率100，可以把3平方千米换算成 300 公顷，再加上 12 公顷即可.\n【详解】 6×100 = 600 (公顷),6 平方千米 = 600 公顷。920×10000 = 9200000 (平方米),920 公顷 = 9200000 平方米。8×10000 = 80000 (平方米),80000 平方米 + 80 平方米 = 80080 平方米,8 公顷 80 平方米 = 80080 平方米。 3×100 = 300 (公顷),300 公顷+12 公顷 = 312 公顷,3 平方千米 12 公顷 = 312 公顷。6 平方千米 = 600 公顷 920 公顷 = 9200000 平方米 8 公顷 80 平方米 = 80080平方米 3 平方千米 12 公顷 = 312 公顷",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·湖南娄底·期中"
+    },
+    {
+        "id":  "66",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "我国的领土面积约是 9600000( ),北京故宫的占地面积是 72( ),合()平方米。",
+        "options":  [
+
+                    ],
+        "answer":  "平方千米;公顷;720000",
+        "solution":  "【分析】我国领土面积：领土面积属于超大范围的计量，常用平方千米作为单位,9600000平方千米是符合我国领土面积的实际数值的;北京故宫占地面积：建筑占地面积的计量常用公顷，故宫作为大型宫殿建筑群,72公顷是其实际占地面积的常用表述，又因为1公顷 = 10000 平方米，所以 72 公顷=720000 平方米.\n【详解】1 公顷 = 10000 平方米72×10000 = 720000 (平方米)所以，我国的领土面积约是9600000平方千米，北京故宫的占地面积是 72 公顷，合 720000 平方米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 四年级上·河北张家口·期中"
+    },
+    {
+        "id":  "67",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "400000 平方米=公顷 \n\n31 公顷 = 平方米",
+        "options":  [
+
+                    ],
+        "answer":  "40;310000",
+        "solution":  "【分析】根据1公顷 = 10000 平方米进行单位换算。400000平方米里面有40个10000平方米，也就是有40个1公顷，所以 400000 平方米 = 40 公顷;31公顷里面有31个1公顷，也就是有31个10000平方米，所以31公顷=310000平方米。据此分析解答.\n【详解】400000平方米 = 40 公顷\n31 公顷= 310000 平方米",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·河北张家口·期中"
+    },
+    {
+        "id":  "68",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "7050 克=()千克( )克 7 公顷-40000 平方米 = ( )公顷 6 平方千米 5 公顷=( )公顷",
+        "options":  [
+
+                    ],
+        "answer":  "7;50;3;605",
+        "solution":  "【分析】7050克 = 7000 克+50克,1千克 = 1000 克，那么7000 克 = 7 千克，把 7 千克和 50 克合起来，即可把 7050 克换算成几千克几克。1公顷 = 10000 平方米，那么40000平方米 = 4 公顷，再算出 7 公顷-4 公顷的差即可。1 平方千米 = 100 公顷，那么 6 平方千米 = 600 公顷,600 公顷加上 5 公顷，即可算出 6 平方千米 5 公顷是几公顷.\n【详解】7050 克 = 7000 克 + 50 克,7000 克 = 7 千克,7 千克 + 50 克 = 7 千克 50 克,7050 克 = 7 千克50克。40000 平方米 = 4 公顷,7公顷-4公顷 = 3 公顷,7公顷-40000 平方米 = 3 公顷.\n6 平方千米 = 600 公顷,600 公顷 + 5 公顷 = 605 公顷,6 平方千米 5 公顷 = 605 公顷。7050 克 = 7 千克 50 克 7 公顷-40000 平方米 = 3 公顷 6 平方千米 5 公顷 = 605 公顷",
+        "knowledge_points":  [
+                                 "克，千克之间的换算与比较",
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·内蒙古鄂尔多斯·期中"
+    },
+    {
+        "id":  "69",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "填上合适的单位.\n\n我们的教室面积大约是50( );伊旗母亲公园的占地面积大约是 133( );一片森林的占地面积大约是 35( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方米;公顷;平方千米",
+        "solution":  "【分析】1平方米是边长为1米的正方形的面积的大小，教室面积单位用平方米比较合理;边长是 100 米的正方形的面积是 1 公顷,1 公顷大约是一个400米跑道围起来部分的面积，所以计量伊旗母亲公园的占地面积单位用公顷;1平方千米=1000000平方米，计量较大的土地面积，通常用平方千米，一片森林的占地面积很大，所以计量单位用平方千米.\n【详解】我们的教室面积大约是50平方米;伊旗母亲公园的占地面积大约是133公顷;一片森林的占地面积大约是35平方千米。",
+        "knowledge_points":  [
+                                 "平方厘米，平方分米，平方米的认识",
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 四年级上·内蒙古鄂尔多斯·期中"
+    },
+    {
+        "id":  "70",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上合适的数.\n\n50000 平方米=( )公顷 \n\n500 平方千米=( )公顷\n\n4 平方千米=( )公顷=( )平方米",
+        "options":  [
+
+                    ],
+        "answer":  "5;50000;400;4000000",
+        "solution":  "【分析】根据1公顷 = 10000 平方米和1平方千米 = 100 公顷直接换算即可.\n【详解】1 公顷 = 10000 平方米，所以 50000 平方米 = 5 公顷。1 平方千米 = 100 公顷，所以 500 平方千米 = 50000 公顷。1 平方千米 = 100 公顷，所以 4 平方千米 = 400 公顷;1 公顷= 10000 平方米，所以 400 公顷 = 4000000 平方米，所以 4平方千米 = 400 公顷 = 4000000 平方米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "24-25 四年级上·内蒙古乌海·期末"
+    },
+    {
+        "id":  "71",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "读下面一段话，在括号里填上合适的面积单位.\n\n历史悠久的甘肃，是古丝绸之路的重要通道。它见证了中国古代的繁荣与辉煌，承载着中华民族的历史记忆。全省总面积约 425800( ),截至 2023 年末，常住人口为2465.48 万人。欣欣在甘肃的某所小学上学，学校的占地面积约是 4( ),校内有一个面积约为 400( )的篮球场。",
+        "options":  [
+
+                    ],
+        "answer":  "平方千米;公顷;平方米",
+        "solution":  "【分析】根据生活经验对面积单位和数据大小的认识，计量比较大的土地面积，常用平方千米作单位，因此计量甘肃的总面积用平方千米作单位比较合适；计量土地的面积，可以用公顷作单位，所以计量一所学校的占地面积用公顷作单位比较合适；计量一般大小的面积用平方米作单位，所以计量一个篮球场的占地面积用平方米作单位比较合适。`r`n【详解】根据分析可得：历史悠久的甘肃，是古丝绸之路的重要通道。它见证了中国古代的繁荣与辉煌，承载着中华民族的历史记忆。全省总面积约425800平方千米，截至2023年末，常住人口为2465.48万人。欣欣在甘肃的某所小学上学，学校的占地面积约是4公顷，校内有一个面积约为400平方米的篮球场。",
+        "knowledge_points":  [
+                                 "平方厘米，平方分米，平方米的认识",
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "24-25 四年级上·甘肃兰州·期末"
+    },
+    {
+        "id":  "72",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "8000000 平方米 = 平方千米 120000 平方米 = 公顷",
+        "options":  [
+
+                    ],
+        "answer":  "8;12",
+        "solution":  "【分析】根据1平方千米 = 1000000 平方米,1公顷 = 10000 平方米进行单位换算即可.\n【详解】8000000 平方米里面有 8 个 1000000 平方米，所以8000000 平方米 = 8 平方千米;120000 平方米里面有 12 个 10000 平方米，所以 120000 平方米 = 12 公顷。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "24-25 四年级上·新疆吐鲁番·期末"
+    },
+    {
+        "id":  "73",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上合适的面积单位.\n\n(1)边长是 100 米的正方形操场，面积是 1( )。 \n(2)我国领土面积约 960 万( )。 \n(3)一间教室的占地面积约 50( )。",
+        "options":  [
+
+                    ],
+        "answer":  "公顷;平方千米;平方米",
+        "solution":  "【分析】(1)根据正方形的面积 = 边长×边长计算出正方形操场的面积，而 10000 平方米 = 1 公顷，所以填\"公顷\"作单位;(2)鄱阳湖的面积大约是3500平方千米，所以我国领土的面积用\"平方千米\"作单位比较合适;(3)篮球场的面积约420平方米，所以一间教室的占地面积用\"平方米\"作单位比较合适。据此解答.\n【详解】(1) 100×100 = 10000 (平方米)10000 平方米 = 1 公顷，所以边长是100米的正方形操场，面积是1公顷。(2)我国领土面积约960万平方千米。 (3)一间教室的占地面积约 50 平方米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "24-25四年级上·新疆克孜勒苏·期末"
+    },
+    {
+        "id":  "74",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "某县2025年建成一座现代化智慧农业产业园，实现智能温控和无人化种植。该项目总投资一亿五千零六十万元，占地400000平方米。横线上的数写作( ),省略\"亿\"位后面的尾数约是( )亿。400000 平方米合( )公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "150600000;2;40",
+        "solution":  "【分析】整数的写法：从高位到低位，一级一级地写，哪个数位上一个单位也没有，就在那个数位上写 0;省略亿位后面的尾数求近似数，也就是去掉亿位后面的尾数，对千万位上的数进行四舍五入，然后在数的后面写上单位\"亿\";根据1公顷 = 10000 平方米，据此将400000平方米换算成公顷为单位.\n【详解】一亿五千零六十万元写作 150600000.150600000≈2 亿400000 平方米里面有 40 个 10000 平方米，所以 400000 平方米 = 40 公顷。因此该项目总投资一亿五千零六十万元，占地400000平方米。横线上的数写作150600000，省略\"亿\"位后面的尾数约是2亿。400000平方米合40公顷。",
+        "knowledge_points":  [
+                                 "亿以上数的读，写法",
+                                 "公顷，平方千米的进率与换算",
+                                 "亿以上数的近似数"
+                             ],
+        "source":  "25-26四年级上·河南南阳·期中"
+    },
+    {
+        "id":  "75",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上“\u003e”“\u003c”或“=”。`r`n`r`n7802000( )8512000`r`n`r`n888888888( )9亿`r`n`r`n2平方千米( )199公顷`r`n`r`n三十七万( )370000",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003c;\u003e;=",
+        "solution":  "【分析】(1)整数比较大小的方法：位数相同的两个数比较大小，从最高位比起，最高位上的数大的那个数就大，如果最高位上的数相同，就比较下一个数位上的数，一直到比较出大小为止；位数不同的两个数比较大小，位数多的数就大。(2)先将9亿改写成以“个”为单位的数，然后再按照整数比较大小的方法比较。(3)比较2平方千米和199公顷，因为1平方千米 = 100公顷，所以2平方千米换算为公顷是200公顷，因为200 \u003e 199，所以2平方千米 \u003e 199公顷。(4)先将三十七万改写成以“个”为单位的数，然后再按照整数比较大小的方法比较。`r`n【详解】(1)两边都是7位数，右边最高位的8更大，所以7802000\u003c8512000；(2)9亿=900000000，两边都是9位数，左边最高位是8，右边最高位是9，所以888888888\u003c9亿；(3)2平方千米 = 200公顷，200公顷\u003e199公顷，所以2平方千米\u003e199公顷；(4)三十七万是370000，所以三十七万=370000。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "公顷，平方千米的进率与换算",
+                                 "亿以内数的改写",
+                                 "亿以上数的改写"
+                             ],
+        "source":  "24-25 四年级上·甘肃临夏·期末"
+    },
+    {
+        "id":  "76",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上合适的数.\n\n8000 公顷 = ( )平方千米 \n\n9 公顷=( )平方米\n\n15000000 平方米=( )公顷=( )平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "80;90000;1500;15",
+        "solution":  "【分析】根据1平方千米 = 100公顷，1公顷 = 10000平方米进行单位换算即可。`r`n【详解】8000÷100=80（平方千米），即8000公顷=80平方千米；9×10000=90000（平方米），即9公顷=90000平方米；15000000÷10000=1500（公顷），1500÷100=15（平方千米），即15000000平方米=1500公顷=15平方千米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26四年级上·广东珠海·期中"
+    },
+    {
+        "id":  "77",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一个长400米，宽100米的长方形鱼塘，如果长和宽都扩大到原来的2倍，那么鱼塘的面积增加了多少公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "12",
+        "solution":  "【分析】根据题意，先用400乘2,100乘2算出扩大后的长方形的长和宽。长方形的面积=长×宽。分别代入现在的和原来的长和宽，分别算出现在和原来的长方形的面积。用现在的面积减去原来的面积就是增加的面积。再根据1公顷=10000平方米，转换成公顷作单位.\n【详解】400×2=800(米)100×2=200(米)400×100=40000(平方米)800×200=160000(平方米)160000-40000=120000(平方米)120000平方米=12公顷那么鱼塘的面积增加了12公顷。",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26四年级上·江西宜春·期中"
+    },
+    {
+        "id":  "78",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在下面的括号里填上合适的面积单位。足球场的面积约是 7000( ) 国家体育场的占地面积约为20( )",
+        "options":  [
+
+                    ],
+        "answer":  "平方米;公顷",
+        "solution":  "【分析】边长是1米的正方形的面积是1平方米，所以足球场的面积可以用平方米作单位。测量土地的面积可以用“公顷”作单位。400米跑道围起来的部分面积大约是1公顷，所以计量体育场的面积用“公顷”比较合适。`r`n【详解】根据生活实际，足球场的面积约是7000平方米。国家体育场的占地面积约为20公顷。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26四年级上·江西宜春·期中"
+    },
+    {
+        "id":  "79",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "80000 平方米=( )公顷 4000 公顷 = ( )平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "8;40",
+        "solution":  "【分析】将80000平方米换算成公顷,1公顷 = 10000 平方米,80000 平方米里面有 8 个 10000 平方米，即 8 公顷;将4000公顷换算成平方千米,1平方千米 = 100 公顷,4000公顷里面有40个100公顷，即40平方千米.\n【详解】80000 平方米 = 8 公顷;4000 公顷 = 40 平方千米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·江西宜春·期中"
+    },
+    {
+        "id":  "80",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "6 平方千米=( )公顷 80000 平方米=( )公顷\n\n300 公顷=( )平方千米=( )平方米",
+        "options":  [
+
+                    ],
+        "answer":  "600;8;3;3000000",
+        "solution":  "【分析】1平方千米 = 100 公顷，用6乘100换算单位。10000 平方米 = 1 公顷，用80000除以10000换算单位。用300除以100换算成\"平方千米\"作单位，用300乘10000 换算成\"平方米\"作单位.\n【详解】 6×100 = 600 (公顷),故 6 平方千米 = 600 公顷。80000÷10000 = 8 (公顷),故 80000 平方米 = 8 公顷。300÷100 = 3 (平方千米),300公顷 = 3 平方千米;300×10000 = 3000000 (平方米),300公顷 = 3000000 平方米，即 300 公顷 = 3 平方千米 = 3000000 平方米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·湖南邵阳·期中"
+    },
+    {
+        "id":  "81",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "排序题",
+        "difficulty":  0.65,
+        "question":  "把 1 平方分米,1 公顷,1平方厘米,1平方千米,1平方米按从大到小的顺序用\"\u003e\"排列是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "1平方千米\u003e1公顷\u003e1平方米\u003e1平方分米\u003e1平方厘米",
+        "solution":  "【分析】一个指甲盖的大小大约是1平方厘米，一个手掌的大小大约是1平方分米，一扇窗户的大小大约是1平方米。边长是100米的正方形的面积是10000平方米，即1公顷。边长为1000米的正方形的面积是1平方千米，即100公顷。据此解答.\n【详解】由分析可知，把 1 平方分米,1 公顷,1 平方厘米,1 平方千米,1 平方米按从大到小的顺序用\"\u003e\"排列是:1平方千米\u003e1 公顷\u003e1 平方米\u003e1 平方分米\u003e1 平方厘米。",
+        "knowledge_points":  [
+                                 "平方厘米，平方分米，平方米的认识",
+                                 "公顷，平方千米的认识"
+                             ],
+        "source":  "25-26 四年级上·福建福州·期中"
+    },
+    {
+        "id":  "82",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "北京到上海铁路长1223 平方千米。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】在判断该说法是否正确时，需依据长度和面积单位的定义及适用场景。铁路长度指两地间的距离，应使用长度单位(如千米)。平方千米是面积单位。题干中使用平方千米描述铁路长度，不符合单位定义.\n【详解】铁路长度应使用长度单位千米表示，如\"北京到上海铁路长 1223 千米\"。题干中使用面积单位平方千米，表述错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "千米的认识"
+                             ],
+        "source":  "25-26 四年级上·湖北咸宁·期中"
+    },
+    {
+        "id":  "83",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "40000 米与 4 公顷大小相等。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】米表示长度的单位，而公顷表示面积的单位。长度单位和面积单位是不同类型的单位，不能直接比较大小。据此解答.\n【详解】40000米表示长度的大小，而4公顷表示面积的大小，由于单位类型不相同，不能比较大小;原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "米的认识",
+                                 "公顷，平方千米的认识"
+                             ],
+        "source":  "24-25 四年级上·河北衡水·期末"
+    },
+    {
+        "id":  "84",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "面积单位间的进率都是 100。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】判断面积单位间的进率是否都是100，需考虑所有常见面积单位之间的进率关系。常见的面积单位包括平方厘米，平方分米，平方米，公顷等。其中，平方米与平方分米，平方分米与平方厘米的进率为100，但平方米与公顷的进率为10000，即相邻的两个面积单位之间的进率是100，不相邻的两个面积单位之间的进率不是100，据此解答.\n【详解】由分析可知，相邻的两个面积单位之间的进率是100，且面积单位间的进率并不都是 100。例如,1 公顷=10000平方米，公顷和平方米之间的进率是10000，而非100;原题干说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "24-25 四年级上·云南文山·期末"
+    },
+    {
+        "id":  "85",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "香港特别行政区的陆地总面积约是 1106 平方千米。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】计量比较大的土地面积，常用\"平方千米\"( k m 2 ) 作单位，边长是1千米的正方形的面积是1平方千米。再结合生活实际判断香港特别行政区的陆地总面积即可.\n【详解】根据生活实际，香港特别行政区的陆地总面积约为1106平方千米。故答案为:√",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识"
+                             ],
+        "source":  "24-25 四年级上·江西宜春·期末"
+    },
+    {
+        "id":  "86",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "判断题",
+        "difficulty":  0.94,
+        "question":  "乐乐的房间面积约12 公顷。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】公顷是较大的面积单位,1公顷等于10000平方米，通常用于测量土地，公园等大面积。房间的面积一般较小，约为10至30平方米。12公顷相当于120000平方米，远大于普通房间的面积，因此该说法错误.\n【详解】因为公顷是较大的面积单位，而房间的面积较小，通常用平方米表示,12公顷不符合实际房间面积的大小，所以判断为错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "24-25 四年级上·江西萍乡·期末"
+    },
+    {
+        "id":  "87",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "一个果园的占地面积是 1 公顷，这个果园一定是边长为 100 米的正方形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】这个说法是错误的。首先,1公顷等于10000平方米，边长100米的正方形面积确实是1公顷，但面积是10000平方米的图形不只有边长100米的正方形，像长200米，宽50米的长方形，面积也是10000平方米，还有梯形，平行四边形等形状也能达到这个面积，所以面积为1公顷的果园不一定是边长100米的正方形.\n【详解】1公顷虽等于边长100米的正方形面积，但面积为1公顷的图形可以是长方形等其他形状，因此这个果园不一定是边长 100 米的正方形，该说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "正方形的面积",
+                                 "公顷，平方千米的认识",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25四年级上·江西上饶·期末"
+    },
+    {
+        "id":  "88",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "判断题",
+        "difficulty":  0.94,
+        "question":  "1 平方千米-57 公顷 = 43 公顷。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】将1平方千米换算成100公顷后，再减去57公顷，计算差值是否等于43公顷。由此可做出判断.\n【详解】因为 1 平方千米 = 100 公顷，所以 1 平方千米-57公顷 = 100 公顷-57 公顷 = 43 公顷。故答案为:√",
+        "knowledge_points":  [
+                                 "两，三位数的连续退位减法",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·天津蓟州·期末"
+    },
+    {
+        "id":  "89",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "一个篮球场的面积是 420 平方千米。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】根据生活经验对面积单位和数据大小的认识，计量比较大的土地的面积，常用平方千米(( k m 2 ))作单位，我国的国土面积大约是960万平方千米，计量一般大小的面积用平方米(( m 2 ))作单位，一间教室的面积大约是50平方米，一个篮球场的面积用\"平方千米\"作单位太大了，应该用\"平方米\"作单位比较合适，据此判断即可.\n【详解】一个篮球场的面积是420平方米，原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "24-25 四年级上·新疆阿克苏·期末"
+    },
+    {
+        "id":  "90",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "4 平方千米比500000 平方米小。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】比较两个面积的大小，需要统一单位。根据面积单位换算,1平方千米=1000000平方米，将单位统一成平方米再比较大小.\n【详解】4 平方千米 = 4000000 平方米,4000000 平方米\u003e500000 平方米，所以 4 平方千米比 500000 平方米大。故答案为: ×",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·甘肃临夏·期末"
+    },
+    {
+        "id":  "91",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "张叔叔的养殖场有 20 个同样大池塘，每个池塘都是长 40 米，宽 25 米的长方形。这些池塘的占地面积一共是多少公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "2 公顷",
+        "solution":  "【分析】首先根据长方形面积计算公式：面积 = 长×宽，计算单个池塘的面积，再乘以池塘的数量即可得到总面积，最后将面积单位从平方米换算成公顷(1公顷 = 10000 平方米).\n【详解】 40×25 = 1000 (平方米)1000×20 = 20000 (平方米)1 公顷 = 10000 平方米,20000 平方米 = 2 公顷。答：这些池塘的占地面积一共是2公顷。",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·黑龙江绥化·期末"
+    },
+    {
+        "id":  "92",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "一块边长为400米的正方形菜地，每公顷可以收获 3 吨白菜，这块菜地可以收获多少吨白菜。",
+        "options":  [
+
+                    ],
+        "answer":  "48 吨",
+        "solution":  "【分析】根据题意，正方形的面积 = 边长×边长，代入计算出正方形菜地的面积。再根据1公顷=10000平方米，转换成公顷作单位。再用菜地的面积乘每公顷可以收获的白菜的吨数，就是这块菜地可以收获多少吨白菜.\n【详解】 400×400 = 160000 (平方米)160000 平方米 = 16 公顷;3×16 = 48 (吨)。答：这块菜地可以收获 48 吨白菜。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "正方形的面积"
+                             ],
+        "source":  "25-26四年级上·江西吉安·期中"
+    },
+    {
+        "id":  "93",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "学校劳动实践基地有一块长方形水稻田，专门种植袁爷爷团队培育的\"超优千号\"品种。这块田长 700 米，宽 400 米，平均每公顷产水稻 16 吨，这块水稻田一共能收获水稻多少吨。",
+        "options":  [
+
+                    ],
+        "answer":  "448 吨",
+        "solution":  "【分析】这块田长700米，宽400米，根据长×宽求出水稻田的总面积，再根据 1 公顷 = 10000 平方米，把单位换算为公顷，平均每公顷产水稻16吨，用总面积乘16即可求出收获水稻的总产量.\n【详解】 700×400 = 280000 (平方米)280000 平方米 = 28 公顷;28×16 = 448 (吨)。答：这块水稻田一共能收获水稻 448 吨。",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26四年级上·浙江宁波·期中"
+    },
+    {
+        "id":  "94",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一块长方形花田长800 米，宽 350 米，这个花田的面积是多少平方米。合多少公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "280000 平方米;28 公顷",
+        "solution":  "【分析】长方形的面积 = 长×宽。由题意得，一块长方形花田长800米，宽350米，直接将数据代入可以算出这个花田的面积，然后根据10000平方米 = 1 公顷来转化单位即可.\n【详解】 800×350 = 280000 (平方米);10000 平方米 = 1 公顷，所以 280000 平方米 = 28 公顷。答：这个花田的面积是280000平方米，合28公顷。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "长方形的面积"
+                             ],
+        "source":  "25-26 四年级上·湖南常德·期中"
+    },
+    {
+        "id":  "95",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一个长方形鱼塘，长 400 米，宽 200 米，如果每公顷养鱼 1200 条，这个鱼塘共养鱼多少条。",
+        "options":  [
+
+                    ],
+        "answer":  "9600 条",
+        "solution":  "【分析】长方形面积 = 长×宽，把数据代入可以算出这个鱼塘有多少平方米。1公顷 = 10000 平方米，据此把这个鱼塘的面积换算成几公顷。这个鱼塘每公顷养鱼条数乘这个鱼塘面积，即可算出这个鱼塘共养鱼多少条.\n【详解】 400×200 = 80000 (平方米)80000 平方米 = 8 公顷;1200×8 = 9600 (条)。答：这个鱼塘共养鱼9600条。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "长方形的面积"
+                             ],
+        "source":  "24-25 四年级上·云南玉溪·期末"
+    },
+    {
+        "id":  "96",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "为一块长 300 米，宽 200 米的农田喷洒农药，每公顷需要农药 15 千克，这块农田共需要农药多少千克。",
+        "options":  [
+
+                    ],
+        "answer":  "90 千克",
+        "solution":  "【分析】根据\"长方形的面积 = 长×宽\"，把数据代入公式求出这块农田的面积是多少平方米，再根据\"10000平方米= 1 公顷\"将结果换算为公顷，然后乘每公顷用农药的质量即可.\n【详解】 300×200 = 60000 (平方米)60000 平方米 = 6 公顷;15×6 = 90 (千克)。答：这块农田共需要农药90千克。",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "24-25 四年级上·天津津南·期末"
+    },
+    {
+        "id":  "97",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一个长方形牧场长8千米，王叔叔骑自行车绕这个牧场一周，每小时骑15千米，正好需要2小时。这个牧场的面积是多少平方千米。",
+        "options":  [
+
+                    ],
+        "answer":  "56平方千米",
+        "solution":  "【分析】根据题意，王叔叔绕牧场一周需要2小时，速度为每小时15千米，可根据速度×时间=路程，代入数据用乘法求出牧场的周长;根据长方形的周长=(长+宽)×2，可知长方形的宽=周长/2-长，代入数据得出这个牧场的宽;最后根据长方形面积公式计算面积，长方形的面积=长×宽.\n【详解】15×2=30(千米)30÷2-8=15-8=7(千米)8×7=56(平方千米)。答：这个牧场的面积是56平方千米。",
+        "knowledge_points":  [
+                                 "除加，除减混合运算",
+                                 "基础行程问题",
+                                 "长方形的面积，长，正方形周长的应用"
+                             ],
+        "source":  "25-26四年级上·内蒙古鄂尔多斯·期中"
+    },
+    {
+        "id":  "98",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一片长方形的白皮松树林，长是 5 千米，宽是 4 千米。这片白皮松树林的面积是多少平方千米。合多少公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "20 平方千米;2000 公顷",
+        "solution":  "【分析】用长乘宽，计算出这片白皮松树林面积的平方千米数。根据 1 平方千米 = 100 公顷，在这片白皮松树林面积的平方千米数末尾添上 2 个 0，换算成以公顷为单位。据此解答.\n【详解】 5×4 = 20 (平方千米)20 平方千米 = 2000 公顷。答：这片白皮松树林的面积是 20 平方千米，合 2000 公顷。",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·新疆吐鲁番·期末"
+    },
+    {
+        "id":  "99",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "某实验学校为推广校园体育活动，新建了一批同样的长方形羽毛球场。每个羽毛球场长 20 米，宽 10 米。多少个这样的球场的总面积正好是 1 公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "50 个",
+        "solution":  "【分析】根据长方形的面积 = 长×宽，又已知每个羽毛球场长20米，宽10米，代入数据，即可求出一个羽毛球场的面积;要求多少个这样的球场的总面积正好是1公顷，根据1公顷 = 10000 平方米，看10000平方米里有几个200平方米，就有几个这样的球场的总面积正好是1公顷。据此解答.\n【详解】 20×10 = 200 (平方米);1公顷 = 10000 平方米,10000 平方米里有 50 个 200 平方米。答:50个这样的羽毛球场的总面积是1公顷。",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26四年级上·河南南阳·期中"
+    },
+    {
+        "id":  "100",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "某农场有一块麦地，长250米，宽200米，共收小麦30吨。这块麦地平均每公顷收小麦多少吨。",
+        "options":  [
+
+                    ],
+        "answer":  "6吨",
+        "solution":  "【分析】先求麦地的面积，长250米，宽200米，面积=长×宽。再根据1公顷=10000平方米，将面积换算成公顷。题目给出了总产量30吨，平均每公顷的产量=总产量/公顷数.\n【详解】250×200=50000(平方米)=5(公顷)30÷5=6(吨)。答：这块麦地平均每公顷收小麦6吨。",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26四年级上·广东珠海·期中"
+    },
+    {
+        "id":  "101",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下列说法中，正确的是( )。",
+        "options":  [
+                        "A. 边长为 1 千米的正方形的面积是 1 公顷。",
+                        "B. 一个长方形地的占地面积是 2000 平方米,5 块这样的地的占地面积是 1 公顷。",
+                        "C. 一间教室的占地面积大约是1公顷。"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】边长是100米的正方形的面积为1公顷，边长是1千米的正方形的面积是1平方千米。1公顷 = 10000 平方米。据此解答.\n【详解】A。边长为1千米的正方形的面积是1平方千米。该选项说法错误。B。一个长方形地的占地面积是 2000 平方米, 2000×5 =10000(平方米),10000 平方米 = 1 公顷，即 5 块这样的地的占地面积是 1 公顷。该选项说法正确。C。一间教室的占地面积较大，大约是几十平方米，肯定没有 1 公顷。该选项说法错误。故答案为:B",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "公顷，平方千米的认识"
+                             ],
+        "source":  "25-26四年级上·山西忻州·期中"
+    },
+    {
+        "id":  "102",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.94,
+        "question":  "下面所给的单位中，不是面积单位的是( )。",
+        "options":  [
+                        "A. 千米",
+                        "B. 公顷",
+                        "C. 平方米",
+                        "D. 平方千米"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】由题目可知，平方千米，平方米以及公顷是面积单位，它可用来度量物体的面或封闭图形的大小;千米是长度单位，它只能用来度量长度，不能度量面积的大小，据此即可解题.\n【详解】根据分析可知：千米不是面积单位。故答案为:A",
+        "knowledge_points":  [
+                                 "千米的认识",
+                                 "平方厘米，平方分米，平方米的认识",
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 四年级上·海南省直辖县级单位·期中"
+    },
+    {
+        "id":  "103",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "黄河是中华文明最主要的发源地，人们称其为\"母亲河\"。黄河发源于青藏高原，干流流经9个省，自治区，流域面积约达80000000公顷。黄河流域面积约合( )万平方千米。",
+        "options":  [
+                        "A. 8",
+                        "B. 80",
+                        "C. 800"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】1 平方千米 = 100 公顷，先去掉 80000000 公顷末尾的2个0换算成平方千米为单位，再根据整数的改写，改写成以万为单位的数则去掉末尾4个0，并添上\"万\"字，据此选择即可.\n【详解】80000000 公顷 = 800000 平方千米 = 80 万平方千米，黄河流域面积约合 80 万平方千米。故答案为:B",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "亿以内数的改写"
+                             ],
+        "source":  "25-26四年级上·广东东莞·月考"
+    },
+    {
+        "id":  "104",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "有两块长方形花圃，第一块的面积是 10 公顷，第二块长 400 米，宽 250 米，这两块花圃的面积比较,( )。",
+        "options":  [
+                        "A. 第一块大",
+                        "B. 第二块大",
+                        "C. 一样大"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】第二块长400米，宽250米，长方形的面积 = 长×宽，即可求出第二块的面积。再与第一块的面积进行比较即可.\n【详解】第二块的面积为 400×250 = 100000 (平方米),10000 平方米 = 1 公顷,100000 平方米 = 10 公顷,10 公顷 = 10 公顷，所以这两块花圃的面积一样大。故答案为:C",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "长方形的面积"
+                             ],
+        "source":  "25-26 四年级上·湖南郴州·期中"
+    },
+    {
+        "id":  "105",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "1公顷的土地大约能种1000棵果树，那么1平方千米的土地大约能种( )棵果树。",
+        "options":  [
+                        "A. 10000",
+                        "B. 100000",
+                        "C. 1000000",
+                        "D. 1000000"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据 1 平方千米 = 100 公顷进行单位换算,1 平方千米的土地就是100公顷的土地,1公顷的土地大约能种1000棵果树，用100乘上1000，即可得知1平方千米的土地大约能种多少棵果树.\n【详解】根据分析可知:1平方千米 = 100 公顷,100×1000 = 100000 (棵),1平方千米的土地大约能种100000棵果树。故答案为:B",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "因数中间或末尾是0的乘法"
+                             ],
+        "source":  "25-26四年级上·广西南宁·期中"
+    },
+    {
+        "id":  "106",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "《孟子·梁惠王上》中说：\"海内之地，方千里者九.\" \"方千里\"指纵横各一千里，即正方形的边长是 1000 里。如果 1 里 = 500 米，那么\"方一里\"是( )。",
+        "options":  [
+                        "A. 250平方米",
+                        "B. 25公顷",
+                        "C. 25平方米",
+                        "D. 250 公顷"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】\"方一里\"指的是边长500米的正方形的面积，正方形的面积 = 边长×边长，然后依据 1 公顷 = 10000 平方米，进行单位换算即可解答.\n【详解】 500×500 = 250000 (平方米),1 公顷 = 10000 平方米,250000 平方米中有 25 个 10000 平方米，也就是 25 个 1 公顷;250000 平方米 = 25 公顷.\"方一里\"是25公顷。故答案为:B",
+        "knowledge_points":  [
+                                 "因数中间或末尾是0的乘法",
+                                 "正方形的面积",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26四年级上·浙江杭州·期中"
+    },
+    {
+        "id":  "107",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "学校准备建一个操场，下面是四块地的面积，选面积为( )的地比较合适。",
+        "options":  [
+                        "A. 2000 平方分米",
+                        "B. 200 平方米",
+                        "C. 2 公顷",
+                        "D. 2 平方千米"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】1平方米 = 100 平方分米,1公顷 = 10000 平方米,1平方千米 = 1000000 平方米。将各个选项的面积统一换算成平方米，结合实际操场大小判断哪个选项合适.\n【详解】A。2000平方分米 = 20 平方米,20平方米面积过小，该选项错误。B。200平方米，可以理解成2个边长10米的正方形的面积，面积过小，该选项错误。C。2公顷 = 2×10000 = 20000 平方米,20000平方米面积大小适中，该选项正确。D。2 平方千米 = 2×1000000 = 2000000 平方米,2000000 平方米面积太大，该选项错误。学校准备建一个操场，下面是四块地的面积，选面积为2公顷的地比较合适。故答案为:C",
+        "knowledge_points":  [
+                                 "平方厘米，平方分米，平方米之间的进率与换算",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26四年级上·河南驻马店·期中"
+    },
+    {
+        "id":  "108",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "妈妈的卧室面积约为( )。",
+        "options":  [
+                        "A. 18平方厘米",
+                        "B. 18平方米",
+                        "C. 18平方千米"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据生活经验对面积单位和数据大小的认识，计量非常小的面积用平方厘米 (cm2) 作单位，数学书封面的面积大约是40平方厘米;计量一般大小的面积用平方米( m 2 ) 作单位，一间教室的面积大约是40平方米;计量比较大的土地的面积，常用平方千米 ( k m 2 ) 作单位，我国的国土面积大约是 960 万平方千米，据此选择即可.\n【详解】A。18平方厘米比数学书封面还要小，不符合题意;B。妈妈的卧室面积约为18平方米，符合题意;C。18平方千米的面积太大了，不符合题意。卧室面积约为 18 平方米。故答案为:B",
+        "knowledge_points":  [
+                                 "平方厘米，平方分米，平方米的认识",
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 四年级上·广东阳江·期中"
+    },
+    {
+        "id":  "109",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "一块正方形菜地，边长是 10 米,( )块这样的菜地面积是 1 公顷。",
+        "options":  [
+                        "A. 10",
+                        "B. 100",
+                        "C. 1000",
+                        "D. 10000"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据题意分析，正方形的面积 = 边长×边长，边长是10米的正方形面积 = 10×10 = 100 (平方米),1公顷=10000 平方米,10000 平方米里面有 100 个 100 平方米，所以一块正方形菜地，边长是10米,100块这样的菜地面积是1公顷。据此解答即可.\n【详解】一块正方形菜地，边长是10米,100块这样的菜地面积是 1 公顷。故答案为:B",
+        "knowledge_points":  [
+                                 "正方形的面积",
+                                 "公顷，平方千米的认识"
+                             ],
+        "source":  "25-26四年级上·河南驻马店·期中"
+    },
+    {
+        "id":  "110",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.94,
+        "question":  "我国的陆地国土面积大约是 9600000( )。",
+        "options":  [
+                        "A. 平方千米",
+                        "B. 米",
+                        "C. 平方米",
+                        "D. 千米"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】计量我国的陆地国土面积，应用面积单位平方千米作单位;【详解】我国的陆地国土面积大约是9600000平方千米;故答案为:A",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26四年级上·河南驻马店·期中"
+    },
+    {
+        "id":  "111",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "学校操场的面积是8000 平方米，再扩建( )平方米，面积就可以达到 1 公顷。",
+        "options":  [
+                        "A. 200",
+                        "B. 2000",
+                        "C. 20000"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】1 公顷 = 10000 平方米，用 10000 减去 8000，即可求出再扩建多少平方米，面积就可以达到 1 公顷.\n【详解】1公顷 = 10000 平方米,10000-8000=2000(平方米)。学校操场的面积是8000平方米，再扩建2000平方米，面积就可以达到1公顷。故答案为:B",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26四年级上·河南南阳·期中"
+    },
+    {
+        "id":  "112",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "国家游泳中心(水立方)的占地面积约 6 万( )。",
+        "options":  [
+                        "A. 平方米",
+                        "B. 公顷",
+                        "C. 平方千米"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】根据生活经验以及对面积单位的认识：边长为1米的正方形，面积是1平方米，生活中，教师讲桌的桌面面积大约是1平方米，平方米我们常用来描述建筑的面积，房屋的面积等;边长是100米的正方形的面积是1公顷，所以公顷适合计量稍大的土地面积，如计量天安门广场面积，校园面积等;边长为1千米的正方形，面积是1平方千米，生活中，我们计量国土面积，城市面积等特别大的土地面积，常用平方千米为单位;结合数据的大小可知，国家游泳中心(水立方)的占地面积用平方米作单位比较合适.\n【详解】由分析可知，国家游泳中心(水立方)的占地面积约 6 万平方米。故答案为:A",
+        "knowledge_points":  [
+                                 "面积单位的选择",
+                                 "公顷，平方千米的认识",
+                                 "平方厘米，平方分米，平方米的认识"
+                             ],
+        "source":  "25-26 四年级上·河南南阳·期中"
+    },
+    {
+        "id":  "113",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面说法中，能描述 1 平方千米有多大的是( )。",
+        "options":  [
+                        "A. 一个篮球场的面积",
+                        "B. 5 个占地面积约为 20 公顷的\"鸟巢\"的面积",
+                        "C. 20 间小学教室的面积",
+                        "D. 一个长 200 米，宽 100 米的操场的面积"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】计量比较大的土地面积，常用平方千米作单位，边长是 1 千米的正方形的面积是 1 平方千米,1 平方千米=1000000平方米;测量较大物体的面积，通常用平方米作单位，边长1米的正方形，面积是1平方米，如：方桌桌面的面积约 1 平方米，所以计量一个篮球场的面积用\"平方米\"作单位比较合适;用一个\"鸟巢\"的占地面积乘 5，求出 5个\"鸟巢\"的占地总面积，再根据 1 平方千米 = 100 公顷，换算成用平方千米作单位即可判断;一间小学教室大约50平方米，用一个教室的面积乘20，求出20间小学教室的面积;根据长方形面积 = 长×宽，求出这个长方形操场的面积，再根据1公顷=10000平方米，换算成用公顷作单位即可判断;【详解】A。根据生活实际可知，一个标准的篮球场的面积约是420平方米，所以不能描述1平方千米有多大;B. 20×5 = 100 (公顷),100公顷 = 1 平方千米，所以5个占地面积约为 20 公顷的\"鸟巢\"的面积能描述 1 平方千米有多大;C。20 间小学教室面积为: 50×20 = 1000 (平方米),所以 20 间小学教室面积不能描述 1 平方千米有多大;D. 200×100 = 20000 (平方米),20000 平方米 = 2 公顷，所以一个长 200 米，宽 100 米的操场的面积不能描述 1 平方千米有多大。故答案为:B",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 四年级上·河北沧州·期中"
+    },
+    {
+        "id":  "114",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "一个正方形的动物园，周长是 1200 米，它的面积是( )公顷。",
+        "options":  [
+                        "A. 3",
+                        "B. 6",
+                        "C. 9"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】先用1200除以4，求出正方形的动物园的边长，再用边长乘边长，求出它的面积，最后根据 10000 平方米=1 公顷进行单位的换算，解答此题即可.\n【详解】 1200÷4 = 300 (米),300×300 = 90000 (平方米),90000 平方米 = 9 公顷。一个正方形的动物园，周长是 1200 米，它的面积是 9 公顷。故答案为:C",
+        "knowledge_points":  [
+                                 "正方形的周长",
+                                 "正方形的面积",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·山东济宁·期中"
+    },
+    {
+        "id":  "115",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "白洋淀被誉为\"华北明珠\"，是野生鸟类在华北中部的重要栖息地，总面积 366000000 平方米，合( )平方千米。",
+        "options":  [
+                        "A. 366",
+                        "B. 3660",
+                        "C. 36600",
+                        "D. 366000"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】由题意得，白洋淀的总面积为366000000平方米，可以根据10000平方米 = 1 公顷将其转化为多少公顷，然后再根据100公顷 = 1 平方千米将得数转化为多少平方千米.\n【详解】10000 平方米 = 1 公顷，所以 366000000 平方米=36600 公顷。100 公顷 = 1 平方千米，所以 36600 公顷 = 366 平方千米，即白洋淀的面积为 366 平方千米。故答案为:A",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26四年级上·重庆沙坪坝·期中"
+    },
+    {
+        "id":  "116",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一个长方形的果园，长500米，宽400米。( )个这样的果园占地面积是1平方千米。",
+        "options":  [
+                        "A. 2",
+                        "B. 5",
+                        "C. 50",
+                        "D. 500"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据长方形面积 = 长×宽，求出一个长方形果园的面积，再根据1平方千米=1000000平方米，把一个长方形果园的面积换算成用平方千米作单位，计算1平方千米是几个这样的果园，即可解答.\n【详解】 500×400 = 200000 (平方米),1 平方千米 = 1000000 平方米,1000000÷200000 = 5 (个),所以 5 个这样的果园占地面积是 1 平方千米。故答案为:B",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·广东广州·期中"
+    },
+    {
+        "id":  "117",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一段公路长 40 千米，宽 10 米。这段公路的占地面积是( )。",
+        "options":  [
+                        "A. 40公顷",
+                        "B. 400平方米",
+                        "C. 400平方千米",
+                        "D. 400000 千米"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】公路的占地面积 = 长×宽，求面积时长和宽的单位要统一，根据单位的换算,1千米 = 1000 米，高级单位换算成低级单位，乘单位间的进率。40千米 = 40000 米,1公顷= 10000 平方米，解答此题即可.\n【详解】40千米 = 40000 米。长:40 千米 = 40000 米，宽:10 米，公路的占地面积 = 长×宽，所以 40000×10 = 400000 (平方米)。1公顷 = 10000 平方米，所以400000平方米 = 40 公顷。故答案为:A",
+        "knowledge_points":  [
+                                 "千米和米之间的进率与换算",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·广东广州·期中"
+    },
+    {
+        "id":  "118",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "从化儿童公园占地面积约 5( )。",
+        "options":  [
+                        "A. 平方米",
+                        "B. 公顷",
+                        "C. 平方千米",
+                        "D. 米"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据生活经验以及对面积单位和数据的大小的认识进行分析:5 平方米大概是一个小房间的大小，远小于儿童公园的实际占地面积;边长是 100 米的正方形的面积是 1 公顷,5 公顷符合小型儿童公园的常见规模;边长是1千米的正方形的面积是1平方千米，计量比较大的土地面积用\"平方千米\"作单位,5平方千米面积过大，远超普通儿童公园的规模;\"米\"是长度单位，不是面积单位，无法表示占地面积.\n【详解】根据分析，从化儿童公园占地面积约 5 公顷。故答案为:B",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26四年级上·广东广州·期中"
+    },
+    {
+        "id":  "119",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "一块长方形麦田的长是 300 米，长是宽的 3 倍，它占地( )。",
+        "options":  [
+                        "A. 300000 平方米",
+                        "B. 3公顷",
+                        "C. 27 公顷",
+                        "D. 30公顷"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】一块长方形麦田的长是300米，长是宽的3倍，先用长300米除以3得到宽为100米;根据长方形面积 = 长×宽，代入数据求出这个长方形麦田的面积，再根据1公顷=10000平方米，将单位化为公顷即可.\n【详解】 300÷3 = 100 (米),300×100 = 30000 (平方米),30000 平方米 = 3 公顷，它占地3公顷。故答案为:B",
+        "knowledge_points":  [
+                                 "整十，整百，整千数除以一位数的除法",
+                                 "长方形的面积",
+                                 "面积单位间的进率及换算",
+                                 "已知一个数的几倍是多少，求这个数"
+                             ],
+        "source":  "25-26 四年级上·河南南阳·期中"
+    },
+    {
+        "id":  "120",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面的面积，最大的是( )。",
+        "options":  [
+                        "A. 5500 平方米",
+                        "B. 5 公顷",
+                        "C. 0.55 平方千米"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】根据1平方千米 = 100 公顷,1公顷 = 10000 平方米，将单位换算成相同的单位，比较大小即可选择.\n【详解】5公顷 = 50000 平方米,0.55平方千米 = 55 公顷 = 550000 平方米,5500 平方米\u003c50000 平方米\u003c550000平方米，所以面积最大的是 0.55 平方千米。故答案为:C",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "24-25五年级上·河北邢台·期末"
+    },
+    {
+        "id":  "121",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "一块长方形菜地，长 200 米，宽 50 米，这块菜地的面积是( )平方米，合()公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "10000;1",
+        "solution":  "【分析】根据长方形的面积 = 长×宽即可求这块菜地的面积;根据1公顷 = 10000 平方米，将菜地的面积换算为公顷即可.\n【详解】 200×50 = 10000 (平方米),10000 平方米 = 1 公顷，即这块菜地的面积是10000平方米，合1公顷。",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·河南信阳·期中"
+    },
+    {
+        "id":  "122",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "南湾湖是信阳著名的风景区，湖面面积约 75 平方千米，合( )公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "7500",
+        "solution":  "【分析】根据1平方千米 = 100 公顷，用75乘进率100即可换算.\n【详解】 75×100 = 7500 (公顷),即湖面面积约 75 平方千米，合 7500 公顷。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·河南信阳·期中"
+    },
+    {
+        "id":  "123",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "9 平方千米 =​ ()公顷\n\n15000000 平方米=( )公顷=( )平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "900;1500;15",
+        "solution":  "【分析】1平方千米 = 100 公顷,1公顷 = 10000 平方米，据此进行单位换算.\n【详解】1 平方千米 = 100 公顷,9 平方千米是 9 个 1 平方千米，即 900 公顷，因此 9 平方千米 = 900 公顷。1 平方千米 = 100 公顷,1 公顷 = 10000 平方米,15000000平方米是 1500 个 10000 平方米即 1500 公顷,1500 公顷是15 个 100 公顷即 15 平方千米，因此 15000000 平方米 = 1500 公顷 = 15 平方千米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·河南驻马店·期中"
+    },
+    {
+        "id":  "124",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "2400 平方厘米=()平方分米\n\n56 平方米=( )平方分米\n\n10 平方千米=( )公顷=( )平方米",
+        "options":  [
+
+                    ],
+        "answer":  "24;5600;1000;10000000",
+        "solution":  "【分析】1平方分米 = 100 平方厘米，用2400除以进率100即可换算成平方分米为单位;1 平方米 = 100 平方分米，用56乘进率100即可换算成平方分米为单位;1平方千米 = 100 公顷，用10乘进率100即可换算成公顷为单位;1平方千米 = 1000000 平方米，用10乘进率1000000即可换算成平方米为单位.\n【详解】2400÷100 = 24,2400 平方厘米 = 24 平方分米;56×100 = 5600,56 平方米 = 5600 平方分米;10×100 = 1000,10×100000 = 10000000,10 平方千米 = 1000 公顷=10000000 平方米。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "公顷，平方千米的进率与换算",
+                                 "平方厘米，平方分米，平方米之间的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·江西赣州·期中"
+    },
+    {
+        "id":  "125",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果 1 平方米能种植4株牡丹,1公顷能种植( )株,1平方千米能种植()株。",
+        "options":  [
+
+                    ],
+        "answer":  "40000;4000000",
+        "solution":  "【分析】1公顷 = 10000 平方米，用1平方米能种植牡丹的株数乘10000，计算出1公顷能种植牡丹的株数;1平方千米 = 1000000 平方米，用 1 平方米能种植牡丹的株数乘1000000，计算出 1 平方千米能种植牡丹的株数。据此解答.\n【详解】1 公顷 = 10000 平方米,4×10000 = 40000 (株);1平方千米 = 1000000 平方米,4×1000000 = 4000000 (株)。所以如果1平方米能种植4株牡丹,1公顷能种植40000株,1平方千米能种植4000000株。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·河南三门峡·期中"
+    },
+    {
+        "id":  "126",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在□里填“\u003e”“\u003c”或“=”，在( )里填合适的数。`r`n`r`n610000平方米 □ 61公顷`r`n340000000平方米=( )平方千米`r`n500公顷 □ 50000平方千米`r`n8平方千米60公顷=( )公顷",
+        "options":  [
+
+                    ],
+        "answer":  "=;340;\u003c;860",
+        "solution":  "【分析】根据 1 平方千米 = 100 公顷,1 公顷 = 10000 平方米,1 平方千米 = 1000000 平方米，先统一单位，再进行比较.\n【详解】1 公顷 = 10000 平方米,61 公顷=610000 平方米,610000平方米 = 61 公顷，所以 610000平方米和61公顷相等;1 平方千米 = 1000000 平方米,340000000 平方米 = 340 平方千米;50000 平方千米 = 5000000 公顷,500 公顷\u003c5000000 公顷，所以 500 公顷\u003c50000 平方千米;1 平方千米 = 100 公顷,8 平方千米 60 公顷 = 800 公顷 + 60 公顷 = 860 公顷。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·山西忻州·期中"
+    },
+    {
+        "id":  "127",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "11 公顷=( )平方米 800 平方千米 = ( )公顷",
+        "options":  [
+
+                    ],
+        "answer":  "110000;80000",
+        "solution":  "【分析】1 公顷 = 10000 平方米,11 里面有 11 个 1，所以 11公顷=110000平方米;1平方千米 = 100 公顷,800里面有800个1，所以800平方千米 = 80000 公顷。据此解答.\n【详解】根据分析:11 公顷=110000 平方米;800平方千米 = 80000 公顷。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·广西玉林·期中"
+    },
+    {
+        "id":  "128",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上合适的数或单位名称.\n\n课桌的面积大约是 60( ).\n教室的面积大约是 80( ).\n3 平方千米=( )公顷=( )平方米",
+        "options":  [
+
+                    ],
+        "answer":  "平方分米;平方米;300;3000000",
+        "solution":  "【分析】生活中1平方米大约是1个餐桌的大小,1平方分米大约是一个手掌的大小,1平方厘米大约是1个指甲盖的大小;据此根据面积单位的认识，以及生活经验进行填空。1平方千米 = 100 公顷=1000000平方米，高级单位化低级单位要乘进率，低级单位化高级单位要除以进率，平方千米化为公顷要乘 100;公顷化为平方米要乘 10000.\n【详解】课桌的面积大约是 60 平方分米。教室的面积大约是 80 平方米。3 平方千米 = 300 公顷 = 3000000 平方米。",
+        "knowledge_points":  [
+                                 "面积单位的选择",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25 五年级上·重庆云阳·期末"
+    },
+    {
+        "id":  "129",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "比一比，在括号里填上\"\u003e\"\"\u003c\"或 \"=\".\n\n2000 平方米( )2 公顷 \n400 公顷( )40000 平方米 \n70000 平方米( )7 公顷",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003e;=",
+        "solution":  "【分析】先统一单位，再比较即可。根据1公顷 = 10000 平方米，高级单位化低级单位乘进率，低级单位化高级单位除以进率，将公顷全部化为平方米要乘10000.\n【详解】2 公顷 = 20000 平方米,2000 平方米\u003c20000 平方米，即2000平方米\u003c2 公顷。400 公顷 = 4000000 平方米,4000000 平方米\u003e40000 平方米，即400 公顷\u003e40000 平方米。70000 平方米 = 7 公顷。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25 五年级上·重庆云阳·期末"
+    },
+    {
+        "id":  "130",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在( )里填上\"\u003e\"\"\u003c\"或\"=\" .\n\n9 公顷( )1 平方千米 \n500 公顷( )400 平方千米 \n6 公顷( )600 平方米 \n2 平方千米( )2000 公顷",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003c;\u003e;\u003c",
+        "solution":  "【分析】比较 9 公顷和 1 平方千米，先根据 1 平方千米=100 公顷，把 1 平方千米换算成公顷再比较;比较 500 公顷和 400 平方千米，先把公顷换算平方千米再比较，因为 1 平方千米 = 100 公顷,500 公顷里面有 5 个 100公顷，即 5 平方千米;比较 6 公顷和 600 平方米，先把 6 公顷换算成平方米再比较，因为 1 公顷 = 10000 平方米,6 公顷里面有 6 个 10000 平方米，即 60000 平方米;比较2平方千米和2000公顷，先把2平方千米换算成公顷再比较，因为1平方千米 = 100 公顷,2平方千米里面有2个 100 公顷，即 200 公顷.\n【详解】1平方千米 = 100 公顷，因为 9 公顷\u003c100 公顷，所以 9 公顷\u003c1 平方千米;500 公顷 = 5 平方千米，因为 5 平方千米\u003c400 平方千米，所以 500 公顷\u003c400 平方千米;6公顷 = 60000 平方米，因为 60000 平方米\u003e600 平方米，所以6公顷\u003e600 平方米;2平方千米 = 200 公顷，因为 200 公顷\u003c2000 公顷，所以 2 平方千米\u003c2000 公顷。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·广东东莞·月考"
+    },
+    {
+        "id":  "131",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "单位换算闯关.\n\n( )平方千米 = 600 公顷;8 公顷+12 公顷 = ( )平方米;9000000 平方米=( )公顷 = ( )平方千米。",
+        "options":  [
+
+                    ],
+        "answer":  "6;200000;900;9",
+        "solution":  "【分析】根据1平方千米 = 100 公顷,1公顷 = 10000 平方米,1平方千米 = 1000000 平方米，进行单位换算即可.\n【详解】(1)1 平方千米 = 100 公顷，所以 600 公顷 = 6 平方千米。(2)1 公顷 = 10000 平方米,8 公顷+12 公顷 = 20 公顷,20 公顷 = 200000 平方米，所以 8 公顷+12 公顷 = 200000 平方米。(3)10000 平方米 = 1 公顷,9000000 平方米 = 900 公顷。1 平方千米 = 1000000 平方米,9000000 平方米 = 9 平方千米。所以 9000000 平方米 = 900 公顷 = 9 平方千米。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·河南郑州·期中"
+    },
+    {
+        "id":  "132",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在( )里填合适的单位.\n\n郑州奥体中心总建筑面积约 58( ).\n教室的门高约2( ),面积约 2( ).\n龙湖金融岛环岛周长约 10( ).\n一张课桌桌面面积约 24()。",
+        "options":  [
+
+                    ],
+        "answer":  "公顷;米;平方米;千米;平方分米",
+        "solution":  "【分析】公顷适合计量稍大的土地面积,400米跑道围起来的部分的面积大约是1公顷，所以根据实际数据可知，计量郑州奥体中心总建筑面积用\"公顷\"作单位比较合适;小学生双手张开，手掌之间的距离大约是 1 米，所以根据实际数据可知，计量教室的门高用\"米\"作单位比较合适;1平方米大约是1米的正方形瓷砖的面积，所以根据实际数据可知，计量教室的门的面积用\"平方米\"作单位比较合适;1千米大约是一个学校操场跑道两圈半的长度，所以根据实际数据可知，计量龙湖金融岛环岛的周长用\"千米\"作单位比较合适;1平方分米大约是一个手掌面的面积，所以根据实际数据可知，计量一张课桌桌面的面积用\"平方分米\"作单位比较合适.\n【详解】郑州奥体中心总建筑面积约58公顷。教室的门高约2米，面积约2平方米。龙湖金融岛环岛周长约10千米。一张课桌桌面面积约24平方分米。",
+        "knowledge_points":  [
+                                 "面积单位的选择",
+                                 "公顷，平方千米的认识",
+                                 "长度单位的选择"
+                             ],
+        "source":  "25-26 四年级上·河南郑州·期中"
+    },
+    {
+        "id":  "133",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "郑州中国国际学校秋季运动会有 8642 名师生参与，这个数读作( ),省略千位后面的尾数约是( )千;郑州高新区某公园总面积约 28公顷，相当于( )个长 70米，宽 40 米的标准篮球场。",
+        "options":  [
+
+                    ],
+        "answer":  "八千六百四十二;9;100",
+        "solution":  "【分析】根据题意，对于数字8642，按照中文读数的规则，从高位读起，每一位数字读作相应的中文数字并加上位名(如千，百，十),注意没有零的情况。省略千位后面的尾数时，看百位数字，百位是 6，大于等于 5，向千位进一，所以约是9千。对于面积换算，先将公顷转换为平方米(1公顷 = 10000 平方米),然后计算一个标准篮球场的面积(长×宽),最后用公园总面积除以篮球场面积，得到个数。以此答题即可.\n【详解】8642 读作：八千六百四十二。8642≈9000.28 公顷=280000 平方米,70×40 = 2800 (平方米),280000÷2800 = 100 (个)。郑州中国国际学校秋季运动会有8642名师生参与，这个数读作八千六百四十二，省略千位后面的尾数约是9千;郑州高新区某公园总面积约28公顷，相当于100个长70米，宽40 米的标准篮球场。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "亿以内数的读，写法"
+                             ],
+        "source":  "25-26 四年级上·河南郑州·期中"
+    },
+    {
+        "id":  "134",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在横线里面填上适当的单位.\n\n(1)学校操场的面积大约 9800;(2)桂阳文化园包括古郡城，环湖游园和东塔公园，其占地面积约是 144;(3)我国陆地面积世界第三，大约是 960 万 。",
+        "options":  [
+
+                    ],
+        "answer":  "平方米;公顷;平方千米",
+        "solution":  "【分析】常用的面积单位有平方厘米，平方分米，平方米。我们知道，指甲盖的面积大约1平方厘米，手掌的面积大约1平方分米，一块地板砖的面积大约1平方米。计量较大的土地面积，常用公顷和平方千米作单位。边长是100米的正方形的面积为1公顷，边长是1千米的正方形的面积是1平方千米。(1)由题意得,400米跑道的操场面积大约是1公顷。1公顷 = 10000 平方米，所以学校操场的面积大约是10000平方米。题目中给出的数据是9800,9800与10000较为接近，所以横线上填平方米比较合适。(2)由题意得，桂阳文化园的面积很大。计量公园的面积，常用公顷作单位。(3)由题意得，我国的陆地面积特别大。计量国家的陆地的面积，常用平方千米作单位.\n【详解】(1)学校操场的面积大约 9800 平方米。(2)桂阳文化园包括古郡城，环湖游园和东塔公园，其占地面积约是144公顷。(3)我国陆地面积世界第三，大约是 960 万平方千米。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "面积单位的选择",
+                                 "公顷，平方千米的认识"
+                             ],
+        "source":  "25-26四年级上·湖南郴州·期中"
+    },
+    {
+        "id":  "135",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\"\u003e\"\"\u003c\"或\"=\".\n\n6 公顷( )59000 平方米 \n\n2 平方千米( )200 公顷\n\n30000 平方米( )3 平方千米 \n\n1010100( )1101000",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;=;\u003c;\u003c",
+        "solution":  "【分析】根据1公顷 = 10000 平方米将单位统一成平方米再比较大小;根据 1 平方千米 = 100 公顷将单位统一成公顷再比较大小;根据1平方千米 = 1000000 平方米将单位统一成平方米再比较大小;整数比较大小，先比较位数，位数多的数就大，如果位数相同，则从高位开始比较数字大小，直到比较出大小为止.\n【详解】6 公顷 = 60000 平方米,60000\u003e59000，所以 6 公顷\u003e59000 平方米;2 平方千米 = 200 公顷;3 平方千米 = 3000000 平方米,30000\u003c3000000，所以30000 平方米\u003c3 平方千米;1010100\u003c1101000。因此 6 公顷\u003e59000 平方米;2 平方千米 = 200 公顷;30000平方米\u003c3 平方千米;1010100\u003c1101000。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·河南南阳·期中"
+    },
+    {
+        "id":  "136",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上适当的单位.\n\n教室的面积大约是 50( ).\n\n河南省的面积大约是 16.7 万().\n\n一个标准足球场的面积大约是 7000( ),10个这样的足球场面积大约是 7( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方米;平方千米;平方米;公顷",
+        "solution":  "【分析】面积的常用较大单位有平方千米，公顷，平方米等,1平方米的面积大小是边长为1米的正方形大小,1平方千米的面积相当于一个边长为1000米的正方形面积。般国家，城市，省份，岛屿等用平方千米;大型广场，公园，田地等用公顷;房屋，足球场，操场等用平方米;根据实际情况及数值大小选择合适单位;根据1公顷 = 10000 平方米进行单位换算.\n【详解】教室的面积大约是50平方米;河南省的面积大约是16.7万平方千米;一个标准足球场的面积大约是7000平方米;7000×10 = 70000 (平方米) = 7 公顷。",
+        "knowledge_points":  [
+                                 "面积单位的选择",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·河南南阳·期中"
+    },
+    {
+        "id":  "137",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\"\u003e\"\"\u003c\"或\"=\".\n\n3573269( )3573629\n\n6 公顷( )5990 平方米 \n\n430×3( )43×30\n\n7506000000( )7 亿 \n\n960000 平方米( )96 平方千米\n\n260×4( )206×4",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003e;=;\u003e;\u003c;\u003e",
+        "solution":  "【分析】整数比较大小，首先要看它们的位数，如果位数不同，那么位数多的数就大；如果位数相同，则从高位开始比较数字大小，直到比较出大小为止。据此比较(1)小题即可。(2)小题，1公顷 = 10000平方米，6公顷就是6个10000平方米，据此将6公顷的单位换算为平方米，再比较大小即可；两位数乘两位数的计算方法：先用第二个因数的个位上的数与第一个因数相乘，所得的积末尾与个位对齐；然后用第二个因数的十位上的数与第一个因数相乘，所得的积末尾与十位对齐，最后把两次乘得的积相加；当乘数末尾有0时，先算0前面的数，再在积的末尾添加对应个数的0。三位数乘一位数，相同数位对齐，从个位起，用一位数依次乘三位数的每一位数，乘到哪一位，就把积写在哪一位上。哪一位上乘得的积满几十就向前一位进几。据此计算出(3)(6)小题的结果，再比较大小即可；(4)小题，把7亿后面的亿字去掉，再在7的后面添上8个0，即可把7亿改写，然后再比较两数的大小即可；(5)小题，1平方千米 = 1000000平方米，96平方千米就是96个1000000平方米，据此将96平方千米的单位换算为平方米，再比较大小即可。`r`n【详解】3573269与3573629的百万位、十万位、万位和千位上数字都相同，百位上的数字2\u003c6，因此3573269\u003c3573629。6个10000平方米是60000平方米，60000平方米\u003e5990平方米，即6公顷\u003e5990平方米。430×3 = 1290，43×30 = 1290，即430×3 = 43×30。7亿=700000000，7506000000是十位数，而700000000是九位数，因此7506000000\u003e700000000，即7506000000\u003e7亿。96个1000000平方米是96000000平方米，960000平方米\u003c96000000平方米，即960000平方米\u003c96平方千米。260×4 = 1040，206×4 = 824，1040\u003e824，即260×4\u003e206×4。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "两位数乘整十数的口算乘法",
+                                 "因数中间或末尾是0的乘法",
+                                 "大数的比较"
+                             ],
+        "source":  "25-26四年级上·河南南阳·期中"
+    },
+    {
+        "id":  "138",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上适当的面积单位.\n\n国家游泳中心(\"水立方\")的占地面积约 6( )\n\n香港特别行政区的陆地面积约 1106( )\n\n一间教室的面积约 63( )\n\n一部手机的屏幕面积大约是 60( )",
+        "options":  [
+
+                    ],
+        "answer":  "公顷;平方千米;平方米;平方厘米",
+        "solution":  "【分析】根据生活经验，对面积单位和数据大小的认识可知：边长是100米的正方形土地的面积是10000平方米，也即1公顷,400米跑道围起来的部分的面积大约是1公顷。测量土地的面积可以用\"公顷\"做单位，所以计量国家游泳中心(\"水立方\")的占地面积，结合数据6用\"公顷\"作单位比较合适。边长是1千米的正方形土地的面积是1平方千米，也即100公顷，计量比较大的土地面积常用\"平方千米\"做单位，所以计量香港特别行政区的陆地面积，结合数据1106用\"平方千米\"作单位比较合适。边长1米的正方形，面积是1平方米;所以计量一间教室的面积，再结合数据63用\"平方米\"作单位比较合适。边长1厘米的正方形的面积是1平方厘米;生活中手指甲的面积接近1平方厘米;所以计量一部手机的屏幕面积，再结合数据60用\"平方厘米\"作单位比较合适.\n【详解】国家游泳中心(\"水立方\")的占地面积约6公顷。香港特别行政区的陆地面积约1106平方千米。一间教室的面积约 63 平方米。一部手机的屏幕面积大约是60平方厘米。",
+        "knowledge_points":  [
+                                 "平方厘米，平方分米，平方米的认识",
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26四年级上·河南南阳·期中"
+    },
+    {
+        "id":  "139",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "南阳武侯祠是三国时期诸葛亮\"躬耕于南阳\"的历史遗址，是国家级4A级旅游景区，它的占地面积约为15公顷，北京颐和园的占地面积是南阳武侯祠的 20 倍，北京颐和园的占地面积为( )平方千米。",
+        "options":  [
+
+                    ],
+        "answer":  "3",
+        "solution":  "【分析】南阳武侯祠的占地面积约为 15 公顷，北京颐和园的占地面积是南阳武侯祠的20倍，北京颐和园的占地面积= 南阳武侯祠的占地面积*20.1平方千米 = 100 公顷，低级单位换算成高级单位，除以单位间的进率，据此可得出.\n【详解】由分析可得出，北京颐和园的占地面积为:15×20 = 300 (公顷),1平方千米 = 100 公顷，所以 300÷100 = 3(平方千米)。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26四年级上·河南南阳·期中"
+    },
+    {
+        "id":  "140",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "国家体育馆\"鸟巢\"的占地面积约20( ),( )个鸟巢的占地面积约1平方千米。",
+        "options":  [
+
+                    ],
+        "answer":  "公顷;5",
+        "solution":  "【分析】根据生活经验，计量大型建筑的占地面积用公顷作单位比较合适，如一个中学操场的面积大约是1公顷，据此填写面积单位;1平方千米 = 100 公顷，先统一单位，看100公顷中有多少个20公顷，鸟巢的个数就有多少个，据此作答.\n【详解】根据分析可得：大型建筑物的占地面积用公顷作单位比较合适，再结合数据 20，可知国家体育馆\"鸟巢\"的占地面积约 20 公顷;因为 1 平方千米 = 100 公顷,20 + 20 + 20 + 20 + 20 = 100(公顷),故5个鸟巢的占地面积约为1平方千米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26四年级上·浙江温州·期中"
+    },
+    {
+        "id":  "141",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在( )里填上合适的数.\n\n2800000 平方米=( )公顷 \n\n7 平方千米 = ( )公顷\n\n1 公顷-4000 平方米 = ( )平方米 \n\n80 公顷+20 公顷=( )平方千米\n\n15 平方千米=( )公顷=( )平方米",
+        "options":  [
+
+                    ],
+        "answer":  "280;700;6000;1;1500;15000000",
+        "solution":  "【分析】1公顷等于10000平方米,1平方千米等于100公顷,1平方千米等于1000000平方米，大单位化小单位乘进率，小单位化大单位除以进率.\n【详解】2800000 平方米 = (2800000÷10000)公顷 = 280 公顷;7 平方千米 = (7×100)公顷 = 700 公顷;1公顷-4000平方米 = 10000 平方米-4000平方米 = 6000平方米;80 公顷+20 公顷 = 100 公顷 = (100÷100)平方千米 = 1 平方千米;15 平方千米 = (15×100)公顷 = 1500 公顷=((1500×10000))平方米=15000000 平方米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26四年级上·浙江温州·期中"
+    },
+    {
+        "id":  "142",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "3公顷( )平方米 \n\n12平方千米( )公顷\n\n400000000平方米( )公顷( )平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "30000;1200;40000;400",
+        "solution":  "【分析】1公顷=10000平方米,1平方千米=100公顷,1平方千米=1000000平方米，高级单位换算成低级单位乘进率，低级单位换算成高级单位除以进率.\n【详解】3公顷=3×10000=30000平方米。12平方千米=12×100=1200公顷。400000000平方米=400000000÷10000=40000公顷,40000公顷=40000÷100=400平方千米。所以3公顷=30000平方米,12平方千米=1200公顷;400000000平方米=40000公顷=400平方千米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26四年级上·河南南阳·期中"
+    },
+    {
+        "id":  "143",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "4000 公顷=( )平方千米 \n\n5 公顷 = ( )平方米",
+        "options":  [
+
+                    ],
+        "answer":  "40;50000",
+        "solution":  "【分析】1平方千米 = 100 公顷,1公顷 = 10000 平方米，大单位化小单位乘进率，小单位化大单位除以进率，据此解答.\n【详解】4000 公顷里面有 40 个 100 公顷，所以 4000 公顷=40 平方千米。5 公顷里面有 5 个 1 公顷，所以 5 公顷 = 50000 平方米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·山东济宁·期中"
+    },
+    {
+        "id":  "144",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "6 平方米=( )平方分米;23平方千米=( )公顷;10 公顷 40 平方米=()平方米。",
+        "options":  [
+
+                    ],
+        "answer":  "600;2300;100040",
+        "solution":  "【分析】1 平方米 = 100 平方分米,6 平方米就是 6 个 100 平方分米，是 600 平方分米。1 平方千米 = 100 公顷,23 平方千米是 23 个 100 公顷，是2300 公顷。1公顷 = 10000 平方米,10公顷是10个10000平方米，是100000平方米。再和40平方米合起来即可.\n【详解】6平方米就是6个100平方分米，是600平方分米。所以6平方米 = 600 平方分米。23 平方千米是 23 个 100 公顷，是 2300 公顷。所以 23 平方千米 = 2300 公顷。10 公顷是 10 个 10000 平方米，是 100000 平方米。100000+40 = 100040 (平方米)。所以 10 公顷 40 平方米=100040平方米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "平方厘米，平方分米，平方米之间的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·江西九江·期中"
+    },
+    {
+        "id":  "145",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "北京故宫是世界上最大的宫殿，占地面积约是\u003cu\u003e720000\u003c/u\u003e平方米，横线上的数是( )位数，改写成用\"万\"作单位的数是( )万平方米，合( )公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "六;72;72",
+        "solution":  "【分析】①位数判断:720000是一个整数，从最高位开始数，有6个数字，因此是六位数。②改写为\"万\"作单位：根据整数改写，将一个数改写成用\"万\"作单位的数，需要抹掉万位后面的0，并加上\"万\"字。③单位换算：根据公顷和平方米的进率,1公顷 =10000平方米，因此将平方米数除以10000即可得到公顷数.\n【详解】根据分析可得:①720000是一个六位数;②720000改写成\"万\"作单位，需要抹掉万位后面的0，并加上\"万\"字，即72万平方米;③因为1公顷=10000平方米,720000平方米化成公顷作单位，除以进率10000，即720000÷10000=72(公顷)。所以横线上的数是六位数，改写成用\"万\"作单位的数是72万，合72公顷。",
+        "knowledge_points":  [
+                                 "亿以内数的改写",
+                                 "公顷，平方千米的进率与换算",
+                                 "整数的数级，数位和计数单位的认识"
+                             ],
+        "source":  "25-26四年级上·江西九江·期中"
+    },
+    {
+        "id":  "146",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "3 平方米=( )平方分米 \n\n1400 平方分米=( )平方米\n\n5 公顷=( )平方米 \n\n3 平方千米=( )公顷",
+        "options":  [
+
+                    ],
+        "answer":  "300;14;50000;300",
+        "solution":  "【分析】把3平方米换算为平方分米，因为1平方米 = 100 平方分米，所以3平方米里面有3个100平方分米，即300平方分米;把1400平方分米换算为平方米,1平方米 = 100 平方分米，因为1400平方分米里面有14个100平方分米，即14平方米。把5公顷换算为平方米，因为1公顷 = 10000 平方米，所以5公顷里面有5个10000平方米，即50000平方米。3平方千米换算为公顷，因为1平方千米 = 100 公顷,3平方千米里面有3个100公顷，即300公顷.\n【详解】3 平方米 = 300 平方分米;1400 平方分米 = 14 平方米;5公顷 = 50000 平方米;3平方千米 = 300 公顷。",
+        "knowledge_points":  [
+                                 "平方厘米，平方分米，平方米之间的进率与换算",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·广东江门·期中"
+    },
+    {
+        "id":  "147",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "我们学校占地面积约2( )，余江区的面积约是933( )。`r`n`r`n47000000平方米=( )公顷=( )平方千米`r`n`r`n( )平方米=4公顷500平方米",
+        "options":  [
+
+                    ],
+        "answer":  "公顷;平方千米;4700;47;40500",
+        "solution":  "【分析】(1)1公顷是边长为100米的正方形面积，计量学校的面积用公顷作单位比较合适。(2)1平方千米是边长为1000米的正方形面积，计量余江区的面积用平方千米作单位比较合适。(3)1公顷 = 10000平方米，1平方千米 = 1000000平方米，高级单位化低级单位要乘进率，低级单位化高级单位要除以进率。(4)1公顷 = 10000平方米，先把4公顷换算成平方米，再加上500平方米。`r`n【详解】我们学校占地面积约2公顷；余江区的面积约是933平方千米。1公顷 = 10000平方米，47000000÷10000 = 4700（公顷），4700÷100 = 47（平方千米），所以47000000平方米 = 4700公顷 = 47平方千米。1公顷 = 10000平方米，4×10000 = 40000（平方米），40000+500 = 40500（平方米），所以40500平方米 = 4公顷500平方米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 四年级上·江西鹰潭·期中"
+    },
+    {
+        "id":  "148",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "括号里填上\"\u003e\"\"\u003c\"或 \"=\".\n\n99999999( )9 亿 \n\n3031988888( )3030988888\n\n8 平方千米( )800 公顷 \n\n1000 平方米( )1 公顷",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003e;=;\u003c",
+        "solution":  "【分析】本题考查大数的比较和面积单位的换算。对于第一个比较，需要将“9亿”转换为数字900000000，再与99999999比较；第二个比较直接比较两个十位数的大小，从高位到低位逐位比较；第三个和第四个比较需要先进行单位换算，1平方千米 = 100公顷，1公顷 = 10000平方米，换算后再比较数值大小。`r`n【详解】(1)99999999\u003c900000000，所以99999999\u003c9亿，填“\u003c”；(2)3031988888和3030988888都是十位数，前三位数字相同，第四位数字分别是1和0，因为1\u003e0，所以3031988888\u003e3030988888，填“\u003e”；(3)1平方千米 = 100公顷，所以8平方千米 = 800公顷，因此8平方千米=800公顷，填“=”；(4)1公顷 = 10000平方米，1000平方米\u003c10000平方米，所以1000平方米\u003c1公顷，填“\u003c”。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26四年级上·江西抚州·期中"
+    },
+    {
+        "id":  "149",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "5 公顷=( )平方米;350000000 平方米=( )平方千米。",
+        "options":  [
+
+                    ],
+        "answer":  "50000;350",
+        "solution":  "【分析】1公顷 = 10000 平方米,5里面有5个1，所以5公顷 = 50000 平方米;1平方千米 = 1000000 平方米,350000000 里面有 350 个 1000000，所以 350000000 平方米= 350 平方千米。据此解答.\n【详解】根据分析可知,5公顷 = 50000 平方米;350000000平方米 = 350 平方千米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·湖北十堰·期中"
+    },
+    {
+        "id":  "150",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "8 公顷=( )平方米,( )公顷 = 2 平方千米。",
+        "options":  [
+
+                    ],
+        "answer":  "80000;200",
+        "solution":  "【分析】根据 1 平方千米 = 100 公顷,1 公顷 = 10000 平方米进行单位换算，从高级单位到低级单位，乘上进率;从低级单位到高级单位，除以进率.\n【详解】1 公顷 = 10000 平方米,8×10000 = 80000 (平方米);1平方千米 = 100 公顷,2×100 = 200(公顷)。所以8公顷 = 80000 平方米,200公顷 = 2 平方千米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·广西南宁·期中"
+    },
+    {
+        "id":  "151",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "5 平方米=( )平方厘米 \n\n64 平方千米=( )公顷\n\n23 公顷=( )平方米 \n\n50 公顷 20 平方米=( )平方米",
+        "options":  [
+
+                    ],
+        "answer":  "50000;6400;230000;500020",
+        "solution":  "【分析】根据面积单位之间的进率:1平方千米 = 100 公顷,1公顷 = 10000 平方米,1平方米 = 10000 平方厘米，从高级单位名数换算成低级单位名数乘进率，从低级单位名数换算成高级单位名数除以进率.\n【详解】5平方米里面有5个1平方米，也就是有5个10000平方厘米，所以5平方米 = 50000 平方厘米。1 平方千米 = 100 公顷，所以 64 平方千米 = 6400 公顷。1 公顷 = 10000 平方米，所以 23 公顷=230000 平方米。先把 50 公顷换算成平方米:50 公顷 = 500000 平方米;再加上 20 平方米:500000 平方米 + 20 平方米 = 500020 平方米，即 50 公顷 20 平方米 = 500020 平方米。",
+        "knowledge_points":  [
+                                 "平方厘米，平方分米，平方米之间的进率与换算",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·浙江杭州·期中"
+    },
+    {
+        "id":  "152",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "5 平方千米=( )公顷 \n\n50000 平方米=( )公顷",
+        "options":  [
+
+                    ],
+        "answer":  "500;5",
+        "solution":  "【分析】根据 1 平方千米 = 100 公顷,1 公顷 = 10000 平方米，据此单位换算.\n【详解】5平方千米中有5个1平方千米，也就是5个100公顷，所以 5 平方千米 = 500 公顷;50000 平方米中有 5 个 10000 平方米，也就是 5 个 1 公顷，所以 50000 平方米 = 5 公顷。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·广东阳江·期中"
+    },
+    {
+        "id":  "153",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号填上合适的单位.\n\n杭州西湖的水域面积约为 6.39( )。周末，小欣一家坐环湖观光车游览，观光车线路全程约 14( ),用时约70分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "平方千米;千米",
+        "solution":  "【分析】(1)计量较大的土地面积，常用公顷和平方千米作单位。边长是100米的正方形的面积为1公顷，边长是1千米的正方形的面积是1平方千米。由题意得，第一个括号里应该填较大的面积单位。杭州西湖的水域面积特别大，选择平方千米作单位比较合适。(2)常用的长度单位有厘米，分米，米和千米。我们知道，食指宽大约1厘米，一支笔的长度约1分米，小学生展开双臂，两臂的距离大约是1米，连续步行10分钟所走的路程 大概就是1千米。由题意得，括号里应该填长度单位。观光车线路全程较长，选择千米作单位比较合适.\n【详解】杭州西湖的水域面积约为6.39平方千米。周末，小欣一家坐环湖观光车游览，观光车线路全程约14千米，用时约 70 分钟。",
+        "knowledge_points":  [
+                                 "面积单位的选择",
+                                 "公顷，平方千米的认识",
+                                 "长度单位的选择",
+                                 "千米的认识"
+                             ],
+        "source":  "25-26四年级上·浙江杭州·期中"
+    },
+    {
+        "id":  "154",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "地球的海洋总面积约是 361000000( )(填上合适的单位名称),把横线上的数改写成用\"万\"做单位的数是( )万。",
+        "options":  [
+
+                    ],
+        "answer":  "平方千米;36100",
+        "solution":  "【分析】根据生活经验，对面积单位和数据大小的认识可知：边长是1千米的正方形土地的面积是1平方千米，也即100公顷，计量比较大的土地面积常用\"平方千米\"做单位，所以计量地球的海洋总面积，结合数据361000000用\"平方千米\"作单位比较合适。把一个整万数改写成用\"万\"作单位的数，把个级里4个0去掉同时在后面写上\"万\"字即可.\n【详解】地球的海洋总面积约是361000000(平方千米)(填上合适的单位名称),把横线上的数改写成用\"万\"做单位的数是(36100)万。",
+        "knowledge_points":  [
+                                 "亿以上数的改写",
+                                 "面积单位的选择",
+                                 "公顷，平方千米的认识"
+                             ],
+        "source":  "25-26四年级上·河南驻马店·期中"
+    },
+    {
+        "id":  "155",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上合适的数.\n\n80000 平方米 = ( )公顷\n\n3 平方千米 = ( )公顷=( )平方米",
+        "options":  [
+
+                    ],
+        "answer":  "8;300;3000000",
+        "solution":  "【分析】根据10000平方米 = 1 公顷,1平方千米 = 100 公顷=1000000 平方米，低级单位转换为高级单位除以进率，高级单位转换为低级单位乘上进率。据此换算填空即可.\n【详解】80000 平方米 = 8 公顷;3×100 = 300 , 3×1000000 = 3000000 ,3 平方千米 = 300 公顷 = 3000000 平方米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26四年级上·山东日照·期中"
+    },
+    {
+        "id":  "156",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "单位换算.\n\n3 平方千米=( )公顷 \n\n50000 平方米=( )公顷\n\n700 公顷=( )平方千米 \n\n2 公顷=( )平方米",
+        "options":  [
+
+                    ],
+        "answer":  "300;5;7;20000",
+        "solution":  "【分析】1 平方千米 = 100 公顷 = 1000000 平方米,1 公顷=10000 平方米，据此进行换算解答.\n【详解】(1)1平方千米 = 100 公顷,3平方千米中有3个1平方千米，也就是3个100公顷，所以3平方千米 = 300 公顷;(2)1 公顷 = 10000 平方米,50000 平方米中有 5 个 10000平方米，也就是 5 个 1 公顷，所以 50000 平方米 = 5 公顷;(3)1平方千米 = 100 公顷,700公顷中有7个100公顷，也就是 7 个 1 平方千米，所以 700 公顷 = 7 平方千米;(4)1公顷 = 10000 平方米,2公顷中有2个1公顷，也就是 2 个 10000 平方米，所以 2 公顷 = 20000 平方米。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·河南驻马店·期中"
+    },
+    {
+        "id":  "157",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上合适的单位名称.\n\n北京市的面积约是 16410( );我们学校的占地面积约是 2( );我们班黑板的面积约是 5( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方千米;公顷;平方米",
+        "solution":  "【分析】根据生活经验以及面积单位和数据大小的认识，面积的常用单位有平方千米，公顷，平方米，平方分米等;1平方千米大约是小型湖泊的湖面面积,1公顷比一个标准足球场稍大一点,1平方米大约1张方桌面的大小,1平方分米大约1个成人手掌面的大小，拇指指甲盖的面积大约是1平方厘米;北京市作为首都，管辖面积非常大，应选用大的面积单位，结合所给数据，应选用\"平方千米\"作单位;学校的占地面积应该略大于足球场，结合所给数据，应选用\"公顷\"作单位;黑板的面积，结合所给数据，应选用\"平方米\"作单位，据此即可解答.\n【详解】北京市的面积约是 16410 平方千米;我们学校的占地面积约是 2 公顷;我们班黑板的面积约是 5 平方米。",
+        "knowledge_points":  [
+                                 "面积单位的选择",
+                                 "公顷，平方千米的认识",
+                                 "平方厘米，平方分米，平方米的认识"
+                             ],
+        "source":  "25-26四年级上·河南驻马店·期中"
+    },
+    {
+        "id":  "158",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上合适的单位.\n\n(1)我国的国土面积是 960 万( )。 \n(2)糖糖的一天，早上 7( )起床，称体重 35( ),量身高145( ),吃完早饭，看一会书，书的封面面积约是 6( ),然后走路去 1( )外的舞蹈室跳舞，舞蹈室面积约 100( ),下午去农田体验劳动的乐趣，农田面积约 2( ),傍晚回家。",
+        "options":  [
+
+                    ],
+        "answer":  "平方千米;时;千克;厘米;平方分米;千米;平方米;公顷",
+        "solution":  "【分析】计量比较大的土地的面积，常用平方千米(( k m 2 ))作单位，两个半故宫的面积大约是1平方千米，因此计量我国的国土面积以\"平方千米\"为单位比较合适;起床时间的单位为\"时\";称一般物体的重量用千克(kg)作单位，一只小狗的重量大约是10千克，所以糖糖的体重用\"千克\"作单位比较合适;一把直尺大约是15厘米，但因为单位前面的数字是145，所以糖糖的身高用\"厘米\"作单位比较合适;成人的手掌面积大约1平方分米所以书的封面面积用\"平方分米\"作单位比较合适;操场的一圈是400米，两圈半是1千米，所以计量去舞蹈室的路程用\"千米\"作单位比较合适;正方形地板砖面积是 1 平方米，所以舞蹈室面积用\"平方米\"作单位比较合适;结合数据2，农田的面积用公顷作单位比较合适.\n【详解】(1)我国的国土面积是 960 万平方千米。(2)糖糖的一天，早上7时起床，称体重35千克，量身高145厘米，吃完早饭，看一会书，书的封面面积约是6平方分米，然后走路去1千米外的舞蹈室跳舞，舞蹈室面积约100平方米，农田面积约2公顷，傍晚回家。",
+        "knowledge_points":  [
+                                 "时间单位的选择",
+                                 "质量单位的选择",
+                                 "长度单位的选择",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26四年级上·江西抚州·期中"
+    },
+    {
+        "id":  "159",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上合适的面积单位.\n\n一个篮球场的面积是 420( );某地公园占地约 3( );香港特别行政区的陆地面积约 1106( );课桌面的面积约 40( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方米;公顷;平方千米;平方分米",
+        "solution":  "【分析】根据生活经验，对面积单位的认识可知，边长是 1 米的正方形的面积是 1 平方米。双人课桌面的面积大约是1平方米，所以计量一个篮球场的面积用\"平方米\"作单位比较合适;边长是100米的正方形的面积是1公顷,400米跑道围起来的部分的面积大约是1公顷，所以计量学校，公园占地面积用\"公顷\"作单位比较合适;边长是1千米的正方形的面积是1平方千米，国土，地区面积常用平方千米作单位，所以计量香港特别行政区的陆地面积用\"平方千米\"作单位比较合适;课桌面的面积常用\"平方分米\"作单位.\n【详解】由分析可知，一个篮球场的面积是 420 平方米，某地公园占地约 3 公顷，香港特别行政区的陆地面积约 1106平方千米，课桌面的面积约 40 平方分米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26四年级上·湖北十堰·期中"
+    },
+    {
+        "id":  "160",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "300000000 平方米 =( )公顷 = ( )平方千米\n\n80 公顷=( )平方米 \n\n600 平方千米=( )公顷",
+        "options":  [
+
+                    ],
+        "answer":  "30000;300;800000;60000",
+        "solution":  "【分析】根据 1 平方千米=1000000 平方米,1 公顷 = 10000平方米,1平方千米 = 100 公顷进行单位换算即可.\n【详解】300000000 平方米里面有 30000 个 10000 平方米，有 300 个 1000000 平方米。则 300000000 平方米 = 30000 公顷 = 300 平方千米;1 公顷 = 10000 平方米,80 公顷里面有 800000 个平方米,80 公顷 = 800000 平方米;1 平方千米 = 100 公顷,600 平方千米 60000 个公顷,600 平方千米 = 60000 公顷。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·山东日照·期中"
+    },
+    {
+        "id":  "161",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上适当的面积单位.\n\n广州市的面积约 7434( )\n\n手指甲的面积约是 1( )\n\n教室地面的面积是 96( )\n\n国家体育场的占地面积约为 20( )",
+        "options":  [
+
+                    ],
+        "answer":  "平方千米;平方厘米;平方米;公顷",
+        "solution":  "【分析】平方千米适合计量特别大的土地面积，边长是1千米的正方形的面积是1平方千米，一个足球场的面积大约是7000平方米,140个这样的足球场的面积和大约是1平方千米，所以计量广州市的面积用\"平方千米\"作单位比较合适;1平方厘米大约是计算机键盘上一个正方形按键面的面积，所以计量手指甲的面积用\"平方厘米\"作单位比较合适;1平方米大约是1米的正方形瓷砖的面积，所以计量教室地面的面积用\"平方米\"作单位比较合适;公顷适合计量稍大的土地面积，边长是100米的正方形的面积是1公顷,400米跑道围起来的部分的面积大约是1公顷，所以计量国家体育场的占地面积用\"公顷\"作单位比较合适.\n【详解】广州市的面积约7434平方千米手指甲的面积约是1平方厘米教室地面的面积是96平方米\n\n国家体育场的占地面积约为20公顷",
+        "knowledge_points":  [
+                                 "平方厘米，平方分米，平方米的认识",
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 四年级上·山东日照·期中"
+    },
+    {
+        "id":  "162",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在横线上填上适当的数.\n\n6 公顷= 平方米 \n\n40000 平方米= 公顷",
+        "options":  [
+
+                    ],
+        "answer":  "60000;4",
+        "solution":  "【分析】1公顷 = 10000 平方米，大单位化小单位乘进率，小单位化大单位除以进率，据此解答.\n【详解】6 公顷里面有 6 个 1 公顷，所以 6 公顷 = 60000 平方米;40000 平方米里面有 4 个 1 公顷，所以 40000 平方米 = 4 公顷。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·广东汕尾·期中"
+    },
+    {
+        "id":  "163",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "如果 1 平方米能摆放 8 盆花,1 公顷能摆放( )盆花,1 平方千米能摆放( )盆花。",
+        "options":  [
+
+                    ],
+        "answer":  "80000;8000000",
+        "solution":  "【分析】1公顷 = 10000 平方米，用1平方米能摆放的盆数乘10000，计算出1公顷能摆放的盆数;1平方千米=1000000 平方米，用 1 平方米能摆放的盆数乘 1000000，计算出 1 平方千米能摆放的盆数;据此解答.\n【详解】1 公顷 = 10000 平方米8×10000 = 80000 (盆)1 平方千米 = 1000000 平方米8×1000000 = 8000000 (盆)所以如果1平方米能摆放8盆花,1公顷能摆放80000盆花,1 平方千米能摆放 8000000 盆花。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "公顷，平方千米的进率与换算",
+                                 "整十，整百，整千数与一位数的乘法"
+                             ],
+        "source":  "25-26 四年级上·山东日照·期中"
+    },
+    {
+        "id":  "164",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在( )里填上适当的数.\n\n28 公顷=( )平方米 \n\n600 平方千米=( )公顷",
+        "options":  [
+
+                    ],
+        "answer":  "280000;60000",
+        "solution":  "【分析】根据1平方千米 = 100 公顷,1公顷 = 10000 平方米，低级单位转换为高级单位除以进率，高级单位转换为低级单位乘进率，换算单位解答即可.\n【详解】28×10000 = 280000，所以 28 公顷 = 280000 平方米;600×100 = 60000，所以 600 平方千米 = 60000 公顷。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26四年级上·河南驻马店·期中"
+    },
+    {
+        "id":  "165",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "国家速滑馆\"冰丝带\"的建筑面积约 8 公顷，那么( )个这样的\"冰丝带\"面积大约是2平方千米。",
+        "options":  [
+
+                    ],
+        "answer":  "25",
+        "solution":  "【分析】根据1平方千米 = 100 公顷进行换算,2平方千米= 200 公顷，用200除以8即可得知需要几个这样的\"冰丝带\" .\n【详解】2平方千米 = 200 公顷200÷8 = 25 (个)所以国家速滑馆\"冰丝带\"的建筑面积约 8 公顷，那么 25个这样的\"冰丝带\"面积大约是 2 平方千米。",
+        "knowledge_points":  [
+                                 "整十，整百，整千数除以一位数的除法",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·河南南阳·期中"
+    },
+    {
+        "id":  "166",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\"\u003e\"\"\u003c\"或\"=\".\n\n500 公顷( )50 平方千米 \n\n308 公顷( )3 平方千米+8公顷 \n\n50 万( )499800",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;=;\u003e",
+        "solution":  "【分析】1平方千米 = 100 公顷,500公顷里面有几个100公顷，就是几平方千米;据此先统一单位，再比较大小即可;1平方千米 = 100 公顷,3平方千米就是3个100公顷，据此先将3平方千米的单位换算为公顷，再与8公顷相加，求出和，再比较大小即可;将 50 万改写为个的单位，去掉万字，在数的末尾加上 4 个0;据此先统一单位，再比较大小即可.\n【详解】500公顷里面有5个100公顷,500公顷 = 5 平方千米,5平方千米\u003c50平方千米，即500公顷 \u003c 50 平方千米3 个 100 公顷是 300 公顷,3 平方千米 = 300 公顷,300 公顷+ 8 公顷 = 308 公顷，即308公顷 = 3 平方千米 + 8 公顷50 万=500000,500000\u003e499800，即 50 万\u003e499800",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "公顷，平方千米的进率与换算",
+                                 "亿以内数的改写"
+                             ],
+        "source":  "25-26 四年级上·河南南阳·期中"
+    },
+    {
+        "id":  "167",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\"\u003e\"\"\u003c\"或\"=\".\n\n204 万( )20450000 \n\n999999( )1000000\n\n9 平方千米( )90 公顷 \n\n20 公顷( )20000 平方米",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003c;\u003e;\u003e",
+        "solution":  "【分析】整数大小比较方法：位数不相同的两个数，位数多的数较大。位数相同的两个数，从最高位比起，最高位上的数大的那个数较大。如果最高位上的数相同，就比较下一个数位上的数，以此类推。把204万改写成2040000，然后按照整数大小比较方法进行比较;先统一单位，再比较大小,1平方千米 = 100 公顷,1公顷 = 10000 平方米，级单位换算成低级单位，就乘以单位间的进率，低级单位换算成高级单位，就除以单位间的进率。由此进行比较.\n【详解】204 万=2040000，所以 2040000\u003c20450000;999999 是 6 位数,1000000 是 7 位数，所以 999999\u003c1000000;1平方千米 = 100 公顷,9平方千米 = 900 公顷,900 { \u003e } 90，所以 9 平方千米 { tt \u003e } 90 公顷;1 公顷 = 10000 平方米,20 公顷 = 200000 平方米,200000\u003e20000，所以 20 公顷\u003e20000 平方米。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26四年级上·新疆乌鲁木齐·期中"
+    },
+    {
+        "id":  "168",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上合适的单位.\n\n(1)我国陆地领土面积约为 960 万( )。 \n(2)明明家客厅的面积约 40( )。 \n(3)国家游泳中心\"水立方\"的占地面积约 6( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方千米;平方米;公顷",
+        "solution":  "【分析】常见的用于描述建筑的面积单位有平方千米，公顷，平方米等,1 平方千米 = 100 公顷,1 公顷 = 10000 平方米，平方千米一般用于很大的面积，如上海市的面积约6340 平方千米，公顷一般用于较大的面积，如\"鸟巢\"的占地面积约 20 公顷，平方米用于较小的面积，如商品房的面积约 100 平方米.\n【详解】(1)平方千米一般用于很大的面积，我国陆地领土面积很大，结合所给数据 960 万，单位用平方千米比较合适，因此我国陆地领土面积约 960 万平方千米.\n\n(2)平方米用于较小的面积，一般商品房的面积约 100 平方米，客厅是商品房的一部分，结合所给数据 40，单位用平方米比较合适，因此明明家客厅的面积约 40 平方米.\n\n(3)公顷一般用于较大的面积，\"水立方\"是重要的运动场馆，面积较大，结合所给数据 6，单位用公顷比较合适，因此国家游泳中心\"水立方\"的占地面积约 6 公顷。",
+        "knowledge_points":  [
+                                 "平方厘米，平方分米，平方米的认识",
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 四年级上·新疆乌鲁木齐·期中"
+    },
+    {
+        "id":  "169",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "6 公顷=( )平方米 1700 公顷 = ( )平方千米\n\n5 平方千米=( )平方米=( )公顷 \n\n8 公顷 80 平方米 = ( )平方米",
+        "options":  [
+
+                    ],
+        "answer":  "60000;17;5000000;500;80080",
+        "solution":  "【分析】单位换算需熟记进率:1 公顷 = 10000 平方米,1平方千米 = 100 公顷 = 1000000 平方米.\n【详解】6 公顷 = 6×10000 = 60000 平方米;1700 公顷 = 1700÷100 = 17 平方千米;5平方千米 = 5×1000000 = 5000000 平方米;5平方千米 = 5×100 = 500 公顷;8公顷80平方米 = 8×10000 + 80 = 80000 + 80 = 80080 平方米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·新疆乌鲁木齐·期中"
+    },
+    {
+        "id":  "170",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "沧州市区的总面积约为 198( );而广州城市新中轴线上的核心景点--花城广场，它的占地面积约为 56( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方千米;公顷",
+        "solution":  "【分析】1 平方千米相当于边长为 1 千米的正方形面积，计量沧州市区的总面积用平方千米作单位比较合适。1公顷相当于边长为100米的正方形面积，计量花城广场的面积用公顷作单位比较合适.\n【详解】沧州市区的总面积约为198平方千米;而广州城市新中轴线上的核心景点--花城广场，它的占地面积约为56 公顷。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26四年级上·河北沧州·期中"
+    },
+    {
+        "id":  "171",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "6 平方千米=( )公顷,80000平方米=( )公顷;嵩明县的区域面积大约是825( )。",
+        "options":  [
+
+                    ],
+        "answer":  "600;8;平方千米",
+        "solution":  "【分析】根据题意，明确单位间的进率,1平方千米 = 100 公顷,10000平方米 = 1 公顷，由此根据公顷和平方千米之间的进率换算单位。根据实际情况结合单位的大小可知，平方千米表示大的土地的面积，如我国领土的面积是960万平方千米，可以判断嵩明县的区域面积大约是825平方千米。以此答题即可.\n【详解】根据分析可知:6平方千米 = 600 公顷,80000平方米 = 8 公顷;嵩明县的区域面积大约是 825 平方千米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·云南昆明·期中"
+    },
+    {
+        "id":  "172",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "昆明翠湖被誉为\"城市绿宝石\"，总面积约 20 公顷,( )个翠湖的占地面积约 1 平方千米。",
+        "options":  [
+
+                    ],
+        "answer":  "5",
+        "solution":  "【分析】首先统一面积单位，将平方千米换算成公顷，根据1平方千米 = 100 公顷，再利用四舍五入法将翠湖的面积看成是20公顷，用总面积1平方千米换算后的公顷数，除以翠湖的占地面积，就能得出几个翠湖的占地面积约1平方千米。据此解答即可.\n【详解】1平方千米 = 100 公顷,100÷20 = 5 (个)。故可得出5个翠湖的占地面积约1平方千米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·云南昆明·期中"
+    },
+    {
+        "id":  "173",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "在括号里填上合适的单位名称.\n\n合川中学占地面积约 18( );教室的面积约是 60( );合川因嘉陵江，渠江，涪江汇流而得名，占地面积约 2344()。",
+        "options":  [
+
+                    ],
+        "answer":  "公顷;平方米;平方千米",
+        "solution":  "【分析】1 公顷是一个边长为 100 米正方形的面积，计量学校的占地面积用\"公顷\"作单位比较合适;1平方米是一个边长为1米正方形的面积，计量教室的面积用\"平方米\"作单位比较合适;1平方千米是一个边长为1000米正方形的面积，计量合川的占地面积用\"平方千米\"作单位比较合适.\n【详解】合川中学占地面积约18公顷;教室的面积约是60平方米;合川因嘉陵江，渠江，涪江汇流而得名，占地面积约2344平方千米。",
+        "knowledge_points":  [
+                                 "面积单位的选择"
+                             ],
+        "source":  "24-25五年级上·重庆合川·期末"
+    },
+    {
+        "id":  "174",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一片原始森林长 85 千米，宽 2 千米，这片原始森林的面积是( )平方千米，合( )公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "170;17000",
+        "solution":  "【分析】根据题意，明确长方形的面积 = 长×宽，用 85 乘 2，求出原始森林的面积，再根据 1 平方千米 = 100 公顷，把平方千米换算成公顷即可.\n【详解】根据分析可知:85×2 = 170 (平方千米)。1平方千米 = 100 公顷,170 平方千米 = 17000 公顷。一片原始森林长85千米，宽2千米，这片原始森林的面积是170平方千米，合17000公顷。",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·湖北省直辖县级单位·期中"
+    },
+    {
+        "id":  "175",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上合适的面积单位.\n\n山东省的陆地面积约 16 万( ).\n\n北京\"水立方\"的占地面积约 6( ) 。",
+        "options":  [
+
+                    ],
+        "answer":  "平方千米;公顷",
+        "solution":  "【分析】根据生活经验以及对面积单位和数据大小的认识，选择合适的计量单位：计量比较大的土地面积，常用平方千米作单位，边长是1千米的正方形的面积是1平方千米，一个足球场的面积大约是7000平方米,140个这样的足球场的总面积大约是1平方千米，所以计量山东省的陆地面积应用\"平方千米\"作单位比较合适;公顷适合计量稍大的土地面积，边长是100米的正方形的面积是1公顷,400米跑道围起来的部分的面积大约是1公顷，所以计量北京\"水立方\"的占地面积应用\"公顷\"作单位比较合适.\n【详解】根据分析可知：山东省的陆地面积约 16 万平方千米;北京\"水立方\"的占地面积约 6 公顷。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 四年级上·山东济宁·期中"
+    },
+    {
+        "id":  "176",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上合适的数字.\n\n70 公顷=( )平方米 ( )公顷 = 8 平方千米\n\n300000 米 = ( )千米 20 平方米=( )平方分米",
+        "options":  [
+
+                    ],
+        "answer":  "700000;800;300;2000",
+        "solution":  "【分析】1公顷 = 10000 平方米,1平方千米 = 100 公顷,1千米 = 1000 米,1平方米 = 100 平方分米，高级单位化成低级单位就乘它们之间的进率，低级单位化成高级单位就除以它们之间的进率，依此换算.\n【详解】70 公顷是 70 个 10000 平方米，即 70 公顷=700000平方米。8 平方千米是 8 个 100 公顷，即 800 公顷 = 8 平方千米。300000 米是 300 个 1000 米，即 300000 米 = 300 千米。20 平方米是 20 个 100 平方分米，即 20 平方米 = 2000 平方分米。",
+        "knowledge_points":  [
+                                 "千米和米之间的进率与换算",
+                                 "平方厘米，平方分米，平方米之间的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·河南南阳·期中"
+    },
+    {
+        "id":  "177",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上适当的单位.\n\n\"水立方\"占地面积约 6( )。香港特别行政区的面积约是1100( )。",
+        "options":  [
+
+                    ],
+        "answer":  "公顷;平方千米",
+        "solution":  "【分析】1公顷等于10000平方米，像学校，公园，湿地等的面积一般用公顷作单位，所以\"水立方\"占地面积用公顷作单位比较合适;100公顷等于1平方千米，像国家，省，市，县的面积一般用平方千米作单位，所以香港特别行政区的面积用平方千米作单位比较合适.\n【详解】\"水立方\"占地面积约 6 公顷。香港特别行政区的面积约是1100平方千米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 四年级上·四川遂宁·期中"
+    },
+    {
+        "id":  "178",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "1000 公顷=( )平方千米 \n\n5 公顷 70 平方米 = ( )平方米",
+        "options":  [
+
+                    ],
+        "answer":  "10;50070",
+        "solution":  "【分析】1公顷 = 10000 平方米,1平方千米 = 100 公顷，高级单位化成低级单位就乘它们之间的进率，低级单位化成高级单位就除以它们之间的进率，依此换算.\n【详解】1000 公顷是 10 个 100 公顷，即 10 个 1 平方千米，则 1000 公顷 = 10 平方千米。5 公顷 = 50000 平方米，即 5 公顷 70 平方米 = 50070 平方米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·四川遂宁·期中"
+    },
+    {
+        "id":  "179",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号内填上合适的面积单位.\n\n(1)我家客厅的面积大约是 30( )。 \n(2)綦江区是重庆的卫星城，幅员面积约 2748( )。 \n(3)北京天安门广场是世界上最大的城市广场，占地面积约44( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方米;平方千米;公顷",
+        "solution":  "【分析】根据生活经验，对面积单位和数据大小的认识,1平方米大约能站12个小朋友;1公顷大约是学校操场的面积;1平方千米大约是2.3个天安门广场那么大;所以计量我家客厅的占地面积用\"平方米\"作单位，计量綦江区的面积用\"平方千米\"作单位，计量天安门广场是世界上最大的城市中心广场，用\"公顷\"作单位;据此得解.\n【详解】(1)我家客厅的面积大约是30平方米。(2)綦江区是重庆的卫星城，幅员面积约 2748 平方千米。(3)北京天安门广场是世界上最大的城市广场，占地面积约 44 公顷。",
+        "knowledge_points":  [
+                                 "面积单位的选择"
+                             ],
+        "source":  "24-25 五年级上·重庆綦江·期末"
+    },
+    {
+        "id":  "180",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在( )里填上适当的面积单位.\n\n(1)一枚 1 元硬币面积约为 5( )。 \n(2)天安门广场占地面积约 44( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方厘米;公顷",
+        "solution":  "【分析】1个指甲盖的大小大约是1平方厘米，故一枚1元硬币面积用平方厘米作单位;1公顷比一个标准足球场的面积稍大一点，故天安门广场占地面积用公顷作单位.\n【详解】(1)一枚 1 元硬币面积约为 5 平方厘米。(2)天安门广场占地面积约44公顷。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 四年级上·河南南阳·期中"
+    },
+    {
+        "id":  "181",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "排序题",
+        "difficulty":  0.65,
+        "question":  "把下面各数从大到小排列.\n\n5000平方厘米 5公顷 5平方千米 500平方分米 50平方米\n\n( )\u003e( )\u003e( )\u003e( )\u003e( )",
+        "options":  [
+
+                    ],
+        "answer":  "5 平方千米;5 公顷;50 平方米;500 平方分米;5000 平方厘米",
+        "solution":  "【分析】根据1平方千米 = 100 公顷,1公顷 = 10000 平方米,1平方米 = 100 平方分米,1平方分米 = 100 平方厘米，据此可以统一单位后比较即可.\n【详解】5000平方厘米 = 50 平方分米;5公顷 = 50000 平方米;5平方千米 = 5000000 平方米;500平方分米 = 5 平方米。5000000 平方米\u003e50000 平方米\u003e50 平方米\u003e5 平方米\u003e50 平方分米。5 平方千米\u003e5 公顷\u003e50 平方米\u003e500 平方分米\u003e5000 平方厘米。",
+        "knowledge_points":  [
+                                 "平方厘米，平方分米，平方米之间的进率与换算",
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26四年级上·广东东莞·月考"
+    },
+    {
+        "id":  "182",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "排序题",
+        "difficulty":  0.65,
+        "question":  "把 53 平方千米,900000平方米,700公顷,600公顷90平方米按从大到小的顺序排列是\n\n( )\u003e( )\u003e( )\u003e( )。",
+        "options":  [
+
+                    ],
+        "answer":  "53 平方千米 700 公顷 600 公顷 90 平方米 900000 平方米",
+        "solution":  "【分析】比较不同单位的量，需要先统一单位，大单位换小单位乘进率，小单位换大单位除以进率。根据1平方千米= 100 公顷,1公顷 = 10000 平方米,1平方千米=1000000平方米，把题目中的单位都统一为中间单位，也就是\"公顷\"，再进行从大到小的排序即可.\n【详解】53 平方千米 = 5300 公顷;900000 平方米 = 90 公顷。单位统一后的数据:5300公顷,90公顷,700公顷,600公顷 90 平方米，从大到小排序为:5300 公顷\u003e700 公顷\u003e600 公顷 90 平方米\u003e90 公顷。所以按从大到小的顺序排列是:53平方千米\u003e700公顷\u003e600 公顷 90 平方米\u003e900000 平方米。",
+        "knowledge_points":  [
+                                 "万以内数的大小比较",
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·河南驻马店·期中"
+    },
+    {
+        "id":  "183",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "信阳平桥区的占地面积约为 1886 平方米。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】1平方米是边长1米的正方形面积的大小，像大门，操场，篮球场等的面积一般用平方米作单位,1公顷等于10000平方米，像学校，公园，湿地等的面积一般用公顷作单位;100公顷等于1平方千米，像国家，省，市，县的面积一般用平方千米作单位，据此即可解.\n【详解】根据分析可知，信阳平桥区的占地面积约为1886平方千米，原说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识"
+                             ],
+        "source":  "25-26 四年级上·河南信阳·期中"
+    },
+    {
+        "id":  "184",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "我国拥有湿地面积超过 6600 万公顷，合 66 平方千米。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】根据1平方千米 = 100 公顷，去掉6600万公顷中6600末尾的2个0即可换算成平方千米为单位，据此判断即可.\n【详解】6600 万公顷 = 66 万平方千米。我国拥有湿地面积超过6600万公顷，合66万平方千米，原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26四年级上·河南开封·期中"
+    },
+    {
+        "id":  "185",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "判断题",
+        "difficulty":  0.94,
+        "question":  "5 平方千米比 5 千米大。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平方千米是面积单位，千米是长度单位，面积和长度是不同类的量，不能直接比较大小.\n【详解】平方千米是面积单位，用于测量面积;千米是长度单位，用于测量长度。由于面积和长度是不同类的量，不能直接比较大小。所以,5平方千米和5千米无法比较大小。故答案为: ×",
+        "knowledge_points":  [
+                                 "千米的认识",
+                                 "公顷，平方千米的认识"
+                             ],
+        "source":  "25-26 四年级上·广东江门·期中"
+    },
+    {
+        "id":  "186",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "平方厘米，平方分米，平方米和公顷，每相邻两个面积单位之间的进率都是 10。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】根据1公顷 = 10000 平方米,1平方米 = 100 平方分米,1 平方分米 = 100 平方厘米，可知原题说法错误。据此判断.\n【详解】由分析可知，平方米和公顷之间的进率不是10，是10000。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·河南驻马店·期中"
+    },
+    {
+        "id":  "187",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "8 个长 50 米，宽 25 米的游泳池的面积是 1 公顷。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】根据长方形的面积公式计算一个游泳池的面积，再计算8个游泳池的总面积，并与1公顷进行比较。1公顷等于10000平方米，若总面积等于10000平方米，则说法正确.\n【详解】一个游泳池的面积: 50×25 = 1250 (平方米)。8 个游泳池的总面积: 8×1250 = 10000 (平方米)。10000 平方米 = 1 公顷。因为10000平方米等于1公顷，所以8个这样的游泳池的面积是1公顷;原题干说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "公顷，平方千米的认识",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·湖北省直辖县级单位·期中"
+    },
+    {
+        "id":  "188",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "如果 1 平方米能摆 6盆花，那么 1 公顷就能摆 6000 盆花。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】已知 1 平方米能摆放 6 盆花,1 公顷 = 10000 平方米，用每平方米的盆数乘平方米数，即可求出1公顷能摆的盆数，据此判断.\n【详解】1 公顷 = 10000 平方米,6×10000 = 60000 (盆)。6000!=60000，因此该说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·山东济宁·期中"
+    },
+    {
+        "id":  "189",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "说理题.\n\n将\"30000 平方米\"转化为\"公顷\"，请说明你是如何迁移面积单位间的进率完成换算的，并写出算法。算理：迁移\"公顷与平方米的进率\":1 公顷=( )平方米，将低级单位 \"平方米\"转化为高级单位\"公顷\",( )进率即可。算法：列式计算：结果:",
+        "options":  [
+
+                    ],
+        "answer":  "10000 除以 30000÷10000 = 330000 平方米 = 3 公顷",
+        "solution":  "【分析】依据单位换算得出，算理：迁移\"公顷与平方米的进率\":1公顷=10000平方米，将低级单位\"平方米\"转化为高级单位\"公顷\"，除以单位间进率即可。据此可解得.\n【详解】1公顷 = 10000 平方米，平方米换算成公顷，除以单位间的进率，可列式计算: 30000÷10000 = 3 。结果:30000 平方米 = 3 公顷。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·河南驻马店·期中"
+    },
+    {
+        "id":  "190",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "李叔叔承包了一个面积为2公顷的桃园，如果每4平方米可种1株桃树，那么这个桃园一共可种桃树多少株。",
+        "options":  [
+
+                    ],
+        "answer":  "5000 株",
+        "solution":  "【分析】先进行单位换算,1公顷 = 10000 平方米,2里面有2个1，所以2公顷=20000平方米。再用20000除以4，即可求出这个桃园一共可种桃树多少株.\n【详解】2 公顷 = 20000 平方米,20000÷4 = 5000 (株)。答：这个桃园一共可种桃树5000株。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26四年级上·山东菏泽·期中"
+    },
+    {
+        "id":  "191",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一块长方形稻田，长 600 米，宽 200 米.\n\n(1)这块稻田的面积是多少平方米。合多少公顷。 \n(2)如果每公顷收稻谷 8 吨，这块田可收稻谷多少吨。",
+        "options":  [
+
+                    ],
+        "answer":  "120000;12;96",
+        "solution":  "【分析】(1)长方形的面积 = 长×宽，已知长方形稻田的长是600米，宽是200米，先求出稻田的面积，再根据1公顷 = 10000 平方米，把稻田的面积换算成公顷;(2)每公顷收稻谷8吨，根据\"单产量×数量 = 总产量\"即可求出这块田可收稻谷的总量.\n【详解】(1) 600×200 = 120000 (平方米),120000 平方米 = 12 公顷。答：这块稻田的面积是120000平方米，合12公顷。(2) 8×12 = 96 (吨)。答：这块田可收稻谷96吨。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "长方形的面积"
+                             ],
+        "source":  "25-26 四年级上·浙江温州·期中"
+    },
+    {
+        "id":  "192",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一个长方形的玉米农场，长是 500 米，宽是 400 米。如果每公顷需要施肥 300千克，这个农场一共需要施肥多少千克。",
+        "options":  [
+
+                    ],
+        "answer":  "6000",
+        "solution":  "【分析】根据题意可知，长方形的玉米农场，长是500米，宽是400米，根据长方形的面积 = 长×宽，用 500 乘 400，即可求出玉米农场的面积;再根据1公顷 = 10000 平方米，将单位换算成公顷，每公顷需要施肥300千克，再乘300，即可求出一共需要施肥多少千克.\n【详解】 500×400 = 200000 (平方米),200000 平方米 = 20 公顷,20×300 = 6000 (千克)。答：这个农场一共需要施肥6000千克。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "长方形的面积"
+                             ],
+        "source":  "25-26 四年级上·湖南郴州·期中"
+    },
+    {
+        "id":  "193",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "王叔叔承包了一块土地，用来开设\"家庭农场\"。农场中有一块边长 400 米的正方形稻田，如果每公顷稻田能收稻谷 7 吨，这块稻田能收多少吨稻谷。",
+        "options":  [
+
+                    ],
+        "answer":  "112吨",
+        "solution":  "【分析】正方形稻田的面积 = 正方形稻田的边长×正方形稻田的边长，计算出面积,1 公顷 = 10000 平方米，再将单位化为公顷;最后用面积乘单位面积的产量，即可解答此题.\n【详解】 400×400 = 160000 (平方米),160000 平方米 = 16 公顷,16×7 = 112 (吨)。答：每公顷稻田能收稻谷7吨，这块稻田能收112吨稻谷。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "正方形的面积"
+                             ],
+        "source":  "25-26 四年级上·河南南阳·期中"
+    },
+    {
+        "id":  "194",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一个正方形农场的占地面积为 9 公顷，现要退耕还林，农场主将农场各边均缩减了 100 米，缩减后农场的面积减少了多少公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "5",
+        "solution":  "【分析】根据题意,1 公顷=10000 平方米,9 公顷就是90000平方米。正方形的面积 = 边长×边长，由此可知原来的边长是多少米。再用原来的边长减去缩小的 100 米，就是现在的边长。再代入正方形的面积公式，算出缩减后的面积。根据 1 公顷 = 10000 平方米，转换成公顷作单位。再用9 公顷减去缩减后的公顷数就是减少多少公顷.\n【详解】9 公顷 = 90000 平方米,300×300 = 90000 (平方米),300 - 100 = 200 (米),200×200 = 40000 (平方米),40000 平方米 = 4 公顷,9 - 4 = 5 (公顷)。答：缩减后农场的面积减少了 5 公顷。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "正方形的面积"
+                             ],
+        "source":  "25-26 四年级上·江西宜春·期中"
+    },
+    {
+        "id":  "195",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "中国是世界上最大的小麦生产国。王叔叔承包了一块长方形的麦田，长625米，宽80米。由于他的种植技术和管理水平高，平均每公顷产量达到8吨。这块麦田的总产量是多少吨。",
+        "options":  [
+
+                    ],
+        "answer":  "40 吨",
+        "solution":  "【分析】首先需要根据长方形面积公式算出麦田面积，长方形面积 = 长×宽，再将面积单位从平方米换算成公顷，最后用每公顷产量乘公顷数得到总产量.\n【详解】 625×80 = 50000 (平方米)。1 公顷 = 10000 平方米,50000 平方米里面有 5 个 10000 平方米，即 50000 = 5 公顷。5×8 = 40 (吨)。答：这块麦田的总产量是 40 吨。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "长方形的面积"
+                             ],
+        "source":  "25-26四年级上·浙江杭州·期中"
+    },
+    {
+        "id":  "196",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "张叔叔计划修建一个总面积是35000平方米的养殖场，养殖场的规划如下:\n\n①鸡场的面积是800平方米;②牛场的面积是2公顷;③猪舍的面积是8500平方米;④其他场地的面积是6520平方米.\n\n张叔叔的规划是否合理。请说明理由。",
+        "options":  [
+
+                    ],
+        "answer":  "张叔叔的规划不合理;规划的面积超过了养殖场的总面积",
+        "solution":  "【分析】根据1公顷=10000平方米,2公顷就是2个10000平方米，据此将牛场的面积单位换算成平方米;然后再把鸡场，牛场，猪舍和其他场地的面积相加，求出一共有多少平方米，再与张叔叔计划修建养殖场的总面积比较，如果面积相等，则规划合理，如果不相等，则规划不合理.\n【详解】2公顷=20000平方米。800+20000+8500+6520=35820(平方米)。35820\u003e35000。答：张叔叔的规划不合理;因为规划的面积超过了养殖场的总面积。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26四年级上·河南驻马店·期中"
+    },
+    {
+        "id":  "197",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一块长方形土地，长 500 米，宽 300 米，若把它的宽增加 100 米，长不变，现在的面积多少公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "20公顷",
+        "solution":  "【分析】首先计算出宽增加100米后的宽度，再根据长方形的面积 = 长×宽，即可求出长方形的面积，再进行单位换算即可.\n【详解】 300 + 100 = 400 (米),500×400 = 200000 (平方米),200000 平方米 = 20 公顷。答：现在的面积是20公顷。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "长方形的面积"
+                             ],
+        "source":  "25-26 四年级上·河南驻马店·期中"
+    },
+    {
+        "id":  "198",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "一架直升机在一片长 6 千米，宽 4 千米的长方形树林上空喷洒药水，这片树林的面积是多少平方千米。合多少公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "24 平方千米;2400 公顷",
+        "solution":  "【分析】根据长方形的面积 = 长×宽，代入数据即可求出面积。1 平方千米 = 100 公顷，据此将单位换算为公顷即可.\n【详解】 6×4 = 24 (平方千米),24平方千米 = 2400 公顷。答：这片树林的面积是 24 平方千米，合 2400 公顷。",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26四年级上·河南驻马店·期中"
+    },
+    {
+        "id":  "199",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "欢欢是个爱锻炼的孩子，她绕着一个长方形公园走一圈用了 10 分钟，她每分钟走50米。已知长方形公园长200米，它的占地面积是多少公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "1 公顷",
+        "solution":  "【分析】绕着一个长方形公园走一圈用了10分钟，她每分钟走50米，用50乘上10，算出这个长方形公园周长，再根据长方形周长 = (长+宽)×2，用周长除以2再减去200，算出长方形的宽，再根据长方形面积 = 长×宽，计算出该长方形面积，根据1公顷 = 10000 平方米，进行单位换算即可解答.\n【详解】 50×10 = 500 (米)。500÷2 - 200 = 250 - 200 = 50 (米)。200×50 = 10000 (平方米)。1公顷 = 10000 平方米。答：它的占地面积是1公顷。",
+        "knowledge_points":  [
+                                 "长方形的周长",
+                                 "长方形的面积",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26四年级上·河南南阳·期中"
+    },
+    {
+        "id":  "200",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "某新能源汽车制造厂计划新建一个长方形的电池生产厂区，用于扩大生产。已知厂区的长为 800 米，宽为 50 米。那么厂区的占地面积是多少公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "4公顷",
+        "solution":  "【分析】根据长方形的面积公式:S=ab,把数据代入公式求出它的面积是多少平方米，然后换算成用公顷作单位即可.\n【详解】 800×50 = 40000 (平方米),40000 平方米 = 4 公顷。答：厂区的占地面积是 4 公顷。",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·河南郑州·期中"
+    },
+    {
+        "id":  "201",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列选项中，面积最接近 1 公顷的是( )。",
+        "options":  [
+                        "A. 2间卧室的地面",
+                        "B. 200张学生课桌的桌面",
+                        "C. 200间普通教室的地面"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】根据生活经验，对面积单位和数据大小的认识可知，一间卧室的面积约为10平方米，一张课桌面的面积约为50平方分米;一间教室的面积约为50平方米，根据选项逐一排查.\n【详解】根据分析:A。2 间卧室的面积大约是 20 平方米，不符合题意;B。200 张学生课桌的桌面约是 100 平方米，不符合题意;C。200 间普通教室的地面约是 1 公顷，符合题意;故答案为:C",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·山东菏泽·期中"
+    },
+    {
+        "id":  "202",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "关于面积大小的比较，下面正确的是( )。",
+        "options":  [
+                        "A. 400 公顷\u003c40 平方千米",
+                        "B. 2 平方千米 30 公顷 = 2030公顷",
+                        "C. 2 公顷\u003c2900 平方米",
+                        "D. 20000 平方米 = 2 平方千米"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】1公顷 = 10000 平方米,1平方千米 = 100 公顷,1平方米 = 100 平方分米,1000000 平方米 = 1 平方千米，高级单位化成低级单位就乘它们之间的进率，低级单位化成高级单位就除以它们之间的进率，依此换算并比较即可解答.\n【详解】A。40平方千米 = 4000 公顷，即:400公顷\u003c40平方千米，是正确的;B。2平方千米30公顷 = 230 公顷，所以,2平方千米30公顷 != 2030 公顷，因此此项错误;C。2 公顷 = 20000 平方米，所以,2 公顷\u003e2900 平方米，因此此项错误;D。20000 平方米 = 2 公顷，所以,20000平方米 != 2 平方千米，因此此项错误。故答案为:A",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26四年级上·福建三明·期中"
+    },
+    {
+        "id":  "203",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面说法中正确的是( )。",
+        "options":  [
+                        "A. 四(1)班教室的占地面积大约是1公顷",
+                        "B. 飞机飞行的高度达 10000 平方米",
+                        "C. 北京湿地总面积达 51400 公顷，合 514 平方千米",
+                        "D. 1 平方千米比 1 公顷大 9999 公顷"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】1公顷 = 10000 平方米,1平方千米 = 100 公顷。结合生活常识和单位换算，逐项判断选项的正确性.\n【详解】根据分析:A。1公顷 = 10000 平方米，一个教室的面积通常为几十平方米，远小于1公顷，此选项错误。B。高度应使用长度单位(如米),平方米是面积单位，用于高度不合适，此选项错误。C。51400 公顷 / 100 = 514 平方千米，换算正确，此选项正确。D。1 平方千米 = 100 公顷，比 1 公顷大 99 公顷，而不是 9999 公顷，此选项错误。故答案为:C",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26四年级上·福建三明·期中"
+    },
+    {
+        "id":  "204",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "有两块长方形的花圃，第一块花圃的面积是8公顷，第二块花圃的长是320米，宽是 25 米。这两块花圃的面积相比,( )。",
+        "options":  [
+                        "A. 两块花圃面积相等",
+                        "B. 第一块花圃面积大",
+                        "C. 第二块花圃面积大",
+                        "D. 无法确定"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】先根据长方形面积公式 = 长×宽，算出面积，再根据 1 公顷 = 10000 平方米，换算成一样的单位进行比较得出答案.\n【详解】320×25 = 8000(平方米),8公顷= 80000平方米,8000平方米\u003c80000平方米，所以第一块花圃面积大。故答案为:B",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "长方形的面积"
+                             ],
+        "source":  "25-26 四年级上·江西赣州·期中"
+    },
+    {
+        "id":  "205",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "2025 年 10 月 27 日\"仲子文化节.\"在泗水仲庙隆重举办。泗水仲子文化广场占地面积约为 6( )。",
+        "options":  [
+                        "A. 平方分米",
+                        "B. 平方米",
+                        "C. 公顷",
+                        "D. 平方千米"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】1平方分米大约是1个成人手掌的面积,1平方米大约是1张方桌桌面的大小,1公顷比一个标准足球场的面积稍大一点,1平方千米大约是小型湖泊的湖面面积;结合生活经验和题中数据 6 可知，泗水仲子文化广场占地面积用公顷作单位较合适.\n【详解】根据分析可知：仲子文化广场占地面积约为6公顷。故答案为:C",
+        "knowledge_points":  [
+                                 "面积单位的选择",
+                                 "公顷，平方千米的认识"
+                             ],
+        "source":  "25-26 四年级上·山东济宁·期中"
+    },
+    {
+        "id":  "206",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面( )场所的面积最接近 1 公顷。",
+        "options":  [
+                        "A. 一间教室",
+                        "B. 一个篮球场",
+                        "C. 北京天安门广场",
+                        "D. 一个标准游泳池"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】一间教室的面积通常在 50 ~ 85 平方米范围内;一个篮球场的面积约为500平方米;天安门广场的面积为44公顷;一个标准游泳池的面积为1250平方米。再由1公顷=10000 平方米，即可做出判断哪个场所的面积最接近 1 公顷.\n【详解】一个标准游泳池的面积最接近 1 公顷。故答案为:D",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 四年级上·山东菏泽·期中"
+    },
+    {
+        "id":  "207",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "房山区素有\"人之源\"\"城之源\"\"都之源\"的美誉，总面积约 2019( )。",
+        "options":  [
+                        "A. 平方分米",
+                        "B. 平方米",
+                        "C. 公顷",
+                        "D. 平方千米"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】边长是 1 千米的正方形的面积是 1 平方千米。计量地区面积一般用平方千米作单位。所以计量房山区占地面积用\"平方千米\"单位比较恰当.\n【详解】房山区总面积约 2019 平方千米。故答案为:D",
+        "knowledge_points":  [
+                                 "面积单位的选择",
+                                 "公顷，平方千米的认识"
+                             ],
+        "source":  "24-25 五年级上·北京房山·期末"
+    },
+    {
+        "id":  "208",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "南宁青秀山占地 78()。",
+        "options":  [
+                        "A. 平方米",
+                        "B. 公顷",
+                        "C. 平方千米"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据生活经验以及对面积单位和数据大小的认识，鸟巢的占地面积约20公顷，所以南宁青秀山用\"公顷\"作单位比较合适。据此解答.\n【详解】根据分析可知，南宁青秀山占地78公顷。故答案为:B",
+        "knowledge_points":  [
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 四年级上·广西玉林·期中"
+    },
+    {
+        "id":  "209",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "小亮家的房子约是100 平方米,( )个这样的房子约是 1 公顷。",
+        "options":  [
+                        "A. 10",
+                        "B. 100",
+                        "C. 1000",
+                        "D. 10000"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】1公顷 = 10000 平方米，先把1公顷换算为10000平方米，再看10000平方米里面有几个100平方米，就需要几个这样的房子.\n【详解】1 公顷 = 10000 平方米,10000平方米里面有100个100平方米，则约需要100个这样的房子。故答案为:B",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "公顷，平方千米的认识"
+                             ],
+        "source":  "25-26 四年级上·河南开封·期中"
+    },
+    {
+        "id":  "210",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "杭州西湖湖面面积约为 6 平方千米。扬州瘦西湖的湖面面积约为 360000 平方米，嘉兴南湖的湖面面积约为58公顷。( )湖面面积最小。",
+        "options":  [
+                        "A. 杭州西湖",
+                        "B. 扬州瘦西湖",
+                        "C. 嘉兴南湖"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】先把单位进行换算，即 1 平方千米 = 1000000 平方米,1 公顷 = 10000 平方米，然后进行比较即可.\n【详解】6平方千米=6000000平方米,58公顷 = 580000 平方米,360000 \u003c 580000 \u003c 6000000，所以扬州瘦西湖湖面面积最小。故答案为:B",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·甘肃平凉·期中"
+    },
+    {
+        "id":  "211",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一个正方形的居民活动中心，周长是 1600 米，它的面积是( )。",
+        "options":  [
+                        "A. 1600 公顷",
+                        "B. 16000 平方米",
+                        "C. 16 公顷"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】已知周长1600米，正方形边长 = 1600÷4;面积 = 边长×边长;再根据1公顷 = 10000 平方米换算.\n【详解】1600÷4 = 400(米),400×400 = 160000(平方米),160000平方米 = 16公顷。故答案为:C",
+        "knowledge_points":  [
+                                 "正方形的周长",
+                                 "正方形的面积",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·河北唐山·期中"
+    },
+    {
+        "id":  "212",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "济南被称为\"泉城\"。趵突泉是济南著名的景点，趵突泉的面积大约是10( )。",
+        "options":  [
+                        "A. 公顷",
+                        "B. 平方米",
+                        "C. 平方千米",
+                        "D. 千米"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】计量公园，景点等占地面积常用公顷;平方米太小，平方千米太大;千米是长度单位.\n【详解】根据生活经验可知，趵突泉面积比10平方米大得多但比10平方千米小得多，千米又是长度单位，所以选择公顷作单位比较合适。故答案为:A",
+        "knowledge_points":  [
+                                 "面积单位的选择",
+                                 "公顷，平方千米的认识"
+                             ],
+        "source":  "25-26 四年级上·山东济南·期中"
+    },
+    {
+        "id":  "213",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一个标准足球场的占地面积约是 ( )。",
+        "options":  [
+                        "A. 平方千米",
+                        "B. 平方米",
+                        "C. 公顷",
+                        "D. 平方分米"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】根据生活经验可知，计量一个足球场的占地面积用\"公顷\"作单位较合适.\n【详解】一个标准足球场的占地面积约是 1 公顷。故答案为:C",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识"
+                             ],
+        "source":  "25-26 四年级上·广东潮州·期中"
+    },
+    {
+        "id":  "214",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "体育馆看台的总面积大约是 1 公顷，被划分成 5 块大小相同的区域，每块区域的面积大约是( )平方米。",
+        "options":  [
+                        "A. 20",
+                        "B. 50",
+                        "C. 200",
+                        "D. 2000"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】1公顷 = 10000 平方米，先换算再平均分.\n【详解】10000÷5 = 2000(平方米)。故答案为:D",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·河北邢台·期中"
+    },
+    {
+        "id":  "215",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "计算慈利县的城市面积时，一般要用( )作单位。",
+        "options":  [
+                        "A. 平方千米",
+                        "B. 公顷",
+                        "C. 平方米"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】国家，城市，省份等面积一般用平方千米作单位.\n【详解】计算慈利县的城市面积时，一般用平方千米作单位。故答案为:A",
+        "knowledge_points":  [
+                                 "面积单位的选择",
+                                 "公顷，平方千米的认识"
+                             ],
+        "source":  "25-26 四年级上·湖南张家界·期中"
+    },
+    {
+        "id":  "216",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "洞庭湖水域面积约为 2700( )。",
+        "options":  [
+                        "A. 平方千米",
+                        "B. 公顷",
+                        "C. 平方米"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】洞庭湖是大型湖泊，面积非常广阔，宜用平方千米作单位.\n【详解】洞庭湖水域面积约为 2700 平方千米。故答案为:A",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识"
+                             ],
+        "source":  "25-26 四年级上·湖南岳阳·期中"
+    },
+    {
+        "id":  "217",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列说法正确的是( )。",
+        "options":  [
+                        "A. 大于 90 circ 的角是钝角。",
+                        "B. 在 10 的后面添上 8 个 0，这个数就变成了 1 亿。",
+                        "C. 溪溪家房子的面积是 105 公顷。",
+                        "D. 最小的自然是 0，没有最大的自然数。"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】大于 90 circ 小于 180 circ 的角是钝角;在10后面添8个0得到1000000000=10亿;房屋面积应为平方米量级;自然数包含0且无穷无尽，没有最大自然数.\n【详解】A错误;B错误;C错误;D正确。故答案为:D",
+        "knowledge_points":  [
+                                 "自然数的认识",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "公顷，平方千米的认识",
+                                 "亿以上数的改写"
+                             ],
+        "source":  "25-26 四年级上·福建泉州·期中"
+    },
+    {
+        "id":  "218",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "某社区需要一块面积约为 1 公顷的地来作消防演练的场地，你觉得( )比较合适。",
+        "options":  [
+                        "A. 两个面积为 480 平方米的篮球场。",
+                        "B. 边长 200 米的正方形停车场。",
+                        "C. 面积 8000 平方米的市民广场。",
+                        "D. 长 200 米，宽 50 米的中心广场。"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】1公顷 = 10000 平方米，分别计算各选项面积，看哪个接近10000平方米.\n【详解】A:960平方米;B:40000平方米;C:8000平方米;D:200×50 = 10000平方米，最合适。故答案为:D",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "正方形的面积",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26四年级上·广东汕尾·期中"
+    },
+    {
+        "id":  "219",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "填上合适的面积单位和数。我们的校园面积约为 3( ).\n\n6 平方千米=( )公顷\n\n北京市的面积大约为 16410( ).\n\n250000 平方米=( )公顷",
+        "options":  [
+
+                    ],
+        "answer":  "公顷;600;平方千米;25",
+        "solution":  "【分析】校园占地常用公顷;1平方千米=100公顷;北京市面积用平方千米;1公顷=10000平方米.\n【详解】校园约3公顷;6平方千米=600公顷;北京市约16410平方千米;250000平方米=25公顷。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位的选择",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·河南郑州·期中"
+    },
+    {
+        "id":  "220",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "在括号内填上适当的面积单位.\n\n北京故宫博物院的占地面积约 72( )\n\n一间教室的面积约 63( )\n\n我国陆地领土面积约 960 万( )\n\n一块橡皮的面积约 10( )",
+        "options":  [
+
+                    ],
+        "answer":  "公顷;平方米;平方千米;平方厘米",
+        "solution":  "【分析】故宫占地较大用公顷;教室用平方米;国土用平方千米;橡皮用平方厘米.\n【详解】故宫约72公顷;教室约63平方米;我国约960万平方千米;橡皮约10平方厘米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26四年级上·山东菏泽·期中"
+    },
+    {
+        "id":  "221",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "15000000 平方米=()公顷=( )平方千米。",
+        "options":  [
+
+                    ],
+        "answer":  "1500;15",
+        "solution":  "【分析】1公顷=10000平方米,1平方千米=100公顷.\n【详解】15000000÷10000=1500(公顷);1500÷100=15(平方千米)。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26四年级上·山东菏泽·期中"
+    },
+    {
+        "id":  "222",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上适当的单位.\n\n(1)河南省科技馆是华中地区最大的科技馆，整个场馆占地面积约 10( )。 \n(2)河北省是全国唯一兼有草原，海滨，沙漠，湖泊，高原，平原，山地，林地等地貌的省份，面积约 188800( )。",
+        "options":  [
+
+                    ],
+        "answer":  "公顷;平方千米",
+        "solution":  "【分析】科技馆占地较大用公顷;河北省面积很大用平方千米.\n【详解】(1)约10公顷;(2)约188800平方千米。",
+        "knowledge_points":  [
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26四年级上·重庆沙坪坝·期中"
+    },
+    {
+        "id":  "223",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在下面的( )里填上\"\u003e\"\"\u003c\"或\"=\".\n\n70700( )70070\n\n369870( )37 万\n\n8 亿( )80000万\n\n2 平方千米( )196公顷",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;\u003c;=;\u003e",
+        "solution":  "【分析】数的大小比较与单位改写;面积单位换算:1平方千米=100公顷.\n【详解】70700\u003e70070;369870\u003c370000(37万);8亿=80000万;2平方千米=200公顷\u003e196公顷。",
+        "knowledge_points":  [
+                                 "亿以上数的改写",
+                                 "亿以内数的改写",
+                                 "公顷，平方千米的进率与换算",
+                                 "大数的比较"
+                             ],
+        "source":  "25-26 四年级上·山东菏泽·期中"
+    },
+    {
+        "id":  "224",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "5 平方千米=( )平方米=( )公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "5000000;500",
+        "solution":  "【分析】1平方千米=1000000平方米=100公顷.\n【详解】5×1000000=5000000(平方米);5×100=500(公顷)。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·山东菏泽·期中"
+    },
+    {
+        "id":  "225",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在 96 后面添上()个 0，这个数读作九百六十万，我国的领土面积是九百六十万( )。(填面积单位)",
+        "options":  [
+
+                    ],
+        "answer":  "5;平方千米",
+        "solution":  "【分析】九百六十万是7位数,96后添5个0得9600000;我国国土面积单位用平方千米.\n【详解】添5个0;九百六十万平方千米。",
+        "knowledge_points":  [
+                                 "亿以内数的读，写法",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 四年级上·江西宜春·期中"
+    },
+    {
+        "id":  "226",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果每平方米能种3棵花苗，那么 1 公顷能种( )棵花苗,1 平方千米能种( )棵花苗。",
+        "options":  [
+
+                    ],
+        "answer":  "30000;3000000",
+        "solution":  "【分析】1公顷=10000平方米,1平方千米=1000000平方米.\n【详解】3×10000=30000(棵);3×1000000=3000000(棵)。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26四年级上·山东菏泽·期中"
+    },
+    {
+        "id":  "227",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "6 平方千米=( )公顷 80000 平方米=( )公顷\n\n300 公顷=( )平方千米=( )平方米",
+        "options":  [
+
+                    ],
+        "answer":  "600;8;3;3000000",
+        "solution":  "【分析】1平方千米=100公顷,1公顷=10000平方米.\n【详解】6×100=600(公顷);80000÷10000=8(公顷);300÷100=3(平方千米);300×10000=3000000(平方米)。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 五年级上·湖南邵阳·期中"
+    },
+    {
+        "id":  "228",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "25 平方千米=()公顷 6 公顷=( )平方米\n\n55 平方千米=( )平方米 27000 公顷=( )平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "2500;60000;55000000;270",
+        "solution":  "【分析】1平方千米=100公顷,1公顷=10000平方米.\n【详解】25平方千米=2500公顷;6公顷=60000平方米;55平方千米=55000000平方米;27000公顷=270平方千米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·河南三门峡·期中"
+    },
+    {
+        "id":  "229",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个长方形广场，长是 100 米，宽是 50 米,( )个这样的广场的面积是 1 公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "2",
+        "solution":  "【分析】长方形面积=长×宽;1公顷=10000平方米.\n【详解】100×50=5000(平方米);10000÷5000=2。",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·山东临沂·期中"
+    },
+    {
+        "id":  "230",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "填上合适的单位.\n\n山东省国土面积约是 15.8 万( ).\n\n东东家住房的面积约是 135( ).\n\n鸟巢占地面积约是 20( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方千米;平方米;公顷",
+        "solution":  "【分析】省域面积用平方千米;住房面积用平方米;鸟巢占地用公顷.\n【详解】山东省约15.8万平方千米;住房约135平方米;鸟巢约20公顷。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 四年级上·山东临沂·期中"
+    },
+    {
+        "id":  "231",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "800 公顷 = ( )平方米 230000 公顷=( )平方千米\n\n8000000 平方米=( )公顷=( )平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "8000000;2300;800;8",
+        "solution":  "【分析】1公顷=10000平方米,1平方千米=100公顷.\n【详解】800公顷=8000000平方米;230000公顷=2300平方千米;8000000平方米=800公顷=8平方千米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·山东临沂·期中"
+    },
+    {
+        "id":  "232",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "4 公顷=( )平方米 1500 公顷=( )平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "40000;15",
+        "solution":  "【分析】1公顷=10000平方米;100公顷=1平方千米.\n【详解】4公顷=40000平方米;1500公顷=15平方千米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·湖南长沙·期中"
+    },
+    {
+        "id":  "233",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在( )里填上适当的面积单位.\n\n(1)一个篮球场的面积是 420( )。 \n(2)动物园的占地面积约是 3( )。 \n(3)东北三省总面积约是 80 万( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方米;公顷;平方千米",
+        "solution":  "【分析】根据实际大小选择单位：篮球场用平方米，动物园用公顷，省域总面积用平方千米.\n【详解】(1)用平方米合适;(2)用公顷合适;(3)用平方千米合适。",
+        "knowledge_points":  [
+                                 "平方厘米，平方分米，平方米的认识",
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 四年级上·山东菏泽·期中"
+    },
+    {
+        "id":  "234",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "在括号里填上适当的面积单位.\n\n(1)一个篮球场的面积是 420( ).\n(2)国家游泳中心\"水立方\"的占地面积约 6( ).\n(3)香港特别行政区的陆地面积约 1100( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方米;公顷;平方千米",
+        "solution":  "【分析】根据生活经验，面积单位和数据大小的认识进行分析解答。(1)1 平方米是边长为 1 米的正方形面积大小，所以一个篮球场的面积用\"平方米\"作单位合适;(2)1公顷大约是边长为100米的正方形面积的大小，所以计量国家游泳中心\"水立方\"的占地面积用\"公顷\"作单位合适;(3)边长是1千米的正方形的面积是1平方千米，计量比较大的土地的面积常用平方千米为单位，因此计量香港特别行政区陆地的占地面积以\"平方千米\"作单位合适.\n【详解】(1)一个篮球场的面积是 420 平方米;(2)国家游泳中心\"水立方\"的占地面积约 6 公顷;(3)香港特别行政区的陆地面积约 1100 平方千米。",
+        "knowledge_points":  [
+                                 "公顷",
+                                 "平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26四年级上·山东菏泽·期中"
+    },
+    {
+        "id":  "235",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "900000 平方米=( )公顷,40 平方千米=( )公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "90;4000",
+        "solution":  "【分析】1 公顷 = 10000 平方米,1 平方千米 = 100 公顷，据此解答.\n【详解】900000 平方米里面包含 90 个 10000 平方米，所以 900000 平方米 = 90 公顷;40 平方千米里面包含 40 个 100 公顷，所以 40 平方千米 = 4000 公顷。",
+        "knowledge_points":  [
+                                 "公顷",
+                                 "平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·山东菏泽·期中"
+    },
+    {
+        "id":  "236",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "7 公顷=( )平方米\n\n162 公顷+138 公顷=( )平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "70000;3",
+        "solution":  "【分析】1 公顷 = 10000 平方米,1 平方千米 = 100 公顷，据此解答.\n【详解】7公顷里面包含7个1公顷，所以 7 公顷=70000 平方米;162 公顷+138 公顷 = 300 公顷,100 公顷 = 1 平方千米，所以 300 公顷 = 3 平方千米。",
+        "knowledge_points":  [
+                                 "公顷",
+                                 "平方千米的认识",
+                                 "公顷",
+                                 "平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·山东菏泽·期中"
+    },
+    {
+        "id":  "237",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "( )公顷=500000平方米\n\n8000 公顷=( )平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "50;80",
+        "solution":  "【分析】1公顷 = 10000 平方米,1 平方千米 = 100 公顷。据此换算.\n【详解】500000 平方米有 50 个 10000 平方米，所以是 50 公顷;8000 公顷有 80 个 100 公顷，所以是 80 平方千米。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "公顷",
+                                 "平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·河南南阳·期中"
+    },
+    {
+        "id":  "238",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "广西龙脊梯田规模宏大，其中一块梯田的面积约 2000 平方米,( )块这样的梯田面积合 1 公顷;1 公顷=( )平方米。",
+        "options":  [
+
+                    ],
+        "answer":  "5;10000",
+        "solution":  "【分析】1 公顷 = 10000 平方米，求多少块 2000 平方米的梯田合 1 公顷，就是求\"10000 平方米里有几个 2000 平方米\".\n【详解】10000÷2000 = 5(块)。所以 5 块这样的梯田面积合 1 公顷;1 公顷=10000 平方米。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26四年级上·广西玉林·期中"
+    },
+    {
+        "id":  "239",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "填空.\n\n(1)颐和园的占地面积是 3000000( ),合( )公顷.\n(2)澳门的面积约为 33.3( ).\n(3)西藏的面积是 1228000( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方米;300;平方千米;平方千米",
+        "solution":  "【分析】(1)结合颐和园的实际面积，适合用平方米作单位，并用进率1公顷 = 10000 平方米换算成公顷。(2)(3)计量澳门，西藏等地区面积用平方千米作单位比较合适.\n【详解】(1)3000000 平方米 / 10000 = 300，合 300 公顷。(2)澳门的面积约为 33.3 平方千米。(3)西藏的面积是 1228000 平方千米。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "公顷",
+                                 "平方千米的认识",
+                                 "平方厘米，平方分米，平方米的认识"
+                             ],
+        "source":  "25-26 四年级上·山东菏泽·期中"
+    },
+    {
+        "id":  "240",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "7 公顷=( )平方米\n\n600000 平方米=( )公顷\n\n8 平方千米=( )公顷\n\n19000000 平方米=( )平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "70000;60;800;19",
+        "solution":  "【分析】掌握1公顷=10000平方米,1平方千米=100公顷,1平方千米=1000000平方米，根据进率进行乘除换算.\n【详解】7 公顷 = 7×10000 = 70000 平方米;600000 平方米 = 600000÷10000 = 60 公顷;8 平方千米 = 8×100 = 800 公顷;19000000 平方米 = 19000000÷1000000 = 19 平方千米。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "公顷",
+                                 "平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·山东菏泽·期中"
+    },
+    {
+        "id":  "241",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上合适的数.\n\n(1)250000 平方米=( )公顷\n(2)1000 公顷=( )平方千米\n(3)3 平方千米=( )公顷=( )平方米\n(4)80000000 平方米=( )平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "25;10;300;3000000;80",
+        "solution":  "【分析】根据1公顷=10000平方米,1平方千米=1000000平方米=100公顷进行单位换算.\n【详解】(1)250000÷10000=25(公顷);(2)1000÷100=10(平方千米);(3)3平方千米=3×100=300(公顷),3×1000000=3000000(平方米);(4)80000000÷1000000=80(平方千米)。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "公顷",
+                                 "平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·新疆乌鲁木齐·期中"
+    },
+    {
+        "id":  "242",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上合适的面积单位.\n\n(1)一个篮球场的面积大约是 500( ).\n(2)一个体育馆的占地面积约为 10( ).\n(3)我国领土面积约为 960 万( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方米;公顷;平方千米",
+        "solution":  "【分析】国家，城市，省份等用平方千米;大型广场，公园，田地等用公顷;房屋，球场，操场等用平方米，根据实际情况选择合适单位.\n【详解】(1)500平方米;(2)10公顷;(3)960万平方千米。",
+        "knowledge_points":  [
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26四年级上·湖南长沙·期中"
+    },
+    {
+        "id":  "243",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "5 公顷=( )平方米\n\n250000 平方米=( )公顷\n\n14000000 平方米=( )平方千米=( )公顷",
+        "options":  [
+
+                    ],
+        "answer":  "50000;25;14;1400",
+        "solution":  "【分析】1平方千米=100公顷=1000000平方米,1公顷=10000平方米，按进率进行换算.\n【详解】5公顷=50000平方米;250000平方米=25公顷;14000000平方米=14平方千米=1400公顷。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "公顷",
+                                 "平方千米的进率与换算"
+                             ],
+        "source":  "25-26 六年级上·山西晋中·期中"
+    },
+    {
+        "id":  "244",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上合适的单位.\n\n一间教室的面积约 63( ).\n\n香港特别行政区的陆地面积约 1106( ).\n\n动物园的占地面积约是 3( ).\n\n一个篮球场的面积是 420( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方米;平方千米;公顷;平方米",
+        "solution":  "【分析】面积常用单位有公顷，平方千米，平方米等;教室，篮球场用平方米;香港陆地面积用平方千米;动物园占地用公顷.\n【详解】教室约63平方米;香港约1106平方千米;动物园约3公顷;篮球场约420平方米。",
+        "knowledge_points":  [
+                                 "公顷",
+                                 "平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 四年级上·山东临沂·期中"
+    },
+    {
+        "id":  "245",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "6 公顷=( )平方米\n\n8000000 平方米=( )平方千米\n\n5000000 平方米=( )公顷=( )平方千米\n\n8030000 公顷=( )平方米=( )平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "60000;8;500;5;80300000000;80300",
+        "solution":  "【分析】掌握进率:1公顷=10000平方米,1平方千米=100公顷=1000000平方米，根据进率乘除换算.\n【详解】(1)6公顷=60000平方米;(2)8000000平方米=8平方千米;(3)5000000平方米=500公顷=5平方千米;(4)8030000公顷=80300000000平方米=80300平方千米。",
+        "knowledge_points":  [
+                                 "公顷",
+                                 "平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·山东临沂·期中"
+    },
+    {
+        "id":  "246",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "边长是( )的正方形的面积是 1 公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "100 米",
+        "solution":  "【分析】1公顷=10000平方米，正方形面积=边长×边长，找边长使面积为10000.\n【详解】100×100=10000(平方米),即1公顷，所以边长是100米。",
+        "knowledge_points":  [
+                                 "公顷",
+                                 "平方千米的进率与换算",
+                                 "正方形的面积"
+                             ],
+        "source":  "25-26 四年级上·山东临沂·期中"
+    },
+    {
+        "id":  "247",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上合适的数或面积单位.\n\n(1)一所学校的占地面积约 4( ).\n(2)北京奥林匹克公园总占地面积 11590000( ),合( )公顷.\n(3)43000 公顷=( )平方千米。",
+        "options":  [
+
+                    ],
+        "answer":  "公顷;平方米;1159;430",
+        "solution":  "【分析】(1)学校占地用公顷较合适。(2)11590000应以平方米计量，并用1公顷=10000平方米换算。(3)1平方千米=100公顷，将43000公顷换算成平方千米.\n【详解】(1)约4公顷。(2)11590000平方米=1159公顷。(3)43000公顷=430平方千米。",
+        "knowledge_points":  [
+                                 "面积单位的选择",
+                                 "公顷",
+                                 "平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·湖北武汉·期中"
+    },
+    {
+        "id":  "248",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上适当的单位.\n\n武汉大学建筑面积约 252( )\n\n小丽家的房屋面积约100( )\n\n武汉东湖湖面面积约 33( )\n\n武汉东湖绿道全长101.98( )",
+        "options":  [
+
+                    ],
+        "answer":  "公顷;平方米;平方千米;千米",
+        "solution":  "【分析】建筑面积较大用公顷;家庭房屋面积用平方米;湖面面积用平方千米;道路长度用千米.\n【详解】武汉大学约252公顷;房屋约100平方米;东湖湖面约33平方千米;绿道全长约101.98千米。",
+        "knowledge_points":  [
+                                 "长度单位的选择",
+                                 "公顷",
+                                 "平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "24-25四年级上·湖北武汉·期中"
+    },
+    {
+        "id":  "249",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上合适的单位.\n\n(1)大田县是三明市的一个行政县，土地面积是 2233( ).\n(2)大田县大仙峰自然保护区总面积为 6893 公顷，核心区面积为 1848( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方千米;公顷",
+        "solution":  "【分析】县域土地面积通常用平方千米;保护区内部面积常用公顷.\n【详解】(1)2233平方千米;(2)1848公顷。",
+        "knowledge_points":  [
+                                 "面积单位的选择",
+                                 "公顷",
+                                 "平方千米的认识"
+                             ],
+        "source":  "25-26四年级上·福建三明·期中"
+    },
+    {
+        "id":  "250",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个体育场长 200米，宽 50 米，它的面积是( )平方米，合( )公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "10000;1",
+        "solution":  "【分析】长方形面积=长×宽，先求面积(平方米),再用10000平方米=1公顷换算.\n【详解】200×50 = 10000(平方米),10000平方米=1公顷，所以合1公顷。",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·湖北黄冈·期中"
+    },
+    {
+        "id":  "251",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "边长是 3 千米的正方形土地，它的面积是( )公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "900",
+        "solution":  "【分析】正方形面积=边长×边长，先得平方千米，再用1平方千米=100公顷换算.\n【详解】3×3 = 9(平方千米),9×100 = 900(公顷)。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "正方形的面积"
+                             ],
+        "source":  "24-25 五年级上·河北保定·期末"
+    },
+    {
+        "id":  "252",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在( )里填上适当的面积单位。国庆节假期，妈妈带果果从广州出发去北京旅游。她们来到占地面积约 2094( )的广州白云国际机场，发现机场跑道最长的有 3800 米。她们坐的飞机大约飞行了 3 小时，最终到达了占地面积为 16400( )的北京市，然后参观了占地面积约 720000( )的故宫博物院。",
+        "options":  [
+
+                    ],
+        "answer":  "公顷;平方千米;平方米",
+        "solution":  "【分析】机场占地较大用公顷;北京市面积用平方千米;故宫博物院面积用平方米.\n【详解】广州白云国际机场约2094公顷;北京市约16400平方千米;故宫博物院约720000平方米。",
+        "knowledge_points":  [
+                                 "公顷",
+                                 "平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 四年级上·山东济南·期中"
+    },
+    {
+        "id":  "253",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "40000 平方米=( )公顷\n\n3 平方千米=( )公顷=( )平方米",
+        "options":  [
+
+                    ],
+        "answer":  "4;300;3000000",
+        "solution":  "【分析】1 公顷=10000平方米;1平方千米=100公顷;1公顷=10000平方米.\n【详解】40000平方米=4公顷;3平方千米=300公顷=3000000平方米。",
+        "knowledge_points":  [
+                                 "公顷",
+                                 "平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·山东济南·期中"
+    },
+    {
+        "id":  "254",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上合适的数或单位.\n\n480000 平方米=( )公顷;( )公顷 = 6 平方千米;一个篮球场的面积约是 400( ),( )个这样的篮球场总面积约为 2 公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "48;600;平方米;50",
+        "solution":  "【分析】480000平方米换算公顷：除以10000;6平方千米换算公顷：乘100;计量篮球场面积用平方米;2公顷=20000平方米,20000÷400=50.\n【详解】480000平方米=48公顷;600公顷=6平方千米;篮球场约400平方米;50个这样的篮球场总面积约2公顷。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "面积单位的选择",
+                                 "公顷",
+                                 "平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·云南楚雄·期中"
+    },
+    {
+        "id":  "255",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "填上合适的面积单位.\n\n(1)一间教室的面积约是 40( ).\n(2)\"鸟巢\"占地面积约是 20( ).\n(3)我国陆地领土面积约为 960 万( ).\n(4)教材封面的面积大约是 3( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方米;公顷;平方千米;平方分米",
+        "solution":  "【分析】教室用平方米;鸟巢占地用公顷;国土面积用平方千米;教材封面用平方分米.\n【详解】(1)40平方米;(2)20公顷;(3)960万平方千米;(4)3平方分米。",
+        "knowledge_points":  [
+                                 "平方厘米，平方分米，平方米的认识",
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 四年级上·广东汕头·期中"
+    },
+    {
+        "id":  "256",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上合适的面积单位.\n\n唐山市位于河北省东部，土地总面积为 13472( ).\n\n颐和园世界文化遗产区面积约 297( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方千米;公顷",
+        "solution":  "【分析】唐山市是较大的行政区域，用平方千米;遗产区面积用公顷较合适.\n【详解】唐山市约13472平方千米;颐和园世界文化遗产区约297公顷。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26四年级上·河北唐山·期中"
+    },
+    {
+        "id":  "257",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "在横线上填上合适的单位.\n\n一个教室的面积大约是 63( ).\n\n北京故宫的占地面积是 72( ).\n\n新疆维吾尔自治区是我国陆地面积最大的省级行政区，它的面积约是 166 万( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方米;公顷;平方千米",
+        "solution":  "【分析】教室面积用平方米;故宫占地面积用公顷;省级行政区面积用平方千米.\n【详解】63平方米;72公顷;166万平方千米。",
+        "knowledge_points":  [
+                                 "平方厘米，平方分米，平方米的认识",
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 四年级上·山东济南·期中"
+    },
+    {
+        "id":  "258",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在横线里填上适当的单位.\n\n(1)北京圆明园的占地面积约是 350( ).\n(2)澳门特别行政区的面积约是 33( ).\n(3)一枚北京2022年冬奥会闭幕式纪念章的面积约是3( ).\n(4)明明家客厅的面积约是 30( ).\n(5)一张 A4 纸的面积大约是 6( )。",
+        "options":  [
+
+                    ],
+        "answer":  "公顷;平方千米;平方厘米;平方米;平方分米",
+        "solution":  "【分析】公园占地用公顷;城市/行政区面积用平方千米;纪念章面积小用平方厘米;客厅面积用平方米;A4纸面积用平方分米.\n【详解】(1)350公顷;(2)33平方千米;(3)3平方厘米;(4)30平方米;(5)6平方分米。",
+        "knowledge_points":  [
+                                 "面积单位的选择",
+                                 "公顷，平方千米的认识",
+                                 "平方厘米，平方分米，平方米的认识"
+                             ],
+        "source":  "25-26四年级上·内蒙古乌海·期中"
+    },
+    {
+        "id":  "259",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "暑假期间小诺和家人打卡济南，她在旅行手账里记录：济南市的总面积约10244( ),我们去的趵突泉景区占地面积约 158( ),景区内观澜亭的占地面积约20( ),最后在景区文创店买了一张明信片，明信片的面积约 150( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方千米;公顷;平方米;平方厘米",
+        "solution":  "【分析】城市面积用平方千米;景区占地用公顷;亭子占地用平方米;明信片面积用平方厘米.\n【详解】济南市约10244平方千米;趵突泉景区约158公顷;观澜亭约20平方米;明信片约150平方厘米。",
+        "knowledge_points":  [
+                                 "面积单位的选择",
+                                 "公顷，平方千米的认识",
+                                 "平方厘米，平方分米，平方米的认识"
+                             ],
+        "source":  "25-26四年级上·山东济南·期中"
+    },
+    {
+        "id":  "260",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "20000000 平方米=( )公顷=( )平方千米。",
+        "options":  [
+
+                    ],
+        "answer":  "2000;20",
+        "solution":  "【分析】1平方千米=100 公顷=1000000 平方米;20000000 平方米换算成公顷去掉末尾4个0，换算成平方千米去掉末尾6个0.\n【详解】20000000 平方米 = 2000 公顷 = 20 平方千米。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·山东济南·期中"
+    },
+    {
+        "id":  "261",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "53 米=( )分米( )厘米\n\n7 公顷=( )平方米\n\n1200 公顷=( )平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "530;5300;70000;12",
+        "solution":  "【分析】1米=10分米=100厘米;1公顷=10000平方米;1平方千米=100公顷，据此换算.\n【详解】53米=530分米=5300厘米;7公顷=70000平方米;1200公顷=12平方千米。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "公顷，平方千米的进率与换算",
+                                 "毫米，厘米，分米，米之间的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·河南驻马店·期中"
+    },
+    {
+        "id":  "262",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "括号里填上合适的面积单位.\n\n(1)一间教室的面积约 54( ).\n(2)国家游泳中心\"水立方\"的占地面积约 6( ).\n(3)广西壮族自治区的陆地面积约为 23.76 万( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方米;公顷;平方千米",
+        "solution":  "【分析】常用面积单位有平方厘米，平方分米，平方米，公顷，平方千米;教室用平方米，\"水立方\"用公顷，省级行政区面积用平方千米.\n【详解】(1)54平方米;(2)6公顷;(3)23.76万平方千米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 四年级上·广西河池·期中"
+    },
+    {
+        "id":  "263",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在( )里填上适当的面积单位.\n\n(1)一块橡皮的占地面积约是 6( );(2)河南省总面积约为 17 万( );(3)北京圆明园的面积约是 350( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方厘米;平方千米;公顷",
+        "solution":  "【分析】橡皮面积小用平方厘米;省域面积大用平方千米;圆明园等公园面积用公顷.\n【详解】(1)6平方厘米;(2)17万平方千米;(3)350公顷。",
+        "knowledge_points":  [
+                                 "平方厘米，平方分米，平方米的认识",
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 四年级上·河南三门峡·期中"
+    },
+    {
+        "id":  "264",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在( )里填上合适的面积单位.\n\n(1)音乐广场的占地面积约是 4.5( ).\n(2)学校会议室的面积约是 150( ).\n(3)电视机屏幕的面积大约是 110( ).\n(4)河南省的面积大约是 167 万( )。",
+        "options":  [
+
+                    ],
+        "answer":  "公顷;平方米;平方分米;平方千米",
+        "solution":  "【分析】广场占地较大用公顷;会议室用平方米;电视屏幕用平方分米;省域面积用平方千米.\n【详解】(1)4.5公顷;(2)150平方米;(3)110平方分米;(4)167万平方千米。",
+        "knowledge_points":  [
+                                 "面积单位的选择",
+                                 "公顷，平方千米的认识",
+                                 "平方厘米，平方分米，平方米的认识"
+                             ],
+        "source":  "25-26 五年级上·河南洛阳·期中"
+    },
+    {
+        "id":  "265",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上合适的单位.\n\n一张邮票的面积约 6( ).\n\n公园的占地面积约是 25( ).\n\n教室黑板的面积约 4( ).\n\n香港特别行政区的面积约 1100( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方厘米;公顷;平方米;平方千米",
+        "solution":  "【分析】邮票面积小用平方厘米;公园占地用公顷;黑板面积用平方米;香港面积用平方千米.\n【详解】邮票约6平方厘米;公园约25公顷;黑板约4平方米;香港约1100平方千米。",
+        "knowledge_points":  [
+                                 "平方厘米，平方分米，平方米的认识",
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 四年级上·广东潮州·期中"
+    },
+    {
+        "id":  "266",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "8 平方分米=( )平方厘米\n\n10 米=( )厘米\n\n7000000 平方米=( )公顷=( )平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "800;1000;700;7",
+        "solution":  "【分析】1平方分米=100平方厘米,1米=100厘米,1公顷=10000平方米,1平方千米=100公顷=1000000平方米，按进率换算.\n【详解】8平方分米=800平方厘米;10米=1000厘米;7000000平方米=700公顷=7平方千米。",
+        "knowledge_points":  [
+                                 "毫米，厘米，分米，米之间的进率与换算",
+                                 "平方厘米，平方分米，平方米之间的进率与换算",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·广东潮州·期中"
+    },
+    {
+        "id":  "267",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\"\u003e\"\"\u003c\"或\"=\".\n\n5 公顷( )49000 平方米\n\n20 吨( )200 千克\n\n270 万( )2700000",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;\u003e;=",
+        "solution":  "【分析】单位统一后比较:1公顷=10000平方米,1吨=1000千克;\"万\"表示10000，把270万改写成以\"个\"为单位.\n【详解】5公顷=50000平方米,50000\u003e49000，所以填\"\u003e\";20吨=20000千克,20000\u003e200，所以填\"\u003e\";270万=2700000，所以填\"=\"。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "吨，千克之间的换算与比较",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·广东潮州·期中"
+    },
+    {
+        "id":  "268",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在( )里填上合适的单位.\n\n(1)某城市的面积是 5800( ).\n(2)鸟巢的占地面积约为 20( ).\n(3)颐和园的占地面积约 300( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方千米;公顷;公顷",
+        "solution":  "【分析】城市面积用平方千米;大型场馆，公园占地面积常用公顷.\n【详解】(1)5800平方千米;(2)20公顷;(3)300公顷。",
+        "knowledge_points":  [
+                                 "面积单位的选择",
+                                 "公顷，平方千米的认识"
+                             ],
+        "source":  "25-26四年级上·山东临沂·期中"
+    },
+    {
+        "id":  "269",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在下面的横线上填上合适的面积单位.\n\n郑州是河南省的省会，是著名的历史名城，面积约为7446( ).\n\n龙子湖公园的绿地面积约为103( ).\n\n郑州东站候车大厅占地面积是 12000( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方千米;公顷;平方米",
+        "solution":  "【分析】很大的地域面积(城市)常用平方千米;较大的土地面积(公园绿地)常用公顷;建筑内场地面积常用平方米.\n【详解】郑州面积约7446平方千米;龙子湖公园绿地约103公顷;郑州东站候车大厅约12000平方米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26四年级上·河北邢台·期中"
+    },
+    {
+        "id":  "270",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "微山湖是中国北方最大的淡水湖，湖区面积约 1266 平方千米，合( )公顷;微山湖国家湿地公园被誉为\"亚洲第一湿地\"，其中湿地面积约 650 公顷，合( )平方米。",
+        "options":  [
+
+                    ],
+        "answer":  "126600;6500000",
+        "solution":  "【分析】根据 1 平方千米 = 100 公顷,1 公顷 = 10000 平方米，高级单位换算成低级单位乘进率，解答即可.\n【详解】1 平方千米 = 100 公顷,1266 个 100 公顷是 126600公顷，所以 1266 平方千米=126600 公顷;1 公顷 = 10000 平方米,650 个 10000 平方米是 6500000 平方米，所以650公顷=6500000平方米。微山湖是中国北方最大的淡水湖，湖区面积约1266平方千米，合126600公顷;微山湖国家湿地公园被誉为\"亚洲第一湿地\"，其中湿地面积约650公顷，合6500000平方米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·山东济宁·期中"
+    },
+    {
+        "id":  "271",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上合适的单位名称.\n\n微山县面积约 1780( ) 一所学校占地面积约 4( )\n\n一间教室面积约 80( ) 一张身份证的面积约 50( )",
+        "options":  [
+
+                    ],
+        "answer":  "平方千米;公顷;平方米;平方厘米",
+        "solution":  "【分析】根据题意，选择单位时，要结合生活实际和单位大小,1平方厘米大约是指甲盖表面大小，所以计量一张身份证的面积用\"平方厘米\"作单位比较合适;1平方米大约是一张方桌的大小，所以计量一间教室地面的面积用\"平方米\"作单位比较合适;微山县是一个较大的行政区域(县),面积通常较大，所以微山县面积约1780平方千米;学校包括操场，教学楼等设施，占地面积较大，一般都用公顷作单位，以此答题即可.\n【详解】根据分析可知：微山县面积约1780平方千米一所学校占地面积约4公顷一间教室面积约 80 平方米 一张身份证的面积约 50 平方厘米",
+        "knowledge_points":  [
+                                 "平方厘米，平方分米，平方米的认识",
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26四年级上·山东济宁·期中"
+    },
+    {
+        "id":  "272",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在横线上填上合适的面积单位.\n\n(1)学校操场的面积大约是 9600 .\n(2)一个国际机场的占地面积约为 3 .\n(3)一家小型工厂的面积约为 2 。",
+        "options":  [
+
+                    ],
+        "answer":  "平方米;平方千米;公顷",
+        "solution":  "【分析】联系生活实际，根据对面积单位的认识和数据的大小可知:(1)结合数据9600可知，计量操场的面积用\"平方米\"作单位比较合适;(2)结合数据3可知，一个国际机场的占地面积用\"平方千米\"作单位比较合适;(3)结合数据2可知，计量一家小型工厂的面积用\"公顷\"作单位比较合适.\n【详解】由分析可知:(1)学校操场的面积大约是 9600 平方米;(2)一个国际机场的占地面积约为 3 平方千米;(3)一家小型工厂的面积约为2公顷。",
+        "knowledge_points":  [
+                                 "面积单位的选择",
+                                 "公顷，平方千米的认识"
+                             ],
+        "source":  "25-26 四年级上·河北唐山·期中"
+    },
+    {
+        "id":  "273",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "8 平方千米=( )公顷\n\n600000 平方米=( )公顷\n\n90000000 平方米=( )平方千米=( )公顷",
+        "options":  [
+
+                    ],
+        "answer":  "800;60;90;9000",
+        "solution":  "【分析】1平方千米 = 100 公顷 = 1000000 平方米,1公顷=10000 平方米，据此将单位换算即可.\n【详解】8 平方千米= 800 公顷;600000 平方米 = 60 公顷;90000000 平方米 = 90 平方千米 = 9000 公顷。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·河北唐山·期中"
+    },
+    {
+        "id":  "274",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "6 公顷=( )平方米\n\n12 平方千米=( )公顷",
+        "options":  [
+
+                    ],
+        "answer":  "60000;1200",
+        "solution":  "【分析】1 公顷 = 10000 平方米,1 平方千米 = 100 公顷，据此换算.\n【详解】6 公顷 = 60000 平方米;12 平方千米 = 1200 公顷。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·湖南岳阳·期中"
+    },
+    {
+        "id":  "275",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "5 平方千米=( )公顷\n\n70000 平方米=( )公顷",
+        "options":  [
+
+                    ],
+        "answer":  "500;7",
+        "solution":  "【分析】1平方千米 = 100 公顷;1 公顷=10000 平方米，据此换算.\n【详解】5 平方千米 = 500 公顷;70000 平方米 = 7 公顷。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·湖南长沙·期中"
+    },
+    {
+        "id":  "276",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上\"\u003e\"\"\u003c\"或\"=\".\n\n3403400( )30403400\n\n6 亿 2 千万( )620000000\n\n50000 平方米( )5 公顷\n\n7030030( )704 万\n\n10020000( )1 千万\n\n30 平方千米( )300 公顷",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;=;=;\u003c;\u003e;\u003e",
+        "solution":  "【分析】整数比较大小看位数与数位;把\"万\"\"亿\"统一为以\"个\"为单位后再比较;公顷与平方米，平方千米与公顷先统一单位再比较.\n【详解】3403400\u003c30403400;6 亿 2 千万=620000000;50000平方米=5公顷;7030030\u003c7040000(即704万);10020000\u003e10000000(即1千万);30平方千米=3000公顷\u003e300公顷。",
+        "knowledge_points":  [
+                                 "亿以上数的改写",
+                                 "亿以内数的改写",
+                                 "公顷，平方千米的进率与换算",
+                                 "大数的比较"
+                             ],
+        "source":  "25-26 四年级上·湖北鄂州·期中"
+    },
+    {
+        "id":  "277",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "有\"九寨归来不看水\"之说的九寨沟号称\"水景之王\"，其总面积约为 620( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方千米",
+        "solution":  "【分析】计量大的土地面积常用平方千米，结合数据 620 判断单位用平方千米更合适.\n【详解】九寨沟总面积约为620平方千米。",
+        "knowledge_points":  [
+                                 "面积单位的选择",
+                                 "公顷，平方千米的认识"
+                             ],
+        "source":  "25-26 四年级上·湖北鄂州·期中"
+    },
+    {
+        "id":  "278",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上合适的单位.\n\n某小学占地面积约 3( );泉州市陆地面积约 11015( )。",
+        "options":  [
+
+                    ],
+        "answer":  "公顷;平方千米",
+        "solution":  "【分析】小学占地面积用公顷;城市陆地面积用平方千米.\n【详解】某小学占地面积约3公顷;泉州市陆地面积约11015平方千米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26四年级上·福建泉州·期中"
+    },
+    {
+        "id":  "279",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上合适的面积单位.\n\n江西省总面积约为 16.69 万( ).\n\n天安门广场占地面积 4( ).\n\n上海市的面积约是 6340( ).\n\n足球场的面积约是 7000( ).\n\n北京天坛公园的占地面积约是 273( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方千米;公顷;平方千米;平方米;公顷",
+        "solution":  "【分析】国家，省，市，县面积一般用平方千米;较大的场地如广场，公园常用公顷;足球场面积常用平方米.\n【详解】江西省约16.69万平方千米;天安门广场约4公顷;上海市约6340平方千米;足球场约7000平方米;北京天坛公园约273公顷。",
+        "knowledge_points":  [
+                                 "面积单位的选择",
+                                 "公顷，平方千米的认识"
+                             ],
+        "source":  "25-26四年级上·黑龙江哈尔滨·期中"
+    },
+    {
+        "id":  "280",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上合适的单位.\n\n(1)湖南省的面积约 21.18 万( )\n\n(2)一个果园的占地面积约是 25( )\n\n(3)一间教室的面积约是 40( )\n\n(4)数学书封面的面积约是 3( )",
+        "options":  [
+
+                    ],
+        "answer":  "平方千米;公顷;平方米;平方分米",
+        "solution":  "【分析】省域面积用平方千米;果园占地用公顷;教室面积用平方米;数学书封面面积用平方分米.\n【详解】(1)21.18万平方千米;(2)25公顷;(3)40平方米;(4)3平方分米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "平方厘米，平方分米，平方米的认识"
+                             ],
+        "source":  "25-26 五年级上·湖南邵阳·期中"
+    },
+    {
+        "id":  "281",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "25 公顷=( )平方米\n\n300 厘米=( )米",
+        "options":  [
+
+                    ],
+        "answer":  "250000;3",
+        "solution":  "【分析】1公顷 = 10000 平方米;1米 = 100 厘米，据此换算.\n【详解】25 公顷 = 250000 平方米;300厘米 = 3 米。",
+        "knowledge_points":  [
+                                 "厘米和米之间的进率与换算",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·广东汕尾·期中"
+    },
+    {
+        "id":  "282",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在横线上填上合适的单位.\n\n(1)上海迪士尼乐园，是中国内地首座迪士尼主题乐园，也是一座融合迪士尼经典元素与中国文化特色的主题乐园，面积约是 116 .\n\n(2)圆明园，中国清代大型皇家园林，堪称人类文化的宝藏之一，占地面积约 3500000，合 350 。",
+        "options":  [
+
+                    ],
+        "answer":  "公顷;平方米;公顷",
+        "solution":  "【分析】(1)边长为 100 米的正方形的面积是 1 公顷，所以上海迪士尼乐园面积约是 116 公顷;(2)圆明园占地面积约3500000平方米,1公顷=10000平方米，据此将3500000平方米换算成350公顷.\n【详解】(1)上海迪士尼乐园面积约116公顷。(2)圆明园占地面积约3500000平方米，合350公顷。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 四年级上·河北石家庄·期中"
+    },
+    {
+        "id":  "283",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "帮阳阳改数学日记。春节前夕我们搬新家了，我的卧室可大，可漂亮了，面积有 12 平方千米( ),后院还有一个小花园，面积是 100平方分米( )。春节期间，爸爸妈妈放假了，带我去了北京大学本部的校园参观，北京大学的足球场真大呀，估计约有 7000 公顷( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方米;平方米;平方米",
+        "solution":  "【分析】常用的面积单位有平方厘米，平方分米，平方米，一个指甲盖约1平方厘米，一个手掌约1平方分米，一扇窗户约1平方米。结合生活实际判断单位大小是否合理.\n【详解】卧室面积应改为12平方米;小花园100平方分米=1平方米过小，应改为100平方米;足球场面积应改为7000平方米。故三处都应改为平方米。",
+        "knowledge_points":  [
+                                 "面积单位的选择",
+                                 "公顷，平方千米的进率与换算",
+                                 "平方厘米，平方分米，平方米之间的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·河北石家庄·期中"
+    },
+    {
+        "id":  "284",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "中国地大物博，人口众多，是名副其实的大国。请根据相关数据填空.\n\n(1)第七次人口普查结果：全国人口为十四亿四千三百四十九万七千三百七十八人。横线上的数写作:( ),省略\"亿\"位后的尾数，约是( )亿.\n\n(2)中国陆地面积约为9600000平方千米，横线上的数读作:( )。改写成用\"万\"位作单位的数是( )万.\n\n(3)我国海域面积约为 473 万( ),北京故宫面积约为 106( ),一个篮球场的面积约为 420( ).\n\n(4)布达拉宫占地面积约为 36 公顷，合( )平方米;青海湖是我国最大的咸水湖，面积约为 4650 平方千米，合( )公顷.\n\n(5)第七次人口普查中，江苏有 84748016 人。横线上的数是( )位数，最高位是( )位;数中的7表示( ),从左数第1 个 4 表示( );从左数起，第一个8是第二个8的( )倍。",
+        "options":  [
+
+                    ],
+        "answer":  "1443497378;14;九百六十万;960;平方千米;公顷;平方米;360000;465000;8;千万;7个十万;4个百万;10000",
+        "solution":  "【分析】(1)按数位顺序写数;省略\"亿\"位后的尾数按千万位四舍五入取近似数。(2)按整数读法读数;改写成用\"万\"为单位去掉末尾4个0并加\"万\"。(3)根据实际大小选择面积单位。(4)1公顷 = 10000 平方米;1平方千米 = 100 公顷。(5)整数有几个数字就是几位数;最高位看最左边数字所在数位;理解数位表示的计数单位并比较倍数.\n【详解】(1)十四亿四千三百四十九万七千三百七十八写作:1443497378;省略\"亿\"位后的尾数约是14亿。(2)9600000读作：九百六十万;改写成用\"万\"为单位是960万。(3)我国海域面积约为473万平方千米;北京故宫面积约为106公顷;一个篮球场的面积约为420平方米。(4)36公顷=360000平方米;4650平方千米=465000公顷。(5)84748016是8位数，最高位是千万位;7表示7个十万;第1个4表示4个百万;第一个8(8千万)是第二个8(8千)的10000倍。",
+        "knowledge_points":  [
+                                 "亿以上数的近似数",
+                                 "公顷，平方千米的进率与换算",
+                                 "亿以上数的读，写法",
+                                 "亿以内数的读，写法"
+                             ],
+        "source":  "25-26 四年级上·江西赣州·期中"
+    },
+    {
+        "id":  "285",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上\"\u003c\"\"\u003e\"或\"=\".\n\n653401( )635401\n\n99999( )10 万\n\n8 亿( )80000000\n\n500000( )50 万\n\n5 公顷( )600 平方千米\n\n2000000 平方米( )2 平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;\u003c;\u003e;=;\u003c;=",
+        "solution":  "【分析】整数比较先比位数再比数位;把\"万\"\"亿\"改写成以\"个\"为单位;公顷与平方千米，平方米与平方千米先统一单位再比较.\n【详解】653401\u003e635401;10万=100000，所以99999\u003c10万;8亿=800000000\u003e80000000;50万=500000，所以500000=50万;600平方千米=60000公顷，所以5公顷\u003c600平方千米;2000000平方米=2平方千米，所以2000000平方米=2平方千米。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "2025四年级上·广东广州·专题练习"
+    },
+    {
+        "id":  "286",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "鸟巢的占地面积约为 20 公顷,5 个\"鸟巢\"的占地面积约为 1 平方千米。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】5个鸟巢占地面积约为 20×5 = 100 公顷，再根据 100 公顷=1平方千米换算即可.\n【详解】20×5 = 100 公顷 = 1 平方千米，故答案为:√",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·江西宜春·期中"
+    },
+    {
+        "id":  "287",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "一个标准篮球场的占地面积约为 1 平方千米。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】1平方千米=1000000平方米。标准篮球场面积约为420平方米，远小于1000000平方米，因此题干错误.\n【详解】标准篮球场面积约 28×15 = 420(平方米),而1平方千米 = 1000000 平方米,420远小于1000000，故答案为: ×",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 四年级上·山东菏泽·期中"
+    },
+    {
+        "id":  "288",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "判断题",
+        "difficulty":  0.94,
+        "question":  "四(一)班教室里课桌面的面积有 1 公顷。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】1公顷=10000平方米。教室里课桌面总面积通常只有几十平方米，远小于1公顷，因此题干错误.\n【详解】1公顷 = 10000 平方米。按常见课桌面面积和课桌数量估算，课桌面总面积远小于10000平方米，故答案为: ×",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识"
+                             ],
+        "source":  "25-26四年级上·山东菏泽·期中"
+    },
+    {
+        "id":  "289",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "花园小学阅览室的面积约是100平方米,10个这样的阅览室的面积约是1公顷。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】1公顷=10000平方米。10个阅览室面积约10×100=1000平方米，不是10000平方米，因此题干错误.\n【详解】10个阅览室面积约1000平方米，而1公顷=10000平方米，应为100个这样的阅览室约1公顷，故答案为: ×",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·河北唐山·期中"
+    },
+    {
+        "id":  "290",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "状状家的客厅面积很大，约 28 公顷。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】客厅面积应使用平方米。公顷用于计量较大土地面积,28公顷过大，不符合生活实际.\n【详解】状状家的客厅面积应约为28平方米，原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "面积单位的选择",
+                                 "公顷，平方千米的认识"
+                             ],
+        "source":  "25-26 四年级上·湖北鄂州·期中"
+    },
+    {
+        "id":  "291",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "天安门广场，位于北京市中心，地处北京市东城区东长安街，北起天安门，南至正阳门。东起中国国家博物馆，西至人民大会堂，南北长880米。东西宽500米，天安门广场占地面积为多少公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "44公顷",
+        "solution":  "【分析】天安门广场近似长方形，面积=长×宽，先求平方米，再用1公顷=10000平方米换算成公顷.\n【详解】880×500 = 440000(平方米),440000÷10000 = 44(公顷)。答：占地面积为44公顷。",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26四年级上·山东菏泽·期中"
+    },
+    {
+        "id":  "292",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "中国文字博物馆的总占地面积约为470亩，约合31公顷。文字博物馆的总占地面积相当于多少个边长100米的正方形。",
+        "options":  [
+
+                    ],
+        "answer":  "31 个",
+        "solution":  "【分析】边长100米的正方形面积=100×100=10000平方米=1公顷;用总公顷数除以1即可.\n【详解】100×100 = 10000(平方米),10000平方米=1公顷,31÷1 = 31(个)。答：相当于31个边长100米的正方形。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "正方形的面积"
+                             ],
+        "source":  "25-26 四年级上·河南三门峡·期中"
+    },
+    {
+        "id":  "293",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "劳动社团到蔬菜基地学习种菜，基地的一块正方形菜地里种植了菠菜，菜地的边长为200米，若每公顷收菠菜4000千克，则这块菜地可以收菠菜多少千克。",
+        "options":  [
+
+                    ],
+        "answer":  "16000 千克",
+        "solution":  "【分析】正方形面积=边长×边长，先求平方米，再换算成公顷(除以10000),最后用每公顷产量乘公顷数.\n【详解】200×200 = 40000(平方米),40000平方米=4公顷,4000×4 = 16000(千克)。答：可收菠菜16000千克。",
+        "knowledge_points":  [
+                                 "正方形的面积",
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26四年级上·山东临沂·期中"
+    },
+    {
+        "id":  "294",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "现有1块长方形水稻田，长100米，宽40米。如果平均每公顷产稻谷 7100 千克，那么 5 块这样的水稻田一共能产稻谷多少千克。",
+        "options":  [
+
+                    ],
+        "answer":  "14200 千克",
+        "solution":  "【分析】先求1块面积，再乘5得到总面积(平方米),换算成公顷后乘每公顷产量.\n【详解】100×40×5 = 20000(平方米),20000平方米=2公顷,7100×2 = 14200(千克)。答：一共能产稻谷14200千克。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "长方形的面积"
+                             ],
+        "source":  "25-26四年级上·新疆乌鲁木齐·期中"
+    },
+    {
+        "id":  "295",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "橘子园里一共种了20000棵橘子树。如果每 4 平方米种一棵橘子树，那么这个橘子园的面积是多少平方米。合多少公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "80000 平方米;8 公顷",
+        "solution":  "【分析】用橘子树总棵数乘每棵占地面积得总面积(平方米),再用1公顷=10000平方米换算.\n【详解】20000×4 = 80000(平方米),80000平方米=8公顷。答：面积80000平方米，合8公顷。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "整十，整百，整千数与一位数的乘法"
+                             ],
+        "source":  "24-25 四年级上·湖北武汉·期中"
+    },
+    {
+        "id":  "296",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "小华家有一块周长800米的正方形苹果园。一共收了48吨苹果，这块苹果园的面积是多少公顷。平均每公顷收苹果多少吨。",
+        "options":  [
+
+                    ],
+        "answer":  "4 公顷;12 吨",
+        "solution":  "【分析】正方形边长=周长/4;面积=边长×边长(平方米);再换算成公顷;平均每公顷产量=总吨数/公顷数.\n【详解】800÷4 = 200(米),200×200 = 40000(平方米),40000平方米=4公顷,48÷4 = 12(吨)。答：面积4公顷，平均每公顷12吨。",
+        "knowledge_points":  [
+                                 "正方形的周长",
+                                 "正方形的面积",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·云南楚雄·期中"
+    },
+    {
+        "id":  "297",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一块长方形土地的长是 900 米，宽是 500 米，农民用播种机播种，一共用了 3天播完，平均每天播种多少公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "15 公顷",
+        "solution":  "【分析】先求土地总面积(平方米),换算成公顷后再除以3天.\n【详解】900×500 = 450000(平方米),450000平方米=45公顷,45÷3 = 15(公顷)。答：平均每天播种15公顷。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "长方形的面积"
+                             ],
+        "source":  "25-26 四年级上·河北唐山·期中"
+    },
+    {
+        "id":  "298",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "王爷爷一家将一片沙化地开发为一个面积为 2 公顷的苹果园。如果每 5 平方米可种 1 棵苹果树，那么这个苹果园一共可种苹果树多少棵。",
+        "options":  [
+
+                    ],
+        "answer":  "4000 棵",
+        "solution":  "【分析】2公顷先换算成平方米(1公顷=10000平方米),再用总平方米数除以5求棵数.\n【详解】2公顷=20000平方米,20000÷5 = 4000(棵)。答：一共可种苹果树4000棵。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26四年级上·山东济南·期中"
+    },
+    {
+        "id":  "299",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "一块正方形玉米地的边长是 800 米，这块玉米地的面积是多少平方米。合多少公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "640000 平方米;64 公顷",
+        "solution":  "【分析】正方形面积=边长×边长;再用1公顷=10000平方米换算成公顷.\n【详解】800×800 = 640000(平方米),640000平方米里面有64个10000平方米，所以合64公顷。答：面积640000平方米，合64公顷。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "正方形的面积"
+                             ],
+        "source":  "25-26 四年级上·海南省直辖县级单位·期中"
+    },
+    {
+        "id":  "300",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "塔克拉玛干沙漠是我国面积最大的沙漠。为了防风固沙，援疆某志愿队要开垦3公顷的沙地进行植树造林。现已开垦出一块长200米，宽 100 米的长方形沙地，还要继续开垦多少公顷沙地。",
+        "options":  [
+
+                    ],
+        "answer":  "1 公顷",
+        "solution":  "【分析】先求已开垦长方形沙地面积(平方米),换算成公顷，再用3公顷减去已开垦公顷数.\n【详解】200×100 = 20000(平方米),20000平方米=2公顷,3 - 2 = 1(公顷)。答：还要继续开垦1公顷。",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·广东汕尾·期中"
+    },
+    {
+        "id":  "301",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "比5公顷少5000平方米是( )。",
+        "options":  [
+                        "A. 45000 平方米",
+                        "B. 4 公顷",
+                        "C. 55000 平方米"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】先统一单位，根据1公顷 = 10000 平方米，把5公顷换算成50000平方米。用换算后5公顷对应的平方米数减去5000平方米，再据此判断选项.\n【详解】1 公顷 = 10000 平方米5×10000 = 50000(平方米)所以5公顷 = 50000 平方米50000-5000=45000(平方米)故答案为:A",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26五年级上·湖南邵阳·期中"
+    },
+    {
+        "id":  "302",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "幸福村计划挖一个占地16公顷的正方形池塘，如果它的边长缩短100米，那么这个池塘的面积就减少( )公顷。",
+        "options":  [
+                        "A. 1",
+                        "B. 7",
+                        "C. 9"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】把16公顷换算成160000平方米;由正方形面积=边长×边长求原边长400米;缩短100米后边长300米，求新面积并换算成公顷;用原面积减新面积.\n【详解】16公顷=160000平方米，因为 400×400 = 160000(平方米),所以边长是400米。400 - 100 = 300(米),300×300 = 90000(平方米),90000平方米=9公顷,16 - 9 = 7(公顷)。故答案为:B",
+        "knowledge_points":  [
+                                 "正方形的面积",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26五年级上·河北石家庄·期中"
+    },
+    {
+        "id":  "303",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "奥林匹克森林公园占地面积约 6.8( )。",
+        "options":  [
+                        "A. 平方米",
+                        "B. 平方千米",
+                        "C. 公顷",
+                        "D. 千米"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据生活经验与单位大小判断：大型公园占地适合用平方千米.\n【详解】平方米过小;6.8公顷过小;千米是长度单位;6.8平方千米符合实际。故答案为:B",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26五年级上·湖南邵阳·期中"
+    },
+    {
+        "id":  "304",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下列选项中，面积为1 公顷的场所是( )。",
+        "options":  [
+                        "A. 学校室内篮球场的面积。",
+                        "B. 一间教室的占地面积。",
+                        "C. 400米跑道围成的面积。",
+                        "D. 一栋教学楼的占地面积。"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】边长为 100 米的正方形面积是1公顷,400米跑道围起来的部分面积约为1公顷.\n【详解】A约400平方米;B约50平方米;D几百平方米;C最接近1公顷。故答案为:C",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识"
+                             ],
+        "source":  "25-26 四年级上·福建福州·期中"
+    },
+    {
+        "id":  "305",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面最接近1平方千米的是( )。",
+        "options":  [
+                        "A. 一间卧室",
+                        "B. 一个篮球场",
+                        "C. 一所中等规模的小学",
+                        "D. 一座大型城市公园"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】1平方千米=100公顷=1000000平方米，比较各选项的典型面积大小.\n【详解】卧室约20平方米，篮球场约420平方米，小学约几万平方米，大型城市公园约数百公顷，最接近1平方千米。故答案为:D",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26四年级上·广东东莞·期中"
+    },
+    {
+        "id":  "306",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "莆田市的陆地面积大约是 4100( )。",
+        "options":  [
+                        "A. 千米",
+                        "B. 平方米",
+                        "C. 公顷",
+                        "D. 平方千米"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】地级市陆地面积通常用平方千米表示,4100平方千米符合实际.\n【详解】千米是长度单位;平方米太小;4100公顷=41平方千米偏小;4100平方千米合适。故答案为:D",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26四年级上·福建莆田·期中"
+    },
+    {
+        "id":  "307",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "杭州西湖的湖面面积约为6平方千米，扬州瘦西湖的湖面面积约为500000平方米，武汉东湖的湖面面积约为33平方千米，嘉兴南湖的湖面面积约为 100 公顷。( )的湖面面积最小。",
+        "options":  [
+                        "A. 杭州西湖",
+                        "B. 扬州瘦西湖",
+                        "C. 武汉东湖",
+                        "D. 嘉兴南湖"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】把平方米，平方千米都化成公顷(1平方千米=100公顷,1公顷=10000平方米)再比较.\n【详解】6平方千米=600公顷;500000平方米=50公顷;33平方千米=3300公顷;南湖=100公顷。最小为50公顷，对应扬州瘦西湖。故答案为:B",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·福建厦门·期中"
+    },
+    {
+        "id":  "308",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "一块占地2公顷的果园中，种了 5000 棵果树，平均每棵树占地( )平方米。",
+        "options":  [
+                        "A. 400",
+                        "B. 4000",
+                        "D. 4"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】2公顷=20000平方米，用总面积除以棵数.\n【详解】20000÷5000=4(平方米)。故答案为:D",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26四年级上·福建福州·期中"
+    },
+    {
+        "id":  "309",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "学校足球场的占地面积大约是 8000( )。",
+        "options":  [
+                        "A. 平方米",
+                        "B. 公顷",
+                        "C. 平方千米",
+                        "D. 千米"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】足球场面积一般在几千平方米量级,8000更符合平方米.\n【详解】学校足球场占地面积约8000平方米。故答案为:A",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26四年级上·广东东莞·期中"
+    },
+    {
+        "id":  "310",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "一块长方形的草坪长 50 米，宽 20 米,( )个这样的草坪面积是 1 公顷。",
+        "options":  [
+                        "A. 10",
+                        "B. 100",
+                        "C. 20"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】先求一个草坪面积，再用1公顷=10000平方米求需要块数.\n【详解】50×20 = 1000(平方米),10000÷1000=10(块)。故答案为:A",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·江西赣州·期中"
+    },
+    {
+        "id":  "311",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "拙政园占地面积约5公顷,( )个这样的苏州园林是1平方千米。",
+        "options":  [
+                        "A. 2",
+                        "B. 20",
+                        "C. 200"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】1平方千米=100公顷，用100÷5.\n【详解】100 ÷ 5 = 20(个)。故答案为:B",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·河北邢台·期中"
+    },
+    {
+        "id":  "312",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一个正方形果园，边长是 100 米，它的面积是( )。",
+        "options":  [
+                        "A. 1 公顷",
+                        "B. 100 平方米",
+                        "C. 1 平方千米",
+                        "D. 1000 平方米"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】正方形面积=边长×边长,100×100=10000平方米=1公顷.\n【详解】面积是1公顷。故答案为:A",
+        "knowledge_points":  [
+                                 "正方形的面积",
+                                 "公顷，平方千米的认识",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "23-24 四年级上·贵州遵义·期末"
+    },
+    {
+        "id":  "313",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "( )个边长 100 米的正方形面积和是 1 平方千米。",
+        "options":  [
+                        "A. 10",
+                        "B. 100",
+                        "C. 1000",
+                        "D. 10000"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】边长100米的正方形面积=1公顷,1平方千米=100公顷.\n【详解】需要100个。故答案为:B",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·河南开封·期中"
+    },
+    {
+        "id":  "314",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "厦门市的油罐公园占地面积约为10万平方米,( )油罐公园的占地面积约为1平方千米。",
+        "options":  [
+                        "A. 1",
+                        "B. 10",
+                        "C. 100",
+                        "D. 1000"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】1平方千米=1000000平方米,10万平方米=100000平方米.\n【详解】1000000里有10个100000，所以约需要10个。故答案为:B",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·福建厦门·期中"
+    },
+    {
+        "id":  "315",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "32个小朋友手拉手围成一个正方形，面积大约是100平方米,( )个这样的正方形的面积大约是1平方千米。",
+        "options":  [
+                        "A. 10",
+                        "B. 100",
+                        "C. 1000",
+                        "D. 10000"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】1平方千米=1000000平方米，求多少个100平方米组成1000000平方米.\n【详解】1000000÷100=10000。故答案为:D",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·福建厦门·期中"
+    },
+    {
+        "id":  "316",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面( )的面积最接近 1 公顷。",
+        "options":  [
+                        "A. 30 块黑板",
+                        "B. 50张课桌",
+                        "C. 100间教室",
+                        "D. 20 个篮球场"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】1公顷=10000平方米，估算各项总面积后比较.\n【详解】30块黑板约120平方米;50张课桌约50平方米;100间教室约5000平方米;20个篮球场约10000平方米=1公顷。故答案为:D",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识"
+                             ],
+        "source":  "24-25 四年级上·福建厦门·期中"
+    },
+    {
+        "id":  "317",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面最接近1公顷的是( )。",
+        "options":  [
+                        "A. 10 个标准篮球场的面积",
+                        "B. 一个标准游泳池的面积",
+                        "C. 200间教室的面积",
+                        "D. 湖北省的面积"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】1公顷=10000平方米，结合篮球场，游泳池，教室的典型面积判断.\n【详解】10个篮球场约5000平方米;游泳池约1000平方米;200间教室约50×200=10000平方米=1公顷;湖北省面积远大。故答案为:C",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "24-25四年级上·湖北孝感·期中"
+    },
+    {
+        "id":  "318",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面各组中，两个面积单位进率最大的是( )。",
+        "options":  [
+                        "A. 平方米和平方千米",
+                        "B. 平方米和公顷",
+                        "C. 公顷和平方千米",
+                        "D. 平方分米和平方米"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】比较各组相邻单位换算进率大小:1平方千米=1000000平方米;1公顷=10000平方米;1平方千米=100公顷;1平方米=100平方分米.\n【详解】进率分别为1000000,10000,100,100，最大的是平方米和平方千米。故答案为:A",
+        "knowledge_points":  [
+                                 "平方厘米，平方分米，平方米之间的进率与换算",
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "24-25 四年级上·河南许昌·期中"
+    },
+    {
+        "id":  "319",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "张叔叔要修建一个大型的农家乐，选择占地面积为( )的土地修建比较合适。",
+        "options":  [
+                        "A. 100平方米",
+                        "B. 1 公顷",
+                        "C. 1 平方千米",
+                        "D. 100平方千米"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】结合不同面积单位的实际大小:100平方米太小;1平方千米,100平方千米过大;1公顷较适中.\n【详解】1公顷=10000平方米，适合大型农家乐用地规模。故答案为:B",
+        "knowledge_points":  [
+                                 "平方厘米，平方分米，平方米的认识",
+                                 "公顷，平方千米的认识",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·广西柳州·期中"
+    },
+    {
+        "id":  "320",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "张叔叔承包了一片土地来种植刺梨，这片土地大约有 5 个足球场的大小，这片土地的面积最有可能是( )。",
+        "options":  [
+                        "A. 50000 平方米",
+                        "B. 50平方米",
+                        "C. 5平方千米",
+                        "D. 50 公顷"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】一个足球场约1公顷,5个足球场约5公顷=50000平方米.\n【详解】50000平方米=5公顷符合;其余选项不符合实际。故答案为:A",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积的估测",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "24-25 四年级上·贵州六盘水·期中"
+    },
+    {
+        "id":  "321",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上合适的单位名称.\n\n①舞蹈室的面积大约是120( );②金华小学的占地面积大约4( );③湖南省的面积大约是21.18万( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方米;公顷;平方千米",
+        "solution":  "【分析】舞蹈室面积用平方米;学校占地用公顷;省域面积用平方千米.\n【详解】①120平方米;②4公顷;③21.18万平方千米。",
+        "knowledge_points":  [
+                                 "平方厘米，平方分米，平方米的认识",
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26五年级上·湖南邵阳·期中"
+    },
+    {
+        "id":  "322",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "北京天安门广场的占地面积大约是 44( ),山东省的占地面积大约是157900( )。",
+        "options":  [
+
+                    ],
+        "answer":  "公顷;平方千米",
+        "solution":  "【分析】天安门广场用公顷;省域面积用平方千米.\n【详解】44公顷;157900平方千米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 五年级上·河北石家庄·期中"
+    },
+    {
+        "id":  "323",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "操场跑道一圈长400( ),小明的课桌面积约是40( ),我国陆地领土面积约是960万( ),国家游泳中心\"水立方\"的建筑面积约为 8( )。",
+        "options":  [
+
+                    ],
+        "answer":  "米;平方分米;平方千米;公顷",
+        "solution":  "【分析】跑道长度用米;课桌面积常用平方分米;国土面积用平方千米;大型建筑占地用公顷.\n【详解】400米,40平方分米,960万平方千米,8公顷。",
+        "knowledge_points":  [
+                                 "长度单位的选择",
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 四年级上·湖北襄阳·期中"
+    },
+    {
+        "id":  "324",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "填上合适的数或单位名称.\n\n(1)福建省占地面积约是 12 万( );(2)我国著名园林北京圆明园的占地面积约 350( );(3)福州西湖公园总面积约为 50 公顷，合( )平方米,( )个这样的西湖公园面积大约是 1 平方千米。",
+        "options":  [
+
+                    ],
+        "answer":  "平方千米;公顷;500000;2",
+        "solution":  "【分析】省域面积用平方千米;大型园林用公顷;1公顷=10000平方米;1平方千米=100公顷.\n【详解】(1)12万平方千米。(2)350公顷。(3)50公顷=500000平方米;2个50公顷合100公顷=1平方千米。",
+        "knowledge_points":  [
+                                 "面积单位的选择",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·福建福州·期中"
+    },
+    {
+        "id":  "325",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号中填上合适的面积单位。海口市的总面积约为 2297( ).\"鸟巢\"的占地面积大约是 20( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方千米;公顷",
+        "solution":  "【分析】海口市总面积较大用平方千米;鸟巢占地用公顷.\n【详解】海口市约2297平方千米;鸟巢约20公顷。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 四年级上·海南省直辖县级单位·期中"
+    },
+    {
+        "id":  "326",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上合适的数.\n\n70000 平方米=( )公顷\n\n7 平方米=( )平方分米\n\n68 平方千米=( )公顷=( )平方米",
+        "options":  [
+
+                    ],
+        "answer":  "7;700;6800;68000000",
+        "solution":  "【分析】1公顷=10000平方米;1平方米=100平方分米;1平方千米=100公顷=1000000平方米.\n【详解】70000平方米=7公顷;7平方米=700平方分米;68平方千米=6800公顷=68000000平方米。",
+        "knowledge_points":  [
+                                 "平方厘米，平方分米，平方米之间的进率与换算",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·海南省直辖县级单位·期中"
+    },
+    {
+        "id":  "327",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "700 平方厘米=( )平方分米\n\n5 公顷=( )平方米\n\n6000000 平方米=( )公顷=( )平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "7;50000;600;6",
+        "solution":  "【分析】平方厘米与平方分米进率100;公顷与平方米进率10000;平方米与公顷进率10000;公顷与平方千米进率100.\n【详解】700平方厘米=7平方分米;5公顷=50000平方米;6000000平方米=600公顷=6平方千米。",
+        "knowledge_points":  [
+                                 "平方厘米，平方分米，平方米之间的进率与换算",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·广东东莞·期中"
+    },
+    {
+        "id":  "328",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "9 公顷=800 平方米+( )平方米;64000000 平方米=( )公顷=( )平方千米。",
+        "options":  [
+
+                    ],
+        "answer":  "89200;6400;64",
+        "solution":  "【分析】9公顷=90000平方米，用90000-800求第一空;平方米换公顷除以10000;公顷换平方千米除以100.\n【详解】9公顷=90000平方米,90000-800=89200;64000000平方米=6400公顷=64平方千米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·河北保定·期中"
+    },
+    {
+        "id":  "329",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "东莞市人民公园的占地面积大约是 23 公顷，合( )平方米;广东丹霞山国家级自然保护区面积为 292 平方千米，合( )公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "230000;29200",
+        "solution":  "【分析】1公顷=10000平方米;1平方千米=100公顷.\n【详解】23公顷=230000平方米;292平方千米=29200公顷。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·广东东莞·期中"
+    },
+    {
+        "id":  "330",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上适当的面积单位.\n\n(1)国家游泳中心(\"水立方\")的占地面积约 6( ).\n(2)标准的足球场面积约 7140( ).\n(3)湖北省的土地面积约 19700( )。",
+        "options":  [
+
+                    ],
+        "answer":  "公顷;平方米;平方千米",
+        "solution":  "【分析】水立方占地用公顷;足球场面积用平方米;省域面积用平方千米.\n【详解】(1)6公顷;(2)7140平方米;(3)19700平方千米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 四年级上·湖北襄阳·期中"
+    },
+    {
+        "id":  "331",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "60000000 平方米=( )公顷=( )平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "6000;60",
+        "solution":  "【分析】1公顷=10000平方米,1平方千米=100公顷.\n【详解】60000000÷10000=6000(公顷);6000÷100=60(平方千米)。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·广东潮州·期中"
+    },
+    {
+        "id":  "332",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上合适的单位名称.\n\n(1)定安文笔峰盘古文化旅游区的核心景区占地面积大约是80( ).\n(2)一个乒乓球馆的占地面积约 600( ).\n(3)定安县行政区域陆地面积大约为 1196( )。",
+        "options":  [
+
+                    ],
+        "answer":  "公顷;平方米;平方千米",
+        "solution":  "【分析】景区占地较大用公顷;乒乓球馆用平方米;县域面积用平方千米.\n【详解】80公顷;600平方米;1196平方千米。",
+        "knowledge_points":  [
+                                 "平方厘米，平方分米，平方米的认识",
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 五年级上·海南省直辖县级单位·期中"
+    },
+    {
+        "id":  "333",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上合适的单位.\n\n北京故宫的占地面积约是 72( )。一枚邮票的面积约是6( )。我国陆地领土面积约是 960 万( )。",
+        "options":  [
+
+                    ],
+        "answer":  "公顷;平方厘米;平方千米",
+        "solution":  "【分析】故宫占地用公顷;邮票面积小用平方厘米;国土面积大用平方千米.\n【详解】72公顷;6平方厘米;960万平方千米。",
+        "knowledge_points":  [
+                                 "平方厘米，平方分米，平方米的认识",
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26四年级上·河北邯郸·期中"
+    },
+    {
+        "id":  "334",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上适当的面积单位.\n\n(1)颐和园的占地面积约为 300( ).\n\n(2)香港特别行政区陆地面积约 1106( ).\n(3)1 间教室的面积约为 63( )。",
+        "options":  [
+
+                    ],
+        "answer":  "公顷;平方千米;平方米",
+        "solution":  "【分析】(1)边长是 1 米的正方形面积是 1 平方米，一般小学的操场大约是10000平方米，因为1公顷 = 10000 平方米，所以一般小学的操场面积大约是1公顷，而颐和园的占地面积相对于操场大得多，根据这里的数据300应选择公顷作单位。(2)香港特别行政区陆地面积与颐和园的占地面积相比较，香港特别行政区陆地面积比颐和园的占地面积大很多很多，根据这里的数据1106，选择平方千米作单位合适。(3)1间教室的面积一般不会超过100平方米，且一般大于 50 平方米，根据这里的数据 63，选择平方米作单位合适.\n【详解】(1)颐和园的占地面积约为 300 公顷。(2)香港特别行政区陆地面积约1106平方千米。(3)1 间教室的面积约为 63 平方米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26四年级上·辽宁鞍山·期中"
+    },
+    {
+        "id":  "335",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上合适的数.\n\n(1)12 公顷=( )平方米.\n(2)80000 公顷=( )平方千米.\n(3)5000000 平方米=( )公顷 = ( )平方千米。",
+        "options":  [
+
+                    ],
+        "answer":  "120000;800;500;5",
+        "solution":  "【分析】高级单位化低级单位乘进率;低级单位化高级单位除以进率。(1)1 公顷 = 10000 平方米，据此将 12 公顷换算成以\"平方米\"为单位的数;(2)1平方千米 = 100 公顷，据此将80000公顷换算成以\"平方千米\"为单位的数;(3)1 平方千米 = 100 公顷 = 1000000 平方米，据此解题.\n【详解】(1)12 公顷=120000 平方米。(2)80000 公顷=800 平方千米。(3)5000000 平方米=500 公顷=5 平方千米。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·新疆阿克苏·期中"
+    },
+    {
+        "id":  "336",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上合适的单位.\n\n北京故宫的占地面积约 72( ).\n\n教室的面积大约是 56( ).\n\n新疆维吾尔自治区的面积大约是 166 万( )。",
+        "options":  [
+
+                    ],
+        "answer":  "公顷;平方米;平方千米",
+        "solution":  "【分析】400米跑道围起来的部分的面积大约是1公顷，所以北京故宫的占地面积约72公顷比较合理;学生桌面面积大约是1平方米，所以教室的面积大约是56平方米比较合理;计量比较大的土地面积，常用\"平方千米\"作单位，所以新疆维吾尔自治区的面积大约是166万平方千米.\n【详解】北京故宫的占地面积约72公顷。教室的面积大约是56平方米。新疆维吾尔自治区的面积大约是166万平方千米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 四年级上·新疆阿克苏·期中"
+    },
+    {
+        "id":  "337",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "学校操场的面积大约是 2( )。",
+        "options":  [
+
+                    ],
+        "answer":  "公顷",
+        "solution":  "【分析】常用较大的面积单位有平方千米，公顷，平方米等,1平方米的面积大小是边长为1米的正方形大小，一般国家，城市，省份，岛屿等用平方千米，大型广场，公园，田地等用公顷，房屋，足球场等用平方米，根据实际情况选择合适单位.\n【详解】学校操场一般有跑道，长宽大约有1到200米，那么面积大约1到4万平方米，也就是1到4公顷，所以学校操场的面积大约是2公顷。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 四年级上·内蒙古呼伦贝尔·期中"
+    },
+    {
+        "id":  "338",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "50000平方米=( )公顷\n\n3平方千米=( )公顷\n\n6 公顷=( )平方米\n\n4 平方千米=( )平方米",
+        "options":  [
+
+                    ],
+        "answer":  "5;300;60000;4000000",
+        "solution":  "【分析】根据10000平方米 = 1 公顷进行单位换算;根据 1 平方千米 = 100 公顷进行单位换算;根据1公顷 = 10000 平方米进行单位换算;根据 1 平方千米 = 1000000 平方米进行单位换算.\n【详解】50000 里有 5 个 10000，所以 50000 平方米=5 公顷;3 平方千米=300 公顷;6 公顷=60000 平方米;4 平方千米=4000000 平方米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26四年级上·内蒙古呼伦贝尔·期中"
+    },
+    {
+        "id":  "339",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上合适的单位名称.\n\n一个标准足球场的面积约是 7140( ).\n\n一所希望小学的占地面积约是 2( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方米;公顷",
+        "solution":  "【分析】计量一个标准足球场的面积用平方米比较合适;测量土地的面积常用公顷.\n【详解】一个标准足球场的面积约是7140平方米。一所希望小学的占地面积约是2公顷。",
+        "knowledge_points":  [
+                                 "平方厘米，平方分米，平方米的认识",
+                                 "公顷，平方千米的认识"
+                             ],
+        "source":  "25-26四年级上·河北邢台·期中"
+    },
+    {
+        "id":  "340",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "请你为下面三条信息选择合适的计量单位.\n\n内蒙古自治区的面积约为 1183000( ).\n\n鄂尔多斯市体育中心占地面积约为 70.88( ).\n\n一间教室占地面积约为 56( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方千米;公顷;平方米",
+        "solution":  "【分析】内蒙古自治区作为省级行政区，面积非常广阔,1183000这个巨大数字应选用\"平方千米\"。体育中心属于较大的建筑区域，用公顷更合适;教室是较小空间，用平方米合适.\n【详解】内蒙古自治区约1183000平方千米;鄂尔多斯市体育中心约70.88公顷;一间教室约56平方米。",
+        "knowledge_points":  [
+                                 "平方厘米，平方分米，平方米的认识",
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 四年级上·内蒙古鄂尔多斯·期中"
+    },
+    {
+        "id":  "341",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在( )里填上适当的单位.\n\n(1)莆田市总面积约是 4200( ).\n(2)飞机 1 分钟飞行 12( ).\n(3)故宫博物院的占地面积 72( ).\n(4)操场一周的长是 200( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方千米;千米;公顷;米",
+        "solution":  "【分析】莆田市总面积用平方千米;飞机1分钟飞行距离用千米;故宫占地用公顷;操场一周长度用米.\n【详解】(1)4200平方千米;(2)12千米;(3)72公顷;(4)200米。",
+        "knowledge_points":  [
+                                 "长度单位的选择",
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26四年级上·福建莆田·期中"
+    },
+    {
+        "id":  "342",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "( )公顷 = 300000",
+        "options":  [
+
+                    ],
+        "answer":  "30",
+        "solution":  "【分析】1公顷 = 10000 平方米,300000末尾去掉4个0即可化为公顷.\n【详解】300000平方米=30公顷，所以填30。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·福建莆田·期中"
+    },
+    {
+        "id":  "343",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\"\u003e\"\"\u003c\"或\"=\".\n\n78798( )78789\n\n800000000( )8 亿\n\n360920( )37 万\n\n12 平方千米( )120 公顷\n\n4 平方千米( )4000 公顷\n\n110000 平方米( )10 公顷",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;=;\u003c;\u003e;\u003c;\u003e",
+        "solution":  "【分析】按整数大小比较方法比较;并把8亿,37万改写成以\"个\"为单位;平方千米和公顷，平方米和公顷先统一单位再比较.\n【详解】78798\u003e78789;800000000=8亿;360920\u003c370000(37万);12平方千米=1200公顷\u003e120公顷;4平方千米=400公顷\u003c4000公顷;10公顷=100000平方米,110000平方米\u003e100000平方米，所以110000平方米\u003e10公顷。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "公顷，平方千米的进率与换算",
+                                 "亿以内数的改写",
+                                 "亿以上数的改写"
+                             ],
+        "source":  "25-26四年级上·内蒙古通辽·月考"
+    },
+    {
+        "id":  "344",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "边长是( )米的正方形，面积是 1 平方千米，也就是( )公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "1000;100",
+        "solution":  "【分析】边长1千米的正方形面积是1平方千米;1平方千米=100公顷.\n【详解】边长1000米的正方形面积是1平方千米，也就是100公顷。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·湖南永州·期中"
+    },
+    {
+        "id":  "345",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "20000 平方米=( )公顷\n\n12 平方千米=( )公顷",
+        "options":  [
+
+                    ],
+        "answer":  "2;1200",
+        "solution":  "【分析】1公顷=10000平方米,1平方千米=100公顷.\n【详解】20000平方米=2公顷;12平方千米=1200公顷。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·湖南永州·期中"
+    },
+    {
+        "id":  "346",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在下面的括号里填上合适的单位.\n\n(1)湖南省的占地面积大约是 211800( ).\n(2)箴龙学校的占地面积约是 6( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方千米;公顷",
+        "solution":  "【分析】省域面积用平方千米；学校占地用公顷。`r`n【详解】(1)湖南省的占地面积大约是211800平方千米。(2)箴龙学校的占地面积约是6公顷。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "24-25 四年级上·湖南益阳·期中"
+    },
+    {
+        "id":  "347",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "36 公顷=( )平方米\n\n5000 公顷=( )平方千米\n\n14 公顷=( )平方米\n\n9 公顷 800 平方米=( )平方米\n\n820000000 平方米=( )公顷=( )平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "360000;50;140000;90800;82000;820",
+        "solution":  "【分析】根据1公顷=10000平方米,1平方千米=100公顷=1000000平方米进行换算.\n【详解】36公顷=360000平方米;5000公顷=50平方千米;14公顷=140000平方米;9公顷800平方米=90800平方米;820000000平方米=82000公顷=820平方千米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·山东日照·期中"
+    },
+    {
+        "id":  "348",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上\"\u003e\"\"\u003c\"或\"=\".\n\n4 平方千米( )400 公顷\n\n350000( )530000\n\n10999999( )9850000\n\n296400400( )3 亿",
+        "options":  [
+
+                    ],
+        "answer":  "=;\u003c;\u003e;\u003c",
+        "solution":  "【分析】(1)1平方千米=100公顷，先统一单位;(2)(3)按整数大小比较;(4)把3亿改写成300000000再比较.\n【详解】4平方千米=400公顷，所以\"=\";350000\u003c530000;10999999\u003e9850000;3亿=300000000,296400400\u003c300000000，所以296400400\u003c3亿。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "公顷，平方千米的进率与换算",
+                                 "亿以上数的改写"
+                             ],
+        "source":  "25-26 四年级上·云南昭通·期中"
+    },
+    {
+        "id":  "349",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "9 公顷=( )平方米\n\n45000000 平方米=( )公顷=( )平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "90000;4500;45",
+        "solution":  "【分析】1公顷=10000平方米,1平方千米=100公顷.\n【详解】9公顷=90000平方米;45000000平方米=4500公顷=45平方千米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·云南昭通·期中"
+    },
+    {
+        "id":  "350",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "中国园林是世界园林艺术的瑰宝，其中承德避暑山庄是世界上现存最大的皇家园林，占地面积约为564( ),苏州留园的占地面积约为23300( )。横线上的数由( )个万和( )个一组成，用\"四舍五入\"法省略万位后面的尾数是( ),这个数中左边的\"3\"表示( ),右边的\"3\"表示( )。",
+        "options":  [
+
+                    ],
+        "answer":  "公顷;平方米;2;3300;2万;3个千;3个百",
+        "solution":  "【分析】结合实际选择面积单位;理解\"万\"和\"个一\"的组成;省略万位后面的尾数看千位四舍五入;理解数位上数字表示的计数单位.\n【详解】承德避暑山庄约564公顷;苏州留园约23300平方米。23300由2个万和3300个一组成，省略万位后面的尾数约是2万;左边的3表示3个千，右边的3表示3个百。",
+        "knowledge_points":  [
+                                 "亿以内数的组成",
+                                 "整数的数级，数位和计数单位的认识",
+                                 "公顷，平方千米的认识"
+                             ],
+        "source":  "24-25 四年级上·河南开封·期中"
+    },
+    {
+        "id":  "351",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在( )里填上合适的面积单位。颐和园的占地面积约为 293( );一枚邮票的面积约是 6( );一张课桌桌面的面积大约是 40( );一间卧室占地约 25( )。",
+        "options":  [
+
+                    ],
+        "answer":  "公顷;平方厘米;平方分米;平方米",
+        "solution":  "【分析】根据生活实际选择合适的面积单位：较大区域用公顷，较小物体用平方厘米/平方分米，房间用平方米.\n【详解】颐和园约293公顷;邮票约6平方厘米;课桌面约40平方分米;卧室约25平方米。",
+        "knowledge_points":  [
+                                 "平方厘米，平方分米，平方米的认识",
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "24-25 四年级上·山西长治·期中"
+    },
+    {
+        "id":  "352",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "380000 平方米=( )公顷\n\n3 平方千米 5 公顷=( )公顷\n\n56 平方千米=( )公顷\n\n12 公顷=( )平方米",
+        "options":  [
+
+                    ],
+        "answer":  "38;305;5600;120000",
+        "solution":  "【分析】1公顷=10000平方米,1平方千米=100公顷，按进率换算.\n【详解】380000平方米=38公顷;3平方千米5公顷=300+5=305公顷;56平方千米=5600公顷;12公顷=120000平方米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "24-25 四年级上·山西长治·期中"
+    },
+    {
+        "id":  "353",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "15 平方千米=( )平方米=( )公顷\n\n500 公顷=( )平方千米\n\n10 公顷=( )平方米",
+        "options":  [
+
+                    ],
+        "answer":  "15000000;1500;5;100000",
+        "solution":  "【分析】1平方千米=1000000平方米=100公顷;100公顷=1平方千米;1公顷=10000平方米.\n【详解】15平方千米=15000000平方米=1500公顷;500公顷=5平方千米;10公顷=100000平方米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "24-25 四年级上·湖北武汉·期中"
+    },
+    {
+        "id":  "354",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "5 平方千米=( )公顷\n\n3 平方千米=( )平方米\n\n520000=( )万",
+        "options":  [
+
+                    ],
+        "answer":  "500;3000000;52",
+        "solution":  "【分析】1平方千米=100公顷=1000000平方米;把520000改写成用\"万\"作单位去掉末尾4个0.\n【详解】5平方千米=500公顷;3平方千米=3000000平方米;520000=52万。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "亿以内数的改写"
+                             ],
+        "source":  "24-25 四年级上·广东江门·期中"
+    },
+    {
+        "id":  "355",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上合适的单位.\n\n有\"鹭岛\"美称的厦门市面积约 158( );于近期正式开业的厦门大悦城占地面积约 10( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方千米;公顷",
+        "solution":  "【分析】城市面积用平方千米;商业建筑占地用公顷.\n【详解】厦门市约158平方千米;厦门大悦城占地约10公顷。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "24-25 四年级上·福建厦门·期中"
+    },
+    {
+        "id":  "356",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上合适的单位名称.\n\n课桌桌面的面积约 40( )\n\n四川省的面积约 485000( )\n\n教室的面积约 60( )\n\n故宫博物院的占地面积约 72( )",
+        "options":  [
+
+                    ],
+        "answer":  "平方分米;平方千米;平方米;公顷",
+        "solution":  "【分析】结合生活经验与单位大小选择：课桌面用平方分米;省域面积用平方千米;教室用平方米;故宫占地用公顷.\n【详解】课桌桌面约40平方分米;四川省约485000平方千米;教室约60平方米;故宫博物院约72公顷。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 六年级上·河北唐山·期中"
+    },
+    {
+        "id":  "357",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "5 公顷=( )平方米\n\n6000 公顷=( )平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "50000;60",
+        "solution":  "【分析】1公顷=10000平方米;1平方千米=100公顷.\n【详解】5公顷=50000平方米;6000公顷=60平方千米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 五年级上·海南省直辖县级单位·期中"
+    },
+    {
+        "id":  "358",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "48 分=( )时\n\n1.02 公顷=( )平方米",
+        "options":  [
+
+                    ],
+        "answer":  "0.8;10200",
+        "solution":  "【分析】1时=60分;1公顷=10000平方米.\n【详解】48÷60=0.8(时);1.02×10000=10200(平方米)。",
+        "knowledge_points":  [
+                                 "时，分的认识",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 五年级上·福建福州·期中"
+    },
+    {
+        "id":  "359",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "狮山镇的占地面积约是 336( );佛出梅沙双语学校的占地面积是6.5( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方千米;公顷",
+        "solution":  "【分析】镇域面积较大用平方千米;学校占地用公顷.\n【详解】狮山镇约336平方千米;佛出梅沙双语学校约6.5公顷。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "24-25 四年级上·广东佛山·期中"
+    },
+    {
+        "id":  "360",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在( )里填上适当的面积单位.\n\n(1)鼓浪屿是一个面积为 178( )的小岛.\n(2)73 公顷 500 平方米=( )平方米.\n(3)5000 公顷=( )平方千米=( )平方米。",
+        "options":  [
+
+                    ],
+        "answer":  "公顷;730500;50;50000000",
+        "solution":  "【分析】(1)结合数据选择合适单位;(2)1公顷=10000平方米;(3)1平方千米=100公顷,1公顷=10000平方米.\n【详解】(1)鼓浪屿约178公顷;(2)73公顷=730000平方米,730000+500=730500平方米;(3)5000公顷=50平方千米=50000000平方米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位的选择"
+                             ],
+        "source":  "24-25四年级上·福建厦门·期中"
+    },
+    {
+        "id":  "361",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "杭州西湖的面积约是 5.6( ),嘉善缪家村的碧水绕村河道全长 10.4( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方千米;千米",
+        "solution":  "【分析】计量湖泊面积用平方千米;计量河道长度用千米.\n【详解】杭州西湖约5.6平方千米;河道全长约10.4千米。",
+        "knowledge_points":  [
+                                 "长度单位的选择",
+                                 "面积单位的选择"
+                             ],
+        "source":  "24-25 五年级上·浙江宁波·期中"
+    },
+    {
+        "id":  "362",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "6 平方千米=( )公顷\n\n45 公顷=( )平方米\n\n12000000 平方米=( )公顷=( )平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "600;450000;1200;12",
+        "solution":  "【分析】1平方千米=100公顷;1公顷=10000平方米;10000平方米=1公顷,100公顷=1平方千米.\n【详解】6平方千米=600公顷;45公顷=450000平方米;12000000平方米=1200公顷=12平方千米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "24-25 四年级上·广东江门·期中"
+    },
+    {
+        "id":  "363",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上适当的单位.\n\n(1)北京圆明园的占地面积约是 350( ).\n(2)一所学校占地面积约 2( ).\n(3)澳门特别行政区的面积约是 33( ).\n(4)一间卧室的面积约 20( ).\n(5)一块橡皮的面积约是 6( )。",
+        "options":  [
+
+                    ],
+        "answer":  "公顷;公顷;平方千米;平方米;平方厘米",
+        "solution":  "【分析】根据物体/区域大小选择面积单位：园林和学校用公顷;行政区用平方千米;卧室用平方米;橡皮用平方厘米.\n【详解】(1)350公顷;(2)2公顷;(3)33平方千米;(4)20平方米;(5)6平方厘米。",
+        "knowledge_points":  [
+                                 "平方厘米，平方分米，平方米的认识",
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "24-25 四年级上·重庆秀山·期中"
+    },
+    {
+        "id":  "364",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号填上\"\u003e\"\"\u003c\"或\"=\".\n\n368432156( )37206200\n\n5 公顷( )5000 平方米\n\n364500( )365400\n\n6800000( )680 万\n\n2000000 平方米( )4 公顷\n\n10 平方千米( )10000000 平方米",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;\u003e;\u003c;=;\u003e;=",
+        "solution":  "【分析】按整数大小比较;并把公顷，平方千米换算成平方米再比较;把\"万\"改写成以\"个\"为单位后比较.\n【详解】368432156\u003e37206200;5公顷=50000平方米\u003e5000平方米;364500\u003c365400;6800000=680万;4公顷=40000平方米,2000000平方米\u003e40000平方米;10平方千米=10000000平方米，所以\"=\"。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "平方厘米，平方分米，平方米之间的进率与换算",
+                                 "公顷，平方千米的进率与换算",
+                                 "亿以内数的改写"
+                             ],
+        "source":  "24-25四年级上·重庆大渡口·期中"
+    },
+    {
+        "id":  "365",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "5 平方千米=( )公顷=( )平方米\n\n36000000 平方米=( )公顷=( )平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "500;5000000;3600;36",
+        "solution":  "【分析】1平方千米=100公顷=1000000平方米,1公顷=10000平方米.\n【详解】5平方千米=500公顷=5000000平方米;36000000平方米=3600公顷=36平方千米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "24-25 四年级上·广东潮州·期中"
+    },
+    {
+        "id":  "366",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "( )平方米=200 公顷=( )平方千米。",
+        "options":  [
+
+                    ],
+        "answer":  "2000000;2",
+        "solution":  "【分析】1公顷=10000平方米，所以200公顷=200×10000=2000000平方米;1平方千米=100公顷，所以200公顷=2平方千米.\n【详解】2000000平方米=200公顷=2平方千米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "24-25四年级上·宁夏吴忠·期中"
+    },
+    {
+        "id":  "367",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在( )里填上\"\u003e\"\"\u003c\"或\"=\".\n\n50 平方米( )500 平方分米\n\n90000 平方米( )9 公顷\n\n3 公顷( )300 平方千米\n\n10 平方千米( )1000000 平方米",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;=;\u003c;\u003e",
+        "solution":  "【分析】1平方米=100平方分米;1公顷=10000平方米;1平方千米=100公顷=1000000平方米，先统一单位再比较.\n【详解】50平方米=5000平方分米\u003e500平方分米;90000平方米=9公顷;3公顷\u003c1平方千米\u003c300平方千米;10平方千米\u003e1平方千米=1000000平方米。",
+        "knowledge_points":  [
+                                 "平方厘米，平方分米，平方米之间的进率与换算",
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "24-25四年级上·宁夏吴忠·期中"
+    },
+    {
+        "id":  "368",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上\"\u003e\"\"\u003c\"或\"=\".\n\n4630000( )463 万\n\n5 平方千米( )498 公顷\n\n3403400( )30403400\n\n100000 平方米( )1 平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "=;\u003e;\u003c;\u003c",
+        "solution":  "【分析】(1)把4630000改写为463万;(2)1平方千米=100公顷;(3)按位数比较;(4)1平方千米=1000000平方米.\n【详解】4630000=463万;5平方千米=500公顷\u003e498公顷;3403400是7位数,30403400是8位数，所以3403400\u003c30403400;100000平方米\u003c1000000平方米，所以100000平方米\u003c1平方千米。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算",
+                                 "亿以内数的改写"
+                             ],
+        "source":  "24-25 四年级上·河南许昌·期中"
+    },
+    {
+        "id":  "369",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上合适的面积单位.\n\n鸟巢的占地面积约 20( ).\n\n一间教室的面积约 63( ).\n\n我国领土的面积约为 960 万( )。",
+        "options":  [
+
+                    ],
+        "answer":  "公顷;平方米;平方千米",
+        "solution":  "【分析】鸟巢占地较大用公顷;教室面积用平方米;国土面积用平方千米.\n【详解】鸟巢约20公顷;教室约63平方米;我国领土约960万平方千米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "24-25四年级上·河南许昌·期中"
+    },
+    {
+        "id":  "370",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一块长方形绿地长250 米，宽 80 米，这块绿地的面积是( )平方米，合( )公顷;如果原来这块绿地的长不变。宽增加到 240 米，这时绿地的面积是( )公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "20000;2;6",
+        "solution":  "【分析】根据长方形的面积公式，长方形面积 = 长×宽;根据平方米和公顷之间的进率为10000，低级单位换算成高级单位除以进率.\n【详解】10000 平方米 = 1 公顷。250×80 = 20000(平方米),20000 平方米 = 2 公顷。250×240 = 60000(平方米),60000 平方米 = 6 公顷。答：面积是20000平方米，合2公顷;宽增加到240米时面积是6公顷。",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·浙江宁波·期中"
+    },
+    {
+        "id":  "371",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "6 平方千米=( )公顷\n\n78 公顷=( )平方米\n\n8 公顷 80 平方米=( )平方米\n\n3平方千米12公顷=( )公顷",
+        "options":  [
+
+                    ],
+        "answer":  "600;780000;80080;312",
+        "solution":  "【分析】1平方千米=100公顷;1公顷=10000平方米;先统一单位后计算.\n【详解】6平方千米=600公顷;78公顷=780000平方米;8公顷80平方米=80080平方米;3平方千米12公顷=300+12=312公顷。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "24-25 四年级上·浙江宁波·期中"
+    },
+    {
+        "id":  "372",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "单位转换.\n\n12 公顷=( )平方米\n\n100 平方米=( )平方分米\n\n10000 公顷=( )平方米\n\n3000000 平方米=( )平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "120000;10000;100000000;3",
+        "solution":  "【分析】1公顷=10000平方米;1平方米=100平方分米;1平方千米=1000000平方米.\n【详解】12公顷=120000平方米;100平方米=10000平方分米;10000公顷=100000000平方米;3000000平方米=3平方千米。",
+        "knowledge_points":  [
+                                 "平方厘米，平方分米，平方米之间的进率与换算",
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "24-25 四年级上·湖南益阳·期中"
+    },
+    {
+        "id":  "373",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "3 平方千米=( )公顷\n\n6 公顷=( )平方米\n\n45 平方千米=( )平方米\n\n4 平方千米 12 公顷=( )公顷",
+        "options":  [
+
+                    ],
+        "answer":  "300;60000;45000000;412",
+        "solution":  "【分析】1平方千米=100公顷=1000000平方米;1公顷=10000平方米.\n【详解】3×100=300公顷;6×10000=60000平方米;45×1000000=45000000平方米;4平方千米12公顷=400+12=412公顷。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "24-25 四年级上·广西柳州·期中"
+    },
+    {
+        "id":  "374",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "4010000=( )万\n\n2749090000≈( )亿\n\n5 公顷=( )平方米\n\n4 平方千米=( )公顷",
+        "options":  [
+
+                    ],
+        "answer":  "401;27;50000;400",
+        "solution":  "【分析】整万数改写：去掉末尾4个0;省略亿后尾数看千万位四舍五入;1公顷=10000平方米;1平方千米=100公顷.\n【详解】4010000=401万;2749090000≈27亿;5公顷=50000平方米;4平方千米=400公顷。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "亿以内数的改写",
+                                 "亿以上数的近似数"
+                             ],
+        "source":  "24-25 四年级上·湖南益阳·期中"
+    },
+    {
+        "id":  "375",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "( )平方米=8 公顷\n\n6000000 平方米=( )公顷=( )平方千米\n\n3 平方千米+50 公顷=( )公顷\n\n( )平方米=8500 公顷=( )平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "80000;600;6;350;85000000;85",
+        "solution":  "【分析】1公顷=10000平方米;1平方千米=100公顷;按进率换算并统一单位计算.\n【详解】8公顷=80000平方米;6000000平方米=600公顷=6平方千米;3平方千米+50公顷=300+50=350公顷;8500公顷=85000000平方米=85平方千米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "24-25 四年级上·广东潮州·期中"
+    },
+    {
+        "id":  "376",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "单位换算.\n\n600 公顷=( )平方千米\n\n8000000 平方米=( )平方千米\n\n520000 平方米=( )公顷\n\n4 平方千米+40000 平方米=( )公顷",
+        "options":  [
+
+                    ],
+        "answer":  "6;8;52;404",
+        "solution":  "【分析】100公顷=1平方千米;1000000平方米=1平方千米;10000平方米=1公顷;统一为公顷再计算.\n【详解】600公顷=6平方千米;8000000平方米=8平方千米;520000平方米=52公顷;4平方千米=400公顷,400+4=404公顷。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "24-25 四年级上·广东佛山·期中"
+    },
+    {
+        "id":  "377",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "3800000平方米=( )公顷\n\n56 平方千米=( )平方米",
+        "options":  [
+
+                    ],
+        "answer":  "380;56000000",
+        "solution":  "【分析】1公顷=10000平方米;1平方千米=1000000平方米.\n【详解】3800000÷10000=380公顷;56×1000000=56000000平方米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·海南省直辖县级单位·期中"
+    },
+    {
+        "id":  "378",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上合适的面积单位.\n\n(1)一间教室的面积约 60( ).\n(2)广东省的面积大约是 190000( ).\n(3)\"鸟巢\"的占地面积约为 20( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方米;平方千米;公顷",
+        "solution":  "【分析】教室面积用平方米;省域面积用平方千米;鸟巢占地用公顷.\n【详解】(1)60平方米;(2)190000平方千米;(3)20公顷。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "24-25 四年级上·广东佛山·期中"
+    },
+    {
+        "id":  "379",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "250023 平方米=( )公顷( )平方米\n\n300000 平方米=( )公顷",
+        "options":  [
+
+                    ],
+        "answer":  "25;23;30",
+        "solution":  "【分析】1公顷=10000平方米，拆分或除以10000求公顷.\n【详解】250023平方米=25公顷23平方米;300000平方米=30公顷。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·浙江宁波·期中"
+    },
+    {
+        "id":  "380",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一辆喷洒车要给一个占地面积为6公顷的长方形广场进行冲洗。已知它每分钟行驶 250 米，喷洒宽度是 20 米,10 分钟( )完成冲洗工作。(填\"能\"或\"不能\")",
+        "options":  [
+
+                    ],
+        "answer":  "不能",
+        "solution":  "【分析】6公顷=60000平方米。喷洒车10分钟行驶距离=250×10=2500米，喷洒面积=2500×20=50000平方米，与广场面积比较判断.\n【详解】250×10 = 2500(米),2500×20 = 50000(平方米)。6公顷=60000平方米,60000\u003e50000，所以10分钟不能完成冲洗工作。",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26四年级上·甘肃庆阳·期中"
+    },
+    {
+        "id":  "381",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "1 平方千米比 1 千米大,1 公顷比 1 平方米大。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平方千米是面积单位，千米是长度单位，不能比较大小;1公顷=10000平方米，确实比1平方米大，因此整句话错误.\n【详解】\"1平方千米比1千米大\"错误;\"1公顷比1平方米大\"正确，所以该说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·河北邯郸·期中"
+    },
+    {
+        "id":  "382",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "1平方米可以站8个学生，那么 2 平方千米可以站 16 万个学生。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】2平方千米=2000000平方米，人数=8×2000000=16000000=1600万，不是16万.\n【详解】2平方千米=2000000平方米,8×2000000=16000000(个)=1600万，原题错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "亿以内数的改写"
+                             ],
+        "source":  "25-26六年级上·河北唐山·期中"
+    },
+    {
+        "id":  "383",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "判断题",
+        "difficulty":  0.94,
+        "question":  "边长是 1000 米的正方形面积是 1 公顷。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】边长1000米正方形面积=1000000平方米=100公顷=1平方千米，不是1公顷.\n【详解】1000×1000=1000000(平方米),1000000平方米=100公顷,1平方千米=1000000平方米，所以应为1平方千米，原题错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "正方形的面积",
+                                 "公顷，平方千米的认识",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·广东江门·期中"
+    },
+    {
+        "id":  "384",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "某小学操场的面积大约是 500 平方千米。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平方千米用于非常大的区域，小学操场面积通常用平方米或公顷,500平方千米远超实际.\n【详解】该说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "24-25 四年级上·广东潮州·期中"
+    },
+    {
+        "id":  "385",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "天安门广场的面积约是 44 公顷。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】资料中天安门广场面积约440000平方米,1公顷=10000平方米,440000平方米=44公顷.\n【详解】44万平方米=44公顷，所以说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识"
+                             ],
+        "source":  "25-26 四年级上·云南曲靖·月考"
+    },
+    {
+        "id":  "386",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "王伯伯家有一片正方形玉米地，土地边长500米，这片玉米地的面积是多少公顷。每公顷土地收获玉米 6000 千克，这片玉米地能收获多少千克玉米。",
+        "options":  [
+
+                    ],
+        "answer":  "25 公顷;150000 千克",
+        "solution":  "【分析】正方形面积=边长×边长，先求平方米再换算公顷;收获量=公顷数×每公顷产量.\n【详解】500×500=250000(平方米),250000平方米=25公顷;25×6000=150000(千克)。答：面积25公顷，收获150000千克。",
+        "knowledge_points":  [
+                                 "正方形的面积",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·海南省直辖县级单位·期中"
+    },
+    {
+        "id":  "387",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "北京奥林匹克公园是国家5A级旅游景区，公园中有宽 100 米，长 2400 米的树阵景观带。(1)景观带的面积是多少平方米。合多少公顷。(2)若将宽扩大到原来的2倍，长保持不变，那么景观带面积增加了多少公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "240000 平方米;24 公顷;24 公顷",
+        "solution":  "【分析】(1)长方形面积=长×宽，换算公顷;(2)宽变为200米，重新求面积并与原面积作差.\n【详解】(1)2400×100=240000(平方米)=24公顷。 (2)宽变为200米:200×2400=480000(平方米)=48公顷，增加48-24=24公顷。答：面积240000平方米(24公顷);增加24公顷。",
+        "knowledge_points":  [
+                                 "因数中间或末尾是0的乘法",
+                                 "长方形的面积",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·广东东莞·期中"
+    },
+    {
+        "id":  "388",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "现有一块长方形的湿地，长为 20 千米，管理员开车绕湿地外围巡视一圈共行驶了 60 千米(车与湿地边缘的距离忽略不计)。这块湿地的面积是多少公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "20000",
+        "solution":  "【分析】周长=(长+宽)×2，已知周长60千米，长20千米，先求宽;再求面积(平方千米)并换算公顷(1平方千米=100公顷).\n【详解】60÷2 - 20 = 30 - 20 = 10(千米)。面积:20×10=200(平方千米),200平方千米=20000公顷。答：面积20000公顷。",
+        "knowledge_points":  [
+                                 "长方形的周长",
+                                 "长方形的面积",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·广东东莞·期中"
+    },
+    {
+        "id":  "389",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一个正方形果园的周长是 2000 米，如果每公顷栽 80 棵果树，这个果园一共能栽多少棵果树。",
+        "options":  [
+
+                    ],
+        "answer":  "2000 棵",
+        "solution":  "【分析】边长=周长/4;面积=边长×边长(平方米),换算公顷后乘80.\n【详解】2000÷4=500(米),500×500=250000(平方米)=25公顷,25×80=2000(棵)。答：能栽2000棵。",
+        "knowledge_points":  [
+                                 "正方形的周长",
+                                 "正方形的面积",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·湖北襄阳·期中"
+    },
+    {
+        "id":  "390",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "原候车厅的面积约达 15000 平方米。新的约是旧的 2 倍大。你能帮他们算一算新候车厅的面积是多少平方米。合多少公顷吗。",
+        "options":  [
+
+                    ],
+        "answer":  "30000 平方米;3 公顷",
+        "solution":  "【分析】新候车厅面积=旧面积*2;再用1公顷=10000平方米换算.\n【详解】15000×2=30000(平方米),30000平方米=3公顷。答:30000平方米，合3公顷。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·山西晋中·期中"
+    },
+    {
+        "id":  "391",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "度尾文旦柚是莆田四大名果之一，现有一块长方形文旦柚果园，长 200 米，宽150 米，按每公顷产量 8 吨计算，这片果园可收柚子多少吨。",
+        "options":  [
+
+                    ],
+        "answer":  "24 吨",
+        "solution":  "【分析】果园面积=长×宽(平方米),换算成公顷后乘每公顷产量.\n【详解】200×150=30000(平方米)=3公顷,3×8=24(吨)。答：可收24吨。",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·福建莆田·期中"
+    },
+    {
+        "id":  "392",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "某水稻种植基地有一块水稻田长 400 米，宽 200 米，去年这块田共收获水稻 56 吨，平均每公顷收水稻多少吨。",
+        "options":  [
+
+                    ],
+        "answer":  "7 吨",
+        "solution":  "【分析】先求稻田面积并换算公顷，再用总吨数/公顷数.\n【详解】400×200=80000(平方米)=8公顷,56÷8=7(吨)。答：平均每公顷收7吨。",
+        "knowledge_points":  [
+                                 "用 7 ~ 9 的乘法口诀求商",
+                                 "长方形的面积",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26五年级上·海南省直辖县级单位·期中"
+    },
+    {
+        "id":  "393",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "某果园的占地面积约5公顷，多少个果园的占地面积约是1平方千米。",
+        "options":  [
+
+                    ],
+        "answer":  "20 个",
+        "solution":  "【分析】1平方千米=100公顷，用100÷5.\n【详解】100 ÷ 5=20(个)。答:20个。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·新疆阿克苏·期中"
+    },
+    {
+        "id":  "394",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "李叔叔用一台收割机收割小麦，收割机的作业宽度是4米，每小时前进5千米，这台收割机3小时可以收割完这块小麦地。小麦地的总面积是多少公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "6 公顷",
+        "solution":  "【分析】前进距离=5千米*3=15千米=15000米;面积=15000×4=60000平方米;60000平方米=6公顷.\n【详解】5×3=15(千米)=15000米,15000×4=60000(平方米)=6公顷。答：总面积6公顷。",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26四年级上·福建莆田·期中"
+    },
+    {
+        "id":  "395",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一个长方形公园，长 360 米，宽 250 米，现在把它扩建为一个边长 400 米的正方形公园，面积增加了多少公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "7 公顷",
+        "solution":  "【分析】分别求原长方形与扩建后正方形面积(平方米),换算公顷后作差.\n【详解】原面积:360×250=90000平方米=9公顷;扩建后:400×400=160000平方米=16公顷;增加16-9=7公顷。答：增加7公顷。",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "正方形的面积",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·湖南永州·期中"
+    },
+    {
+        "id":  "396",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "某工业园区规划面积是 6 平方千米，已经开发了 3 个经开区，每个经开区占地面积是 150 公顷，还剩多少公顷未开发。",
+        "options":  [
+
+                    ],
+        "answer":  "150 公顷",
+        "solution":  "【分析】6平方千米=600公顷;已开发面积=3×150=450公顷;未开发=600-450.\n【详解】6平方千米=600公顷,600-3×150=600-450=150(公顷)。答：还剩150公顷未开发。",
+        "knowledge_points":  [
+                                 "乘加，乘减混合运算",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26四年级上·湖南永州·期中"
+    },
+    {
+        "id":  "397",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "一个长方形果园，宽是250米，长是宽的4倍。这个果园的占地面积是多少平方米。合多少公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "250000 平方米;25 公顷",
+        "solution":  "【分析】长=宽*4;面积=长×宽;再用1公顷=10000平方米换算.\n【详解】长:250×4=1000(米);面积:1000×250=250000(平方米)=25公顷。答:250000平方米，合25公顷。",
+        "knowledge_points":  [
+                                 "求一个数的几倍是多少",
+                                 "长方形的面积",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·湖南永州·期中"
+    },
+    {
+        "id":  "398",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "园区内现有一块长600 米，宽 400 米的长方形土地将用作郁金香花展，如果每公顷能种 80 万棵郁金香，那么这块土地能种多少万棵郁金香。",
+        "options":  [
+
+                    ],
+        "answer":  "1920 万棵",
+        "solution":  "【分析】先求土地面积并换算公顷;总棵数=公顷数*80万;再化成\"万棵\".\n【详解】面积:600×400=240000平方米=24公顷。24×80万=1920万棵。答：能种1920万棵。",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "24-25 四年级上·福建厦门·期中"
+    },
+    {
+        "id":  "399",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "黄老师每天沿着一个正方形草坪的边缘走 5 圈，正好是 2 千米。这个草坪的面积是多少公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "1 公顷",
+        "solution":  "【分析】2千米=2000米;1圈周长=2000÷5=400米;边长=400÷4=100米;面积=100×100=10000平方米=1公顷.\n【详解】2千米=2000米,2000÷5=400米,400÷4=100米,100×100=10000平方米=1公顷。答：面积1公顷。",
+        "knowledge_points":  [
+                                 "正方形的周长",
+                                 "正方形的面积",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·广东潮州·期中"
+    },
+    {
+        "id":  "400",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一架飞机在一片长125 千米，宽 16 米的防护林带上空喷洒农药。这片防护林带的面积是多少平方米。合多少公顷。合多少平方千米。",
+        "options":  [
+
+                    ],
+        "answer":  "2000000 平方米;200 公顷;2 平方千米",
+        "solution":  "【分析】125千米=125000米;面积=125000×16=2000000平方米;换算:2000000平方米=200公顷=2平方千米.\n【详解】125千米=125000米,125000×16=2000000平方米;2000000平方米=200公顷;200公顷=2平方千米。答:2000000平方米，合200公顷，合2平方千米。",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "千米和米之间的进率与换算",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25四年级上·广东佛山·期中"
+    },
+    {
+        "id":  "401",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一个长方形果园长200 米，宽 100 米，扩建后长增加到原来的 2 倍，宽不变，扩建后的面积是( )公顷。",
+        "options":  [
+                        "A. 4",
+                        "B. 2",
+                        "C. 40000"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】长方形果园长200米，宽100米，扩建后长增加到原来的2倍，则扩建后果园的长为原来的200米乘2，宽为100米，根据长方形的面积 = 长×宽，即可计算长方形果园扩建后的面积.\n【详解】 200×2 = 400 (米) \n400×100 = 40000 (平 方 米)\n40000 平方米 = 4​ 公顷，扩建后的面积是 4 公顷。故答案为:A",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "长方形的面积"
+                             ],
+        "source":  "25-26 四年级上·黑龙江鹤岗·期中"
+    },
+    {
+        "id":  "402",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面说法正确的是()。",
+        "options":  [
+                        "A. 边长是 100 米的正方形的面积是 1 平方千米",
+                        "B. 平方分米和平方米，公顷和平方千米之间的进率都是100",
+                        "C. 1 平方千米比 1 公顷大 9999 公顷",
+                        "D. 81000 平方米 = 81 公顷"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据正方形的面积 = 边长×边长，计算出边长为100 米的正方形的面积，再根据 1 平方千米 = 1000000 平方米，将求出的面积的单位换算为平方千米，然后再判断即可;1平方米 = 100 平方分米,1平方千米 = 100 公顷，据此判断即可;1平方千米 = 100 公顷，用100减去1，求出1平方千米比1公顷大多少公顷，再进一步判断即可;1 公顷 = 10000 平方米,81 公顷就是 81 个 10000 平方米，是810000 平方米;据此先统一单位，再比较大小即可.\n【详解】A。 100×100 = 10000，即边长是 100 米的正方形的面积是10000平方米,1平方千米 = 1000000 平方米, 1000000\u003e10000,1 平方千米\u003e10000 平方米，所以边长是100米的正方形的面积小于1平方千米，原题说法错误，不符合题意;B。1平方米 = 100 平方分米,1平方千米 = 100 公顷，因此平方分米和平方米，公顷和平方千米之间的进率都是 100;原题说法正确，符合题意;C。1 平方千米 = 100 公顷, 100 - 1 = 99，即 1 平方千米比 1公顷大 99 公顷;原题说法错误，不符合题意;D。81 公顷 = 810000 平方米,810000 平方米\u003e81000 平方米，即81000平方米 != 81 公顷;原题说法错误，不符合题意。故答案为:B",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·辽宁鞍山·月考"
+    },
+    {
+        "id":  "403",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.94,
+        "question":  "我们教室的占地面积是 72( )。",
+        "options":  [
+                        "A. 平方米",
+                        "B. 公顷",
+                        "C. 平方千米",
+                        "D. 平方分米"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】根据我们的生活经验，边长是 1 米的正方形的面积是1平方米，双人课桌面的面积大约是1平方米;边长是100米的正方形的面积是1公顷,1公顷比标准足球场大一些,400米跑道围起来的部分的面积大约是是1公顷;边长是1千米的正方形的面积是1平方千米;1平方分米大约是个手掌面的大小;结合数据72，所以计量我们教室的占地面积用平方米作单位比较合适。据此解答.\n【详解】由分析可知，我们教室的占地面积是72平方米。故答案为:A",
+        "knowledge_points":  [
+                                 "面积单位的选择",
+                                 "公顷，平方千米的认识",
+                                 "平方厘米，平方分米，平方米的认识"
+                             ],
+        "source":  "24-25 四年级上·湖南永州·期中"
+    },
+    {
+        "id":  "404",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "李大伯家的长方形苗圃面积大约有 8 公顷，它的长是 400 米，宽是( )米。",
+        "options":  [
+                        "A. 200",
+                        "B. 2",
+                        "C. 2000",
+                        "D. 20"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】根据长方形面积公式，面积 = 长×宽，已知面积和长，求宽需用面积除以长。注意单位换算,1公顷 = 10000 平方米，将8公顷转换为80000平方米后计算.\n【详解】8 公顷 = 8×10000 平方米 = 80000 平方米;根据长方形面积公式，面积 = 长×宽，已知面积和长;长方形苗圃的宽: 80000÷400 = 200 (米)。故答案为:A",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25四年级上·浙江宁波·期中"
+    },
+    {
+        "id":  "405",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "1□5平方千米\u003c16600 公顷,□里可以填( )。",
+        "options":  [
+                        "A. 0~4",
+                        "B. 5~9",
+                        "C. 1~9",
+                        "D. 0~6"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】根据1平方千米 = 100 公顷可知,16600公顷 = 166 平方千米,1□5和166都是三位数，百位上都是1，所以要比较十位上的数，当十位上的数小于6时，符合1□5\u003c166;当十位上是 6 时,1□5 的个位上是 5,166 个位上是6, 5 \u003c 6，符合 1□5\u003c166;所以□里可以填 0 ~ 6;据此解答.\n【详解】根据分析可知:1□5平方千米\u003c16600公顷,□里可以填 0~6 故答案为:D",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25四年级上·福建福州·期中"
+    },
+    {
+        "id":  "406",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一块长方形菜地长200米，宽50米，如果每公顷施化肥120千克，共需要化肥( )千克。",
+        "options":  [
+                        "A. 120",
+                        "B. 12",
+                        "C. 120000",
+                        "D. 1200000"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】根据长方形的面积 = 长×宽，计算出菜地的面积。再根据 1 公顷 = 10000 平方米，将菜地的面积转换成公顷。最后用每公顷施化肥的千克数乘菜地的面积，即可求出化肥的总重量.\n【详解】 200×50 = 10000 (平方米)10000 平方米 = 1 公顷120×1 = 120 (千克)故答案为:A",
+        "knowledge_points":  [
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "长方形的面积",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·浙江杭州·阶段练习"
+    },
+    {
+        "id":  "407",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面选项中，最接近1 公顷的是( )。",
+        "options":  [
+                        "A. 杭州市的占地面积",
+                        "B. 12 个小朋友手拉手围成的正方形的面积",
+                        "C. 一个标准游泳池的面积",
+                        "D. 200间教室的面积"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】根据题意，常用的面积单位有平方千米，公顷，平方米，平方分米，平方厘米等。1 公顷 = 10000 平方米;例如:400米跑道围起来的面积大约是1公顷,1平方厘米大约是一个手指甲的面积,1平方分米大约是一个手掌面的大小;1 平方米大约是一个电视机的面积,100 个标准足球场的面积大约是1平方千米;根据生活实际，计量杭州市的占地面积用平方千米作单位,12 个小朋友手拉手围成正方形的面积用平方米作单位合适;游泳池大约 200 平方米;200 间教室的面积大约是 1 公顷;以此选择即可.\n【详解】根据分析可知:A。杭州市的占地面积一般用平方千米作单位，不符合题意。 B。12个小朋友手拉手围成正方形的面积，用平方米作单位，不符合题意。 C。一个标准游泳池面积，大约是200平方米，不符合题意。D。200 间教室面积大约是 1 公顷，符合题意。故答案为:D",
+        "knowledge_points":  [
+                                 "面积单位的选择",
+                                 "公顷，平方千米的认识",
+                                 "平方厘米，平方分米，平方米的认识"
+                             ],
+        "source":  "24-25四年级上·浙江杭州·期中"
+    },
+    {
+        "id":  "408",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "一块正方形地的边长是 200 米,( )块这样的正方形地的占地面积是 1 平方千米。",
+        "options":  [
+                        "A. 3",
+                        "B. 25",
+                        "C. 250"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据正方形的面积公式 = 边长×边长，代入数据即可求出一块正方形地的面积。因为1平方千米=1000000平方米，求几块这样的地的面积正好是1平方千米，用除法解答.\n【详解】 200×200 = 40000 (平方米)1 平方千米 = 1000000 平方米1000000÷40000 = 25 (块)块正方形地的边长是 200 米,25 块这样的正方形地的占地面积是 1 平方千米。故答案为:B",
+        "knowledge_points":  [
+                                 "正方形的面积",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "24-25 四年级上·甘肃金昌·期中"
+    },
+    {
+        "id":  "409",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "有两块正方形土地，第一块的边长是 200 米，第二块的面积是 3 公顷，这两块土地的面积相比,( )。",
+        "options":  [
+                        "A. 第一块大",
+                        "B. 第二块大",
+                        "C. 一样大"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】正方形的面积 = 边长×边长。由题意得，第一块正方形土地的边长是200米，直接将数据代入可以算出第一块正方形土地的面积，然后根据10000平方米 = 1 公顷转化为多少公顷。第二块土地的面积是 3 公顷，然后直接比较两块土地的面积大小即可.\n【详解】 200×200 = 40000 (平方米)10000 平方米 = 1 公顷，所以 40000 平方米 = 4 公顷。4 公顷 { \u003e } 3 公顷，所以第一块土地的面积更大。故答案为:A",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "正方形的面积"
+                             ],
+        "source":  "25-26 四年级上·山东临沂·阶段练习"
+    },
+    {
+        "id":  "410",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.94,
+        "question":  "公顷和平方米这两个面积单位之间的进率是( )。",
+        "options":  [
+                        "A. 10000",
+                        "B. 100",
+                        "C. 1000"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】根据 1 公顷 = 10000 平方米，据此作答.\n【详解】公顷和平方米这两个面积单位之间的进率是10000。故答案为:A",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·广西柳州·期中"
+    },
+    {
+        "id":  "411",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面面积中最大的是( )。",
+        "options":  [
+                        "A. 9000000 平方分米",
+                        "B. 789 公顷",
+                        "C. 8000 平方米",
+                        "D. 8平方千米"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】1平方米 = 100 平方分米,1平方千米 = 100 公顷,1公顷 = 10000 平方米,1平方千米 = 1000000 平方米。根据面积单位间的进率，先将选项中的面积单位都统一为平方米，然后再进行比较，即可解答。大数的比较，先比较数的位数，位数多的数就大，位数相同，再从高位向低位依次比较。据此解答.\n【详解】A。100平方分米 = 1 平方米,9000000平方分米=90000 平方米;B。1 公顷 = 10000 平方米,789 公顷 = 7890000 平方米;C。8000 平方米的单位是平方米，不需换算;D。1 平方千米 = 1000000 平方米,8 平方千米 = 8000000 平方米。8000000 和 7890000 都是七位数, 8 \u003e 7，所以 8000000\u003e7890000;90000 是五位数,8000 是四位数，所以这四个数的大小关系为: 8000000 \u003e 7890000 \u003e 90000 \u003e 80000，即 8 平方千米\u003e789 公顷\u003e9000000 平方分米\u003e8000 平方米。故答案为:D",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "公顷，平方千米的进率与换算",
+                                 "大数的比较"
+                             ],
+        "source":  "24-25 四年级上·云南楚雄·期中"
+    },
+    {
+        "id":  "412",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "某学校有一块面积为 4 公顷的长方形劳动基地，这块劳动基地的长为 400 米，那么它的周长是( )。",
+        "options":  [
+                        "A. 800 米",
+                        "B. 200 米",
+                        "C. 1千米",
+                        "D. 1600 米"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】先将公顷转换为平方米,1公顷 = 10000 平方米，再利用长方形面积公式：面积 = 长×宽，反推出宽的长度，即宽 = 面积/长。再将长和宽代入周长公式：周长 = 2×(长+宽),得到最终结果.\n【详解】4公顷 = 40000 平方米宽: 40000÷400 = 100 (米)周长: 2×( 400 + 100 ) \\ = 2×500 = 1000 (米)1000 米 = 1 千米故答案为:C",
+        "knowledge_points":  [
+                                 "长方形的周长",
+                                 "长方形的面积",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "24-25 四年级上·贵州六盘水·期中"
+    },
+    {
+        "id":  "413",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "世界最大的广场是我国首都的天安门广场，占地大约 40( )。",
+        "options":  [
+                        "A. 平方千米",
+                        "B. 平方米",
+                        "C. 公顷"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】常见的面积单位中，平方米适用于较小场所，公顷适用于较大区域如广场，平方千米适用于城市等极大范围，据此选择.\n【详解】根据分析可得：根据面积单位的实际应用，天安门广场作为大型广场，应选用公顷作为单位比较合适，所以天安门广场占地约40公顷。故答案为:C",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 四年级上·陕西安康·期中"
+    },
+    {
+        "id":  "414",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "学校准备建一个足球场，下面三块地的面积选( )比较合适。",
+        "options":  [
+                        "A. 100 平方分米",
+                        "B. 100 平方米",
+                        "C. 1公顷"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】面积的常用单位有公顷，平方千米，平方米，平方分米等,1公顷比一个标准足球场稍大一点,1平方千米大约是小型湖泊的湖面面积,1平方米大约1张方桌面的大小,1平方分米大约1个成人手掌面的大小，据此解答.\n【详解】A。100 平方分米也就是 100 个成人手掌面的大小，足球场的面积要大于100平方分米，不符合题意;B。100平方米也就是100张方桌面的大小，足球场的面积要大于100平方米，不符合题意;C。1公顷也就是一个标准足球场的大小，学校足球场的面积选择面积为1公顷的地比较合适，符合题意。选1公顷比较合适。故答案为:C",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "24-25 四年级上·江西九江·期中"
+    },
+    {
+        "id":  "415",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "边长是1000米的正方形土地面积是1( )。",
+        "options":  [
+                        "A. 平方米",
+                        "B. 公顷",
+                        "C. 平方千米"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】1000米 = 1 千米，根据正方形的面积 = 边长×边长求出土地面积.\n【详解】1000 米 = 1 千米1×1 = 1 (平方千米)所以边长是1000米的正方形土地面积是1平方千米。故答案为:C",
+        "knowledge_points":  [
+                                 "正方形的面积",
+                                 "千米和米之间的进率与换算",
+                                 "公顷，平方千米的认识"
+                             ],
+        "source":  "24-25四年级上·湖南益阳·期中"
+    },
+    {
+        "id":  "416",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一所小学的占地面积大约是( )。",
+        "options":  [
+                        "A. 30平方米",
+                        "B. 3公顷",
+                        "C. 3000平方千米"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】面积的常用单位有公顷，平方千米，平方米，平方分米等,1公顷比一个标准足球场稍大一点,1平方千米大约是小型湖泊的湖面面积,1 平方米大约 1 张方桌面的大小,1平方分米大约1个成人手掌面的大小，据此解答.\n【详解】A。30平方米也就是30张方桌面的大小，一所小学的占地面积大于30平方米，不符合题意;B。3公顷也就是比3个标准足球场稍大一点，一所小学的占地面积大约是3公顷，符合题意;C。3000平方千米也就是3000个小型湖泊的湖面面积，一所小学的占地面积没有3000平方千米大，不符合题意。占地面积大约是3公顷。故答案为:B",
+        "knowledge_points":  [
+                                 "面积单位的选择",
+                                 "公顷，平方千米的认识"
+                             ],
+        "source":  "24-25四年级上·贵州铜仁·期中"
+    },
+    {
+        "id":  "417",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "中南大学湘雅医院的占地面积约为 140000( )。",
+        "options":  [
+                        "A. 公顷",
+                        "B. 平方分米",
+                        "C. 平方米",
+                        "D. 平方千米"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】根据生活经验，以及对面积单位和数据大小的认识，可知：测量稍小图形的面积，通常用平方分米作单位，边长1分米的正方形，面积是1平方分米，如：一个手掌面的面积约1平方分米;测量较大图形的面积，通常用平方米作单位，边长1米的正方形，面积是1平方米，如：方桌桌面的面积约1平方米;公顷适合计量稍大的土地面积，如计量天安门广场面积，校园面积等;边长是100米的正方形的面积是1公顷,400米跑道围起来的部分的面积大约是1公顷;计量比较大的土地面积，常用平方千米作单位，如计量国土面积，城市面积等;边长是 1 千米的正方形的面积是 1 平方千米，一个足球场的面积大约是7000平方米,140个这样的足球场的面积和大约是1平方千米;所以结合数据140000可知，计量中南大学湘雅医院的占地面积应用\"平方米\"作单位比较合适;据此解答.\n【详解】据分析可得：中南大学湘雅医院的占地面积约为140000平方米。故答案为:C",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "24-25四年级上·湖南长沙·期中"
+    },
+    {
+        "id":  "418",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "李明沿着面积是 9公顷的正方形公园跑了一圈，他跑了( )米。",
+        "options":  [
+                        "A. 300",
+                        "B. 1200",
+                        "C. 900"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据题意，明确平方米与公顷之间的进率是10000，先把 9 公顷换算成 90000 平方米，根据正方形的面积 = 边长×边长，据此求出正方形公园的边长，然后根据正方形的周长 = 边长 * 4，求出公园的周长即可.\n【详解】根据分析可知:1公顷 = 10000 平方米9公顷 = 90000 平方米因为 300×300 = 90000 (平方米)正方形公园的边长为300米。300×4 = 1200 (米)李明沿着面积是9公顷的正方形公园跑了一圈，他跑了1200 米。故答案为:B",
+        "knowledge_points":  [
+                                 "正方形的周长",
+                                 "正方形的面积",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·陕西咸阳·期中"
+    },
+    {
+        "id":  "419",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "有两块长方形地，第一块地的面积是 6 公顷，第二块地的长是 120 米，宽是40 米，这两块地的面积相比较,( )。",
+        "options":  [
+                        "A. 面积相等",
+                        "B. 第一块地面积大",
+                        "C. 第二块地面积大"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】比较两块地的面积，需统一单位。第一块地面积已知为 6 公顷，第二块地面积需通过长×宽计算出面积，再根据1公顷 = 10000 平方米将单位换算统一后再比较.\n【详解】6 公顷 = 60000 平方米 第二块地的面积: 120×40 = 4800 (平方米)60000 平方米\u003e4800 平方米这两块地的面积相比较，第一块地的面积大。故答案为:B",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·陕西咸阳·期中"
+    },
+    {
+        "id":  "420",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "8 平方千米=( )公顷 130000 平方米=( )公顷 6000 公顷=( )平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "800;13;60",
+        "solution":  "【分析】根据1平方千米 = 100 公顷进行单位换算;根据 10000 平方米 = 1 公顷进行单位换算;根据 100 公顷 = 1 平方千米进行单位换算.\n【详解】8 平方千米 = 800 公顷;130000 里有 13 个 10000，所以 130000 平方米 = 13 公顷;6000 里有 60 个 100，所以 6000 公顷 = 60 平方千米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·甘肃庆阳·期中"
+    },
+    {
+        "id":  "421",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "我国某自然保护区总面积约为120000公顷，合( )平方千米;某湿地公园占地面积约 3 平方千米，合( )平方米。",
+        "options":  [
+
+                    ],
+        "answer":  "1200;3000000",
+        "solution":  "【分析】由题意得,100公顷 = 1 平方千米，求120000公顷=( )平方千米，可以先用除法求出 120000 公顷里面有多少个100公顷，那么就对应着多少个1平方千米;1平方千米 = 1000000 平方米，求3平方千米 = ( )平方米，可以想3平方千米里面有多少个1平方千米，那么就对应着多少个 1000000 平方米.\n【详解】 120000÷100 = 1200，所以 120000 公顷 = 1200 平方千米;1平方千米 = 1000000 平方米，所以3平方千米 = 3000000 平方米。我国某自然保护区总面积约为 120000 公顷，合 1200 平方千米;某湿地公园占地面积约 3 平方千米，合 3000000 平方米。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26四年级上·黑龙江鹤岗·期中"
+    },
+    {
+        "id":  "422",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "650 公顷 = ( )平方米 30 平方千米=( )公顷800000 平方米 = ( )公顷 9000 公顷=( )平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "6500000;3000;80;90",
+        "solution":  "【分析】根据题意，明确单位间的进率,1公顷 = 10000 平方米,1平方千米 = 100 公顷，换算时注意单位大小：大单位转小单位时乘进率，小单位转大单位时除进率。以此答题即可.\n【详解】根据分析可知:650 公顷 = 6500000 平方米30平方千米 = 3000 公顷800000 平方米 = 80 公顷 9000 公顷 = 90 平方千米",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期中"
+    },
+    {
+        "id":  "423",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个长方形果园，长是 400 米，宽是 250 米，这个果园占地面积是( )平方米，合( )公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "100000;10",
+        "solution":  "【分析】长方形的面积 = 长×宽。由题意得，一个长方形果园的长是400米，宽是250米，直接将数据代入可以算出长方形果园的面积，然后根据10000平方米 = 1 公顷将其转化为多少公顷。`r`n【详解】400×250 = 100000（平方米），10000 平方米 = 1 公顷，所以100000平方米 = 10 公顷。一个长方形果园，长是400米，宽是250米，这个果园占地面积是100000平方米，合10公顷。",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·湖南永州·月考"
+    },
+    {
+        "id":  "424",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填合适的单位.\n\n(1)学校球场占地面积是 2( )。 \n(2)永州市占地面积 22441( )。",
+        "options":  [
+
+                    ],
+        "answer":  "公顷;平方千米",
+        "solution":  "【分析】计量较大的土地面积，常用公顷和平方千米做单位。边长是100米的正方形的面积为1公顷，边长是1千米的正方形的面积是 1 平方千米。(1)计量公园或者球场的面积，常用公顷做单位。由题意得，学校球场的占地面积很大，所以括号里填公顷比较合适。 (2)计量一个市或一个省的面积，常用平方千米做单位。由题意得，永州市的占地面积特别大，所以括号里填平方千米比较合适.\n【详解】(1)学校球场占地面积是 2 公顷。(2)永州市占地面积 22441 平方千米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 四年级上·湖南永州·月考"
+    },
+    {
+        "id":  "425",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "2 平方千米=( )公顷=( )平方米40000 平方米=( )公顷 600 公顷=( )平方千米。",
+        "options":  [
+
+                    ],
+        "answer":  "200;2000000;4;6",
+        "solution":  "【分析】1平方千米 = 100 公顷 = 1000000 平方米。2平方千米是 2 个 100 公顷，是 2 个 1000000 平方米。1公顷 = 10000 平方米,40000平方米是4个1公顷。1平方千米 = 100 公顷,600公顷是6个1平方千米.\n【详解】2平方千米是2个100公顷，是200公顷。也是2个 1000000 平方米，是 2000000 平方米。所以 2 平方千米=200 公顷=2000000 平方米。 40000 平方米是 4 个 1 公顷，是 4 公顷。所以 40000 平方米= 4 公顷。 600公顷是6个1平方千米，是6平方千米。所以600公顷= 6 平方千米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·湖南永州·月考"
+    },
+    {
+        "id":  "426",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "单位换算。 \n3 公顷=( )平方米 6 平方千米-78 公顷=( )公顷 \n8 平方千米=( )平方米 9 平方千米+45 公顷=( )公顷 \n6 平方米+3 公顷=( )平方米 7 公顷 500 平方米=()平方米",
+        "options":  [
+
+                    ],
+        "answer":  "30000;522;8000000;945;30006;70500",
+        "solution":  "【分析】1 公顷 = 10000 平方米，在 3 公顷末尾添上 4 个 0即可换算成平方米为单位;1平方千米 = 100 公顷，在6平方千米末尾添上2个0换算成公顷为单位再计算即可;1平方千米 = 1000000 平方米，在8平方千米末尾添上6个0即可换算成平方米为单位;在9平方千米末尾添上2个0换算成公顷为单位再计算即可;在 3 公顷末尾添上 4 个 0 换算成平方米为单位再计算即可;在 7 公顷末尾添上 4 个 0 换算成平方米为单位再加上 500 平方米即可.\n【详解】3 公顷 = 30000 平方米;6 平方千米-78 公顷 = 600 公顷-78 公顷 = 522 公顷;8 平方千米 = 8000000 平方米;9平方千米+45公顷 = 900 公顷+45公顷 = 945 公顷;6平方米 + 3 公顷 = 6 平方米+30000平方米 = 30006 平方米;7公顷500平方米 = 70000 平方米+500平方米 = 70500 平方米。 3公顷 = 30000 平方米;6平方千米-78公顷 = 522 公顷;8平方千米 = 8000000 平方米;9平方千米+45公顷 = 945 公顷;6平方米 + 3 公顷 = 30006 平方米;7公顷500平方米 = 70500 平方米。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·黑龙江佳木斯·期中"
+    },
+    {
+        "id":  "427",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "单位换算。8000000 平方米=( )公顷 600 公顷=( )平方千米( )平方千米 = 500 公顷 20 平方千米 = ( )平方米",
+        "options":  [
+
+                    ],
+        "answer":  "800;6;5;20000000",
+        "solution":  "【分析】1公顷 = 10000 平方米,1平方千米 = 100 公顷,1平方千米 = 1000000 平方米，据此进行单位换算即可.\n【详解】8000000 平方米 = 800 公顷 600 公顷 = 6 平方千米 5 平方千米 = 500 公顷 20 平方千米 = 20000000 平方米",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·黑龙江齐齐哈尔·期中"
+    },
+    {
+        "id":  "428",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上\"\u003e\"\"\u003c\"或 \" = \" 。 \n6 公顷( )5900 平方米 4 平方千米( )404 公顷 \n3 公顷 4 平方米( )34000 平方米 2900 平方米( )30公顷",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;\u003c;\u003c;\u003c",
+        "solution":  "【分析】1公顷 = 10000 平方米,1平方千米 = 100 公顷，据此将不同单位换算成同一单位后再进行比较即可.\n【详解】6 公顷 = 60000 平方米,60000 平方米\u003e5900 平方米，所以 6 公顷\u003e5900 平方米;4 平方千米 = 400 公顷, 400 \u003c 404，所以 4 平方千米\u003c404公顷;3公顷4平方米 = 30000 平方米 + 4 平方米 = 30004 平方米,30004 \u003c 34000，所以 3 公顷 4 平方米\u003c34000 平方米;30 公顷 = 300000 平方米, 2900 \u003c 300000，所以 2900 平方米\u003c30公顷。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26四年级上·黑龙江齐齐哈尔·期中"
+    },
+    {
+        "id":  "429",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一台小麦收割机每小时行驶4千米，收割宽度是2米，这台收割机工作 5 小时，可以收割小麦( )公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "4",
+        "solution":  "【分析】1 千米 = 1000 米，所以 4 千米 = 4000 米，收割机收割的面积是长方形，据此先用1小时收割的长度×宽度计算出1小时收割的面积，然后再乘5，最后根据1公顷=10000 平方米进行单位换算.\n【详解】4千米 = 4000 米\n 4000×2×5 = 8000×5 = 40000 (平 方 米) \n40000 平方米 = 4 公顷\n\n一台小麦收割机每小时行驶4千米，收割宽度是2米，这台收割机工作5小时，可以收割小麦4公顷。",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "千米和米之间的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "24-25 四年级上·湖北省直辖县级单位·期中"
+    },
+    {
+        "id":  "430",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "20 公顷=( )平方米 500 公顷=( )平方千米46000000 平方米=( )平方千米 850000 平方米=( )公顷",
+        "options":  [
+
+                    ],
+        "answer":  "200000;5;46;85",
+        "solution":  "【分析】1 公顷 = 10000 平方米,1 平方千米 = 100 公顷,1平方千米 = 1000000 平方米，高级单位化成低级单位就乘它们之间的进率，低级单位化成高级单位就除以它们之间的进率，依此换算.\n【详解】(1) 20 times10000 = 200000，即 20 公顷 = 200000 平方米;(2) 500÷100 = 5 ·s 5，即 500 公顷 = 5 平方千米;(3) 46000000÷1000000 = 46，即 46000000 平方米 = 46 平方千米;(4) 850000÷10000 = 85，即 850000 平方米 = 85 公顷。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·湖南永州·期中"
+    },
+    {
+        "id":  "431",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "8 公顷=( )平方米 5500 公顷=( )平方千米160000 平方米=( )公顷 300 平方千米=( )公顷",
+        "options":  [
+
+                    ],
+        "answer":  "80000;55;16;30000",
+        "solution":  "【分析】公顷换算成平方米，大单位换算成小单位，乘进率 10000, 8 times10000 = 80000，所以 8 公顷 = 80000 平方米;根据1平方千米 = 100 公顷可知,5500公顷里面有55个100公顷，即5500公顷里面有55个1平方千米，所以5500公顷 = 55 平方千米;根据1公顷 = 10000 平方米可知,160000平方米里面有16个10000平方米，即160000平方米里面有16个1公顷，所以 160000 平方米 = 16 公顷;根据1平方千米 = 100 公顷可知,300平方千米里面有300个1平方千米，即300平方千米里面有300个100公顷，就是30000公顷，所以300平方千米 = 30000 公顷.\n【详解】根据分析可知:8 公顷 = 80000 平方米5500 公顷 = 55 平方千米160000 平方米 = 16 公顷300 平方千米 = 30000 公顷",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·宁夏银川·期中"
+    },
+    {
+        "id":  "432",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在( )里填上适当的面积单位.\n\n(1)海南省陆地总面积约 35400( )。 \n(2)天坛公园是中国现存最大的古代祭祀性建筑群，占地面积约 273( )。 \n(3)青海湖是世界上海拔最高的湖泊之一，水体总面积约4626( )。 \n(4)一间舞蹈教室的面积约 120( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方千米;公顷;平方千米;平方米",
+        "solution":  "【分析】(1)根据生活经验，对面积单位和数据大小的认识，省份的面积比较大，一般用平方千米作单位，因此海南省陆地总面积约35400用平方千米作单位比较合适;(2)根据生活经验，对面积单位和数据大小的认识，生活中边长为100米的正方形面积是1公顷，因此天坛公园占地面积约 273 用公顷作单位比较合适;(3)根据生活经验，对面积单位和数据大小的认识，青海湖是世界上海拔最高的湖泊之一，因此水体总面积约4626用平方千米作单位比较合适;(4)根据生活经验，对面积单位和数据大小的认识，生活中的方桌面积大约是1平方米，因此一间舞蹈教室的面积约120用平方米作单位比较合适.\n【详解】(1)海南省陆地总面积约 35400 平方千米。(2)天坛公园是中国现存最大的古代祭祀性建筑群，占地面积约273公顷。(3)青海湖是世界上海拔最高的湖泊之一，水体总面积约4626平方千米。 (4)一间舞蹈教室的面积约 120 平方米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "24-25四年级上·宁夏银川·期中"
+    },
+    {
+        "id":  "433",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "8000000 平方米 = ( )平方千米;300 公顷+900 公顷 = ( )平方千米。",
+        "options":  [
+
+                    ],
+        "answer":  "8;12",
+        "solution":  "【分析】根据分析可得,8000000平方米 = 8 平方千米300 公顷+900 公顷 = 12 平方千米\n【详解】(1)低级单位平方米化高级单位平方千米除以进率 1000000。 (2)300 公顷+900 公顷 = 1200 公顷，再把公顷除以进率100 化成平方千米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·宁夏石嘴山·期中"
+    },
+    {
+        "id":  "434",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "黑龙江扎龙国家级自然保护区是世界上面积最大的野生丹顶鹤栖息繁殖地，它的总面积约 210000 公顷。210000 公顷=( )平方千米=( )平方米",
+        "options":  [
+
+                    ],
+        "answer":  "2100;2100000000",
+        "solution":  "【分析】根据单位进率进行计算:1平方千米 = 100 公顷,1公顷 = 10000 平方米。将公顷换算成平方千米时，需要除以进率 100;将公顷换算成平方米时，需要乘以进率 10000。以此答题即可.\n【详解】根据分析可知:1 平方千米 = 100 公顷 1 公顷 = 10000 平方米 210000 公顷 = 2100 平方千米=2100000000 平方米",
+        "knowledge_points":  [
+                                 "公顷",
+                                 "平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "24-25 四年级上·江西南昌·期中"
+    },
+    {
+        "id":  "435",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里面填上合适的数.\n\n50 平方千米=( )公顷=( )平方米 780 公顷=( )平方千米( )公顷 6 平方千米 3 公顷=( )公顷",
+        "options":  [
+
+                    ],
+        "answer":  "5000;50000000;7;80;603",
+        "solution":  "【分析】把50平方千米换算成公顷和平方米，因为1平方千米 = 100 公顷,50平方千米里面有50个100公顷，即5000 公顷;因为 1 平方千米 = 1000000 平方米,50 平方千米里面有 50 个 1000000 平方米，即 50000000 平方米;把780公顷换算成平方千米和公顷，因为1平方千米 = 100 公顷,700 公顷里面有 7 个 100 公顷，即 7 平方千米;把 6 平方千米 3 公顷换算成公顷，因为 1 平方千米 = 100 公顷,6 平方千米里面有 6 个 100 公顷，即 600 公顷，再加上3公顷是603公顷.\n【详解】50 平方千米 = 5000 公顷 = 50000000 平方米;780 公顷 = 7 平方千米 80 公顷;6 平方千米 3 公顷 = 603 公顷。",
+        "knowledge_points":  [
+                                 "公顷",
+                                 "平方千米的进率与换算"
+                             ],
+        "source":  "24-25四年级上·福建福州·期中"
+    },
+    {
+        "id":  "436",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上合适的数。12 平方千米=( )公顷7 公顷=( )平方米9000000 平方米=( )公顷=( )平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "1200;70000;900;9",
+        "solution":  "【分析】根据1平方千米 = 100 公顷,1公顷 = 10000 平方米，高级单位转换成低级单位乘进率，低级单位转换成高级单位除以进率，据此解答即可.\n【详解】1平方千米 = 100 公顷,12个100公顷是1200公顷，所以 12 平方千米 = 1200 公顷;7 公顷 = 7×10000 = 70000 平方米;9000000 平方米里面含有 900 个 10000 平方米,9000000 平方米里面含有9个1000000平方米，所以9000000平方米=900 公顷 = 9 平方千米。",
+        "knowledge_points":  [
+                                 "公顷",
+                                 "平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·新疆喀什·期中"
+    },
+    {
+        "id":  "437",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "填上适当的数.\n\n30 平方米=( )平方分米 3 公顷 800 平方米=( )平方米\n\n120000000 平方米=( )平方千米 4000000 平方米=()公顷=( )平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "3000;30800;120;400;4",
+        "solution":  "【分析】1平方米 = 100 平方分米,30平方米就是30个100平方分米;1 公顷 = 10000 平方米,3 公顷就是 3 个 10000 平方米，是30000 平方米，再加上 800 平方米，即可解答;1平方千米 = 1000000 平方米,120000000平方米里面有几个 1000000 平方米，就是几平方千米;1 公顷 = 10000 平方米,4000000 平方米里面有几个 10000平方米，就是几公顷;1平方千米=1000000平方米,4000000平方米里面有几个1000000平方米，就是几平方千米.\n【详解】30 个 100 平方分米是 3000，即 30 平方米 = 3000 平方分米;3 公顷 = 30000 平方米, 30000 + 800 = 30800，即 3 公顷 800平方米 = 30800 平方米;120000000 平方米里面有 120 个 1000000 平方米，即120000000 平方米 = 120 平方千米;4000000 平方米里面有 400 个 10000 平方米,4000000 平方米 = 400 公顷,4000000 平方米里面有 4 个 1000000 平方米,4000000 平方米 = 4 平方千米;即 4000000 平方米 = 400 公顷 = 4 平方千米。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "公顷",
+                                 "平方千米的进率与换算",
+                                 "平方厘米",
+                                 "平方分米",
+                                 "平方米之间的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·广东汕尾·期中"
+    },
+    {
+        "id":  "438",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上合适的单位.\n\n西湖的面积大约 650( ) 一间教室的面积大约是54( )\n\n绕操场一周大约是 400( ) 余杭区的面积大约 942( )",
+        "options":  [
+
+                    ],
+        "answer":  "公顷;平方米;米;平方千米",
+        "solution":  "【分析】常用的面积单位有平方厘米，平方分米，平方米。我们知道，指甲盖的面积大约1平方厘米，手掌的面积大约1平方分米，一块地板砖的面积大约1平方米。计量较大的土地面积，常用公顷和平方千米做单位。边长是100米的正方形的面积为1公顷，边长是1千米的正方形的面积是1平方千米;常用的长度单位有毫米，厘米，分米，米和千米。我们知道，银行卡的厚度大约1毫米，食指宽大约1厘米，支笔的长度约1分米，小学生展开双臂大约1米，连续步行 10 分钟所走的路程大概就是 1 千米。(1)由题意得，西湖的面积很大。计量公园的面积，常用公顷做单位。(2)由题意得，一间教室的面积较大，比一块地砖的面积大很多，所以选择平方米做单位比较合适。 (3)由题意得，绕操场一周的距离较长，但不到1千米，所以选择米做单位比较合适。 (4)由题意得，余杭区的面积特别大。计量行政区域的面积，常用平方千米做单位.\n【详解】由分析可得：西湖的面积大约650公顷;一间教室的面积大约是 54 平方米绕操场一周大约是 400 米;余杭区的面积大约 942 平方千米",
+        "knowledge_points":  [
+                                 "长度单位的选择",
+                                 "公顷",
+                                 "平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 四年级上·浙江杭州·阶段练习"
+    },
+    {
+        "id":  "439",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "在( )填合适的单位.\n\n(1)开福区的面积为 188( )。 \n(2)长沙烈士公园的面积为 153.3( )。 \n(3)阳光小学的操场面积大约是 3000( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方千米;公顷;平方米",
+        "solution":  "【分析】根据生活经验和对面积单位的认识可知,1平方千米相当于2-3个普通乡镇的核心区域面积，适用于衡量城市，区县，湖泊等大范围区域的面积;1公顷相当于2个大型超市的占地面积，主要用于计量公园，农场，校园等中等规模的场地;1平方米相当于1张标准单人课桌的桌面面积，常用于衡量房间，桌面，小型场地等小范围空间，据此选择合适的面积单位即可.\n【详解】(1)行政区域(如区，县)面积范围大，通常用平方千米作单位。开福区的面积为188平方千米。(2)大型公园，景区面积常用公顷计量,1公顷 = 10000 平方米,153.3 公顷约 1533000 平方米，契合城市大型公园的面积体量。长沙烈士公园的面积为 153.3 公顷。(3)学校操场属于小范围场地，平方米是日常小型场地的常用面积单位。阳光小学的操场面积大约是 3000 平方米。",
+        "knowledge_points":  [
+                                 "公顷",
+                                 "平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "24-25 四年级上·湖南长沙·期中"
+    },
+    {
+        "id":  "440",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "40 平方千米=()公顷 120 公顷=( )平方米12000000 平方米=( )公顷=( )平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "4000;1200000;1200;12",
+        "solution":  "【分析】根据 1 平方千米 = 100 公顷,1 公顷 = 10000 平方米进行单位间的换算，大单位换算成小单位，乘进率，单位换算成大单位，除以进率;据此即可得出结果.\n【详解】40 平方千米 = 4000 公顷 120 公顷 = 1200000 平方米12000000 平方米 = 1200 公顷 = 12 平方千米",
+        "knowledge_points":  [
+                                 "公顷",
+                                 "平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "24-25 四年级上·湖南长沙·期中"
+    },
+    {
+        "id":  "441",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "5 公顷=( )平方米 3 平方千米=( )公顷\n\n4000000 平方米=( )平方千米=( )公顷",
+        "options":  [
+
+                    ],
+        "answer":  "50000;300;4;400",
+        "solution":  "【分析】1 公顷 = 10000 平方米，在 5 公顷末尾添上 4 个 0即可换算成平方米为单位;1平方千米 = 100 公顷，在3平方千米末尾添上 2 个 0 即可换算成公顷为单位;去掉4000000平方米末尾的6个0即可换算成平方千米为单位，去掉末尾的 4 个 0 即可换算成公顷为单位.\n【详解】5 公顷 = 50000 平方米;3 平方千米 = 300 公顷;4000000 平方米 = 4 平方千米 = 400 公顷。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "公顷",
+                                 "平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·山东临沂·阶段练习"
+    },
+    {
+        "id":  "442",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上合适的数。`r`n`r`n24 平方千米=( )公顷`r`n0.78 公顷=( )平方米`r`n7.5 公顷=( )公顷( )平方米",
+        "options":  [
+
+                    ],
+        "answer":  "2400;7800;7;5000",
+        "solution":  "【分析】1平方千米 = 100 公顷，24平方千米换算为公顷是：24×100 = 2400（公顷）。1公顷 = 10000 平方米，0.78公顷换算为平方米是：0.78×10000 = 7800（平方米）。7.5公顷的整数部分7就是7公顷，小数部分0.5公顷换算为平方米。因为1公顷 = 10000 平方米，所以0.5×10000 = 5000 平方米，即7.5公顷 = 7 公顷 5000 平方米。`r`n【详解】24 平方千米 = 2400 公顷；0.78 公顷 = 7800 平方米；7.5 公顷 = 7 公顷 5000 平方米。",
+        "knowledge_points":  [
+                                 "公顷",
+                                 "平方千米的进率与换算"
+                             ],
+        "source":  "24-25五年级上·河南洛阳·期中"
+    },
+    {
+        "id":  "443",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "8 公顷 20 平方米=()平方米 36 平方千米 = ( )公顷",
+        "options":  [
+
+                    ],
+        "answer":  "80020;3600",
+        "solution":  "【分析】1公顷 = 10000 平方米，先把8公顷用乘法转化为80000平方米，再加上20平方米;1平方千米 = 100 公顷，高级单位换算低级单位用乘法，用题目中的数据乘进率，即 36 times100，据此解答.\n【详解】8 公顷 = 8×10000 = 80000 平方米，则 8 公顷 20 平方米 = 80000 + 20 = 80020 平方米;36 平方千米 = 36×100 = 3600 公顷。综上所述,8 公顷 20 平方米 = 80020 平方米,36 平方千米= 3600 公顷。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "公顷",
+                                 "平方千米的进率与换算"
+                             ],
+        "source":  "24-25 五年级上·河南商丘·期中"
+    },
+    {
+        "id":  "444",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在( )里填上“\u003e”“\u003c”或“=”。`r`n`r`n8087809( )8081099`r`n5 亿( )589000000`r`n300 公顷( )3000 平方米`r`n2000000 平方米( )2 平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;\u003c;\u003e;=",
+        "solution":  "【分析】比较数的大小，先看位数的多少，位数多的数就大，位数相同，从高位比较，高位上的数大则这个数大，高位上的数相同，就比较下一位，以此类推，据此比较即可；根据整数的改写，将5亿改写成500000000然后比较即可；1公顷 = 10000 平方米，在300公顷末尾添上4个0换算成平方米为单位再比较；2000000平方米和2平方千米都可以换算成公顷后再比较。`r`n【详解】8087809和8081099都是七位数，百万位、十万位、万位上的数都相同，千位上7 \u003e 1，所以8087809 \u003e 8081099；5亿 = 500000000，500000000 和 589000000 都是九位数，亿位上都是5，千万位上0 \u003c 8，所以500000000 \u003c 589000000，即5亿 \u003c 589000000；300公顷 = 3000000平方米，3000000 \u003e 3000，所以300公顷 \u003e 3000平方米；2000000平方米 = 200公顷，2平方千米 = 200公顷，所以2000000平方米 = 2平方千米。",
+        "knowledge_points":  [
+                                 "亿以上数的改写",
+                                 "面积单位间的进率及换算",
+                                 "大数的比较"
+                             ],
+        "source":  "24-25 四年级上·江西吉安·期中"
+    },
+    {
+        "id":  "445",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "请帮明明完善日记中的单位。今天，我和奶奶去自家的菜园收菜。从家到菜园的路程大约是 1000，菜园不大，面积只有 20 。收完这些菜，可把我累坏了。晚上吃了饭，我和家人去公园散步，爸爸说这个公园的面积大约有 3，也就是 30000 。",
+        "options":  [
+
+                    ],
+        "answer":  "米;平方米;公顷;平方米",
+        "solution":  "【分析】计量较长的物体一般用米作单位，二年级学生双臂展开的长度大约是1米;测量较大物体的面积，通常用平方米作单位，边长1米的正方形，面积是 1 平方米;公顷适合计量稍大的土地面积，边长是100米的正方形的面积是1公顷,400米跑道围起来的部分的面积大约是1公顷,1公顷 = 10000 平方米.\n【详解】今天，我和奶奶去自家的菜园收菜。从家到菜园的路程大约是1000米，菜园不大，面积只有20平方米。收完这些菜，可把我累坏了。晚上吃了饭，我和家人去公园散步，爸爸说这个公园的面积大约有3公顷，也就是30000平方米。",
+        "knowledge_points":  [
+                                 "长度单位的选择",
+                                 "面积单位的选择"
+                             ],
+        "source":  "24-25 四年级上·湖北黄石·期中"
+    },
+    {
+        "id":  "446",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "20 平方千米=( )公顷 90000 平方米 = ( )公顷",
+        "options":  [
+
+                    ],
+        "answer":  "2000;9",
+        "solution":  "【分析】根据1平方千米 = 100 公顷,1公顷 = 10000 平方米，据此单位换算即可.\n【详解】20平方千米中有20个1平方千米，即20个100公顷，也就是 2000 公顷，所以 20 平方千米 = 2000 公顷;90000平方米中有9个10000平方米，即9个1公顷，也就是 9 公顷，所以 90000 平方米 = 9 公顷。",
+        "knowledge_points":  [
+                                 "公顷",
+                                 "平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·广西柳州·期中"
+    },
+    {
+        "id":  "447",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "第19届亚运会于2023年9月23日在杭州开幕，其中，杭州奥体博览城规划总面积约为 6( ),分场馆温州奥体中心二期占地面积约为23( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方千米;公顷",
+        "solution":  "【分析】边长是1000米的正方形的面积是1平方千米,1平方千米=1000000平方米，计量较大的土地面积，通常用平方千米，杭州奥体博览城是较大的区域，用平方千米合适;边长是100米的正方形的面积是1公顷,1公顷等于10000平方米，公顷通常用于计量稍大的土地面积，温州奥体中心二期面积相对小一些，用公顷合适.\n【详解】第19届亚运会于2023年9月23日在杭州开幕，其中，杭州奥体博览城规划总面积约为6平方千米，分场馆温州奥体中心二期占地面积约为23公顷。",
+        "knowledge_points":  [
+                                 "公顷",
+                                 "平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "24-25四年级上·云南楚雄·期中"
+    },
+    {
+        "id":  "448",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "1公顷的水稻大约可以产出 9 吨粮食,5 平方千米的水稻大约可以产( )吨粮食。",
+        "options":  [
+
+                    ],
+        "answer":  "4500",
+        "solution":  "【分析】先进行单位换算,1平方千米 = 100 公顷,5平方千米 = 500 公顷，已知1公顷大约可以产出9吨粮食，那么500公顷就是500个9吨，利用乘法计算可解答.\n【详解】5 平方千米 = 500 公顷500×9 = 4500 (吨)5 平方千米的水稻大约可以产 4500 吨。",
+        "knowledge_points":  [
+                                 "公顷",
+                                 "平方千米的进率与换算"
+                             ],
+        "source":  "24-25四年级上·云南楚雄·期中"
+    },
+    {
+        "id":  "449",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上合适的计量单位.\n\n一张单人课桌桌面的面积约是 24( ).\n\n一头大白鲨约重 5( )\n\n绕学校操场走一圈大约 400( )\n\n天安门广场的面积约是 44( ).\n\n一间教室地面的面积约是 54( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方分米;吨;米;公顷;平方米",
+        "solution":  "【分析】常用的面积单位有平方厘米，平方分米，平方米。我们知道，指甲盖的面积大约 1 平方厘米，手掌的面积大约1 平方分米，一块地板砖的面积大约 1 平方米。计量较大的土地面积，常用公顷和平方千米做单位。边长是 100 米的正方形的面积为1公顷，边长是1千米的正方形的面积是1平方千米。计量大型场馆或公园的面积，常用公顷做单位。计量城市或行政区的面积，常用平方千米做单位;常用的长度单位有毫米，厘米，分米，米和千米。我们知道，银行卡的厚度大约1毫米，食指宽大约1厘米，一支笔的长度约1分米，小学生展开双臂，两臂的距离大约是1米，连续步行10分钟所走的路程大概就是1千米;常用的质量单位有克，千克和吨。我们知道，两枚回形针大约重1克，两瓶矿泉水大约重 1 千克，两头牛的重量大约是 1 吨。(1)由题意得，一张单人课桌桌面的面积较大，但比一块地砖的面积小一些，所以括号里填平方分米比较合适。 (2)由题意得，一头大白鲨特别重，所以括号里填吨比较合适。(3)由题意得，绕学校操场走一圈的距离较长，但不到1千米，所以括号里填米比较合适。 (4)由题意得，天安门广场的面积特别大，选择公顷做单位比较合适。 (5)由题意得，一间教室地面的面积较大，比一块地砖的面积大很多，所以括号里填平方米比较合适.\n【详解】一张单人课桌桌面的面积约是 24 平方分米。一头大白鲨约重 5 吨。绕学校操场走一圈大约400米。天安门广场的面积约是44公顷。一间教室地面的面积约是 54 平方米。",
+        "knowledge_points":  [
+                                 "质量单位的选择",
+                                 "长度单位的选择",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26四年级上·新疆乌鲁木齐·月考"
+    },
+    {
+        "id":  "450",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上合适的单位.\n\n巩义市总面积 1043( )。天安门广场是世界上最大的广场，占地约 40( )。一个会议室的面积大约是 420( )。一所小学的占地面积是 4( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方千米;公顷;平方米;公顷",
+        "solution":  "【分析】边长1米的正方形面积是1平方米。1平方厘米大约是1个大拇指甲盖这么大，而1平方分米大约是1个手掌大小,100个手掌大小大约是1平方米。常用的面积单位有平方米，平方分米，平方厘米，比较大的面积单位有公顷，平方千米。第1题，因为是给面积的数据后面填单位，所以用的是面积单位。巩义市的面积是比较大，根据这里的数据1043用平方米或者公顷作单位，都太小了，不合适，应选择平方千米作单位。第2题，天安门广场是非常大的，不可能是4000个手掌大小，结合生活常识，这里应选择公顷作单位。第3题，会议室的面积相对比较大，与边长是1米的正方形比较，相当于几百个边长是1米的正方形，根据这里的数据420，应选择用平方米作单位。第4题，学校的占地面积相对是比较大的，但比天安门广场小很多，结合实际情况，以及数据4，用平方米作单位显然小了，应选择公顷作单位比较合适.\n【详解】巩义市总面积1043平方千米。天安门广场是世界上最大的广场，占地约 40 公顷。一个会议室的面积大约是 420 平方米。一所小学的占地面积是4公顷。",
+        "knowledge_points":  [
+                                 "平方厘米",
+                                 "平方分米",
+                                 "平方米的认识",
+                                 "公顷",
+                                 "平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "24-25四年级上·河南郑州·期中"
+    },
+    {
+        "id":  "451",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "如果1平方米收小麦 1000 克，那么 1 公顷收小麦( )千克。",
+        "options":  [
+
+                    ],
+        "answer":  "10000",
+        "solution":  "【分析】根据 1 千克 = 1000 克，那么 1 平方米就可以收小麦1千克，再根据1公顷 = 10000 平方米，可知1公顷中有10000 个 1 平方米，那么 1 公顷就可以收小麦 10000 个 1 千克，即10000千克。据此作答.\n【详解】根据分析可知:1平方米收小麦1000克，即1平方米收小麦1千克，而1公顷 = 10000 平方米，那么 1 公顷收小麦10000 千克。",
+        "knowledge_points":  [
+                                 "克",
+                                 "千克之间的换算与比较",
+                                 "公顷",
+                                 "平方千米的进率与换算"
+                             ],
+        "source":  "24-25四年级上·河南郑州·期中"
+    },
+    {
+        "id":  "452",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上“\u003e”“\u003c”或“=”。`r`n`r`n6000800( )6008000`r`n100390000( )九千七百万`r`n800 公顷( )80 平方千米`r`n10个一百万( )一千万",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003e;\u003c;=",
+        "solution":  "【分析】大数的比较：先比较两个数的位数，位数多的数就大，如果两个数的位数相同，就从最高位比起，最高位上的数字大的那个数就大，如果最高位上的数字相同，就依次比较下一个数位上的数字，直到比出大小为止。大数的写法，从高位写起：按照数位顺序，从高位到低位，一级一级地写。哪个数位上一个计数单位也没有，就在那个数位上写0占位。根据1平方千米 = 100 公顷，将80平方千米换算为公顷，再进行比较。根据数的认识，相邻单位之间的进率是10，10个一百万就是一千万，据此比较大小。`r`n【详解】根据分析可知：6000800\u003c6008000；九千七百万写作97000000，100390000\u003e97000000，所以100390000\u003e九千七百万；80平方千米 = 8000公顷，800公顷\u003c8000公顷，所以800公顷\u003c80平方千米；10个一百万 = 一千万。",
+        "knowledge_points":  [
+                                 "公顷",
+                                 "平方千米的进率与换算",
+                                 "大数的比较",
+                                 "十进制计数法",
+                                 "亿以内数的读",
+                                 "写法"
+                             ],
+        "source":  "25-26四年级上·云南文山·阶段练习"
+    },
+    {
+        "id":  "453",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果1平方米可以种 4 棵树苗，则 1 公顷土地可以种( )棵树苗,1 平方千米可以种( )棵树苗。",
+        "options":  [
+
+                    ],
+        "answer":  "40000;4000000",
+        "solution":  "【分析】1公顷 = 10000 平方米,1平方千米 = 1000000 平方米,1 平方米种 4 棵树苗，则 10000 平方米种 4×10000 棵树苗;1000000 平方米种 4×1000000 棵树苗.\n【详解】1 公顷 = 10000 平方米4×10000 = 40000 (棵)1 平方千米 = 1000000 平方米\n4×1000000 mathrm = 4000000 (棵)1 公顷土地可以种 40000 棵树苗,1 平方千米可以种4000000 棵树苗。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26四年级上·云南文山·阶段练习"
+    },
+    {
+        "id":  "454",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上合适的单位.\n\n暑假我和爸爸妈妈去昆明游玩，昆明真大呀，我用手机搜索后，知道了昆明市的面积是 21012( ),第一天我们游玩了翠湖公园，翠湖公园占地面积约 22( ),参观了陆军讲武堂，陆军讲武堂位于翠湖公园里面，占地面积70000( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方千米;公顷;平方米",
+        "solution":  "【分析】根据题意，面积的常用单位有公顷，平方千米，平方米等,1公顷比一个标准足球场稍大一点,1平方千米大约是小型湖泊的湖面面积,1张方桌面的大小大约1平方米;可以判断计量昆明市的面积用\"平方千米\";计量翠湖公园占地面积用\"公顷\";计量陆军讲武堂占地面积用\"平方米\"。以此答题即可.\n【详解】根据分析可知：暑假我和爸爸妈妈去昆明游玩，昆明真大呀，我用手机搜索后，知道了昆明市的面积是21012平方千米，第一天我们游玩了翠湖公园，翠湖公园占地面积约22公顷，参观了陆军讲武堂，陆军讲武堂位于翠湖公园里面，占地面积70000平方米。",
+        "knowledge_points":  [
+                                 "公顷",
+                                 "平方千米的认识"
+                             ],
+        "source":  "25-26四年级上·云南文山·阶段练习"
+    },
+    {
+        "id":  "455",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上合适的数.\n\n5 公顷=( )平方米 90 平方千米=( )公顷",
+        "options":  [
+
+                    ],
+        "answer":  "50000;9000",
+        "solution":  "【分析】根据1公顷 = 10000 平方米,1平方千米 = 100 公顷进行单位换算.\n【详解】5公顷里面有5个1公顷，所以5公顷 = 50000 平方米;90平方千米里面有90个1平方千米，所以90平方千米=9000 公顷。",
+        "knowledge_points":  [
+                                 "公顷",
+                                 "平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·云南文山·阶段练习"
+    },
+    {
+        "id":  "456",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "盘州市是贵州省辖县级市，由六盘水市代管。位于贵州省西部，六盘水市西南部，东邻普安县，南接兴义市，西连云南省富源县，宣威市，北邻水城区，它的总面积约 4057 平方千米，合公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "405700",
+        "solution":  "【分析】1 平方千米 = 100 公顷，据此进行单位换算.\n【详解】1平方千米 = 100 公顷,4057平方千米 = 405700 公顷，所以它的总面积约 4057 平方千米，合 405700 公顷。",
+        "knowledge_points":  [
+                                 "公顷",
+                                 "平方千米的进率与换算"
+                             ],
+        "source":  "24-25四年级上·贵州六盘水·期中"
+    },
+    {
+        "id":  "457",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在下面的括号里填上合适的数.\n\n8 公顷=( )平方米 600 平方千米=( )公顷\n\n1 平方千米=( )平方米 800000 平方米=( )公顷",
+        "options":  [
+
+                    ],
+        "answer":  "80000;60000;1000000;80",
+        "solution":  "【分析】把 8 公顷换算为平方米，因为 1 公顷 = 10000 平方米，所以 8 公顷换算为平方米是 80000 平方米;把600平方千米换算为公顷，因为1平方千米 = 100 公顷，所以 600 平方千米换算为公顷是 60000 公顷;把1平方千米换算成平方米，因为1平方千米 = 1000000 平方米，所以1平方千米 = 1000000 平方米。把 800000 平方米换算为公顷;因为 1 公顷 = 10000 平方米,800000 平方米里面有 80 个 10000 平方米，即 80 公顷;【详解】8 公顷 = 80000 平方米;600平方千米 = 60000 公顷;1 平方千米 = 1000000 平方米;800000 平方米 = 80 公顷。",
+        "knowledge_points":  [
+                                 "公顷",
+                                 "平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "24-25四年级上·湖北孝感·期中"
+    },
+    {
+        "id":  "458",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上合适的单位.\n\n(1)明明家客厅的面积约是 30( )。 \n(2)北京圆明园的面积约是 350( )。 \n(3)广东省的占地面积约是 179725( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方米;公顷;平方千米",
+        "solution":  "【分析】常用的面积单位有平方千米，公顷，平方米，平方分米和平方厘米等，计量很大的土地面积，如国家，城市的占地面积常用平方千米作单位;计量较大的土地面积，如公园，学校的面积常用公顷作单位;计量建筑或房间的面积常用平方米作单位;计量较小的物体表面的面积，如课桌面，课本封面的面积常用平方分米或平方厘米作单位。(1)明明家客厅的面积约是 30 平方米比较合适;(2)北京圆明园的面积约是 350 公顷比较合适;(3)广东省的占地面积约是179725平方千米比较合适。据此解答.\n【详解】(1)根据分析可知：明明家客厅的面积约是 30 平方米。(2)根据分析可知：北京圆明园的面积约是 350 公顷。 (3)根据分析可知：广东省的占地面积约是179725平方千米。",
+        "knowledge_points":  [
+                                 "面积单位的选择",
+                                 "公顷",
+                                 "平方千米的认识"
+                             ],
+        "source":  "24-25 四年级上·广东潮州·期中"
+    },
+    {
+        "id":  "459",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "6 平方千米=( )公顷 30000 平方米=( )公顷 200 公顷=( )平方千米=( )平方米",
+        "options":  [
+
+                    ],
+        "answer":  "600;3;2;2000000",
+        "solution":  "【分析】1平方千米 = 100 公顷,10000平方米 = 1 公顷，据此进行单位的换算.\n【详解】(1)6 平方千米里面有 6 个 1 平方千米，也就是 6个 100 公顷，即 600 公顷;(2)30000 平方米里面有 3 个 10000 平方米，也就是 3 个 1公顷，即 3 公顷;(3)200 公顷里面有 2 个 100 公顷，也就是 2 个 1 平方千米，即2平方千米;200公顷里面有200个1公顷，也就是200个 10000 平方米，即 2000000 平方米。6 平方千米 = 600 公顷 30000 平方米 = 3 公顷 200 公顷 = 2 平方千米 = 2000000 平方米",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "公顷",
+                                 "平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·安徽合肥·期中"
+    },
+    {
+        "id":  "460",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "65 公顷=( )平方米 14 平方千米=( )公顷",
+        "options":  [
+
+                    ],
+        "answer":  "650000;1400",
+        "solution":  "【分析】根据面积单位之间的进率,1公顷 = 10000 平方米,1平方千米 = 100 公顷，进行单位换算，大单位化小单位乘进率，小单位化大单位除以进率，据此解答.\n【详解】根据分析可得：因为1公顷 = 10000 平方米,65公顷 = 65 times10000 = 650000 平方米，故65公顷 = 650000 平方米。因为1平方千米 = 100 公顷,14平方千米 = 14×100 = 1400 公顷，故14平方千米 = 1400 公顷。所以65公顷 = 650000 平方米,14平方千米 = 1400 公顷。",
+        "knowledge_points":  [
+                                 "公顷",
+                                 "平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·陕西安康·期中"
+    },
+    {
+        "id":  "461",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在横线上填上合适的单位名称.\n\n数学课本封面的面积约 5 。宁夏回族自治区的面积约 66400 。",
+        "options":  [
+
+                    ],
+        "answer":  "平方分米;平方千米",
+        "solution":  "【分析】常用面积单位有平方千米，平方米，平方分米，平方厘米等,1平方米的面积大小是边长为1米的正方形大小，一般国家，城市，省份，岛屿等用平方千米,1平方分米的大小大约是一个手掌的大小,1平方厘米大约是一个指甲盖的大小，根据实际情况选择合适的单位.\n【详解】数学课本封面大约几个手掌的大小，所以单位用平方分米，数学课本封面的面积约5平方分米;宁夏回族自治区的面积较大，用平方千米，所以宁夏回族自治区的面积约66400平方千米。",
+        "knowledge_points":  [
+                                 "面积单位的选择"
+                             ],
+        "source":  "24-25四年级上·宁夏吴忠·期中"
+    },
+    {
+        "id":  "462",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "填上合适的面积单位和数.\n\n我们的校园约 3( ) 浙江省的面积大约是 105500( )\n\n900 平方千米=( )公顷 120000 平方米=( )公顷",
+        "options":  [
+
+                    ],
+        "answer":  "公顷;平方千米;90000;12",
+        "solution":  "【分析】计量较大的面积单位有公顷和平方千米;边长为100米的正方形，面积是1公顷;100个标准足球场的面积大约是1平方千米;根据实际数据可知，公顷可以用来计量公园等占地面积;国土面积比较大，常用平方千米为单位;1平方千米 = 100 公顷,1公顷 = 10000 平方米。由此进行单位换算;据此解答.\n【详解】我们的校园约 3 公顷。浙江省的面积约是 105500 平方千米。 1 平方千米 = 100 公顷,900 平方千米 = 90000 公顷。 10000 平方米 = 1 公顷,120000 平方米 = 12 公顷。",
+        "knowledge_points":  [
+                                 "面积单位的选择",
+                                 "公顷",
+                                 "平方千米的进率与换算",
+                                 "公顷",
+                                 "平方千米的认识"
+                             ],
+        "source":  "24-25 四年级上·浙江杭州·期中"
+    },
+    {
+        "id":  "463",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "5 平方千米=()平方米 4 公顷=( )平方米\n\n350000 平方米=( )公顷 5000 公顷=( )平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "5000000;40000;35;50",
+        "solution":  "【分析】(1)1平方千米 = 1000000 平方米，求5平方千米=( )平方米，可以想5平方千米里面有多少个1平方千米，那么就对应着多少个 1000000 平方米。(2)1 公顷 = 10000 平方米，求 4 公顷 = ( )平方米，可以想 4 公顷里面有多少个 1 公顷，那么就对应着多少个10000 平方米(3)10000 平方米 = 1 公顷，求 350000 平方米 = ( )公顷，可以想 350000 平方米里面有几个10000 平方米，那么就对应着几公顷。 (4)100 公顷 = 1 平方千米，求5000公顷 = ( )平方千米，可以想5000公顷里面有几个100公顷，那么就对应着几平方千米.\n【详解】(1)1 平方千米 = 1000000 平方米，所以 5 平方千米 = 5000000 平方米。(2)1 公顷 = 10000 平方米，所以 4 公顷 = 40000 平方米。 (3)10000 平方米 = 1 公顷，所以 350000 平方米 = 35 公顷。 (4)100 公顷 = 1 平方千米，所以5000公顷 = 50 平方千米。5平方千米 = 5000000 平方米;4公顷 = 40000 平方米350000 平方米 = 35 公顷;5000公顷 = 50 平方千米",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "公顷",
+                                 "平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·湖南长沙·期中"
+    },
+    {
+        "id":  "464",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个近似长方形的人工湖，长 2 千米，宽 3 千米，绕湖走一圈，一共要走()千米，这个人工湖的面积大约是( )平方千米。",
+        "options":  [
+
+                    ],
+        "answer":  "10;6",
+        "solution":  "【分析】长方形的周长 = (长+宽)×2，长方形的面积=长×宽，代入计算即可.\n【详解】 ( 2 + 3 ) \\ * 2\n= 5×2\n= 10 (千米)3×2 = 6 (平方千米)一个近似长方形的人工湖，长2千米，宽3千米，绕湖走圈，一共要走10千米，这个人工湖的面积大约是6平方千米。",
+        "knowledge_points":  [
+                                 "长方形的周长",
+                                 "长方形的面积",
+                                 "公顷",
+                                 "平方千米的认识"
+                             ],
+        "source":  "24-25 四年级上·江西九江·期中"
+    },
+    {
+        "id":  "465",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "3 公顷=( )平方米 9 平方千米=( )公顷\n\n102000 平方米=( )公顷( )平方米",
+        "options":  [
+
+                    ],
+        "answer":  "30000;900;10;2000",
+        "solution":  "【分析】(1)1 公顷 = 10000 平方米,3 公顷里面有 3 个 1公顷，也就是 3 公顷里面有 3 个 10000 平方米;(2)1平方千米 = 100 公顷,9平方千米里面有9个1平方千米，也就是 9 平方千米里面有 9 个 100 公顷;(3)1 公顷 = 10000 平方米,102000 平方米 = 100000 平方米+2000 平方米,100000 平方米 = 10 公顷.\n【详解】(1)1 公顷 = 10000 平方米，所以 3 公顷 = 30000 平方米;(2)1 平方千米 = 100 公顷，所以 9 平方千米 = 900 公顷;(3)1 公顷 = 10000 平方米，所以 102000 平方米 = 10 公顷2000 平方米。",
+        "knowledge_points":  [
+                                 "公顷",
+                                 "平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "24-25 四年级上·江西九江·期中"
+    },
+    {
+        "id":  "466",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号填上\"\u003e\" \"\u003c\"或 \" = \" .\n\n56 万( )5686705 平方千米( )499 公顷 10 公顷( )10 万平方米",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003e;=",
+        "solution":  "【分析】(1)大数比较看数位：位数越多，数越大。位数相同，从最高位比起，最高位上的数大的那个数就越大，如果最高位数相同，就依次比较下一个数位上的数，直到比较出大小。(2)1 平方千米 = 100 公顷，将 5 平方千米换算成公顷，再比较大小即可;(3)1 公顷 = 10000 平方米，将公顷换算成平方米，再比较大小即可.\n【详解】(1)56 万=560000,560000\u003c568670，所以 56万\u003c568670;(2)1 平方千米 = 100 公顷,5 平方千米 = 500 公顷,500公顷\u003e499 公顷，所以 5 平方千米\u003e499 公顷;(3)1公顷 = 10000 平方米,10公顷 = 100000 平方米,100000 平方米 = 10 万平方米，所以10公顷 = 10 万平方米。56 万\u003c568670;5 平方千米\u003e499 公顷;10 公顷 = 10 万平方米。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "公顷",
+                                 "平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "24-25四年级上·江西九江·期中"
+    },
+    {
+        "id":  "467",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上合适的面积单位.\n\n一个篮球场的面积大约是 500( ) 一个果园的面积大约是 1( ) 我国领土面积大约是 960 万( )",
+        "options":  [
+
+                    ],
+        "answer":  "平方米;公顷;平方千米",
+        "solution":  "【分析】面积的常用单位有公顷，平方千米，平方米，平方分米等,1公顷比一个标准足球场稍大一点,1平方千米大约是小型湖泊的湖面面积,1平方米大约是1张方桌面的大小,1 平方分米大约是 1 个成人手掌面的大小，据此解答.\n【详解】一个篮球场的面积大约是500平方米;一个果园的面积大约是 1 公顷;我国领土面积大约是 960 万平方千米。",
+        "knowledge_points":  [
+                                 "公顷",
+                                 "平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "24-25四年级上·江西九江·期中"
+    },
+    {
+        "id":  "468",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在( )里填上适当的单位.\n\n我国陆地领土面积约为 960 万( ) 旗杆高约 15( )一间教室的面积约为 60( )鸟巢的占地面积约为 20( )",
+        "options":  [
+
+                    ],
+        "answer":  "平方千米;米;平方米;公顷",
+        "solution":  "【分析】根据生活经验对面积单位，长度单位和数据大小的认识，计量比较大的土地的面积，常用平方千米( ( k m 2 ) )作单位，江苏省的面积大约是10万平方千米，所以计量我国陆地的领土面积用\"平方千米\"作单位比较合适;计量般的高度用米(m)作单位，门的高度大约是 2 米，所以计量旗杆的高度用\"米\"作单位比较合适;计量一般大小的面积用平方米 ( m 2 ) )作单位，一扇门的面积大约是2平方米，所以计量一间教室的面积用\"平方米\"作单位比较合适;计量土地的面积，可以用公顷为单位，纽约博物馆的占地面积约8公顷，所以鸟巢的占地面积用\"公顷\"作单位比较合适.\n【详解】我国陆地领土面积约为 960 万平方千米;旗杆高约 15 米;一间教室的面积约为 60 平方米;鸟巢的占地面积约为 20 公顷。",
+        "knowledge_points":  [
+                                 "长度单位的选择",
+                                 "面积单位的选择"
+                             ],
+        "source":  "24-25四年级上·广东潮州·期中"
+    },
+    {
+        "id":  "469",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上合适的单位.\n\n一所学校的操场面积约是 1( ) 温州市陆域面积约12110( )",
+        "options":  [
+
+                    ],
+        "answer":  "公顷;平方千米",
+        "solution":  "【分析】根据生活经验，以及对面积单位和数据大小的认识，可知：公顷适合计量稍大的土地面积，边长是100米的正方形的面积是1公顷,400米跑道围起来的部分的面积大约是1公顷;计量比较大的土地面积，常用平方千米作单位，例如：国土面积，城市面积;据此解答.\n【详解】据分析可得：一所学校的操场面积约是1公顷 温州市陆域面积约12110 平方千米",
+        "knowledge_points":  [
+                                 "公顷",
+                                 "平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "24-25 四年级上·浙江·期中"
+    },
+    {
+        "id":  "470",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "6 公顷=( )平方米 1000 公顷=( )平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "60000;10",
+        "solution":  "【分析】根据1平方千米 = 100 公顷,1公顷=10000平方米，解答此题即可。把高级单位的名数换算成低级单位的名数，就乘单位间的进率，反之，则除以进率.\n【详解】6 公顷 = (6×10000)平方米 = 60000 平方米;因为 1000 公顷里面含有 10 个 100 公顷，所以 1000 公顷=10 平方千米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·浙江·期中"
+    },
+    {
+        "id":  "471",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个长 500 米，宽80米的长方形鱼塘，如果长和宽都扩大到原来的2倍，那么鱼塘的面积增加了( )公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "12",
+        "solution":  "【分析】已知原鱼塘长500米，宽80米，根据长方形面积= 长×宽，计算出原鱼塘面积;长和宽都扩大到原来的2倍，新长为 500 times2 = 1000 (米),新宽为 80×2 = 160 (米),计算出扩大后鱼塘的面积，再根据增加的面积 = 新面积-原面积，计算出鱼塘扩建后增加的面积，最后根据1公顷=10000平方米把结果换算成公顷即可.\n【详解】( ( 500×2 ) ~ * ~ ( 80×2 ) )-500×80\n\n\n= 1000×160 - 500×80\n\n = 160000 - 40000 = 120000 (平 方 米) 120000 = 12 公 顷 \n\n\n所以，鱼塘的面积增加了 12 公顷。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "长方形的面积"
+                             ],
+        "source":  "24-25 四年级上·甘肃兰州·期中"
+    },
+    {
+        "id":  "472",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "3 公顷=( )平方米 70000 平方米=( )公顷\n\n60 平方千米=( )公顷 9 平方千米=( )平方米",
+        "options":  [
+
+                    ],
+        "answer":  "30000;7;6000;9000000",
+        "solution":  "【分析】1 公顷 = 10000 平方米,3 乘 10000 换算成平方米作单位;看 70000 里面有几个 10000 就有几公顷;1 平方千米 = 100 公顷,60 乘 100 换算成公顷作单位;1平方千米 = 1000000 平方米,9乘1000000换算成平方米作单位.\n【详解】 3×10000 = 30000 (平方米),所以 3 公顷 = 30000 平方米70000 里面有 7 个 10000，所以 70000 平方米 = 7 公顷60×100 = 6000 (公顷),所以 60 平方千米 = 6000 公顷\n\n9×1000000 mathrm = 9000000 (平方米),所以 9 平方千米 =9000000 平方米",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·甘肃兰州·期中"
+    },
+    {
+        "id":  "473",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "60000 平方米=()公顷 5 平方千米=( )公顷 12000000 平方米=( )公顷=( )平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "6;500;1200;12",
+        "solution":  "【分析】根据 10000 平方米 = 1 公顷进行单位换算;根据 1 平方千米 = 100 公顷进行单位换算;根据 10000 平方米 = 1 公顷,100公顷 = 1 平方千米进行单位换算.\n【详解】60000 平方米 = 6 公顷;5 平方千米 = 500 公顷;12000000 里有 1200 个 10000，所以 12000000 平方米 = 1200公顷,1200 里有 12 个 100，所以 1200 公顷 = 12 平方千米12000000 平方米 = 1200 公顷 = 12 平方千米。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·黑龙江绥化·期中"
+    },
+    {
+        "id":  "474",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "括号里填上合适的单位.\n\n颐和园的占地面积约为 300( ).\n\n某科技馆的占地面积约是 4( ).\n\n我国领土面积约是 960 万( ).\n\n大树约高 16( )。",
+        "options":  [
+
+                    ],
+        "answer":  "公顷;公顷;平方千米;米",
+        "solution":  "【分析】计量较大的土地面积，常用公顷和平方千米作单位。边长是100米的正方形的面积为1公顷，边长是1千米的正方形的面积是1平方千米。计量公园或者体育场馆的面积，常用公顷做单位。计量县，市，省的土地面积，常用平方千米做单位;常用的长度单位有毫米，厘米，分米，米和千米。我们知道，银行卡的厚度大约1毫米，食指宽大约1厘米，一支笔的长度约1分米，小学生展开双臂大约1米，连续步行 10 分钟所走的路程大概就是 1 千米。(1)由题意得，颐和园的面积很大，所以括号里填公顷比较合适。 (2)由题意得，科技馆的占地面积很大，所以括号里填公顷比较合适。 (3)由题意得，我国的领土面积特别大。计量全国的面积，括号里填平方千米比较合适。(4)由题意得，大树的高度和几层楼的高度差不多，所以括号里填米比较合适.\n【详解】由分析可得：颐和园的占地面积约为300公顷。某科技馆的占地面积约是4公顷。我国领土面积约是960万平方千米。大树约高 16 米。",
+        "knowledge_points":  [
+                                 "长度单位的选择",
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26 四年级上·黑龙江绥化·期中"
+    },
+    {
+        "id":  "475",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上合适的面积单位.\n\n(1)一个篮球场的面积约是 420( )。 \n(2)我国陆地领土面积约为 960 万( )。 \n(3)2023 年亚运会在杭州市举行，杭州市的面积是 16850()。",
+        "options":  [
+
+                    ],
+        "answer":  "平方米;平方千米;平方千米",
+        "solution":  "【分析】需要根据实际物体的面积大小选择合适的单位。常见面积单位有平方厘米，平方分米，平方米，公顷，平方千米等。小型方桌桌面大约是 1 平方米,1.4 个标准足球场的大小大约是 1 公顷,140 个标准足球场的面积总和大约是1平方千米。篮球场的面积较大，但相对于领土和城市面积较小，标准篮球场面积约为420平方米。我国陆地领土面积非常大，通常以平方千米为单位，常识中为960万平方千米。杭州市作为大城市，面积较大,16850 这个数值适合用平方千米为单位.\n【详解】(1)一个篮球场的面积约是420平方米。(2)我国陆地领土面积约为960万平方千米。 (3)2023年亚运会在杭州市举行，杭州市的面积是16850平方千米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "25-26四年级上·陕西咸阳·期中"
+    },
+    {
+        "id":  "476",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上\"\u003e\"\"\u003c\"或\"=\".\n\n30000 平方米( )29999 平方米 90 公顷( )900000 平方米5 公顷 40 平方米( )54000 平方米 9 平方千米 7 公顷()907 公顷",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;=;\u003c;=",
+        "solution":  "【分析】(1)直接比较数值大小。(2)1 公顷 = 10000 平方米，用 90 times10000 换算成平方米为单位后比较即可。 (3)1公顷 = 10000 平方米，用 5 times10000 div 40 换算成平方米为单位后比较即可。 (4)1平方千米 = 100 公顷，用 9 times100 + 7 换算成公顷为单位后比较即可。据此解答.\n【详解】(1)30000\u003e29999所以 30000 平方米(\u003e)29999 平方米。(2)1公顷 = 10000 平方米90×10000 = 900000 (平方米)所以 90 公顷( ( = ) )900000 平方米。(3)1公顷 = 10000 平方米\n 5×10000 + 40 = 50000 + 40 = 50040 (平 方 米) \n\n50040 \u003c 54000\n所以 5 公顷 40 平方米(\u003c)54000 平方米。(4)1平方千米 = 100 公顷\n 9×100 + 7 = 900 + 7 = 907 (公 顷) \n所以 9 平方千米 7 公顷( [ = ] )907 公顷。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "大数的比较"
+                             ],
+        "source":  "25-26 四年级上·陕西咸阳·期中"
+    },
+    {
+        "id":  "477",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "7 平方千米=( )公顷 300000 平方米=( )公顷",
+        "options":  [
+
+                    ],
+        "answer":  "700;30",
+        "solution":  "【分析】根据 1 平方千米 = 100 公顷,1 公顷 = 10000 平方米，据此单位换算即可.\n【详解】7个平方千米中有7个1平方千米，即7个100公顷，也就是 700 公顷，所以 7 平方千米 = 700 公顷;300000 平方米中有 30 个 10000 平方米，即 30 个 1 公顷，也就是 30 公顷，所以 300000 平方米 = 30 公顷。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·陕西咸阳·期中"
+    },
+    {
+        "id":  "478",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "排序题",
+        "difficulty":  0.65,
+        "question":  "把 2 平方千米,300公顷,4000000平方米和425平方米按面积从小到大的顺序排列:( )。",
+        "options":  [
+
+                    ],
+        "answer":  "425 平方米 \u003c 2 平方千米\u003c300 公顷\u003c4000000 平方米",
+        "solution":  "【分析】1 平方千米=1000000 平方米,1 公顷 = 10000 平方米，据此把2平方千米,300公顷都换算成平方米作单位，再与 4000000 平方米,425 平方米比较大小.\n【详解】2 平方千米 = 2000000 平方米300 公顷 = 3000000 平方米425 平方米\u003c2000000 平方米\u003c3000000 平方米\u003c4000000 平方米所以 425 平方米\u003c2 平方千米\u003c300 公顷\u003c4000000 平方米。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·江西赣州·期中"
+    },
+    {
+        "id":  "479",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "排序题",
+        "difficulty":  0.65,
+        "question":  "按从小到大的顺序排列下面各数.\n\n(1)793847794837793807794767 \n(2)1 平方千米 102 公顷 10000 平方米 10 万平方米",
+        "options":  [
+
+                    ],
+        "answer":  "793807\u003c793847\u003c794767\u003c794837;10000 平方米\u003c10 万平方米 \u003c 1 平方千米\u003c102 公顷",
+        "solution":  "【分析】(1)整数比较大小，先比较位数，位数多的数就大，如果位数相同，则从高位开始比较数字大小，直到比较为止。 (2)根据1平方千米 = 100 公顷,1公顷 = 10000 平方米将单位统一成公顷再比较.\n【详解】(1)根据分析，从小到大排列如下:\n793807 \u003c 793847 \u003c 794767 \u003c 794837;(2)1 平方千米 = 100 公顷;10000 平方米 = 1 公顷;10 万平方米 = 10 公顷，所以排序如下:10000 平方米 \u003c 10 万平方米 \u003c 1 平方千米\u003c102 公顷。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·陕西咸阳·期中"
+    },
+    {
+        "id":  "480",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "面积 1 公顷的正方形边长必为 100 米。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】根据面积单位公顷的定义,1公顷等于边长为100米的正方形的面积，即 100×100 = 10000 平方米。因此，面积为1公顷的正方形，其边长必然为100米，这里要注意另外一句话，面积是1公顷的图形一定是边长100米的正方形，这句话是错误的，例如长 1000 米，宽 10 米的长方形的面积是 1000×10 = 10000 平方米 = 1 公顷.\n【详解】根据分析可知：面积1公顷的正方形边长必为100米，原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "正方形的面积",
+                                 "公顷，平方千米的认识"
+                             ],
+        "source":  "25-26 四年级上·湖南永州·月考"
+    },
+    {
+        "id":  "481",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "20 公顷 = 20 万平方米。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】把一个整万数改写成用\"万\"作单位的数，只需要找到万位，把万位后面的4个0去掉，最后在数的后面写上\"万\"字。由题意得,1公顷 = 10000 平方米，求20公顷=( )平方米，就是求20公顷里面有多少个1公顷，那么就对应几个10000平方米。然后将得数转化为以\"万\"作单位的数即可.\n【详解】1 公顷 = 10000 平方米，所以 20 公顷 = 200000 平方米。200000 平方米 = 20 万平方米，原题说法正确.\n故答案为:√",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "亿以内数的改写"
+                             ],
+        "source":  "25-26 四年级上·甘肃临夏·期中"
+    },
+    {
+        "id":  "482",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "四年级教室的面积约 1 公顷。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】1公顷 = 10000 平方米，而普通教室的长和宽通常在10米和8米左右，面积约为80平方米，远小于1公顷.\n【详解】1 公顷 = 10000 平方米。一般教室的长为 10 米，宽为8米，面积为 10 times8 = 80 (平方米),80平方米与1公顷相差极大。故答案为: ×",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识"
+                             ],
+        "source":  "24-25 四年级上·吉林松原·期中"
+    },
+    {
+        "id":  "483",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "判断题",
+        "difficulty":  0.94,
+        "question":  "4000 公顷 = 40 平方千米。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】根据面积单位换算，1平方千米 = 100 公顷，因此将公顷转换为平方千米时，需要除以进率100，也就是看4000里面有几个100就有几平方千米。`r`n【详解】4000里面有40个100，所以4000公顷 = 40 平方千米。题干中等式成立。故答案为:√",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·宁夏石嘴山·期中"
+    },
+    {
+        "id":  "484",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "一个农场的占地面积是 50 公顷,2 个这样的农场有 1 平方千米。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】先计算2个农场的总占地面积，再根据100公顷=1平方千米将公顷换算成平方千米，判断是否等于1平方千米.\n【详解】 50×2 = 100 (公顷)1平方千米 = 100 公顷 100 公顷 = 1 平方千米所以 2 个这样的农场有 1 平方千米。故答案为:√",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25四年级上·湖北孝感·期中"
+    },
+    {
+        "id":  "485",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "48平方千米 = 4800 平方米。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】1平方千米 = 100 公顷 = 1000000 平方米,48平方千米里面有 48 个 1000000 平方米。据此转换之后看是不是4800 平方米即可.\n【详解】48 平方千米里面有 48 个 1000000 平方米，是48000000 平方米。所以，题目中48平方千米 = 4800 平方米的换算错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25四年级上·贵州铜仁·期中"
+    },
+    {
+        "id":  "486",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "公顷和平方千米都是常用的面积单位,1 平方千米 = 100 公顷。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】计量较大的土地面积，常用公顷和平方千米作单位。边长是100米的正方形的面积为1公顷，边长是1千米的正方形的面积是1平方千米。1平方千米 = 100 公顷。据此解答.\n【详解】由分析得，公顷和平方千米都是常用的面积单位,1 平方千米 = 100 公顷。原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·黑龙江绥化·期中"
+    },
+    {
+        "id":  "487",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "袁隆平爷爷的杂交水稻试验田是一块长方形地，长 800 米，宽 500 米。这块试验田的面积是多少公顷。如果每公顷产水稻12吨，这块地一共能产水稻多少吨。",
+        "options":  [
+
+                    ],
+        "answer":  "40公顷;480吨",
+        "solution":  "【分析】根据长方形的面积 = 长×宽，把数代入即可求解，再根据1公顷=10000平方米，将长方形的面积换算成以公顷为单位，再用这块田地的面积乘每公顷产水稻的重量，就是总共产的水稻的重量.\n【详解】 800×500 = 400000 (平方米)400000 平方米 = 40 公顷40×12 = 480 (吨)\n答：这块试验田的面积是40公顷，这块地一共能产水稻480 吨。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "长方形的面积"
+                             ],
+        "source":  "25-26 四年级上·黑龙江鹤岗·期中"
+    },
+    {
+        "id":  "488",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "有一个正方形林场，如果用每小时步行 3 千米的速度绕林场一周，需要 8 小时，林场的面积是多少平方千米。合多少公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "36 平方千米;3600 公顷",
+        "solution":  "【分析】路程 = 速度×时间，用3乘8可算出林场一周是多少千米，根据正方形周长 = 边长 * 4，用周长除以4可算出边长，再根据边长×边长可算出正方形的面积，最后根据1平方千米 = 100 公顷进行单位换算.\n【详解】 3×8÷4\n= 24÷4\n= 6 (千米)6×6 = 36 (平方千米)36平方千米 = 3600 公顷\n\n答：林场的面积是 36 平方千米;合 3600 公顷。",
+        "knowledge_points":  [
+                                 "正方形的周长",
+                                 "基础行程问题",
+                                 "正方形的面积",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·辽宁鞍山·月考"
+    },
+    {
+        "id":  "489",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "劳动社团到蔬菜基地学习种菜，基地的一块正方形菜地里种植了油菜，菜地的边长为 200 米，若每公顷收油菜 3000 千克，则这块菜地可以收油菜多少千克。",
+        "options":  [
+
+                    ],
+        "answer":  "12000千克",
+        "solution":  "【分析】根据正方形面积 = 边长×边长，已知正方形菜地的边长是 200 米，计算出正方形菜地的面积;因为每公顷收油菜3000千克，因此根据10000平方米 = 1 公顷，先把平方米换算为公顷，再乘 3000，即可计算出收油菜的质量;据此解答.\n【详解】 200×200 = 40000 (平方米)40000 平方米 = 4 公顷3000×4 = 12000 (千克)\n答：这块菜地可以收油菜 12000 千克。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "正方形的面积"
+                             ],
+        "source":  "25-26 四年级上·黑龙江齐齐哈尔·期中"
+    },
+    {
+        "id":  "490",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "齐齐哈尔市深入践行\"绿水青山就是金山银山\"理念，将在 4 公顷的山上种满松树，如果每 5 平方米可种一棵松树，这座山一共可种松树多少棵。",
+        "options":  [
+
+                    ],
+        "answer":  "8000 棵",
+        "solution":  "【分析】根据 1 公顷 = 10000 平方米进行单位换算，再除以每种一棵松树需要的面积即可求出棵数.\n【详解】4 公顷 = 40000 平方米40000÷5 = 8000 (棵)答：这座山一共可种松树 8000 棵。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·江西南昌·期中"
+    },
+    {
+        "id":  "491",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一个长方形农业观光园长 7 千米，李叔叔驾驶汽车以 44 千米/时的速度绕观光园一周需要半个小时。这个观光园的面积是多少公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "2800公顷",
+        "solution":  "【分析】根据题意，李叔叔驾驶汽车以44千米/时的速度绕观光园一周需要半个小时，先用 44÷2 求出绕一圈走的长度，已知长是 7 千米，根据长方形周长 = (长+宽)×2，用一圈的长度除以2再减去长，即可求出这个观光园的宽是多少千米，根据长方形面积 = 长×宽，计算出面积后根据1平方千米 = 100 公顷，据此换算成公顷为单位.\n【详解】 44÷2 = 22 (千米)\n 22÷2 - 7 = 11 - 7 = 4 (千 米) \n7×4 = 28 (平方千米)28平方千米 = 2800 公顷 答：这个观光园的面积是2800公顷。",
+        "knowledge_points":  [
+                                 "长方形的周长",
+                                 "长方形的面积",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·福建福州·期中"
+    },
+    {
+        "id":  "492",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "某市新建了一个小区。文文陪妈妈一起去看房子，他们在小区门口看到一则广告：美丽家园理想居所，小区环境优美，景色宜人，是本市绿化示范小区，占地面积11公顷。其中游泳馆，儿童游乐场，健身房，网球场，道路等公共设施的占地面积是 2 公顷，绿化面积达 6 公顷。文文在小区里走了一圈，发现该小区共新建住宅楼25幢。文文估计了一下，每幢楼长约80米，宽约10米，请你和文文一起算一算，小区门口的广告是否真实。",
+        "options":  [
+
+                    ],
+        "answer":  "不真实",
+        "solution":  "【分析】根据长方形的面积 = 长×宽，代入数据求出每幢楼的占地面积，再乘25，求出25幢楼的占地面积;再根据10000 平方米 = 1 公顷，换算单位;最后住宅楼总占地面积，加上公共设施和绿化面积，若总和等于广告中的11公顷则真实，否则不真实.\n【详解】每幢楼占地面积: 80×10 = 800 (平方米)25 幢楼总占地面积: 800×25 = 20000 (平方米)20000 平方米 = 2 公顷2+6+2=8+2=10(公顷)10\u003c11答：广告中总占地面积为 11 公顷，实际计算为 10 公顷，故广告不真实。",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26四年级上·浙江杭州·阶段练习"
+    },
+    {
+        "id":  "493",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一块长方形玉米地，长450米，宽200米。如果平均每公顷收玉米8吨，这块玉米地能收玉米多少吨。",
+        "options":  [
+
+                    ],
+        "answer":  "72 吨",
+        "solution":  "【分析】根据题意，先计算出长方形玉米地的面积，再将面积的单位\"平方米\"化成\"公顷\"作单位，最后乘以每公顷的产量，即可算得这块玉米地可以收的玉米量.\n【详解】计算长方形面积:450×200 = 90000 (平方米)因为 1 公顷 = 10000 平方米90000 平方米有 9 个 10000 平方米，所以 90000 平方米 = 9 公顷计算总产量:9×8 = 72 (吨)答：这块玉米地能收玉米72吨。",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·湖南长沙·期中"
+    },
+    {
+        "id":  "494",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "一块长 600 米，宽400米的长方形试验田，每公顷大约收小麦12吨。这块试验田共收小麦多少吨。",
+        "options":  [
+
+                    ],
+        "answer":  "288 吨",
+        "solution":  "【分析】根据长方形面积 = 长×宽，先求出这个长方形试验田的面积，再根据1公顷 = 10000 平方米，将单位换算为公顷，再用面积乘每公顷收小麦的质量，即可求出这块试验田共收小麦的质量.\n【详解】 600×400 = 240000 (平方米)240000 平方米 = 24 公顷24×12 = 288 (吨)答：这块试验田共收小麦288吨。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的实际问题",
+                                 "公顷，平方千米的进率与换算",
+                                 "长方形的面积"
+                             ],
+        "source":  "24-25 四年级上·河南郑州·期中"
+    },
+    {
+        "id":  "495",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "小军家有一块长800米，宽50米的长方形土地，现在小军的爸爸准备将这块土地全部种上刺梨.\n\n(1)这块土地的面积是多少公顷。 \n(2)如果平均每公顷栽刺梨350棵，一共可以栽多少棵刺梨。",
+        "options":  [
+
+                    ],
+        "answer":  "4 公顷;1400 棵",
+        "solution":  "【分析】长方形的面积 = 长×宽，求出土地的面积。平方米和公顷之间的进率是10000，据此将土地的面积换算成公顷。再用土地的面积乘每公顷土地种刺梨树棵数，求出种刺梨树总棵数.\n【详解】(1) 800×50 = 40000 (平方米) = 4 公顷答：这块土地的面积是 4 公顷.\n(2) 4×350 = 1400 (棵)答：一共可以栽 1400 棵刺梨。",
+        "knowledge_points":  [
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "长方形的面积",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·贵州六盘水·期中"
+    },
+    {
+        "id":  "496",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "张爷爷有一块长方形麦地，长 600 米，宽 150 米，平均每公顷能收 3 吨小麦，那么这块麦地一共能收多少吨小麦。",
+        "options":  [
+
+                    ],
+        "answer":  "27 吨",
+        "solution":  "【分析】根据题意，明确长方形的面积 = 长×宽，先计算长方形麦地的面积，根据1公顷 = 10000 平方米，将平方米转换为公顷，再用公顷数乘每公顷产量，得到总产量.\n【详解】根据分析可知:\n600×150 = 90000 (平 方 米)\n1公顷 = 10000 平方米90000 平方米 = 9 公顷\n9×3 = 27 (吨)\n答：这块麦地一共能收27吨小麦。",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·湖北孝感·期中"
+    },
+    {
+        "id":  "497",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "王叔叔承包了一块长方形稻田，长 250 米，宽 80 米。这块稻田的面积是多少公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "2公顷",
+        "solution":  "【分析】根据长方形的面积 = 长×宽求出稻田的面积，再根据 1 公顷 = 10000 平方米换算单位.\n【详解】 250×80 = 20000 (平方米)20000 平方米 = 2 公顷答：这块稻田的面积是2公顷。",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25四年级上·浙江杭州·期中"
+    },
+    {
+        "id":  "498",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "小明沿正方形花坛走了 4 圈，共行 4800 米，这个花坛的面积是多少平方米。合多少公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "90000 平方米;9 公顷",
+        "solution":  "【分析】沿正方形花坛走了4圈，共行4800米，用4800米除以4圈，则计算出走一圈的长度，一圈的长度就是正方形的周长，根据正方形的周长 = 边长 * 4，用走一圈的长度除以4，即可求得正方形花坛的边长，然后根据正方形的面积= 边长 boldsymbol times 边长，即可求得这个花坛的面积，根据1公顷=10000 平方米，最后换算成公顷为单位，据此解答.\n【详解】 4800÷4 = 1200 (米)1200÷4 = 300 (米)300×300 = 90000 (平方米)90000 平方米 = 9 公顷答：这个花坛的面积是 90000 平方米，合 9 公顷。",
+        "knowledge_points":  [
+                                 "正方形的周长",
+                                 "正方形的面积",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "24-25 四年级上·江西九江·期中"
+    },
+    {
+        "id":  "499",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一个周长为1200米的正方形果园，占地多少公顷。如果每公顷地施肥200千克，这个果园一共要施肥多少千克。",
+        "options":  [
+
+                    ],
+        "answer":  "9 公顷;1800 千克",
+        "solution":  "【分析】根据题意，已知一个周长为1200米的正方形果园，首先根据正方形周长公式求出边长，再用边长计算面积,1公顷 = 10000 平方米，并换算为公顷。接着用面积乘每公顷施肥量，得到总施肥量.\n【详解】根据分析可知:1200÷4 = 300 (米)300×300 = 90000 (平方米)1公顷 = 10000 平方米90000 平方米 = 9 公顷9×200 = 1800 (千克)\n\n答：占地 9 公顷。这个果园一共要施肥 1800 千克。",
+        "knowledge_points":  [
+                                 "正方形的周长",
+                                 "正方形的面积",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26四年级上·陕西咸阳·期中"
+    },
+    {
+        "id":  "500",
+        "unit":  "第二单元 平方千米",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一种播种机的播种宽度是 3 米，播种机每小时行 4 千米，照这样计算,5 小时可以播种多少公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "6 公顷",
+        "solution":  "【分析】播种机每小时行驶 4 千米,5 小时行 4 times5 = 20 (千米),播种的面积为一个长方形，长为 20 千米也就是20000米，宽为3米，根据长方形面积 = 长×宽计算，最后根据 10000 平方米 = 1 公顷进行单位换算即可.\n【详解】 4×5 = 20 (千米)20 千米 = 20000 米20000×3 = 60000 (平方米) = 6 公顷答:5 小时可以播种 6 公顷。",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·陕西咸阳·期中"
+    }
+]
diff --git "a/\345\233\233\345\271\264\347\272\247\351\242\230\351\233\206/\345\233\233\345\271\264\347\272\247\344\270\212/\345\233\233\345\271\264\347\272\247\344\270\212\347\254\254\344\272\224\345\215\225\345\205\203\345\271\263\350\241\214\345\233\233\350\276\271\345\275\242\345\222\214\346\242\257\345\275\242.json" "b/\345\233\233\345\271\264\347\272\247\351\242\230\351\233\206/\345\233\233\345\271\264\347\272\247\344\270\212/\345\233\233\345\271\264\347\272\247\344\270\212\347\254\254\344\272\224\345\215\225\345\205\203\345\271\263\350\241\214\345\233\233\350\276\271\345\275\242\345\222\214\346\242\257\345\275\242.json"
new file mode 100644
index 0000000000000000000000000000000000000000..0ab9942ff7a723b02f455969524f2fe3641031f4
--- /dev/null
+++ "b/\345\233\233\345\271\264\347\272\247\351\242\230\351\233\206/\345\233\233\345\271\264\347\272\247\344\270\212/\345\233\233\345\271\264\347\272\247\344\270\212\347\254\254\344\272\224\345\215\225\345\205\203\345\271\263\350\241\214\345\233\233\350\276\271\345\275\242\345\222\214\346\242\257\345\275\242.json"
@@ -0,0 +1,6644 @@
+[
+    {
+        "id":  "1",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "长方形是特殊的()。",
+        "options":  [
+                        "A. 三角形",
+                        "B. 正方形",
+                        "C. 梯形",
+                        "D. 平行四边形"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】正方形，长方形，平行四边形和梯形都是四边形，正方形的四条边都相等，四个角都是直角;长方形的对边相等，有四个直角;平行四边形的对边互相平行且相等，只有一组对边平行的四边形叫做梯形;依此选择。熟练掌握四边形的分类及关系，是解答此题的关键.\n【详解】正方形是特殊的长方形，正方形和长方形都是特殊的平行四边形(如下图所示)。![](images/a18080b5adb0f875e8f954ec5306b4433eb34e61c097d00afd4c18483889ccdd.jpg)故答案为:D",
+        "knowledge_points":  [
+                                 "四边形的分类及关系"
+                             ],
+        "source":  "20-21 四年级上·广东肇庆·期末"
+    },
+    {
+        "id":  "2",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面说法错误的是()。",
+        "options":  [
+                        "A. 正方形相邻的两条边互相垂直",
+                        "B. 平行四边形具有稳定性",
+                        "C. 平行四边形和梯形都有无数条高"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】正方形的四个角是直角，因此相邻的两条边互相垂直;三角形具有稳定性，平行四边形易变形，具有不稳定性;从平行四边形的一条边上的任意一点都可以向对边作垂直线段，即是平行四边形的高，平行四边形有无数条高;从梯形一条底边上的一点到它对边的垂直线段叫做梯形的高，梯形有 2 条底，有无数条高;据此即可解答。本题考查了正方形，平行四边形及梯形的特征，要熟练掌握.\n【详解】A。根据分析可知，正方形相邻的两条边互相垂直，原说法正确;B。根据分析可知，平行四边形具有不稳定性，原说法错误;C。根据分析可知，平行四边形和梯形都有无数条高，原说法正确。故答案为:B",
+        "knowledge_points":  [
+                                 "正方形的概念及特点，梯形的高及画法，平行 四边形的高及画法，平行四边形的不稳定性及应用"
+                             ],
+        "source":  "21-22四年级上·四川德阳·期末"
+    },
+    {
+        "id":  "3",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面说法正确的是()。",
+        "options":  [
+                        "A. 每两个计数单位之间的进率是 10。",
+                        "B. 直线比射线长。",
+                        "C. 只有一组对边平行的四边形叫做梯形。"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】先对每个选项中的说法进行分析并判断，然后选择正确的一项即可。A。每相邻两个计数单位之间的进率是10。B。由于无法测量出直线和射线的长度，因此直线和射线的长度无法比较。C。只有一组对边平行的四边形叫做梯形。此题考查了十进制计数法，直线和射线的特点，以及梯形的特点，应逐项进行分析并判断.\n【详解】A。每相邻两个计数单位之间的进率是10，即原说法错误。B。无法比较直线和射线的长短，即原说法错误。C。只有一组对边平行的四边形叫做梯形，即原说法正确。故答案为:C",
+        "knowledge_points":  [
+                                 "梯形的概念及特点，线段，直线，射线的认识及特征，十进制计数法"
+                             ],
+        "source":  "22-23四年级上·广东中山·期末"
+    },
+    {
+        "id":  "4",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列说法错误的是()。",
+        "options":  [
+                        "A. 电线杆与地面互相垂直",
+                        "B. 直线可以无限延伸",
+                        "C. 过直线a外一点可以作无数条与a平行的直线"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】两条直线相交成直角时，这两条直线互相垂直，其中一条直线是另一条直线的垂线，这两条直线的交点叫做垂足;把线段的两端无限延长，得到一条直线，直线没有端点，是可以无限延伸的;根据平行的特征可知，过直线 a 外一点可以作一条与 a 平行的直线;据此解答即可。本题考查了垂直与平行及直线的特征，结合题意分析解答即可.\n【详解】A。根据垂直的特征可知，电线杆与地面互相垂直，正确;B。根据直线的特征，直线可以无限延伸，正确;C。过直线a外一点可以作一条与a平行的直线，所以此项说法错误。故答案为:C.",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征，垂直的特征，平行的特征及性质"
+                             ],
+        "source":  "21-22 四年级上·广东东莞·期末"
+    },
+    {
+        "id":  "5",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "延长( ),这两条直线永不相交。",
+        "options":  [
+                        "A. 梯形的两腰 下底",
+                        "B. 梯形的高和腰",
+                        "C. 梯形的上底与下底"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】同一平面内，永不相交的两条直线叫做平行线，组成平行线的两条直线互相平行;只有一组对边平行的四边形叫做梯形;依此选择。熟练掌握平行和梯形的特点，是解答此题的关键.\n【详解】根据分析可知，延长梯形的上底与下底，这两条直线永不相交。故答案为:C.",
+        "knowledge_points":  [
+                                 "平行的特征及性质，梯形的概念及特点"
+                             ],
+        "source":  "21-22四年级上·广东江门·期末"
+    },
+    {
+        "id":  "6",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面说法正确的是()。",
+        "options":  [
+                        "A. 两条平行线都是 5 厘米",
+                        "B. 大于90°的角都是钝角",
+                        "C. 长方形是特殊的平行四边形"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】两条平行线是两条互相平行的直线，直线无限长;大于 90°,小于 180° 的角是钝角;长方形和正方形都是特殊的平行四边形;据此即可解答。本题主要考查学生对平行线的定义，角的分类和平行四边形的特征及性质的掌握.\n【详解】A。平行线都是直线，无限长，原说法错误。B。大于 90°,小于 180° 的角是钝角，原说法错误。C。长方形，正方形都是特殊的平行四边形，原说法正确。故答案为:C.",
+        "knowledge_points":  [
+                                 "角的分类及换算，平行的特征及性质，平行四边形的概念及特点"
+                             ],
+        "source":  "21-22 四年级上·广东中山·期末"
+    },
+    {
+        "id":  "7",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "关于梯形，下列说法错误的是( )。",
+        "options":  [
+                        "A. 梯形只有一组对边互相平行",
+                        "B. 直角梯形中一定有两个直角",
+                        "C. 等腰梯形的两组对边分别相等",
+                        "D. 上底与下底之间的垂线段是梯形的高"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】只有一组对边平行的四边形叫梯形;如果梯形的两腰相等，这样的梯形叫做等腰梯形。一腰垂直于底的梯形叫直角梯形。从梯形一条底边上的一点到它对边的垂直线段叫做梯形的高。据此即可解答。本题主要考查学生对梯形的定义，特征和分类知识的掌握.\n【详解】A。根据梯形的定义可知，梯形只有一组对边互相平行，原说法正确。B。直角梯形有一个腰垂直于上下底，所以直角梯形中一定有两个直角，原说法正确。C。等腰梯形的两腰相等，并不是两组对边分别相等，原说法错误。D。根据梯形高的定义可知，上底与下底之间的垂线段是梯形的高，原说法正确。故答案为:C.",
+        "knowledge_points":  [
+                                 "直角梯形和等腰梯形的概念及特点，梯形的概念及特点，梯形的高及画法"
+                             ],
+        "source":  "21-22 四年级上·湖南长沙·期末"
+    },
+    {
+        "id":  "8",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面说法中，有()个是正确的.\n①一个八位数，它的最高位是亿位.\n②两条平行线长都是4千米.\n③梯形只有一组对边是平行的.\n④1平方千米=100公顷=1000000平方米。",
+        "options":  [
+                        "A. 1",
+                        "B. 2",
+                        "C. 3",
+                        "D. 4"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】先对每个说法进行分析并判断，然后再选择即可。①整数的数位从右往左依次为：个位，十位，百位，千位，万位，十万位，百万位，千万位,。;②同一平面内永不相交的两条直线叫做平行线，组成平行线的两条直线互相平行。③只有一组对边平行的四边形叫做梯形。④1平方千米=100公顷=1000000平方米。此题考查了对整数的数位的认识，对平行线的认识，梯形的特点，以及面积单位之间的换算.\n【详解】①根据数位顺序表，一个八位数，它的最高位是千万位，即原说法错误。②因为直线是无限延伸的，所以不能测量出两条平行线之间的长度，即原说法错误。③根据梯形的特征可知，梯形只有一组对边平行，即原说法正确。④1平方千米=100公顷=1000000平方米，即原说法正确。由此可知说法正确的只有2个。故答案为:B.",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算，梯形的概念及特点，平行的特征及性质，整数的数级，数位和计数单位的认识"
+                             ],
+        "source":  "19-20 四年级上·湖南长沙·期末"
+    },
+    {
+        "id":  "9",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "两条直线相交成直角，就说这两条直线相互垂直，这两条直线的交点叫做()。",
+        "options":  [
+                        "A. 垂直",
+                        "B. 垂线",
+                        "C. 垂足",
+                        "D. 顶点"
+                    ],
+        "answer":  "C",
+        "solution":  "【详解】两条直线相交所成的四个角中，有一个角是直角时，就说这两条直线互相垂直，其中一条直线叫做另一条直线的垂线，它们的交点叫做垂足。如图：点 O 即是垂足。![](images/aa64a4b20f00c54b524f802c1540590b.jpg)故答案为:C",
+        "knowledge_points":  [
+                                 "垂直的特征"
+                             ],
+        "source":  "19-20 四年级上·湖南长沙·期末"
+    },
+    {
+        "id":  "10",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "在两条平行线之间作了四条垂直线段，这四条垂直线段的长度( )。",
+        "options":  [
+                        "A. 都相等",
+                        "B. 不相等",
+                        "C. 不一定都相等"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】两条平行线中可以画无数条垂线段，这些线段的长度都相等，依此选择。此题考查的是平行线之间的距离，熟练掌握平行与垂直的特点是解答此题的关键.\n【详解】根据分析可知，在两条平行线之间作了四条垂直线段，这四条垂直线段的长度都相等(如下图所示)![](images/b817879793af2bfe9519ec13a9c298d9d5b8c622e92cd85f228b8b8156f6c90a.jpg)故答案为:A",
+        "knowledge_points":  [
+                                 "平行的特征及性质，垂直的特征"
+                             ],
+        "source":  "14-15 四年级·全国·期末"
+    },
+    {
+        "id":  "11",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "一个梯形中只有()组对边平行。",
+        "options":  [
+                        "A. 1",
+                        "B. 2",
+                        "C. 3",
+                        "D. 4"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】只有一组对边平行的四边形叫梯形，互相平行的组对边分别是梯形的上底和下底，不平行的一组对边是梯形的腰，据此即可解答。本题主要考查学生对梯形的概念和特征的掌握.\n【详解】根据分析可知，一个梯形中只有一组对边平行。故答案为:A.",
+        "knowledge_points":  [
+                                 "梯形的概念及特点"
+                             ],
+        "source":  "19-20四年级上·湖南娄底·期末"
+    },
+    {
+        "id":  "12",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "在同一平面上,a 和b 两条直线都和 c 直线垂直，则直线 a 和直线 b()。",
+        "options":  [
+                        "A. 互相平行",
+                        "B. 互相垂直",
+                        "C. 无法确定"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】在同一平面内两条直线相交成直角，这两条直线互相垂直;同一平面内永不相交的两条直线叫做平行线，组成平行线的两条直线互相平行;依此画图并选择即可。熟练掌握平行与垂直的特点是解答此题的关键.\n【详解】画图如下:![](images/5eb3c3249f1cf5e74887414f73616b374067966bdd1307bdd9d5439842c025dd.jpg)由此可知，直线a和直线b互相平行。故答案为:A",
+        "knowledge_points":  [
+                                 "垂直的特征，平行的特征及性质"
+                             ],
+        "source":  "20-21 四年级上·湖南株洲·期末"
+    },
+    {
+        "id":  "13",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面说法正确的是( ).\nA。最小的自然数是 1，没有最大的自然数.\nB。两个锐角之和一定大于钝角.\nC。把一个平行四边形沿高剪开，可以得到两个梯形。",
+        "options":  [
+                        "A. 最小的自然数是 1，没有最大的自然数。",
+                        "B. 两个锐角之和一定大于钝角。",
+                        "C. 把一个平行四边形沿高剪开，可以得到两个梯形。"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】先对每个选项中的说法进行分析并判断，然后选择正确的一项即可。A。最小的自然数是0，没有最大的自然数，依此判断。B。小于 90° 的角是锐角，等于 90° 的角是直角，大于 90° 小于 180° 的角是钝角，依此判断。C。两组对边分别平行的四边形，叫做平行四边形，从平行四边形一条边上的一点向对边引一条垂线，这点和垂足之间的线段叫做平行四边形的高，垂足所在的边叫做平行四边形的底;只有一组对边平行的四边形叫做梯形;依此判断。此题考查了对自然数的认识，锐角，钝角的特点，以及平行四边形的高，平行四边形和梯形的特点.\n【详解】A。最小的自然数是0，没有最大的自然数，即原说法错误。B。45°+30°=75°,75°\u003c90°;89°+89°=178°,178°\u003c180°,因此两个锐角之和可能比钝角小，也可能是钝角，但一定不会比钝角大，即原说法错误。C。![](images/38fdcee9e3577f34b6ab61457fe5d10c49f4d6192d4349bc41a20b41f5a6a651.jpg)由此可知，把一个平行四边形沿高剪开，可以得到两个梯形，即原说法正确。故答案为:C.",
+        "knowledge_points":  [
+                                 "梯形的概念及特点，直角，钝角，锐角的认识及特征，自然数的认识，平行四边形的概念及特点"
+                             ],
+        "source":  "21-22 四年级上·云南昭通·期末"
+    },
+    {
+        "id":  "14",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下列图形中具有稳定性的是( )。",
+        "options":  [
+                        "A. 三角形",
+                        "B. 正方形",
+                        "C. 平行四边形"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】根据三角形的特性：三角形具有稳定性;据此选择。此题考查的是对三角形稳定性的知识的掌握情况.\n【详解】正方形和平行四边形具有不稳定性，具有稳定性的图形是三角形。故答案为:A.",
+        "knowledge_points":  [
+                                 "正方形的概念及特点，平行四边形的不稳定性及应用，三角形的稳定性及应用"
+                             ],
+        "source":  "9-10四年级下·河北张家口·期末"
+    },
+    {
+        "id":  "15",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "已知在同一平面内，直线 a 和直线 b 都与直线 c 平行，那么直线 a 和直线 b( )。",
+        "options":  [
+                        "A. 互相平行",
+                        "B. 互相垂直",
+                        "C. 相交",
+                        "D. 无法确定"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】根据平行的性质得出：在同一个平面内，平行于同一直线的两条直线一定平行。据此解答。本题主要考查学生对平行性质的掌握。平行线具有传递性.\n【详解】由分析可知，当直线a平行于直线c,直线b平行于直线c,那么直线a一定和直线b互相平行。选项A符合。故答案为:A.",
+        "knowledge_points":  [
+                                 "平行的特征及性质"
+                             ],
+        "source":  "22-23 四年级上·山东济南·期末"
+    },
+    {
+        "id":  "16",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下列说法正确的是( )。",
+        "options":  [
+                        "A. 最小的自然数是 1",
+                        "B. 过一点只能画一条直线",
+                        "C. 梯形只有一条高",
+                        "D. 平行四边形具有容易变形的特性"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】(1)最小的自然数是0。(2)过两点只能画一条直线，过一点能画无数条直线。(3)从梯形一条底边上的一点到它对边的垂直线段叫做梯形的高。则梯形有无数条高，这些高的长度相等。(4)平行四边形具有不稳定性，容易变形。本题考查的知识点较多，但都比较基础，熟练掌握即可.\n【详解】A。最小的自然数是0，说法错误;B。过一点能画无数条直线，说法错误;C。梯形有无数条高，说法错误;D。平行四边形具有容易变形的特性，说法正确;故答案为:D.",
+        "knowledge_points":  [
+                                 "平行四边形的不稳定性及应用，线段，直线，射线的认识及特征，自然数的认识，梯形的高及画法"
+                             ],
+        "source":  "19-20 四年级上·湖南岳阳·期末"
+    },
+    {
+        "id":  "17",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "聪聪在同一平面上画了一条直线的三条垂线，这三条垂线的关系是( )。",
+        "options":  [
+                        "A. 相交",
+                        "B. 互相平行",
+                        "C. 互相垂直"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】在同一平面内两条直线相交成直角，这两条直线互相垂直;同一平面内永不相交的两条直线叫做平行线，组成平行线的两条直线互相平行，由此可知，垂直于同条直线的垂线互相平行，依此选择。熟练掌握平行与垂直的特点是解答此题的关键.\n【详解】根据题意，画图如下:![](images/647d7c956a19c24bf671543445b3b4314b9faf47db30aaaf48eaae001d82c97b.jpg)由此可知，这三条垂线的关系是互相平行。故答案为:B.",
+        "knowledge_points":  [
+                                 "垂直的特征，平行的特征及性质"
+                             ],
+        "source":  "22-23 四年级上·河北保定·期末"
+    },
+    {
+        "id":  "18",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面说法错误的是( )。",
+        "options":  [
+                        "A. 平行四边形不容易变形",
+                        "B. 正方形相邻的两条边互相垂直",
+                        "C. 长方形是特殊的平行四边形"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】A。平行四边形易变形，具有不稳定性;B。正方形的四个角都是直角，因此相邻的两条边互相垂直;C。两组对边分别平行的四边形叫做平行四边形，长方形具有平行四边形的所有性质，因此长方形是特殊的平行四边形。此题主要考查正方形，长方形，平行四边形的特征及平行四边形的易变形的特点.\n【详解】A。平行四边形容易变形，所以原题说法错误;B。正方形相邻的两条边互相垂直，此说法正确;C。长方形是特殊的平行四边形，此说法正确。故答案为:A.",
+        "knowledge_points":  [
+                                 "平行四边形的概念及特点，长方形的概念及特点，正方形的概念及特点，平行四边形的不稳定性及应用"
+                             ],
+        "source":  "22-23 四年级上·湖南岳阳·期末"
+    },
+    {
+        "id":  "19",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一个四边形的两组对边分别平行，这个四边形不可能是( )。",
+        "options":  [
+                        "A. 正方形",
+                        "B. 长方形",
+                        "C. 梯形"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】平行四边形的两组对边分别平行;正方形和长方形是特殊的平行四边形;梯形只有一组对边平行，据此解答。此题根据正方形，长方形，平行四边形，梯形的特征进行解答.\n【详解】一个四边形的两组对边分别平行，这个四边形不可能是(梯形)。故答案为:C.",
+        "knowledge_points":  [
+                                 "梯形的概念及特点，正方形的概念及特点，长方形的概念及特点，平行四边形的概念及特点"
+                             ],
+        "source":  "21-22 四年级上·广东广州·期末"
+    },
+    {
+        "id":  "20",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "两个高相等的平行四边形拼在一起( )是平行四边形。",
+        "options":  [
+                        "A. 一定",
+                        "B. 可能",
+                        "C. 不可能"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】两组对边分别平行的四边形，叫做平行四边形;从平行四边形一条边上的一点向对边引一条垂线，这点和垂足之间的线段叫做平行四边形的高，垂足所在的边叫做平行四边形的底，依此画图并选择即可。此题考查的是平面图形的拼接，熟练掌握平行四边形的高和平行四边形的特点是解答此题的关键.\n【详解】画图如下(平行四边形的高都是 2 厘米):![](images/4890281d96f76d8bfaa4016f65fc1968b8b3b520650a8ff4015b809a13fcfd7f.jpg)\n由此可知，两个高相等的平行四边形拼在一起可能是平行四边形。故答案为:B.",
+        "knowledge_points":  [
+                                 "平面图形的拼接，平行四边形的高及画法，平行四边形的概念及特点"
+                             ],
+        "source":  "21-22 四年级上·广东广州·期末"
+    },
+    {
+        "id":  "21",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面说法正确的是( )。",
+        "options":  [
+                        "A. 两个锐角的和一定大于直角",
+                        "B. 伸缩门是利用平行四边形容易变形的特性",
+                        "C. 黄老师家有两室两厅，面积约是 90 平方千米。",
+                        "D. 一个梯形可以分成一个平行四边形和一个等腰梯形"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据角的概念，平行四边形的特性以及面积单位的大小和梯形的特征等知识，分析解答即可。本题考查了角的概念，平行四边形的特性以及面积单位的大小和梯形的特征等知识，结合题意分析解答即可.\n【详解】A。两个锐角的和不一定大于直角，例如 40°+40°=80°,80°是锐角，所以本选项原说法错误;B。伸缩门是利用平行四边形容易变形的特性，正确;C。黄老师家有两室两厅，面积约是90平方米，所以本选项原说法错误;D。一个梯形可以分成一个平行四边形和一个三角形，所以本选项原说法错误。故答案为:B.",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征，平行四边形的不稳定性及应用"
+                             ],
+        "source":  "21-22 四年级上·广东广州·期末"
+    },
+    {
+        "id":  "22",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "以下说法错误的是( )。",
+        "options":  [
+                        "A. 梯形有两组对边平行",
+                        "B. 平行四边形对边平行且相等",
+                        "C. 平行四边形具有不稳定性",
+                        "D. 梯形有无数条高"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】先对每个选项中的说法进行分析并判断，然后选择错误的一项即可。A。只有一组对边平行的四边形叫做梯形。B。平行四边形对边平行且相等。C。平行四边形容易变形，具有不稳定性。D。梯形上底到下底的距离是梯形的高，依此判断。解答此题的关键是要熟练掌握梯形和平行四边形的特点，平行四边形的不稳定性，以及梯形的高及画法.\n【详解】A。梯形只有一组对边平行，原说法错误。B。平行四边形对边平行且相等，原说法正确。C。平行四边形具有不稳定性，原说法正确。D。梯形有无数条高，原说法正确。故答案为:A.",
+        "knowledge_points":  [
+                                 "平行四边形的概念及特点，梯形的高及画法，平行四边形的不稳定性及应用，梯形的概念及特点"
+                             ],
+        "source":  "22-23 四年级上·广东广州·期末"
+    },
+    {
+        "id":  "23",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面的说法，错误的是( )。",
+        "options":  [
+                        "A. 平行四边形的对边平行且相等。",
+                        "B. 平行四边形不容易变形。",
+                        "C. 长方形和正方形都是特殊的平行四边形。"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】先对每个选项中的说法进行分析并判断，然后选择错误的一项即可。A。两组对边分别平行的四边形，叫做平行四边形，平行四边形的对边分别相等。B。平行四边形容易变形。C。长方形的两组对边分别互相平行，两组对边分别相等，四个角都是直角;正方形的两组对边分别互相平行，相邻的两条边相互垂直，四条边都相等，四个角都是直角;依此判断。解答此题的关键是要熟练掌握平行四边形的特点，四边形的分类及其关系，四边形的不稳定性.\n【详解】A。平行四边形的对边平行且相等，即原说法正确。B。平行四边形不容易变形，即原说法错误。C。长方形和正方形都是特殊的平行四边形，即原说法正确。故答案为:B.",
+        "knowledge_points":  [
+                                 "平行四边形的不稳定性及应用，四边形的分类及关系，平行四边形的概念及特点"
+                             ],
+        "source":  "22-23四年级上·广东广州·期末"
+    },
+    {
+        "id":  "24",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列说法中错误的是( )。",
+        "options":  [
+                        "A. 平行四边形是特殊的长方形",
+                        "B. 长方形是特殊的平行四边形",
+                        "C. 梯形是特殊的四边形",
+                        "D. 正方形是特殊的长方形"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】平行四边形，梯形，长方形，正方形都是四边形，长方形和正方形是特殊的平行四边形，正方形是特殊的长方形，据此即可解答。本题主要考查学生对四边形分类知识的掌握.\n【详解】A。长方形是特殊的平行四边形，原说法错误。B。长方形，正方形是特殊的平行四边形，原说法正确。C。梯形是特殊的四边形，原说法正确。D。正方形是特殊的长方形，原说法正确。故答案为:A.",
+        "knowledge_points":  [
+                                 "四边形的分类及关系"
+                             ],
+        "source":  "21-22四年级上·湖北黄冈·期末"
+    },
+    {
+        "id":  "25",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面说法正确的是( )。",
+        "options":  [
+                        "A. 平行四边形也是长方形",
+                        "B. 平行四边形是特殊的梯形",
+                        "C. 梯形是特殊的平行四边形",
+                        "D. 两个完全相同的梯形一定可以拼成一个平行四边形"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】长方形对边相等，有四个直角;平行四边形两组对边分别平行且相等，有四个角;梯形只有一组对边平行，且不相等，有四个角;根据长方形，平行四边形和梯形的特征，将四个选项分析，即可解答。本题考查学生对四边形之间关系的掌握。熟练掌握各种四边形的特征是解决此题的关键.\n【详解】A。根据长方形和平行四边形的特征可知，当平行四边形的四个角都是直角时，这个四边形就是长方形，所以长方形是特殊的平行四边形，选项说法错误;B。根据平行四边形和梯形的特征可知，平行四边形和梯形是四边形的两种，两者之间没有关系，所以选项说法错误;C。根据平行四边形和梯形的特征可知，平行四边形和梯形是四边形的两种，两者之间没有关系，所以选项说法错误;D。两个完全相同的梯形，将其中一个梯形的上底和另一个梯形的下底拼在一起，能够组成一个平行四边形，选项说法正确。故答案为:D.",
+        "knowledge_points":  [
+                                 "梯形的概念及特点，四边形的分类及关系，长方形的概念及特点，平行四边形的概念及特点"
+                             ],
+        "source":  "21-22四年级上·天津河北·期末"
+    },
+    {
+        "id":  "26",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列说法错误的是( )。",
+        "options":  [
+                        "A. 长方形相邻的两条边互相垂直",
+                        "B. 平行四边形有无数条高，梯形只有1条",
+                        "C. 同一平面内，如果直线 a//c,直线 b//c,那么直线 a//b"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】长方形两组对边相等，有四个直角;从平行四边形一条边上的一点向对边引一条垂线，这点和垂足之间的线段叫做平行四边形的高，垂足所在的边叫做平行四边形的底;从梯形上底的一点向对边引一条垂线，这点和垂足之间的线段叫做梯形的高;平行于同一条直线的两条直线互相平行;根据长方形的特征，平行四边形和梯形的高的知识以及平行知识，结合选项，逐一分析解答即可。本题考查了长方形的特征，平行四边形和梯形的高的知识以及平行知识，结合题意分析解答即可.\n【详解】A。长方形相邻的两条边互相垂直，长方形的内角都是直角，选项说法正确;B。根据平行四边形高的定义和梯形高的定义可知，平行四边形有无数条高，梯形也有无数条高，选项说法错误;C。直线a∥c,直线b∥c,直线a,b都平行于直线c,由分析可知，同平面内，如果直线a∥c,直线b∥c,那么直线a∥b,选项说法正确。故答案为:B.",
+        "knowledge_points":  [
+                                 "平行的特征及性质，垂直的特征，平行四边形的高及画法，梯形的高及画法"
+                             ],
+        "source":  "21-22 四年级上·山东临沂·期末"
+    },
+    {
+        "id":  "27",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "同一平面内，两条直线相互垂直，可以组成( )个直角。",
+        "options":  [
+                        "A. 2",
+                        "B. 3",
+                        "C. 4"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】两条直线相交成直角，就说这两条直线互相垂直，其中一条直线叫做另一条直线的垂线，这两条直线的交点叫做垂足。根据垂直的定义，即可解答。本题主要考查学生对垂直特性的掌握。熟练掌握垂直的定义，结合图形，能够快速解决此题.\n【详解】![](images/9f805eb232470f32222163154c9b6e85bbdbd4683b72672ac1fff7a093983017.jpg)直线可以无限延伸。如图所示，同一平面内，两条直线互相垂直，可以组成 4 个直角。选项 C 符合题意。故答案为:C.",
+        "knowledge_points":  [
+                                 "垂直的特征"
+                             ],
+        "source":  "21-22 四年级上·广西玉林·期末"
+    },
+    {
+        "id":  "28",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "丽丽在同一平面上画了一条直线的 5 条垂线，这 5 条垂线的关系是( )。",
+        "options":  [
+                        "A. 可能相交",
+                        "B. 互相垂直",
+                        "C. 互相平行"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】平行线的性质：在同一平面内，如果两条直线都与另一条直线互相垂直，那么这两条直线互相平行。此题考查了平行的特征和性质，要熟练掌握.\n【详解】丽丽在同一平面上画了一条直线的5条垂线，这5条垂线的关系是互相平行。故答案为:C",
+        "knowledge_points":  [
+                                 "平行的特征及性质"
+                             ],
+        "source":  "21-22 四年级上·河北保定·期末"
+    },
+    {
+        "id":  "29",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "把一张平行四边形纸片剪一刀，剪成的图形可能是( )。",
+        "options":  [
+                        "A. 一个正方形和一个梯形",
+                        "B. 一个三角形和一个梯形",
+                        "C. 两个长方形"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】如下图，把一张平行四边形纸片剪一刀，可以分成两个三角形，或两个梯形，或两个平行四边形，或一个三角形和一个梯形，但不能分成一个正方形和一个梯形，或两个长方形，据此即可解答。本题主要考查了平行四边形的分割，动手画一下就能清楚地知道可以分成哪些图形。![](images/cecdb4676e6ae03b273fcfe00d9697f5b74183f59b395d061e6e7dd9b379324d.jpg)\n【详解】根据分析可知，把一张平行四边形纸片剪一刀，剪成的图形可能是一个三角形和一个梯形，但不能分成一个正方形和一个梯形，或两个长方形。故答案为:B",
+        "knowledge_points":  [
+                                 "平行四边形的切拼"
+                             ],
+        "source":  "21-22 四年级上·山东济南·期末"
+    },
+    {
+        "id":  "30",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列说法错误的是( )。",
+        "options":  [
+                        "A. 平行四边形是特殊的梯形",
+                        "B. 平行四边形容易变形",
+                        "C. 梯形只有一组对边平行",
+                        "D. 过一点能画无数条直线"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】先对每个选项中的说法进行分析并判断，然后选择错误的一项即可。A。两组对边分别平行的四边形，叫做平行四边形;只有一组对边平行的四边形叫做梯形;依此判断。B。平行四边形具有不稳定性，容易变形。C。只有一组对边平行的四边形叫做梯形。D。过一点能画无数条直线。解答此题的关键是要熟练掌握梯形的特点，平行四边形的不稳定性，四边形的分类及其关系，以及过一点画直线的条数.\n【详解】A。平行四边形不是特殊的梯形，即原说法错误。B。平行四边形容易变形，即原说法正确。C。梯形只有一组对边平行，即原说法正确。D。过一点能画无数条直线，即原说法正确。故答案为:A.",
+        "knowledge_points":  [
+                                 "四边形的分类及关系，线段，直线，射线的认识及特征，平行四边形的不稳定性及应用，梯形的概念及特点"
+                             ],
+        "source":  "21-22 四年级上·贵州六盘水·期末"
+    },
+    {
+        "id":  "31",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "梯形有( )条高。",
+        "options":  [
+                        "A. 1",
+                        "B. 2",
+                        "C. 4",
+                        "D. 无数"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】梯形上底到下底的距离是梯形的高;梯形的高：在梯形的上底上任意找一点，过这个点向下底作垂线，这个点到垂足之间的线段就是梯形的高，依此画图并选择。熟练掌握梯形的高及画法是解答此题的关键.\n【详解】![](images/cf9a4b0af9508f696df44e2743d952251310aef4d78df2fbb8e7afaeb1f8d7f0.jpg)由此可知，梯形有无数条高。故答案为:D.",
+        "knowledge_points":  [
+                                 "梯形的高及画法"
+                             ],
+        "source":  "21-22 四年级上·贵州六盘水·期末"
+    },
+    {
+        "id":  "32",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "学校的电动伸缩门是利用了平行四边形的( )。",
+        "options":  [
+                        "A. 稳定性",
+                        "B. 容易变形的特点",
+                        "C. 节省材料"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】平行四边形容易变形，具有不稳定性，依此选择。熟练掌握平行四边形的不稳定性及应用是解答此题的关键.\n【详解】根据分析可知，学校的电动伸缩门是利用了平行四边形容易变形的特点。故答案为:B",
+        "knowledge_points":  [
+                                 "平行四边形的不稳定性及应用"
+                             ],
+        "source":  "21-22 四年级上·西藏林芝·期末"
+    },
+    {
+        "id":  "33",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "两个完全一样的正方形，只能拼成一个( )。",
+        "options":  [
+                        "A. 长方形",
+                        "B. 平行四边形",
+                        "C. 正方形"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】正方形的四条边都相等，有4个直角;长方形的对边相等，有4个直角;平行四边形的对边相互平行且相等，依此画图并选择。此题考查的是平面图形的拼接，熟练掌握正方形和长方形的特点是解答此题的关键.\n【详解】![](images/c47a5592dad448e29ca0f52faa1b08619b644751a4bbe7703845ed369d0f0219.jpg)即两个完全一样的正方形，只能拼成一个长方形。故答案为:A.",
+        "knowledge_points":  [
+                                 "平面图形的拼接，正方形的概念及特点，长方形的概念及特点，平行四边形的概念及特点"
+                             ],
+        "source":  "21-22三年级上·山东临沂·期末"
+    },
+    {
+        "id":  "34",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列说法中正确的是( )。",
+        "options":  [
+                        "A. 周角是一条射线",
+                        "B. 两个锐角一定能拼成一个钝角",
+                        "C. 伸缩门应用了平行四边形易变形的特性",
+                        "D. 有一组对边平行的四边形是梯形"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】A。根据角的含义\u0027由一点引出的两条射线所围成的图形，叫做角\u0027可知：周角的特点是两条边重合成射线。但不能说成周角是一条射线，因为角和线是两个不同的概念，两者不能混淆;据此判断即可;B。锐角大于0度小于90度，据此举例解答;C。根据平行四边形不具有稳定性，进行判断即可;D。根据梯形的特征，只有一组对边平行的四边形是梯形，进行判断。熟练掌握平行四边形的周长的计算是解答此题的关键.\n【详解】A。根据角的含义可知：周角的特点是两条边重合成射线。但不能说成周角是一条射线，因为角和线是两个不同的概念，两者不能混淆，所以原题说法错误;B。两个锐角可能拼成一个锐角或一个钝角或一个直角，如10度+10度=20度，是个锐角,35度+55度=90度，是个直角,75度+30度=105度，是个钝角;所以原题说法错误;C。伸缩门应用了平行四边形易变形的特性，此说法正确;D。只有一组对边平行的四边形是梯形，所以原题说法错误。故答案为:C",
+        "knowledge_points":  [
+                                 "角的分类及换算",
+                                 "平行四边形的不稳定性及应用",
+                                 "梯形的概念及特点",
+                                 "平面图形的拼接"
+                             ],
+        "source":  "21-22四年级上·天津南开·期末"
+    },
+    {
+        "id":  "35",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面说法中不正确的有( )个.\n①两条平行线长都是4厘米.\n②已知客车3小时行的路程是120千米，可以求出它的平均速度.\n③两个完全相同的梯形一定能拼成一个长方形。",
+        "options":  [
+                        "A. 3",
+                        "B. 2",
+                        "C. 1"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据题意,①根据平行线的定义：在同一平面内，不相交的两条直线叫做平行线;由此可知：两条平行线都是直线，无限长;据此判断;②已知客车3小时行的路程是120千米，根据速度=路程/时间，可以求出它的平均速度;③两个完全相同的梯形一定能拼成一个平行四边形;只有两个完全相同的直角梯形才能拼成一个长方形;据此判断。本题考查了平行线，速度=路程/时间以及梯形的拼组知识，结合题意分析解答即可.\n【详解】①两条平行线长都是4厘米，说法错误，因为平行线都是直线，无限长。②已知客车3小时行的路程是120千米，根据速度=路程/时间，可以求出它的平均速度，列式为:120÷3=40(千米/小时),所以原题说法正确。③两个完全相同的梯形一定能拼成一个平行四边形，不一定能拼成长方形，只有两个完全相同的直角梯形才能拼成一个长方形，所以原题说法错误。所以上面说法中不正确的有2个。故答案为:B.",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "平行的特征及性质",
+                                 "梯形的概念及特点",
+                                 "平面图形的拼接",
+                                 "用同样的图形拼接"
+                             ],
+        "source":  "21-22 四年级上·天津和平·期末"
+    },
+    {
+        "id":  "36",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一条直线既垂直于直线a也垂直于直线b,那么直线a和直线b( )。",
+        "options":  [
+                        "A. 相交",
+                        "B. 平行",
+                        "C. 垂直",
+                        "D. 无法确定"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据垂直的性质：垂直于同一条直线的两条直线平行;进行解答即可。解题的关键是熟练掌握垂直的性质，本题是一个基础题.\n【详解】根据分析可知，根据垂直和平行的特征可知：一条直线既垂直于直线a也垂直于直线b,那么直线a和直线b平行;故答案为:B.",
+        "knowledge_points":  [
+                                 "垂直的特征",
+                                 "平行的特征及性质"
+                             ],
+        "source":  "20-21 四年级上·湖北鄂州·期末"
+    },
+    {
+        "id":  "37",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "平行四边形的高是一条( )。",
+        "options":  [
+                        "A. 垂线",
+                        "B. 线段",
+                        "C. 直线"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】从平行四边形一条边上的一点向对边引一条垂线，这点和垂足之间的线段叫做平行四边形的高，垂足所在的边叫做平行四边形的底，依此选择。熟练掌握平行四边形的高及画法是解答此题的关键.\n【详解】平行四边形的高是一条线段。故答案为:B.",
+        "knowledge_points":  [
+                                 "平行四边形的高及画法"
+                             ],
+        "source":  "20-21四年级上·湖北鄂州·期末"
+    },
+    {
+        "id":  "38",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "捏住长方形的两个对角，向相反方向拉，拉完之后图形的周长( )。",
+        "options":  [
+                        "A. 不变",
+                        "B. 变长",
+                        "C. 变短"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】用手捏住长方形的两个对角，向相反方向拉，把长方形拉成平行四边形，四个边的长度没变，则其周长不变。此题主要考查了长方形和平行四边形周长公式的灵活运用.\n【详解】根据分析可知，用手捏住长方形的两个对角，向相反方向拉，把长方形拉成平行四边形，拉完之后图形的周长不变。故答案为:A.",
+        "knowledge_points":  [
+                                 "平行四边形的概念及特点",
+                                 "长方形的周长",
+                                 "周长的认识和大小比较",
+                                 "平行四边形的周长"
+                             ],
+        "source":  "21-22 四年级上·黑龙江绥化·期末"
+    },
+    {
+        "id":  "39",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "两个完全一样的三角形不可能拼成一个( )。",
+        "options":  [
+                        "A. 平行四边形",
+                        "B. 三角形",
+                        "C. 梯形"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】两个完全一样的三角形不可能拼成一个梯形，因为根据三角形的面积推导过程;两个完全一样的三角形可以拼出一个平行四边形;两个完全一样的直角三角形可以拼出一个正方形或等腰三角形，据此解答。本题考查平行四边形，梯形的特征，熟练掌握并灵活运用.\n【详解】A。![](images/b905755b8df9f76277a1dbdbeb8c6a57faa8ff436e9509c7428a05baf43bf0db.jpg)由分析可得，该选项正确;B。![](images/00a4ef801d40a027659e6c3d59b4fd6bc8207ec6300a0604af7023a98e59079b.jpg)由分析可得，该选项正确;C。![](images/7afc345b849de9190a922ce00dcba6d1e30edb2583e50f3e28976cdf34e5c32d.jpg)由分析可得，该选项错误。两个完全一样的三角形不可能拼成一个(梯形)。故答案为:C",
+        "knowledge_points":  [
+                                 "平行四边形的概念及特点",
+                                 "梯形的概念及特点"
+                             ],
+        "source":  "21-22 四年级上·山西阳泉·期末"
+    },
+    {
+        "id":  "40",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "把一张正方形纸对折两次，展开后折痕( )。",
+        "options":  [
+                        "A. 一定互相垂直",
+                        "B. 一定互相平行",
+                        "C. 互相垂直或平行",
+                        "D. 以上都不对"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】把一张正方形纸对折两次，有两种折法，一种是纵对折再对折，此时形成的两条折痕互相平行;一种是横对折再纵对折，此时形成的两条折痕互相垂直。本题是考查简图形的折叠问题，此题不难理解，只要动手操作一下即可解决问题.\n【详解】如图:![](images/f6dd8addf3ec01b52ea4751900378ded.jpg)所以，把一张正方形纸对折两次，展开后折痕互相平行或垂直。故答案为:C.",
+        "knowledge_points":  [
+                                 "平行的特征及性质",
+                                 "垂直的特征",
+                                 "图形的折叠问题",
+                                 "正方形的概念及特点"
+                             ],
+        "source":  "21-22 四年级上·河南三门峡·期末"
+    },
+    {
+        "id":  "41",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列说法正确的是( )。",
+        "options":  [
+                        "A. 过直线外一点可以作无数条垂线。",
+                        "B. 如果a∥b,b∥c,则a⊥b。",
+                        "C. 两个完全一样的梯形一定能拼成平行四边形。"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】根据过直线外一点可以作一条垂线;如果a∥b,b∥c,则a∥b;两个完全一样的梯形一定能拼成平行四边形，解答即可。本题考查了过直线外一点可以作一条垂线，平行的特征及两个完全一样的梯形一定能拼成平行四边形知识，结合题意分析解答即可.\n【详解】A。过直线外一点可以作一条垂线，所以原选项说法错误;B。如果a∥b,b∥c,则a∥b,所以原选项说法错误;C。两个完全一样的梯形一定能拼成平行四边形，所以原选项说法正确。故答案为:C.",
+        "knowledge_points":  [
+                                 "平行的特征及性质",
+                                 "画垂线",
+                                 "平行四边形的切拼"
+                             ],
+        "source":  "21-22 四年级上·江西吉安·期末"
+    },
+    {
+        "id":  "42",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "过直线外两点分别作这条直线的垂线，这两条垂线的位置关系是( )。",
+        "options":  [
+                        "A. 垂直",
+                        "B. 平行",
+                        "C. 相交"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】把三角板的一条直角边与已知直线重合，使三角板的另一条直角边和已知直线上任意一点重合，过这点沿三角板的直角边，向已知直线画直线即可画出已知直线的垂线，用同样的方法画出即可。观察画的这两条垂线发现：一条直线的垂线有无数条，这些垂线相互平行。本题考查了学生垂线的认识，培养学生的作图能力，一条直线的垂线有无数条，这些垂线相互平行.\n【详解】如图:![](images/a0c769127854e9ca4336a2b432718a279875779b82888c9463cb95aa7026932f.jpg)观察画的这两条垂线，这两条垂线的位置关系是相互平行。故答案为:B.",
+        "knowledge_points":  [
+                                 "垂直的特征",
+                                 "画垂线"
+                             ],
+        "source":  "21-22 四年级上·江西吉安·期末"
+    },
+    {
+        "id":  "43",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列说法错误的是( )。",
+        "options":  [
+                        "A. 正方形相邻的两条边互相垂直",
+                        "B. 平行四边形和梯形都有无数条高",
+                        "C. 不相交的两条直线一定互相平行"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】正方形的两组对边分别互相平行，相邻的两条边相互垂直，四条边都相等，四个角都是直角;在同一平面内两条直线相交成直角，这两条直线互相垂直;从平行四边形一条边上的一点向对边引一条垂线，这点和垂足之间的线段叫做平行四边形的高，垂足所在的边叫做平行四边形的底;梯形上底到下底的距离是梯形的高;在同一平面内，永不相交的两条直线叫做平行线，组成平行线的两条直线互相平行，依此选择。此题考查的是垂直与相交的特点，以及平行四边形和梯形的高，应熟练掌握.\n【详解】A。正方形相邻的两条边互相垂直，即原说法正确;B。平行四边形和梯形都有无数条高，即原说法正确;C。在同一平面内，永不相交的两条直线一定互相平行，即原说法错误。故答案为:C.",
+        "knowledge_points":  [
+                                 "平行的特征及性质",
+                                 "垂直的特征",
+                                 "平行四边形的高及画法",
+                                 "梯形的高及画法"
+                             ],
+        "source":  "21-22 四年级上·江西赣州·期末"
+    },
+    {
+        "id":  "44",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面说法中，有()个是正确的.\n①一个八位数，它的最高位是亿位.\n②两条平行线长都是4千米.\n③梯形只有一组对边是平行的.\n④1平方千米=100公顷=10000平方米。",
+        "options":  [
+                        "A. 1",
+                        "B. 2",
+                        "C. 3"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】①根据整数数位顺序表，一个八位数，它的最高位是千万位，据此判断即可;②根据平行线的定义：在同一平面内，不相交的两条直线叫做平行线;由此可知：两条平行线都是直线，无限长;据此判断即可;③只有一组对边平行的四边形叫做梯形，据此判断;④根据1平方千米=100公顷,1公顷=10000平方米即可求解。此题主要考查了整数数位顺序表的应用，以及梯形的特征和面积单位间的进率以及换算的方法，要熟练掌握.\n【详解】①根据整数数位顺序表，一个八位数，它的最高位是千万位，所以原题说法不正确;②因为直线是无限延伸的，所以两条平行线长都是4千米的说法不正确;③只有一组对边平行的四边形叫做梯形，所以原题的说法正确;④因为1平方千米=100公顷=1000000平方米，所以原题的说法不正确;由此可知，上面说法中，有1个是正确的。故答案为:A.",
+        "knowledge_points":  [
+                                 "整数的数级，数位和计数单位的认识",
+                                 "线段，直线，射线的认识及特征",
+                                 "梯形的概念及特点",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "20-21四年级上·广东江门·期末"
+    },
+    {
+        "id":  "45",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列描述中，正确的是( )。",
+        "options":  [
+                        "A. 直线上两点之间的一段叫做线段",
+                        "B. 小明画了一条长5厘米的射线",
+                        "C. 当两条直线相交成直角时，这两条直线互相垂直",
+                        "D. 两点之间所有连线中线段最短"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】一根拉紧的线，绷紧的弦，都可以看作线段，线段有两个端点，可以测量出长度;把线段向一端无限延伸，就得到一条射线，射线只有一个端点，不可以测量出长度;在同一平面内两条直线相交成直角，这两条直线互相垂直，依此选择。解答此题的关键是要熟练掌握线段，射线的特点，以及垂直的特点.\n【详解】A。直线上两点之间的一段叫做线段，即原说法正确;B。小明画了一条长5厘米的线段，即原说法错误;C。当两条直线相交成直角时，这两条直线互相垂直，即原说法错误;D。两点之间所有连线中线段最短，即原说法错误。故答案为:A.",
+        "knowledge_points":  [
+                                 "垂直的特征",
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "20-21四年级上·浙江金华·期末"
+    },
+    {
+        "id":  "46",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "平行四边形中画一条线段，不可能把它分割成( )。",
+        "options":  [
+                        "A. 两个完全一样的梯形",
+                        "B. 两个完全一样的三角形",
+                        "C. 两个完全一样的平行四边形",
+                        "D. 两个完全一样的长方形"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】如图:![](images/04655c4fb835e686b9416fe067461c930ca618fdc701ccd635375ec03adcbcc5.jpg),在平行四边形中画一条线段，可以把它分割成两个完全一样的三角形，梯形和平行四边形，但不能分割成两个完全一样的长方形。熟练掌握平行四边形，梯形，三角形和长方形的特征，是解答此题的关键.\n【详解】根据分析可知,A。平行四边形中画一条线段，可以把它分割成两个完全一样的梯形;B。平行四边形中画一条线段，可以把它分割成两个完全一样的三角形;C。平行四边形中画一条线段，可以把它分割成两个完全一样的平行四边形;D。平行四边形中画一条线段，不可能把它分割成两个完全一样的长方形。故答案为:D.",
+        "knowledge_points":  [
+                                 "梯形的概念及特点",
+                                 "平行四边形的切拼",
+                                 "长方形的概念及特点",
+                                 "平行四边形的概念及特点"
+                             ],
+        "source":  "2022 四年级上·全国·专题练习"
+    },
+    {
+        "id":  "47",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面说法正确的是()。",
+        "options":  [
+                        "A. 平行线就是不相交的两条直线",
+                        "B. 两条直线相交就一定垂直",
+                        "C. 一个梯形的上底和下底之间的距离处处相等",
+                        "D. 梯形中的两组对边可以互相平行"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】选项A,在同一平面内，不相交的两条直线是平行线;选项B,在同一个平面上，当两条直线相交成直角时，这两条直线就互相垂直;由此判断即可;选项C,因为梯形的上底和下底互相平行，在梯形上底上任取一点，过这一点向下底作垂线段即为梯形的高;选项D,只有一组对边平行的四边形叫做梯形。据此解答。本题主要考查了平行的特征及性质等。解题关键是正确理解梯形的特征:\u0027梯形的两底平行\u0027,另外还要理解平行线的特征:\u0027两平行线间的距离处处相等\u0027.\n【详解】A。根据平行的定义，在同一平面内，不相交的两条直线是平行线;所以原题说法错误。B。在同一个平面上，当两条直线相交成直角时，这两条直线就互相垂直;由此可知：在同一平面内，垂直一定相交，但相交所成的角不一定是直角，所以相交不一定垂直;所以原题说法错误。C。因为梯形的上底和下底互相平行，在梯形上底上任取点，过这一点向下底作垂线段即为梯形的高。这样的线段可以作无数条，又因为梯形的上底和下底互相平行，因而这些高都相等。所以一个梯形的上底和下底之间的距离处处相等，此说法正确。D。只有一组对边平行的四边形叫做梯形。所以此项说法错误。故答案为:C.",
+        "knowledge_points":  [
+                                 "梯形的概念及特点",
+                                 "平行的特征及性质",
+                                 "垂直的特征"
+                             ],
+        "source":  "2022 四年级上·全国·专题练习"
+    },
+    {
+        "id":  "48",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "用完全相同的两副三角板可以拼成( )。",
+        "options":  [
+                        "A. 长方形",
+                        "B. 正方形",
+                        "C. 平行四边形",
+                        "D. 长方形，正方形，三角形，平行四边形"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】用两副完全相同的三角板可以拼成长方形，正方形，三角形和平行四边形，据此解题即可。解答此题时可以用三角板动手拼一拼，操作一下更直观.\n【详解】如图:![](images/07184107f0fd7ece0b16981302ebdf65.jpg)所以，用完全相同的两副三角板可以拼成长方形，正方形，三角形，平行四边形。故答案为:D.",
+        "knowledge_points":  [
+                                 "长方形的概念及特点",
+                                 "平行四边形的切拼",
+                                 "三角形的概念及表示方法",
+                                 "平面图形的拼接",
+                                 "用同样的图形拼接"
+                             ],
+        "source":  "2022 四年级上·全国·专题练习"
+    },
+    {
+        "id":  "49",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "张强在纸上画了两条直线，两条直线都垂直于纸上另一条直线，这两条直线()。",
+        "options":  [
+                        "A. 互相垂直",
+                        "B. 互相平行",
+                        "C. 相交"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】平行线的性质：在同一平面内，如果两条直线都与另一条直线互相垂直，那么这两条直线互相平行。此题考查了平行的特征和性质，要熟练掌握.\n【详解】如图所示:![](images/b3b601c74f210ee4f378901a3398cb9fb5598204605d15aebcbb20bd0e8889cf.jpg)张强在纸上画了两条直线，两条直线都垂直于纸上另一条直线，这两条直线互相平行。故答案为:B.",
+        "knowledge_points":  [
+                                 "垂直的特征",
+                                 "平行的特征及性质"
+                             ],
+        "source":  "2022四年级上·全国·专题练习"
+    },
+    {
+        "id":  "50",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "两个完全一样的直角梯形可以拼成一个( )。",
+        "options":  [
+                        "A. 长方形",
+                        "B. 等腰梯形",
+                        "C. 平行四边形",
+                        "D. 以上三种都有可能"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】两个完全一样的直角梯形可以拼成如下图形，据此解答。![](images/a5fe431c67c706a8d88642007aedc047.jpg)\n本题考查了长方形，等腰梯形，平行四边形的认识以及平面图形的切拼.\n【详解】由分析可知，两个完全一样的直角梯形可以拼成长方形，平行四边形和等腰梯形。故答案为:D.",
+        "knowledge_points":  [
+                                 "平行四边形的概念及特点",
+                                 "长方形的概念及特点",
+                                 "直角梯形和等腰梯形的概念及特点",
+                                 "平面图形的拼接"
+                             ],
+        "source":  "20-21 五年级上·湖南邵阳·期中"
+    },
+    {
+        "id":  "51",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一个平行四边形的高有( )。",
+        "options":  [
+                        "A. 1条",
+                        "B. 2条",
+                        "C. 4条",
+                        "D. 无数条"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】从平行四边形一条边上的一点到它的对边引一条垂线，这点和垂足之间的线段叫做平行四边形的高，从一条边的任一点都可以向对边做垂线段，所以平行四边形有无数条高。本题考查了平行四边形的高，明确平行四边形的高的定义是解题的关键.\n【详解】由分析可知：一个平行四边形的高有无数条。故答案为:D.",
+        "knowledge_points":  [
+                                 "平行四边形的高及画法"
+                             ],
+        "source":  "20-21五年级上·吉林长春·期中"
+    },
+    {
+        "id":  "52",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在一张纸上，任意画两条直线，这两条直线要么( ),要么( )。",
+        "options":  [
+
+                    ],
+        "answer":  "互相平行;相交",
+        "solution":  "【分析】在同一平面内的两条直线可能互相平行，也可能相交;同一平面内，永不相交的两条直线叫做平行线，组成平行线的两条直线互相平行;依此填空。熟练掌握平行与相交的特点是解答此题的关键.\n【详解】根据分析可知，在一张纸上，任意画两条直线，这两条直线要么互相平行，要么相交。",
+        "knowledge_points":  [
+                                 "平行的特征及性质"
+                             ],
+        "source":  "21-22四年级上·广东江门·期末"
+    },
+    {
+        "id":  "53",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "两条直线互相垂直可以得到( )个直角;捏住一个长方形两个对角向相反方向拉，得到的平行四边形的周长( )原来长方形的周长。(填\u0027大于\u0027,\u0027小于\u0027或\u0027等于\u0027)",
+        "options":  [
+
+                    ],
+        "answer":  "4/四;等于",
+        "solution":  "【分析】在同一平面内，两条直线相交成直角，这两条直线互相垂直，依此画图并填空。捏住一个长方形两个对角向相反方向拉，拉成一个平行四边形，即长方形的四条边的长度没有改变，根据对周长的认识进行填空即可。解答此题的关键是应熟练掌握是对周长的认识，以及垂直的特点.\n【详解】![](images/8f70fb233043a7076ec1e4ffdaef054a8dbad7abe8d7c54404bafc988ea409ff.jpg)由此可知，两条直线互相垂直可以得到4个直角;捏住一个长方形两个对角向相反方向拉，围成图形一周的长度不变，所以得到的平行四边形的周长等于原长方形的周长。",
+        "knowledge_points":  [
+                                 "平行四边形的概念及特点",
+                                 "长方形的概念及特点",
+                                 "垂直的特征",
+                                 "周长的认识和大小比较"
+                             ],
+        "source":  "21-22 四年级上·广东东莞·期末"
+    },
+    {
+        "id":  "54",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个平行四边形，有两条边分别是8cm和5cm,一条高是3cm,它的周长是()cm。",
+        "options":  [
+
+                    ],
+        "answer":  "26",
+        "solution":  "【分析】平行四边形的两组对边分别平行且相等，由题目可知，这个平行四边形的一组邻边分别是8cm和5cm,根据周长的定义，求出平行四边形一组邻边之和，再乘2，即可算出这个平行四边形的周长是多少。据此解答。本题主要考查平行四边形的特征和周长，属于基础知识，要熟练掌握.\n【详解】(8+5)×2=13×2=26(cm),一个平行四边形，有两条边分别是8cm和5cm,一条高是3cm,它的周长是26cm。",
+        "knowledge_points":  [
+                                 "平行四边形的周长",
+                                 "平行四边形的概念及特点"
+                             ],
+        "source":  "21-22 四年级上·湖南长沙·期末"
+    },
+    {
+        "id":  "55",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "桌面上摆了3根小棒，分别是a⊥b,b⊥c,那么a和c这两根小棒互相( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平行",
+        "solution":  "【分析】同一平面内永不相交的两条直线叫做平行线，组成平行线的两条直线互相平行;在同一平面内两条直线相交成直角，这两条直线互相垂直，因此垂直于同一直线的两条直线互相平行;依此画图并填空即可。熟练掌握平行与垂直的特点是解答此题的关键.\n【详解】画图如下:![](images/50d284f089a4a3dc9a658ce16a5ae8bd1c6a2b85197e01613e76371c7da380ee.jpg)由此可知,a和c这两根小棒互相平行。",
+        "knowledge_points":  [
+                                 "垂直的特征",
+                                 "平行的特征及性质"
+                             ],
+        "source":  "20-21 四年级上·湖南株洲·期末"
+    },
+    {
+        "id":  "56",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "两条直线相交成直角时，这两条直线叫做互相垂直，其中的一条叫做另一条的( ),它们的交点叫做( )。",
+        "options":  [
+
+                    ],
+        "answer":  "垂线;垂足",
+        "solution":  "【详解】在同一平面内的两条不重合的直线，只有两种位置关系，不是相交就是平行，垂直是相交的特殊情况。两条直线相交成直角时，这两条直线互相垂直，其中一条直线是另一条直线的垂线，这两条直线的交点叫做垂足。",
+        "knowledge_points":  [
+                                 "垂直的特征"
+                             ],
+        "source":  "21-22 四年级上·湖南娄底·期末"
+    },
+    {
+        "id":  "57",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰梯形的周长是29厘米，已知它的上底是4厘米，下底是7厘米，它的其中一条腰是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "9",
+        "solution":  "【分析】等腰梯形的两腰相等，因此用等腰梯形的周长减去上底和下底的长度之和，然后再除以2即可，依此计算。熟练掌握等腰梯形的特点是解答此题的关键.\n【详解】4+7=11(厘米),29-11=18(厘米),18÷2=9(厘米),即它的其中一条腰是9厘米。",
+        "knowledge_points":  [
+                                 "直角梯形和等腰梯形的概念及特点",
+                                 "梯形的周长"
+                             ],
+        "source":  "21-22四年级上·云南玉溪·期末"
+    },
+    {
+        "id":  "58",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果一个平行四边形两邻边之和是9分米，这个平行四边形的周长是( )分米。",
+        "options":  [
+
+                    ],
+        "answer":  "18",
+        "solution":  "【分析】平行四边形的对边相等，且它两条邻边之和是9分米，所以它的周长用9乘2即可。平行四边形的周长等于相邻两边之和乘2，平行四边形的对边平行且相等.\n【详解】9×2=18(分米),这个平行四边形的周长是18分米。",
+        "knowledge_points":  [
+                                 "平行四边形的周长",
+                                 "周长的认识和大小比较",
+                                 "平行四边形的概念及特点"
+                             ],
+        "source":  "20-21三年级上·北京朝阳·期末"
+    },
+    {
+        "id":  "59",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小亮想要验证a∥b,就从直线a上任取一些点并分别向直线b作垂线，通过测量这些垂直的线段相等即可判断a∥b,这样做的依据是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平行线之间的距离处处相等",
+        "solution":  "【分析】两直线互相平行时，从一条直线上任意一点向另条直线作垂线，所得的平行线间的垂直线段的长度，叫做平行线间的距离。平行线之间的距离处处相等;据此解题即可。熟练掌握平行的特征及性质，是解答此题的关键.\n【详解】根据分析可知，小亮想要验证a∥b,就从直线a上任取一些点并分别向直线b作垂线，通过测量这些垂直的线段相等即可判断a∥b,这样做的依据是平行线之间的距离处处相等。",
+        "knowledge_points":  [
+                                 "平行的特征及性质"
+                             ],
+        "source":  "21-22 四年级上·山东临沂·期末"
+    },
+    {
+        "id":  "60",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在一张梯形纸上剪下一个最大的平行四边形，剩下的部分一定是( )形。",
+        "options":  [
+
+                    ],
+        "answer":  "三角",
+        "solution":  "【分析】过梯形的上底的一个顶点，向一条腰作平行线，这条平行线把梯形分成一个平行四边形和一个三角形;过梯形上底一点，作一条腰的平行线，可以把这个梯形分成一个平行四边形和一个梯形。在一张梯形纸上剪一刀，剪下一个平行四边形，剩下的图形可能是三角形，梯形。在一张梯形纸上剪下一个最大的平行四边形，剩下的部分一定是三角形。解答此题的关键是掌握梯形，平行四边形的特征，画图更容易直观理解.\n【详解】如图:![](images/c724a714a890e39ce071adfe99d2381b.jpg)在一张梯形纸上剪下一个最大的平行四边形，剩下的部分一定是三角形。",
+        "knowledge_points":  [
+                                 "平面图形的分割",
+                                 "梯形的概念及特点",
+                                 "平行四边形的概念及特点"
+                             ],
+        "source":  "21-22 四年级上·山东临沂·期末"
+    },
+    {
+        "id":  "61",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在四边形中，只有一组对边平行的四边形，叫做( );两组对边平行且相等，有四个直角的四边形，叫做( )。",
+        "options":  [
+
+                    ],
+        "answer":  "梯形;长方形",
+        "solution":  "【分析】只有一组对边平行的四边形叫做梯形;两组对边平行且相等，且有四个直角的四边形叫做长方形。此题考查的目的是理解掌握梯形和长方形的特征.\n【详解】在四边形中，只有一组对边平行的四边形，叫做梯形;两组对边平行且相等，有四个直角的四边形，叫做长方形。",
+        "knowledge_points":  [
+                                 "正方形的概念及特点",
+                                 "梯形的概念及特点"
+                             ],
+        "source":  "21-22 四年级上·广西玉林·期末"
+    },
+    {
+        "id":  "62",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "当一个梯形的两腰相等时，这是一个( )梯形，当梯形中含有一个直角时，这是一个( )梯形，梯形有( )条高。",
+        "options":  [
+
+                    ],
+        "answer":  "等腰;直角;无数",
+        "solution":  "【分析】只有一组对边平行的四边形叫做梯形;互相平行的一组对边分别是梯形的上底和下底，不平行的一组对边是梯形的腰;从梯形一条底边上的一点到它对边的垂直线段叫做梯形的高，梯形有无数条高;如果梯形的两腰相等，这样的梯形叫做等腰梯形;一腰垂直于底的梯形叫直角梯形。正确理解梯形，等腰梯形和直角梯形的定义，是解答此题的关键.\n【详解】根据分析可知，当一个梯形的两腰相等时，这是一个等腰梯形，当梯形中含有一个直角时，这是一个直角梯形，梯形有无数条高。",
+        "knowledge_points":  [
+                                 "梯形的高及画法",
+                                 "直角梯形和等腰梯形的概念及特点"
+                             ],
+        "source":  "21-22四年级上·江西赣州·期末"
+    },
+    {
+        "id":  "63",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个等腰梯形的周长是22米，其中上底是4米，下底是6米，这个等腰梯形的一条腰长( )米。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】等腰梯形的周长减去上下两底的长度，等于两腰的长度和，再除以2即等于一条腰的长度。等腰梯形的两条腰相等，这是解答本题的关键.\n【详解】(22-4-6)÷2=12÷2=6(米),这个梯形的一条腰长6米。",
+        "knowledge_points":  [
+                                 "直角梯形和等腰梯形的概念及特点",
+                                 "梯形的周长"
+                             ],
+        "source":  "21-22 四年级上·天津和平·期末"
+    },
+    {
+        "id":  "64",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "聪聪和明明都用两根长7厘米和两根长5厘米的小棒，摆了一个平行四边形，他们摆的图形的( )一定相等，是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "周长;24",
+        "solution":  "【分析】平行四边形对边平行且相等，用两根长7厘米和两根长5厘米的小棒摆平行四边形，随着平行四边形内角度数的变化，平行四边形的形状可能不同，因此面积可能不同，但它们的周长都等于这4根小棒的总长度，因此周长一样，据此即可解答。此题主要考查了平行四边形的特征以及周长的意义.\n【详解】(7+5)×2=12×2=24(厘米),聪聪和明明都用两根长7厘米和两根长5厘米的小棒，摆了一个平行四边形，他们摆的图形的周长一定相等，是24厘米。",
+        "knowledge_points":  [
+                                 "平行四边形的周长",
+                                 "平行四边形的概念及特点"
+                             ],
+        "source":  "21-22 四年级上·山西阳泉·期末"
+    },
+    {
+        "id":  "65",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "( )分别平行的四边形叫做平行四边形。",
+        "options":  [
+
+                    ],
+        "answer":  "两组对边",
+        "solution":  "【详解】两组对边分别平行的四边形叫做平行四边形。如下图所示:![](images/99d58353f51c87f9c00684499e28d7aa7407e153829c036e59d6741c340457fd.jpg)",
+        "knowledge_points":  [
+                                 "平行四边形的概念及特点"
+                             ],
+        "source":  "21-22 四年级上·云南楚雄·期末"
+    },
+    {
+        "id":  "66",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "平行四边形有( )条高，具有( )性。",
+        "options":  [
+
+                    ],
+        "answer":  "无数;不稳定",
+        "solution":  "【分析】从平行四边形一条边上的一点向对边引一条垂线，这点和垂足之间的线段叫做平行四边形的高，垂足所在的边叫做平行四边形的底，依此填空;平行四边形容易变形，具有不稳定性，依此填空。解答此题的关键是要熟练掌握平行四边形的高及画法，以及平行四边形的不稳定性.\n【详解】根据分析可知，平行四边形有无数条高，具有不稳定性。![](images/5096c27866bed3ebb0715fd16dce998cd9ab8cf5a7e71bf1b7bb1ea32163238f.jpg)",
+        "knowledge_points":  [
+                                 "平行四边形的不稳定性及应用",
+                                 "平行四边形的高及画法"
+                             ],
+        "source":  "21-22 四年级上·江西赣州·期末"
+    },
+    {
+        "id":  "67",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "伸缩门是利用了平行四边形的( );直角梯形中有( )个直角，等腰梯形中有( )个锐角。",
+        "options":  [
+
+                    ],
+        "answer":  "不稳定性;2;2",
+        "solution":  "【分析】平行四边形具有不稳定性;只有一个腰垂直于底边的梯形叫做直角梯形;两腰相等的梯形叫做等腰梯形，据此解答。本题考查直角梯形和等腰梯形的特征，熟练掌握并正确记忆.\n【详解】![](images/9c3edd4f6e8d7767b81964f1b848e73f.jpg)伸缩门是利用了平行四边形的(不稳定性);直角梯形中有(2)个直角，等腰梯形中有(2)个锐角。",
+        "knowledge_points":  [
+                                 "直角梯形和等腰梯形的概念及特点"
+                             ],
+        "source":  "2022 四年级上·全国·专题练习"
+    },
+    {
+        "id":  "68",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "等腰梯形的周长是18cm,腰长3cm,这个梯形的上下底之和是( )cm。",
+        "options":  [
+
+                    ],
+        "answer":  "12",
+        "solution":  "【分析】因为等腰梯形的两条腰相等，将围成等腰梯形的所有线段的长度加起来，就是其周长，已知周长和腰长，求上下底之和，只要用周长减去两条腰长即可。本题主要考查等腰梯形的性质的应用以及平面图形周长的概念.\n【详解】18-3×2=18-6=12(cm),所以，这个梯形的上下底之和是12cm。",
+        "knowledge_points":  [
+                                 "梯形的周长",
+                                 "直角梯形和等腰梯形的概念及特点"
+                             ],
+        "source":  "21-22 四年级下·四川达州·期末"
+    },
+    {
+        "id":  "69",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰梯形的周长是40cm,上底长10cm,比下底短4cm,这个等腰梯形的腰长是( )cm。",
+        "options":  [
+
+                    ],
+        "answer":  "8",
+        "solution":  "【分析】根据题意可知，下底比10cm的上底多4cm,所以用10加4先求出下底的长，再用梯形的周长减去上底和下底，求出梯形两腰的和，等腰梯形的两条腰相等，所以再除以2求出腰长即可。熟练掌握等腰梯形的特征，是解答此题的关键.\n【详解】10+4=14(cm),40-14-10=16(cm),16÷2=8(cm),这个等腰梯形的腰长是8cm。",
+        "knowledge_points":  [
+                                 "梯形的周长",
+                                 "直角梯形和等腰梯形的概念及特点"
+                             ],
+        "source":  "21-22 四年级下·四川自贡·期末"
+    },
+    {
+        "id":  "70",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "下面的说法对吗。对的在( )里画\u0027√\u0027。(1)长方形也是平行四边形。( ) (2)平行四边形是特殊的梯形。( ) (3)两个完全相同的梯形可以拼成一个长方形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√;×;×",
+        "solution":  "【分析】根据长方形的特点：两组对边平行且相等，四个角都是直角;平行四边形的特点：两组对边平行且相等;梯形的特点：只有一组对边平行的四边形;长方形也是平行四边形，平行四边形不是特殊的梯形，两个完全相同的梯形一定可以拼成一个平行四边形，但不一定能拼成一个长方形，据此判断即可。本题考查了长方形，平行四边形和梯形的特点.\n【详解】(1)长方形也是平行四边形。(√)(2)平行四边形是特殊的梯形，说法错误。(×)(3)两个完全相同的梯形不一定可以拼成一个长方形，说法错误。(×)",
+        "knowledge_points":  [
+                                 "平行四边形的概念及特点",
+                                 "长方形的概念及特点",
+                                 "梯形的概念及特点"
+                             ],
+        "source":  "22-23 四年级上·全国·课前预习"
+    },
+    {
+        "id":  "71",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "平行四边形很容易变形，平行四边形是特殊的梯形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平行四边形的特性：平行四边形易变形，具有不稳定性;根据平行四边形的特征，平行四边形的两组对边分别平行且相等，而梯形只有一组对边平行，并且平行的一组对边不相等;据此判断即可。本题主要考查平行四边形及梯形的意义及性质.\n【详解】据分析可知：平行四边形很容易变形，具有不稳定性;平行四边形不是特殊的梯形，所以原题的说法错误。故答案为: ×。",
+        "knowledge_points":  [
+                                 "平行四边形的概念及特点",
+                                 "平行四边形的不稳定性及应用",
+                                 "梯形的概念及特点"
+                             ],
+        "source":  "20-21四年级上·广东肇庆·期末"
+    },
+    {
+        "id":  "72",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "互相垂直的两条直线，有4个交点。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】在同一平面内，两条直线相交成直角，这两条直线互相垂直，其中一条直线叫做另一条直线的垂线，这两条直线的交点叫做垂足，依此画图并判断即可。熟练掌握垂直的特点是解答此题的关键.\n【详解】![](images/9cdc1e4d07115d52a67ffd274c53ed5913e9d4fd8679c3e2e05719921fe902be.jpg)如图，互相垂直的两条直线，有1个交点。故答案为: ×。",
+        "knowledge_points":  [
+                                 "垂直的特征"
+                             ],
+        "source":  "21-22 四年级上·广东江门·期末"
+    },
+    {
+        "id":  "73",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "长方形是特殊的平行四边形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】当平行四边形的一个内角是90°时，则该平行四边形是长方形，据此进行判断即可。解答此题的关键是理解平行四边形和长方形的关系.\n【详解】由分析可知：长方形是特殊的平行四边形。所以原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "平行四边形的概念及特点",
+                                 "长方形的概念及特点"
+                             ],
+        "source":  "2014 三年级·全国·课后作业"
+    },
+    {
+        "id":  "74",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "长方形和正方形都是特殊的梯形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】根据长方形，正方形及梯形的特征：长方形和正方形都是两组对边平行且相等，梯形只有一组对边平行，所以长方形和正方形都不是梯形。据此判断即可。本题主要考查长方形，正方形及梯形的特征，需熟练掌握.\n【详解】长方形和正方形都是特殊的梯形，说法错误。故答案为: ×。",
+        "knowledge_points":  [
+                                 "长方形的概念及特点",
+                                 "正方形的概念及特点",
+                                 "梯形的概念及特点"
+                             ],
+        "source":  "21-22 四年级上·云南昭通·期末"
+    },
+    {
+        "id":  "75",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "钟面上午9时整，分针和时针互相垂直。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】钟面上,6时整时，时针与分针之间的夹角是180°,有6个大格，因此每个大格是:180°/6=30°,钟面上9时整，时针和分针之间的较小角有3个大格，因此用30°乘3即可;依此计算并根据垂直的特点进行判断。此题考查的是垂直的特点，熟练掌握对钟面时间的认识，以及平角的特点，是解答此题的关键.\n【详解】180°/6=30°,30°*3=90°,即钟面上午9时整，分针和时针互相垂直。故答案为:√",
+        "knowledge_points":  [
+                                 "整点时间与钟面指针的位置",
+                                 "平角，周角的认识及特征",
+                                 "垂直的特征"
+                             ],
+        "source":  "19-20 四年级上·湖南郴州·期末"
+    },
+    {
+        "id":  "76",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "过直线外一点只能画一条直线平行于已知直线。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】根据平行的性质：过直线外一点画已知直线的平行线，有且只有一条直线与已知直线平行;据此解答即可。此题考查的知识点是平行的性质，关键是正确理解和运用.\n【详解】过直线外一点只能画一条直线平行于已知直线，原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "平行的特征及性质"
+                             ],
+        "source":  "2014 四年级·全国·课后作业"
+    },
+    {
+        "id":  "77",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "直角梯形仍然只有两条高。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【详解】根据梯形高的含义：梯形虽然只有一组对边平行，但是，在这组对边里，也可以画无数条垂直线段，所以也有无数条高。故答案为: ×",
+        "knowledge_points":  [
+                                 "梯形的高及画法"
+                             ],
+        "source":  "22-23 五年级上·浙江金华·期末"
+    },
+    {
+        "id":  "78",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "数学课本封面相邻的两边互相垂直，相对的两边互相平行。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】两条直线相交所成的四个角中，有一个角是直角时，就说这两条直线互相垂直;在同一平面内，不相交的两条直线叫做平行线，据此解答。此题考查在同一平面内，两条直线的位置关系以及特征.\n【详解】数学课本封面相邻的两边互相垂直，相对的两边互相平行。故答案为:√。",
+        "knowledge_points":  [
+                                 "垂直的特征",
+                                 "平行的特征及性质"
+                             ],
+        "source":  "21-22 四年级上·广东广州·期末"
+    },
+    {
+        "id":  "79",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "在直线外一点A,向直线画垂线，可以画无数条。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】根据垂线的特征和性质可知，过直线外一点，有且只有一条直线与已知直线垂直，据此即可解答。本题主要考查学生对垂线特征和性质的掌握.\n【详解】根据分析可知，在直线外一点A,向直线画垂线，只能画一条，原说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "画垂线",
+                                 "垂直的特征"
+                             ],
+        "source":  "2014 四年级·全国·课后作业"
+    },
+    {
+        "id":  "80",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "电动伸缩门是利用平行四边形具有不稳定性的特点制成的。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】电动伸缩门可以开，也可以回缩起来，此处应用了四边形的不稳定性。三角形具有稳定性，四边形易变性.\n【详解】根据分析，这句话是对的。故答案为:√。",
+        "knowledge_points":  [
+                                 "平行四边形的不稳定性及应用"
+                             ],
+        "source":  "21-22 四年级上·陕西安康·期末"
+    },
+    {
+        "id":  "81",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "两底之和与高分别相等的两个梯形一定可以拼成一个平行四边形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】上底和下底的和与高都分别相等的梯形，并不能代表是完全相同的梯形，比如，甲乙两梯形，上底与下底的和为8厘米，甲梯形上底可以为2厘米，下底为6厘米，乙梯形上底为3厘米，下底为5厘米，由此可见上底与下底的和相等，并不代表上底和下底分别相等，由此判断即可。本题考查了梯形的拼接问题，只有完全一样的两个梯形才能保证一定能拼成平行四边形.\n【详解】由分析可得：下图两个梯形上底与下底的和与高都分别相等，但是形状不同，就不能拼成平行四边形.\n![](images/9b5d62e0b88d332c0fbf25f71d5d32c2f9dff3ee63d6f65073353964e290aaeb.jpg)故答案为: ×",
+        "knowledge_points":  [
+                                 "平面图形的拼接",
+                                 "梯形的概念及特点",
+                                 "梯形的面积"
+                             ],
+        "source":  "21-22五年级上·陕西商洛·期末"
+    },
+    {
+        "id":  "82",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "同一平面内，两条线段平行，它们一定相等。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平行线的含义：在同一平面内，不相交的两条直线叫做平行线;同样在同一平面内，不相交的两条线段也是平行的，但是没有说长度相等。据此解答即可。此题考查了平行线的含义及性质，应注意基础知识的积累和理解.\n【详解】同一平面内，不相交的两条线段也是平行的，但是长度不一定相等，例如梯形的上底和下底是平行的，但是不相等。所以两条线段互相平行，它们也一定相等，这个说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平行的特征及性质",
+                                 "线段，直线，射线的认识及特征"
+                             ],
+        "source":  "21-22 四年级上·广西玉林·期末"
+    },
+    {
+        "id":  "83",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "互相平行的两条线段，它们的长度相等。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】同一平面内，将线段的两端无限延长，如果这两条直线永不相交，就说明这两条线段互相平行，依此画图并判断。熟练掌握平行的特点是解答此题的关键.\n【详解】由此可知，互相平行的两条线段，它们的长度不一定相等。故答案为: ×",
+        "knowledge_points":  [
+                                 "平行的特征及性质"
+                             ],
+        "source":  "20-21 四年级上·重庆黔江·期末"
+    },
+    {
+        "id":  "84",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "两条直线相交一定不垂直。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】在同一个平面内，两条直线相交所成的四个角中，有一个角是直角时，就说这两条直线互相垂直。此题考查了垂直的特征和性质，要熟练掌握.\n【详解】两条直线相交不一定垂直。故原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "垂直的特征"
+                             ],
+        "source":  "21-22 四年级上·河北保定·期末"
+    },
+    {
+        "id":  "85",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "平面上的两条直线的关系有平行，垂直和相交三种。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【详解】在同一平面内的两条直线，只有两种位置关系，不是相交就是平行，垂直是相交的特殊情况。所以，平面上的两条直线的关系有平行，垂直和相交三种，说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "垂直的特征",
+                                 "平行的特征及性质"
+                             ],
+        "source":  "20-21四年级上·湖北鄂州·期末"
+    },
+    {
+        "id":  "86",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "正方形是特殊的长方形，平行四边形是特殊的梯形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平行四边形是两组对边分别平行且相等的四边形;长方形有四个直角，有两组分别相等的对边;正方形有四个直角，四条边都相等;梯形是只有一组对边平行，而另一组对边不平行的四边形;这几种图形既有联系又有区别，根据定义即可判断出。本题考查平行四边形，长方形，正方形，梯形的联系与区别，正确了解它们的特征是解题的关键.\n【详解】由分析可知，长方形是特殊的平行四边形，正方形是特殊的长方形;平行四边形是两组对边分别平行的四边形，而梯形只有一组对边平行，另一组对边不平行，所以平行四边形不是特殊的梯形。此题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "四边形的分类及关系"
+                             ],
+        "source":  "21-22 四年级上·河南三门峡·期末"
+    },
+    {
+        "id":  "87",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "两个完全相同的三角形一定能拼成一个梯形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】如图：两个完全相同的三角形拼成了一个长方形，正方形，平行四边形，但是拼不出梯形，梯形分为等腰梯形，直角梯形和一般的梯形，但是任何一种梯形也无法拆成两个完全相同的三角形，据此判断即可。本题重点考查平面图形的拼接，掌握图形的特征是解题的关键.\n![](images/ab239fd4b79be7755e5fdbf4d5237b51.jpg)\n【详解】由分析可知：两个完全相同的三角形不能拼成一个梯形，原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "梯形的概念及特点",
+                                 "平面图形的拼接"
+                             ],
+        "source":  "18-19 四年级下·浙江·课后作业"
+    },
+    {
+        "id":  "88",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "n个同样的平行四边形一定能拼成一个较大的平行四边形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】根据平行四边形的性质可得：完全一样的两个(或多个)平行四边形能拼成一个较大的平行四边形，据此判断即可。本题考查了图形的拼组及平行四边形的特征.\n【详解】如图:![](images/7dce5a46eb78bf7744054aa651570bf7.jpg)\n所以，几个同样的平行四边形一定能拼成一个较大的平行四边形，说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "平行四边形的切拼"
+                             ],
+        "source":  "22-23 五年级上·黑龙江双鸭山·期末"
+    },
+    {
+        "id":  "89",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "在同一平面内，两条平行线延长后不可能相交。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【详解】如图:AB//CD,即使把这两条平行线延长后，它们仍然平行。![](images/2eeef43702625f77ebef5c65bf3411fcfaf1081284a53a0b23cd32bb33f15378.jpg)同一平面内不相交的两条直线互相平行，其中一条直线是另一条直线的平行线;据此可知，在同一平面内，两条平行线延长后不可能相交。故答案为:√",
+        "knowledge_points":  [
+                                 "平行的特征及性质"
+                             ],
+        "source":  "2022四年级上·全国·专题练习"
+    },
+    {
+        "id":  "90",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "平行四边形有无数条高。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【详解】从平行四边形一边上的一点向对边引垂线，这点到垂足之间的线段叫做平行四边形的高。由于平行四边形的一条边上可以确定无数个点，就可以向对边引无数条垂线，所以有无数条高。故答案为:√",
+        "knowledge_points":  [
+                                 "平行四边形的高及画法"
+                             ],
+        "source":  "2014 三年级·全国·课后作业"
+    },
+    {
+        "id":  "91",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "在梯形里，平行的一组对边叫做梯形的腰。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】梯形是一个四边形，由四条线段围成。梯形有一组对边平行，这组对边叫梯形的底，长的那条叫梯形的下底，短的那条叫梯形的上底。梯形另外两条不平行的边叫做梯形的腰。据此解答。本题考查学生对梯形各部分名称的掌握。熟练理解梯形各边名称是解决此题的关键.\n【详解】由梯形的各部分名称可知，梯形平行的一组对边叫梯形的底，梯形另外两条不平行的边叫梯形的腰。所以题目说法不正确。故答案为: ×",
+        "knowledge_points":  [
+                                 "梯形的概念及特点"
+                             ],
+        "source":  "21-22四年级下·广西贵港·期末"
+    },
+    {
+        "id":  "92",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "只有一组对边平行的四边形叫做梯形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】根据梯形的定义：只有一组对边平行的四边形叫做梯形。如果两组对边都平行，则是平行四边形;如果没有对边平行，则是一般四边形。据此判断.\n【详解】由梯形的定义可知，必须满足两个条件:①是四边形;②只有一组对边平行。题目描述完全符合梯形的定义。故答案为:√",
+        "knowledge_points":  [
+                                 "梯形的概念及特点"
+                             ],
+        "source":  "23-24 四年级上·江苏南京·期末"
+    },
+    {
+        "id":  "93",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "判断题",
+        "difficulty":  0.7,
+        "question":  "梯形的高一定比它的腰短。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】梯形的高是指两底之间的垂直距离。在一般梯形中，高确实比腰短(因为垂线段最短);但在直角梯形中，其中一条腰垂直于底边，这条腰的长度就等于梯形的高。因此,\u0027一定比腰短\u0027的说法是错误的.\n【详解】1。在普通梯形中，从上底一点向下底作垂线(高),这条垂线段与腰构成直角三角形，高是直角边，腰是斜边，此时高\u003c腰。2。在直角梯形中，垂直于底边的那条腰，其长度本身就等于高，此时高=腰。综上所述，梯形的高可能等于腰，不一定比腰短。故答案为: ×",
+        "knowledge_points":  [
+                                 "梯形的高及特点",
+                                 "直角梯形的特征"
+                             ],
+        "source":  "23-24 四年级上·湖北武汉·期中"
+    },
+    {
+        "id":  "94",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "判断题",
+        "difficulty":  0.75,
+        "question":  "两个完全一样的梯形一定能拼成一个平行四边形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】将两个完全一样的梯形，把其中一个倒置(旋转180度),使它们相等的腰重合，或者让上底与下底分别对接。由于梯形有一组对边平行，拼接后新图形的一组对边长度相等且平行(原上底+原下底),另一组对边也平行且相等，因此一定能拼成一个平行四边形.\n【详解】设梯形的上底为a,下底为b,高为h。将两个完全相同的梯形拼在一起，新图形的一组对边长度均为a+b,且互相平行(因为原梯形的上下底平行)。根据平行四边形的判定定理(一组对边平行且相等的四边形是平行四边形),拼成的图形一定是平行四边形。这也是推导梯形面积公式S=(a+b)h/2的基础方法。故答案为:√",
+        "knowledge_points":  [
+                                 "梯形的拼接",
+                                 "平行四边形的特征"
+                             ],
+        "source":  "23-24 四年级上·四川成都·单元测试"
+    },
+    {
+        "id":  "95",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一块平行四边形的菜地，相邻两边的长分别是30米和25米，这块菜地的周长是多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "110米",
+        "solution":  "【分析】根据\u0027对边平行且相等的四边形是平行四边形\u0027可知，平行四边形的周长等于相邻两边的长度和乘2，据此求出这块菜地的周长即可。正确理解周长的意义，是解答此题的关键.\n【详解】(30+25)×2=55×2=110(米)答：这块菜地的周长是110米。",
+        "knowledge_points":  [
+                                 "平行四边形的周长"
+                             ],
+        "source":  "22-23四年级上·广东珠海·期末"
+    },
+    {
+        "id":  "96",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "小明说：梯形也是平行四边形。你认为对吗。请说明理由。",
+        "options":  [
+
+                    ],
+        "answer":  "不对;平行四边形的两组对边互相平行，梯形只有一组对边互相平行，所以梯形不是平行四边形。",
+        "solution":  "【分析】根据平行四边形的定义和梯形的定义进行解答即可。本题考查了平行四边形的定义和梯形的定义.\n【详解】两组对边互相平行的四边形叫平行四边形;一组对边平行，另一组对边不平行的四边形叫梯形。所以不对。",
+        "knowledge_points":  [
+                                 "梯形的概念及特点",
+                                 "平行四边形的概念及特点"
+                             ],
+        "source":  "21-22四年级上·广东广州·期末"
+    },
+    {
+        "id":  "97",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一个平行四边形的菜地，周长是42米，其中一边长13米，另外三条边分别长多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "8米;13米;8米",
+        "solution":  "【分析】两组对边分别平行的四边形，叫做平行四边形，平行四边形的对边相等，因此用平行四边形的周长减去2个13米，然后再除以2即可计算出13米相邻一边的边长，依此解答。熟练掌握平行四边形的周长的计算是解答此题的关键.\n【详解】13×2=26(米),42-26=16(米),16÷2=8(米)答：另外三条边分别长8米,13米,8米。",
+        "knowledge_points":  [
+                                 "平行四边形的周长"
+                             ],
+        "source":  "21-22 四年级上·河北保定·期末"
+    },
+    {
+        "id":  "98",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "等腰梯形的周长为25厘米，上，下底分别长3厘米和6厘米，它的一条腰长多少厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "8厘米",
+        "solution":  "【分析】等腰梯形的周长减去上，下底的长度等于两腰长度和，再除以2即等于腰长。熟练掌握等腰梯形的定义和特征是解答本题的关键.\n【详解】(25-3-6)÷2=16÷2=8(厘米)答：它的一条腰长8厘米。",
+        "knowledge_points":  [
+                                 "直角梯形和等腰梯形的概念及特点",
+                                 "梯形的周长"
+                             ],
+        "source":  "21-22 四年级上·山西阳泉·期末"
+    },
+    {
+        "id":  "99",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一个等腰梯形，上底长3厘米，下底长5厘米，它的周长是18厘米，这个梯形的腰长是多少厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "5厘米",
+        "solution":  "【分析】等腰梯形的两个腰相等，用周长分别减去上底，下底的长度，再除以2即可解答。此题主要考查学生对等腰梯形性质的认识与应用.\n【详解】(18-3-5)÷2=10÷2=5(厘米)答：这个等腰梯形的腰长是5厘米。",
+        "knowledge_points":  [
+                                 "直角梯形和等腰梯形的概念及特点",
+                                 "梯形的周长"
+                             ],
+        "source":  "2022 四年级上·全国·专题练习"
+    },
+    {
+        "id":  "100",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一个等腰梯形，上底长度13米，下底比上底2倍少4米，腰长8米，梯形的周长是多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "51米",
+        "solution":  "【分析】等腰梯形的两条腰相等，先求出这个等腰梯形的下底，再将等腰梯形的上底，下底，两条腰加起来，即可算出这个梯形的周长是多少米。据此解答。本题考查学生对等腰梯形特征的掌握。牢记等腰梯形两腰相等是解决此题的关键.\n【详解】13×2-4=26-4=22(米),13+22+8×2=35+16=51(米)答：梯形的周长是51米。",
+        "knowledge_points":  [
+                                 "梯形的周长",
+                                 "直角梯形和等腰梯形的概念及特点"
+                             ],
+        "source":  "2022 四年级上·全国·专题练习"
+    },
+    {
+        "id":  "101",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "平行四边形的两组对边( )。",
+        "options":  [
+                        "A. 分别平行",
+                        "B. 分别垂直",
+                        "C. 相交",
+                        "D. 以上答案都不对"
+                    ],
+        "answer":  "A",
+        "solution":  "【详解】平面上两条直线的位置关系有两种相交或平行，垂直是相交的一种，两条直线相交成直角，说明这两条直线互相垂直，在同一平面内不相交的两条直线叫做平行线，而平行四边形的两组对边分别平行，如下图所示:![](images/4598e40070f8873dffe71f6cb6058a05da66985a8f7a87e2091d2324c1b8fc62.jpg)故答案为:A",
+        "knowledge_points":  [
+                                 "平行四边形的概念及特点"
+                             ],
+        "source":  "21-22六年级下·云南玉溪·期末"
+    },
+    {
+        "id":  "102",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面说法正确的是()。",
+        "options":  [
+                        "A. 同一平面内，两条直线不是相交就是平行",
+                        "B. 同一平面内，两条直线垂直但可能不相交",
+                        "C. 两条平行线之间可能有交点"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】利用平行线的判定与定理及有关定义，定理逐一判断即可。此题考查了平行线的性质与判定，熟记有关定义，定理是解题的关键.\n【详解】A。同一平面内，两条不重合的直线不平行就相交，此说法正确;B。同一平面内，两条直线垂直但可能不相交，此说法错误;C。两条平行线之间可能有交点，故此说法错误。故选:A.",
+        "knowledge_points":  [
+                                 "垂直的特征，平行的特征及性质"
+                             ],
+        "source":  "2021 四年级上·全国·专题练习"
+    },
+    {
+        "id":  "103",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面说法正确的是()。",
+        "options":  [
+                        "A. 正方形有四组垂线",
+                        "B. 正方形有四组平行线",
+                        "C. 长方形有两组垂线",
+                        "D. 长方形有四组平行线"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】依据垂直与平行的定义，以及正方形和长方形的特点，即可逐个进行分析解答。此题主要依据：垂直与平行的定义，以及正方形，长方形的特点解决问题.\n【详解】由题意可知：在正方形和长方形中分别有两组平行线四组垂线,A。正方形有四组垂线，说法正确;B。正方形有四组平行线，说法错误;C。长方形有两组垂线，说法错误;D。长方形有四组平行线，说法错误。故选:A.",
+        "knowledge_points":  [
+                                 "垂直的特征，平行的特征及性质"
+                             ],
+        "source":  "2021 四年级上·全国·专题练习"
+    },
+    {
+        "id":  "104",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "在同一平面内,a平行于b,a平行于c,那么b( )c。",
+        "options":  [
+                        "A. 平行于",
+                        "B. 垂直于",
+                        "C. 无法确定"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】在同一平面内不相交的直线叫做平行线，先依此画图并选择即可。熟练掌握平行的特点是解答此题的关键.\n【详解】![](images/91fb56fc950c74783645eab70b3e465df3857e26fae001845c9a80d6e6f0a60d.jpg)因此：在同一平面内,a平行于b,a平行于c,那么b平行于c。故答案为:A.",
+        "knowledge_points":  [
+                                 "平行的特征及性质"
+                             ],
+        "source":  "2021 四年级上·全国·专题练习"
+    },
+    {
+        "id":  "105",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面说法正确的是()。",
+        "options":  [
+                        "A. 同一平面内，两条直线不是垂直就是平行",
+                        "B. 相交的两条直线一定互相垂直",
+                        "C. 垂直的两条直线一定不互相平行"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】在同一平面内，两直线的位置关系有2种：平行，相交，根据以上结论判断即可.\n【详解】A。∵在同一平面内，两直线的位置关系是平行，相交,2种,∴在同一平面内，两直线的位置关系是平行，相交(相交不一定垂直),故本选项错误;B。在同一平面内，不平行的两条直线一定相交，但不一定互相垂直，故本选项错误;C。在同一平面内，垂直的两条直线一定不互相平行，故本选项正确。故选:C.\n[点睛]本题考查了对平行线，垂直的理解和运用，注意:①在同一平面内，两直线的位置关系有2种：平行，相交,②相交不一定垂直。",
+        "knowledge_points":  [
+                                 "垂直的特征，平行的特征及性质"
+                             ],
+        "source":  "2021 四年级上·全国·专题练习"
+    },
+    {
+        "id":  "106",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "两个形状，大小完全相同的直角三角形不可能拼成( )。",
+        "options":  [
+                        "A. 梯形",
+                        "B. 平行四边形",
+                        "C. 三角形"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】如下图，两个形状，大小完全相同的直角三角形可以拼成一个大的三角形和长方形，但不能拼成梯形，据此即可解答。![](images/38d8f6be5e587a73320b3906909209b8b267052be846eb466e39cb4abfc23239.jpg)本题主要考查学生的动手能力.\n【详解】根据分析可知，两个形状，大小完全相同的直角三角形可以拼成一个大的三角形和平行四边形，但不能拼成梯形。故答案为:A.",
+        "knowledge_points":  [
+                                 "用同样的图形拼接，平面图形的拼接"
+                             ],
+        "source":  "2021 四年级上·全国·专题练习"
+    },
+    {
+        "id":  "107",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面选项中正确的是( )。",
+        "options":  [
+                        "A. 梯形只有一条高",
+                        "B. 平行四边形有无数条高",
+                        "C. 任意两个直角梯形可以组成一个长方形"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】(1)从平行四边形的一条边上的任意一点都可以向对边作高，平行四边形的每一条边都可以看做是底，所以平行四边形有4条底，有无数条高。(2)从梯形一条底边上的一点到它对边的垂直线段叫做梯形的高。梯形有2条底，有无数条高。(3)一腰垂直于底的梯形叫直角梯形。正确理解平行四边形的高，梯形的高的概念，是解答此题的关键.\n【详解】A。梯形有无数条高，故此项错误;B。平行四边形有无数条高，正确。C。任意两个直角梯形可以组成一个长方形，是错误的。故答案为:B.",
+        "knowledge_points":  [
+                                 "长方形的概念及特点，梯形的高及画法，平行四边形的高及画法，直角梯形和等腰梯形的概念及特点"
+                             ],
+        "source":  "2021 四年级上·全国·专题练习"
+    },
+    {
+        "id":  "108",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面选项中正确的是( )。",
+        "options":  [
+                        "A. 长方形是梯形",
+                        "B. 正方形是平行四边形",
+                        "C. 梯形的对边相互平行"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】长方形的两组对边相互平行，并且两组对边分别相等，四个角都是直角;只有一组对边平行的四边形叫做梯形;正方形的四条边都相等，并且四个角都是直角;两组对边分别平行的四边形，叫做平行四边形，依此选择。此题考查的是对长方形，正方形，梯形，平行四边形的认识，熟练掌握它们各自的特点是解答此题的关键.\n【详解】A。长方形不是梯形，因此此项说法错误;B。正方形是特殊的平行四边形，因此此项正确;C。只有一组对边平行的四边形叫做梯形，因此此项说法错误;故答案为:B.",
+        "knowledge_points":  [
+                                 "平行四边形的概念及特点，长方形的概念及特点，正方形的概念及特点，梯形的概念及特点"
+                             ],
+        "source":  "2021 四年级上·全国·专题练习"
+    },
+    {
+        "id":  "109",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一个梯形不可以分成( )。",
+        "options":  [
+                        "A. 两个梯形",
+                        "B. 两个平行四边形",
+                        "C. 一个梯形和一个平行四边形"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】只有一组对边平行的四边形叫做梯形;两组对边分别平行的四边形，叫做平行四边形;先画图再依此选择即可。熟练掌握平行四边形，梯形的特点是解答此题的关键.\n【详解】A。![](images/bb3b83e2886a251e9f74dc767f734ac94ad00090cc6bd3037dcbbd7bcb5ff496.jpg);因此一个梯形可以分成两个梯形;B。梯形只有一组对边相互平行，而平行四边形有2组对边相互平行，因此一个梯形不可以分成两个平行四边形;C。![](images/670bf93b0e4fa73a1e8c4300f33b01bc4c808e7916a009adbcf874117b3a6ff1.jpg);因此一个梯形可以分成一个梯形和一个平行四边形;故答案为:B",
+        "knowledge_points":  [
+                                 "平行四边形的概念及特点，梯形的概念及特点"
+                             ],
+        "source":  "2021 四年级上·全国·专题练习"
+    },
+    {
+        "id":  "110",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面哪句话是正确的。( )",
+        "options":  [
+                        "A. 梯形是四边形",
+                        "B. 平行四边形只要有一组对边平行就可以",
+                        "C. 梯形至少要有一组对边平行",
+                        "D. 平行四边形非常稳定"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】根据平行边形和梯形的特征及概念进行作答。本题考查梯形和平行四边形的认识，熟记两者的特征是解答此题的关键.\n【详解】A,梯形是有一组对边平行的四边形，所以梯形是四边形，故选项说法正确;B,平行四边形对边平行且相等，故选说法项错误;C,梯形是有一组对边平行的四边形，有且只有一组对边平行，故选项说法错误;D,平行四边形不具有稳定性，故选项错误。故答案为:A.",
+        "knowledge_points":  [
+                                 "平行四边形的概念及特点，梯形的概念及特点"
+                             ],
+        "source":  "2021 四年级上·全国·专题练习"
+    },
+    {
+        "id":  "111",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "已知直线a与直线c互相平行，直线b与直线c互相平行。那么，直线a与直线b()。",
+        "options":  [
+                        "A. 互相平行",
+                        "B. 互相垂直",
+                        "C. 无法确定"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】在同一平面内不相交的直线叫做平行线，也可以说这两条直线互相平行，依此选择即可。熟练掌握平行的特点是解答此题的关键.\n【详解】已知直线a与直线c互相平行，直线b与直线c互相平行。那么，直线a与直线b互相平行。故答案为:A",
+        "knowledge_points":  [
+                                 "平行的特征及性质"
+                             ],
+        "source":  "21-22 四年级上·全国·期末"
+    },
+    {
+        "id":  "112",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面说法错误的是( )。",
+        "options":  [
+                        "A. 每相邻两个计数单位之间的进率是10",
+                        "B. 一条直线长6厘米",
+                        "C. 垂直于平行线间的线段都是相等的"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】对各个选项进行判断，找出错误的选项即可解答。本题主要考查学生对整数的计数单位，直线和平行线特征的掌握.\n【详解】A。根据整数的数位，计数单位知识可知，每相邻两个计数单位之间的进率是10，原说法正确。B。直线没有端点，无限长，原说法错误。C。两直线互相平行时，从一条直线上任意一点向另一条直线作垂线，所得的平行线间的垂直线段的长度，叫做平行线间的距离。平行线之间的距离处处相等，原说法正确。故答案为:B.",
+        "knowledge_points":  [
+                                 "平行的特征及性质，线段，直线，射线的认识及特征，整数的数级，数位和计数单位的认识"
+                             ],
+        "source":  "19-20四年级上·河南信阳·期末"
+    },
+    {
+        "id":  "113",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "如果在纸上画甲乙两条直线都与第三条直线相交成直角，那么甲乙两条直线就( )。",
+        "options":  [
+                        "A. 相互平行",
+                        "B. 相交",
+                        "C. 互相垂直"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】在同一平面内，垂直于同一条直线的两条直线互相平行;进而得出结论。此题应根据平行和垂直的特征及性质进行解答.\n【详解】根据垂直和平行的性质可知：如果在纸上画甲乙两条直线都与第三条直线相交成直角，那么甲乙两条直线就互相平行。故答案为:A.",
+        "knowledge_points":  [
+                                 "平行的特征及性质，垂直的特征"
+                             ],
+        "source":  "19-20 四年级上·四川眉山·期末"
+    },
+    {
+        "id":  "114",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "长方形相邻的两条边( )。",
+        "options":  [
+                        "A. 互相垂直",
+                        "B. 互相平行",
+                        "C. 无法确定"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】两条直线相交成直角时，这两条直线互相垂直，其中一条直线是另一条直线的垂线，这两条直线的交点叫做垂足。据此解答即可。本题考查垂直的性质，需熟练掌握.\n【详解】根据垂直的性质可知，长方形相邻两条边互相垂直。故答案为:A.",
+        "knowledge_points":  [
+                                 "垂直的特征"
+                             ],
+        "source":  "20-21 四年级上·广东深圳·期末"
+    },
+    {
+        "id":  "115",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面的说法，正确的是( )。",
+        "options":  [
+                        "A. 平行四边形是特殊的梯形。",
+                        "B. 平行四边形不容易变形。",
+                        "C. 在同一平面内的两条直线，如果不是相交，就一定是互相平行。"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】对各个选项逐个判断，找出正确的选项即可解答。本题主要考查学生对平行四边形，梯形和平行与相交等知识的掌握.\n【详解】A。平行四边形有两组对边平行，梯形只有一组对边平行，所以原说法错误。B。平行四边形具有易变性，容易变形，所以原说法错误。C。在同一平面内不相交的两条直线互相平行，所以原说法正确。故答案为:C.",
+        "knowledge_points":  [
+                                 "梯形的概念及特点，平行的特征及性质，四边形的分类及关系，平行四边形的概念及特点"
+                             ],
+        "source":  "20-21四年级上·广东广州·期末"
+    },
+    {
+        "id":  "116",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列哪一句话是错误的。( )",
+        "options":  [
+                        "A. 平行线延长也可能相交",
+                        "B. 两腰相等的梯形叫做等腰梯形",
+                        "C. 平行四边形的两组对边分别相等"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】同一平面内不相交的两条直线互相平行，其中条直线是另一条直线的平行线;平行线之间的距离处处相等。故选项A错误;如果梯形的两腰相等，这样的梯形叫做等腰梯形，故选项B正确;平行四边形的两组对边分别平行且相等，故选项C正确。本题主要考查了平行，等腰梯形，平行四边形的定义，是解答此题是关键.\n【详解】A。平行线之间的距离处处相等，故此项错误。B。两腰相等的梯形叫做等腰梯形，正确。C。平行四边形的两组对边分别相等，正确。故答案为:A.",
+        "knowledge_points":  [
+                                 "梯形的概念及特点，平行的特征及性质，平行四边形的概念及特点"
+                             ],
+        "source":  "21-22 四年级上·全国·单元测试"
+    },
+    {
+        "id":  "117",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在同一平面内,a∥b,b⊥c,那么a( )c。",
+        "options":  [
+
+                    ],
+        "answer":  "⊥",
+        "solution":  "【分析】根据题意可知：直线a与直线b相互平行，直线b与直线c相互垂直，依此画出这三条线，然后填空即可。熟练掌握垂直与平行的特点是解答此题的关键.\n【详解】根据题意可画图为:![](images/65dcaaf437bea0a90123c29c4e99996a3ee1bf45847e186231c4260a5451740b.jpg)因此a⊥c",
+        "knowledge_points":  [
+                                 "垂直的特征，平行的特征及性质"
+                             ],
+        "source":  "2021 四年级上·全国·专题练习"
+    },
+    {
+        "id":  "118",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "过一个点可以向一条直线做( )条垂线。",
+        "options":  [
+
+                    ],
+        "answer":  "1",
+        "solution":  "【分析】过直线外一点有且仅有一条直线与已知直线垂直，据此解答。此题考查了过直线外一点有且仅有一条直线与已知直线垂直知识的掌握情况.\n【详解】由分析可知：过一个点可以向一条直线做1条垂线。",
+        "knowledge_points":  [
+                                 "画垂线"
+                             ],
+        "source":  "2021四年级上·全国·专题练习"
+    },
+    {
+        "id":  "119",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "平行四边形的( )组对边分别平行，正方形有( )条边相等。",
+        "options":  [
+
+                    ],
+        "answer":  "两;四",
+        "solution":  "【详解】正方形的定义：四条边相等，四个角都是直角的四边形叫做正方形;平行四边形，是在同一个二维平面内，由两组平行线段组成的闭合图形;即平行四边形的两组对边分别平行且相等。",
+        "knowledge_points":  [
+                                 "正方形的概念及特点，平行四边形的概念及特点"
+                             ],
+        "source":  "19-20 四年级上·黑龙江牡丹江·期末"
+    },
+    {
+        "id":  "120",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "两条直线互相垂直，形成( )个直角。",
+        "options":  [
+
+                    ],
+        "answer":  "4",
+        "solution":  "【分析】两条直线相交，所形成的角是直角，我们就说这两条直线互相垂直，据此解答。本题考查的是垂直意义的运用，回想垂直的定义.\n【详解】两条直线互相垂直，形成4个直角。",
+        "knowledge_points":  [
+                                 "垂直的特征"
+                             ],
+        "source":  "21-22 四年级上·全国·单元测试"
+    },
+    {
+        "id":  "121",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "直线a与直线b互相垂直，记作( ),读作( )。",
+        "options":  [
+
+                    ],
+        "answer":  "a⊥b;a垂直于b",
+        "solution":  "【分析】垂直，是指一条线与另一条线成直角，这两条直线互相垂直。通常用符号\u0027⊥\u0027表示;依此填空。熟练掌握垂直的特点及掌握垂直的符号是解答此题的关键.\n【详解】![](images/3e6e991c89ae1faa86812b489b9c6fcfbaac68897010739bbd8a8456551e5b2e.jpg)直线a与直线b互相垂直，记作a⊥b,读作:a垂直于b。",
+        "knowledge_points":  [
+                                 "垂直的特征"
+                             ],
+        "source":  "21-22 四年级上·全国·单元测试"
+    },
+    {
+        "id":  "122",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "可以画出( )条端点分别在两条平行线上，且与平行线垂直的线段，这些线段的长度都( )。",
+        "options":  [
+
+                    ],
+        "answer":  "无数;相等",
+        "solution":  "【分析】两直线互相平行时，从一条直线上任意一点向另条直线作垂线，所得的平行线间的垂直线段的长度，叫做平行线间的距离。平行线之间的距离处处相等。本题主要考查学生对平行线特征的掌握和灵活运用.\n【详解】根据分析可知，可以画出无数条端点分别在两条平行线上，且与平行线垂直的线段，这些线段的长度都相等。",
+        "knowledge_points":  [
+                                 "平行的特征及性质"
+                             ],
+        "source":  "21-22 四年级上·全国·单元测试"
+    },
+    {
+        "id":  "123",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰梯形的周长是49米，上底长10米，下底长15米，这个梯形的一条腰长( )米。",
+        "options":  [
+
+                    ],
+        "answer":  "12",
+        "solution":  "【分析】用周长减去上，下底的长度，即可得出梯形的两条腰的和，等腰梯形两条腰的长度相等，再用梯形的两条腰的和除以2就是一条腰的长度，据此计算即可解答问题。明确梯形的周长是四条边的长度和，以及等腰梯形两条腰的长度相等是解答本题的关键.\n【详解】(49-10-15)÷2=24÷2=12(米),一个等腰梯形的周长是49米，上底长10米，下底长15米，这个梯形的一条腰长(12)米。",
+        "knowledge_points":  [
+                                 "直角梯形和等腰梯形的概念及特点，梯形的周长"
+                             ],
+        "source":  "16-17 四年级上·浙江杭州·期末"
+    },
+    {
+        "id":  "124",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "观察你的数学试卷，它的左右两边互相( ),上边和左边互相( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平行;垂直",
+        "solution":  "【分析】两条直线相交成直角时，这两条直线互相垂直，其中一条直线是另一条直线的垂线，这两条直线的交点叫做垂足。同一平面内不相交的两条直线互相平行，其中条直线是另一条直线的平行线。据此解答即可。本题考查平行和垂直的特征，同一平面的两条直线不是相交就是平行，夹角是直角是两条直线垂直.\n【详解】观察你的数学试卷，它的左右两边互相平行，上边和左边互相垂直。",
+        "knowledge_points":  [
+                                 "垂直的特征，平行的特征及性质"
+                             ],
+        "source":  "19-20 四年级上·河南三门峡·期末"
+    },
+    {
+        "id":  "125",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一张长方形纸对折两次后展开，折痕互相( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平行或垂直",
+        "solution":  "【分析】如果两次都朝一个方向折叠，折痕互相平行;如果两次都朝两个方向折叠即先上下折，然后再左右折，折痕互相垂直，据此解答.\n【详解】根据分析，一张长方形纸对折两次后展开，折痕互相平行或垂直。",
+        "knowledge_points":  [
+                                 "图形的折叠问题，垂直的特征，平行的特征及性质"
+                             ],
+        "source":  "25-26四年级上·福建漳州·期中"
+    },
+    {
+        "id":  "126",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "通过动手拉平行四边形框架的活动，我们知道，平行四边形有( )的特性，生活中( )运用了这一特性。",
+        "options":  [
+
+                    ],
+        "answer":  "易变形;伸缩门",
+        "solution":  "【分析】两组对边分别平行且相等的四边形，叫做平行四边形;平行四边形具有易变形的特性，伸缩门利用了平行四边形易变形的特性。正确理解平行四边形的不稳定性，是解答此题的关键.\n【详解】根据分析可知，通过动手拉平行四边形框架的活动，我们知道，平行四边形有易变形的特性，生活中伸缩门运用了这一特性。(答案不唯一)",
+        "knowledge_points":  [
+                                 "平行四边形的不稳定性及应用"
+                             ],
+        "source":  "21-22四年级上·河南郑州·期末"
+    },
+    {
+        "id":  "127",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "从直线外一点到这条直线所画的( )最短，它的长度叫做这点到直线的( )。",
+        "options":  [
+
+                    ],
+        "answer":  "垂直线段;距离",
+        "solution":  "【详解】如图所示:![](images/162c72c649d3309e7184f4bc613e39f63af00bd3ba82e4d0070ae999addbe790.jpg)从直线外一点到这条直线所画的垂直线段最短，它的长度叫作点到直线的距离。",
+        "knowledge_points":  [
+                                 "画垂线，垂直的特征"
+                             ],
+        "source":  "2014 四年级·全国·课后作业"
+    },
+    {
+        "id":  "128",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "学生课桌桌面的邻边互相( ),对边互相( )。",
+        "options":  [
+
+                    ],
+        "answer":  "垂直;平行",
+        "solution":  "【分析】长方形的特征：对边平行且相等,4个角都是直角，因为课桌的桌面是一个长方形，课桌桌面的邻边就是长方形的一条长和一条宽，互相垂直;对边就是长方形的两条宽或者两条长，互相平行;据此解答即可。此题应根据长方形的特征进行解答即可.\n【详解】由分析可得：学生课桌桌面的邻边互相垂直，对边互相平行。",
+        "knowledge_points":  [
+                                 "长方形的概念及特点，垂直的特征，平行的特征及性质"
+                             ],
+        "source":  "20-21 四年级上·贵州遵义·期末"
+    },
+    {
+        "id":  "129",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "同一平面内两条直线都与同一条直线垂直，这两条直线的位置关系是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "互相平行",
+        "solution":  "【分析】根据垂直和平行的特征：同一平面内两条直线都垂直于同一条直线，那么这两条直线互相平行;进而解答即可。此题考查了垂直和平行的特征及性质.\n【详解】如果两条直线都垂直于同一条直线，那么这两条直线互相平行。",
+        "knowledge_points":  [
+                                 "平行的特征及性质，垂直的特征"
+                             ],
+        "source":  "19-20 四年级上·云南曲靖·期中"
+    },
+    {
+        "id":  "130",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "过直线外一点向直线作的( )最短。",
+        "options":  [
+
+                    ],
+        "answer":  "垂直线段",
+        "solution":  "【分析】从直线外一点向直线可以作无数条线段，其中垂直线段最短。如图分析观察:![](images/b94c60d677c5b0be6d6414523bcc652ecca7eeba8cdf18445720aceadf15d031.jpg)本题考查了垂线的性质.\n【详解】过直线外一点向直线作的垂直线段最短。",
+        "knowledge_points":  [
+                                 "画垂线"
+                             ],
+        "source":  "19-20 四年级上·云南曲靖·期中"
+    },
+    {
+        "id":  "131",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "过直线外一点可以画( )条平行线。",
+        "options":  [
+
+                    ],
+        "answer":  "1",
+        "solution":  "【分析】在同一平面内永不相交的两条直线叫做平行线，依此画图并填空即可。此题考查的是过直线外一点画平行线，熟练掌握平行的特点是解答此题的关键.\n【详解】![](images/738c543daf32f2ea6d5c01546b305126f531da2ada055b3eace0a67588a2d487.jpg)\n如图所示，过直线外一点可以画1条平行线。",
+        "knowledge_points":  [
+                                 "画平行线"
+                             ],
+        "source":  "21-22 四年级上·湖北荆门·期末"
+    },
+    {
+        "id":  "132",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰梯形有两个相等的( )角和两个相等的( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "锐;钝",
+        "solution":  "【详解】根据等腰梯形的特征可知;一个等腰梯形有两个相等的锐角和两个相等的钝角。",
+        "knowledge_points":  [
+                                 "直角梯形和等腰梯形的概念及特点"
+                             ],
+        "source":  "2022六年级下·全国·专题练习"
+    },
+    {
+        "id":  "133",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个长方形中有( )组互相垂直的线段。",
+        "options":  [
+
+                    ],
+        "answer":  "4",
+        "solution":  "【分析】两条直线相交成直角时，这两条直线互相垂直，其中一条直线是另一条直线的垂线，这两条直线的交点叫做垂足。据此解答即可。本题考查垂直的性质，关键是看哪两条线段的夹角是直角.\n【详解】一个长方形中有4个直角，有4组互相垂直的线段。",
+        "knowledge_points":  [
+                                 "垂直的特征"
+                             ],
+        "source":  "20-21 四年级上·辽宁本溪·期中"
+    },
+    {
+        "id":  "134",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "电动伸缩门由许多个平行四边形组成，这是利用了平行四边形( )的特点。",
+        "options":  [
+
+                    ],
+        "answer":  "易变形或不稳定",
+        "solution":  "【详解】平行四边形有易变形的特点，所以任意拉动平行四边形就![](images/b0defd073da0135baea1bdaa5db42a189a0c81892b69c60bdfeebb268f2dd80b.jpg)能得到一个新的平行四边形。电动伸缩门由许多个平行四边形组成，这是利用了平行四边形(易变形)的特点。",
+        "knowledge_points":  [
+                                 "平行四边形的不稳定性及应用"
+                             ],
+        "source":  "20-21四年级上·河南信阳·期末"
+    },
+    {
+        "id":  "135",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "从直线外一点到这条直线所画的( )线段最短，它的( )叫做这点到直线的距离。",
+        "options":  [
+
+                    ],
+        "answer":  "垂直;长度",
+        "solution":  "【详解】从直线外一点到这条直线所画的(垂直)线段最短，它的(长度)叫做这点到直线的距离。如图所示:![](images/fdc010c626d884743b66bcee238bcb904944f6e0f5f8c25e22cb44a16c052fd6.jpg)",
+        "knowledge_points":  [
+                                 "画垂线，垂直的特征"
+                             ],
+        "source":  "20-21四年级上·河南信阳·期末"
+    },
+    {
+        "id":  "136",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在同一平面内直线a与直线c互相垂直，直线b也与直线c互相垂直，那么，直线a与直线b互相( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平行",
+        "solution":  "【分析】在同一平面内，垂直于同一条直线的两条直线互相平行，据此即可解答。本题主要考查学生对垂直和平行概念和特征的掌握和灵活运用.\n【详解】根据分析可知，在同一平面内直线a与直线c互相垂直，直线b也与直线c互相垂直，那么，直线a与直线b互相平行。",
+        "knowledge_points":  [
+                                 "平行的特征及性质，垂直的特征"
+                             ],
+        "source":  "21-22 四年级上·新疆克拉玛依·期末"
+    },
+    {
+        "id":  "137",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "两个完全一样的梯形，它们的上底是10厘米，下底是5厘米，高是8厘米，把这两个梯形拼成一个平行四边形，则平行四边形的底是( )厘米，高是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "15;8",
+        "solution":  "【分析】先根据题意画出这样的两个梯形，然后将它们拼在一起，最后再根据所拼成的图形计算出平行四边形的底和高即可填空。此题考查的是平面图形的拼组，应熟练掌握梯形和平行四边形的特点.\n【详解】![](images/acde41b6f49f17e7e87c132e9b350f6645a4d2078d0abdd74fc75327c03adbde.jpg)10+5=15(厘米)根据画图可知，平行四边形的底是15厘米，高是8厘米。",
+        "knowledge_points":  [
+                                 "平行四边形的概念及特点，平行四边形的高及画法，梯形的概念及特点，平面图形的拼接，用同样的图形拼接"
+                             ],
+        "source":  "21-22 四年级上·山西太原·期末"
+    },
+    {
+        "id":  "138",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "两条直线相交成( )角时，这两条直线互相垂直。两条平行线之间的距离是5厘米，在这两条平行线之间画一条垂线，这条垂线段的长是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "直;5",
+        "solution":  "【分析】在同一平面内两条直线相交成直角，这两条直线互相垂直;两条平行线间的垂线段的长度为这两条平行线之间的距离，依此填空。熟练掌握平行与垂直的特点是解答此题的关键.\n【详解】根据分析可知，两条直线相交成直角时，这两条直线互相垂直。两条平行线之间的距离是5厘米，在这两条平行线之间画一条垂线，这条垂线段的长是5厘米。",
+        "knowledge_points":  [
+                                 "垂直的特征，平行的特征及性质"
+                             ],
+        "source":  "21-22 四年级上·山西太原·期末"
+    },
+    {
+        "id":  "139",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "长方形相邻的两条边互相( ),平行四边形相对的两条边互相( )。",
+        "options":  [
+
+                    ],
+        "answer":  "垂直;平行",
+        "solution":  "【分析】两条直线相交成直角时，这两条直线互相垂直，其中一条直线是另一条直线的垂线，这两条直线的交点叫做垂足。同一平面内不相交的两条直线互相平行，其中条直线是另一条直线的平行线。据此解答即可。熟练掌握平行和垂直的性质，是解答此题的关键.\n【详解】长方形相邻的两条边互相垂直，平行四边形相对的两条边互相平行。",
+        "knowledge_points":  [
+                                 "平行的特征及性质，垂直的特征"
+                             ],
+        "source":  "19-20 四年级上·四川宜宾·期末"
+    },
+    {
+        "id":  "140",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "伸缩门在我们的生活中有广泛的应用，它利用了平行四边形( )的特点。",
+        "options":  [
+
+                    ],
+        "answer":  "不稳定",
+        "solution":  "【分析】平行四边形易变形，具有不稳定性，据此解答。此题主要考查了平行四边形的不稳定性，要熟练掌握.\n【详解】伸缩门在我们的生活中有广泛的应用，它利用了平行四边形不稳定的特点。",
+        "knowledge_points":  [
+                                 "平行四边形的不稳定性及应用"
+                             ],
+        "source":  "21-22 四年级上·四川乐山·期末"
+    },
+    {
+        "id":  "141",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "两根长8厘米和两根长6厘米的小棒，小聪摆了一个平行四边形，小明摆了一个长方形，他们摆的图形的( )一定相等。",
+        "options":  [
+
+                    ],
+        "answer":  "周长",
+        "solution":  "【分析】因为平行四边形对边平行且相等，长方形的对边平行且相等，又已知四根小棒的长度分别是8厘米,8厘米,6厘米,6厘米，根据周长的意义，围成平面图形所有边的长度和就是这个图形的周长，所以平行四边形和长方形的周长是相等的，都是这4根小棒的总长度，只不过是形状变了，据此即可进行解答。解答此题的关键是：小棒的长度一定，则围成的长方形和平行四边形的周长是一样的.\n【详解】两根长8厘米和两根长6厘米的小棒，小聪摆了一个平行四边形，小明摆了一个长方形，他们摆的图形的周长一定相等。",
+        "knowledge_points":  [
+                                 "平行四边形的周长，长方形的周长，长方形的概念及特点，平行四边形的概念及特点"
+                             ],
+        "source":  "21-22 四年级上·江西上饶·期末"
+    },
+    {
+        "id":  "142",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个四边形，只有一组对边互相平行，这个四边形是( );两个完全一样的梯形一定能拼成一个( )。",
+        "options":  [
+
+                    ],
+        "answer":  "梯形;平行四边形",
+        "solution":  "【分析】梯形是只有一组对边平行的四边形，据此可知，这个四边形是梯形。本题考查梯形的定义以及图形的拼组，梯形只有组对边平行，平行四边形有两组对边平行。![](images/bd8298668136ad3bfbdedf518754760b6396961948ec3663e52cd9546161bc96.jpg)如图所示，一个平行四边形可以被分成两个完全一样的梯形，则两个完全一样的梯形可以拼成一个平行四边形.\n【详解】一个四边形，只有一组对边互相平行，这个四边形是梯形;两个完全一样的梯形一定能拼成一个平行四边形。",
+        "knowledge_points":  [
+                                 "平行四边形的概念及特点，梯形的概念及特点，平面图形的拼接，用同样的图形拼接"
+                             ],
+        "source":  "20-21 四年级上·湖南张家界·期末"
+    },
+    {
+        "id":  "143",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个平行四边形框架，相邻两条边的长度分别是6厘米和4厘米，将它拉成长方形，那么长方形的周长是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "20",
+        "solution":  "【分析】平行四边形相邻两条边的长度和乘2等于平行四边形的周长，也是长方形的周长，据此即可解答。平行四边形的周长和长方形的周长相等，这是解答本题的关键.\n【详解】(6+4)×2=10×2=20(厘米)",
+        "knowledge_points":  [
+                                 "平行四边形的周长，长方形的周长"
+                             ],
+        "source":  "20-21 四年级上·福建福州·期末"
+    },
+    {
+        "id":  "144",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在同一个平面内，两条直线相交成( )角时，就说这两条直线互相垂直。",
+        "options":  [
+
+                    ],
+        "answer":  "直",
+        "solution":  "【详解】在同一个平面内，两条直线相交成直角时，这两条直线互相垂直，其中一条直线是另一条直线的垂线，这两条直线的交点叫做垂足。",
+        "knowledge_points":  [
+                                 "垂直的特征"
+                             ],
+        "source":  "21-22 四年级上·贵州黔西·期末"
+    },
+    {
+        "id":  "145",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一般情况下，两条路互相( )的路口叫十字路口，行人在过马路时应走由互相( )的白线组成的斑马线。",
+        "options":  [
+
+                    ],
+        "answer":  "垂直;平行",
+        "solution":  "【分析】在同一平面内两条直线相交成直角，这两条直线互相垂直;同一平面内永不相交的两条直线叫做平行线，组成平行线的两条直线互相平行，依此填空。熟练掌握平行与垂直的特点是解答此题的关键.\n【详解】根据分析可知，一般情况下，两条路互相垂直的路口叫十字路口，行人在过马路时应走由互相平行的白线组成的斑马线。",
+        "knowledge_points":  [
+                                 "垂直的特征，平行的特征及性质"
+                             ],
+        "source":  "21-22 四年级上·贵州黔南·期末"
+    },
+    {
+        "id":  "146",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用四根吸管串成一个长方形，然后用两手捏住长方形的两个对角，向相反方向拉,( )不变,( )变了。",
+        "options":  [
+
+                    ],
+        "answer":  "吸管长度;形状",
+        "solution":  "【分析】观察图形可知，用两手捏住长方形的两个对角，向相反方向拉，长方形变成了平行四边形，即吸管的长度没有改变，长方形的形状发生改变。本题考查的是长方形与平行四边形的关系，也可以说长方形是特殊的平行四边形。本题可以动手操作尝试完成.\n【详解】根据分析画图如下:![](images/8cdc3ba2b5dab0bd4cfbc7082b6adb0246def79f4c84829b79ffc1c78c11068a.jpg)所以，用四根吸管串成一个长方形，然后用两手捏住长方形的两个对角，向相反方向拉，吸管长度不变，形状变了。",
+        "knowledge_points":  [
+                                 "平行四边形的概念及特点，长方形的概念及特点"
+                             ],
+        "source":  "20-21四年级上·福建福州·期末"
+    },
+    {
+        "id":  "147",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个直角梯形的下底是上底的5倍，是高的3倍;如果上底延长12cm,就变成了一个长方形;这个梯形的上底是( )厘米，下底是( )厘米，高是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "3;15;5",
+        "solution":  "【分析】已知一个直角梯形的下底是上底的5倍，是高的3倍，如果上底延长12cm,就变成了一个长方形。由此可知:12厘米相当于梯形上底的(5-1)倍，即可求出直角梯形的上底，上底的长度*5=下底的长度，下底的长度/3=直角梯形的高;据此解答.\n【详解】12÷(5-1)=12÷4=3(厘米),下底:3×5=15(厘米),高:15÷3=5(厘米).\n[点睛]本题考查直角梯形的特征与长方形的特征，先求出直角梯形的上底是解题关键。",
+        "knowledge_points":  [
+                                 "直角梯形和等腰梯形的概念及特点，长方形的概念及特点"
+                             ],
+        "source":  "21-22四年级上·河南南阳·阶段练习"
+    },
+    {
+        "id":  "148",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个平行四边形的两条边分别是10厘米和15厘米，已知这个平行四边形的一条高是12厘米，那么这条高对应的底边长( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "10",
+        "solution":  "【分析】从底边对边的一个顶点向底边作一条垂线，这点和垂足之间的距离就是平行四边形的高，与高垂直的边是这条高的底边;相对应底边上的高要比不同的底要矮一些，因为高是底与其对边的最短距离，所以高一定小于其所在底的邻边，因此15厘米底边对应的高小于10厘米,12厘米大于10厘米，这条高对应的底边长度为10厘米;据此解答。本题考查了平行四边形底和高的互相对应的关系，也可以通过画图更清晰直观.\n【详解】据分析可知：一个平行四边形的两条边分别是10厘米和15厘米，已知这个平行四边形的一条高是12厘米，那么这条高对应的底边长10厘米。",
+        "knowledge_points":  [
+                                 "平行四边形的高及画法"
+                             ],
+        "source":  "21-22 四年级上·福建莆田·期末"
+    },
+    {
+        "id":  "149",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "平行四边形具有( )的特性。",
+        "options":  [
+
+                    ],
+        "answer":  "易变形",
+        "solution":  "【分析】平行四边形容易变形，具有不稳定性，依此填空。熟练掌握平行四边形的特性是解答此题关键.\n【详解】平行四边形具有易变形的特性，例如：生活中的电动伸缩门就是应用了平行四边形易变形的特性。",
+        "knowledge_points":  [
+                                 "平行四边形的不稳定性及应用"
+                             ],
+        "source":  "14-15 四年级·全国·期末"
+    },
+    {
+        "id":  "150",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "黑板的上下两边互相( ),相邻两边互相( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平行;垂直",
+        "solution":  "【分析】两条直线相交成直角时，这两条直线互相垂直，其中一条直线是另一条直线的垂线，这两条直线的交点叫做垂足。同一平面内不相交的两条直线互相平行，其中一条直线是另一条直线的平行线。据此解答即可。熟练掌握平行和垂直的性质是解决本题的关键.\n【详解】黑板的上下两边互相平行，相邻两边互相垂直。",
+        "knowledge_points":  [
+                                 "垂直的特征，平行的特征及性质"
+                             ],
+        "source":  "17-18 四年级上·江苏·单元测试"
+    },
+    {
+        "id":  "151",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "黑板左右两条边互相( ),下面的边和左边的边互相( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平行;垂直",
+        "solution":  "【分析】黑板的形状是长方形，根据长方形的特征：相对的两组对边分别平行且相等，相邻两边互相垂直;解答即可。熟悉长方形的特征是解答此题的关键.\n【详解】黑板左右两条边互相(平行),下面的边和左边的边互相(垂直)。",
+        "knowledge_points":  [
+                                 "长方形的概念及特点，垂直的特征，平行的特征及性质"
+                             ],
+        "source":  "21-22 四年级上·湖北武汉·期末"
+    },
+    {
+        "id":  "152",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "正方形相邻的两条边互相( ),( )的两条边互相平行。",
+        "options":  [
+
+                    ],
+        "answer":  "垂直;相对",
+        "solution":  "【分析】正方形的性质:(1)正方形的四条边相等;(2)正方形的四个角都是直角;(3)正方形的对边互相平行，相邻的两条边互相垂直，据此解答即可。此题考查正方形的特征，平时注意基础知识的积累.\n【详解】据分析可得：正方形相邻的两条边互相垂直，相对的两条边互相平行。",
+        "knowledge_points":  [
+                                 "正方形的概念及特点，垂直的特征，平行的特征及性质"
+                             ],
+        "source":  "21-22 四年级上·湖南岳阳·期末"
+    },
+    {
+        "id":  "153",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "只有一组对边平行的四边形叫做( )。平行四边形的两组对边分别( )而且( ),伸缩门的设计利用了平行四边形( )的特点。",
+        "options":  [
+
+                    ],
+        "answer":  "梯形;平行;相等;容易变形",
+        "solution":  "【详解】如图所示:![](images/c8286364aacd9e63c9a69b0b205628f214aceed8e6fd4464618b9a656bd4e912.jpg)根据梯形的定义以及平行四边形的特征可知：只有一组对边平行的四边形叫做梯形。平行四边形的两组对边分别平行而且相等，平行四边形具有不稳定性，所以容易变形，如生活中常见的电动伸缩门，升降机等，这是应用了平行四边形容易变形进行制作的，便于伸缩。所以伸缩门的设计利用了平行四边形容易变形的特点。",
+        "knowledge_points":  [
+                                 "平行四边形的概念及特点，平行四边形的不稳定性及应用，梯形的概念及特点"
+                             ],
+        "source":  "2022 四年级上·四川德阳·专题练习"
+    },
+    {
+        "id":  "154",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "等腰梯形周长35cm,上底5cm,一条腰长7cm,它的下底长是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "16",
+        "solution":  "【分析】根据等腰梯形的两腰相等，解答此题即可。熟练掌握等腰梯形的特征，是解答此题的关键.\n【详解】35-5-7×2=30-14=16(cm)",
+        "knowledge_points":  [
+                                 "梯形的周长，直角梯形和等腰梯形的概念及特点"
+                             ],
+        "source":  "20-21 四年级下·四川雅安·期末"
+    },
+    {
+        "id":  "155",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一张长42cm,宽4cm的长方形纸。(1)用这张纸最多能剪出( )个边长为2cm的正方形。(2)用这张纸最多能剪出( )个边长为4cm的正方形。(3)用这张纸最多能剪出( )个两条直角边都是4cm的直角三角形。(4)用这张纸最多能剪出( )个上底3cm,下底6cm,高4cm的直角梯形。",
+        "options":  [
+
+                    ],
+        "answer":  "42;10;20;9",
+        "solution":  "【分析】(1)(2)分别求长方形的长和宽分别含有多少个正方形的边长，再把个数相乘即可;(3)分别求长方形的长和宽分别含有多少个边长是4厘米的正方形，根据(2)的结果，再乘2，就是直角三角形的个数;(4)两个上底3cm,下底6cm,高4cm的直角梯形可以拼成一个长9厘米、宽4厘米的长方形，42厘米里可以先拼4组这样的长方形，剩下6厘米宽的部分还可以再剪1个直角梯形。本题主要考查图形的剪拼，关键是求各边含有小图形的边的个数.\n【详解】(1)(42÷2)×(4÷2)=21×2=42(个);(2)(42÷4)×(4÷4)=10×1=10(个);(3)10×2=20(个);(4)42÷9=4(组)……6(厘米),4×2+1=9(个)。",
+        "knowledge_points":  [
+                                 "长方形的概念及特点，正方形的概念及特点，梯形的概念及特点，平面图形的分割"
+                             ],
+        "source":  "20-21 五年级下·四川雅安·期末"
+    },
+    {
+        "id":  "156",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "长方形的对边互相( ),相邻的两条边互相( )。(填\u0027平行\u0027或\u0027垂直\u0027)",
+        "options":  [
+
+                    ],
+        "answer":  "平行;垂直",
+        "solution":  "【分析】根据长方形的特征，对边平行且相等,4个角都是直角，可知，长方形的两组对边互相平行，相邻的两条边互相垂直，据此解答即可。解答此题的主要依据是：长方形的特征及性质.\n【详解】据分析可知：长方形的对边互相平行，相邻的两条边互相垂直。",
+        "knowledge_points":  [
+                                 "长方形的概念及特点，垂直的特征，平行的特征及性质"
+                             ],
+        "source":  "21-22 四年级上·河北石家庄·期末"
+    },
+    {
+        "id":  "157",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个直角梯形的下底是6厘米，将下底减少2厘米后它就变成了正方形，这个梯形的高是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "4",
+        "solution":  "【分析】下底是6厘米，将下底减少2厘米后它就变成了正方形，可知上底是4厘米;正方形边长都相等，因此这个梯形的高也是4厘米。此题的关键是明确正方形边长都相等，然后再进一步解答.\n【详解】6-2=4(厘米),这个梯形的高是4厘米。",
+        "knowledge_points":  [
+                                 "正方形的概念及特点，梯形的概念及特点"
+                             ],
+        "source":  "21-22 四年级上·湖北武汉·期末"
+    },
+    {
+        "id":  "158",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "两条直线相交，其中有一个角是直角，则其余3个角都是( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "直",
+        "solution":  "【详解】两条直线相交成直角时，这两条直线互相垂直，相交形成的4个角都是直角。所以如果两条直线相交，其中有一个角是直角，那么就说明这两条直线互相垂直，其余3个角都是直角。",
+        "knowledge_points":  [
+                                 "垂直的特征"
+                             ],
+        "source":  "21-22 四年级上·云南曲靖·期末"
+    },
+    {
+        "id":  "159",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在同一平面内画与已知直线相距2厘米的平行线可以画( )条。",
+        "options":  [
+
+                    ],
+        "answer":  "2/两",
+        "solution":  "【分析】利用画图直观的看一看，在同一平面内，画与已知直线相距2厘米的平行线有2条，在已知直线的两边各有1条。本题考查了平行线的知识，结合题意，画出图示，是解答此题的关键.\n【详解】如图所示:![](images/05c3e6a1fa5570dd52c4f44284cfed0d647702ba8e6dec1dab634787e8272a7d.jpg)在同一平面内画与已知直线相距2厘米的平行线可以画2条。",
+        "knowledge_points":  [
+                                 "平行的特征及性质"
+                             ],
+        "source":  "20-21 四年级上·山西忻州·期末"
+    },
+    {
+        "id":  "160",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "明明用33厘米的铁丝做了一个等腰梯形，它的上底和下底分别是9厘米和12厘米，它的腰是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】33厘米是等腰梯形的周长，所以用33-9-12就等于两条腰的长度和，注意两条腰的长度相等，再除以2即等于腰长。本题主要考查学生对等腰梯形特点的掌握和灵活运用.\n【详解】(33-9-12)÷2=12÷2=6(厘米)",
+        "knowledge_points":  [
+                                 "梯形的周长，直角梯形和等腰梯形的概念及特点"
+                             ],
+        "source":  "20-21四年级上·山西忻州·期末"
+    },
+    {
+        "id":  "161",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "两条直线a,b相交成( )角时，这两条直线可记作:a⊥b。",
+        "options":  [
+
+                    ],
+        "answer":  "直",
+        "solution":  "【分析】两条直线相交成直角时，这两条直线互相垂直，其中一条直线是另一条直线的垂线，这两条直线的交点叫做垂足。本题主要考查学生对垂线定义和特征的掌握.\n【详解】两条直线a,b相交成直角时，这两条直线互相垂直，这两条直线可记作:a⊥b。",
+        "knowledge_points":  [
+                                 "垂直的特征"
+                             ],
+        "source":  "19-20 四年级上·湖北黄冈·期末"
+    },
+    {
+        "id":  "162",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在同一平面内，与已知直线平行的直线有( )条，过直线外一点作已知直线的垂线，可以作( )条。",
+        "options":  [
+
+                    ],
+        "answer":  "无数;1或一",
+        "solution":  "【分析】在同一平面内，和一条直线平行的直线可以画无数条;过直线外一点，作已知直线的垂线，点到直线的垂直线段只有一条，据此解答。本题考查了垂直与平行的性质.\n【详解】在同一平面内，与已知直线平行的直线有无数条，过直线外一点作已知直线的垂线，可以作1条。",
+        "knowledge_points":  [
+                                 "画垂线，平行的特征及性质"
+                             ],
+        "source":  "21-22 四年级上·湖北随州·期末"
+    },
+    {
+        "id":  "163",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "平行四边形有( )条高，把它拉成一个长方形后，它的( )不变。",
+        "options":  [
+
+                    ],
+        "answer":  "无数;周长",
+        "solution":  "【分析】从平行四边形一条边上的一点向对边引一条垂线，这点和垂足之间的线段叫做平行四边形的高;把平行四边形拉成一个长方形后，它每条边的长度没变，依此填空。此题考查的是平行四边形的高，对平行四边形的周长的认识，应熟练掌握.\n【详解】平行四边形有无数条高;平行四边形一周的长度是它的周长，因此把它拉成一个长方形后，它的周长不变。",
+        "knowledge_points":  [
+                                 "平行四边形的高及画法，平行四边形的周长"
+                             ],
+        "source":  "21-22 四年级上·云南玉溪·期末"
+    },
+    {
+        "id":  "164",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "两条平行线之间的距离是8厘米，在这两条平行线之间作一条垂线段，这条垂线的长是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "8",
+        "solution":  "【分析】从直线外一点到这条直线的线段中，垂直线段最短，这条垂直线段的长度叫做点到直线的距离。据此可知，这条垂线的长度就是两条平行线之间的距离，即8厘米。本题考查平行线之间的距离的定义，需熟练掌握.\n【详解】由分析得：这条垂线长是8厘米。",
+        "knowledge_points":  [
+                                 "平行的特征及性质"
+                             ],
+        "source":  "19-20四年级上·河南许昌·期末"
+    },
+    {
+        "id":  "165",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用两个完全相同的等腰梯形拼成一个平行四边形，这个平行四边形的周长是100厘米，已知一个等腰梯形的周长是60厘米，等腰梯形一条腰的长是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "10",
+        "solution":  "【分析】如图所示，等腰梯形中上底与下底的和等于平行四边形的一条边，平行四边形中相邻两条边的长度和是周长/2，则用梯形的周长减去平行四边形中相邻两条边的长度和，即可求出梯形中一条腰的长度。本题考查等腰梯形和平行四边形的性质，关键是明确上底与下底的和等于平行四边形的一条边.\n![](images/ddfe9b939f9660193478e98253c07065237f827d54ac172552e5775e2ae10f92.jpg)\n【详解】60-100÷2=60-50=10(厘米),则等腰梯形一条腰的长是10厘米。",
+        "knowledge_points":  [
+                                 "梯形的周长，平行四边形的切拼，平行四边形的周长"
+                             ],
+        "source":  "21-22四年级上·河南三门峡·期末"
+    },
+    {
+        "id":  "166",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "经过一点可以画( )条直线，经过两点可以画( )条线段，梯形有( )条高。",
+        "options":  [
+
+                    ],
+        "answer":  "无数;1;无数",
+        "solution":  "【详解】经过一点可以画无数条直线，经过两点可以画1条线段。从梯形一条底边上的一点到它对边的垂直线段叫做梯形的高。则梯形的高有无数条。",
+        "knowledge_points":  [
+                                 "梯形的高及画法，线段，直线，射线的认识及特征"
+                             ],
+        "source":  "21-22 四年级上·湖北随州·期末"
+    },
+    {
+        "id":  "167",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "同一平面内，两条直线不是平行就是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "相交",
+        "solution":  "【分析】根据平行的含义：在同一平面内，不相交的两条直线叫做平行线，组成平行线的两条直线相互平行;据此判断即可。此题考查了垂直与平行的特征和性质，要熟练掌握.\n【详解】在同一平面内的两条直线不是平行就是相交。",
+        "knowledge_points":  [
+                                 "平行的特征及性质"
+                             ],
+        "source":  "21-22 四年级上·甘肃金昌·期末"
+    },
+    {
+        "id":  "168",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在同一个平面内( )的两条直线叫做平行线。只有( )平行的四边形叫做梯形。",
+        "options":  [
+
+                    ],
+        "answer":  "不相交;一组对边",
+        "solution":  "【详解】在同一平面内，两条永不相交的直线叫做平行线，如果同一平面内这样的两条直线不平行，说明它们一定相交。两组对边分别平行的四边形叫做平行四边形。只有一组对边平行的四边形叫做梯形。",
+        "knowledge_points":  [
+                                 "平行的特征及性质，梯形的概念及特点"
+                             ],
+        "source":  "21-22 四年级上·贵州六盘水·期末"
+    },
+    {
+        "id":  "169",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰梯形的周长100厘米，上底17厘米，下底33厘米，它的一条腰长( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "25",
+        "solution":  "【分析】等腰梯形的周长减去上下底的长度，再除以2，即等于一条腰的长度。等腰梯形的两条腰相等，这是解答本题的关键.\n【详解】(100-17-33)÷2=50÷2=25(厘米)",
+        "knowledge_points":  [
+                                 "梯形的周长，直角梯形和等腰梯形的概念及特点"
+                             ],
+        "source":  "21-22 四年级上·四川德阳·期末"
+    },
+    {
+        "id":  "170",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "过直线外一点可以作( )条直线与已知直线平行;从直线外一点到已知直线可以画( )条线段，其中( )最短，它的长度叫点到直线的( )。",
+        "options":  [
+
+                    ],
+        "answer":  "1或一;无数;垂线段;距离",
+        "solution":  "【分析】根据平行与垂直的知识填空即可。此题涉及的知识点较多，但都比较简单，属于基础题，只要认真，容易完成，注意平时基础知识的积累.\n【详解】过直线外一点可以作一条直线与已知直线平行，从直线外一点到已知直线可以画无数条线段，其中垂线段最短，它的长度叫做点到直线的距离。",
+        "knowledge_points":  [
+                                 "垂直的特征，平行的特征及性质"
+                             ],
+        "source":  "2022 四年级上·四川德阳·专题练习"
+    },
+    {
+        "id":  "171",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个直角梯形的上底长是4厘米，如果将上底延长到原来的4倍就变成了一个正方形，那么这个梯形的高是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "16",
+        "solution":  "【分析】将这个直角梯形的上底延长到原来的4倍后变成正方形，而正方形的四条边相等，则直角梯形的高是上底的4倍。上底是4厘米，上底的4倍是4×4=16厘米。也就是梯形的高是16厘米。正方形的四条边相等，直角梯形的直角边是梯形的一条高，则这个梯形的高是上底的4倍.\n【详解】4×4=16(厘米),那么这个梯形的高是16厘米。",
+        "knowledge_points":  [
+                                 "正方形的概念及特点，直角梯形和等腰梯形的概念及特点"
+                             ],
+        "source":  "21-22 四年级上·四川广元·期末"
+    },
+    {
+        "id":  "172",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰梯形，它的周长是43厘米，上底和下底分别是5厘米和8厘米，它的一条腰长是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】因为梯形的周长=两腰长度+上底+下底，又根据等腰梯形的特点，两腰相等，所以一条腰的长度=(周长-上底-下底)÷2，代入数据计算即可。此题考查的目的是理解掌握等腰梯形的特征以及梯形周长公式的灵活运用.\n【详解】(43-5-8)÷2=30÷2=15(厘米)。",
+        "knowledge_points":  [
+                                 "梯形的周长，直角梯形和等腰梯形的概念及特点"
+                             ],
+        "source":  "21-22四年级上·四川凉山·期末"
+    },
+    {
+        "id":  "173",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个四边形，如果它的一组对边平行且长度都是8厘米，它一定是( )形;如果它的一组对边平行且长度分别是6厘米和10厘米，它一定是( )形。",
+        "options":  [
+
+                    ],
+        "answer":  "平行四边形;梯形",
+        "solution":  "【分析】如果四边形的一组对边平行且长度都是8厘米，则这个四边形的一组对边平行且相等，这个四边形是平行四边形。如果四边形的一组对边平行且长度分别是6厘米和10厘米，则这个四边形的一组对边平行且不相等，这个四边形是梯形。本题考查平行四边形和梯形的性质，两组对边都平行的四边形是平行四边形，只有一组对边平行的四边形是梯形.\n【详解】一个四边形，如果它的一组对边平行且长度都是8厘米，它一定是平行四边形;如果它的一组对边平行且长度分别是6厘米和10厘米，它一定是梯形。",
+        "knowledge_points":  [
+                                 "平行四边形的概念及特点，梯形的概念及特点"
+                             ],
+        "source":  "21-22 四年级上·辽宁鞍山·期末"
+    },
+    {
+        "id":  "174",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "三条直线相交，最多有( )个交点，最少有( )个交点。",
+        "options":  [
+
+                    ],
+        "answer":  "3;1",
+        "solution":  "【分析】三条直线相交，有三种情况，即：两条直线平行，被第三条直线所截，有两个交点;三条直线经过同一个点，有一个交点;三条直线两两相交且不经过同一点，有三个交点，故可得答案。解决本题的关键是画出三条直线相交时的三种情况，找出交点.\n【详解】由分析可知，三条线相交情况如下:![](images/5b90880fe27a9e6f55a4d3846c47473426e8c27018a9be2e76d50c4c2550924e.jpg)![](images/b917d9da7f27fc24d0372c1496bed74a44e7806cae8899d6456a77d3db722711.jpg)所以三条直线相交，最多有3个交点，最少有1个交点。",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征，平行的特征及性质"
+                             ],
+        "source":  "21-22 四年级上·湖南长沙·期末"
+    },
+    {
+        "id":  "175",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在两条直线相交组成的4个角中，一个角是直角，另外三个角都是( )度。",
+        "options":  [
+
+                    ],
+        "answer":  "90",
+        "solution":  "【分析】在同一平面内两条直线相交成直角，这两条直线互相垂直，依此填空。熟练掌握垂直的特点是解答此题的关键.\n【详解】因此在两条直线相交组成的4个角中，一个角是直角，另外三个角都是90°。",
+        "knowledge_points":  [
+                                 "垂直的特征"
+                             ],
+        "source":  "20-21 四年级上·云南文山·期末"
+    },
+    {
+        "id":  "176",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "平行四边形的对边互相( ),长方形的两条邻边互相( );伸缩门是应用了平行四边形( )的特性;线段有( )端点，把线段的两端无限延长就得到一条( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平行;垂直;容易变形;2个;直线",
+        "solution":  "【分析】两组对边分别平行的四边形，叫做平行四边形;长方形的两组对边分别互相平行，两组对边分别相等，四个角都是直角，在同一平面内两条直线相交成直角，这两条直线互相垂直;平行四边形容易变形，因此平行四边形具有不稳定性;一根拉紧的线，绷紧的弦，都可以看作线段，线段有两个端点;把线段向两端无限延伸，就得到一条直线，直线没有端点。此题考查的是平行四边形和长方形的特点，平行四边形的特性，以及直线与线段的特点，应熟练掌握.\n【详解】根据分析可知：平行四边形的对边互相平行，长方形的两条邻边互相垂直;伸缩门是应用了平行四边形容易变形的特性;线段有2个端点，把线段的两端无限延长就得到一条直线。",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征，垂直的特征，平行的特征及性质，平行四边形的不稳定性及应用"
+                             ],
+        "source":  "21-22四年级上·四川凉山·期末"
+    },
+    {
+        "id":  "177",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个平行四边形的周长是36厘米，其中一条边的长度是12厘米，与它相邻的另一条边的长度是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】根据\u0027平行四边形的周长=相邻两条边长度和乘2\u0027可得，相邻两条边的长度和是:36÷2=18(厘米),再减去12厘米，即可求出与它相邻的另一条边的长度。要明确，平行四边形相邻两条边的长度和是平行四边形周长的一半，是解答此题的关键.\n【详解】36÷2-12=18-12=6(厘米),所以，与它相邻的另一条边的长度是6厘米。",
+        "knowledge_points":  [
+                                 "平行四边形的概念及特点，平行四边形的周长"
+                             ],
+        "source":  "21-22四年级上·内蒙古通辽·期末"
+    },
+    {
+        "id":  "178",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在同一平面内有a,b,c三条直线,a∥b,b∥c,那么a与c的关系是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "互相平行",
+        "solution":  "【分析】在同一平面内，平行于同一条直线的两条直线互相平行。此题考查了平行的特征及性质。在同一个平面内，不相交的两条直线叫做平行线.\n【详解】根据分析可知：在同一平面内有a,b,c三条直线,a∥b,b∥c,那么a与c的关系是互相平行。",
+        "knowledge_points":  [
+                                 "平行的特征及性质"
+                             ],
+        "source":  "21-22 四年级上·重庆永川·期末"
+    },
+    {
+        "id":  "179",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在同一平面内( )的两条直线叫做平行线，如果将其中的一条直线旋转90°,那么这两条直线互相( )。",
+        "options":  [
+
+                    ],
+        "answer":  "不相交;垂直",
+        "solution":  "【分析】在同一平面内不相交的两条直线叫做平行线，相交成直角的两条直线互相垂直，据此即可解答。本题主要考查学生对平行和垂直定义的掌握.\n【详解】在同一平面内不相交的两条直线叫做平行线，如果将其中的一条直线旋转90°,则两条直线相交成直角，这两条直线互相垂直。",
+        "knowledge_points":  [
+                                 "垂直的特征，平行的特征及性质"
+                             ],
+        "source":  "20-21 四年级上·河南郑州·期末"
+    },
+    {
+        "id":  "180",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个梯形中最多有( )个直角。",
+        "options":  [
+
+                    ],
+        "answer":  "2",
+        "solution":  "【分析】根据梯形的特征及四边形的内角和是360°,梯形只有一组对边平行，所以梯形中最多有两个直角;据此解答.\n【详解】根据分析可知，一个梯形中最多有2个直角.\n[点睛]此题考查的目的是理解掌握梯形的特征，以及四边形的内角和是360°。",
+        "knowledge_points":  [
+                                 "梯形的概念及特点，直角梯形和等腰梯形的概念及特点"
+                             ],
+        "source":  "19-20 四年级上·全国·单元测试"
+    },
+    {
+        "id":  "181",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "平行四边形有( )条高，它的对边互相平行且( )。",
+        "options":  [
+
+                    ],
+        "answer":  "无数;相等",
+        "solution":  "【详解】根据平行四边形的意义：在同一平面内有两组对边分别平行的四边形叫做平行四边形，平行四边形的对边平行且相等;从平行四边形的一条边上的任意一点都可以向对边作垂直线段，即是平行四边形的高，平行四边形的每一条边都可以看做是底，所以平行四边形有4条底，有无数条高。",
+        "knowledge_points":  [
+                                 "平行四边形的概念及特点，平行四边形的高及画法"
+                             ],
+        "source":  "21-22 四年级上·广西河池·期末"
+    },
+    {
+        "id":  "182",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "( )平行的四边形叫梯形，长方形和正方形是特殊的( ),长方形可以拉伸成平行四边形，说明平行四边形具有( )的特性。这个长方形无论怎么拉,( )始终不变。",
+        "options":  [
+
+                    ],
+        "answer":  "只有一组对边;平行四边形;易变形;四条边",
+        "solution":  "【分析】梯形只有一组对边平行，平行四边形的两组对边平行。长方形的两组对边平行，四个角是直角，则长方形是特殊的平行四边形。正方形的两组对边平行，四条边相等，四个角是直角，则正方形是特殊的长方形。平行四边形的不稳定性即易变形性，是指平行四边形边长确定，其形状，大小不能完全确定。则长方形拉成平行四边形时，四个角变了，四条边不变.\n【详解】一组对边平行的四边形叫梯形，长方形和正方形是特殊的平行四边形，长方形可以拉伸成平行四边形，说明平行四边形具有易变形的特性。这个长方形无论怎么拉，四条边始终不变.\n[点睛]正方形是特殊的长方形，长方形是特殊的平行四边形。梯形只有一组对边平行。平行四边形具有易变形性。",
+        "knowledge_points":  [
+                                 "四边形的分类及关系，平行四边形的不稳定性及应用，梯形的概念及特点"
+                             ],
+        "source":  "21-22四年级上·河南新乡·期末"
+    },
+    {
+        "id":  "183",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "画已知直线的垂线可以用工具( )和( )。",
+        "options":  [
+
+                    ],
+        "answer":  "三角板;直尺",
+        "solution":  "【详解】已知直线的垂线和已知直线是互相垂直的，成90度角。所以可以利用三角板上的直角以及直尺画图即可。画已知直线的垂线可以用工具三角板和直尺.\n如图:![](images/34689991981bfd8e25cb4262e9a58a2c538d96b43b4c88ebb498722ba63fe039.jpg)",
+        "knowledge_points":  [
+                                 "画垂线"
+                             ],
+        "source":  "21-22 四年级上·河南安阳·期末"
+    },
+    {
+        "id":  "184",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "从3根7厘米的小棒和3根8厘米的小棒中选出4根围成一个平行四边形，这个平行四边形的周长是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "30",
+        "solution":  "【分析】根据平行四边形的特点可知,2根7厘米的小棒和2根8厘米的小棒可以组成1个平行四边形，平行四边形周的长度是平行四边形的周长，依此计算。此题考查的是平行四边形的周长的计算，熟练掌握平行四边形的特点是解答此题的关键.\n【详解】7×2+8×2=14+16=30(厘米)。",
+        "knowledge_points":  [
+                                 "平行四边形的概念及特点，平行四边形的周长"
+                             ],
+        "source":  "21-22四年级上·河南开封·期末"
+    },
+    {
+        "id":  "185",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个长为160厘米，宽为60厘米的长方形木框架，平放在地面上，它占地面积是( )平方厘米;把它拉成一个平行四边形，平行四边形高为40厘米，这时，平行四边形的周长是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "9600;440",
+        "solution":  "【分析】根据长方形面积=长×宽代入计算占地面积;把它拉成一个平行四边形，平行四边形的周长不变，和长方形周长一样，根据长方形周长=(长+宽)×2代入解答即可。动手做一做更容易理解，在拉的过程中四条边的长度不变，而高度变长.\n【详解】占地面积:160×60=9600(平方厘米);平行四边形周长:(160+60)×2=220×2=440(厘米)。",
+        "knowledge_points":  [
+                                 "长方形的面积，平行四边形的周长"
+                             ],
+        "source":  "21-22 四年级上·四川乐山·期末"
+    },
+    {
+        "id":  "186",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个平行四边形的周长是44cm,一条边长是14cm,邻边长是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "8厘米",
+        "solution":  "【分析】根据\u0027对边平行且相等\u0027可知，平行四边形的周长等于两条邻边的和乘2，据此解题即可。熟记：对边平行且相等，是解答此题的关键.\n【详解】44÷2-14=22-14=8(cm),所以，一个平行四边形的周长是44cm,一条边长是14cm,邻边长是8厘米。",
+        "knowledge_points":  [
+                                 "平行四边形的周长"
+                             ],
+        "source":  "21-22四年级上·江西吉安·期末"
+    },
+    {
+        "id":  "187",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "\u0027梯形\u0027是只有一组对边( )的\u0027四边形\u0027;\u0027长方形\u0027是有4个( )的\u0027平行四边形\u0027;\u0027正方形\u0027是4条边都( )的\u0027长方形\u0027。",
+        "options":  [
+
+                    ],
+        "answer":  "平行;直角;相等",
+        "solution":  "【分析】长方形，平行四边形，正方形，梯形都是属于四边形，正方形和长方形是特殊的平行四边形，正方形是特殊的长方形，再根据它们的特点填空即可。此题考查的是四边形的分类及其关系，熟练掌握它们的特点是解答此题的关键.\n【详解】只有一组对边平行的四边形叫做梯形，即\u0027梯形\u0027是只有一组对边平行的\u0027四边形\u0027;长方形的对边相等，并且四个角都是直角，即\u0027长方形\u0027是有4个直角的\u0027平行四边形\u0027;正方形的四条边都相等，并且四个角都是直角，即\u0027正方形\u0027是4条边都相等的\u0027长方形\u0027。",
+        "knowledge_points":  [
+                                 "长方形的概念及特点，正方形的概念及特点，四边形的分类及关系，梯形的概念及特点"
+                             ],
+        "source":  "21-22 四年级上·四川德阳·期末"
+    },
+    {
+        "id":  "188",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "两条直线相交组成了一个直角，其余三个角是( )°,这两条直线( )。",
+        "options":  [
+
+                    ],
+        "answer":  "90;互相垂直",
+        "solution":  "【分析】根据垂直的定义：两条直线相交成直角时，这两条直线互相垂直，其中一条直线是另一条直线的垂线。其中一个角是直角时，其余的三个角都是直角，即90°。此题考查了垂直的含义，注意对一些基础概念和性质的理解.\n【详解】两条直线相交组成了一个直角，其余三个角是90°,这两条直线互相垂直。",
+        "knowledge_points":  [
+                                 "垂直的特征"
+                             ],
+        "source":  "20-21四年级上·河北承德·期末"
+    },
+    {
+        "id":  "189",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "同一平面内，如果a//b,b//c,那么a与c互相( ),记作:( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平行;a//c",
+        "solution":  "【分析】从平行线的角度出发，根据平行线的性质及平行线的判定判断结论.\n【详解】据分析可知：同一平面内，如果a//b,b//c,那么a与c互相平行，记作:a//c。",
+        "knowledge_points":  [
+                                 "平行的特征及性质"
+                             ],
+        "source":  "21-22 四年级上·四川德阳·期末"
+    },
+    {
+        "id":  "190",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "用一条线把一个平行四边形任意分割成两个梯形，这两个梯形中高总是相等。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】平行四边形的两组对边是平行的，它的高有无数条且都是相等的，据此判断。解答本题的关键是要熟练掌握平行四边形的性质及特征.\n【详解】由于平行四边形有无数条高且都是相等的，所以把一个平行四边形任意分割成两个梯形，这两个梯形的高还是等于原来平行四边形的高，所以这两个梯形中高总是相等，原题干说法是正确的。故答案为:√",
+        "knowledge_points":  [
+                                 "梯形的概念及特点，平行四边形的高及画法，梯形的高及画法，平行四边形的切拼"
+                             ],
+        "source":  "16-17五年级·北京·课后作业"
+    },
+    {
+        "id":  "191",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "长方形有四组平行线。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】长方形的两组对边相互平行，两组对边分别相等，四个角都是直角;依此判断。熟练掌握长方形的特点是解答此题的关键.\n【详解】根据长方形的特点可知：长方形有两组平行线。故答案为: ×。",
+        "knowledge_points":  [
+                                 "长方形的概念及特点"
+                             ],
+        "source":  "2021 四年级上·全国·专题练习"
+    },
+    {
+        "id":  "192",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "只给一条5厘米长的线段，无法画出正方形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】正方形的两组对边分别平行，四条边都相等，相邻的两条边相互垂直，依此判断。此题考查的是画指定边长的正方形，熟练掌握正方形的特点是解答此题的关键.\n【详解】根据正方形的特点可知：只给一条5厘米长的线段，可以画出正方形。故答案为: ×。",
+        "knowledge_points":  [
+                                 "正方形的概念及特点，画指定长，宽(边长)的长方形，正方形"
+                             ],
+        "source":  "2021 四年级上·全国·专题练习"
+    },
+    {
+        "id":  "193",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "梯形不属于平行四边形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】只有一组对边平行的四边形叫做梯形;两组对边分别平行的四边形，叫做平行四边形;依此判断。熟练掌握梯形和平行四边形的特点是解答此题的关键.\n【详解】根据梯形和平行四边形的特点可知：梯形不属于平行四边形。故答案为:√",
+        "knowledge_points":  [
+                                 "平行四边形的概念及特点，梯形的概念及特点"
+                             ],
+        "source":  "2021 四年级上·全国·专题练习"
+    },
+    {
+        "id":  "194",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "有直角的四边形一定不是平行四边形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】两组对边分别平行的四边形，叫做平行四边形，依此判断。熟练掌握平行四边形和长方形的特点是解答此题的关键.\n【详解】长方形的两组对边分别平行，两组对边分别相等，并且有四个直角，但长方形是特殊的平行四边形。故答案为: ×",
+        "knowledge_points":  [
+                                 "长方形的概念及特点，平行四边形的概念及特点"
+                             ],
+        "source":  "2021 四年级上·全国·专题练习"
+    },
+    {
+        "id":  "195",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "正方形不是平行四边形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】依据平行四边形的定义：两组对边分别平行的四边形叫做平行四边形。正方形的两组对边分别平行，符合平行四边形的特征，因此正方形是特殊的平行四边形。据此判断原题说法错误.\n【详解】两组对边分别平行的四边形叫平行四边形，长方形和正方形是特殊的平行四边形，所以判断错误。",
+        "knowledge_points":  [
+                                 "平行四边形的概念及特点，正方形的概念及特点"
+                             ],
+        "source":  "21-22 四年级上·全国·期末"
+    },
+    {
+        "id":  "196",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "有一组对边平行的四边形叫做平行四边形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】本题考查平行四边形的定义。平行四边形是指两组对边分别平行的四边形;而只有一组对边平行的四边形叫做梯形。据此判断原题说法错误.\n【详解】根据平行四边形的概念可知，两组对边分别平行的四边形叫平行四边形，平行四边形的两组对边分别相等，所以判断错误。",
+        "knowledge_points":  [
+                                 "平行四边形的概念及特点"
+                             ],
+        "source":  "2014 三年级·全国·课后作业"
+    },
+    {
+        "id":  "197",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "两条平行线都是长4米。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平行线的定义：在同一平面内，不相交的两条直线叫做平行线。由此可知，两条平行线都是直线。把线段的两端无限延长，得到一条直线，直线没有端点，是无限长的。据此判断即可。熟练掌握平行线的定义，明确平行线是直线，无限长，这是解决本题的关键.\n【详解】根据分析可知，两条平行线是直线，是不能度量的。故答案为: ×",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征，平行的特征及性质"
+                             ],
+        "source":  "19-20四年级上·黑龙江牡丹江·期末"
+    },
+    {
+        "id":  "198",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "把一个长方形木质框架拉成一个平行四边形，这个平行四边形的周长和原长方形的周长一样长。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】把一个长方形的框架拉成一个平行四边形，四条边的长度没变，也就是它们的和没变，即周长没变，据此判断即可。本题的关键是理解变化前后框架长度没有发生变化，即图形的周长没有变.\n【详解】把一个长方形框架拉成一个平行四边形，这个平行四边形的周长和原长方形的周长一样。故题干说法正确。",
+        "knowledge_points":  [
+                                 "周长的认识和大小比较"
+                             ],
+        "source":  "19-20 四年级上·全国·单元测试"
+    },
+    {
+        "id":  "199",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "两个等腰梯形可以拼成一个平行四边形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】等腰梯形只有一组对边平行，两条腰相等。平行四边形的两组对边平行。两个等腰梯形不一定可以拼成一个平行四边形。画图举例解答。本题考查等腰梯形和平行四边形的性质，亲自动手拼一拼，即可得出结论.\n【详解】![](images/3e29b494889c04161e06632c9d500e6a.jpg)如图所示，两个等腰梯形不一定可以拼成一个平行四边形。故答案为: ×",
+        "knowledge_points":  [
+                                 "平行四边形的概念及特点，直角梯形和等腰梯形的概念及特点，平面图形的拼接"
+                             ],
+        "source":  "19-20 四年级上·河南三门峡·期末"
+    },
+    {
+        "id":  "200",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "有一块周长为30米的等腰梯形菜地，腰长8米，上底比下底短4米，这块菜地的上底和下底各是多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "上底是5米;下底是9米",
+        "solution":  "【分析】等腰梯形的特点是两条腰都相等，梯形的周长=上底的长度+下底的长度+2条腰的长度，依此先计算出上底和下底的长度之和，再根据和差问题的计算方法计算出上底和下底的长度即可。此题考查的是等腰梯形周长的计算，熟练掌握等腰梯形的特点以及和差问题的计算方法是解答此题的关键.\n【详解】30-8×2=30-16=14(米),下底-上底=4米，下底+上底=14米;下底:(14+4)÷2=18÷2=9(米),上底:14-9=5(米)。答：这块菜地的上底是5米，下底是9米。",
+        "knowledge_points":  [
+                                 "直角梯形和等腰梯形的概念及特点，梯形的周长"
+                             ],
+        "source":  "21-22 四年级上·全国·单元测试"
+    },
+    {
+        "id":  "201",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在同一平面内，直线a垂直于直线b,直线b垂直于直线c(三条直线均不重合),那么直线a与直线c的位置关系是互相( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平行",
+        "solution":  "【分析】![](images/c14f93e44808e5daba606a77dd2b30f5ec7c6114f66a253a18186208b77a20b0.jpg)根据垂直和平行的性质可知，同一平面内,a⊥b,b⊥c,则a∥c。即同一平面内，垂直于同一条直线的两条直线互相平行.\n【详解】在同一平面内，直线a垂直于直线b,直线b垂直于直线c(三条直线均不重合),那么直线a与直线c的位置关系是互相平行.\n[点睛]熟练掌握平行和垂直的定义是解决本题的关键，画图能帮助学生更好的理解。",
+        "knowledge_points":  [
+                                 "垂直的特征，平行的特征及性质"
+                             ],
+        "source":  "21-22四年级上·湖北黄冈·期末"
+    },
+    {
+        "id":  "202",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "正方形相对的两条边互相( ),相邻的两条边互相( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平行;垂直",
+        "solution":  "【分析】正方形的性质:(1)正方形的四条边相等;(2)正方形的四个角都是直角;(3)正方形的对边互相平行，相邻的两条边互相垂直，据此解答即可。此题考查正方形的特征，平时注意基础知识的积累.\n【详解】据分析可得：正方形相对的两条边互相平行，相邻的两条边互相垂直。",
+        "knowledge_points":  [
+                                 "长方形的概念及特点，垂直的特征"
+                             ],
+        "source":  "18-19四年级上·山东菏泽·期末"
+    },
+    {
+        "id":  "203",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "平行四边形和梯形都有( )条高，平行四边形有( )组对边分别平行，梯形有( )组对边平行。",
+        "options":  [
+
+                    ],
+        "answer":  "无数;两或2;一",
+        "solution":  "【分析】![](images/0efd28c4d7cac81566b133bf53955b1f.jpg)从梯形一条底边上的一点到它对边的垂直线段叫做梯形的高。从平行四边形一条边上的一点到对边引一条垂线，这点和垂足之间的线段叫做平行四边形的高。则平行四边形和梯形均有无数条高。平行四边形有两组对边平行，而梯形只有一组对边平行。据此解答。本题考查平行四边形和梯形的性质，平行四边形有两组对边平行，梯形有一组对边平行，它们都有无数条高。\n【详解】平行四边形和梯形都有无数条高，平行四边形有两组对边分别平行，梯形有一组对边平行。",
+        "knowledge_points":  [
+                                 "梯形的概念及特点，平行四边形的高及画法，梯形的高及画法，平行四边形的概念及特点"
+                             ],
+        "source":  "21-22 四年级上·湖南怀化·期末"
+    },
+    {
+        "id":  "204",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在公路上有三条小路通往池塘，它们的长度分别为248米,312米,283米，其中有一条小路与公路是垂直的，那么这条小路的长度是( )米。",
+        "options":  [
+
+                    ],
+        "answer":  "248",
+        "solution":  "【分析】根据点到直线的距离垂线段最短，由此进行解答即可。本题主要考查最短路线问题，解题关键是了解点到直线的距离垂线段最短.\n【详解】因为这条小路与公路是垂直的，垂线段最短。312\u003e283\u003e248，所以在公路上有三条小路通往池塘，它们的长度分别为248米,312米,283米，其中有一条小路与公路是垂直的，那么这条小路的长度是248米。",
+        "knowledge_points":  [
+                                 "画垂线，垂直的特征"
+                             ],
+        "source":  "21-22四年级上·湖南邵阳·期末"
+    },
+    {
+        "id":  "205",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "黑板的两条邻边互相( ),两组对边分别( )。",
+        "options":  [
+
+                    ],
+        "answer":  "垂直;平行且相等",
+        "solution":  "【分析】根据题意分析可知，黑板的四个角都是直角，为一个长方形，长方形每相邻的两条边互相垂直，并且两组对边分别平行且相等，解答此题即可.\n【详解】据分析可知：黑板的两条邻边互相垂直，两组对边分别平行且相等.\n[点睛]熟练掌握长方形的特征，是解答此题的关键。",
+        "knowledge_points":  [
+                                 "平行的特征及性质，垂直的特征，长方形的概念及特点"
+                             ],
+        "source":  "19-20 四年级上·福建福州·期末"
+    },
+    {
+        "id":  "206",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "把一个长6厘米，宽5厘米的长方形拉成一个高是4厘米的平行四边形，这个平行四边形的周长是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "22",
+        "solution":  "【分析】形状改变，但周长不变，用长方形长加宽的和乘2即可解答。长方形拉成平行四边形，形状发生变化，但周长不变.\n【详解】(6+5)×2=11×2=22(厘米)",
+        "knowledge_points":  [
+                                 "平行四边形的周长，长方形的周长"
+                             ],
+        "source":  "21-22 四年级上·福建厦门·期末"
+    },
+    {
+        "id":  "207",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "把一张长方形纸向同方向连续对折两次，所折出的折痕互相( );把一张正方形纸按两个相反方向分别对折一次，折出的两条折痕互相( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平行;垂直",
+        "solution":  "【分析】![](images/43e3a0efd14f96bef57155b4e1174234.jpg)如图所示，如果两次都朝一个方向折叠，折痕互相平行;如果两次都朝两个方向折叠即先上下折，然后再左右折，折痕互相垂直，据此解答。本题最好的解决办法是找一张纸亲自动手操作下，问题就能迎刃而解.\n【详解】把一张长方形纸向同方向连续对折两次，所折出的折痕互相平行;把一张正方形纸按两个相反方向分别对折一次，折出的两条折痕互相垂直。",
+        "knowledge_points":  [
+                                 "垂直的特征，平行的特征及性质"
+                             ],
+        "source":  "21-22 四年级上·吉林白山·期末"
+    },
+    {
+        "id":  "208",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "过直线外一点到已知直线可以画( )条垂线;所有端点分别在两条平行线上的垂直线段的长度都( )。",
+        "options":  [
+
+                    ],
+        "answer":  "1;相等",
+        "solution":  "【分析】根据垂线的特征可知，过一点有且只有一条直线与已知直线垂直。两直线互相平行时，从一条直线上任意点向另一条直线作垂线，所得的平行线间的垂直线段的长度，叫做平行线间的距离。平行线之间的距离处处相等。本题主要考查学生对垂直和平行的特征的掌握.\n【详解】根据分析可知，过直线外一点到已知直线可以画一条垂线;所有端点分别在两条平行线上的垂直线段的长度都相等。",
+        "knowledge_points":  [
+                                 "垂直的特征，平行的特征及性质"
+                             ],
+        "source":  "21-22 四年级上·吉林白山·期末"
+    },
+    {
+        "id":  "209",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在同一平面内两条直线分别和第三条直线垂直，那么这两条直线( )。",
+        "options":  [
+
+                    ],
+        "answer":  "互相平行",
+        "solution":  "【分析】根据垂线段的性质，平行线的性质，平行线的判定定理判断即可。本题考查的是线段的性质，平行线的性质，平行线的判定定理.\n【详解】据分析可知：在同一平面内两条直线都与第三条直线垂直，那么这两条直线互相平行。",
+        "knowledge_points":  [
+                                 "平行的特征及性质"
+                             ],
+        "source":  "21-22 四年级上·吉林四平·期末"
+    },
+    {
+        "id":  "210",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "与已知直线平行的直线有( )条;与已知直线垂直的直线有( )条;一个正方形，相邻的两边互相( )。",
+        "options":  [
+
+                    ],
+        "answer":  "无数;无数;垂直",
+        "solution":  "【分析】在同一平面内，一条直线与另一条直线相交并有一个角是直角，那么这两条直线互相垂直;在平面上的两条直线与一平面之间没有任何公共点时，称它们平行，据此解答。熟练掌握垂直以及平行的特征并灵活应用是解答本题的关键.\n【详解】与已知直线平行的直线有(无数)条;与已知直线垂直的直线有(无数)条;一个正方形，相邻的两边互相(垂直)。",
+        "knowledge_points":  [
+                                 "垂直的特征，平行的特征及性质"
+                             ],
+        "source":  "18-19 四年级上·河南商丘·期中"
+    },
+    {
+        "id":  "211",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "延长平行四边形的一组对边，它们永远不会( )。",
+        "options":  [
+
+                    ],
+        "answer":  "相交",
+        "solution":  "【分析】根据平行四边形的特征：对边分别平行且长度相等;所以延长平行四边形的一组对边仍然平行，据此解答即可。熟练掌握平行四边形的特征，是解答本题的关键.\n【详解】根据平行四边形的特征可知：平行四边形的对边平行;所以延长平行四边形的一组对边，它们永远不会相交。",
+        "knowledge_points":  [
+                                 "平行四边形的概念及特点，平行的特征及性质"
+                             ],
+        "source":  "21-22四年级上·全国·期末"
+    },
+    {
+        "id":  "212",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "亮亮玩摆小棒的游戏，他在桌面上摆了三根小棒，其中第一根垂直于第二根，第一根也垂直于第三根，那么第二根和第三根互相( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平行",
+        "solution":  "【分析】根据平行的性质：同一平面内两条直线同时垂直于一条直线，那么，这两条直线相互平行;据此解答。此题考查了垂直和平行的性质，应注意积累和理解.\n【详解】据分析可得：亮亮玩摆小棒的游戏，他在桌面上摆了三根小棒，其中第一根垂直于第二根，第一根也垂直于第三根，那么第二根和第三根互相平行。",
+        "knowledge_points":  [
+                                 "平行的特征及性质"
+                             ],
+        "source":  "21-22四年级上·湖北武汉·期末"
+    },
+    {
+        "id":  "213",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "长方形的邻边互相( );长方形有( )组对边互相平行。",
+        "options":  [
+
+                    ],
+        "answer":  "垂直;两",
+        "solution":  "【详解】长方形有4个直角，即长方形的邻边互相垂直，长方形有两(2)组对边互相平行，如图所示:![](images/be0618fc35e36634ed3aa8840cfdda57b4d98c70fd31b9dd8a8594d4cff3db46.jpg)",
+        "knowledge_points":  [
+                                 "长方形的概念及特点，垂直的特征，平行的特征及性质"
+                             ],
+        "source":  "21-22 四年级上·四川广元·期末"
+    },
+    {
+        "id":  "214",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把一张长方形纸沿着一组对边折两次，则三条折痕互相( );把一张正方形纸沿着对角线对折两次，则两条折痕互相( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平行;垂直",
+        "solution":  "【分析】同一平面内永不相交的两条直线叫做平行线，组成平行线的两条直线互相平行;在同一平面内两条直线相交成直角，这两条直线互相垂直;依此画图并填空。此题考查的是平行与垂直的特点，熟练掌握图形的折叠问题是解答此题的关键.\n【详解】![](images/f5f83cc46984ea8fbb39eb7681b476df.jpg)根据画图可知，把一张长方形纸沿着一组对边折两次，则三条折痕互相平行;把一张正方形纸沿着对角线对折两次，则两条折痕互相垂直。",
+        "knowledge_points":  [
+                                 "平行的特征及性质，垂直的特征，图形的折叠问题"
+                             ],
+        "source":  "2021四年级上·江西抚州·专题练习"
+    },
+    {
+        "id":  "215",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "从直线外一点到这条直线所画的( )最短。平行线间的( )处处相等。",
+        "options":  [
+
+                    ],
+        "answer":  "垂线段;距离",
+        "solution":  "【详解】根据点到直线的距离的含义：从直线外的一点向这条直线所画的垂直线段的长度，叫做这点到直线的距离;从直线外一点到这条直线所画的垂直线段最短;因为平行线之间的距离就是两条平行线的垂线段的长度，所以平行线间的距离处处相等。",
+        "knowledge_points":  [
+                                 "垂直的特征，平行的特征及性质"
+                             ],
+        "source":  "2022 四年级上·全国·专题练习"
+    },
+    {
+        "id":  "216",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个平行四边形的周长是80厘米，其中一条边是30厘米，那么它邻边的长度是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "10厘米",
+        "solution":  "【分析】平行四边形的周长是相邻两条边的长度和的2倍，则邻边的长度是周长/2-一条边。熟练掌握平行四边形的周长公式是解决本题的关键.\n【详解】80÷2-30=40-30=10(厘米),那么它邻边的长度是10厘米。",
+        "knowledge_points":  [
+                                 "平行四边形的周长"
+                             ],
+        "source":  "21-22 四年级上·山东临沂·期末"
+    },
+    {
+        "id":  "217",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "过直线外的一点可以画( )条平行线。",
+        "options":  [
+
+                    ],
+        "answer":  "1",
+        "solution":  "【分析】根据平行线的性质，过直线外一点作已知直线的平行线，能作且只能作一条。本题考查了平行的性质，是基础题，熟记性质是解题的关键.\n【详解】过直线外的一点，画已知直线的平行线，这样的平行线可以画1条。",
+        "knowledge_points":  [
+                                 "画平行线"
+                             ],
+        "source":  "21-22 四年级上·广西玉林·期末"
+    },
+    {
+        "id":  "218",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个等腰梯形的周长是30厘米。已知上底与下底的和是20厘米，它的每条腰都是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "5",
+        "solution":  "【分析】根据梯形周长的意义，梯形的周长是指围成这个图形的4条边的长度总和。用梯形的周长减去上下底的和，再除以2即可求出一条腰的长度。此题考查的目的是理解掌握等腰梯形的特征以及梯形周长公式的灵活运用.\n【详解】(30-20)÷2=10÷2=5(厘米)",
+        "knowledge_points":  [
+                                 "直角梯形和等腰梯形的概念及特点，梯形的周长"
+                             ],
+        "source":  "18-19 四年级上·浙江杭州·期末"
+    },
+    {
+        "id":  "219",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个平行四边形的框架，把它拉成长方形，长方形的面积( )平行四边形的面积。(填\u0027大于\u0027,\u0027小于\u0027或\u0027等于\u0027)",
+        "options":  [
+
+                    ],
+        "answer":  "大于",
+        "solution":  "【分析】平行四边形活动框架拉成长方形之后，原来平行四边形的高比现在的长方形的高要小，但是对应的底的长度不变，又因为长方形是特殊的平行四边形，根据面积计算公式，平行四边形的面积=底×高，所以平行四边形的面积比长方形的面积要小，所以一个平行四边形活动框架拉成长方形，长方形的面积大于平行四边形。此题考查的是平行四边形的特性的应用，解决本题的关键是熟悉前后两个图形的主要变化：边长不变，把一个平行四边形活动框架拉成长方形后，高变大.\n【详解】平行四边形活动框架拉成长方形之后，每条边的长度不变，所以周长不变;平行四边形活动框架拉成长方形之后，原来平行四边形的高比现在的长方形的高要小，但对应的底长度不变，又因为长方形是特殊的平行四边形，根据面积计算公式，平行四边形的面积=底×高，所以长方形的面积大于平行四边形。",
+        "knowledge_points":  [
+                                 "平行四边形的周长，长方形的面积，平行四边形的面积"
+                             ],
+        "source":  "18-19 五年级上·浙江杭州·期末"
+    },
+    {
+        "id":  "220",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "把一张长方形纸对折，再对折，两条折痕的关系是( )或( )。",
+        "options":  [
+
+                    ],
+        "answer":  "互相垂直;互相平行",
+        "solution":  "【分析】根据前后两次对折方向是否一致来分别判断两条折痕的关系。此题的关键在于知道两次对折方向上不同就会得到不同的结果.\n【详解】如果两次对折都是朝同一方向对折，则得到的两条折痕是互相平行的;如果两次对折，一次是左右对折，另一次是上下对折，则得到的两条折痕是互相垂直的。所以根据对折的不同情况，两条折痕的关系是互相垂直或互相平行。",
+        "knowledge_points":  [
+                                 "垂直的特征，平行的特征及性质"
+                             ],
+        "source":  "21-22 四年级上·湖北省直辖县级单位·期末"
+    },
+    {
+        "id":  "221",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "平行四边形相邻两条边的和是23厘米，那么这个平行四边形的周长是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "46",
+        "solution":  "【分析】平行四边形的周长就是四条边长度的和，平行四边形的对边相等，据此解答。本题主要考查了学生对平行四边形周长计算方法的掌握情况.\n【详解】根据题意可知:23×2=46(厘米),所以平行四边形相邻两条边的和是23厘米，那么这个平行四边形的周长是46厘米。",
+        "knowledge_points":  [
+                                 "平行四边形的周长"
+                             ],
+        "source":  "19-20 四年级上·河南郑州·期末"
+    },
+    {
+        "id":  "222",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "等腰梯形的周长是67厘米，已知腰长13厘米，上底长12厘米，下底长( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "29",
+        "solution":  "【分析】梯形的周长，是由上下底边及两腰组成。所以用梯形周长减去两腰及上底边的长度，即为下底边的长度，由此解答即可。此题主要考查了对梯形周长的了解掌握.\n【详解】67-(13+13+12)=67-38=29(厘米),所以梯形下底边为29厘米。",
+        "knowledge_points":  [
+                                 "梯形的周长"
+                             ],
+        "source":  "21-22 四年级上·湖北黄冈·期末"
+    },
+    {
+        "id":  "223",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "与两条平行线互相垂直的线段的长度都( )。",
+        "options":  [
+
+                    ],
+        "answer":  "相等",
+        "solution":  "【分析】根据\u0027在两条平行线之间的线段中，垂直两条平行线的线段最短，这条线段的长叫做平行线之间的距离\u0027可知：在两条平行线之间再画几条和平行线垂直的线段，这些线段的长度都相等;据此解答即可。此题考查了垂直和平行的特征以及性质，注意基础知识的灵活运用.\n【详解】与两条平行线互相垂直的线段的长度都相等。",
+        "knowledge_points":  [
+                                 "垂直的特征，平行的特征及性质"
+                             ],
+        "source":  "21-22 四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "224",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "我们学校的大门可以伸缩，它是利用了四边形( )原理制成的。",
+        "options":  [
+
+                    ],
+        "answer":  "不稳定性",
+        "solution":  "【分析】根据平行四边形的性质：平行四边形具有容易变形的特性，据此即可解答。本题考查了平行四边形的性质：具有不稳定性，容易变形.\n【详解】我们学校的大门可以伸缩，它是利用了四边形不稳定性原理制成的。",
+        "knowledge_points":  [
+                                 "平行四边形的不稳定性及应用"
+                             ],
+        "source":  "22-23 三年级上·全国·假期作业"
+    },
+    {
+        "id":  "225",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "四边形有( )条边,( )个角。写出三个你知道名字的四边形分别是( ),( ),( )。对边相等的有( ),( )。",
+        "options":  [
+
+                    ],
+        "answer":  "4;4;梯形;正方形;平行四边形;正方形;平行四边形",
+        "solution":  "【分析】根据四边形的含义：由四条线段首尾顺次相连围成的图形叫四边形;它有4条边,4个角，进而解答即可。熟练掌握四边形的定义及分类，是解答此题的关键.\n【详解】四边形有4条边,4个角。写出三个你知道名字的四边形分别是：梯形，正方形，平行四边形;对边相等的有正方形，平行四边形。",
+        "knowledge_points":  [
+                                 "四边形的分类及关系，四边形，五边形及多边形的初步认识"
+                             ],
+        "source":  "22-23 三年级上·全国·假期作业"
+    },
+    {
+        "id":  "226",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "两条平行线之间的距离( ),正方形相邻的两条边互相( )。",
+        "options":  [
+
+                    ],
+        "answer":  "相等;垂直",
+        "solution":  "【分析】(1)因为平行线之间的距离都是两条平行线的垂线段，所以两条平行线之间的距离处处相等;(2)根据正方形的特征,4个角都是直角,4条边的长度都相等，因此正方形的邻边互相垂直。本题考查了平行线之间的距离和正方形的特性，应灵活掌握和运用.\n【详解】两条平行线之间的距离相等，正方形相邻的两条边互相垂直。",
+        "knowledge_points":  [
+                                 "正方形的概念及特点，垂直的特征，平行的特征及性质"
+                             ],
+        "source":  "21-22四年级上·甘肃天水·期末"
+    },
+    {
+        "id":  "227",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个平行四边形的三条边分别是4厘米,5厘米,4厘米，第四条边是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "5厘米",
+        "solution":  "【分析】两组对边分别平行且相等的四边形，叫做平行四边形;据此可知，一个平行四边形的三条边分别是4厘米,5厘米,4厘米，第四条边是5厘米。熟练掌握平行四边形的意义和特征，是解答此题的关键.\n【详解】根据平行四边形的定义可知，一个平行四边形的三条边分别是4厘米,5厘米,4厘米，第四条边是5厘米。",
+        "knowledge_points":  [
+                                 "平行四边形的概念及特点"
+                             ],
+        "source":  "21-22 四年级上·黑龙江七台河·期末"
+    },
+    {
+        "id":  "228",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "梯形上，下底之间的距离是梯形的( )。",
+        "options":  [
+
+                    ],
+        "answer":  "高",
+        "solution":  "【分析】只有一组对边平行的四边形叫梯形。梯形两底间的距离叫做梯形的高，梯形有无数条高，通常过上底的个顶点作下底的垂线，用三角板的直角可以画出梯形的条高。熟悉梯形高的定义是解答此题的关键.\n【详解】梯形上，下底之间的距离是梯形的(高)。",
+        "knowledge_points":  [
+                                 "梯形的高及画法"
+                             ],
+        "source":  "21-22 四年级上·广东汕尾·期末"
+    },
+    {
+        "id":  "229",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "如果一个平行四边形的四个角都变成直角，就变成了( )形或( )形。",
+        "options":  [
+
+                    ],
+        "answer":  "长方形;正方形",
+        "solution":  "【分析】两组对边分别平行且相等的四边形，叫做平行四边形;平行四边形具有不稳定性，所以，如果一个平行四边形的四个角都变成直角，就变成了长方形或正方形。两组对边分别平行，四个角都是直角的四边形叫做长方形;四条边相等，四个角都是直角的四边形叫做正方形.\n【详解】根据分析可知，如果一个平行四边形的四个角都变成直角，就变成了长方形或正方形。",
+        "knowledge_points":  [
+                                 "平行四边形的概念及特点，长方形的概念及特点，正方形的概念及特点，平行四边形的不稳定性及应用"
+                             ],
+        "source":  "21-22 四年级上·广东汕尾·期末"
+    },
+    {
+        "id":  "230",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "折叠椅利用了平行四边形的( )性，在不使用时收起来可以节省空间。",
+        "options":  [
+
+                    ],
+        "answer":  "不稳定",
+        "solution":  "【分析】平行四边形容易变形，具有不稳定性，依此填空。此题考查的是平行四边形的不稳定性及其应用，应熟练掌握.\n【详解】根据分析可知，折叠椅利用了平行四边形的不稳定性，在不使用时收起来可以节省空间。",
+        "knowledge_points":  [
+                                 "平行四边形的不稳定性及应用"
+                             ],
+        "source":  "21-22 四年级上·广东广州·期末"
+    },
+    {
+        "id":  "231",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "学校伸缩门是运用了( )形的易变形特点。",
+        "options":  [
+
+                    ],
+        "answer":  "平行四边形",
+        "solution":  "【分析】三角形具有稳定性，不容易变形;平行四边形与三角形不同，容易变形，也就是具有不稳定性;这种不稳定性在实践中有广泛的应用，学校伸缩门是利用平行四边形具有不稳定性的特点制成的，进行解答即可。本题是考查平行四边形的特征及性质.\n【详解】由分析可知：学校伸缩门通常是运用了平行四边形的不稳定性或易变形的特性。",
+        "knowledge_points":  [
+                                 "平行四边形的不稳定性及应用"
+                             ],
+        "source":  "21-22 四年级上·河南洛阳·期末"
+    },
+    {
+        "id":  "232",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "黑板上下两条边是互相( )的，三角尺中的两条直角边是互相( )的。梯形中最多有( )个直角。",
+        "options":  [
+
+                    ],
+        "answer":  "平行;垂直;2",
+        "solution":  "【分析】因为黑板是一个长方形，所以根据长方形的特征：对边平行且相等,4个角都是直角，可知，长方形相邻的两条边互相垂直，相对的两边互相平行;根据三角尺的特征可知：三角尺有两条边互相垂直。根据梯形的特征及四边形的内角和是360°,梯形只有一组对边平行，所以梯形中最多有两个直角;据此解答。解答此题的主要依据是：长方形的邻边互相垂直，对边互相平行。梯形的特征以及四边形的内角和是360°.\n【详解】黑板上下两条边是互相平行的，三角尺中的两条直角边是互相垂直的。梯形中最多有2个直角。",
+        "knowledge_points":  [
+                                 "垂直的特征，平行的特征及性质，梯形的概念及特点"
+                             ],
+        "source":  "21-22 四年级上·河南洛阳·期末"
+    },
+    {
+        "id":  "233",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "从平行四边形的一条边上的一点向对边可以引( )条垂线，这条垂线就是平行四边形的( )。",
+        "options":  [
+
+                    ],
+        "answer":  "1;高",
+        "solution":  "【分析】在平行四边形中，从一条边上的任意一点向对边作垂线，这点与垂足间的距离叫做以这条边为底的平行四边形的高，即平行四边形一条边上的一点到它的对边的垂直线段就是平行四边形的高。此题主要是考查平行四边形高的意义。平行四边形的高是指确定的底到与它相对边的距离，也就是垂直线段.\n【详解】从平行四边形的一条边上的一点向对边可以引1条垂线，这条垂线就是平行四边形的高。",
+        "knowledge_points":  [
+                                 "平行四边形的高及画法"
+                             ],
+        "source":  "18-19四年级上·吉林延边·期末"
+    },
+    {
+        "id":  "234",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "用手捏住长方形框架的两个对角向相反方向拉，长方形就变成了( ),两个图形的周长( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平行四边形;相等",
+        "solution":  "【分析】平行四边形易变形，具有不稳定性，但图形的周长不变。本题主要考查学生对平行四边形不稳定性知识的掌握.\n【详解】用手捏住长方形框架的两个对角向相反方向拉，长方形就变成了平行四边形，两个图形的周长相等。",
+        "knowledge_points":  [
+                                 "平行四边形的周长，平行四边形的不稳定性及应用"
+                             ],
+        "source":  "21-22四年级上·黑龙江佳木斯·期末"
+    },
+    {
+        "id":  "235",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小红在用计算器计算一道题时，把被除数的前两位数按反了，结果除以25得到的商是650，那么正确的商应该是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "2450",
+        "solution":  "【分析】用除数25乘商650求出错误的被除数，再把错误的被除数的前两位交换位置求得正确的被除数，然后再除以25即得正确的商。熟练掌握整数除法各部分间的关系是解答本题的关键.\n【详解】650×25=16250，正确的被除数为61250;61250÷25=2450。",
+        "knowledge_points":  [
+                                 "乘，除法的意义和各部分间的关系"
+                             ],
+        "source":  "20-21四年级上·湖南怀化·单元测试"
+    },
+    {
+        "id":  "236",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个数除以62，把除数看成60来试商,\u0027试商\u0027可能( );一个数除以37，把除数看成40来试商,\u0027试商\u0027可能( )。(填\u0027偏大\u0027或\u0027偏小\u0027)",
+        "options":  [
+
+                    ],
+        "answer":  "偏大;偏小",
+        "solution":  "【分析】除法计算中，把除数看大了，商就容易偏小，调商时要调大一点，把除数看小了，商就容易偏大，调商时要调小一点，据此解答。熟练掌握整数除法的试商方法是解答本题的关键.\n【详解】一个数除以62，把除数看成60来试商,\u0027试商\u0027可能偏大;一个数除以37，把除数看成40来试商,\u0027试商\u0027可能偏小。",
+        "knowledge_points":  [
+                                 "多位数除以两位数的试商"
+                             ],
+        "source":  "20-21 四年级上·湖南怀化·单元测试"
+    },
+    {
+        "id":  "237",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "将平行四边形拉成长方形，它的周长( ),面积( )。",
+        "options":  [
+
+                    ],
+        "answer":  "不变;变大",
+        "solution":  "【分析】根据将平行四边形拉成长方形后，由于边的长度没有变化，所以周长不变;由于长方形的长等于平行四边形的底，长方形的宽大于原来平行四边形的高，所以面积变大，据此解答即可。本题主要考查了对平行四边形和长方形的周长和面积的理解和理解运用情况.\n【详解】据分析可得：将一个平行四边形拉成长方形，它的周长不变，面积变大。",
+        "knowledge_points":  [
+                                 "平行四边形的周长，周长的认识和大小比较，平行四边形的面积"
+                             ],
+        "source":  "20-21 四年级上·湖南怀化·单元测试"
+    },
+    {
+        "id":  "238",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "A/35=18。B,B最大是( ),这时A是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "34;664",
+        "solution":  "【分析】余数小于除数,B小于35,B最大是34，再根据\u0027被除数=商×除数+余数\u0027求出A。熟练掌握除法各部分间的关系是解答本题的关键.\n【详解】A/35=18。B,B小于35,B最大是34;当B=34时,A=18×35+34=630+34=664。",
+        "knowledge_points":  [
+                                 "乘，除法的意义和各部分间的关系，余数与除数的关系"
+                             ],
+        "source":  "20-21 四年级上·湖南怀化·单元测试"
+    },
+    {
+        "id":  "239",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "要使367÷3□的商是两位数,□里最大可填( );□67÷44要使商是一位数,□里的数最大可填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "6;3",
+        "solution":  "【分析】三位数除以两位数，如果被除数的前两位大于等于除数，商是两位数，否则商是一位数。熟练掌握整数除法的试商方法是解答本题的关键.\n【详解】要使367÷3□的商是两位数，则36\u003e=3□,□里最大可填6;□67÷44要使商是一位数，则□6\u003c44,□里的数最大可填3。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "19-20四年级上·全国·课后作业"
+    },
+    {
+        "id":  "240",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在384÷96中，如果除数除以8，要使商不变，被除数应该( )。",
+        "options":  [
+
+                    ],
+        "answer":  "除以8",
+        "solution":  "【分析】商的变化规律：被除数乘(或除)几，商就乘(或除以)几;除数乘(或除以)几，商就除以(或乘)几;被除数和除数同时乘(或除以)几，商不变。熟练掌握商的变化规律是解答本题的关键.\n【详解】在384÷96中，如果除数除以8，要使商不变，被除数应该除以8。",
+        "knowledge_points":  [
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "20-21 四年级上·湖南怀化·单元测试"
+    },
+    {
+        "id":  "241",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "平行四边形的( )组对边分别平行。梯形里一组互相平行的对边分别叫做梯形的( )和( )。",
+        "options":  [
+
+                    ],
+        "answer":  "两;上底;下底",
+        "solution":  "【详解】平行四边形的两组对边分别平行且相等，平行四边形易变形，具有不稳定性;只有一组对边平行的四边形叫梯形，平行的这组边叫做梯形的上底和下底。",
+        "knowledge_points":  [
+                                 "梯形的概念及特点，平行四边形的概念及特点"
+                             ],
+        "source":  "20-21 四年级上·湖南怀化·单元测试"
+    },
+    {
+        "id":  "242",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "长方形拉成平行四边形后( )不变。",
+        "options":  [
+
+                    ],
+        "answer":  "周长",
+        "solution":  "【分析】长方形的周长=(长+宽)×2;平行四边形的周长=相邻两边长度之和*2;根据长方形和平行四边形的特点填空即可。熟练掌握长方形，平行四边形的周长的计算是解答此题的关键.\n【详解】根据长方形和平行四边形的特点可知：长方形拉成平行四边形后周长不变。",
+        "knowledge_points":  [
+                                 "平行四边形的周长，长方形的周长，长方形的概念及特点，平行四边形的概念及特点"
+                             ],
+        "source":  "21-22 四年级上·山东菏泽·阶段练习"
+    },
+    {
+        "id":  "243",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个平行四边形的周长是48厘米，其中一条边的长是16厘米，与它相邻的一条边长是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "8",
+        "solution":  "【分析】封闭图形一周的长度就是周长，两组对边分别平行的四边形是平行四边形，据此解答。熟知周长的定义以及平行四边形的定义是解答本题的关键.\n【详解】48-16×2=48-32=16(厘米),16÷2=8(厘米)。一个平行四边形的周长是48厘米，其中一条边的长是16厘米，与它相邻的一条边长是8厘米。",
+        "knowledge_points":  [
+                                 "平行四边形的周长"
+                             ],
+        "source":  "21-22 四年级上·黑龙江佳木斯·期末"
+    },
+    {
+        "id":  "244",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个四边形，一组对边平行但长度不相等，另一组对边长度相等但不平行，这个四边形是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "等腰梯形",
+        "solution":  "【分析】根据梯形的特征，只有一组对边平行的四边形叫做梯形，如果在图形中一组对边平行但不相等，另一组对边相等但不平行，那么这个梯形就是等腰梯形，据此填空即可。此题考查了梯形的应用，关键是明确梯形的特点即可.\n【详解】一个四边形，一组对边平行但长度不相等，另一组对边长度相等但不平行，这个四边形是等腰梯形。",
+        "knowledge_points":  [
+                                 "梯形的概念及特点"
+                             ],
+        "source":  "20-21 四年级上·河南安阳·期末"
+    },
+    {
+        "id":  "245",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在同一平面内如果两条直线都与一条直线平行，则这两条直线互相( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平行",
+        "solution":  "【分析】同一平面内永不相交的两条直线叫做平行线，组成平行线的两条直线互相平行，依此画图并填空。熟练掌握平行的特点是解答此题的关键.\n【详解】如图所示：在同一平面内如果两条直线都与一条直线平行，则这两条直线互相平行。",
+        "knowledge_points":  [
+                                 "平行的特征及性质"
+                             ],
+        "source":  "19-20四年级上·全国·单元测试"
+    },
+    {
+        "id":  "246",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "数学书封面的左右两条边互相( ),上方的边与左方的边互相( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平行;垂直",
+        "solution":  "【分析】在同一平面内，不相交的两条直线叫做平行线;两条直线相交成直角时，这两条直线互相垂直，据此解答。熟知平行与垂直的定义是解答本题的关键.\n【详解】数学书封面的左右两条边互相平行，上方的边与左方的边互相垂直。",
+        "knowledge_points":  [
+                                 "垂直的特征，平行的特征及性质"
+                             ],
+        "source":  "20-21 四年级上·广东韶关·期末"
+    },
+    {
+        "id":  "247",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个平行四边形的周长34厘米，其中一条边的长是10厘米，与它相邻的一条边的长是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "7",
+        "solution":  "【分析】平行四边形的周长=两条相邻的边的边长之和*2，因此用周长除以2后，再减去10厘米即可，依此计算。熟练掌握平行四边形的周长的计算是解答此题的关键.\n【详解】34÷2=17(厘米),17-10=7(厘米)。",
+        "knowledge_points":  [
+                                 "平行四边形的周长"
+                             ],
+        "source":  "18-19四年级上·四川绵阳·期末"
+    },
+    {
+        "id":  "248",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "平行四边形具有( )的特性，拉伸( )变了,( )没变。",
+        "options":  [
+
+                    ],
+        "answer":  "不稳定;形状;周长",
+        "solution":  "【分析】平行四边形具有不稳定性，易变形，但周长不变，据此即可解答。本题主要考查学生对平行四边形特性的掌握.\n【详解】平行四边形具有不稳定的特性，拉伸形状变了，周长没变。",
+        "knowledge_points":  [
+                                 "平行四边形的概念及特点，平行四边形的周长"
+                             ],
+        "source":  "21-22 四年级上·全国·期末"
+    },
+    {
+        "id":  "249",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "壮壮有一根长1米9分米的铁丝，如果要用它围成相邻两条边分别是5厘米,4厘米的平行四边形，最多可以围( )个，还剩( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "10;10",
+        "solution":  "【分析】首先根据平行四边形的周长=(长+宽)×2，求出围成一个平行四边形需要铁丝的长度，然后根据\u0027包含\u0027除法的意义，用除法解答，注意单位的统一.\n【详解】1米9分米=190厘米,(5+4)×2=9×2=18(厘米),190÷18=10(个)。10(厘米),所以最多可以围10个，还剩下10厘米.\n[点睛]此题主要考查平行四边形周长公式的灵活运用，以及\u0027包含\u0027除法的意义及应用，关键是熟记公式。",
+        "knowledge_points":  [
+                                 "平行四边形的周长"
+                             ],
+        "source":  "21-22 四年级上·湖北黄冈·阶段练习"
+    },
+    {
+        "id":  "250",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个等腰梯形的周长是36米，腰长是8米，下底是上底的3倍，上底是( )米，下底是( )米。",
+        "options":  [
+
+                    ],
+        "answer":  "5;15",
+        "solution":  "【分析】根据下底长是上底长的3倍，可设梯形的上底长为x米，则下底长为3x米，梯形的周长是梯形四条边的长度和，可列方程:x+3x+8+8=36，解方程即可得出答案。解决本题的关键是明确梯形的周长=两腰长度+上底+下底，另外还有理解等腰梯形两腰相等的性质.\n【详解】解：设梯形的上底长为x米，则下底长为3x米，根据题意得:x+3x+8+8=36,4x+16=36,4x=20,x=5。下底长为:5×3=15(米)。梯形的上底长为5米，下底长为15米。",
+        "knowledge_points":  [
+                                 "梯形的周长"
+                             ],
+        "source":  "21-22 四年级上·湖北黄冈·阶段练习"
+    },
+    {
+        "id":  "251",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "平行四边形有( )条高，这些高可以分为( )组。",
+        "options":  [
+
+                    ],
+        "answer":  "无数;两",
+        "solution":  "【分析】从平行四边形的一条边上的任意一点都可以向对边作垂直线段，即平行四边形有无数条高。平行四边形有4条底，这四条底可以分为两组，则有两组高。本题考查平行四边形的高的定义。从平行四边形条边上的一点到对边引一条垂线，这点和垂足之间的线段叫做平行四边形的高.\n【详解】![](images/b831d3cefc452a0d5baa2ee7ed96e1e9c0e43a8e3c47ac8aa29c8d041e09c15a.jpg)如图所示，平行四边形有无数条高，这些高可以分为两组。",
+        "knowledge_points":  [
+                                 "平行四边形的高及画法"
+                             ],
+        "source":  "21-22 四年级上·全国·期末"
+    },
+    {
+        "id":  "252",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "从直线外一点到这条直线所画的垂直线段最短，它的长度叫做这点到直线的( )。",
+        "options":  [
+
+                    ],
+        "answer":  "距离",
+        "solution":  "【详解】从直线外一点到这条直线所画的垂直线段最短，它的长度叫做这点到直线的距离。如图所示：点到直线的距离![](images/ecd1646129f514caac4b5313faba00c8369c23534b78eb84085a051fadfc4dbd.jpg)",
+        "knowledge_points":  [
+                                 "画垂线"
+                             ],
+        "source":  "19-20 四年级上·云南文山·期末"
+    },
+    {
+        "id":  "253",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "两组对边( )的四边形叫做平行四边形;如果以平行四边形的一边为底，能作出( )条高，这些高的长度都( );如果两条平行线之间的距离是9厘米，在这两条平行线之间作一条垂线段，这条垂线段的长是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "分别平行;无数;相等;9",
+        "solution":  "【分析】两组对边分别平行的四边形叫平行四边形，从平行四边形的一条边上的任意一点都可以向对边作高，平行四边形的每一条边都可以看做是底，所以平行四边形有4条底，有无数条高。两直线互相平行时，从一条直线上任意点向另一条直线作垂线，所得的平行线间的垂直线段的长度，叫做平行线间的距离。平行线之间的距离处处相等。本题主要考查学生对平行线和平行四边形的定义和特点的掌握.\n【详解】两组对边分别平行的四边形叫做平行四边形;如果以平行四边形的一边为底，能作出无数条高，这些高的长度都相等;如果两条平行线之间的距离是9厘米，在这两条平行线之间作一条垂线段，这条垂线段的长是9厘米。",
+        "knowledge_points":  [
+                                 "垂直的特征，平行四边形的高及画法，平行四边形的概念及特点"
+                             ],
+        "source":  "20-21 四年级上·湖南怀化·单元测试"
+    },
+    {
+        "id":  "254",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "只有一组对边平行的四边形叫做( );( )等腰梯形;从梯形上底的一点向下底引一条垂线，这点与垂足间的线段叫做梯形的( )。",
+        "options":  [
+
+                    ],
+        "answer":  "梯形;两腰相等的梯形叫做;高",
+        "solution":  "【详解】只有一组对边平行的四边形叫做梯形;互相平行的一组对边分别是梯形的上底和下底，不平行的一组对边是梯形的腰。两腰相等的梯形叫等腰梯形;一腰垂直于底的梯形叫直角梯形;从梯形上底的一点向下底引一条垂线，这点与垂足间的线段叫做梯形的高。",
+        "knowledge_points":  [
+                                 "直角梯形和等腰梯形的概念及特点，梯形的概念及特点，梯形的高及画法"
+                             ],
+        "source":  "20-21四年级上·湖南怀化·单元测试"
+    },
+    {
+        "id":  "255",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "长方形的长和宽是互相( );在( )内，不相交的两条直线叫做平行线;平行线间的距离处处( )。",
+        "options":  [
+
+                    ],
+        "answer":  "垂直;同一平面;相等",
+        "solution":  "【分析】根据垂直的性质可知，长方形的长和宽的夹角是直角，则长和宽互相垂直。同一平面内不相交的两条直线互相平行，其中一条直线是另一条直线的平行线。两直线互相平行时，从一条直线上任意一点向另一条直线作垂线，所得的平行线间的垂直线段的长度，叫做平行线间的距离。平行线之间的距离处处相等。据此解答即可。本题考查平行和垂直的性质，同一平面内两条直线的位置关系不是平行就是相交，垂直是特殊的相交.\n【详解】长方形的长和宽是互相垂直;在同一平面内，不相交的两条直线叫做平行线;平行线间的距离处处相等。",
+        "knowledge_points":  [
+                                 "平行的特征及性质，垂直的特征"
+                             ],
+        "source":  "20-21 四年级上·湖南怀化·单元测试"
+    },
+    {
+        "id":  "256",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "只有一组对边平行的四边形叫做( )。长方形和正方形都可以看作特殊的( )。",
+        "options":  [
+
+                    ],
+        "answer":  "梯形;平行四边形",
+        "solution":  "【详解】根据梯形的概念：只有一组对边平行的四边形是梯形;根据平行四边形的特征：两组对边平行且相等;则得出：长方形，正方形两组对边平行且相等，有四个角是直角，所以是特殊的平行四边形。",
+        "knowledge_points":  [
+                                 "梯形的概念及特点，平行四边形的概念及特点"
+                             ],
+        "source":  "20-21 四年级上·四川绵阳·期末"
+    },
+    {
+        "id":  "257",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在梯形里，平行的一组对边叫做梯形的( )和( ),不平行的一组对边叫做梯形的( )。",
+        "options":  [
+
+                    ],
+        "answer":  "上底;下底;腰",
+        "solution":  "【分析】![](images/0e9b9033b76ff905e31659f1c34ff99dd15ec3e55bcd589ac06ec5f5748f483e.jpg)只有一组对边平行的四边形叫做梯形。互相平行的一组对边分别是梯形的上底和下底，不平行的一组对边是梯形的腰。从梯形一条底边上的一点到它对边的垂直线段叫做梯形的高。本题考查梯形的定义，需熟练掌握.\n【详解】在梯形里，平行的一组对边叫做梯形的上底和下底，不平行的一组对边叫做梯形的腰。",
+        "knowledge_points":  [
+                                 "梯形的概念及特点"
+                             ],
+        "source":  "20-21四年级上·全国·课后作业"
+    },
+    {
+        "id":  "258",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "已知两条平行线间的距离是8厘米，在其中一条线上任选一点向对边画垂直线段，这条垂直线段的长度是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "8",
+        "solution":  "【分析】端点分别在两条平行线上，且与平行线垂直的所有线段的长度都相等。两条平行线之间的距离处处都相等.\n【详解】已知两条平行线间的距离是8厘米，在其中一条线上任选一点向对边画垂直线段，这条垂直线段的长度是8厘米。",
+        "knowledge_points":  [
+                                 "平行的特征及性质"
+                             ],
+        "source":  "21-22四年级上·湖北黄冈·月考"
+    },
+    {
+        "id":  "259",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "两条直线相交成直角，就说这两条直线( ),其中一条直线叫做另一条直线的( ),这两条直线的交点叫做( )。",
+        "options":  [
+
+                    ],
+        "answer":  "互相垂直;垂线;垂足",
+        "solution":  "【详解】根据垂直的定义：两条直线相交成直角，就说这两条直线互相垂直，其中一条直线叫做另一条直线的垂线，这两条直线的交点叫做垂足。例如:![](images/a5cb341f064f03f302183e83856e2002c8555623d0efb805e7cfcf05352b45a8.jpg)",
+        "knowledge_points":  [
+                                 "垂直的特征"
+                             ],
+        "source":  "16-17四年级上·云南楚雄·期末"
+    },
+    {
+        "id":  "260",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰梯形的周长是40米，每条腰长12米，上底与下底的和是( )米。",
+        "options":  [
+
+                    ],
+        "answer":  "16",
+        "solution":  "【分析】等腰梯形的两腰相等，因此用40米减去2个12米即可，依此计算。熟练掌握等腰梯形的特点是解答此题的关键.\n【详解】12×2=24(米),40-24=16(米)。",
+        "knowledge_points":  [
+                                 "直角梯形和等腰梯形的概念及特点，梯形的周长"
+                             ],
+        "source":  "20-21四年级上·湖南怀化·单元测试"
+    },
+    {
+        "id":  "261",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "有一个角是直角的梯形叫做( )。",
+        "options":  [
+
+                    ],
+        "answer":  "直角梯形",
+        "solution":  "【分析】只有一组对边平行的四边形叫做梯形。一腰垂直于底的梯形叫直角梯形。直角梯形有两个直角，据此解答即可。本题考查直角梯形的定义，需熟练掌握.\n【详解】有一个角是直角的梯形叫做直角梯形。",
+        "knowledge_points":  [
+                                 "直角梯形和等腰梯形的概念及特点"
+                             ],
+        "source":  "20-21四年级上·云南德宏·期末"
+    },
+    {
+        "id":  "262",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "如果同一平面内的两条直线不相交，就说这两条直线( )。",
+        "options":  [
+
+                    ],
+        "answer":  "互相平行",
+        "solution":  "【分析】同一平面内永不相交的两条直线叫做平行线，组成平行线的两条直线互相平行，依此填空。熟练掌握平行的特点是解答此题的关键.\n【详解】根据平行的特点可知：如果同一平面内的两条直线不相交，就说这两条直线互相平行。",
+        "knowledge_points":  [
+                                 "平行的特征及性质"
+                             ],
+        "source":  "20-21 四年级上·贵州黔南·期末"
+    },
+    {
+        "id":  "263",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "长方形，正方形相邻的两条边互相( ),相对的两条边互相( )。",
+        "options":  [
+
+                    ],
+        "answer":  "垂直;平行",
+        "solution":  "【分析】两条直线相交所成的四个角中，有一个角是直角时，就说这两条直线互相垂直;在同一平面内，不相交的两条直线就叫做平行线，据此解答。解答此题的关键，熟知正方形，长方形的特点，以及垂直与平行的特征，熟练掌握并灵活应用.\n【详解】长方形，正方形相邻的两条边互相垂直，相对的两条边互相平行。",
+        "knowledge_points":  [
+                                 "长方形的概念及特点，正方形的概念及特点，垂直的特征，平行的特征及性质"
+                             ],
+        "source":  "17-18 四年级上·全国·课后作业"
+    },
+    {
+        "id":  "264",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个平行四边形的周长是80cm,其中一条边长是16cm,则与它相邻的边是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "24",
+        "solution":  "【分析】根据平行四边形的特点，对边相等可得，平行四边形的周长的求解方法与长方形相似，都是相邻两条边的和的2倍，由此先用周长80厘米除以2，求出相邻两边的和，再减去其中的一条边16厘米，即可求出另一条边。熟知平行四边形的特点，找出其周长的计算方法是解答本题的关键.\n【详解】80÷2-16=40-16=24(厘米)。",
+        "knowledge_points":  [
+                                 "平行四边形的概念及特点，平行四边形的周长"
+                             ],
+        "source":  "20-21四年级上·河南漯河·期末"
+    },
+    {
+        "id":  "265",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把符合要求的图形序号填在括号里。(1)两组对边分别平行，有四个直角。( )(2)只有一组对边平行。( )(3)两组对边分别平行。( )",
+        "options":  [
+                        "A. 正方形",
+                        "B. 长方形",
+                        "C. 平行四边形",
+                        "D. 梯形"
+                    ],
+        "answer":  "A,B;D;C",
+        "solution":  "【分析】根据长方形，正方形，平行四边形，梯形的定义选择即可。两组对边分别平行，四条边相等，四个角都是直角的四边形叫做正方形。两组对边分别平行，四个角都是直角的四边形叫做长方形。两组对边分别平行的四边形叫做平行四边形。只有一组对边平行的四边形叫做梯形。熟悉各种四边形的定义是解答此题的关键.\n【详解】(1)两组对边分别平行，有四个直角。(AB)(2)只有一组对边平行。(D)(3)两组对边分别平行。(C)A。正方形 B。长方形 C。平行四边形 D。梯形",
+        "knowledge_points":  [
+                                 "梯形的概念及特点，正方形的概念及特点，长方形的概念及特点，平行四边形的概念及特点"
+                             ],
+        "source":  "21-22 四年级上·全国·期末"
+    },
+    {
+        "id":  "266",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "等腰梯形的( )相等。",
+        "options":  [
+
+                    ],
+        "answer":  "两腰",
+        "solution":  "【详解】两腰相等的梯形叫做等腰梯形，即等腰梯形的两腰相等，如图所示:![](images/93e5194eeaa02acaad68bbc8eff9d3331448d6da8b0a975ad460262d947e7eba.jpg)等腰梯形",
+        "knowledge_points":  [
+                                 "直角梯形和等腰梯形的概念及特点"
+                             ],
+        "source":  "21-22 四年级上·全国·期末"
+    },
+    {
+        "id":  "267",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "两条平行线之间可以画( )条垂直线段，所有的垂直线段的长度都( ),这些垂直线段互相( )。(填\u0027平行\u0027或\u0027垂直\u0027)",
+        "options":  [
+
+                    ],
+        "answer":  "无数;相等;平行",
+        "solution":  "【分析】同一平面内永不相交的两条直线叫做平行线，组成平行线的两条直线互相平行;在同一平面内两条直线相交成直角，这两条直线互相垂直，其中一条直线叫做另条直线的垂线;依此画图并填空。熟练掌握平行与垂直的特点是解答此题的关键.\n【详解】如图所示:![](images/aa1d805086e0ac7b2a0372575dc6a01dc946c994b5ed71de4456c6f710618d51.jpg)即：两条平行线之间可以画无数条垂直线段，所有的垂直线段的长度都相等，这些垂直线段互相平行。",
+        "knowledge_points":  [
+                                 "垂直的特征，平行的特征及性质"
+                             ],
+        "source":  "21-22 四年级上·全国·期末"
+    },
+    {
+        "id":  "268",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "直线a,b,c在同一平面内，且直线a和直线b都与直线c垂直，则直线a和直线b的位置关系是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平行",
+        "solution":  "【分析】在同一平面内，垂直于同一条直线的两条直线互相平行，据此解答。解题的关键应熟练掌握垂直的性质.\n【详解】由垂直的性质可得：在同一个平面内垂直于同一条直线的两条直线一定互相平行。a,b,c在同一平面内，且直线a和直线b都与直线c垂直，则直线a和直线b互相平行。",
+        "knowledge_points":  [
+                                 "平行的特征及性质，垂直的特征"
+                             ],
+        "source":  "19-20 四年级上·河北石家庄·期末"
+    },
+    {
+        "id":  "269",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在一个正方形中可以找到( )组相互垂直的线段。",
+        "options":  [
+
+                    ],
+        "answer":  "4",
+        "solution":  "【分析】正方形的两组对边分别互相平行，相邻的两条边相互垂直，四条边都相等，四个角都是直角;依此填空。熟练掌握正方形的特点是解答此题的关键.\n【详解】正方形有4个直角，因此在一个正方形中可以找到4(四)组相互垂直的线段。",
+        "knowledge_points":  [
+                                 "垂直的特征，正方形的概念及特点"
+                             ],
+        "source":  "20-21 四年级上·广东惠州·期末"
+    },
+    {
+        "id":  "270",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "平行线之间的距离是6厘米，在这两条平行线之间做一条垂直线段，这条垂直线段的长度是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "6厘米",
+        "solution":  "【分析】两直线互相平行时，从一条直线上任意一点向另一条直线作垂线，所得的平行线间的垂直线段的长度，叫做平行线间的距离。则这条垂直线段的长度等于平行线之间的距离。本题考查平行的性质，关键是明确这两条平行线之间垂线段的长度就是平行线间的距离.\n【详解】平行线之间的距离是6厘米，在这两条平行线之间做一条垂直线段，这条垂直线段的长度是6厘米。",
+        "knowledge_points":  [
+                                 "平行的特征及性质"
+                             ],
+        "source":  "14-15 四年级下·山东青岛·单元测试"
+    },
+    {
+        "id":  "271",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "把一个长方形的框架拉成一个平行四边形，这个平行四边形的周长和原来的长方形的周长相比,( )。(填\u0027增大\u0027\u0027不变\u0027或\u0027减小\u0027)",
+        "options":  [
+
+                    ],
+        "answer":  "不变",
+        "solution":  "【分析】把一个长方形的框架拉成一个平行四边形后，四条边的长度没变，也就是它们的和没有发生变化，即它的周长不变;据此解答即可。解答此题的关键是明确四边形的形状虽然发生了变化，但长和宽没有变化，周长就不发生变化.\n【详解】把一个长方形的框架拉成一个平行四边形，这个平行四边形的周长和原来的长方形的周长相比，不变。",
+        "knowledge_points":  [
+                                 "平行四边形的周长，长方形的周长，平行四边形的概念及特点"
+                             ],
+        "source":  "21-22 四年级上·全国·期末"
+    },
+    {
+        "id":  "272",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "从平行四边形一条边上的一点到( )引一条( ),这点到( )之间的线段叫做平行四边形的高。垂足所在的边叫做( )。",
+        "options":  [
+
+                    ],
+        "answer":  "对边;垂线;垂足;平行四边形的底",
+        "solution":  "【详解】从平行四边形一条边上的一点到对边引一条垂线，这点到垂足之间的线段叫做平行四边形的高。垂足所在的边叫做平行四边形的底。平行四边形有4条底，有无数条高。",
+        "knowledge_points":  [
+                                 "平行四边形的高及画法"
+                             ],
+        "source":  "2021 四年级上·全国·专题练习"
+    },
+    {
+        "id":  "273",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在纸上任意画两条直线，会有( )种情况：一种( ),一种( )。",
+        "options":  [
+
+                    ],
+        "answer":  "两;相交;平行",
+        "solution":  "【详解】根据平面内两条直线的位置关系，在同一平面内的两条直线，只有两种位置关系，不是相交就是平行，垂直是相交的特殊情况。",
+        "knowledge_points":  [
+                                 "垂直的特征，平行的特征及性质"
+                             ],
+        "source":  "2021四年级上·全国·专题练习"
+    },
+    {
+        "id":  "274",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在( ),两条直线不是平行就是相交。",
+        "options":  [
+
+                    ],
+        "answer":  "同一平面内",
+        "solution":  "【详解】在同一平面内的两条直线，只有两种位置关系，不是相交就是平行，垂直是相交的特殊情况。",
+        "knowledge_points":  [
+                                 "垂直的特征，平行的特征及性质"
+                             ],
+        "source":  "2021 四年级上·全国·专题练习"
+    },
+    {
+        "id":  "275",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "( )点整和( )点整时，时针与分针所在直线互相垂直。",
+        "options":  [
+
+                    ],
+        "answer":  "3;9",
+        "solution":  "【分析】当两条直线相交成90度时，这两条直线就互相垂直。3点钟时，钟表的时针指向数字3，分针指向12;时针和分针互相垂直;9点钟时，钟表的时针指向数字9，分针指向12;时针和分针互相垂直。熟悉钟表上几个特殊整点时针和分针组成的角是解答此题的关键.\n【详解】3点整和9点整时，时针与分针所在直线互相垂直。",
+        "knowledge_points":  [
+                                 "垂直的特征，整点时间与钟面指针的位置"
+                             ],
+        "source":  "21-22 四年级上·广东东莞·周测"
+    },
+    {
+        "id":  "276",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个梯形的下底是上底的3倍，如果将上底延长8厘米，梯形就成了一个平行四边形，这个梯形的上底是( )厘米，下底是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "4;12",
+        "solution":  "【分析】这个梯形的下底是上底的3倍，即下底是3个上底长;延长8厘米后，变成了平行四边形，也就是延长了3-1=2个上底长，即2个上底长=8厘米;求一个上底是多少，据此求解即可。本题的关键是要清楚变成平行四边形后对边的长度是相等的.\n【详解】根据分析可得：上底长:8÷(3-1)=8÷2=4(厘米);下底长:4×3=12(厘米)。",
+        "knowledge_points":  [
+                                 "和差倍问题，平行四边形的概念及特点，梯形的概念及特点"
+                             ],
+        "source":  "21-22四年级上·全国·期末"
+    },
+    {
+        "id":  "277",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "梯形有( )条高。",
+        "options":  [
+
+                    ],
+        "answer":  "无数",
+        "solution":  "【分析】本题考查梯形高的概念。根据梯形高的定义，从梯形一条底边上的任意一点向对边作垂线，这点与垂足间的距离叫做梯形的高。因为底边上有无数个点，所以可以画出无数条垂线，即梯形有无数条高.\n【详解】在梯形中，从一条底上的任意一点向对边作垂线，这点与垂足间的距离叫做梯形的高，梯形有无数条高，通常过上底的一个顶点作下底的垂线，用三角板的直角可以画出梯形的高。",
+        "knowledge_points":  [
+                                 "梯形的高及画法"
+                             ],
+        "source":  "21-22 四年级上·广东东莞·周测"
+    },
+    {
+        "id":  "278",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "平行四边形相邻的两条边分别是5厘米,4厘米，它的周长是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "18厘米",
+        "solution":  "【分析】平行四边形的周长等于两条邻边长的和的2倍;据此解答即可。解答此题应明确平行四边形的周长的含义及计算方法，注意基础知识的灵活掌握.\n【详解】(5+4)×2=9×2=18(厘米),所以，它的周长是18厘米。",
+        "knowledge_points":  [
+                                 "平行四边形的周长"
+                             ],
+        "source":  "21-22 四年级上·广东东莞·周测"
+    },
+    {
+        "id":  "279",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在梯形上剪一刀，其中一个是平行四边形，另一个是( )形。",
+        "options":  [
+
+                    ],
+        "answer":  "三角或梯",
+        "solution":  "【分析】只有一组对边平行的四边形叫做梯形;两组对边分别平行的四边形，叫做平行四边形，依此画图并填空即可。![](images/99a13bad36ea46248e9ab229ffd90d6a.jpg)此题考查的是平面图形的分割，熟练掌握平行四边形和梯形的特点是解答此题的关键.\n【详解】如图所示，在梯形上剪一刀，其中一个是平行四边形，另一个是三角形或梯形。",
+        "knowledge_points":  [
+                                 "平行四边形的概念及特点，梯形的概念及特点，平面图形的分割"
+                             ],
+        "source":  "20-21 四年级上·江西鹰潭·单元测试"
+    },
+    {
+        "id":  "280",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在平行四边形上剪一刀，其中一个三角形，另一个是( )形。",
+        "options":  [
+
+                    ],
+        "answer":  "梯形、五边形、三角形",
+        "solution":  "【分析】见详解![](images/7a97f46e6b612ffd2b168e71400900216703ceefb001cb8d1da584e1a1b909a0.jpg)此题解答时答案不唯一，分情况考虑全面即可.\n【详解】(梯形)(三角形)(五边形)",
+        "knowledge_points":  [
+                                 "平面图形的分割"
+                             ],
+        "source":  "20-21四年级上·江西鹰潭·单元测试"
+    },
+    {
+        "id":  "281",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "把2根小棒都摆成和第3根小棒垂直，那么这两根小棒互相( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平行",
+        "solution":  "【分析】在同一平面内两条直线相交成直角，这两条直线互相垂直;同一平面内永不相交的两条直线叫做平行线，组成平行线的两条直线互相平行，依此填空。熟练掌握平行与垂直的特点是解答此题的关键.\n【详解】![](images/a0bd6913b73fe39c8b24e00f42316fda1ea4e9657267ed13740edfa39421cb0c.jpg)如图所示，把2根小棒都摆成和第3根小棒垂直，那么这两根小棒互相平行。",
+        "knowledge_points":  [
+                                 "平行的特征及性质，垂直的特征"
+                             ],
+        "source":  "17-18 四年级上·河南郑州·期末"
+    },
+    {
+        "id":  "282",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "长方形中，有( )组平行线，有( )组垂线。",
+        "options":  [
+
+                    ],
+        "answer":  "2;4",
+        "solution":  "【分析】根据长方形的特点：两组对边分别平行且相等，所以有2组平行线;因为长方形有四个直角，所以有4组垂线。此题考查长方形的定义，两组对边分别平行且相等;根据长方形邻边互相垂直，可得到垂线的组数.\n【详解】长方形中，有2组平行线，有4组垂线。",
+        "knowledge_points":  [
+                                 "平行的特征及性质，垂直的特征"
+                             ],
+        "source":  "20-21 四年级上·江西鹰潭·单元测试"
+    },
+    {
+        "id":  "283",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "2个完全一样的三角形一定可以拼成( )形。",
+        "options":  [
+
+                    ],
+        "answer":  "三角形",
+        "solution":  "【分析】根据题意，画2个完全一样的三角形，然后再将它们拼起来即可，根据拼的图形填空即可。此题考查的是平面图形的拼接，熟练掌握平行四边形的特点是解答此题的关键.\n【详解】![](images/d1d3de9eec2ce79b5ff25c4b13d8cd931171a509b8d0bcd39412362e8b009ca1.jpg)如图所示：长方形和正方形是特殊的平行四边形，因此2个完全一样的三角形一定可以拼成平行四边形。",
+        "knowledge_points":  [
+                                 "用同样的图形拼接，平面图形的拼接，平行四边形的概念及特点"
+                             ],
+        "source":  "20-21 四年级上·江西鹰潭·单元测试"
+    },
+    {
+        "id":  "284",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "两条直线相交成( )角时，这两条直线互相垂直，它们的交点叫( )。",
+        "options":  [
+
+                    ],
+        "answer":  "直;垂足",
+        "solution":  "【详解】两条直线相交成直角，就说这两条线相互垂直，其中一条直线叫另一条直线的垂线，这两条直线的交点叫垂足。",
+        "knowledge_points":  [
+                                 "垂直的特征"
+                             ],
+        "source":  "18-19 四年级上·江苏·课后作业"
+    },
+    {
+        "id":  "285",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个平行四边形的周长是40厘米，其中一条边长12厘米，那么与它相邻的一条边长是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "8",
+        "solution":  "【分析】平行四边形的周长=相邻两边的长度之和*2，因此先用周长除以2后，再减去已知的一条边的长度即可。熟练掌握平行四边形的周长的计算是解答此题的关键.\n【详解】40÷2=20(厘米),20-12=8(厘米)。",
+        "knowledge_points":  [
+                                 "平行四边形的周长"
+                             ],
+        "source":  "21-22四年级上·江西赣州·单元测试"
+    },
+    {
+        "id":  "286",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "已知一条直线，可以画( )条直线与它平行;过直线外一点，可以画( )条直线与它垂直。",
+        "options":  [
+
+                    ],
+        "answer":  "无数;一或1",
+        "solution":  "【分析】根据直线的性质以及平行和垂直的性质进行填空即可。此题考查了直线的性质及平行，垂直的性质，注意基础知识的识记和理解.\n【详解】已知一条直线，可以画无数条直线与它平行;过直线外一点，可以画一条直线与它垂直。",
+        "knowledge_points":  [
+                                 "平行的特征及性质，画垂线"
+                             ],
+        "source":  "20-21 四年级上·江西鹰潭·期末"
+    },
+    {
+        "id":  "287",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用四根硬纸条制作成一个长方形，然后用两手捏住长方形的两个对角，向相反方向拉，变成了( ),两组对边的长度( )变化(填\u0027有\u0027或\u0027没有\u0027),平行四边形容易( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平行四边形;没有;变形",
+        "solution":  "【分析】本题考查平行四边形的特性及与长方形的关系。用硬纸条制作的长方形框架，四条边的长度是固定的。当捏住对角向相反方向拉时，四条边的长度没有变化，但角的大小发生了变化，长方形变成了平行四边形。这一现象说明了平行四边形具有容易变形的特性.\n【详解】由实际操作可知：用硬木棍制作的长方形，捏住对角，向相反的方向拉了之后，两组对边平行且相等，就成了平行四边形，平行四边形容易变形。",
+        "knowledge_points":  [
+                                 "平行四边形的概念及特点"
+                             ],
+        "source":  "21-22四年级上·江西赣州·单元测试"
+    },
+    {
+        "id":  "288",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "从直线外一点到这条直线所画的所有线段中，垂线段最短，它的长度叫做点到直线的( )。",
+        "options":  [
+
+                    ],
+        "answer":  "距离",
+        "solution":  "【分析】本题考查点到直线的距离的概念。根据垂线段的性质，从直线外一点到这条直线所画的所有线段中，垂直线段是最短的。数学上规定，这条垂直线段的长度叫做这点到直线的距离.\n【详解】从直线外一点到这条直线所画的所有线段中，垂线段最短，它的长度叫做点到直线的距离。",
+        "knowledge_points":  [
+                                 "画平行线"
+                             ],
+        "source":  "21-22 四年级上·江西赣州·单元测试"
+    },
+    {
+        "id":  "289",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "过直线外一点可以画( )条这条直线的垂线，还可以画( )条这条直线的平行线。",
+        "options":  [
+
+                    ],
+        "answer":  "一;一",
+        "solution":  "【分析】根据平行公理\u0027经过直线外一点，有且只有一条直线与已知直线平行\u0027和垂线的性质\u0027在同一平面内，过一点(直线外或直线上)有且只有一条直线垂直于已知直线\u0027进行解答即可。此题考查了直线的性质及平行，垂直的性质，注意基础知识的熟记和理解.\n【详解】过直线外一点可以画一条这条直线的垂线，还可以画一条这条直线的平行线。",
+        "knowledge_points":  [
+                                 "画平行线，画垂线"
+                             ],
+        "source":  "21-22 四年级上·江西赣州·单元测试"
+    },
+    {
+        "id":  "290",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "两个完全相同的梯形，可以拼成一个平行四边形，它们的( )相等。",
+        "options":  [
+
+                    ],
+        "answer":  "高",
+        "solution":  "【分析】只有一组对边平行的四边形叫做梯形;两组对边分别平行的四边形，叫做平行四边形，平行四边形的对边相等，依此填空。此题考查的是平面图形的拼接，应熟练掌握梯形和平行四边形的特点以及它们的高.\n【详解】根据分析可知，两个完全相同的梯形，可以拼成一个平行四边形，它们的高相等。如图所示:![](images/04339a319d9d39ff4d695f21f9cf9592c4a6f8b27bb3994af18463100aee1499.jpg)",
+        "knowledge_points":  [
+                                 "梯形的概念及特点，平行四边形的高及画法，梯形的高及画法，平行四边形的概念及特点"
+                             ],
+        "source":  "20-21四年级上·江西鹰潭·期末"
+    },
+    {
+        "id":  "291",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "长方形和正方形可以看成特殊的( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平行四边形",
+        "solution":  "【分析】长方形，正方形两组对边平行且相等，有四个角是直角;而平行四边形只需要两组对边平行且相等;长方形和正方形满足平行四边形的定义，所以说它们是特殊的平行四边形。此题考查了平行四边形的特征和性质，应注意基础知识的积累.\n【详解】根据分析可得：长方形和正方形可以看成特殊的平行四边形。",
+        "knowledge_points":  [
+                                 "平行四边形的概念及特点，长方形的概念及特点，正方形的概念及特点"
+                             ],
+        "source":  "2014三年级·全国·课后作业"
+    },
+    {
+        "id":  "292",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在两条平行线中间可以画( )条与平行线垂直的线段，画出的这些线段长度( ),并且互相( )。",
+        "options":  [
+
+                    ],
+        "answer":  "无数;都相等;平行",
+        "solution":  "【分析】在同一平面内不相交的两条直线叫做平行线，两条直线相交成直角，这两条直线互相垂直，在两平行线之间的可以画无数条垂线段，并且所有垂线段都相等，依此填空.\n【详解】在两条平行线中间可以画无数条与平行线垂直的线段，画出的这些线段长度都相等，并且互相平行.\n[点睛]熟练掌握平行与垂直的特点是解答此题的关键。",
+        "knowledge_points":  [
+                                 "垂直的特征，平行的特征及性质"
+                             ],
+        "source":  "20-21四年级上·江西萍乡·期末"
+    },
+    {
+        "id":  "293",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "( )和正方形可以看成特殊的平行四边形。",
+        "options":  [
+
+                    ],
+        "answer":  "长方形",
+        "solution":  "【详解】梯形和平行四边形都属于四边形，长方形和正方形是特殊的平行四边形。",
+        "knowledge_points":  [
+                                 "四边形的分类及关系，平行四边形的概念及特点"
+                             ],
+        "source":  "18-19 四年级上·湖南常德·期末"
+    },
+    {
+        "id":  "294",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "( )一组对边平行的四边形叫做梯形。",
+        "options":  [
+
+                    ],
+        "answer":  "只有",
+        "solution":  "【详解】根据梯形的定义可知，只有一组对边平行的四边形叫做梯形。如果梯形的两腰相等，这样的梯形叫做等腰梯形;一腰垂直于底的梯形叫直角梯形。",
+        "knowledge_points":  [
+                                 "梯形的概念及特点"
+                             ],
+        "source":  "18-19 四年级上·湖南常德·期末"
+    },
+    {
+        "id":  "295",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "长方形和( )可以看成特殊的平行四边形。",
+        "options":  [
+
+                    ],
+        "answer":  "正方形",
+        "solution":  "【分析】两组对边分别平行的四边形叫做平行四边形。两组对边分别平行，四个角都是直角的四边形叫做长方形。两组对边分别平行，四条边相等，四个角都是直角的四边形叫做正方形。则长方形是特殊的平行四边形。正方形是特殊的长方形，也是特殊的平行四边形。据此解答即可。熟练掌握正方形，长方形和平行四边形的定义是解决本题的关键.\n【详解】根据分析可知，长方形和正方形可以看成特殊的平行四边形。",
+        "knowledge_points":  [
+                                 "正方形的概念及特点，长方形的概念及特点，平行四边形的概念及特点"
+                             ],
+        "source":  "17-18 四年级上·湖南常德·期末"
+    },
+    {
+        "id":  "296",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在同一平面内，把两根小棒都摆成和第三根小棒垂直，这两根小棒的位置是互相( )的关系。",
+        "options":  [
+
+                    ],
+        "answer":  "平行",
+        "solution":  "【分析】同一平面内不相交的两条直线叫做平行线，组成平行线的两条直线互相平行;两条直线相交成直角，这两条直线互相垂直，依此填空。熟练掌握平行与垂直的特点是解答此题的关键.\n【详解】![](images/28b122b8be795094dd84aa04bbb23c913254b5ac5c9df1ef5b1f7f78cd8791d1.jpg)如图所示：这两根小棒的位置是互相平行的关系。",
+        "knowledge_points":  [
+                                 "垂直的特征，平行的特征及性质"
+                             ],
+        "source":  "20-21四年级上·新疆·期末"
+    },
+    {
+        "id":  "297",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "平行四边形容易( )的特点，在实际生活中有广泛的应用。",
+        "options":  [
+
+                    ],
+        "answer":  "变形",
+        "solution":  "【分析】平行四边形的不稳定性就是指平行四边形边长确定，其形状，大小不能完全确定。这些特性在生活中有着广泛的应用。比如伸缩衣架，小区门口的电动门等。本题考查学生对平行四边形的不稳定性的掌握情况.\n【详解】平行四边形容易变形的特点，在实际生活中有广泛的应用。",
+        "knowledge_points":  [
+                                 "平行四边形的不稳定性及应用"
+                             ],
+        "source":  "17-18 四年级上·湖南常德·期末"
+    },
+    {
+        "id":  "298",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在同一个平面内( )的两条直线叫做平行线。",
+        "options":  [
+
+                    ],
+        "answer":  "不相交",
+        "solution":  "【分析】本题考查平行线的定义。根据平行线的概念可知，在同一平面内，不相交的两条直线叫做平行线。据此填空.\n【详解】同一平面内不相交的两条直线互相平行，其中一条直线是另一条直线的平行线。据此解答即可。",
+        "knowledge_points":  [
+                                 "平行的特征及性质"
+                             ],
+        "source":  "19-20 四年级上·山东临沂·期末"
+    },
+    {
+        "id":  "299",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "四边形ABCD,已知AB//CD,其中有两个角都是60°。这个四边形可能是( )形;也可能是( )形。",
+        "options":  [
+
+                    ],
+        "answer":  "平行四边;等腰梯",
+        "solution":  "【分析】两组对边分别平行的四边形，叫做平行四边形，平行四边形的对角相等;只有一组对边平行的四边形叫做梯形，等腰梯形的两条腰相等，两个底角也相等;依此填空。熟练掌握等腰梯形，平行四边形的特点是解答此题的关键.\n【详解】根据分析可知：四边形ABCD,已知AB//CD,其中有两个角都是60°。这个四边形可能是平行四边形;也可能是等腰梯形。",
+        "knowledge_points":  [
+                                 "平行四边形的概念及特点，直角梯形和等腰梯形的概念及特点"
+                             ],
+        "source":  "20-21 四年级上·浙江杭州·期末"
+    },
+    {
+        "id":  "300",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个等腰梯形的周长是30厘米。它的上底是5厘米，下底是9厘米。这个梯形的一条腰是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "8",
+        "solution":  "【分析】等腰梯形的腰相等，梯形的周长减去上，下底的长度，再除以2即等于腰的长度。熟练掌握等腰梯形的特点是解答本题的关键.\n【详解】(30-5-9)÷2=16÷2=8(厘米)",
+        "knowledge_points":  [
+                                 "梯形的周长，直角梯形和等腰梯形的概念及特点"
+                             ],
+        "source":  "19-20 四年级上·浙江杭州·期末"
+    },
+    {
+        "id":  "301",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "直线a与直线b互相垂直，记作( )。平行四边形的对边( )。",
+        "options":  [
+
+                    ],
+        "answer":  "a⊥b;平行且相等",
+        "solution":  "【分析】垂直可以用符号\u0027⊥\u0027表示，然后根据平行四边形的特征进行解答.\n【详解】直线a与直线b互相垂直，记作a⊥b;平行四边形的对边平行且相等.\n[点睛]此题主要考查了垂直的基础知识与平行四边形的特征。",
+        "knowledge_points":  [
+                                 "平行四边形的概念及特点，垂直的特征"
+                             ],
+        "source":  "20-21 四年级上·云南玉溪·期末"
+    },
+    {
+        "id":  "302",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "有一块平行四边形的菜地，一条边的长度是8米，与它相邻的另一条边的长度是15米，这块菜地的周长是( )米。",
+        "options":  [
+
+                    ],
+        "answer":  "46",
+        "solution":  "【分析】平行四边形的周长=(一条边的长度+与它相邻的另一条边的长度)×2，依此计算。熟练掌握平行四边形的周长的计算是解答此题的关键.\n【详解】(8+15)×2=23×2=46(米)",
+        "knowledge_points":  [
+                                 "平行四边形的周长"
+                             ],
+        "source":  "20-21四年级上·云南玉溪·期末"
+    },
+    {
+        "id":  "303",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "过两点可以画( )条直线，平行四边形有( )条高。",
+        "options":  [
+
+                    ],
+        "answer":  "1;无数",
+        "solution":  "【分析】根据直线的性质可知，通过两点最多可画1条直线。通过一点可画无数条直线。从平行四边形一条边上的一点到对边引一条垂线，这点和垂足之间的线段叫做平行四边形的高。则平行四边形有两组高，无数条高。本题考查直线的性质和平行四边形高的定义，需熟练掌握.\n【详解】过两点可以画1条直线，平行四边形有无数条高。",
+        "knowledge_points":  [
+                                 "平行四边形的高及画法，线段，直线，射线的认识及特征"
+                             ],
+        "source":  "20-21 四年级上·云南曲靖·期末"
+    },
+    {
+        "id":  "304",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "数学书封面的对边互相( ),邻边互相( )。伸缩门是利用了平行四边形( )的特性制成的。",
+        "options":  [
+
+                    ],
+        "answer":  "平行;垂直;容易变形",
+        "solution":  "【分析】数学书封面是一个长方形，因此根据长方形的特点填空即可;平行四边形具有不稳定性，容易变形，依此填空。此题考查的是平行四边形容易变形的特性，以及长方形的特点，应熟练掌握.\n【详解】数学书封面的对边互相平行，邻边互相垂直。伸缩门是利用了平行四边形容易变形的特性制成的。",
+        "knowledge_points":  [
+                                 "平行四边形的概念及特点，长方形的概念及特点"
+                             ],
+        "source":  "20-21四年级上·重庆开州·期末"
+    },
+    {
+        "id":  "305",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一条直线的垂线有( )条，从直线外一点到这条直线有( )条垂线，它们的交点叫做( )。",
+        "options":  [
+
+                    ],
+        "answer":  "无数;1;垂足",
+        "solution":  "【分析】直线没有端点，是无限长的，两条直线相交成直角，这两条直线互相垂直，其中一条直线叫做另一条直线的垂线，这两条直线的交点叫做垂足;依此填空。熟练掌握直线与垂直的特点是解答此题的关键.\n【详解】一条直线的垂线有无数条，从直线外一点到这条直线有1条垂线，它们的交点叫做垂足。",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征，垂直的特征"
+                             ],
+        "source":  "20-21 四年级上·江西吉安·期末"
+    },
+    {
+        "id":  "306",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "同一平面内，与一条已知直线垂直的直线有( )条;过直线外一点，与这条已知直线垂直的直线有( )条，这条直线叫做已知直线的( )。",
+        "options":  [
+
+                    ],
+        "answer":  "无数;1;垂线",
+        "solution":  "【分析】本题考查垂直的特征及性质。在同一平面内，与一条直线垂直的直线有无数条;但过直线外一点，只能画一条已知直线的垂线。当两条直线相交成直角时，其中一条直线叫做另一条直线的垂线。据此填空.\n【详解】同一平面内，与一条已知直线垂直的直线有无数条;过直线外一点，与这条已知直线垂直的直线有1条，这条直线叫做已知直线的垂线。",
+        "knowledge_points":  [
+                                 "垂直的特征及性质"
+                             ],
+        "source":  "22-23 四年级上·浙江杭州·期末"
+    },
+    {
+        "id":  "307",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在一个平行四边形纸上剪一刀，能剪出两个完全一样的( )。(填一个)",
+        "options":  [
+
+                    ],
+        "answer":  "平行四边形",
+        "solution":  "【分析】平行四边形纸，剪出完全一样的图形，则剪的这条线一定过平行四边形的中心，一种为沿对角线剪，一种为不过对角线剪，以此作答。本题考查图形的分割，可以培养学生的思维能力.\n【详解】根据分析沿对角线剪，剪出的图形为三角形;不沿对角线剪，则为四边形或梯形，如图沿虚线剪开均可得到完全相同的图形，任选一个即可。![](images/a8dd2db5a397cb7539bacda05176f8a7a18b0f9653c125bb09f76112fff105b5.jpg)",
+        "knowledge_points":  [
+                                 "平面图形的分割"
+                             ],
+        "source":  "20-21 四年级上·河南洛阳·期末"
+    },
+    {
+        "id":  "308",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "两组对边分别( )的四边形叫做平行四边形;有一个角是直角的梯形叫做( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平行;直角梯形",
+        "solution":  "【分析】本题考查平行四边形和直角梯形的定义。根据平行四边形的概念，两组对边分别平行的四边形叫做平行四边形;根据直角梯形的概念，有一个角是直角的梯形叫做直角梯形。据此填空.\n【详解】如图所示:![](images/40a853b30dabf5f72768fb96e19b598879668e4ed6b31038b26014d10816c08f.jpg)![](images/902a01db8c5a57a0d85c9673137fa51bce37a5f715f9a017f2f77131d8d06d66.jpg)根据平行四边形，直角梯形的含义及特征可知：两组对边平行的四边形叫做平行四边形;有一个角为直角的梯形叫做直角梯形。",
+        "knowledge_points":  [
+                                 "平行四边形的概念及特点，直角梯形和等腰梯形的概念及特点"
+                             ],
+        "source":  "20-21 四年级上·河南洛阳·期末"
+    },
+    {
+        "id":  "309",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在一个梯形中，只有( )组对边是平行的;平行四边形有( )组对边平行，且相对的两边( )(填\u0027相等\u0027或\u0027不相等\u0027)。",
+        "options":  [
+
+                    ],
+        "answer":  "一;两;相等",
+        "solution":  "【分析】本题考查梯形和平行四边形的特征。根据梯形的定义可知，只有一组对边平行的四边形叫做梯形;根据平行四边形的特征可知，平行四边形两组对边分别平行且相等。据此填空.\n【详解】只有一组对边平行的四边形叫梯形，互相平行的一组对边分别是梯形的上底和下底，不平行的一组对边是梯形的腰。两组对边互相平行的四边形叫平行四边形，平行四边形具有不稳定性，两组对边平行且相等。",
+        "knowledge_points":  [
+                                 "平行四边形的概念及特点，梯形的概念及特点"
+                             ],
+        "source":  "20-21四年级上·山东济宁·期末"
+    },
+    {
+        "id":  "310",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "观察我们的试卷，相对的两条边互相( ),相邻的两边条互相( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平行;垂直",
+        "solution":  "【分析】观察试卷可知试卷是长方形，根据长方形的特征，对边平行且相等,4个角都是直角，可知，长方形相邻的两条边互相垂直，相对的两边互相平行。此题主要考查长方形的特征.\n【详解】根据长方形的特征，相对的两条边互相平行，相邻的两边条互相垂直。",
+        "knowledge_points":  [
+                                 "长方形的概念及特点，垂直的特征，平行的特征及性质"
+                             ],
+        "source":  "2014 四年级·全国·课后作业"
+    },
+    {
+        "id":  "311",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在一张纸上画三条直线a,b,c,其中a//b,c⊥b,那么直线a和直线c的关系是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "a⊥c",
+        "solution":  "【分析】根据题意画出三条直线即可直观判断直线a与c的关系。本题主要考查平行和垂直的概念，熟记平行和垂直的特征是解答此题的关键.\n【详解】如图:![](images/dffc6673179b4214bb89a1dc100883e130d3982b298633018a7e61498b072fa2.jpg)所以直线a与c的关系是a⊥c。",
+        "knowledge_points":  [
+                                 "平行的特征及性质，垂直的特征"
+                             ],
+        "source":  "20-21四年级上·山东济宁·期末"
+    },
+    {
+        "id":  "312",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个梯形，上底是4厘米，如果将上底延长3厘米，这个梯形就变成了正方形，这个梯形的下底是( )厘米，高是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "7;7",
+        "solution":  "【分析】梯形中只有一组对边平行，正方形的两组对边平行,4条边相等。将梯形的上底延长3厘米，变为4+3=7厘米，这个梯形就变成正方形，则正方形的边长为7厘米。这个梯形的下底等于正方形的边长，高也等于正方形的边长。本题考查梯形和正方形的性质，可通过画图帮助更好理解题意.\n【详解】4+3=7(厘米),则这个梯形的下底是7厘米，高是7厘米。",
+        "knowledge_points":  [
+                                 "梯形的概念及特点，正方形的概念及特点"
+                             ],
+        "source":  "20-21 四年级上·山东济宁·期末"
+    },
+    {
+        "id":  "313",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "过同一平面上的任意两点，可以画( )条直线;过直线外一点可以画( )条直线与已知直线垂直。",
+        "options":  [
+
+                    ],
+        "answer":  "1;1",
+        "solution":  "【分析】根据直线的性质可知，过两点可以画一条直线，过一点可以画无数条直线。过一点有且只有一条直线与已知直线垂直。从直线外一点到这条直线上各点所连的线段中，垂线段最短。据此解答即可。本题考查直线和垂直的性质，需熟练掌握.\n【详解】过同一平面上的任意两点，可以画1条直线;过直线外一点可以画1条直线与已知直线垂直。",
+        "knowledge_points":  [
+                                 "线段，直线，射线的认识及特征，垂直的特征"
+                             ],
+        "source":  "20-21 四年级上·山东日照·期末"
+    },
+    {
+        "id":  "314",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把符合描述的图形序号填在括号里。(1)只有一组对边平行。( )(2)两组对边分别平行。( )(3)有四个直角且两组对边分别平行。( )",
+        "options":  [
+                        "A. 正方形",
+                        "B. 长方形",
+                        "C. 平行四边形",
+                        "D. 梯形"
+                    ],
+        "answer":  "D;A、B、C;A、B",
+        "solution":  "【分析】(1)长方形的定义：两组对边分别平行，四个角都是直角的四边形叫做长方形。(2)正方形的定义：四条边相等，四个角都是直角的四边形叫做正方形。(3)平行四边形的定义：平行四边形，是在同一个二维平面内，由两组平行线段组成的闭合图形。(4)只有一组对边平行的四边形叫做梯形。正确理解长方形，正方形，平行四边形和梯形的定义，是解答此题的关键.\n【详解】(1)只有一组对边平行;是D;(2)两组对边分别平行;是A,B,C;(3)有四个直角且两组对边分别平行;是A,B.",
+        "knowledge_points":  [
+                                 "平行四边形的概念及特点，长方形的概念及特点，正方形的概念及特点，梯形的概念及特点"
+                             ],
+        "source":  "20-21 四年级上·山东菏泽·期末"
+    },
+    {
+        "id":  "315",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果直线a垂直于直线c,那么直线a和直线c相交成( )度。",
+        "options":  [
+
+                    ],
+        "answer":  "90",
+        "solution":  "【分析】两条直线相交成直角时，这两条直线互相垂直，其中一条直线是另一条直线的垂线。据此解答即可。本题考查垂直的性质，需熟练掌握.\n【详解】如果直线a垂直于直线c,那么直线a和直线c相交成90度。",
+        "knowledge_points":  [
+                                 "垂直的特征"
+                             ],
+        "source":  "19-20 四年级上·广东中山·期末"
+    },
+    {
+        "id":  "316",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "平行四边形的两组对边互相( )且相等，只有一组对边平行的四边形是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平行;梯形",
+        "solution":  "【分析】本题考查平行四边形和梯形的特征。根据平行四边形的特征可知，平行四边形的两组对边分别平行且相等;根据梯形的定义可知，只有一组对边平行的四边形叫做梯形。据此填空.\n【详解】根据平行四边形和梯形的概念及特点可知，平行四边形的两组对边互相平行且相等，平行四边形具有不稳定性;只有一组对边平行的四边形是梯形，不平行的一组对边是梯形的腰。",
+        "knowledge_points":  [
+                                 "梯形的概念及特点，平行四边形的概念及特点"
+                             ],
+        "source":  "19-20 四年级上·广东中山·期末"
+    },
+    {
+        "id":  "317",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "如果直线a与直线b平行，可以表示为( ),读作( );如果直线a与直线b互相垂直，可以表示为( ),读作( )。",
+        "options":  [
+
+                    ],
+        "answer":  "a//b;a平行于b;a⊥b;a垂直于b",
+        "solution":  "【分析】本题考查平行和垂直的符号表示及读法。根据平行线的表示方法，直线a与直线b平行记作a//b,读作a平行于b;根据垂直线的表示方法，直线a与直线b垂直记作a⊥b,读作a垂直于b。据此填空.\n【详解】在同一平面内的两条直线a与b,如果不相交，那么直线a与b互相平行，记作a//b,读作a平行于b。如果相交成直角，那么直线a与b互相垂直，记作a⊥b,读作a垂直于b。",
+        "knowledge_points":  [
+                                 "垂直的特征，平行的特征及性质"
+                             ],
+        "source":  "20-21四年级上·广东东莞·期末"
+    },
+    {
+        "id":  "318",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在同一个平面内，如果直线a和直线b都与直线c互相垂直，那么直线a和直线b互相( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平行",
+        "solution":  "【分析】根据垂直和平行的特征：两条直线都垂直于同条直线，那么这两条直线平行;进而解答即可。正确理解垂直和平行的特征及性质，是解答此题的关键.\n【详解】根据\u0027如果两条直线都垂直于同一条直线，那么这两条直线互相平行\u0027可知：在同一个平面内，如果直线a和直线b都与直线c互相垂直，那么直线a和直线b互相平行。",
+        "knowledge_points":  [
+                                 "平行的特征及性质，垂直的特征"
+                             ],
+        "source":  "19-20 四年级上·吉林四平·期末"
+    },
+    {
+        "id":  "319",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "已知AB//DC,在直线AB上任选两个不同点，分别向直线DC作垂线，那么这两条垂线的长度( )(相等/不相等)。",
+        "options":  [
+
+                    ],
+        "answer":  "相等",
+        "solution":  "【分析】两条平行线之间画若干条与平行线垂直的线段，这些垂直线段的长度都相等，依此填空。熟练掌握垂直与平行的特点及性质是解答此题的关键.\n【详解】已知AB//DC,在直线AB上任选两个不同点，分别向直线DC作垂线，那么这两条垂线的长度相等。",
+        "knowledge_points":  [
+                                 "平行的特征及性质，垂直的特征"
+                             ],
+        "source":  "2021四年级上·全国·专题练习"
+    },
+    {
+        "id":  "320",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "已知等腰梯形的周长为36厘米，上，下底分别为8厘米,12厘米，则腰长为( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "8",
+        "solution":  "【分析】等腰梯形的两条腰相等，因此用36分别减去上底和下底的长度后，再除以2即可。熟练掌握等腰梯形的特点是解答此题的关键.\n【详解】36-8-12=16(厘米),16÷2=8(厘米)。",
+        "knowledge_points":  [
+                                 "直角梯形和等腰梯形的概念及特点，梯形的周长"
+                             ],
+        "source":  "2021 四年级上·全国·专题练习"
+    },
+    {
+        "id":  "321",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "只有一组对边平行的四边形叫做( )。它有( )条高。",
+        "options":  [
+
+                    ],
+        "answer":  "梯形;无数",
+        "solution":  "【分析】本题考查梯形的定义及高的特征。根据梯形的定义可知，只有一组对边平行的四边形叫做梯形;根据梯形高的定义，从上底任意一点向下底作垂线，这点与垂足间的距离叫做梯形的高，因为上底上有无数个点，所以梯形有无数条高。据此填空.\n【详解】梯形的定义：只有一组对边平行的四边形叫做梯形;梯形高的含义：梯形虽然只有一组对边平行，但是，在这组对边里，也可以画无数条垂直线段，所以有无数条高，所以在梯形内能画出无数条高，且都相等。",
+        "knowledge_points":  [
+                                 "梯形的高及画法，梯形的概念及特点"
+                             ],
+        "source":  "19-20 四年级上·全国·期末"
+    },
+    {
+        "id":  "322",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个平行四边形的周长是24厘米，其中一条边是7厘米，与它相邻的两条边的长度都是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "5",
+        "solution":  "【分析】根据平行四边形的周长公式，首先用周长除以2求出一组邻边的长度和，然后用一组邻边的长度和减去已知的边长即可，据此解答。此题主要考查平行四边形的周长公式的灵活运用，关键是熟记公式.\n【详解】24÷2-7=12-7=5(厘米)。",
+        "knowledge_points":  [
+                                 "平行四边形的周长"
+                             ],
+        "source":  "20-21 四年级上·河北张家口·期末"
+    },
+    {
+        "id":  "323",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "两条直线相交成直角，就说这两条直线互相( ),其中一条直线叫做另一条直线的( ),这两条直线的( )叫做垂足。",
+        "options":  [
+
+                    ],
+        "answer":  "垂直;垂线;交点",
+        "solution":  "【分析】本题考查垂直的定义。根据垂直的概念可知，两条直线相交成直角，就说这两条直线互相垂直，其中一条直线叫做另一条直线的垂线，这两条直线的交点叫做垂足。据此填空.\n【详解】根据垂直的定义可知，两条直线相交成直角，就说这两条直线互相垂直，其中一条直线叫做另一条直线的垂线，这两条直线的交点叫垂足。例如:![](images/da614a0667fff4914e186fbfb8f8438350d98ec0fa3b2d61a7cbdd57c2b2236f.jpg)如图所示，直线AB和CD互相垂直,AB叫做CD的垂线,CD叫做AB的垂线，垂足是点D。",
+        "knowledge_points":  [
+                                 "垂直的特征"
+                             ],
+        "source":  "18-19 四年级上·海南海口·期末"
+    },
+    {
+        "id":  "324",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "等腰梯形,( )相等，它是一个( )图形。",
+        "options":  [
+
+                    ],
+        "answer":  "两腰长;轴对称",
+        "solution":  "【分析】等腰梯形上下两底平行，两腰相等，是轴对称图形，据此解答。本题考查等腰梯形的特征和轴对称图形的判断，属于基础知识，牢记即可.\n【详解】等腰梯形，两腰长相等，它是一个轴对称图形。",
+        "knowledge_points":  [
+                                 "直角梯形和等腰梯形的概念及特点，轴对称的认识及辨认"
+                             ],
+        "source":  "19-20 四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "325",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "平行四边形相邻的两条边分别是8cm和6cm,这个平行四边形的周长是( )cm。",
+        "options":  [
+
+                    ],
+        "answer":  "28",
+        "solution":  "【分析】根据平行四边形的特征，对边平行且相等，平行四边形相邻的两条边分别是8cm和6cm,周长是由两组对边组成，据此解答。本题考查平行四边形的特征和周长，掌握对边平行且相等是解题的关键.\n【详解】(8+6)×2=14×2=28(cm)。",
+        "knowledge_points":  [
+                                 "平行四边形的周长，平行四边形的概念及特点"
+                             ],
+        "source":  "19-20 四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "326",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "数学试卷相邻两条边互相( ),相对的两条边互相平行。",
+        "options":  [
+
+                    ],
+        "answer":  "垂直",
+        "solution":  "【分析】数学书卷是长方形，根据长方形的特征，对边平行且相等,4个角都是直角，据此解答。本题考查垂直的知识，掌握长方形和垂直的特性是解题的关键.\n【详解】如图所示:![](images/960a78c3d1ff2985942632e61f9b3c962e195cbf96e1d5199c8f5ba66c0e6156.jpg)线段L与线段M互相垂直，故数学试卷相邻两条边互相垂直，相对的两条边互相平行。",
+        "knowledge_points":  [
+                                 "垂直的特征，长方形的概念及特点"
+                             ],
+        "source":  "19-20 四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "327",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在两条平行线之间，可以画( )条垂直线段，这些垂直线段的长度( )。垂直于同一条直线的两条直线互相( )。",
+        "options":  [
+
+                    ],
+        "answer":  "无数;相等;平行",
+        "solution":  "【分析】同一平面内不相交的两条直线互相平行，其中一条直线是另一条直线的平行线。两直线互相平行时，从一条直线上任意一点向另一条直线作垂线，所得的平行线间的垂直线段的长度，叫做平行线间的距离。平行线之间的距离处处相等。垂直于同一条直线的两直线互相平行。本题主要考查学生对平行与垂直概念，特征的掌握和灵活运用.\n【详解】在两条平行线之间，可以画无数条垂直线段，这些垂直线段的长度相等。垂直于同一条直线的两条直线互相平行。",
+        "knowledge_points":  [
+                                 "平行的特征及性质，垂直的特征"
+                             ],
+        "source":  "19-20 四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "328",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "( )对边平行的四边形叫梯形。",
+        "options":  [
+
+                    ],
+        "answer":  "只有一组",
+        "solution":  "【详解】只有一组对边平行的四边形叫梯形，常见的梯形有直角梯形，等腰梯形。",
+        "knowledge_points":  [
+                                 "梯形的概念及特点"
+                             ],
+        "source":  "20-21 四年级上·新疆·期末"
+    },
+    {
+        "id":  "329",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "两根小棒都和第三根小棒互相垂直，这两根小棒的位置关系是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "互相平行",
+        "solution":  "【分析】本题考查梯形的定义。根据梯形的概念可知，只有一组对边平行的四边形叫做梯形。据此填空.\n【分析】同一平面内，垂直于同一直线的两条垂线互相平行，据此即可解答。本题主要考查学生对平行定义和特征的掌握和灵活运用.\n【详解】根据分析可知，两根小棒都和第三根小棒互相垂直，这两根小棒的位置关系是互相平行。",
+        "knowledge_points":  [
+                                 "平行的特征及性质"
+                             ],
+        "source":  "20-21 四年级上·新疆·期末"
+    },
+    {
+        "id":  "330",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰梯形上底长12cm,下底长30cm,腰长15cm,这个梯形的周长为( )cm。",
+        "options":  [
+
+                    ],
+        "answer":  "72",
+        "solution":  "【分析】等腰梯形的周长=上底+下底+腰*2，代入数据计算即可。熟练掌握梯形的周长公式是解决本题的关键.\n【详解】12+30+15×2=12+30+30=72(cm),则这个梯形的周长为72cm。",
+        "knowledge_points":  [
+                                 "梯形的周长"
+                             ],
+        "source":  "20-21四年级上·湖南长沙·期末"
+    },
+    {
+        "id":  "331",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个梯形，上底是4厘米，如果将上底延长2厘米，这个梯形就变成了正方形。这个梯形的下底是( )厘米，变成的正方形面积是( )平方厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "6;36",
+        "solution":  "【分析】上底是4厘米，如果将上底延长2厘米，则上底变成了4+2=6(厘米),此时这个梯形就变成了一个正方形，因为正方形的四条边都相等，所以梯形的下底等于正方形的边长，即6厘米，变成正方形后四条边都为6厘米，根据正方形的面积=边长×边长，代入数据进行计算即可。此题考查了梯形的特征以及正方形的面积公式的应用，明确正方形的特征，是解答此题的关键.\n【详解】下底:4+2=6(厘米);正方形的面积:6×6=36(平方厘米)。",
+        "knowledge_points":  [
+                                 "梯形的概念及特点，正方形的面积"
+                             ],
+        "source":  "20-21四年级上·浙江杭州·期末"
+    },
+    {
+        "id":  "332",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个梯形，上底是3厘米，如果将上底延长5厘米，这个梯形就变成了正方形。这个梯形的下底是( )厘米，变成的正方形面积是( )平方厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "8;64",
+        "solution":  "【分析】上底是3厘米，如果将上底延长5厘米，则上底变成了3+5=8(厘米),此时这个梯形就变成了一个正方形，因为正方形的四条边都相等，所以梯形的下底等于正方形的边长，即8厘米，变成正方形后四条边都为8厘米，根据正方形的面积=边长×边长，代入数据进行计算即可。此题考查了梯形的特征以及正方形的面积公式的应用，明确正方形的特征，是解答此题的关键.\n【详解】下底:3+5=8(厘米);正方形的面积:8×8=64(平方厘米)。",
+        "knowledge_points":  [
+                                 "梯形的概念及特点，正方形的面积"
+                             ],
+        "source":  "22-23 四年级上·江苏苏州·期末"
+    },
+    {
+        "id":  "333",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小红在一张纸上画了三条直线a,b,c,其中a⊥c,b⊥c,那么a与b互相( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平行",
+        "solution":  "【分析】![](images/5ca22dfd6d0fd09dc95b614c2fd83eb1d98c62548983ab6f33d128910c31cf47.jpg)根据垂直的性质可知，同一平面内,a⊥c,b⊥c,则a//b。即同一平面内，垂直于同一条直线的两条直线互相平行。据此解答即可。同一平面内，垂直于同一条直线的两条直线互相平行，画图能帮助学生更好的理解.\n【详解】a⊥c,b⊥c,那么a与b互相平行。",
+        "knowledge_points":  [
+                                 "垂直的特征，平行的特征及性质"
+                             ],
+        "source":  "19-20 四年级上·浙江宁波·期末"
+    },
+    {
+        "id":  "334",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "等腰梯形有( )条高，它们的长度都( )。",
+        "options":  [
+
+                    ],
+        "answer":  "无数;相等",
+        "solution":  "【分析】从梯形一条底边上的一点到它对边的垂直线段叫做梯形的高。梯形有2条底，有无数条高。高一般用虚线表示，并画上垂足符号，据此解答。所有的梯形都有无数条高并都相等.\n【详解】如下图:![](images/eb8b4dcec2a0bb4b2835ba7950f0943d7f50eabd6f4df7e63599afda63a48f26.jpg)从梯形一条底边上的一点到它对边的垂直线段是梯形的高，梯形上底和下底之间的距离是固定的，所以这样的高有无数条，并且都相等。",
+        "knowledge_points":  [
+                                 "梯形的高及画法"
+                             ],
+        "source":  "19-20 四年级上·浙江宁波·期末"
+    },
+    {
+        "id":  "335",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "学校的拉伸门利用了平行四边形的( )性质。",
+        "options":  [
+
+                    ],
+        "answer":  "容易变形",
+        "solution":  "【分析】根据平行四边形的不稳定性填空即可。本题考查的是平行四边形的不稳定性。还需注意细心观察生活中的平行四边形.\n【详解】平行四边形具有易变形、不稳定性等特点，学校的拉伸门应用了容易变形的性质，像这样的还有伸缩衣帽架、升降台云梯等等。![](images/de65a27299505bb8e69a8f51efdfdb8ba428599a1ebe0903ac1094689f97867b.jpg)",
+        "knowledge_points":  [
+                                 "平行四边形的不稳定性及应用"
+                             ],
+        "source":  "20-21 四年级上·北京朝阳·期末"
+    },
+    {
+        "id":  "336",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个等腰梯形的周长是48厘米，上底和下底分别是12厘米和14厘米，每条腰长( )。",
+        "options":  [
+
+                    ],
+        "answer":  "11厘米",
+        "solution":  "【分析】因为梯形的周长=两腰长度+上底+下底，又根据等腰梯形的特点，两腰相等，所以一条腰的长度=(周长-上底-下底)÷2，代入数据计算即可。此题考查的目的是理解掌握等腰梯形的特征以及梯形周长公式的灵活运用.\n【详解】(48-12-14)÷2=(36-14)÷2=22÷2=11(厘米)",
+        "knowledge_points":  [
+                                 "梯形的周长，直角梯形和等腰梯形的概念及特点"
+                             ],
+        "source":  "19-20 四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "337",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明在纸上画了一条直线a,然后又画了一条直线b,使得b⊥a,最后画了一条直线c(三条直线均不重合),使得c⊥a。那么直线b和直线c的位置关系是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平行",
+        "solution":  "【分析】根据平行的性质：同一平面内两条直线同时垂直于一条直线，那么，这两条直线相互平行;据此解答。此题考查了垂直和平行的性质，应注意积累和理解.\n【详解】根据同一平面内两条直线的位置关系可知：小明在纸上画了一条直线a,然后又画了一条直线b,使得b⊥a,最后画了一条直线c(三条直线均不重合),使得c⊥a,那么直线b和直线c的位置关系是平行。",
+        "knowledge_points":  [
+                                 "垂直的特征，平行的特征及性质"
+                             ],
+        "source":  "20-21四年级上·湖北武汉·期末"
+    },
+    {
+        "id":  "338",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "只有一组对边平行的四边形叫( )。不平行的一组对边是它的( )。",
+        "options":  [
+
+                    ],
+        "answer":  "梯形;腰",
+        "solution":  "【分析】![](images/cc2cc86444122ecef87f7602c0aece8bfe665fd84c812e6014938764e1c30daa.jpg)只有一组对边平行的四边形叫做梯形。互相平行的一组对边分别是梯形的上底和下底，不平行的一组对边是梯形的腰。从梯形一条底边上的一点到它对边的垂直线段叫做梯形的高。本题考查学生对梯形的定义的认识，需熟练掌握.\n【详解】只有一组对边平行的四边形叫梯形。不平行的一组对边是它的腰。",
+        "knowledge_points":  [
+                                 "梯形的概念及特点"
+                             ],
+        "source":  "20-21 四年级上·湖北武汉·期末"
+    },
+    {
+        "id":  "339",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.4,
+        "question":  "一个梯形的下底长是上底的4倍，如果将上底延长12厘米，就成了一个平行四边形，这个梯形的下底是( )厘米，上底是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "16;4",
+        "solution":  "【分析】一个梯形的下底长是上底的4倍，如果将上底延长12厘米，就成了一个平行四边形，说明上下底长度差为12厘米，倍数差为4-1=3，长度差除以倍数差等于上底，上底乘4等于下底。熟练掌握差倍问题解题方法是解答本题的关键.\n【详解】12÷(4-1)=12÷3=4(厘米),4×4=16(厘米)",
+        "knowledge_points":  [
+                                 "和差倍问题"
+                             ],
+        "source":  "18-19四年级上·广东汕头·期末"
+    },
+    {
+        "id":  "340",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个正方形的周长是32分米，它的边长是( )分米;一个平行四边形的两条邻边分别长10厘米和8厘米，它的周长是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "8;36",
+        "solution":  "【分析】正方形的边长=周长/4，据此求出正方形的边长。平行四边形的周长为相邻两条边长度和乘2，据此解答即可。熟练掌握正方形和平行四边形的周长公式，灵活运用公式解决问题.\n【详解】32÷4=8(分米),(10+8)×2=18×2=36(厘米),则平行四边形的周长是36厘米。",
+        "knowledge_points":  [
+                                 "正方形的周长，平行四边形的周长"
+                             ],
+        "source":  "18-19 三年级上·广东汕头·期末"
+    },
+    {
+        "id":  "341",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用一根长40厘米的铁丝围成一个等腰梯形，这个梯形的上，下底共长26厘米，它的一条腰长为( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "7",
+        "solution":  "【分析】等腰梯形的周长=上底+下底+2*腰，则等腰梯形的腰=[周长-(上底+下底)]/2，代入数据计算即可。等腰梯形的两条腰相等，熟练掌握等腰梯形的周长公式，灵活运用公式解决问题.\n【详解】(40-26)÷2=14÷2=7(厘米),则它的一条腰长为7厘米。",
+        "knowledge_points":  [
+                                 "梯形的周长，直角梯形和等腰梯形的概念及特点"
+                             ],
+        "source":  "19-20四年级上·福建厦门·期末"
+    },
+    {
+        "id":  "342",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "长方形的两条邻边互相( ),两条对边互相( )且( )。",
+        "options":  [
+
+                    ],
+        "answer":  "垂直;平行;相等",
+        "solution":  "【分析】两组对边分别平行，四个角都是直角的四边形叫做长方形，长方形两组对边平行且相等。熟练掌握长方形的定义和特征是解答本题的关键.\n【详解】根据分析可知，长方形的两条邻边互相垂直，两条对边互相平行且相等。",
+        "knowledge_points":  [
+                                 "长方形的概念及特点"
+                             ],
+        "source":  "19-20 四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "343",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "两组对边分别平行且没有直角的四边形是( );只有一组对边平行的四边形是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平行四边形;梯形",
+        "solution":  "【分析】本题考查平行四边形和梯形的定义。根据平行四边形的特征，两组对边分别平行的四边形是平行四边形;根据梯形的定义，只有一组对边平行的四边形是梯形。据此填空.\n【详解】根据平行四边形的定义：两组对边分别平行且没有直角的四边形是平行四边形;根据梯形的定义：只有一组对边平行的四边形是梯形。",
+        "knowledge_points":  [
+                                 "平行四边形的概念及特点，梯形的概念及特点"
+                             ],
+        "source":  "19-20 四年级上·全国·期末"
+    },
+    {
+        "id":  "344",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个平行四边形的相邻边长度和是14厘米，这个平行四边形的周长是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "28",
+        "solution":  "【分析】平行四边形的两组对边分别相等，所以它的周长是相邻两边之和乘2。据此解答即可。熟练掌握平行四边形的周长公式是解决本题的关键.\n【详解】2×14=28(厘米),则这个平行四边形的周长是28厘米。",
+        "knowledge_points":  [
+                                 "平行四边形的周长"
+                             ],
+        "source":  "19-20四年级上·福建福州·期末"
+    },
+    {
+        "id":  "345",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "这份试卷卷面上相邻的边互相( ),相对的边互相( )。",
+        "options":  [
+
+                    ],
+        "answer":  "垂直;平行",
+        "solution":  "【分析】因为这份试卷的卷面是一个长方形，所以根据长方形的特征：对边平行且相等,4个角都是直角可知，长方形相邻的两条边互相垂直，相对的两边互相平行;据此解答。解答此题的主要依据是：长方形的邻边互相垂直，对边互相平行.\n【详解】根据长方形的特征可知：这份试卷卷面上相邻的边互相垂直，相对的边互相平行。",
+        "knowledge_points":  [
+                                 "长方形的概念及特点，垂直的特征，平行的特征及性质"
+                             ],
+        "source":  "20-21 四年级上·福建福州·期末"
+    },
+    {
+        "id":  "346",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "课桌面的形状是长方形，它相邻的两条边互相( ),相对的两条边互相( )。",
+        "options":  [
+
+                    ],
+        "answer":  "垂直;平行",
+        "solution":  "【分析】因为课桌面是长方形，根据长方形的特征：对边平行且相等,4个角都是直角可知，长方形相邻的两条边互相垂直，相对的两条边互相平行;据此解答。解答此题的主要依据是：长方形的邻边互相垂直，对边互相平行.\n【详解】长方形的特征是对边平行且相等，四个角都是直角。因此，长方形相邻的两条边互相垂直，相对的两条边互相平行。",
+        "knowledge_points":  [
+                                 "长方形的概念及特点，垂直的特征，平行的特征及性质"
+                             ],
+        "source":  "21-22 四年级上·江苏南京·期末"
+    },
+    {
+        "id":  "347",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "如果两条直线相交成( ),就说这两条直线互相垂直。这两条直线的交点叫做( )。",
+        "options":  [
+
+                    ],
+        "answer":  "直角;垂足",
+        "solution":  "【分析】本题考查垂直的定义。根据垂直的概念可知，当两条直线相交成直角时，这两条直线互相垂直，它们的交点叫做垂足。据此填空.\n【详解】由互相垂直的定义可知：两条直线相交成直角时，这两条直线互相垂直，其中一条直线是另一条直线的垂线，这两条直线的交点叫做垂足。",
+        "knowledge_points":  [
+                                 "垂直的特征"
+                             ],
+        "source":  "2014 四年级·全国·课后作业"
+    },
+    {
+        "id":  "348",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "同一平面内，垂直于同一条直线的两条直线( );平行于同一条直线的两条直线( )。",
+        "options":  [
+
+                    ],
+        "answer":  "互相平行;互相平行",
+        "solution":  "【分析】![](images/96b1f521f6a8440212192831e81fdb6e2937a74782e2f345afcb85ab5ba26e45.jpg)根据垂直的性质可知，同一平面内,a⊥c,b⊥c,则a//b。即同一平面内，垂直于同一条直线的两条直线互相平行。![](images/89dc93efafa5ce7e9af1103a8f178962314ee12098614fb685f97e10c6c126ee.jpg)根据平行的性质可知，同一平面内,a//c,b//c,则a//b。同一平面内，平行于同一条直线的两条直线互相平行。熟练掌握平行和垂直的定义是解决本题的关键，画图能帮助学生更好的理解.\n【详解】同一平面内，垂直于同一条直线的两条直线互相平行;平行于同一条直线的两条直线互相平行。",
+        "knowledge_points":  [
+                                 "平行的特征及性质，垂直的特征"
+                             ],
+        "source":  "2020四年级上·全国·专题练习"
+    },
+    {
+        "id":  "349",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个平行四边形的相邻边长度和是10厘米，这个平行四边形的周长是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "20厘米",
+        "solution":  "【分析】平行四边形的对边平行且相等，平行四边形的周长等于相邻边长度和乘2，据此解答。本题主要考查了平行四边形的特征和性质，理解相邻边长度和是平行四边形的一条长和一条宽之和是解题关键.\n【详解】10×2=20(厘米)",
+        "knowledge_points":  [
+                                 "平行四边形的周长"
+                             ],
+        "source":  "18-19 四年级上·浙江杭州·期末"
+    },
+    {
+        "id":  "350",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "四边形中是轴对称图形的有( ),( )等。",
+        "options":  [
+
+                    ],
+        "answer":  "长方形;正方形",
+        "solution":  "【分析】一个图形沿一条直线对折，直线两旁的图形完全重合，这样的图形叫做轴对称图形，折痕所在的直线就是对称轴;据此可知，正方形和长方形是轴对称图形，梯形和平行四边形不是轴对称图形。判断一个图案是否是轴对称图形的关键是看在这个图形中能否找到一条直线，使图形沿着这条直线对折后能够完全重合.\n【详解】四边形中是轴对称图形的有长方形，正方形，等腰梯形等。",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "20-21 四年级上·全国·期末"
+    },
+    {
+        "id":  "351",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "过直线外一点可以作( )条直线与已知直线平行。过直线上一点可以作( )条直线与已知直线垂直。",
+        "options":  [
+
+                    ],
+        "answer":  "1;1",
+        "solution":  "【分析】过直线外一点，有且只有一条直线与这条直线平行。这两条直线互相平行，其中一条直线是另一条直线的平行线。过一点有且只有一条直线与已知直线垂直。从直线外一点到这条直线的线段中，垂直线段最短，这条垂直线段的长度叫做点到直线的距离。本题考查平行和垂直的性质，需熟练掌握.\n【详解】过直线外一点可以作1条直线与已知直线平行。过直线上一点可以作1条直线与已知直线垂直。",
+        "knowledge_points":  [
+                                 "平行的特征及性质，垂直的特征"
+                             ],
+        "source":  "2014 四年级·全国·课后作业"
+    },
+    {
+        "id":  "352",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "两组对边分别平行的四边形叫作( ),两腰相等的梯形叫作( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平行四边形;等腰梯形",
+        "solution":  "【分析】本题考查平行四边形和等腰梯形的定义。根据平行四边形的特征可知，两组对边分别平行的四边形叫作平行四边形;根据等腰梯形的概念可知，两腰相等的梯形叫作等腰梯形。据此填空.\n【详解】1，平行四边形的定义：在同一平面内，由两组平行线段组成的闭合图形。2，梯形的定义：只有一组对边平行的四边形叫做梯形。如果梯形的两腰相等，这样的梯形叫做等腰梯形。两组对边分别平行的四边形叫作(平行四边形),两腰相等的梯形叫作(等腰梯形)。",
+        "knowledge_points":  [
+                                 "直角梯形和等腰梯形的概念及特点，平行四边形的概念及特点"
+                             ],
+        "source":  "19-20 四年级上·甘肃陇南·期末"
+    },
+    {
+        "id":  "353",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用一条长76厘米的铁丝围成一个平行四边形，已知其中一条边的长是16厘米，与它相邻的一条边的长是( )厘米;如果把这条铁丝围成上底与下底相加的和是26厘米的等腰梯形，这个梯形的腰长( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "22;25",
+        "solution":  "【分析】根据平行四边形的特性可知：对边平行且相等，周长是76厘米，一条边是16厘米，相邻的一条边=周长/2-16;根据等腰梯形的特性可知：两腰相等，周长是76厘米，上底加下底是26厘米，腰长=(周长-上底加下底)÷2即可解答。本题考查平行四边形与等腰梯形的特征是解题关键.\n【详解】76÷2-16=38-16=22(厘米),(76-26)÷2=50÷2=25(厘米)",
+        "knowledge_points":  [
+                                 "平行四边形的周长，梯形的周长"
+                             ],
+        "source":  "19-20四年级上·浙江宁波·期末"
+    },
+    {
+        "id":  "354",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "通过这学期的学习，你一定知道了长方形和正方形不仅两组对边分别( )并且邻边互相( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平行;垂直",
+        "solution":  "【分析】两组对边分别平行，四个角都是直角的四边形叫做长方形。四条边相等，四个角都是直角的四边形叫做正方形。平行和垂直的定义：两条直线相交成直角时，这两条直线互相垂直。同一平面内不相交的两条直线互相平行。据此解答即可。本题考查学生对平行和垂直定义的理解和掌握.\n【详解】通过这学期的学习，你一定知道了长方形和正方形不仅两组对边分别平行并且邻边互相垂直。",
+        "knowledge_points":  [
+                                 "平行的特征及性质，垂直的特征"
+                             ],
+        "source":  "19-20 四年级上·浙江宁波·期末"
+    },
+    {
+        "id":  "355",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个平行四边形的周长是56厘米，其中一条边长10厘米，与这条边相交的边长( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "18",
+        "solution":  "【分析】平行四边形的周长为两条相邻边的和的2倍，则另一条边的边为56÷2-10厘米。熟练掌握平行四边形的周长公式，灵活运用公式解决问题.\n【详解】56÷2-10=28-10=18(厘米),则与这条边相交的边长18厘米。",
+        "knowledge_points":  [
+                                 "平行四边形的周长"
+                             ],
+        "source":  "19-20四年级上·甘肃兰州·期末"
+    },
+    {
+        "id":  "356",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "梯形有( )组对边互相平行，长方形的邻边互相( )。",
+        "options":  [
+
+                    ],
+        "answer":  "一;垂直",
+        "solution":  "【分析】本题主要考查梯形和长方形的基础几何特征，属于四年级上册平面图形的核心考点。解题需紧扣两种图形的定义：梯形是只有一组对边互相平行的四边形，这是梯形区别于其他四边形的关键特征;长方形是特殊的平行四边形，它的对边互相平行且相等，四个角都是直角，根据直角的定义，相邻的两条边互相垂直，结合这两个核心特征即可得出答案.\n【详解】根据梯形的定义得出在一个梯形中只有一组对边是互相平行的;根据长方形的特征：在长方形中有两组对边是平行的，两条邻边互相垂直。如图所示:![](images/e614b46937c2f40dbe614ed362a1ee614598e1ee0de0cfc975f8bbc90af05313.jpg)![](images/6c992d23f1897a9b62a8826ba2d735a69b5da6e4ad21b4e744e5e60d56fc3383.jpg)",
+        "knowledge_points":  [
+                                 "梯形的概念及特点，长方形的概念及特点"
+                             ],
+        "source":  "19-20 四年级上·新疆·期末"
+    },
+    {
+        "id":  "357",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "梯形有( )组对边互相平行，它有( )条高。",
+        "options":  [
+
+                    ],
+        "answer":  "一;无数",
+        "solution":  "【分析】根据梯形的定义：只有一组对边平行的四边形是梯形，在梯形的上底任意一点作垂直于下底的线段叫高，梯形有无数条高。据此解答.\n【详解】梯形有一组对边互相平行，它有无数条高。",
+        "knowledge_points":  [
+                                 "梯形的高及画法，梯形的概念及特点"
+                             ],
+        "source":  "19-20 四年级上·山东临沂·期末"
+    },
+    {
+        "id":  "358",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "两组对边分别平行，且有四个直角的图形是( )形或( )形。",
+        "options":  [
+
+                    ],
+        "answer":  "长方;正方",
+        "solution":  "【分析】两组对边分别平行，学过的图形中只有平行四边形，而有四个直角说明是特殊的平行四边形长方形或正方形，据此解答。本题考查图形的认知，掌握平行四边形和长方形，正方形的特性是解题的关键.\n【详解】两组对边分别平行，且有四个直角的图形是长方形或正方形。",
+        "knowledge_points":  [
+                                 "正方形的概念及特点，长方形的概念及特点，平行四边形的概念及特点"
+                             ],
+        "source":  "19-20 四年级上·山东临沂·期末"
+    },
+    {
+        "id":  "359",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个平行四边形相邻两边的和是26厘米，它的周长是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "52",
+        "solution":  "【分析】平行四边形的周长是相邻两边的和的2倍，据此解答即可。本题考查学生对平行四边形周长的认识和掌握情况.\n【详解】26×2=52(厘米),则它的周长是52厘米。",
+        "knowledge_points":  [
+                                 "平行四边形的周长"
+                             ],
+        "source":  "19-20 四年级上·山东临沂·期末"
+    },
+    {
+        "id":  "360",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "经过两点可以画( )条直线;两条直线相交成( )角时，这两条直线互相垂直。",
+        "options":  [
+
+                    ],
+        "answer":  "一;直",
+        "solution":  "【分析】根据直线的性质可知，通过两点最多可画1条直线。通过一点可画无数条直线。两条直线相交成直角时，这两条直线互相垂直，其中一条直线是另一条直线的垂线，这两条直线的交点叫做垂足。熟练掌握直线的性质以及垂直的定义.\n【详解】经过两点可以画一条直线;两条直线相交成直角时，这两条直线互相垂直。",
+        "knowledge_points":  [
+                                 "垂直的特征，线段，直线，射线的认识及特征"
+                             ],
+        "source":  "19-20 四年级上·山东临沂·期末"
+    },
+    {
+        "id":  "361",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "长方形相对的两条边( ),正方形的两条对角线( )。",
+        "options":  [
+
+                    ],
+        "answer":  "互相平行;互相垂直",
+        "solution":  "【分析】两组对边分别平行，四个角都是直角的四边形叫做长方形。长方形两组对边平行且相等，两条对角线相等且互相平分，长方形是特殊的平行四边形;四条边相等，四个角都是直角的四边形叫做正方形。正方形是特殊的长方形，也是特殊的平行四边形，正方形的对角线相等且互相垂直。本题主要考查学生对长方形和正方形的概念和特征的掌握.\n【详解】根据分析可知，长方形相对的两条边互相平行，正方形的两条对角线互相垂直。",
+        "knowledge_points":  [
+                                 "长方形的概念及特点，正方形的概念及特点"
+                             ],
+        "source":  "20-21 四年级上·全国·期末"
+    },
+    {
+        "id":  "362",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个平行四边形的周长是36厘米，其中一条边的长是10厘米，与它相邻的一条边的长是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "8",
+        "solution":  "【分析】根据平行四边形的特点，对边相等可得，平行四边形的周长的求解方法与长方形相似，都是相邻两条边的和的2倍，由此先用周长36厘米除以2，求出相邻两边的和，再减去其中的一条边10厘米，即可求出另一条边。熟知平行四边形的特点，找出其周长的计算方法是解答本题的关键.\n【详解】36÷2-10=18-10=8(厘米)",
+        "knowledge_points":  [
+                                 "平行四边形的周长，平行四边形的概念及特点"
+                             ],
+        "source":  "20-21 四年级上·全国·期末"
+    },
+    {
+        "id":  "363",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "直线a,b,c在同一个平面内，若a⊥b,b//c,则a( )c。",
+        "options":  [
+
+                    ],
+        "answer":  "⊥",
+        "solution":  "【分析】![](images/15459f6b70b6f3217b75e2b73f8cdd1cdd83775f2b0151477626a5840f3d58cb.jpg)如图所示:a⊥b,b//c,则∠1=∠2=90°。两条直线相交成直角时，这两条直线互相垂直，则a⊥c。本题考查平行和垂直的性质，需熟练掌握。画图能帮助学生更好理解图形间的关系.\n【详解】直线a,b,c在同一个平面内，若a⊥b,b//c,则a⊥c。",
+        "knowledge_points":  [
+                                 "垂直的特征，平行的特征及性质"
+                             ],
+        "source":  "19-20 四年级上·福建厦门·期末"
+    },
+    {
+        "id":  "364",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "过直线外一点可以画( )条直线与这条直线垂直，可以画( )条直线与这条直线平行。",
+        "options":  [
+
+                    ],
+        "answer":  "一;一",
+        "solution":  "【分析】根据平行公理\u0027经过直线外一点，有且只有一条直线与已知直线平行\u0027和垂线的性质\u0027在同一平面内，过一点(直线外或直线上)有且只有一条直线垂直于已知直线\u0027进行解答即可。此题考查了直线的性质及平行，垂直的性质，注意基础知识的熟记和理解.\n【详解】过直线外一点可以画一条直线与这条直线垂直，可以画一条直线与这条直线平行。",
+        "knowledge_points":  [
+                                 "画垂线，画平行线"
+                             ],
+        "source":  "18-19 四年级上·河北·单元测试"
+    },
+    {
+        "id":  "365",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把一个长方形木框架沿着对角向外拉，可以拉成一个( ),这时图形的( )变了,( )没变。",
+        "options":  [
+
+                    ],
+        "answer":  "平行四边形;面积;周长",
+        "solution":  "【分析】把长方形木框架拉成平行四边形，四个边的长度没变，则其周长不变;但是它的高变短了，所以它的面积就变小了。此题主要考查长方形和平行四边形面积公式的灵活应用.\n【详解】把长方形木框架拉成平行四边形，四个边的长度没变，则其周长不变;长方形的面积=长×宽，平行四边形的面积=底×高，本题中，长=底，宽\u003e高，所以长方形的面积\u003e平行四边形的面积;即：把一个长方形框架沿着对角向外拉，可拉成一个平行四边形，这时图形的面积变了，周长没变。",
+        "knowledge_points":  [
+                                 "平行四边形的周长，长方形的面积，平行四边形的面积，长方形的周长"
+                             ],
+        "source":  "20-21四年级上·全国·单元测试"
+    },
+    {
+        "id":  "366",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "挂衣架，伸缩铁门等通常都是利用了平行四边形( )的特性而设计的。",
+        "options":  [
+
+                    ],
+        "answer":  "易变形",
+        "solution":  "【分析】平行四边形的不稳定性是指平行四边形边长确定，其形状，大小不能完全确定，易变形。据此解答即可。本题考查平行四边形的不稳定性，这些特性在生活中有着广泛的应用，比如千斤顶，折叠座椅等.\n【详解】挂衣架，伸缩铁门等通常都是利用了平行四边形易变形的特性而设计的。",
+        "knowledge_points":  [
+                                 "平行四边形的不稳定性及应用"
+                             ],
+        "source":  "20-21 四年级上·全国·单元测试"
+    },
+    {
+        "id":  "367",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "两把直尺交叉斜放，尺子重叠部分的图形是( )形。",
+        "options":  [
+
+                    ],
+        "answer":  "平行四边",
+        "solution":  "【分析】直尺的对边平行，则由两把直尺重叠形成的图形中，有两组互相平行的对边。根据平行四边形的定义可知，这个重叠部分的图形应是一个平行四边形。熟练掌握平行四边形的定义：平行四边形是在由两组平行线段组成的闭合图形.\n【详解】两把直尺交叉斜放，尺子重叠部分的图形是平行四边形。",
+        "knowledge_points":  [
+                                 "平行四边形的概念及特点"
+                             ],
+        "source":  "19-20 四年级上·福建厦门·期末"
+    },
+    {
+        "id":  "368",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "两条直线相交，有一个角是直角，那么其余三个角一定是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "直角",
+        "solution":  "【分析】两条直线相交成直角时，这两条直线互相垂直，其中一条直线是另一条直线的垂线，这两条直线的交点叫做垂足。据此解答即可。熟练掌握垂直的性质是解决本题的关键.\n【详解】两条直线相交，如果一个角是直角，那么其余的三个角肯定都是直角。",
+        "knowledge_points":  [
+                                 "垂直的特征"
+                             ],
+        "source":  "20-21四年级上·全国·单元测试"
+    },
+    {
+        "id":  "369",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "学过的四边形有哪四个( )。",
+        "options":  [
+
+                    ],
+        "answer":  "正方形、长方形、平行四边形、梯形",
+        "solution":  "【分析】本题考查对常见四边形的认识。根据所学内容，常见的四边形包括长方形，正方形，平行四边形和梯形，它们都具有四条边和四个角，但各自具有不同的特征。据此填空.\n【详解】学过的四边形有正方形，长方形，平行四边形，梯形。其中长方形的对边平行且相等,4个角都是直角。正方形对边平行且相等,4条边相等,4个角都是直角。平行四边形对边平行且相等。梯形只有一组对边平行。",
+        "knowledge_points":  [
+                                 "平行四边形的概念及特点，长方形的概念及特点，正方形的概念及特点，梯形的概念及特点"
+                             ],
+        "source":  "20-21 四年级上·全国·单元测试"
+    },
+    {
+        "id":  "370",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "两组对边分别平行的四边形叫做( );如果它的四个角都是直角，它就是( );如果此时量得它的四条边都相等，它就是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平行四边形;长方形;正方形",
+        "solution":  "【分析】平行四边形的对边平行且相等;长方形的对边平行且相等，邻边互相垂直,4个角都是直角;正方形对边平行且相等，邻边互相垂直且相等,4条边相等,4个角都是直角。据此解答即可。正方形是特殊的长方形，长方形是特殊的平行四边形.\n【详解】两组对边分别平行的四边形叫做平行四边形;如果它的四个角都是直角，它就是长方形;如果此时量得它的四条边都相等，它就是正方形。",
+        "knowledge_points":  [
+                                 "平行四边形的概念及特点，长方形的概念及特点，正方形的概念及特点"
+                             ],
+        "source":  "20-21 四年级上·全国·单元测试"
+    },
+    {
+        "id":  "371",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "从平行四边形的一个顶点出发作一条高(高在平行四边形内)。可把平行四边形分成一个( )和一个( )。",
+        "options":  [
+
+                    ],
+        "answer":  "三角形;直角梯形",
+        "solution":  "【分析】作出图形即可解答:![](images/1a75e44c77d4cad0f3bc0ff5aaa97d4008e3418be8bd71ef4e116338e768d420.jpg)本题考查了平行四边形的切拼，重在考查学生动手操作的能力，通过作图，使学生加深了对图形的理解.\n【详解】从平行四边形的一个顶点出发作一条高(高在平行四边形内)。可以把平行四边形分成一个三角形和一个直角梯形。",
+        "knowledge_points":  [
+                                 "平行四边形的切拼"
+                             ],
+        "source":  "20-21 四年级上·全国·单元测试"
+    },
+    {
+        "id":  "372",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把一张长方形纸对折两次，折痕间的关系是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "互相平行或互相垂直",
+        "solution":  "【分析】把一张长方形的纸对折两次，两次折痕的位置关系，取决于对折的方向，一种情况是沿一条边的同一个方向对折两次，三条折痕是互相平行的;另一种情况是沿两条边的两个方向对折，两次的折痕是互相垂直的;据此即可解答。解答本题的关键在于要从不同的折叠方向考虑.\n【详解】由分析可知：把一张长方形的纸对折两次后，折痕的关系是可能互相平行，也可能互相垂直。",
+        "knowledge_points":  [
+                                 "长方形的概念及特点，平行的特征及性质，垂直的特征"
+                             ],
+        "source":  "20-21 四年级上·全国·单元测试"
+    },
+    {
+        "id":  "373",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "花店里有515枝花，做一个花篮要用23枝花，能做( )个花篮，再添上( )枝花又可以做一个花篮。",
+        "options":  [
+
+                    ],
+        "answer":  "22;14",
+        "solution":  "【分析】用515÷23得到的商即能做多少个花篮，再看余数是多少，余数加上多少等于23，就添上多少枝。本题考查的是除数是两位数有余除法的实际应用，注意相除得到的商和余数在该题中具有什么样的实际意义.\n【详解】515÷23=22(个)。9(枝),23-9=14(枝),所以能做22个花篮，再添上14枝花又可以做一个花篮。",
+        "knowledge_points":  [
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "20-21 四年级上·全国·单元测试"
+    },
+    {
+        "id":  "374",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "我们这张试卷的对边互相( ),邻边互相( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平行;垂直",
+        "solution":  "【分析】两条直线相交成直角时，这两条直线互相垂直，其中一条直线是另一条直线的垂线，这两条直线的交点叫做垂足。同一平面内不相交的两条直线互相平行，其中一条直线是另一条直线的平行线。据此解答即可。熟练掌握平行和垂直的定义是解决本题的关键.\n【详解】我们这张试卷的对边互相平行，邻边互相垂直。",
+        "knowledge_points":  [
+                                 "垂直的特征，平行的特征及性质"
+                             ],
+        "source":  "20-21 四年级上·全国·单元测试"
+    },
+    {
+        "id":  "375",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "两组对边分别平行的四边形叫做( );如果它的四个角都是直角，它就是( );如果此时量得它的四条边都相等，它就是( );如果只有一组对边平行，那它就是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平行四边形;长方形;正方形;梯形",
+        "solution":  "【分析】本题主要考查对平行四边形，长方形，正方形和梯形定义的理解。解题思路是根据各图形的关键特征进行判断：首先依据\u0027两组对边分别平行\u0027确定基础图形为平行四边形;其次根据\u0027四个角都是直角\u0027这一条件推导出长方形;再根据\u0027四条边都相等\u0027这一条件推导出正方形;最后依据\u0027只有一组对边平行\u0027这一特征确定梯形。掌握这些图形的定义及它们之间的联系是解题的关键.\n【详解】两组对边分别平行的四边形叫做平行四边形。两组对边分别平行，四个角都是直角的四边形叫做长方形。四条边相等，四个角都是直角的四边形叫做正方形。只有一组对边平行的四边形叫做梯形。",
+        "knowledge_points":  [
+                                 "梯形的概念及特点，正方形的概念及特点，长方形的概念及特点，平行四边形的概念及特点"
+                             ],
+        "source":  "20-21 四年级上·全国·单元测试"
+    },
+    {
+        "id":  "376",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个平行四边形可以剪成两个相同的( ),也可以剪成两个相同的( ),还可以剪成两个相同的( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平行四边形;三角形;梯形",
+        "solution":  "【分析】作出图形即可解答:![](images/3fbda3e8ee9bf59f107e0bf890f767bb11d25867affdda89a3bb9c7c43897895.jpg)本题考查了平行四边形的切拼，重在考查学生动手操作的能力，通过作图，使学生加深了对图形的理解.\n【详解】一个平行四边形可以剪成两个相同的平行四边形，也可以剪成两个相同的三角形，还可以剪成两个相同的梯形。",
+        "knowledge_points":  [
+                                 "平行四边形的切拼"
+                             ],
+        "source":  "20-21 四年级上·全国·单元测试"
+    },
+    {
+        "id":  "377",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "有3根小棒，分别长8厘米,6厘米,8厘米，要想摆成一个平行四边形，还需要一根长( )厘米的小棒。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】平行四边形中，两组对边互相平行且相等。根据题意可知，这3根小棒中，有相等的两根8厘米的小棒，剩余一根小棒长6厘米，则应再拿一根长6厘米的小棒，即可摆成一个平行四边形。解决本题的关键是明确平行四边形的两组对边相等.\n【详解】根据分析可知，要想摆成一个平行四边形，还需要一根长6厘米的小棒。",
+        "knowledge_points":  [
+                                 "平行四边形的概念及特点"
+                             ],
+        "source":  "20-21四年级上·全国·单元测试"
+    },
+    {
+        "id":  "378",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "元旦节时，张英做了一个平行四边形的贺卡送给老师，她用82厘米长的彩带给这个平行四边形镶了一道花边，其中一条花边的长是18厘米，与它相邻的花边长是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "23",
+        "solution":  "【分析】平行四边形的两条对边相等，平行四边形的周长=(一条边长+与它相邻的另一条边长)×2。则平行四边形的一条边长=周长/2-与它相邻的另一条边长，代入数据计算即可。熟练掌握平行四边形的周长公式，灵活运用公式解决问题.\n【详解】82÷2-18=41-18=23(厘米),所以与它相邻的花边长是23厘米。",
+        "knowledge_points":  [
+                                 "平行四边形的周长"
+                             ],
+        "source":  "20-21四年级上·全国·单元测试"
+    },
+    {
+        "id":  "379",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用四根长分别为7厘米,5厘米,7厘米,5厘米的小棒可以围成一个( ),也可以围成一个( )。",
+        "options":  [
+
+                    ],
+        "answer":  "长方形;平行四边形",
+        "solution":  "【分析】根据长方形的定义：四个内角相等的四边形，即是说所有内角均为直角。从这个定义可以得出矩形两条相对的边等长。平行四边形：在同一个二维平面内，由两组平行线段组成的闭合图形，称为平行四边形。由此可知这四根小棒可以围成一个长方形，也可以围成一个平行四边形。本题主要考查的是长方形，平行四边形的相关知识.\n【详解】用四根长分别为7厘米,5厘米,7厘米,5厘米的小棒可以围成一个长方形，也可以围成一个平行四边形。",
+        "knowledge_points":  [
+                                 "长方形的概念及特点，平行四边形的概念及特点"
+                             ],
+        "source":  "20-21 三年级上·全国·单元测试"
+    },
+    {
+        "id":  "380",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在四边形中,( )的四条边都相等;( )和( )的四个角都是直角;( ),( )和( )的对边相等。",
+        "options":  [
+
+                    ],
+        "answer":  "正方形;长方形;正方形;长方形;正方形;平行四边形",
+        "solution":  "【分析】两组对边分别平行，四个角都是直角的四边形叫做长方形。四条边相等，四个角都是直角的四边形叫做正方形。由两组平行线段组成的闭合图形叫做平行四边形。据此解答即可。熟练掌握正方形，长方形和平行四边形的性质，注意长方形是特殊的平行四边形，正方形是特殊的长方形.\n【详解】在四边形中，正方形的四条边都相等;长方形和正方形的四个角都是直角;长方形，正方形和平行四边形的对边相等。",
+        "knowledge_points":  [
+                                 "正方形的概念及特点，长方形的概念及特点，平行四边形的概念及特点"
+                             ],
+        "source":  "20-21 三年级上·全国·单元测试"
+    },
+    {
+        "id":  "381",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "教室黑板的两组对边分别( ),且长度( ),邻边互相( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平行;相等;垂直",
+        "solution":  "【分析】因为黑板是一个长方形，所以根据长方形的特征：对边平行且相等,4个角都是直角，可知，长方形相邻的两条边互相垂直，相对的两边互相平行;据此解答。此题主要考查长方形的特征，基础知识，要熟练掌握.\n【详解】根据长方形的特征可知：教室黑板面相对的两组对边分别平行且相等，相邻两边互相垂直。",
+        "knowledge_points":  [
+                                 "长方形的概念及特点，平行的特征及性质，垂直的特征"
+                             ],
+        "source":  "2020 四年级上·全国·专题练习"
+    },
+    {
+        "id":  "382",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "把一张长方形纸左右对折两次，那么三条折痕互相( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平行",
+        "solution":  "【分析】把一张长方形纸左右对折后还是长方形，折痕与宽互相平行，再左右对折，仍是一个长方形，第二次的两个折痕与第一次的折痕互相平行，据此即可解答。本题主要考查学生对长方形的特点及平行线知识的掌握和灵活运用.\n【详解】把一张长方形纸左右对折两次，那么三条折痕互相平行。",
+        "knowledge_points":  [
+                                 "平行的特征及性质"
+                             ],
+        "source":  "19-20 四年级上·安徽黄山·期末"
+    },
+    {
+        "id":  "383",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "两条平行线之间的垂线段长度都( )。",
+        "options":  [
+
+                    ],
+        "answer":  "相等",
+        "solution":  "【分析】本题主要考查平行线之间距离的性质。解题关键在于理解\u0027两条平行线之间的垂线段\u0027实际上就是这两条平行线之间的距离。根据平行线的性质可知，平行线间的距离处处相等，因此无论在哪一点作垂线段，其长度都是固定的。由此可得，两条平行线之间的所有垂线段长度都相等.\n【详解】与两条平行线互相垂直的线段的长度都相等。",
+        "knowledge_points":  [
+                                 "垂直的特征，平行的特征及性质"
+                             ],
+        "source":  "2014 四年级·全国·课后作业"
+    },
+    {
+        "id":  "384",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个长方形菜地面积是50平方米，扩建后，长是原来的2倍，宽是原来的3倍，扩建后菜地面积是( )平方米。",
+        "options":  [
+
+                    ],
+        "answer":  "300",
+        "solution":  "【分析】根据长方形的面积公式:S=ab,再根据因数与积的变化规律，积扩大的倍数等于因数扩大倍数的乘积，据此解答。此题主要考查长方形的面积公式的灵活运用，以及因数与积的变化规律的应用.\n【详解】长方形的长扩大到原来的2倍，宽扩大到原来的3倍，那么它的面积就扩大到原来的2×3=6倍。50×6=300(平方米)",
+        "knowledge_points":  [
+                                 "长方形的面积，积的变化规律(小数乘法)"
+                             ],
+        "source":  "20-21 四年级上·福建厦门·期中"
+    },
+    {
+        "id":  "385",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "平行四边形的对边互相( ),长方形的两条邻边互相( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平行;垂直",
+        "solution":  "【分析】本题考查平行四边形和长方形的特征。根据平行四边形的定义可知，平行四边形的两组对边分别平行;根据长方形的特征可知，长方形的四个角都是直角，所以相邻的两条边互相垂直。据此填空.\n【详解】根据平行四边形的特性可知：平行四边形的对边互相平行;根据长方形的特征，对边平行且相等,4个角都是直角，可知，长方形相邻的两条边互相垂直，相对的两边互相平行。",
+        "knowledge_points":  [
+                                 "平行的特征及性质，垂直的特征，长方形的概念及特点"
+                             ],
+        "source":  "2020 四年级上·全国·专题练习"
+    },
+    {
+        "id":  "386",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "两个( )的梯形可以拼成一个平行四边形。梯形的上底与下底的和等于平行四边形的( ),梯形的高等于平行四边形的( ),每个梯形的面积等于平行四边形面积的( )。",
+        "options":  [
+
+                    ],
+        "answer":  "完全相同;底;高;1÷2",
+        "solution":  "【分析】用两个完全相同的梯形可以拼成一个平行四边形，这样有利于推导梯形的面积;通过观察拼成的平行四边形可以看出：平行四边形的底等于梯形的上下底之和，它们的高相等，进而可以推导出梯形的面积.\n【详解】梯形的面积=平行四边形面积/2=平行四边形的底×高/2=(上底+下底)×高/2。两个完全相同的梯形可以拼成一个平行四边形，梯形的上底与下底的和等于平行四边形的底，梯形的高等于平行四边形的高，每个梯形的面积等于拼成的平行四边形面积的1÷2。",
+        "knowledge_points":  [
+                                 "梯形的面积"
+                             ],
+        "source":  "2014五年级·全国·课后作业"
+    },
+    {
+        "id":  "387",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用线把两根6厘米和两根9厘米的吸管串起来，做成一个长方形，长方形的面积是( ),如果把它拉成一个平行四边形，平行四边形的周长是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "54平方厘米;30厘米",
+        "solution":  "【分析】本题考查长方形的面积计算以及平行四边形的周长特征。根据题意可知，长方形的长是9厘米，宽是6厘米，利用长方形的面积公式可求出面积;当把长方形框架拉成平行四边形时，四条边的长度没有变化，所以周长不变，等于原来长方形的周长.\n【详解】长方形的面积:9×6=54(平方厘米);平行四边形的周长:6×2+9×2=12+18=30(厘米)。答：长方形的面积是54平方厘米;平行四边形的周长是30厘米。",
+        "knowledge_points":  [
+                                 "长方形的面积，平行四边形的周长"
+                             ],
+        "source":  "2014 三年级·全国·课后作业"
+    },
+    {
+        "id":  "388",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "平行四边形的两组对边( )且( )。梯形只有一组对边( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平行;相等;平行",
+        "solution":  "【分析】本题考查平行四边形和梯形的特征。根据平行四边形的性质可知，平行四边形的两组对边分别平行且相等;根据梯形的定义可知，只有一组对边平行的四边形叫做梯形。据此填空.\n【详解】根据平行四边形和梯形的特征：平行四边形的两组对边平行且相等。梯形只有一组对边平行。",
+        "knowledge_points":  [
+                                 "平行四边形的概念及特点"
+                             ],
+        "source":  "2014 三年级·全国·课后作业"
+    },
+    {
+        "id":  "389",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把一条长38厘米的细木条钉成一个有一边是12厘米的平行四边形，然后再把它拉成一个长方形，这个长方形的周长是( ),面积是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "38厘米;84平方厘米",
+        "solution":  "【分析】本题考查长方形的周长和面积公式的应用。根据题意可知，把平行四边形框架拉成一个长方形，四条边的长度没有变化，所以长方形的周长等于细木条的总长度。已知长方形的周长和其中一条边的长度，可以根据周长公式求出另一条边的长度，最后利用长方形的面积公式计算出面积.\n【详解】(1)长方形的周长是38厘米;(2)长方形的另一条边:38÷2-12=19-12=7(厘米),长方形的面积:12×7=84(平方厘米)。答：这个长方形的周长是38厘米，面积是84平方厘米。",
+        "knowledge_points":  [
+                                 "长方形的周长，长方形的面积"
+                             ],
+        "source":  "2014三年级·全国·课后作业"
+    },
+    {
+        "id":  "390",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "平行四边形有( )组对边分别( )。梯形只有( )组对边平行,( )叫做等腰梯形。",
+        "options":  [
+
+                    ],
+        "answer":  "两;平行;一;两个腰相等的梯形",
+        "solution":  "【分析】本题主要考查平行四边形，梯形及等腰梯形的定义与特征。解题时需紧扣各图形的关键性质：首先根据平行四边形的定义，明确其拥有两组对边分别平行;其次依据梯形的定义，确定其只有一组对边平行;最后根据等腰梯形的分类标准，可知两腰相等的梯形即为等腰梯形.\n【详解】根据平行四边形，梯形和等腰梯形的含义：两组对边分别平行的四边形叫平行四边形;只有一组对边平行的四边形叫梯形;两个腰相等的梯形叫等腰梯形;据此解答。",
+        "knowledge_points":  [
+                                 "梯形的特征及分类，平行四边形的概念及特点"
+                             ],
+        "source":  "2014 三年级·全国·课后作业"
+    },
+    {
+        "id":  "391",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "判断题",
+        "difficulty":  0.64,
+        "question":  "两条直线相交可以组成4个角，相对的角的大小一定相等( )。",
+        "options":  [
+
+                    ],
+        "answer":  "正确",
+        "solution":  "【分析】本题考查两条直线相交的特征。根据相交线的性质可知，两条直线相交可以组成4个角，其中相对的角大小一定相等。据此判断原题说法正确.\n【详解】如图可知:![](images/e4516a9f5162ab4c8ee3d834e0421a8bc4d683de79cb86970d7fffb0dd697ace.jpg)∠1=∠2,∠3=∠4，两条直线相交可以组成4个角，相对的角的大小一定相等;故答案为正确。",
+        "knowledge_points":  [
+                                 "相交线的性质"
+                             ],
+        "source":  "2014 四年级·全国·课后作业"
+    },
+    {
+        "id":  "392",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个梯形的下底是上底长度的3倍，如果将上底延长8厘米，这个梯形就变成了一个平行四边形。这个梯形的上底是( )厘米，下底是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "4;12",
+        "solution":  "【分析】由题意可知：梯形上底的(3-1)倍是8厘米，根据已知一个数的几倍是多少用除法求出上底的长，进而求出下底的长.\n【详解】上底:8÷(3-1)=4(厘米);下底:4×3=12(厘米)。答：这个梯形的上底是4厘米，下底是12厘米。",
+        "knowledge_points":  [
+                                 "梯形的概念及特点，和差倍问题"
+                             ],
+        "source":  "2014 三年级·全国·课后作业"
+    },
+    {
+        "id":  "393",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.64,
+        "question":  "两条直线垂直时相交成4个( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "直",
+        "solution":  "【分析】本题考查垂直的特征。根据垂直的定义可知，两条直线互相垂直时，它们相交成的4个角都是直角。据此填空.\n【详解】在同一平面内，如果两条直线相交成直角，则我们就说这两条直线互相垂直;所以两条直线互相垂直时相交成4个直角。![](images/0e5bbf2f222d943cfd87568019113ec8de9147c968d5ffcbf00cfc5451e77bf3.jpg)故答案为直。",
+        "knowledge_points":  [
+                                 "垂直的特征"
+                             ],
+        "source":  "2014 四年级·全国·课后作业"
+    },
+    {
+        "id":  "394",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "判断题",
+        "difficulty":  0.64,
+        "question":  "在纸上画一条直线的平行线，可以画无数条( )。",
+        "options":  [
+
+                    ],
+        "answer":  "正确",
+        "solution":  "【分析】本题考查平行线的定义。根据平行的含义可知，在同一平面内，不相交的两条直线叫做平行线，对于一条已知直线，可以画出无数条与之平行的直线。据此判断.\n【详解】根据平行的含义可知：在纸上画一条直线的平行线，可以画无数条，说法正确;故答案为正确。",
+        "knowledge_points":  [
+                                 "平行的特征及性质"
+                             ],
+        "source":  "2014四年级·全国·课后作业"
+    },
+    {
+        "id":  "395",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.64,
+        "question":  "给一条直线作两条垂线，这两条垂线的位置关系是互相( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平行",
+        "solution":  "【分析】本题考查垂直和平行的性质。根据垂直和平行的特征可知，在同一平面内，垂直于同一条直线的两条直线互相平行。据此填空.\n【详解】如图:![](images/756176a3a12338165602e7b841ad87e0d6cb84deb1fd10abe19de29277b75a2e.jpg)这两条垂线互相平行;故答案为平行。",
+        "knowledge_points":  [
+                                 "垂直的特征，平行的特征及性质"
+                             ],
+        "source":  "2014 四年级·全国·课后作业"
+    },
+    {
+        "id":  "396",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.64,
+        "question":  "在同一平面内( )的两条直线叫( )。",
+        "options":  [
+
+                    ],
+        "answer":  "不相交;平行线",
+        "solution":  "【分析】本题考查平行线的定义。根据平行的含义可知，在同一个平面内的不相交的两条直线，叫做平行线。据此填空.\n【详解】由平行的含义可知：在同一个平面内的不相交的两条直线，叫做平行线。",
+        "knowledge_points":  [
+                                 "平行的特征及性质"
+                             ],
+        "source":  "2014 四年级·全国·课后作业"
+    },
+    {
+        "id":  "397",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "两条直线相交，形成4个角，其中一个角是直角，这两条直线是( )的关系。",
+        "options":  [
+
+                    ],
+        "answer":  "垂直",
+        "solution":  "【分析】本题考查垂直的定义。根据垂直的含义可知，如果两条直线相交成直角，就说这两条直线互相垂直。据此填空.\n【详解】根据垂直的定义：如果两条直线相交成直角，其中一条直线叫作另一条直线的垂线;据此判断。",
+        "knowledge_points":  [
+                                 "垂直的特征"
+                             ],
+        "source":  "2014四年级·全国·课后作业"
+    },
+    {
+        "id":  "398",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "判断题",
+        "difficulty":  0.64,
+        "question":  "在同一个平面内不相交的两条线段叫平行线。(判断对错)",
+        "options":  [
+
+                    ],
+        "answer":  "错误",
+        "solution":  "【分析】本题考查平行的含义。根据平行的定义：在同一平面内，不相交的两条直线叫做平行线。题目中说的是两条线段，不是两条直线，所以题目说法错误.\n【详解】平行的含义：在同一平面内，不相交的两条直线叫做平行线;故答案为错误。",
+        "knowledge_points":  [
+                                 "平行的特征及性质"
+                             ],
+        "source":  "2014 四年级·全国·课后作业"
+    },
+    {
+        "id":  "399",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "填空题",
+        "difficulty":  0.64,
+        "question":  "过直线外一点画一条直线的平行线有( )条，垂线有( )条。",
+        "options":  [
+
+                    ],
+        "answer":  "1;1",
+        "solution":  "【分析】本题考查平行公理及垂线的性质。根据平行公理及垂线的性质可知，过直线外一点画已知直线的平行线有1条，垂线也有1条。据此填空.\n【详解】由垂线的性质：过一点作已知直线的垂线，能作且只能作1条;根据平行公理过直线外一点作已知直线的平行线，能作且只能作一条。所以：过直线外一点画一条直线的平行线有1条，垂线有1条。",
+        "knowledge_points":  [
+                                 "画垂线，画平行线"
+                             ],
+        "source":  "2014 四年级·全国·课后作业"
+    },
+    {
+        "id":  "400",
+        "unit":  "第五单元平行四边形和梯形",
+        "question_type":  "判断题",
+        "difficulty":  0.64,
+        "question":  "时钟在三点半时，时针与分针互相平行。(判断对错)",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】本题考查钟面角及直线的位置关系。根据钟面的特征可知，三点半时，时针指向3和4中间，分针指向6，此时两针相交，不互相平行。据此判断.\n【详解】时钟三点半时，时针在3与4的正中间位置，分针在6处，此时时针与分针是相交的关系，据此即可判断。故答案为错误。",
+        "knowledge_points":  [
+                                 "钟面角，相交与垂直"
+                             ],
+        "source":  "2014 四年级·全国·课后作业"
+    }
+]
diff --git "a/\345\233\233\345\271\264\347\272\247\351\242\230\351\233\206/\345\233\233\345\271\264\347\272\247\344\270\212/\345\233\233\345\271\264\347\272\247\344\270\212\347\254\254\345\205\253\345\215\225\345\205\203\346\225\260\345\255\246\345\271\277\350\247\222.json" "b/\345\233\233\345\271\264\347\272\247\351\242\230\351\233\206/\345\233\233\345\271\264\347\272\247\344\270\212/\345\233\233\345\271\264\347\272\247\344\270\212\347\254\254\345\205\253\345\215\225\345\205\203\346\225\260\345\255\246\345\271\277\350\247\222.json"
new file mode 100644
index 0000000000000000000000000000000000000000..6f71241b74a7ec432fe792b721747870054ac5de
--- /dev/null
+++ "b/\345\233\233\345\271\264\347\272\247\351\242\230\351\233\206/\345\233\233\345\271\264\347\272\247\344\270\212/\345\233\233\345\271\264\347\272\247\344\270\212\347\254\254\345\205\253\345\215\225\345\205\203\346\225\260\345\255\246\345\271\277\350\247\222.json"
@@ -0,0 +1,8041 @@
+[
+    {
+        "id":  "1",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈做晚饭的各环节及时间：淘米2分钟，电饭煲煮饭30分钟，洗菜4分钟，切菜5分钟，炒菜6分钟，妈妈做这顿晚饭最快需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "32",
+        "solution":  "【分析】为了最小化总时间，妈妈可以在电饭煲煮饭的同时进行洗菜，切菜和炒菜。淘米是煮饭的前置任务，必须单独完成，耗时2分钟。煮饭需要30分钟，且是自动过程，妈妈无需看守。洗菜，切菜和炒菜必须顺序进行，总时间为4+5+6=15分钟，这些任务可以在煮饭的30分钟内完成。所以用2加30即可求出妈妈做这顿晚饭最快需要几分钟.\n【详解】根据分析可知,4+5+6=15(分钟),15\u003c30,30+2=32(分钟)。所以妈妈做这顿晚饭最快需要32分钟。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "25-26四年级上·辽宁鞍山·期末"
+    },
+    {
+        "id":  "2",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "星期天小红帮妈妈做下面的事情：用洗衣机洗衣服30分钟，收拾房间15分钟，扫地5分钟，晾衣服5分钟。她做完这些事情，至少要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "35",
+        "solution":  "【分析】小红在洗衣机洗衣服的同时做其他任务。洗衣服需要30分钟，晾衣服需要在洗衣服完成后进行，需要5分钟。收拾房间和扫地共需要20分钟(15分钟+5分钟),可以在洗衣服的30分钟内完成，因此总时间由洗衣服和晾衣服的时间决定.\n【详解】30+5=35(分钟)。所以她做完这些事情，至少要35分钟。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25四年级上·重庆璧山·期末"
+    },
+    {
+        "id":  "3",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明早晨要做以下几件事：起床整理被褥3分钟，刷牙3分钟，洗脸2分钟，听英语录音8分钟。如果7时起床，最快( )时( )分做完这些事情。",
+        "options":  [
+
+                    ],
+        "answer":  "7;8",
+        "solution":  "【分析】根据题意，要想最快完成这些事情，可以在听英语录音的同时起床整理被褥、刷牙和洗脸，再根据所用时间求出最快几时几分做完这些事情。\n【详解】3+3+2=8(分钟)，7时+8分钟=7时8分。小明最快7时8分做完这些事情。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "25-26四年级上·甘肃定西·期末"
+    },
+    {
+        "id":  "4",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "随着医疗技术的不断进步和医疗模式的不断创新，校医务室也在逐步完善和提升其服务水平和能力。小易，冬冬，芳芳同时来到校医务室做检查，芳芳检查需要3分钟，冬冬检查需要4分钟，小易检查需要6分钟。要使三人等候的时间总和最少，首先应安排( )就诊，再安排( )就诊，最后安排( )就诊。",
+        "options":  [
+
+                    ],
+        "answer":  "芳芳;冬冬;小易",
+        "solution":  "【分析】为了使三人等候的时间总和最少，需要最小化每个人的等待时间之和。由于三人同时到达，第一个就诊的人等待时间为0，第二个就诊的人等待时间为第一个人的检查时间，第三个就诊的人等待时间为前两个人检查时间之和。因此，总等候时间取决于就诊顺序。为了最小化总等候时间，应安排检查时间最短的人先就诊，这样后续人员的等待时间较短。检查时间芳芳3分钟最短，冬冬4分钟次之，小易6分钟最长，故顺序应为芳芳，冬冬，小易.\n【详解】设就诊顺序为第一个人，第二个人，第三个人。总等候时间=第一个人等待时间+第二个人等待时间+第三个人等待时间。其中，第一个人等待时间为0，第二个人等待时间为第一个人的检查时间，第三个人等待时间为第一个和第二个人检查时间之和。因此，总等候时间=0+第一个人的检查时间+(第一个人的检查时间+第二个人的检查时间)。要最小化这个总和，应使第一个人的检查时间尽可能小，第二个人的检查时间尽可能小。故安排检查时间最短的芳芳先就诊，检查时间次短的冬冬其次就诊，检查时间最长的小易最后就诊。此时，总等候时间为0+3分钟+(3分钟+4分钟)=10分钟，是最小值。其他顺序均大于10分钟，因此首先安排芳芳就诊，再安排冬冬就诊，最后安排小易就诊。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25四年级上·浙江杭州·期末"
+    },
+    {
+        "id":  "5",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "星期六，小轩帮妈妈做家务。淘米2分钟，煮饭10分钟，扫地3分钟，擦桌子2分钟，洗菜3分钟，炒菜5分钟。他做完这些事至少要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】先淘米2分钟，煮饭10分钟，在煮饭的同时，进行扫地，擦桌子，洗菜，炒菜，用这四部分的时间加起来减去10分钟，计算出超过煮饭的10分钟的时间，加上淘米和煮饭的时间，即可求得他做完这些事至少要多少分钟.\n【详解】3+2+3+5-10=13-10=3(分钟),2+10+3=15(分钟)。所以他做完这些事至少要15分钟。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25四年级上·四川凉山·期末"
+    },
+    {
+        "id":  "6",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "彤彤准备冲一杯果汁，洗杯子和汤匙各需要1分钟，烧开水需要10分钟，取出果汁粉冲泡需要2分钟。彤彤至少需要( )分钟才能喝到果汁。",
+        "options":  [
+
+                    ],
+        "answer":  "12",
+        "solution":  "【分析】为了最小化总时间，应合理安排任务顺序。烧开水需要10分钟，且是冲泡的前提条件。洗杯子和汤匙各需1分钟，但可以在烧开水期间同时进行，因此这些任务的时间被覆盖在烧开水时间内。烧开水完成后，冲泡需要2分钟。因此，总时间由烧开水和冲泡的时间决定，即10+2=12(分钟).\n【详解】先开始烧开水(需要10分钟),在烧开水的同时洗杯子和汤匙。烧开水完成后，立即取出果汁粉冲泡(需要2分钟)。10+2=12(分钟)。彤彤至少需要12分钟才能喝到果汁。",
+        "knowledge_points":  [
+                                 "沏茶问题",
+                                 "不进位加法(20以内)"
+                             ],
+        "source":  "25-26四年级上·辽宁盘锦·期末"
+    },
+    {
+        "id":  "7",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈做午饭。洗菜10分钟，切菜5分钟，炒菜10分钟，淘米2分钟，用电饭锅做米饭30分钟，最快( )分钟可以吃上午饭。",
+        "options":  [
+
+                    ],
+        "answer":  "32",
+        "solution":  "【分析】洗菜，切菜，炒菜，淘米，电饭锅做饭这几件事情中，要先洗菜，才能切菜，炒菜;要先淘米，才能电饭锅做饭。电饭锅做饭的时候，能够去洗菜，切菜，炒菜。因此可以先进行淘米，然后用电饭锅做饭，电饭锅做饭的同时去洗菜，切菜，炒菜.\n【详解】10+5+10=25(分钟),25\u003c30，所以总时间为2+30=32(分钟)。妈妈最快32分钟可以吃上午饭。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "25-26四年级上·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "8",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "铝锅煎饼是临夏回族自治州的一种特色美食。李阿姨用平底锅烙饼，每次最多烙两张饼，两面都要烙，每面烙需要3分钟，烙5张饼至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】由题意得，要使烙饼的时间最少，每次烙饼时锅里都应尽量有两张饼。李阿姨烙5张饼时：第一次烙第1张和第2张的正面，用时3分钟；第二次烙第1张的反面和第3张的正面，用时3分钟；第三次烙第2张的反面和第4张的正面，用时3分钟；第四次烙第3张的反面和第5张的正面，用时3分钟；第五次烙第4张和第5张的反面，用时3分钟。\n【详解】3×5=15(分钟)。故李阿姨烙5张饼至少需要15分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "25-26四年级上·甘肃临夏·期末"
+    },
+    {
+        "id":  "9",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "霄霄每天晚上要背诵成语6分钟，烧开水16分钟，用开水温牛奶2分钟，喝牛奶5分钟，那么霄霄在( )的同时可以( ),做完这些事情最少用( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "烧开水;背诵成语;23",
+        "solution":  "【分析】根据题意，霄霄的活动包括背诵成语，烧开水，用开水温牛奶和喝牛奶，其中烧开水需要较长时间且不需要持续看守，可以在烧开水的同时进行背诵成语;用开水温牛奶必须在烧开水之后进行，喝牛奶必须在温牛奶之后进行，所以通过同时进行烧开水和背诵成语，可以最小化总时间.\n【详解】先烧开水16分钟，同时在烧开水期间背诵成语6分钟，然后用开水温牛奶2分钟，最后喝牛奶5分钟。背诵成语的时间被覆盖在烧开水时间内，因此总时间为16+2+5=23(分钟)。答：在烧开水的同时可以背诵成语，做完这些事情最少用23分钟。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "25-26四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "10",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈出门之前要完成以下几件事：整理房间15分钟，把衣服放入洗衣机要1分钟，洗衣机自动洗衣要15分钟，擦鞋要3分钟，那么妈妈最少需要( )分钟可以出门。",
+        "options":  [
+
+                    ],
+        "answer":  "19",
+        "solution":  "【分析】根据题意，把衣服放入洗衣机要1分钟，洗衣机自动洗衣要15分钟，可以同时整理房间15分钟，然后擦鞋要3分钟，即1+15+3=19(分钟).\n【详解】1+15+3=19(分钟)。妈妈最少需要19分钟可以出门。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "25-26四年级上·甘肃天水·期末"
+    },
+    {
+        "id":  "11",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "(1)班参加科技嘉年华活动，开展“VR沉浸体验”和“绿色能源挑战”两个项目体验。同学们分为三组，每个项目每组体验一次，每次体验时间20分钟。从9:00开始，全班最早( )能完成项目体验。",
+        "options":  [
+
+                    ],
+        "answer":  "10:00",
+        "solution":  "【分析】为了最早完成体验，需要优化调度安排。三个组分别记作组1、组2、组3，两个项目记作A(VR沉浸体验)和B(绿色能源挑战)。每个组需体验A和B各一次，每次20分钟，顺序可任意选择。通过合理安排顺序和调度，可使总完成时间最短。\n【详解】一种可行方案为：9:00~9:20，组1体验A，组2体验B；9:20~9:40，组2体验A，组3体验B；9:40~10:00，组1体验B，组3体验A。此时，所有组均在10:00完成体验，总用时60分钟。因此，全班最早10:00能完成项目体验。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "25-26四年级上·北京丰台·期末"
+    },
+    {
+        "id":  "12",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妮妮招待外国小朋友，用一只平底锅煎鱼，每次只能煎2条，煎一面要3分钟，两面都要煎，煎3条鱼至少要用( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "9",
+        "solution":  "【分析】要使煎鱼的总时间最少，那么每一次锅内就应有两条鱼。第一次：第1条鱼的正面，第2条鱼的正面，用时3分钟;第二次：第1条鱼的反面，第3条鱼的正面，用时3分钟;第三次：第3条鱼的反面，第2条鱼的反面，用时3分钟.\n【详解】3×3=9(分钟)。所以煎3条鱼至少要用9分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "25-26四年级上·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "13",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "平底锅每次只能烙2张饼，两面都要烙，每面需2分钟，妈妈要烙3张饼，至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】一共要烙3张饼，先烙第一张和第二张的正面，需要2分钟;再烙第一张的反面和第三张的正面，需要2分钟;最后烙第二张和第三张的反面，需要2分钟，把这3个2分钟相加即可.\n【详解】2+2+2=6(分钟)。妈妈要烙3张饼，至少需要6分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "25-26四年级上·甘肃定西·期末"
+    },
+    {
+        "id":  "14",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈做晚饭需要做的事情有：淘米2分钟，洗炒菜锅2分钟，洗菜6分钟，切菜6分钟，煮饭25分钟，炒菜10分钟。吃上这顿饭至少要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "27",
+        "solution":  "【分析】已知妈妈做晚饭需要做的事情有：淘米2分钟，洗炒菜锅2分钟，洗菜6分钟，切菜6分钟，煮饭25分钟，炒菜10分钟。在煮饭的同时,2+6+6+10=24(分钟),可以洗炒菜锅，洗菜，切菜，炒菜;再加上淘米的2分钟，就是吃上这顿饭至少要25+2=27(分钟).\n【详解】2+6+6+10=24(分钟),25+2=27(分钟)。妈妈吃上这顿饭至少要27分钟。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "25-26四年级上·吉林白城·期末"
+    },
+    {
+        "id":  "15",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "街边的“铁板烧烤”每次同时可以烤2串鱿鱼，两面都要烤，每面烤2分钟。如果要烤3串鱿鱼，至少需要( )分钟；如果要烤7串鱿鱼，至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "6;14",
+        "solution":  "【分析】这是一个类似烙饼的问题。每次尽量让铁板上都有2串鱿鱼在烤，这样最省时间。烤2串及以上的鱿鱼时，最少时间等于串数×每面所需时间。\n【详解】3×2=6(分钟)，7×2=14(分钟)。所以要烤3串鱿鱼，至少需要6分钟；要烤7串鱿鱼，至少需要14分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "25-26四年级上·陕西渭南·期末"
+    },
+    {
+        "id":  "16",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "刘影早上起床，淘米要2分钟，插上开关用电饭锅烧饭要18分钟，背外语要用12分钟，刷牙洗脸要3分钟，吃早饭要8分钟，刘影经过合理安排，起床后( )分钟就能上学了。",
+        "options":  [
+
+                    ],
+        "answer":  "28",
+        "solution":  "【分析】烧饭过程需要18分钟，但在此期间可以同时进行其他活动，如背外语和刷牙洗脸。吃早饭必须在烧饭后进行。通过将背外语和刷牙洗脸安排在烧饭期间，可以最小化总时间。总时间由淘米时间，烧饭时间和吃早饭时间相加得出.\n【详解】2+18+8=28(分钟)。所以刘影经过合理安排，起床后28分钟就能上学了。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "25-26四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "17",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "小明扫地需要8分钟，用电水壶烧开水需要10分钟，用烧开的水沏茶需要1分钟。他完成这三件事至少要用( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "11",
+        "solution":  "【分析】烧开水需要10分钟，且在此期间可以同时进行其他活动，如扫地(需要8分钟)。沏茶必须在烧开水完成后进行，需要1分钟。因此，通过合理安排，在烧开水的同时扫地，总时间最小值为烧开水时间与沏茶时间之和，即10+1=11(分钟).\n【详解】10+1=11(分钟)。小明完成这三件事至少要用11分钟。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "25-26四年级上·辽宁铁岭·期末"
+    },
+    {
+        "id":  "18",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一张饼烙熟要4分钟，正、反面各需要2分钟，每次正好可以烙两张，烙5张饼至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "10",
+        "solution":  "【分析】平底锅每次能烙2张饼，每张饼要烙正、反两面，每面2分钟。尽量让锅里每次都有2张饼在烙，是最省时间的方法。给5张饼编号1、2、3、4、5。第1-2分钟：烙1正面、2正面；第3-4分钟：烙1反面、3正面；第5-6分钟：烙2反面、4正面；第7-8分钟：烙3反面、5正面；第9-10分钟：烙4反面、5反面。\n【详解】共需要烙5次，每次2分钟，2×5=10(分钟)。烙5张饼至少需要10分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "25-26四年级上·辽宁盘锦·期末"
+    },
+    {
+        "id":  "19",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "青青感冒了，妈妈给青青冲感冒药，她需要完成下面几件事：①用水壶烧水4分钟；②找感冒药2分钟；③洗杯子1分钟；④冲感冒药2分钟；⑤洗水壶1分钟。做完这些事至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "7",
+        "solution":  "【分析】要想做完这些事情所花的时间最少，妈妈需要合理安排顺序。可以先洗水壶1分钟，接着用水壶烧水4分钟。水壶烧水的同时，再去找感冒药2分钟和洗杯子1分钟。等水烧好之后，再冲感冒药2分钟。\n【详解】1+4+2=7(分钟)。故妈妈做完这些事至少需要7分钟。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "25-26四年级上·辽宁抚顺·期末"
+    },
+    {
+        "id":  "20",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈淘米用2分钟，用电饭锅煮饭23分钟，洗菜6分钟，切菜4分钟，在煤气上烧热油1分钟，炒菜12分钟，妈妈最快( )分钟可以做好饭菜。",
+        "options":  [
+
+                    ],
+        "answer":  "25",
+        "solution":  "【分析】根据题意，洗菜6分钟，切菜4分钟，烧热油1分钟，再炒菜12分钟，这个时间是6+4+1+12=23分钟。所以妈妈先淘米，然后煮饭。在煮饭23分钟的同时，洗菜，切菜，炒菜，这样用的时间最少.\n【详解】整个时间就是淘米的2分钟，加上煮饭的23分钟。2+23=25(分钟)。所以，最少要25分钟。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "25-26四年级上·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "21",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "平底锅每次只能烙两张饼，两面都要烙，每面需要2分钟。妈妈要烙4张饼，至少需要( )分钟，烙5张饼至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "8;10",
+        "solution":  "【分析】烙4张饼：平底锅每次只能烙两张饼，两面都要烙，每面需要2分钟。4张饼可以分为2组(每组2张),每组需要烙2面，每组时间为每面时间2分钟乘2面，然后用每组时间乘组数2组。烙5张饼：饼编号为1,2,3,4,5。第1次：烙1正面,2正面;第2次：烙1反面,3正面;第3次：烙2反面,3反面;第4次：烙4正面,5正面;第5次：烙4反面,5反面.\n【详解】烙4张饼:2×2=4(分钟),4×2=8(分钟);烙5张饼:2×5=10(分钟)。因此，妈妈要烙4张饼，至少需要8分钟，烙5张饼至少需要10分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "25-26四年级上·黑龙江齐齐哈尔·期末"
+    },
+    {
+        "id":  "22",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "妈妈用平底锅烙饼，两面都要烙，每次最多烙2张，每面3分钟。烙6张饼，最快需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "18",
+        "solution":  "【分析】要使烙饼的总时间最少，每次锅内都应尽量有两张饼。前3张饼可以按烙3张饼的方法用3次烙完，后3张饼也同样用3次烙完，一共需要6次，每次3分钟。\n【详解】3×6=18(分钟)。所以烙好6张饼，最快需要18分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "25-26四年级上·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "23",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "妈妈完成下面几件家务：扫地5分钟，烧开水8分钟，洗茶杯1分钟，找茶叶1分钟，沏茶1分钟。想要尽快喝到茶，至少要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "9",
+        "solution":  "【分析】根据题意可知，想要尽快喝到茶，可以在烧开水的同时去扫地，洗茶杯和找茶叶，这样可以节省时间。求至少需要多少时间，直接把烧开水和沏茶的时间加起来即可.\n【详解】8+1=9(分钟)。所以想要尽快喝到茶，至少要9分钟。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "25-26四年级上·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "24",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "复印机同时可以复印2张稿件，复印一面要30秒，有5张稿件两面都要复印，最少需要( )秒复印完。",
+        "options":  [
+
+                    ],
+        "answer":  "150",
+        "solution":  "【分析】5张稿件两面都要复印，共有5×2=10个面。复印机一次最多复印2个面，所以至少需要10÷2=5次，每次30秒。\n【详解】5×2=10(面)，10÷2=5(次)，5×30=150(秒)。所以最少需要150秒复印完。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "25-26四年级上·黑龙江齐齐哈尔·期末"
+    },
+    {
+        "id":  "25",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "期末体育达标测试有三名同学缺考，后来三人一起补考，一共三项，立定跳远，掷铅球，仰卧起坐，有三名体育老师给他们测试，每项测试都需要2分钟，他们最短需要( )分钟能全部测试完。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】三人一起补考，有三名体育老师给他们测试，立定跳远，掷铅球，仰卧起坐，每项测试需要2分钟。三名同学可以同时分别在三名体育老师处参加不同项目测试，共需要测试3轮.\n【详解】2×3=6(分钟)。所以他们最短需要6分钟能全部测试完。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "25-26四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "26",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小月周末帮妈妈做家务:①整理房间20分钟;②拖地15分钟;③找齐要洗的衣服5分钟;④洗衣机洗好衣服50分钟;⑤晾衣服3分钟。完成这些事她至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "58",
+        "solution":  "【分析】洗衣机洗衣服的过程不需要小月一直等待，她可以在洗衣机工作期间完成其他家务。为了最小化总时间，应先完成洗衣前的准备任务(找齐要洗的衣服),然后启动洗衣机，在洗衣机工作期间完成整理房间和拖地任务，最后在洗衣完成后晾衣服。这样，总时间由洗衣时间加上洗衣前准备和洗衣后晾衣服的时间决定.\n【详解】找齐要洗的衣服5分钟;洗衣机洗好衣服50分钟，在洗衣机工作期间完成整理房间和拖地;晾衣服3分钟。5+50+3=58(分钟)。完成这些事她至少需要58分钟。",
+        "knowledge_points":  [
+                                 "100以内数的连加运算",
+                                 "沏茶问题"
+                             ],
+        "source":  "25-26四年级上·甘肃临夏·期末"
+    },
+    {
+        "id":  "27",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "“铁板烧”可同时烤4条鱿鱼，每条鱿鱼两面都要烤，每面需要烤2分钟，那么烤6条鱿鱼至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】根据“烤鱼的总面数=烤好一条鱼需要烤的面数×鱼的条数”，求出烤好6条鱼需要烤的总面数；再根据“烤好鱼最少需要的次数=总面数÷每次最多能烤的数量”，求出需要烤多少次；总时间=每面烤鱼需要的时间×烤的次数。\n【详解】6×2=12(面)，12÷4=3(次)，3×2=6(分钟)。所以烤6条鱿鱼至少需要6分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "25-26四年级上·吉林·期末"
+    },
+    {
+        "id":  "28",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "乐乐经常帮妈妈做家务，上个周末他洗碗用了10分钟，拖地用了15分钟，用全自动洗衣机洗衣服用了35分钟，晾衣服用了5分钟。他做完这些事情至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "40",
+        "solution":  "【分析】用全自动洗衣机洗衣服用35分钟，同时在这期间洗碗用了10分钟，拖地用了15分钟，共25分钟，可以完成，最后再晾衣服用了5分钟，完成这些事至少需要35+5=40分钟.\n【详解】35+5=40(分钟)。所以他做完这些事情至少需要40分钟。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "25-26四年级上·吉林·期末"
+    },
+    {
+        "id":  "29",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈做家务，擦地8分钟，用洗衣机洗衣服25分钟，晾衣服3分钟，洗碗筷6分钟，整理房间10分钟，她至少需要( )分钟才能做完家务。",
+        "options":  [
+
+                    ],
+        "answer":  "28",
+        "solution":  "【分析】用洗衣机洗衣服的25分钟无需手动操作，期间能同步完成擦地，洗碗，整理房间这三项家务(总耗时24分钟，不超过洗衣时间),最后再单独花3分钟晾衣服，这样安排就能算出完成所有家务的最短时间.\n【详解】在洗衣机洗衣的25分钟内同步完成擦地，洗碗，整理房间，之后再晾衣服3分钟，至少需要25+3=28分钟完成家务。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "25-26四年级上·吉林松原·期末"
+    },
+    {
+        "id":  "30",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用一口平底锅烙饼，每次只能放2张饼，烙一面要3分钟，两面都要烙，烙5张饼至少要用( )分钟，烙9张饼至少要用( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "15;27",
+        "solution":  "【分析】先烙2张饼，同时下锅，一面要3分钟，两面一共需要6分钟。剩下3张饼，用烙3张饼的方法共需3×3=9分钟，把这两部分时间加起来就是烙5张饼的最少时间。同样地，烙9张饼时，每张饼每面都要3分钟，最少时间等于9×3=27分钟。\n【详解】烙5张饼：6+9=15(分钟)；烙9张饼：9×3=27(分钟)。用一口平底锅烙饼，烙5张饼至少要用15分钟，烙9张饼至少要用27分钟。",
+        "knowledge_points":  [
+                                 "3的乘法口诀及应用",
+                                 "用2~5的乘法口诀求商",
+                                 "表内乘加，乘减",
+                                 "烙饼问题"
+                             ],
+        "source":  "24-25四年级上·浙江嘉兴·期末"
+    },
+    {
+        "id":  "31",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "平底锅每次最多只能烙2张饼，两面都要烙，每面烙3分钟。妈妈烙4张饼至少需要( )分钟，烙9张饼至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "12;27",
+        "solution":  "【分析】平底锅每次最多烙2张饼，两面都要烙，每面需要烙3分钟。锅里尽量同时烙2张时，烙2张及以上的饼，最少需要的时间=烙一面需要的时间×张数。\n【详解】3×4=12(分钟)，3×9=27(分钟)。因此，妈妈烙4张饼至少需要12分钟，烙9张饼至少需要27分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "24-25四年级上·重庆大足·期末"
+    },
+    {
+        "id":  "32",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "早上聪聪起床后，刷牙洗脸4分钟，微波炉热早餐2分钟，吃早餐8分钟。他做完这些事至少要花( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "12",
+        "solution":  "【分析】根据题意，刷牙洗脸4分钟，这时可以同时微波炉热早餐2分钟，没有超过4分钟，然后再吃早餐8分钟。最后计算4+8=12(分钟)为做完这些事最少要花的时间.\n【详解】4+8=12(分钟)。早上聪聪起床后，刷牙洗脸4分钟，微波炉热早餐2分钟，吃早餐8分钟。他做完这些事至少要花12分钟。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25四年级上·湖北武汉·期末"
+    },
+    {
+        "id":  "33",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用一个平底锅煎牛排，每次最多煎2块牛排，两面都要煎，每面需要3分钟。要煎5块牛排，至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】共需要煎5块牛排，每次能煎2块，每面需要3分钟，煎2个及以上的牛排最少时间等于煎一面需要的时间乘牛排的个数.\n【详解】3×5=15(分钟)。所以至少需要15分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "24-25四年级上·云南文山·期末"
+    },
+    {
+        "id":  "34",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小华和爸爸，妈妈跳绳。现在只有一根跳绳，可以单人跳，也可以双人跳，每次跳1分钟。他们每人都要跳两次，至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "3",
+        "solution":  "【分析】读题可知，每人要跳2次，每次1分钟，要想用时少，则每次都尽可能进行双人跳，可以先让小华和爸爸跳1次，再让小华和妈妈跳1次，然后爸爸和妈妈跳1次.\n【详解】可以先让小华和爸爸跳1次，再让小华和妈妈跳1次，然后爸爸和妈妈跳1次，需要:1+1+1=3(分钟)。所以至少需要3分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "24-25四年级上·云南曲靖·期末"
+    },
+    {
+        "id":  "35",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一张饼两面都要烙，每面需要3分钟，一只平底锅每次可以烙2张，烙熟7张饼至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "21",
+        "solution":  "【分析】先将前4张饼按每次烙2张的方式进行。每次烙2张饼，每面3分钟，那么烙熟2张饼需要6分钟，所以烙熟4张饼需要12分钟。接着烙剩下的3张饼，用烙3张饼的方法还需要9分钟，总时间为12+9分钟。\n【详解】2×3=6(分钟)，6×(4÷2)=12(分钟)，3×3=9(分钟)，12+9=21(分钟)。一张饼两面都要烙，每面需要3分钟，一只平底锅每次可以烙2张，烙熟7张饼至少需要21分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "24-25四年级上·云南文山·期末"
+    },
+    {
+        "id":  "36",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "楷楷和爸爸，妈妈跳绳。现在只有一根跳绳，可以单人跳，也可以双人跳，每次跳1分钟。他们每人都要跳两次，至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "3",
+        "solution":  "【分析】由于每次跳可以单人跳或双人跳，且每人都要跳两次，总跳次为3人*2次=6次。双人跳一次可以同时完成两个人的跳次，因此为了最小化时间，应尽可能安排双人跳。最多可以安排3次双人跳，覆盖所有跳次，因此至少需要3分钟。安排跳的顺序：第一次双人跳，楷楷和爸爸跳;第二次双人跳，楷楷和妈妈跳;第三次双人跳，爸爸和妈妈跳。这样，每个人均跳了两次，共用了3分钟.\n【详解】3×1=3(分钟)。楷楷和爸爸，妈妈跳绳，每人都要跳两次，至少需要3分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "24-25四年级上·天津南开·期末"
+    },
+    {
+        "id":  "37",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小红给客人沏茶，洗水壶2分钟，接水1分钟，烧水5分钟，洗茶杯3分钟，拿茶叶1分钟，沏茶2分钟。小红合理安排以上事情，最少要( )分钟使客人尽快喝上茶。",
+        "options":  [
+
+                    ],
+        "answer":  "10",
+        "solution":  "【分析】小红先洗水壶2分钟，接水1分钟，烧水5分钟，烧水的同时洗茶杯，拿茶叶，最后沏茶2分钟.\n【详解】2+1+5+2=10(分钟)。所以最少要10分钟使客人尽快喝上茶。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25四年级上·重庆云阳·期末"
+    },
+    {
+        "id":  "38",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "星期天，小红帮妈妈做以下事情：扫地5分钟，擦桌子5分钟，烧水11分钟，沏茶1分钟。她做完这些事至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "12",
+        "solution":  "【分析】烧水的时候可以去扫地和擦桌子，等水烧好后再进行沏茶，这样最节省时间.\n【详解】11+1=12(分钟)。所以她做完这些事至少需要12分钟。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25四年级上·河南安阳·期末"
+    },
+    {
+        "id":  "39",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "烧烤店的王叔叔利用铁板烤鱿鱼，一次可同时烤2条鱿鱼，鱿鱼两面都要烤，每面烤3分钟，烤熟8条鱿鱼需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "24",
+        "solution":  "【分析】先烤熟两条鱿鱼的正面，需要3分钟;再烤熟两条鱿鱼的反面，也需要3分钟。这样烤熟2条鱿鱼需要6分钟。而8条鱿鱼需要分成8÷2=4组来烤熟。用每组6分钟乘4组即可.\n【详解】3+3=6(分钟),8÷2×6=24(分钟)。所以，烤熟8条鱿鱼需要24分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "24-25四年级上·山东济宁·期末"
+    },
+    {
+        "id":  "40",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈做早饭的过程及时间：洗锅1分，淘米2分，熬粥20分，煎鸡蛋5分，拌小菜5分，盛粥1分，妈妈做这顿饭至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "24",
+        "solution":  "【分析】妈妈做早饭的过程包括必须按顺序进行的步骤：洗锅，淘米，熬粥，盛粥;以及可以与其他步骤同时进行的步骤：煎鸡蛋，拌小菜。为了节省时间，在熬粥的同时进行煎鸡蛋和拌小菜。必须顺序进行的步骤总时间为1+2+20+1=24分钟，而煎鸡蛋和拌小菜各5分钟，都可以在熬粥的20分钟内完成.\n【详解】1+2+20+1=24(分钟)。妈妈做这顿饭至少需要24分钟。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25四年级上·江西赣州·期末"
+    },
+    {
+        "id":  "41",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "三个小朋友玩一辆碰碰车，玩一次限时5分钟，可以单人玩，也可以双人玩，每人玩2次，至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】要想省时，尽量双人玩。第一个与第二个小朋友玩一次，第一个与第三个小朋友玩一次，第二个与第三个小朋友玩一次，正好每人玩2次.\n【详解】5×3=15(分钟)。最少要用15分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "24-25四年级上·内蒙古巴彦淖尔·期末"
+    },
+    {
+        "id":  "42",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "有10张复习资料，正反面都要复印，复印机一次最多可以放2张，一次只能复印一面，复印一面需要8秒。全部复印完至少需要( )秒。",
+        "options":  [
+
+                    ],
+        "answer":  "80",
+        "solution":  "【分析】因为有10张资料，每张资料正反面都要复印，所以总面数为10×2=20面。复印机一次最多放2张，一次复印一面，那么复印20面需要20÷2=10次。已知复印一面需要8秒，用复印的次数乘复印一面需要的时间即可.\n【详解】10×2÷2×8=80(秒)。所以，全部复印完至少需要80秒。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "24-25四年级上·江西萍乡·期末"
+    },
+    {
+        "id":  "43",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "李阿姨做家务。擦地5分钟，洗衣机洗衣服25分钟，晾衣服3分钟，整理房间10分钟。她至少需要( )分钟才能做完。",
+        "options":  [
+
+                    ],
+        "answer":  "28",
+        "solution":  "【分析】洗衣机洗衣服25分钟，同时可以擦地，整理房间，但晾衣服时不能与其他几项同时进行。接下来用洗衣服的时间加上晾衣服的时间即可求出她做完这些事情至少需要的时间.\n【详解】25+3=28(分钟)。她至少需要28分钟才能做完。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25四年级上·江西上饶·期末"
+    },
+    {
+        "id":  "44",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈准备晚餐，择菜、洗菜8分钟，炒菜10分钟，淘米2分钟，蒸米饭20分钟，妈妈合理安排以上事情，至少需要( )分。",
+        "options":  [
+
+                    ],
+        "answer":  "22",
+        "solution":  "【分析】妈妈准备晚餐时，可以合理安排顺序以节省时间。蒸米饭需要先淘米，然后才能进行蒸煮。蒸米饭的过程需要20分钟，但在此期间妈妈可以同时进行其他任务。择菜、洗菜和炒菜需要顺序进行，总时间为18分钟，小于蒸米饭的20分钟，因此可以在蒸米饭的同时完成。总时间由淘米和蒸米饭的总时间决定。\n【详解】2+20=22(分钟)。所以，妈妈合理安排以上事情，至少需要22分。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25四年级上·天津蓟州·期末"
+    },
+    {
+        "id":  "45",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "天天用温水给奶奶冲一杯豆奶，需要做以下事情：烧开水7分钟，等开水降温2分钟，取豆奶粉1分钟，洗杯子2分钟，冲豆奶粉1分钟。奶奶至少要等( )分钟才能喝上豆奶。",
+        "options":  [
+
+                    ],
+        "answer":  "10",
+        "solution":  "【分析】要想完成这些事情用时最少，天天需要合理安排顺序。天天可先烧开水，烧开水同时取豆奶粉，洗杯子，然后等开水降温，最后冲豆奶粉.\n【详解】7+2+1=10(分钟)。所以奶奶至少要等10分钟才能喝上豆奶。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25四年级上·江西九江·期末"
+    },
+    {
+        "id":  "46",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "牛排馆烤1块牛排需要6分钟(正、反面各3分钟)，如果一块铁板上最多只能烤2块，那么烤5块牛排至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】先把第一块和第二块的正面同时烤，再把第一块的反面和第三块的正面同时烤，然后再把第二块和第三块的反面同时烤，这样前三块要烤3次；剩下的两块一起烤，又要烤2次，5块牛排总共要烤5次，每次3分钟。\n【详解】3+2=5(次)，3×5=15(分钟)。烤5块牛排至少需要15分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "24-25四年级上·重庆丰都·期末"
+    },
+    {
+        "id":  "47",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "周末华华的爸爸，妈妈在加班，中午她一个人在家自己做饭吃。淘米需要2分钟，用电饭煲煮饭需要20分钟，把妈妈提前做好的几个菜放进微波炉里加热需要8分钟，用电磁炉做一碗鸡蛋汤需要5分钟。华华最快( )分钟后能吃上饭。",
+        "options":  [
+
+                    ],
+        "answer":  "22",
+        "solution":  "【分析】华华可以先淘米，再用电饭煲煮饭。在电饭煲煮饭的同时，把菜放进微波炉里加热，做鸡蛋汤。那么需要的就是淘米的时间和电饭煲煮饭的时间.\n【详解】2+20=22(分钟)。所以，华华最快22分钟后能吃上饭。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25四年级上·重庆秀山·期末"
+    },
+    {
+        "id":  "48",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用一只平底锅煎饼，每次最多只能煎两张饼，两面都要煎，每面煎2分钟。煎5张饼至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "10",
+        "solution":  "【分析】煎5张饼：先同时煎两张，正反面共需2×2=4分钟;再煎后三张，先煎第一张与第二张的正面需2分钟，然后煎第一张的反面与第三张的正面需要2分钟，最后煎第二张的反面与第三张的反面需2分钟，煎完3张共需2×3=6分钟,5张共需4+6=10分钟.\n【详解】根据分析可知：用一只平底锅煎饼，每次最多只能煎两张饼，两面都要煎，每面煎2分钟。煎5张饼至少需要10分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "24-25四年级上·内蒙古乌海·期末"
+    },
+    {
+        "id":  "49",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "李叔叔是一名讲解员，他每天早上的时间安排如下：穿衣服，洗漱7分钟，做早餐10分钟，吃早餐10分钟，听讲解视频20分钟。如果李叔叔6:40起床，他最快在( )可以出门。",
+        "options":  [
+
+                    ],
+        "answer":  "7:07",
+        "solution":  "【分析】先穿衣服、洗漱7分钟。听讲解视频时只用耳朵听，所以可以在听讲解视频20分钟的同时做早餐和吃早餐，这样只需计算穿衣服、洗漱和听讲解视频的时间，再加到6:40上。\n【详解】7+20=27(分钟)，6时40分+27分=7:07。所以如果李叔叔6:40起床，他最快在7:07可以出门。",
+        "knowledge_points":  [
+                                 "时，分，秒时间的推算",
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25四年级上·甘肃兰州·期末"
+    },
+    {
+        "id":  "50",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "笑笑帮妈妈做家务，洗衣机洗衣服30分钟，扫地6分钟，擦家具12分钟，晾衣服3分钟。做完家务至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "33",
+        "solution":  "【分析】洗衣机洗衣服需要30分钟，在此期间笑笑可以同时进行扫地和擦家具，因为这两项任务的总时间为6+12=18分钟，小于30分钟，所以能在洗衣服过程中完成。洗衣服完成后，必须进行晾衣服3分钟。因此，总时间至少为洗衣服时间加上晾衣服时间.\n【详解】30+3=33(分钟)。所以做完家务至少需要33分钟。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25四年级上·甘肃临夏·期末"
+    },
+    {
+        "id":  "51",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "有500个乒乓球运动员通过淘汰赛产生1个冠军，要打( )场比赛。",
+        "options":  [
+
+                    ],
+        "answer":  "499",
+        "solution":  "【分析】淘汰赛中，每场比赛淘汰一名选手，最终产生一名冠军需要淘汰其他所有选手。从500名选手中产生冠军，需淘汰500-1=499名选手，也就是需要比赛的场次.\n【详解】500-1=499(场)。所以有500个乒乓球运动员通过淘汰赛产生1个冠军，要打499场比赛。",
+        "knowledge_points":  [
+                                 "比赛问题"
+                             ],
+        "source":  "2025·湖南长沙·小升初真题"
+    },
+    {
+        "id":  "52",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "乐乐猪会做4种川菜，可可猫会做3种粤菜，如果选择其中一道菜，会有( )种选择。",
+        "options":  [
+
+                    ],
+        "answer":  "7",
+        "solution":  "【分析】选择一道菜的情况分为两类：选乐乐猪做的川菜或选可可猫做的粤菜，分别计算两类的选择数后相加即可.\n【详解】选川菜有4种选择，选粤菜有3种选择。根据分类加法计数原理，总选择数为4+3=7(种)。",
+        "knowledge_points":  [
+                                 "加法原理"
+                             ],
+        "source":  "2024三年级·全国·竞赛"
+    },
+    {
+        "id":  "53",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈给家人烙饼，这口锅只能同时烙两张饼，烙一面需要2分钟，要想一家五口能尽快吃上饼，最少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "10",
+        "solution":  "【分析】妈妈需要烙5张饼且用时最少，所以在烙饼时要保证锅里同时烙2张饼.\n【详解】第一个2分钟：放入2张饼;第二个2分钟：烙第1张饼的另一面和第3张饼的正面;第三个2分钟：烙第2张饼的另一面和第3张饼的另一面;第四，第五个2分钟分别烙剩下2张饼的两面。共需要5个2分钟，即10分钟。所以最少需要10分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "24-25六年级下·贵州黔东南·期末"
+    },
+    {
+        "id":  "54",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一只平底锅上只能煎两条鱼，用它煎一条鱼需要4分钟(正反面各2分钟),那么，煎三条鱼至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】通过合理安排煎鱼的顺序，充分利用平底锅每次能煎两条鱼的条件，来节省时间。首先把三条鱼分别标记为鱼1，鱼2，鱼3。先煎鱼1和鱼2的正面，需要2分钟。然后煎鱼1的反面和鱼3的正面，需要2分钟，此时鱼1已经煎好。最后煎鱼2和鱼3的反面，需要2分钟，此时鱼2和鱼3也煎好.\n【详解】一共煎了3次，每次2分钟,2×3=6(分钟)。答：煎三条鱼至少需要6分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "2014四年级·全国·课后作业"
+    },
+    {
+        "id":  "55",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小波骑在马背上赶马过河，共有甲，乙，丙，丁四匹马，甲马过河需要3分钟，乙马过河需要5分钟，丙马过河需要6分钟，丁马过河需要7分钟，每次最多两匹马同时过河，要把4匹马都有赶到对岸去，最少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "24",
+        "solution":  "【分析】要使时间最少，需让用时少的马多次往返，带动用时多的马过河。先让甲和乙过河，甲返回;再让甲和丁过河，甲返回;最后让甲和丙过河，不再返回，完成过河.\n【详解】第一次：甲和乙过河，用时5分钟，甲返回，用时3分钟，共8分钟。第二次：甲和丁过河，用时7分钟，甲返回，用时3分钟，共10分钟。第三次：甲和丙过河，用时6分钟，不再返回。总时间:8+10+6=24分钟。最少需要24分钟。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25六年级下·湖南长沙·期末"
+    },
+    {
+        "id":  "56",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "烤面包，每次最多只能烤2片面包，两面都要烤，每面2分钟，烤2片至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "4",
+        "solution":  "【分析】每次最多烤2片，每面需要2分钟，将两片面包同时烤正面，再同时烤反面，总时间为2+2=4分钟.\n【详解】将两片面包分别记为A和B。第一次烤:A的正面和B的正面，用时2分钟。第二次烤:A的反面和B的反面，用时2分钟。总时间:2+2=4(分钟)。因此，烤2片至少需要4分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "25-26四年级上·海南海口·单元测试"
+    },
+    {
+        "id":  "57",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "园园给客人烧水沏茶。洗水壶要2分钟，烧开水要6分钟，洗茶杯要2分钟，拿茶叶要1分钟，沏茶要1分钟。我认为最合理安排是( )分钟就能沏好茶。",
+        "options":  [
+
+                    ],
+        "answer":  "9",
+        "solution":  "【分析】先洗水壶，再烧开水，在烧水的过程中，洗茶杯，拿茶叶，最后沏茶;洗水壶需要2分钟，烧水需要6分钟，沏茶需要1分钟，所以至少需要花2+6+1分钟.\n【详解】2+6+1=9(分钟)。答：我认为最合理安排是9分钟就能沏好茶。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "25-26四年级上·全国·课后作业"
+    },
+    {
+        "id":  "58",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个37人的旅行团到大夫山公园游玩并计划租车。现有2人车和3人车可供选择，2人车租金每小时12元，3人车租金每小时16元。若租车时间为1小时，则租( )辆2人车和( )辆3人车最划算，总花费应为( )元。",
+        "options":  [
+
+                    ],
+        "answer":  "2;11;200",
+        "solution":  "【分析】租车方案既要考虑费用，又要保证37人都能坐下。因为同样坐6人时，租2辆3人车要16×2=32(元)，租3辆2人车要12×3=36(元)，所以应尽量多租3人车，再搭配少量2人车。\n【详解】方案1：租12辆3人车可坐36人，还差1人，需再租1辆2人车，总花费12×16+1×12=204(元)。方案2：租11辆3人车可坐33人，还差4人，需再租2辆2人车，总花费11×16+2×12=200(元)。方案3：租10辆3人车可坐30人，还差7人，需再租4辆2人车，总花费10×16+4×12=208(元)。方案4：租9辆3人车可坐27人，还差10人，需再租5辆2人车，总花费9×16+5×12=204(元)。比较可知，租11辆3人车和2辆2人车时，总费用最低，且刚好能坐下37人。因此，租2辆2人车和11辆3人车最划算，总花费应为200元。",
+        "knowledge_points":  [
+                                 "租船问题"
+                             ],
+        "source":  "24-25四年级下·广东广州·期末"
+    },
+    {
+        "id":  "59",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "周末，乐乐帮妈妈做饭：淘米需2分，电饭锅煮饭需25分，洗菜需6分，炒菜需16分，乐乐至少需要( )分才能吃上饭。",
+        "options":  [
+
+                    ],
+        "answer":  "27",
+        "solution":  "【分析】要想尽快吃上饭，乐乐需要合理安排需要做的事情。他应该先淘米，然后再用电饭锅煮饭，在电饭锅煮饭的同时，他还可以洗菜和炒菜，则至少需要2+25分钟.\n【详解】6+16=22(分钟),22\u003c25,2+25=27(分钟)。所以，乐乐至少需要27分才能吃上饭。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25四年级下·湖北十堰·期末"
+    },
+    {
+        "id":  "60",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小红放学回家帮妈妈做家务，完成下面这些工作至少需要( )分钟。用电水壶烧开水15分钟，用电饭煲煮饭20分钟，整理房间6分钟，拖地10分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "20",
+        "solution":  "【分析】烧开水和煮饭都属于可以同时进行的事情，整理房间和拖地也可以安排在煮饭过程中进行。只要这些并行完成的家务总时长不超过煮饭时间，总时间就由最长的关键任务决定.\n【详解】整理房间和拖地共6+10=16(分钟),完成整理房间和拖地需要16分钟,20\u003e16\u003e15，所以在完成煮饭的同时可以完成烧开水，整理房间和拖地。综上可知，完成这些工作至少需要20分钟。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25四年级下·湖北十堰·期末"
+    },
+    {
+        "id":  "61",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "5个鸡蛋同时煮需要8分钟,10个鸡蛋同时煮需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "8",
+        "solution":  "【分析】煮鸡蛋的时间取决于煮熟单个鸡蛋所需的时间，与数量无关，只要同时进行，时间不变.\n【详解】5个鸡蛋同时煮需要8分钟,10个鸡蛋同时煮也需要8分钟。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25四年级上·安徽淮南·期末"
+    },
+    {
+        "id":  "62",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小华想帮妈妈做家务，用洗衣机洗衣服要用20分钟，扫地要6分钟，擦家具要用15分钟，晾衣服要用5分钟，请你帮小华想一想，要尽快完成这些事，最少用( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "26",
+        "solution":  "【分析】要统筹安排时间，使事情能够顺利完成，又不至于相互干扰，可以先用洗衣机洗衣服，同时擦家具和扫地，最后晾衣服，这样最快。\n【详解】先用洗衣机洗衣服20分钟，同时完成擦家具15分钟和扫地6分钟。因为15+6=21(分钟)，比洗衣时间多1分钟，所以做完前面三件事共需21分钟。再晾衣服5分钟，一共需要21+5=26(分钟)。这样最快，最少用26分钟。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25四年级上·新疆乌鲁木齐·期末"
+    },
+    {
+        "id":  "63",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "张老师周末上午9:00开始，要做下面几件事：用洗衣机洗衣服50分钟，拖地20分钟，擦桌子20分钟，晾衣服5分钟，下楼扔垃圾5分钟。如果张老师做完一件事再做另一件，共需要( )分钟;如果合理安排一下，只需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "100;55",
+        "solution":  "【分析】如果一件事做完再做另一件，就把所有时间直接相加。如果合理安排，可以先启动洗衣机洗衣服，在洗衣机工作的50分钟内完成拖地，擦桌子和下楼扔垃圾，最后等衣服洗好后再晾衣服.\n【详解】依次完成需要:50+20+20+5+5=100(分钟)。合理安排时，先用洗衣机洗衣服50分钟，同时完成拖地20分钟，擦桌子20分钟，下楼扔垃圾5分钟，这三项共45分钟，不超过50分钟;衣服洗好后再晾衣服5分钟，所以共需要50+5=55(分钟)。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "23-24三年级下·河北保定·期末"
+    },
+    {
+        "id":  "64",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个锅里每次最多只能烙2张饼，两面都要烙，每面烙2分钟，烙8张饼至少要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "16",
+        "solution":  "【分析】计算烙8张饼需要的次数：已知锅里每次最多只能烙2张饼，现在要烙8张饼，则需要烙的次数为8÷2=4次。计算每次烙2张饼所需的时间：因为每面烙2分钟，一张饼有两面，所以烙好一张饼需要的时间是2×2=4分钟。而每次可以同时烙2张饼，这2张饼是同时进行烙制的，所以每次烙2张饼所需要的时间和烙好一张饼的时间是一样的，即每次烙2张饼需要4分钟。计算烙8张饼至少需要的时间：烙8张饼需要烙4次，每次需要4分钟，那么总共需要的时间为4×4=16分钟.\n【详解】8÷2=4(次),2×2=4(分钟),4×4=16(分钟)。所以一个锅里每次最多只能烙2张饼，两面都要烙，每面烙2分钟，烙8张饼至少要16分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "24-25四年级上·河北保定·期末"
+    },
+    {
+        "id":  "65",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明早饭经常自己做烤面包片，每面需要烤2分钟，两面都要烤，每次只能放两片，小明要烤5片面包需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "10",
+        "solution":  "【分析】这实际是烙饼问题，结合所学知识，一次最多烤2片，一面时间为2分钟，要求所花的时间最少，就要在保证火炉不空着的情况下，要烤5片的最短时间是2×5分钟。当烤2片及以上的面包(两面都要烤),并且锅里只能同时烤2片时，最少需要的时间=烤一面需要的时间×片数.\n【详解】先烤第一，二片面包的正面，需要2分钟。然后烤第一片面包的反面和第三片面包的正面，需要2分钟。接着烤第二，三片面包的反面，需要2分钟。再烤第四，五片面包的正面，需要2分钟。最后烤第四，五片面包的反面，需要2分钟。一共:2×5=10(分钟)。小明要烤5片面包需要10分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "24-25四年级上·河北保定·期末"
+    },
+    {
+        "id":  "66",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "星期天，小明帮奶奶一起做家务。他们择菜5分钟，洗菜3分钟，电饭锅煮饭15分钟，淘米3分钟，切菜3分钟，炒菜5分钟，做完这些至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "19",
+        "solution":  "【分析】要使花费的时间最少，需要考虑哪些家务可以在同一时间内进行，哪些家务是不能少的，合理利用时间，因为电饭锅煮饭不需要一直盯着，在煮饭的15分钟内可以同时做其他家务.\n【详解】首先淘米3分钟，然后电饭锅煮饭15分钟，在煮饭的15分钟内，可以同时进行择菜5分钟，洗菜3分钟，切菜3分钟和炒菜5分钟。这几项家务总共需要5+3+3+5=16分钟，所以煮好饭跟做好这几项家务总共就花了16分钟，加上前面淘米的3分钟，则总共需要:16+3=19(分钟)。所以做完这些至少需要19分钟。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25四年级上·河北保定·期末"
+    },
+    {
+        "id":  "67",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈要烤面包，烤箱每次最多能放两片面包，每片面包的两面都要烤，烤一面需要4分钟，烤3片面包至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "12",
+        "solution":  "【分析】要使烤面包的时间最少，那么每一次烤面包时烤箱里面都必须有2片面包。妈妈烤3片面包时：第一次烤第1片面包的正面和第2片面包的正面，用时4分钟;第二次烤第1片面包的反面和第3片面包的正面，用时4分钟;第三次烤第2片面包的反面和第3片面包的反面，用时4分钟。所以一共需要3个4分钟.\n【详解】3×4=12(分钟)。故妈妈烤3片面包至少需要12分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "24-25四年级上·河北保定·期末"
+    },
+    {
+        "id":  "68",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "煮蛋器有两层，每层一次最多可以煮7个鸡蛋，煮熟一个鸡蛋需要8分钟，那么煮熟28个鸡蛋至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "16",
+        "solution":  "【分析】煮蛋器有两层，每层一次最多可以煮7个鸡蛋,28÷7算出要放4层，然后用4÷2算出需要煮两次，煮一次需要8分钟,2×8即为煮熟28个鸡蛋至少需要几分钟.\n【详解】28÷7÷2=2(次),2×8=16(分钟)。煮熟28个鸡蛋至少需要16分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "24-25四年级上·河北保定·期末"
+    },
+    {
+        "id":  "69",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "星期天妈妈要做以下家务：擦玻璃要20分钟，收拾厨房要15分钟，用全自动洗衣机洗衣服要30分钟，晾衣服要6分钟。妈妈做完这些家务至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "41",
+        "solution":  "【分析】擦玻璃和收拾厨房共需要20+15=35分钟，用全自动洗衣机洗衣服需要30分钟，这30分钟可以和前面的家务同时进行，因此关键时间应看35分钟和30分钟中较长的那个，最后再晾衣服。\n【详解】20+15=35(分钟)，35>30，所以前面三件事至少需要35分钟。再晾衣服6分钟，共需35+6=41(分钟)。所以妈妈做完这些家务至少需要41分钟。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25四年级上·贵州安顺·期末"
+    },
+    {
+        "id":  "70",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "笑笑想给爸爸，妈妈和自己每人煎一个鸡蛋。一个平底锅每次只能煎两个鸡蛋，两面都要煎，每面煎2分钟，最少( )分钟全家人都能吃上煎蛋。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】一共煎3个鸡蛋，先煎第一个与第二个的正面需2分钟，然后煎第一个的反面与第三个的正面需要2分钟，最后煎第二个的反面与第三个的反面需2分钟，煎完3个共需2×3=6分钟.\n【详解】2×3=6(分钟)。笑笑最少6分钟全家人都能吃上煎蛋。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "24-25四年级上·贵州遵义·期末"
+    },
+    {
+        "id":  "71",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个锅里每次最多只能烙2张饼，两面都要烙，每面烙2分钟，烙8张饼至少要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "16",
+        "solution":  "【分析】先算出烙8张饼需要的次数。因为锅里每次最多只能烙2张饼，所以8÷2=4(次);每次烙2张饼，两面都要烙，每面2分钟，所以每次需要2×2=4(分钟).\n【详解】8÷2=4(次),2×2=4(分钟),4×4=16(分钟)。所以烙8张饼至少要16分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "24-25四年级上·贵州黔西·期末"
+    },
+    {
+        "id":  "72",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.4,
+        "question":  "大年初一的主食除水饺外，也可以吃荷包蛋，寓意是圆圆满满。煎荷包蛋要煎两面，正面需要2分钟，反面需要1分钟，锅里每次最多可以同时煎2个荷包蛋。煎3个荷包蛋至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "5",
+        "solution":  "【分析】第一阶段2分钟：同时煎两个荷包蛋的正面。第二阶段2分钟：先煎第一个蛋的反面和第三个蛋的正面1分钟，再取出第一个蛋，放入第二个蛋的反面，同时继续煎第三个蛋的正面1分钟。第三阶段1分钟：单独煎第三个蛋的反面.\n【详解】2+2+1=5(分钟),煎3个荷包蛋至少需要5分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "24-25四年级上·河北保定·期末"
+    },
+    {
+        "id":  "73",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一只平底锅最多能煎两条小黄鱼，用它煎1条小黄鱼需要2分钟(正，反面各一分钟)。那么煎3条小黄鱼至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "3",
+        "solution":  "【分析】开始时可以先放A,B两条鱼,1分钟后可以翻转B,拿出A,放入C;1分钟以后可以拿出B,再把A的反面放入，翻转C,再1分钟即可.\n【详解】1+1+1=3(分钟)。那么，煎3条小黄鱼至少需要3分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "24-25四年级上·贵州铜仁·期末"
+    },
+    {
+        "id":  "74",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用平底锅煎鸡蛋，每次只能放2个鸡蛋，煎一面要2分钟，两面都要煎，那么煎6个鸡蛋要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "12",
+        "solution":  "【分析】根据烙饼问题公式：总时间=饼数*2/每锅可烙的数量×烙每面的时间，把煎鸡蛋看作烙饼，代入数据计算即可.\n【详解】6×2÷2×2=12(分钟)。那么煎6个鸡蛋要12分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "24-25四年级上·河南漯河·期末"
+    },
+    {
+        "id":  "75",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小芳早晨起来是这样安排的：刷牙洗脸4分钟，淘米2分钟，用电饭煲煮饭15分钟，梳头3分钟，吃早饭10分钟，小芳起床后最快( )分钟可以去上学。",
+        "options":  [
+
+                    ],
+        "answer":  "27",
+        "solution":  "【分析】小芳早晨起床后，可以先淘米，再用电饭煲煮饭，煮饭的同时可以刷牙洗脸，梳头，饭煮好后再吃早饭。计算淘米，煮饭，吃早饭的时间之和即可.\n【详解】2+15+10=27(分钟)。小芳起床后最快27分钟可以去上学。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25四年级上·河北张家口·期末"
+    },
+    {
+        "id":  "76",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "琪琪每晚睡觉前都要做几件事：喝牛奶，洗漱，练舞，听英语。完成这些事情所需的时间：喝牛奶2分钟，洗漱14分钟，练舞15分钟，听英语14分钟，琪琪做完这些事，最少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "31",
+        "solution":  "【分析】听英语不需要占用专门的时间去做，可以在洗漱和练舞的时候同时听英语。洗漱14分钟加上练舞15分钟，总共需要29分钟，而听英语的14分钟包含在这29分钟内。再加上喝牛奶单独花费的2分钟即可.\n【详解】14+15=29(分钟),29+2=31(分钟)。所以琪琪做完这些事，最少需要31分钟。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25四年级上·贵州六盘水·期末"
+    },
+    {
+        "id":  "77",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在贵州苗寨的芦笙节活动中，阿花要为客人准备午饭。淘米2分钟，杀鱼5分钟，煮酸汤鱼15分钟，煮饭20分钟。为了让客人尽快吃上午饭，阿花应先( ),在( )的同时可以( ),这样安排只要( )分钟客人就能吃上午饭。",
+        "options":  [
+
+                    ],
+        "answer":  "淘米;煮饭;杀鱼，煮酸汤鱼;22",
+        "solution":  "【分析】要使需要的时间最短，应先淘米用2分钟，然后煮饭用20分钟，在完成煮饭这项任务的同时，可以完成杀鱼和煮酸汤鱼这两项任务，一共需要2+20=22分钟.\n【详解】2+20=22(分钟)。为了让客人尽快吃上午饭，阿花应先淘米，在煮饭的同时可以杀鱼，煮酸汤鱼，这样安排只要22分钟客人就能吃上午饭。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25四年级上·贵州黔南·期末"
+    },
+    {
+        "id":  "78",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一口平底锅能同时煎两条鱼，用它煎一条鱼需要6分钟(正，反面各煎3分钟)。那么煎3条鱼，至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "9",
+        "solution":  "【分析】三条鱼分别为a,b,c;第一次煎a,b的正面;第二次煎a的反面和c的正面;第三次煎b,c的反面。所以煎三条鱼需要3×3分钟.\n【详解】3×3=9(分钟)。那么煎3条鱼，至少需要9分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "24-25四年级上·浙江绍兴·期末"
+    },
+    {
+        "id":  "79",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈早上起来要完成以下几件事：洗水壶要1分钟，烧开水要8分钟，把水灌进水瓶要1分钟，整理卧室需要8分钟，吃早饭要10分钟。把这些事情做好，最少要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "20",
+        "solution":  "【分析】妈妈要想做完这些事情用时最少，需要合理安排顺序。她可以先洗水壶，然后烧开水;在烧开水的同时整理卧室;整理完卧室后，把水灌进水瓶;最后再吃早饭.\n【详解】1+8+1+10=20(分钟)。把这些事情做好，最少要20分钟。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25四年级上·河南漯河·期末"
+    },
+    {
+        "id":  "80",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "放学时王老师，李老师，张老师同时去四(2)班布置作业。王老师需要10分钟，李老师需要7分钟，张老师需要5分钟。要使三位老师等候的时间总和最少，应安排( )先布置，等候的时间总和至少要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "张老师;17",
+        "solution":  "【分析】一个老师布置作业需要的时间越长，其他老师等待的时间就越长。要使三位老师等候的时间总和最少，就安排时间短的老师先布置，需要时间多的老师排在后面布置.\n【详解】张老师先布置5分钟：李老师等待5分钟，王老师等待5分钟;李老师接着布置7分钟：王老师再等待7分钟，累计等待12分钟;王老师最后布置10分钟。等候时间总和为5+5+7=17(分钟)。所以应安排张老师先布置，等候的时间总和至少要17分钟。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25四年级上·浙江台州·期末"
+    },
+    {
+        "id":  "81",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用一只平底锅烙饼，每次能同时烙2张饼，饼的两面都要烙(每面2分钟),那么烙45张饼最少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "90",
+        "solution":  "【分析】先分析烙2个饼的情况：平底锅每次能同时放2个饼，烙一个饼的正，反面各需要2分钟，那么同时烙2个饼，正面2分钟，反面2分钟，共需4分钟。然后烙前42个饼：因为42÷2=21(次),所以烙42个饼可以分21次烙完，每次4分钟，则烙42个饼总共花费84分钟。接着烙剩下的3个饼：第一次正1正2;第二次反1正3;第三次反2反3，共需6分钟。最后将两部分时间相加.\n【详解】21×4+2+2+2=90(分钟)。即用一只平底锅烙饼，每次能同时烙2张饼，烙45张饼最少需要90分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "24-25四年级上·浙江杭州·期末"
+    },
+    {
+        "id":  "82",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一只平底锅一次只能烙两张饼，用它烙好一张饼需要6分钟(正反两面各3分钟),那么烙好7张饼至少需( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "21",
+        "solution":  "【分析】第一次：正1正2;第二次：反1正3;第三次：反2正4;第四次：反3正5;第五次：反4正6;第六次：反5正7;第七次：反6反7。根据烙饼问题公式：总时间=饼数*2/每锅可烙的数量×烙每面的时间.\n【详解】14÷2×3=21(分钟)。一只平底锅一次只能烙两张饼，烙好7张饼至少需21分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "24-25四年级上·安徽安庆·期末"
+    },
+    {
+        "id":  "83",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "家里来了客人，要给客人沏茶，用水壶接水1分钟，水壶烧水需要8分钟，洗茶杯1分钟，沏茶1分钟，合理安排后至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "10",
+        "solution":  "【分析】洗茶杯需要1分钟，烧水需要8分钟，可以在等待烧水的过程中完成洗茶杯的任务，所以接水时间加上烧水时间再加上沏茶的时间就是最少需要的时间.\n【详解】1+8+1=10(分钟)。家里来了客人，要给客人沏茶，合理安排后至少需要10分钟。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25四年级上·浙江绍兴·期末"
+    },
+    {
+        "id":  "84",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "爸爸起床要做完:①整理被褥2分钟,②刷牙3分钟,③洗脸2分钟,④听新闻8分钟这4件事情，至少要用( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "8",
+        "solution":  "【分析】爸爸听新闻8分钟的同时可以依次完成整理被褥2分钟，刷牙3分钟，洗脸2分钟，即2+3+2=7分钟，所以做完这几件事情最少要用的时间就是听新闻的时间8分钟.\n【详解】爸爸起床要做完这4件事情，至少要用8分钟。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25四年级上·浙江绍兴·期末"
+    },
+    {
+        "id":  "85",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈做早饭需要做以下事情：淘米2分钟，煎鸡蛋5分钟，熬米粥20分钟，炒菜10分钟。妈妈做这顿饭至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "22",
+        "solution":  "【分析】妈妈需要完成四项任务，合理安排顺序可缩短总时间。步骤如下：淘米2分钟，必须优先完成，以便开始熬粥;熬米粥20分钟，淘米后立即开始，耗时最长;煎鸡蛋5分钟和炒菜10分钟，在熬粥的20分钟内同时进行.\n【详解】20+2=22(分钟)。妈妈做这顿饭至少需要22分钟。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25四年级上·河南漯河·期末"
+    },
+    {
+        "id":  "86",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "两个油漆工给5块同样的木板正反面刷漆，每面需2分钟，刷完5块木板至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "10",
+        "solution":  "【分析】要使刷完5块木板的时间最少，两个油漆工应同时工作，让工作时间尽可能重叠。把5块木板编号为1,2,3,4,5。第一步：同时刷1号和2号的正面;第二步：同时刷1号和2号的反面;第三步：同时刷3号和4号的正面;第四步：同时刷3号的反面和5号的正面;第五步：同时刷4号和5号的反面.\n【详解】2×5=10(分钟)。两个油漆工给5块同样的木板正反面刷漆，至少需要10分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "24-25四年级上·河南新乡·期末"
+    },
+    {
+        "id":  "87",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈要烙5张饼，锅里每次最多只能烙2张，两面都要烙，每面烙3分钟，至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】当烙2张及以上的饼，并且锅里只能同时烙两张饼时，最少需要的时间=烙一面需要的时间×张数.\n【详解】3×5=15(分钟)。妈妈要烙5张饼，至少需要15分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "2024·广东广州·小升初真题"
+    },
+    {
+        "id":  "88",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "周末，奇奇用自家的烧烤架烤烧饼，烤架上每次能同时放两个烧饼，如果烤熟一个烧饼的正，反面各需要2分钟，那烤熟7个烧饼至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "14",
+        "solution":  "【分析】先烤前4个烧饼：每次同时烤2个，正反两面各2分钟，所以烤熟2个烧饼需要4分钟，烤熟4个烧饼需要8分钟。再烤剩下3个烧饼，可以采用交替烙法：第一次烤两个正面，第二次烤其中一个反面和第三个正面，第三次烤前一个剩余的反面和第三个的反面，共需6分钟.\n【详解】4×2+2+2+2=14(分钟)。所以烤熟7个烧饼至少需要14分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "24-25四年级上·河南郑州·期末"
+    },
+    {
+        "id":  "89",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "淘米2分钟，熬粥18分钟，煎鸡蛋4分钟，拌小菜5分钟，盛粥1分钟。如果妈妈从6:40开始做早饭，最快( )开饭。",
+        "options":  [
+
+                    ],
+        "answer":  "7:01",
+        "solution":  "【分析】淘米需要2分钟，熬粥需要18分钟。煎鸡蛋，拌小菜需要4+5=9分钟，所以妈妈可在等待粥熟的18分钟内完成煎鸡蛋与拌小菜，所以妈妈做这顿饭至少需要2+18+1=21分钟。根据结束时刻=开始时刻+经过时间，求出开饭时间.\n【详解】2+18+1=21(分钟),6:40+21分=7:01。所以如果妈妈从6:40开始做早饭，最快7:01开饭。",
+        "knowledge_points":  [
+                                 "时，分，秒时间的推算",
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25四年级上·湖南岳阳·期末"
+    },
+    {
+        "id":  "90",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈中午下班回家做饭，淘米要2分钟，洗菜要6分钟，煮饭要23分钟，切菜切肉要4分钟，炒菜要12分钟。如果煮饭和炒菜用不同的锅子和炉子，最少要( )分钟才能把饭菜做好。",
+        "options":  [
+
+                    ],
+        "answer":  "25",
+        "solution":  "【分析】洗菜6分钟，切菜切肉4分钟，再炒菜12分钟，这个时间是22分钟。所以妈妈先淘米，然后煮饭。在煮饭23分钟的同时，洗菜，切菜切肉，炒菜，这样用的时间最少.\n【详解】2+23=25(分钟)。所以，最少要25分钟才能把饭菜做好。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25四年级上·广东汕头·期末"
+    },
+    {
+        "id":  "91",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "欢欢在家里烙饼，锅里可最多放2张饼，两面都要烙，每面要烙3分钟。烙5张饼至少要( )分钟，烙9张饼至少要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "15;27",
+        "solution":  "【分析】烙5张饼时，可先烙2张共6分钟，再用烙3张饼的方法再烙9分钟，共15分钟。烙9张饼时，可把9张饼按3张一组分成3组，每组用烙3张饼的方法都需要9分钟，所以一共需要27分钟.\n【详解】3×5=15(分钟),3×9=27(分钟)。所以烙5张饼至少要15分钟，烙9张饼至少要27分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "24-25四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "92",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "如果烙熟一张饼饼需4分钟(每面2分钟),一个平底锅每次最多可以烙2张饼，那么烙熟8张饼至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "16",
+        "solution":  "【分析】先算出8张饼总共的面数，然后根据平底锅每次能烙的面数，得出需要烙的次数，最后结合每次烙饼所需时间，求出烙熟8张饼至少需要的时间.\n【详解】8×2=16(面),16÷2=8(次),2×8=16(分钟)。所以烙熟8张饼至少需要16分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "24-25四年级上·河南三门峡·期末"
+    },
+    {
+        "id":  "93",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一张饼两面都要烙，每面需要3分钟，一个平底锅每次最多可以烙2张，烙熟3张饼至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "9",
+        "solution":  "【分析】烙3张饼时，可以先烙两张饼的正面，再烙其中一张的反面和第三张的正面，最后再烙剩余两面的反面，总共需要3次，每次3分钟.\n【详解】6+3=9(分钟)。张饼两面都要烙，每面需要3分钟，一个平底锅每次最多可以烙2张，烙熟3张饼至少需要9分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "24-25四年级上·河南信阳·期末"
+    },
+    {
+        "id":  "94",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "描金是一种传统工艺美术技艺，有两道工序，先上漆，再描绘花纹，每道工序10分钟。现在两位工匠要为3件漆器描金，最少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "30",
+        "solution":  "【分析】每件漆器有两道工序，上漆和描绘花纹，每道工序10分钟，那么完成一件漆器的描金需要20分钟。有两位工匠同时工作，可以将工序进行合理分配来节省时间。两位工匠先同时给两件漆器上漆，接着其中一位工匠给第一件漆器描绘花纹，另一位工匠给第三件漆器上漆，两位工匠再同时给第二件和第三件漆器描绘花纹.\n【详解】第一步：两位工匠先同时给两件漆器上漆，这需要10分钟。第二步：其中一位工匠给第一件漆器描绘花纹，另一位工匠给第三件漆器上漆，这又需要10分钟。第三步：两位工匠再同时给第二件和第三件漆器描绘花纹，这需要10分钟。总共花费:10+10+10=30(分钟)。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "24-25四年级上·广东佛山·期末"
+    },
+    {
+        "id":  "95",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "妙妙是个安排时间的小能手。周末在家，她不仅可以完成自己的学习任务，还能主动帮忙做家务。洗衣机洗衣服需要30分钟，英语打卡需要10分钟，写作业需要25分钟，晾衣服需要2分钟。妙妙做完这些事至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "37",
+        "solution":  "【分析】本题考查合理安排时间。根据题意，洗衣机洗衣服需要30分钟，英语打卡需要10分钟，写作业需要25分钟。英语打卡和写作业共需要35分钟，可以和洗衣机洗衣服的30分钟部分重合，最后再晾衣服2分钟。\n【详解】10+25=35(分钟)，35>30，所以前面三件事至少需要35分钟。再晾衣服2分钟，共需35+2=37(分钟)。所以妙妙做完这些事至少需要37分钟。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25四年级上·河南信阳·期末"
+    },
+    {
+        "id":  "96",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个小平底锅一次最多可以煎2只鸡翅，两面都要煎，每面需要5分钟，煎6只鸡翅至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "30",
+        "solution":  "【分析】根据题意，每次煎2只鸡翅，每只鸡翅需要煎两面，每面5分钟。煎2只鸡翅的时间：同时煎2只的正面5分钟，然后同时煎2只的反面5分钟，共需10分钟。6只鸡翅需分3组煎，每组耗时10分钟.\n【详解】6÷2=3(组),5×2=10(分),10×3=30(分)。一个小平底锅一次最多可以煎2只鸡翅，煎6只鸡翅至少需要30分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "24-25四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "97",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "公园里有一种射击游戏和一种套圈游戏，每项每次只能一个人玩，玩一次都需要用时5分钟。3位小朋友相约一起来玩，如果每人两种游戏都玩到至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】此类问题中，尽量使两种游戏始终都有小朋友在玩，由此进行合理安排即可解决问题.\n【详解】3位小朋友分别用序号A,B,C表示：第一次,A射击,B套圈，用时5分钟;第二次,A套圈,C射击，用时5分钟;第三次,B射击,C套圈，用时5分钟。5+5+5=15(分钟)。如果每人两种游戏都玩到至少需要15分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "24-25四年级上·湖北荆州·期末"
+    },
+    {
+        "id":  "98",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "小红周六帮妈妈做家务：用自动洗衣机洗衣服需30分钟;洗碗用6分钟;擦地板用16分钟，晾衣服用5分钟，做完这些事至少要用( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "35",
+        "solution":  "【分析】用自动洗衣机洗衣服需30分钟，同时在这期间洗碗用了6分钟，擦地板用16分钟，共22分钟，可以完成，最后再晾衣服用了5分钟，完成这些事至少需要30+5分钟.\n【详解】30+5=35(分钟)。所以做完这些事至少要用35分钟。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "99",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "聪聪要冲一杯牛奶，洗水壶2分钟，洗杯子2分钟，用水壶烧开水5分钟，冲牛奶1分钟，冷却2分钟。聪聪最快( )分钟能喝上牛奶。",
+        "options":  [
+
+                    ],
+        "answer":  "10",
+        "solution":  "【分析】根据沏茶问题，用水壶烧开水的5分钟，可以同时洗杯子2分钟，这样可以节省2分钟，即洗水壶2分钟+烧开水5分钟+冲牛奶1分钟+冷却2分钟.\n【详解】2+5+1+2=10(分钟)。聪聪最快10分钟能喝上牛奶。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25四年级上·湖北荆州·期末"
+    },
+    {
+        "id":  "100",
+        "unit":  "第八章 数学广角",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "星期六，爸爸和明明一起做家务。晾晒洗好的衣服和鞋子6分钟；用洗衣机洗衣服55分钟；刷鞋20分钟；整理书房15分钟；抹桌拖地20分钟。为合理安排这些事情要注意的事项是( )，完成这些事至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "要按一定的顺序做，能同时做的事情可以同时做;61",
+        "solution":  "【分析】这是合理安排时间的问题。要按一定的顺序做，能同时做的事情可以同时做。可以先启动洗衣机洗衣服，在洗衣机工作的55分钟内，同时完成刷鞋、整理书房、抹桌拖地等事情；洗衣完成后再晾晒洗好的衣服和鞋子。\n【详解】用洗衣机洗衣服55分钟时，可以同时完成刷鞋20分钟、整理书房15分钟、抹桌拖地20分钟，这些事情都不超过55分钟；最后再晾晒洗好的衣服和鞋子6分钟。因此总时间为55+6=61(分钟)。所以为合理安排这些事情要注意的事项是：要按一定的顺序做，能同时做的事情可以同时做;完成这些事至少需要61分钟。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "101",
+        "unit":  "第八章数学广角 2",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "战国时，齐王与田忌赛马，双方都有上等马，中等马和下等马三种，且齐王每个等级的马都比田忌马略强。比赛分三场进行，每场各出一马，赢得两场者胜。假设齐王出马的顺序是上等马，中等马，下等马，那么田忌共有( )种应对策略，而田忌赢齐王的策略只有( )种。",
+        "options":  [
+
+                    ],
+        "answer":  "6;1",
+        "solution":  "【分析】根据题意，因为齐王每个等级的马比田忌每个等级的马都略强，但是田忌的上，中等马分别要比齐王的中，下等马强，所以当齐王的马按上，中，下等的顺序出阵时，田忌的马按下，上，中等的顺序出阵，田忌便能取胜，据此解答.\n【详解】根据分析可得：田忌的应对策略如下表\n\n\u003ctable\u003e\u003ctr\u003e\u003ctd\u003e齐王的马\u003c/td\u003e\u003ctd\u003e上中下\u003c/td\u003e\u003ctd\u003e上中下\u003c/td\u003e\u003ctd\u003e上中下\u003c/td\u003e\u003ctd\u003e上中下\u003c/td\u003e\u003ctd\u003e上中下\u003c/td\u003e\u003ctd\u003e上中下\u003c/td\u003e\u003c/tr\u003e\u003ctr\u003e\u003ctd\u003e田忌的马\u003c/td\u003e\u003ctd\u003e上中下\u003c/td\u003e\u003ctd\u003e上下中\u003c/td\u003e\u003ctd\u003e中上下\u003c/td\u003e\u003ctd\u003e中下上\u003c/td\u003e\u003ctd\u003e下上中\u003c/td\u003e\u003ctd\u003e下中上\u003c/td\u003e\u003c/tr\u003e\u003c/table\u003e\n\n由上表可知田忌共有6种应对策略，但是只有一种策略他能赢齐王，即为齐王出上，中，下，田忌出下，上，中。所以那么田忌共有6种应对策略，而田忌赢齐王的策略只有1种。",
+        "knowledge_points":  [
+                                 "田忌赛马问题"
+                             ],
+        "source":  "24-25 四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "102",
+        "unit":  "第八章数学广角 2",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "周末，妈妈交代明明完成以下家务：扫地7分钟，拖地5分钟，烧水10分钟，浇花3分钟。最节省时间的安排方法是( ),共用( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "烧水的同时扫地，然后浇花，最后拖地;15",
+        "solution":  "【分析】根据题意，烧水(10分钟)和扫地(7分钟)可以同时开始。烧水的前7分钟完成扫地。扫地完成后，烧水还剩3分钟，此时进行浇花(3分钟)，刚好在烧水结束时完成。最后单独进行拖地(5分钟)，总时间为10+5=15(分钟)。\n【详解】最节省时间的安排方法是烧水的同时扫地，然后浇花，最后拖地，共用15分钟。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25 四年级上·河南信阳·期末"
+    },
+    {
+        "id":  "103",
+        "unit":  "第八章数学广角 2",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "亮亮有10、8、5三张牌，明明有9、7、4三张牌，规定每人每次出一张牌。如果亮亮先出，明明获胜的策略是10-( )，8-( )，5-( )。(三局两胜制)",
+        "options":  [
+
+                    ],
+        "answer":  "4;9;7",
+        "solution":  "【分析】根据“田忌赛马”的方法，用明明最小的牌对亮亮最大的牌，用明明最大的牌对亮亮次大的牌，用明明次大的牌对亮亮最小的牌，这样明明就会2胜1负，从而取胜。据此解答。\n【详解】10 > 9 > 8 > 7 > 5 > 4，由分析可知，如果亮亮先出，明明获胜的策略是10-4，8-9，5-7。(三局两胜制)",
+        "knowledge_points":  [
+                                 "田忌赛马问题"
+                             ],
+        "source":  "24-25 四年级上·湖北十堰·期末"
+    },
+    {
+        "id":  "104",
+        "unit":  "第八章数学广角 2",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一口锅每次最多只能煎2条鱼,1条鱼两面都要煎，每面煎3分钟，妈妈要煎5条鱼至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】先煎前2条鱼，需要时间(3×2)分钟;再煎后3条鱼，假设剩下的3条鱼分别为a,b,c,第一次煎a的正面,b的正面，第二次煎a的反面,c的正面，第三次煎b的反面,c的反面，则需要:3×3=9(分钟);一共需要:6 + 9 = 15(分钟),这就是至少需要的煎鱼时间。即，一口锅每次最多只能煎2条鱼,1条鱼两面都要煎，每面煎3分钟，妈妈要煎5条鱼至少需要15分钟.\n【详解】计算总时间:=6+9=15(分钟),即，一口锅每次最多只能煎2条鱼,1条鱼两面都要煎，每面煎3分钟，妈妈要煎5条鱼至少需要15分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "24-25 四年级上·湖北十堰·期末"
+    },
+    {
+        "id":  "105",
+        "unit":  "第八章数学广角 2",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "牛排店烤一块牛排需要12分钟(正，反面各6分钟),一块铁板上最多能烤2块，那么烤5块牛排至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "30",
+        "solution":  "【分析】根据题意，每次最多能烤2块，可以列出每一次烤的情况，用烤的次数乘每次烤需要的时间，即可求出至少需要的时间.\n【详解】第一次：第一块牛排正面，第二块牛排正面;第二次：第一块牛排反面，第二块牛排反面;第三次：第三块牛排正面，第四块牛排正面;第四次：第三块牛排反面，第五块牛排正面;第五次：第四块牛排反面，第五块牛排反面。6×5=30(分钟),牛排店烤一块牛排需要12分钟(正，反面各6分钟),一块铁板上最多能烤2块，那么烤5块牛排至少需要30分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "24-25 四年级上·福建龙岩·期末"
+    },
+    {
+        "id":  "106",
+        "unit":  "第八章数学广角 2",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在烤盘里烤面包(一次可以烤两片)时，烤一片面包需要4分钟，正反面各2分钟，那么妈妈要烤5片面包至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "10",
+        "solution":  "【分析】要烤5片面包，每片面包需要烤正反面各2分钟，总共需要4分钟。烤盘一次可以烤两片面包。我们可以按照以下步骤来计算所需的最少时间：第一轮：烤第1片和第2片面包的正面，需要2分钟。第二轮：烤第1片面包的反面，同时开始烤第3片面包的正面，需要2分钟。第三轮：烤第3片面包的反面，同时开始烤第4片面包的正面，需要2分钟。第四轮：烤第4片面包的反面，同时开始烤第5片面包的正面，需要2分钟。第五轮：烤第5片面包的反面，同时烤第2片面包的反面，需要2分钟。因此，总共用时5个2分钟.\n【详解】2×5=10(分钟)。在烤盘里烤面包(一次可以烤两片)时，烤一片面包需要4分钟，正反面各2分钟，那么妈妈要烤5片面包至少需要10分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "24-25 四年级上·河北邯郸·期末"
+    },
+    {
+        "id":  "107",
+        "unit":  "第八章数学广角 2",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "学校进行体质测试,3个同学要去1号教室验视力和2号教室测肺活量，每项检查都要2分钟，他们最短要( )分钟才能全部做完这些检查。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】为了保证做完这些检查用的时间最短，三位同学(A,B,C)要分开检查，如:A验视力，同时B测肺活量，要用2分钟;再B验视力，同时C测肺活量，要用2分钟;最后A测肺活量，同时C测视力，再用2分钟;据此三人做完检查共用了3个2分钟.\n【详解】3×2=6(分钟),他们最短要6分钟才能全部做完这些检查。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "24-25 四年级上·福建漳州·期末"
+    },
+    {
+        "id":  "108",
+        "unit":  "第八章数学广角 2",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "除夕夜，爸爸准备做一道红烧大鲤鱼给家人吃。具体步骤如下：腌制鲤鱼需要20分钟，洗锅需要2分钟，切葱，姜，蒜等配料需要2分钟，热油需要1分钟，烧鱼需要15分钟，装盘需要1分钟。爸爸最快( )分钟可以把这道菜装完盘。",
+        "options":  [
+
+                    ],
+        "answer":  "37",
+        "solution":  "【分析】由题意得，爸爸要想最快把这道菜装完盘，需要合理安排顺序。他可以先腌制鲤鱼，然后在腌制鲤鱼的同时洗锅、切葱、姜、蒜等配料。等鱼腌好后，再热油、烧鱼，最后装盘。据此解答。\n【详解】腌制鲤鱼20分钟(同时洗锅2分钟、切配料2分钟，这两项共4分钟，完全嵌入20分钟内)，热油1分钟，烧鱼15分钟，装盘1分钟。20+1+15+1=37(分钟)，故爸爸最快37分钟可以把这道菜装完盘。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25 四年级上·湖南永州·期末"
+    },
+    {
+        "id":  "109",
+        "unit":  "第八章数学广角 2",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一种电脑小游戏，玩1局要10分钟，可以单人玩，也可以双人玩，小强和小英，小芳一起玩，每人玩两局，至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "30",
+        "solution":  "【分析】要想所用时间最少，那么就按照双人玩来进行，第1个10分钟，小强与小英各玩1局，第2个10分钟，小强与小芳各玩1局，第3个10分钟小芳与小英各玩1局，据此需要3个10分钟，把10与3相乘即可求出需要的最短时间.\n【详解】10分钟：小强1局，小英1局。10分钟：小强1局，小芳1局。10分钟：小芳1局，小英1局。10×3=30(分钟),小强和小英，小芳一起玩，每人玩两局，至少需要30分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "24-25 四年级上·福建福州·期末"
+    },
+    {
+        "id":  "110",
+        "unit":  "第八章数学广角 2",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在《田忌赛马》的故事中，如果以扑克牌的大代表马的竞赛水平高低，例如齐王的马分别是8,6,4，田忌的马分别是7,5,3，请回答以下问题:(1)在下方有序地写出田忌的马的所有出场顺序，数一数共( )种。其中恰能最终赢齐王的出场顺序是( )。 (2)如果从双方的马中挑选一匹赛后与观众见面，有( )种选择。如果从双方的马中各挑选一匹赛后与观众见面，有( )种选择。 (3)如果在故事中的第一局(齐王\"8\"对田忌\"3\")之后，第二局由田忌先确定派出的马，田忌派出\"7\"，那么齐王为了获胜，应当派出( )。说明理由: 。",
+        "options":  [
+
+                    ],
+        "answer":  "6;3、7、5;6;9;4;为保证最终获胜，齐王用4对田忌的7，这样第一局齐王已胜，第三局齐王可用6对田忌的5，能以2胜1负赢得比赛。",
+        "solution":  "【分析】(1)对于田忌的马的所有出场顺序，可以枚举出所有方法有:(7,5,3),(7,3,5),(5,7,3),(5,3,7),(3,7,5),(3,5,7)。所以田忌一共有6种方法安排自己的马的出场顺序。齐王的马是8,6,4，根据田忌赛马的策略，用下等马对上等马，中等马对下等马，上等马对中等马能赢，所以能赢的出场顺序是(3,7,5),即田忌用3对齐王的8，用7对齐王的6，用5对齐王的4。(2)从双方的马中挑选一匹马与观众见面，若只选一匹，双方共3+3=6匹马，所以有6种选择。从双方的马中各挑选一匹马与观众见面，齐王有3种选法，田忌也有3种选法，则共有3×3=9种选择。(3)第一局(齐王\"8\"对田忌\"3\"),第二局田忌派出\"7\"，为了获胜，齐王应该派出4。因为若齐王派出4，虽然这局齐王输，但第三局齐王剩下的6对田忌剩下的5，最终齐王2胜1负;若齐王派出6，虽然这一局齐王赢，但第三局齐王剩下的4对田忌剩下的5，最终齐王1胜2负，所以齐王应派出4，这样能保证最终2胜1负获胜.\n【详解】(1)在下方有序地写出田忌的马的所有出场顺序，数一数共6种。其中恰能最终赢齐王的出场顺序是3,7,5。(2)如果从双方的马中挑选一匹赛后与观众见面，有6种选择。如果从双方的马中各挑选一匹赛后与观众见面，有9种选择。(3)如果在故事中的第一局(齐王\"8\"对田忌\"3\")之后，第二局由田忌先确定派出的马，田忌派出\"7\"，那么齐王为了获胜，应当派出4。说明理由：为保证最终获胜，齐王用4对田忌的7，这样第一局齐王已胜，第三局齐王可用6对田忌的5，能以2胜1负赢得比赛。",
+        "knowledge_points":  [
+                                 "田忌赛马问题"
+                             ],
+        "source":  "24-25 四年级上·福建福州·期末"
+    },
+    {
+        "id":  "111",
+        "unit":  "第八章数学广角 2",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明给爸爸沏茶，接水1分钟，烧水8分钟，洗茶杯2分钟，拿茶叶1分钟，沏茶1分钟。小明合理安排以上事情，至少要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "10",
+        "solution":  "【分析】由题意可知，洗茶杯和拿茶叶一共需要(2+1)分钟，烧水需要8分钟，小明可以在等待烧水的过程中完成洗茶杯和拿茶叶这两项任务，所以接水的时间加上烧水的时间再加上沏茶的时间就是最少需要的时间，据此解答.\n【详解】根据分析可知:1 + 8 + 1 = 10(分钟)。小明给爸爸沏茶，接水1分钟，烧水8分钟，洗茶杯2分钟，拿茶叶1分钟，沏茶1分钟。小明合理安排以上事情，至少要10分钟。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25 四年级上·福建漳州·期末"
+    },
+    {
+        "id":  "112",
+        "unit":  "第八章数学广角 2",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用一个平底锅煎饼，每次都能煎两个饼，如果煎一个饼需要6分钟(正反面各3分钟),煎10个饼最少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "30",
+        "solution":  "【分析】为了节省时间，锅里不能有空余;第一次烙饼：第一块正面，第二块正面;第二次烙饼：第三块正面，第四块正面;第三次烙饼：第五块正面，第六块正面;第四次烙饼：第七块正面，第八块正面;第五次烙饼：第九块正面，第十块正面;第六次烙饼：第一块反面，第二块反面;第七次烙饼：第三块反面，第四块反面;第八次烙饼：第五块反面，第六块反面;第九次烙饼：第七块反面，第八块反面;第十次烙饼：第九块反面，第十块反面;根据烙饼的过程可知，首先求出一共有多少个面;然后求出一共需要烙几次;最后用次数乘一次的时间即可求出总时间;所以总时间 = 饼数*2/每锅的可烙的数量×烙每面的时间，把数代入计算即可.\n【详解】一共需要煎的面数:10×2=20(个);一共需要煎的次数:20÷2=10(次);总时间:10×3=30(分)。所以煎10个饼最少需要30分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "24-25 四年级上·湖南娄底·期末"
+    },
+    {
+        "id":  "113",
+        "unit":  "第八章数学广角 2",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "周末在家，小明完成这些家务：洗水壶2分钟，烧开水15分钟，擦桌子5分钟，洗碗3分钟，拖地2分钟。小明做完这些家务至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "17",
+        "solution":  "【分析】要使用的时间最少，那么能同时做的事情就应同时做，利用烧开水的时间同时做其他家务来计算最少用时。首先洗水壶需要2分钟。然后烧开水需要15分钟，在烧开水的同时，可以擦桌子5分钟，洗碗3分钟和拖地2分钟，因为5 + 3 + 2 = 10分钟,10 \u003c 15，即在烧开水的时间内可以完成这些家务。所以总共花费的时间就是洗水壶的时间加上烧开水的时间，即2 + 15 = 17分钟.\n【详解】5 + 3 + 2 = 8 + 2 = 10(分钟),10 \u003c 15,2 + 15 = 17(分钟)。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25 四年级上·湖南娄底·期末"
+    },
+    {
+        "id":  "114",
+        "unit":  "第八章数学广角 2",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小红每天起床后都要做这些事情：叠被子2分钟，洗漱3分钟，温牛奶3分钟，烤面包1分钟，吃早餐8分钟。她做完这些事情，至少需( )分钟。请用流程图表示出做事的先后顺序。",
+        "options":  [
+
+                    ],
+        "answer":  "13;温牛奶(同时洗漱)→烤面包(同时叠被子)→吃早餐",
+        "solution":  "【分析】由题意得，小红要想做完事情所需时间最少，需要合理安排顺序。她可以先温牛奶，然后在温牛奶的同时去洗漱。接着烤面包，然后在烤面包的同时叠被子，最后再吃早餐。据此解答。前面准备工作最少需要5分钟，再加上吃早餐8分钟，共13分钟。做这些事情的先后顺序为：温牛奶(同时洗漱)→烤面包(同时叠被子)→吃早餐。\n【详解】吃早餐8分钟是固定且无法并行的时间，前面所有准备工作(温牛奶、洗漱、烤面包、叠被子)至少需要5分钟，因此理论最小值为5+8=13(分钟)，此方案已达到最优。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25 四年级上·重庆大渡口·期末"
+    },
+    {
+        "id":  "115",
+        "unit":  "第八章数学广角 2",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "三位同学准备运动会开幕式上用一根跳绳进行表演，他们可以单人跳，也可以双人跳，每人跳两次，每次跳2分钟，三个人都表演完，至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】要使时间最少，可以双人跳，第一位同学和第二位同学跳一次，需要2分钟，第二位同学休息，第一位同学和第三位同学一起跳一次，第一位同学跳了两次，用4分钟;第三位同学再和第二位同学跳一次，需要2分钟;据此求出至少需要的分钟数.\n【详解】据分析可知:2×3=6(分钟)。三位同学准备运动会开幕式上用一根跳绳进行表演，他们可以单人跳，也可以双人跳，每人跳两次，每次跳2分钟，三个人都表演完，至少需要6分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "24-25 四年级上·湖南株洲·期末"
+    },
+    {
+        "id":  "116",
+        "unit":  "第八章数学广角 2",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "小明每天早上起床要做的事有：穿衣服6分钟，整理被褥2分钟，刷牙3分钟，洗脸1分钟，热牛奶3分钟，烤面包2分钟，吃早餐8分钟。做完这些事至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "20",
+        "solution":  "【分析】穿衣服、整理被褥、刷牙、洗脸和吃早餐不能同时进行，但是在进行整理被褥、刷牙、洗脸时可以同时热牛奶和烤面包，所以总时间为6+2+3+1+8=20(分钟)。\n【详解】6+2+3+1+8=20(分钟)。所以小明做完这些事至少需要20分钟。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25 四年级上·湖南常德·期末"
+    },
+    {
+        "id":  "117",
+        "unit":  "第八章数学广角 2",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "玲玲和她的同学两人玩扑克牌比大小的游戏，每人三张牌，每人每次出一张牌(不可重复出牌),赢两次者胜。玲玲拿到的三张牌是5,7,9，她的同学拿到的三张牌是6,8,10。玲玲的同学先出。如果玲玲想要赢，那么当玲玲的同学出10时，玲玲应该出( )。当玲玲的同学出6时，玲玲应该出( )。",
+        "options":  [
+
+                    ],
+        "answer":  "5;7",
+        "solution":  "【分析】本题根据田忌赛马的对策进行解答。玲玲的同学先出，要想玲玲赢，玲玲用最小的牌对同学最大的10，用中间的牌7对同学最小的6，用最大的牌对同学中间的8。据此解答即可.\n【详解】玲玲的同学先出。如果玲玲想要赢，那么当玲玲的同学出10时，玲玲应该出5。当玲玲的同学出6时，玲玲应该出7。",
+        "knowledge_points":  [
+                                 "田忌赛马问题"
+                             ],
+        "source":  "24-25 四年级上·湖南益阳·期末"
+    },
+    {
+        "id":  "118",
+        "unit":  "第八章数学广角 2",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "张红在家做家务。擦地需要9分钟，洗衣机洗衣服需要25分钟，晾衣服需要5分钟，洗碗筷需要7分钟，整理房间需要10分钟。她至少需要( )分钟才能做完。",
+        "options":  [
+
+                    ],
+        "answer":  "31",
+        "solution":  "【分析】由题意得，要想做完这些家务用时最少，张红需要合理安排做这些家务的顺序。她可以先用洗衣机洗衣服，然后在洗衣机洗衣服的同时，去擦地和洗碗筷和整理房间。最后再晾衣服。据此解答.\n【详解】9+7+10 = 16 + 10 = 26(分钟),26 \u003e 25，即张红擦地，洗碗筷和整理房间一共的时间比洗衣机洗衣服的时间要长。26 + 5 = 31(分钟)。故张红至少需要31分钟才能做完全部家务。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25 四年级上·湖南怀化·期末"
+    },
+    {
+        "id":  "119",
+        "unit":  "第八章数学广角 2",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "聪聪每天7时10分起床，整理床铺要4分钟，刷牙3分钟，洗脸2分钟，听英文歌曲15分钟，整理书包2分钟。他最快( )时( )分能做完这些事。",
+        "options":  [
+
+                    ],
+        "answer":  "7;25",
+        "solution":  "【分析】聪聪起床后先听英文歌曲，同时整理床铺、刷牙、洗脸、整理书包，所以最快需要15分钟，再用起床时间加上15分钟即可。\n【详解】7时10分+15分=7时25分，所以他最快7时25分能做完这些事。",
+        "knowledge_points":  [
+                                 "沏茶问题",
+                                 "时，分，秒时间的推算"
+                             ],
+        "source":  "24-25 四年级上·湖南株洲·期末"
+    },
+    {
+        "id":  "120",
+        "unit":  "第八章数学广角 2",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "完成下面几件事：扫地3分钟，洗衣机洗衣服25分钟，拖地8分钟，洗碗6分钟，晾衣服5分钟。做完这些事至少要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "30",
+        "solution":  "【分析】要使用的时间最少，那么能同时做的事情就应同时做，洗衣机洗衣服的同时可以扫地，拖地，洗碗，晾衣服需等洗衣机洗衣服后完成，据此解答.\n【详解】25 + 5 = 30(分钟),所以做完这些事至少要30分钟。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25 四年级上·湖南株洲·期末"
+    },
+    {
+        "id":  "121",
+        "unit":  "第八章数学广角 2",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小红有10,7,4三张扑克牌，小东有3,6,8三张扑克牌，规定每人每次出一张，轮流出牌，各出三次，所出牌数字大的获胜，三局两胜，如果小红按照7-4-10的顺序出牌，小东需按照( )的顺序出牌才能获胜。",
+        "options":  [
+
+                    ],
+        "answer":  "8;6;3",
+        "solution":  "【分析】小红的三张扑克牌中数字最大的是10，小东的三张扑克牌中数字最大的是8，当小红出10时，无论小东出哪张牌都会输。小东可以用最小的牌对小红最大的牌，即小红出10，小东出3，小红胜;小红出7小东出8，小东胜;小红出4小东出6，小东胜.\n【详解】小红有10,7,4三张扑克牌，小东有3,6,8三张扑克牌，规定每人每次出一张，轮流出牌，各出三次，所出牌数字大的获胜，三局两胜，如果小红按照7-4-10的顺序出牌，小东需8-6-3按照的顺序出牌才能获胜。",
+        "knowledge_points":  [
+                                 "田忌赛马问题"
+                             ],
+        "source":  "24-25 四年级上·重庆永川·期末"
+    },
+    {
+        "id":  "122",
+        "unit":  "第八章数学广角 2",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "4个同学排队参加跳远测试。李强排在第一位，他跳完三次花了4分钟，这时他后面的同学一共等了( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "12",
+        "solution":  "【分析】李强排在第一位，他后面的同学有3人。每人需要等李强完成测试才能开始，而李强跳完三次花了4分钟，这意味着每个后面的同学都等了4分钟。求\"一共等了多少分钟\"，即求3个同学等待时间的总和.\n【详解】三个同学，每个同学等4分钟，一共等了3×4=12分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "14-15 三年级上·全国·课后作业"
+    },
+    {
+        "id":  "123",
+        "unit":  "第八章数学广角 2",
+        "question_type":  "填空题",
+        "difficulty":  0.4,
+        "question":  "一名同学擦一块玻璃要用5分钟，五名同学同时擦10块玻璃需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "10",
+        "solution":  "【分析】五名同学同时工作，意味着每次可以同时擦5块玻璃。擦一块玻璃需要5分钟,10块玻璃需要分两轮完成：第一轮5人同时擦5块，第二轮5人同时擦剩下的5块。每轮都是5分钟，且是串行进行.\n【详解】擦10块玻璃，需要分5人擦两次，所以5×2=10分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "14-15 三年级上·全国·课后作业"
+    },
+    {
+        "id":  "124",
+        "unit":  "第八章数学广角 2",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "煮熟一个玉米需要40分钟，一个锅里可以煮6个玉米，那么要煮12个玉米最少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "80",
+        "solution":  "【分析】一个锅一次能煮6个玉米，煮12个玉米需要分两次煮。每次煮都需要40分钟，两次就是80分钟.\n【详解】玉米分两锅煮两次，所以时间一共需要80分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "14-15 三年级上·全国·课后作业"
+    },
+    {
+        "id":  "125",
+        "unit":  "第八章数学广角 2",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "烙一个地瓜片需要4分钟，一个平底锅同时可以烙5个地瓜片，烙5个地瓜片至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "4",
+        "solution":  "【分析】平底锅的容量是5个地瓜片，而正好需要烙5个地瓜片。这意味着5个地瓜可以一次性同时放入平底锅，它们同时开始，同时结束，所需时间与烙一个地瓜片的时间相同，都是4分钟.\n【详解】因为5个地瓜是同时进行的，所以时间没有改变，仍是4分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "14-15 三年级上·全国·课后作业"
+    },
+    {
+        "id":  "126",
+        "unit":  "第八章数学广角 2",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "星期日，老师到兵兵家家访，妈妈做迎接老师的准备。烧开水用15分，洗茶杯2分钟，打扫客厅用5分钟，她至少需要( )分干完这些事。",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】分析各项工作的性质：烧开水需要15分钟，但在这15分钟内不需要一直盯着，可以同时进行其他工作。洗茶杯(2分钟)和打扫客厅(5分钟)都可以与烧开水并行。因此，最优方案是：先开始烧开水，在烧水的15分钟内，穿插完成洗茶杯和打扫客厅，这样总时间由最长的单项工作(烧开水)决定.\n【详解】先烧开水15分钟，烧开水同时洗茶杯和扫客厅，共15分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "14-15 三年级上·全国·课后作业"
+    },
+    {
+        "id":  "127",
+        "unit":  "第八章数学广角 2",
+        "question_type":  "填空题",
+        "difficulty":  0.4,
+        "question":  "某小学组织六年级学生春游，学校买了182瓶汽水分给每个学生。如果每5个空瓶又可换得1瓶汽水，那么这些汽水瓶最多可换得( )瓶汽水。",
+        "options":  [
+
+                    ],
+        "answer":  "45",
+        "solution":  "【分析】根据换汽水的方法一步步计算，直到最后都换成汽水为止.\n【详解】第一次:182÷5=36(瓶)。2(瓶),即可换得36瓶汽水;第二次:36+2=38(瓶),38÷5=7(瓶)。3(瓶),即可换得7瓶汽水;第三次:7+3=10(瓶),10÷5=2(瓶),即可换得2瓶汽水;36+7+2=45(瓶),所以总共可以换得45瓶汽水。",
+        "knowledge_points":  [
+                                 "有余数除法的实际应用"
+                             ],
+        "source":  "2014·全国·小升初模拟"
+    },
+    {
+        "id":  "128",
+        "unit":  "第八章数学广角 2",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "甲，乙，丙三个小朋友问老师题目，他们分别要用5分钟,3分钟,2分钟，要使等候时间总和最少，应安排( )第一个,( )最后一个。",
+        "options":  [
+
+                    ],
+        "answer":  "丙;甲",
+        "solution":  "【分析】这是一个经典的\"排队等待\"优化问题。等候时间总和包括每个人等待前面人完成的时间加上自己问问题的时间。要使总等候时间最少，应该让用时短的人先问，用时长的人后问。按此原则,2分钟(丙)\u003c3分钟(乙)\u003c5分钟(甲),所以顺序应为丙→乙→甲.\n【详解】要使等候的时间最少，需要让用时最少的先问，因为2 \u003c 3 \u003c 5，所以按丙→乙→甲的顺序问问题使等候的时间和最少;问丙问题2分钟，他们3人等候，问乙问题3分钟，有他和小甲2人等候，最后问甲问题5分钟只有她自己等候，据此解答。故答案为丙，甲。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "14-15 四年级·全国·期末"
+    },
+    {
+        "id":  "129",
+        "unit":  "第八章数学广角 2",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈让笑笑烧开水给客人沏茶，洗热水壶要2分钟，烧开水要15分钟，洗茶壶要2分钟，洗茶杯要2分钟，拿茶叶要1分钟。为了让客人早点喝上茶，最少要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "17",
+        "solution":  "【分析】分析各项工作的先后关系和可并行性：必须先洗热水壶(2分钟),才能烧开水(15分钟)。而洗茶壶，洗茶杯，拿茶叶这三项工作可以在烧开水的15分钟内同时进行。因此，最优方案是：先洗热水壶(2分钟),然后开始烧开水(15分钟),在烧水的同时完成洗茶壶，洗茶杯，拿茶叶，总时间为2+15=17分钟.\n【详解】根据题干可知，先洗水壶用2分钟，烧开水15分钟的同时，可以洗茶壶，洗茶杯，拿茶叶，可以节约2 + 2 + 1 = 5分钟，需要一共2 + 15 = 17分钟。答：最少需要17分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "2014·全国·小升初模拟"
+    },
+    {
+        "id":  "130",
+        "unit":  "第八章数学广角 2",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "找规律填数:3,6,7,14,15,30,31,( ),( )。",
+        "options":  [
+
+                    ],
+        "answer":  "62;63",
+        "solution":  "【分析】观察数列3,6,7,14,15,30,31，尝试找出相邻数字之间的运算关系。发现规律：从第一个数开始，先乘2得到第二个数，第二个数加1得到第三个数，然后重复\"乘2，加1\"的循环。验证:3×2=6,6+1=7,7×2=14,14+1=15,15×2=30,30+1=31，完全符合。因此接下来的两个数应该是31×2=62,62+1=63.\n【详解】3×2 = 6,6 + 1 = 7,7×2 = 14,14 + 1 = 15,15×2 = 30,30 + 1 = 31，得出规律：从第一个数开始依次乘2得到第二个数，第二个数加1得到第三个数，然后依次循环，据此得解。31×2=62,62 + 1 = 63，所以数列为3,6,7,14,15,30,31,62,63。",
+        "knowledge_points":  [
+                                 "数字排列的规律"
+                             ],
+        "source":  "15-16 四年级上·全国·阶段练习"
+    },
+    {
+        "id":  "131",
+        "unit":  "第八章数学广角 2",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "平底锅每次最多烙2张饼，两面都要烙，每面烙熟需要4分钟。烙熟3张饼最少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "12",
+        "solution":  "【分析】此类问题中，尽量使每次都有2张饼在烙，由此进行合理安排即可解决问题。考查利用统筹思想进行合理安排事情的能力，抓住锅内始终有2张饼在烙是本题的关键.\n【详解】三张饼分别用序号1,2,3表示：第一次烙:1正,2正,4分钟;第二次烙:1反,3正,4分钟;第三次烙:2反,3反,4分钟;4 + 4 + 4 = 12(分钟),答：烙3张饼至少需要12分钟。",
+        "knowledge_points":  [
+                                 "最优化问题"
+                             ],
+        "source":  "15-16 四年级·全国·期末"
+    },
+    {
+        "id":  "132",
+        "unit":  "第八章数学广角 2",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明给奶奶冲牛奶，他需要烧开水5分钟，取牛奶1分钟，洗杯子2分钟，冲牛奶1分钟，奶奶最少需要( )分钟才能喝上牛奶。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】分析各项工作的先后关系：必须先烧开水，才能冲牛奶。而取牛奶和洗杯子不需要用到开水，可以在烧开水的5分钟内同时进行。冲牛奶必须在烧开水之后，需要1分钟。因此最优方案是：先烧开水(5分钟),同时取牛奶和洗杯子，水开后冲牛奶(1分钟),总时间为5+1=6分钟.\n【详解】等水开需要5分钟，同时可以取牛奶，洗杯子，可以节约2 + 1 = 3分钟，再冲牛奶1分钟，需要5 + 1 = 6(分钟)。答：奶奶最少需要6分钟才能喝上牛奶。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "15-16 四年级上·安徽淮南·期末"
+    },
+    {
+        "id":  "133",
+        "unit":  "第八章数学广角 2",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "平底锅中每次最多放两张饼，一张饼要烙两面，烙熟一面要2分钟，烙熟5张饼最少要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "10",
+        "solution":  "【分析】烙饼问题的关键是充分利用锅的容量，避免空位浪费。5张饼共10个面，每次最多烙2个面(2张饼各一面),每面2分钟。最优策略：先同时烙2张饼(正反面共4分钟);剩下3张饼采用交替烙法：第1次烙饼①正面和饼②正面(2分钟),第2次烙饼①反面和饼③正面(2分钟),第3次烙饼②反面和饼③反面(2分钟),3张饼共6分钟。总计4+6=10分钟.\n【详解】烙5张饼：先同时烙两张，正反面共需2×2=4分钟;再烙后三张，先烙第一张与第二张的正面需2分钟，然后烙第一张的反面与第三张的正面需要2分钟，最后烙第二张的反面与第三张的反面需2分钟，烙完3张共需2×3=6分钟,5张共需4+6=10分钟。答：烙熟5张饼最少需要10分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "15-16 四年级·全国·期末"
+    },
+    {
+        "id":  "134",
+        "unit":  "第八章数学广角 2",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "家里来客人了，你热情地为客人烧水沏茶。其中洗水壶要用 1 分钟，烧开水要用 10 分钟，洗茶杯要用 1 分钟，拿茶叶要用半分钟，用开水泡茶要 1 分钟。为了让客人早点儿喝上茶，请你进行最合理的安排，你至少要用( )分钟就能沏好茶。",
+        "options":  [
+
+                    ],
+        "answer":  "12",
+        "solution":  "【分析】先洗水壶 1 分钟，然后烧开水 10 分钟，在烧水的 10 分钟内可以同时完成洗茶杯和拿茶叶两项工作，最后用开水泡茶 1 分钟.\n【详解】按照\"洗水壶→烧开水(同时洗茶杯，拿茶叶)→泡茶\"的顺序，总时间为 1+10+1=12 分钟，因此至少需要 12 分钟就能沏好茶。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "2015·全国·小升初模拟·重庆"
+    },
+    {
+        "id":  "135",
+        "unit":  "第八章数学广角 2",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "家里来客人了，你热情地为客人烧水沏茶。其中洗水壶要用 1 分钟，烧开水要用 10 分钟，洗茶杯要用 1 分钟，拿茶叶要用半分钟，用开水泡茶要 1 分钟。为了让客人早点儿喝上茶，请你进行最合理的安排，你至少要用( )分钟就能沏好茶。",
+        "options":  [
+
+                    ],
+        "answer":  "12",
+        "solution":  "【分析】本题与上一题情境相同，仍然要充分利用烧水时间，同时完成可以并行的工作，使总时间最少.\n【详解】先洗水壶 1 分钟，再烧开水 10 分钟，同时洗茶杯，拿茶叶，最后用开水泡茶 1 分钟，共 1+10+1=12 分钟，因此至少要用 12 分钟就能沏好茶。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "2015·重庆·小升初模拟"
+    },
+    {
+        "id":  "136",
+        "unit":  "第八章数学广角 2",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一平底锅上能同时煎 4 块牛排，用它煎一块牛排需 6 分钟(正，反各 3 分钟),如果煎 6 块同样的牛排，至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "9",
+        "solution":  "【分析】6 块牛排需要煎 12 个面，每面 3 分钟，一次可以同时煎 4 个面，因此一共需要 3 次，每次 3 分钟.\n【详解】合理安排顺序，使每次锅里都尽量放满 4 块牛排：先煎 4 块正面，再安排部分反面与新牛排正面一起煎，最后把剩余的反面煎完，共 3 次，每次 3 分钟，总时间 9 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "14-15五年级下·山东青岛·课后作业"
+    },
+    {
+        "id":  "137",
+        "unit":  "第八章数学广角 2",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "5 个同学排队打饭，小竹排在第一位，她打饭花了 2 分钟，这时她后面的同学一共等待了( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "8",
+        "solution":  "【分析】小竹后面有 4 个同学，每个人都要等小竹打完 2 分钟后才能开始打饭.\n【详解】4 个同学每人等待 2 分钟，一共等待时间为 4×2=8 分钟，因此她后面的同学一共等待了 8 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "14-15三年级上·全国·课后作业"
+    },
+    {
+        "id":  "138",
+        "unit":  "第八章数学广角 2",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小丽家用高压锅蒸米饭，需要 10 分钟，减压需要 5 分钟，炒菜需要 8 分钟(蒸饭和炒菜用同一煤气灶)。小丽家最快需要( )分钟就可以吃上饭。",
+        "options":  [
+
+                    ],
+        "answer":  "18",
+        "solution":  "【分析】蒸饭 10 分钟用煤气灶，减压 5 分钟一般不占用煤气灶，炒菜 8 分钟要用煤气灶。可以在蒸饭结束后，一边减压一边炒菜，使两项时间重叠.\n【详解】先蒸米饭 10 分钟，然后在减压的 5 分钟内同时炒菜 8 分钟，由于炒菜时间更长，实际从开始减压到炒菜结束共 8 分钟，因此总时间为 10+8=18 分钟，小丽家最快 18 分钟就可以吃上饭。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "14-15三年级上·全国·课后作业"
+    },
+    {
+        "id":  "139",
+        "unit":  "第八章数学广角 2",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "李老师同时找甲，乙，丙三名学生谈话，甲要 12 分钟谈完，乙要 15 分钟谈完，丙要 8 分钟谈完(只能一个一个地谈)。经李老师合理安排，使三人等候的总时间最少是( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "63",
+        "solution":  "【分析】要使等候时间总和最少，应让用时短的同学先谈，按谈话时间从小到大排序为：丙 8 分钟，甲 12 分钟，乙 15 分钟.\n【详解】按\"丙→甲→乙\"的顺序：丙谈话 8 分钟时 3 人等候，甲谈话 12 分钟时 2 人等候，乙谈话 15 分钟时 1 人等候，总等候时间为 8×3+12×2+15×1=24+24+15=63 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "14-15三年级上·全国·课后作业"
+    },
+    {
+        "id":  "140",
+        "unit":  "第八章数学广角 2",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个锅每次只能放两片面包，每烤一面需 2 分钟。小红早点要吃 3 片面包，至少要等( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】3 片面包需要烤 6 个面，每面 2 分钟，一次可以烤 2 个面，应采用交替烤的方法让锅始终满负荷工作.\n【详解】先烤 1,2 片的正面 2 分钟，再烤 1 片反面和 3 片正面 2 分钟，最后烤 2 片反面和 3 片反面 2 分钟，共 2+2+2=6 分钟，所以至少要等 6 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "16-17 四年级上·全国·期末"
+    },
+    {
+        "id":  "141",
+        "unit":  "第八章数学广角 2",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈用高压锅蒸米饭，需要 10 分钟，减压需要 5 分钟，炒菜需要 8 分钟，妈妈最快( )分钟就可以让家人吃上饭。",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】蒸米饭的 10 分钟内可以同时炒菜 8 分钟，因为高压锅工作时不需要一直看管，炒菜时间可以完全嵌入蒸饭时间.\n【详解】先蒸米饭 10 分钟，并在这段时间内完成炒菜 8 分钟，然后再减压 5 分钟，总时间为 10+5=15 分钟，因此妈妈最快 15 分钟就可以让家人吃上饭。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "16-17 四年级上·全国·期末"
+    },
+    {
+        "id":  "142",
+        "unit":  "第八章数学广角 2",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "家里来客人了，你热情地为客人烧水沏茶。其中洗水壶要用 1 分钟，烧开水要用 10 分钟，洗茶杯要用 1 分钟，拿茶叶要用半分钟，用开水泡茶要 1 分钟。为了让客人早点儿喝上茶，请你进行最合理的安排，你至少要用( )分钟就能沏好茶。",
+        "options":  [
+
+                    ],
+        "answer":  "12",
+        "solution":  "【分析】与前面的沏茶问题相同，先完成必须在前的步骤，再把可以并行的工作安排在烧水过程中.\n【详解】先洗水壶 1 分钟，再烧开水 10 分钟，在烧水期间洗茶杯并拿茶叶，最后用开水泡茶 1 分钟，共 1+10+1=12 分钟，所以至少用 12 分钟就能沏好茶。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "2017·四川雅安·小升初模拟"
+    },
+    {
+        "id":  "143",
+        "unit":  "第八章数学广角 2",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用一只平底锅烙饼，每次只能放 2 张饼，烙一面要 4 分钟，两面都要烙，烙 7 张饼至少要用( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "28",
+        "solution":  "【分析】前 4 张饼可以按普通方法两张一组烙完，后 3 张饼需要用交替烙的方法，让锅始终有两张饼在烙，避免空锅浪费时间.\n【详解】先烙 4 张饼，两面都烙完需要 4×2×2=16 分钟;剩下 3 张饼用交替烙法共需 4×3=12 分钟，总时间 16+12=28 分钟，所以至少需要 28 分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题",
+                                 "优化问题"
+                             ],
+        "source":  "16-17 三年级·甘肃庆阳·期末"
+    },
+    {
+        "id":  "144",
+        "unit":  "第八章数学广角 2",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明感冒了，吃完药后要赶快休息。他找杯子倒开水 1 分钟，等开水变温 7 分钟，找感冒药 1 分钟，量体温 5 分钟。小明合理安排以上事情，最少要( )分钟才能尽快休息。",
+        "options":  [
+
+                    ],
+        "answer":  "8",
+        "solution":  "【分析】必须先倒开水再等开水变温，找感冒药和量体温可以在等水变温时同时进行.\n【详解】先倒开水 1 分钟，然后等开水变温 7 分钟，在这 7 分钟内完成找药(1 分钟)和量体温(5 分钟),总时间为 1+7=8 分钟，因此小明最少要 8 分钟才能尽快休息。",
+        "knowledge_points":  [
+                                 "最优化问题",
+                                 "优化问题"
+                             ],
+        "source":  "18-19 四年级上·全国·课后作业"
+    },
+    {
+        "id":  "145",
+        "unit":  "第八章数学广角 2",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈用平底锅煎鱼，每一次只能煎两条鱼，一面要煎 4 分钟，每条鱼要煎两面，煎完 3 条鱼要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "12",
+        "solution":  "【分析】3 条鱼共 6 个面，一次可以煎 2 个面，每面 4 分钟，应采用交替煎法让锅内始终有 2 条鱼.\n【详解】先煎 1,2 号鱼正面 4 分钟，再煎 1 号反面和 3 号正面 4 分钟，最后煎 2 号反面和 3 号反面 4 分钟，共 4+4+4=12 分钟，所以煎完 3 条鱼需要 12 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "2014四年级·全国·课后作业"
+    },
+    {
+        "id":  "146",
+        "unit":  "第八章 数学广角（二） 优化问题",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一只平底锅，只能同时烙两张饼，烙熟饼的一面要 2 分钟，要烙熟 5 张饼最少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "10",
+        "solution":  "【分析】 5 张饼共 10 个面，每次最多烙 2 个面，每面 2 分钟。最优策略：先同时烙 2 张饼(正反面共 4 分钟);剩下 3 张采用交替烙法避免空锅：第 1 次烙饼①正面和饼②正面(2 分钟),第 2 次烙饼①反面和饼③正面(2 分钟),第 3 次烙饼②反面和饼③反面(2 分钟),3 张共 6 分钟。总计 4+6=10 分钟。公式验证：饼数×每面时间 = 5×2=10 分钟.\n【详解】试题分析：烙5张饼：先同时烙两张，正反面共需2×2 = 4 分钟;再烙后三张，先烙第一张与第二张的正面需2分钟，然后烙第一张的反面与第三张的正面需要2分钟，最后烙第二张的反面与第三张的反面需2分钟，烙完3张共需2×3 = 6 分钟,5 张共需 4 + 6 = 10 分钟;据此解答。解：先同时烙两张，正反面共需 2×2 = 4(分钟)再烙后 3 张共需 2×3 = 6(分钟)5 张共需 4 + 6 = 10(分钟)答：烙熟 5 张饼最少需要 10 分钟。故答案为 10。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "15-16四年级上·全国·期末"
+    },
+    {
+        "id":  "147",
+        "unit":  "第八章 数学广角（二） 优化问题",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一只锅，每次只能放2张饼。煎熟一面要2分钟，两面都要煎，煎熟9张饼最少要用( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "18",
+        "solution":  "【分析】 9 张饼共 18 个面，每次煎 2 个面，每面 2 分钟。前 6 张可以 2 张 2 张地煎，每次 4 分钟,3 次共 12 分钟。剩下 3 张采用交替法：第 1 次放①正②正，第 2 次放①反③正，第 3 次放②反③反，每次 2 分钟，共 6 分钟。总计 12+6=18 分钟。公式验证:9×2=18 分钟.\n【详解】试题分析: 9÷ 2 = 4(组)。1(张),前6张可以2张 2 张的煎，煎 3 次共需 4×3 = 12 分钟;剩下的 3 张饼可以这样煎：先煎2张的正面;煎熟后拿出第一张，放入第三张，煎第二张的反面和第三张的正面;煎熟后第二张就熟了，再煎第一张和第三张的反面,2×3 = 6 分钟，这样煎9张饼至少需要 12 + 6 = 18 分钟，据此解答。解：前6张可以2张2张的煎，煎3次共需 4×3 = 12 分钟;剩下 3 张：第一次：放①的正面和②的正面，第二次：放①的反面和③的正面，第三次：放②的反面和③的反面，共用 2×3 = 6 分钟。全部时间: 12 + 6 = 18(分钟);答：煎 9 只饼至少需要 18 分钟。故答案为 18.\n[点评]解决此类问题的方法是使效率最大化，即锅能放满就尽量放满，不做无用工。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "15-16四年级上·全国·期末"
+    },
+    {
+        "id":  "148",
+        "unit":  "第八章 数学广角（二） 优化问题",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个平底锅，每次只能煎 2 个鸡蛋，两面都要煎，每面煎 2 分钟。煎 5 个鸡蛋，至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "10",
+        "solution":  "【分析】 煎鸡蛋问题属于烙饼问题的变形。每次只能煎 2 个鸡蛋，每面需要 2 分钟。煎 5 个鸡蛋，每个鸡蛋需要煎两面，共需煎 10 个面。每次可以煎 2 个面，所以至少需要煎 5 次，每次 2 分钟，总共需要 10 分钟。关键是要合理安排，使平底锅每次都充分利用.\n【详解】设 5 个鸡蛋分别为 1,2,3,4,5，每个鸡蛋有正面和反面。第一次：放 1 的正面和 2 的正面，煎 2 分钟;第二次：放 1 的反面和 2 的反面，煎 2 分钟(鸡蛋 1,2 煎好，用时 4 分钟);第三次：放 3 的正面和 4 的正面，煎 2 分钟;第四次：放 3 的反面和 5 的正面，煎 2 分钟;第五次：放 4 的反面和 5 的反面，煎 2 分钟(鸡蛋 3,4,5 煎好，用时 6 分钟)。全部时间:4 + 6 = 10(分钟)煎 5 个鸡蛋至少需要 10 分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题",
+                                 "优化问题"
+                             ],
+        "source":  "2014 四年级·全国·课后作业"
+    },
+    {
+        "id":  "149",
+        "unit":  "第八章 数学广角（二） 优化问题",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用平底锅烙饼，每次只能烙2张，两面都要烙，烙好每面需要2分钟，烙好9张最少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "18",
+        "solution":  "【分析】当烙2张及以上的饼，并且锅里只能同时烙两张饼时，最少需要的时间 = 烙一面需要的时间×张数;依此计算并填空.\n【详解】 2×9 = 18(分钟)烙9张饼，至少需要18分钟.\n[点睛]熟练掌握优化问题的计算是解答此题的关键。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "2014 四年级·全国·课后作业"
+    },
+    {
+        "id":  "150",
+        "unit":  "第八章 数学广角（二） 优化问题",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小王要给客人烧水沏茶。洗水壶要 1 分钟，烧开一壶水要 15 分钟，洗茶杯要用 2 分钟，洗茶壶要用 1 分钟，拿茶叶要用 1 分钟，总的算起来要20 分钟，合理安排后，最短用( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "16",
+        "solution":  "【分析】由题干可知，先洗水壶用1分钟，烧开水15分钟的同时，可以洗茶壶，洗茶杯，拿茶叶，需要一共需要 1 + 15 = 16 分钟.\n【详解】根据题干分析可设计如下:\n![](images/360e597712e6371e251fd773b8ef7245ebcdea6856df5c9634e8e0c840d9943c.jpg)\n1 + 15 = 16(分钟)答：最短要用16分钟.\n[点睛]此题属于合理安排时间问题，要奔着既节约时间又不使每道程序相互矛盾即可解答。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "2014 四年级·全国·课后作业"
+    },
+    {
+        "id":  "151",
+        "unit":  "第八章 数学广角（二） 优化问题",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在 1 只平底锅上烙薄饼，同时只能放下 2 只薄饼，饼的两面都要烙 1 分钟。如果要烙7 只饼，至少要用( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "7",
+        "solution":  "【分析】第一次先煎2张饼两面需要2分钟，第二次再煎2张饼两面需要 2 分钟，共需要 2×2 = 4 分钟;剩下的3张饼可以这样煎：先煎2张的正面，过1分钟煎熟后拿出第一张，放入第三张，煎第二张的反面和第三张的正面，再过1分钟煎熟后第二张就熟了，最后再煎第一张的反面和第三张的反面，煎熟需要1分钟，这3张煎熟一共需要 1×3 = 3 分钟.\n【详解】 2×2 + 1×3 = 4 + 3 = 7(分钟)所以至少需要 7 分钟.\n[点睛]本题考查优化问题，关键注意每次最多放2张饼，别让锅有空余，最省时间。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "2014 四年级·全国·课后作业"
+    },
+    {
+        "id":  "152",
+        "unit":  "第八章 数学广角（二） 优化问题",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一只平锅上只能煎两只饼，用它煎一只饼需要2分钟(正面，反面各1分钟),煎3 只饼需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "3",
+        "solution":  "【分析】若先把两只饼煎至熟，势必在煎第三张饼时，锅中只有一只饼而造成浪费，所以应把两只饼的两面错开煎.\n【详解】先取两张饼烙1分钟，取出其中一张，另一张的反面和新放入的第三张饼烙1分钟，把烙好的第一张饼取出，剩下两张饼烙反面1分钟;至少需要: 1 + 1 + 1 = 3(分钟);所以，煎 3 只饼需要 3 分钟;[点睛]此题考查了学生的利用统筹思想进行合理安排事情的能力，抓住锅内始终有2张饼在烙是本题的关键。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "2014 四年级·全国·课后作业"
+    },
+    {
+        "id":  "153",
+        "unit":  "第八章 数学广角（二） 优化问题",
+        "question_type":  "填空题",
+        "difficulty":  0.4,
+        "question":  "一次，齐王与大将田忌赛马，每人有四匹马，分为四等。田忌知道齐王这次比赛马的出场顺序依次为一等，二等，三等，四等，而且还知道这八匹马跑的最快的是齐王的一等马，接着依次为自己的一等，齐王的二等，自己的二等，齐王的三等，自己的三等，齐王的四等，自己的四等。那么田忌有( )种方法安排自己的马的出场顺序，保证自己至少能赢两场比赛。",
+        "options":  [
+
+                    ],
+        "answer":  "12",
+        "solution":  "【分析】采用枚举法，找出所有的可能即可.\n【详解】用四位数来表示田忌的出场顺序，可以枚举出所有方法有:1423,2143,2413,3124,3142,3412,3421,4123,4132,4213,4312,4321。所以田忌一共有 12 种方法安排自己的马的出场顺序。田忌有12种方法安排自己的马的出场顺序，保证自己至少能赢两场比赛.\n[点睛]此题主要考查了简单的枚举法应用，根据已知得出所有符合要求的答案是解题关键。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "2014四年级·全国·课后作业"
+    },
+    {
+        "id":  "154",
+        "unit":  "第八章 数学广角（二） 优化问题",
+        "question_type":  "填空题",
+        "difficulty":  0.4,
+        "question":  "某学校组织师生去春游，准备租用以下的两种客车。若租若干辆 45 座的客车，则有 15 人没有座位;若租 60 座的客车，则可少租一辆且恰好全部坐满。按照最省钱的方案租车，租金至少需( )元。(中车:45 座，租金 215 元。大车:60 座，租金 300 元)",
+        "options":  [
+
+                    ],
+        "answer":  "1160",
+        "solution":  "【分析】(1)要求出最省钱的租车方案，需要先求出师生人数共有多少;可以设如果单独租一种车,60座的需要x辆正好座满，根据题意可知：租45座的需要 x+2 辆，还剩下 45-15=30 个空座，抓住人数相等即可列出方程求得总人数;(2)租45座的客车，每人花费约 215÷45approx4.8 元;租60座的客车，每人花费 300÷6=5 元;所以尽量地租45座的车较实惠，但是还要考虑人数和是否座满，保证所选方案最省钱;【详解】(1)设60座的需要 x 辆正好座满，则租45座的需要 x+2 辆，根据题意可得方程:\n45×(x+2)-60x=45-15,\n15x=60,\nx=4，所以师生共有:60×4=240(人),\n(2)根据题干可将所有租车花费情况列出表格如下图所示:\n\u003ctable\u003e\u003ctr\u003e\u003ctd\u003e60座的客车\u003c/td\u003e\u003ctd\u003e4辆\u003c/td\u003e\u003ctd\u003e3辆\u003c/td\u003e\u003ctd\u003e2辆\u003c/td\u003e\u003ctd\u003e1辆\u003c/td\u003e\u003ctd\u003e\u003c/td\u003e\u003c/tr\u003e\u003ctr\u003e\u003ctd\u003e45座的客车\u003c/td\u003e\u003ctd\u003e\u003c/td\u003e\u003ctd\u003e2辆\u003c/td\u003e\u003ctd\u003e3辆\u003c/td\u003e\u003ctd\u003e4辆\u003c/td\u003e\u003ctd\u003e6辆\u003c/td\u003e\u003c/tr\u003e\u003ctr\u003e\u003ctd\u003e话费的钱数(元)\u003c/td\u003e\u003ctd\u003e4×300=1200\u003c/td\u003e\u003ctd\u003e3×300+2×215=1330\u003c/td\u003e\u003ctd\u003e2×300+3×215=1245\u003c/td\u003e\u003ctd\u003e300+4×215=1160\u003c/td\u003e\u003ctd\u003e215×6=1290\u003c/td\u003e\u003c/tr\u003e\u003c/table\u003e\n由上表计算数据可得按照最省钱的方案租车，租金至少需1160 元.\n[点睛]根据题干求得总人数是解决本题的关键，然后根据人数和租车的不同方法计算得出各种情况下的花费情况，从而得出最佳方案解决问题。",
+        "knowledge_points":  [
+                                 "优化问题",
+                                 "方案选择",
+                                 "列方程解决实际问题"
+                             ],
+        "source":  "2014 五年级·全国·课后作业"
+    },
+    {
+        "id":  "155",
+        "unit":  "第八章 数学广角（二） 优化问题",
+        "question_type":  "填空题",
+        "difficulty":  0.4,
+        "question":  "周日妈妈给小明烙鸡蛋饼，如果锅里每次最多可以放两张饼，一张饼需要烙两面，每烙一面需要 2 分钟，要烙五张饼最少需要用( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "10",
+        "solution":  "【分析】先煎前 2 张饼，共需 4 分钟，为了便于说明问题把剩下三个鸡蛋饼分别编号为1,2,3号，可以采用交替煎的办法，先放1,2号,2分钟后把其中的一个取出，比如把2号取出，再把3号放入,1号煎反面;2分钟后,1号熟了取出，再把2号放入;再过2分钟,2,3都熟了;这样后三张一共用了6分钟。所以一共需要:4 + 6 = 10 分钟.\n【详解】 4 + 2×3 = 4 + 6 = 10(分钟);答：煎 5 张饼，至少需要 10 分钟。故答案为 10.\n[点睛]本题需要采用交替煎的办法，这样使锅里始终没有空位，能比先煎两个后煎一个要节省2分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "2014四年级·全国·课后作业"
+    },
+    {
+        "id":  "156",
+        "unit":  "第八章 数学广角（二） 优化问题",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在一个两位数的两个数字之间加上一个 0，所得的新数是原数的 9 倍，原数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "45",
+        "solution":  "【分析】首先设出这个两位数，用十进制计数法表示出来，再利用所得三位数是原来的数的9倍，列方程分析解答即可.\n【详解】解：设原来数为 ab,这样后来的数为 a0b,把数字展开我们可得:100a+b=9×(10a+b),我们可以得到 5a=4b,所以 a=4 ,b=5，因此原来的两位数为45。故答案为45。",
+        "knowledge_points":  [
+                                 "用三角尺画角"
+                             ],
+        "source":  "2014四年级·全国·课后作业"
+    },
+    {
+        "id":  "157",
+        "unit":  "第八章 数学广角（二） 优化问题",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用一只平底锅烙饼，每次只能放2张饼，烙一面要2分钟，两面都要烙，烙11张饼至少要用( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "22",
+        "solution":  "【分析】烙前8张饼按正常方法烙，需要:2×2×(8÷2)=2×2×4=16(分钟);剩下的3张饼可以这样烙：第一次先把 2 张饼同时下锅,2 分钟时，将其中一张盛上来，另外一张反面，同时，再把第三张饼放上去;第二次，第一次放的那张饼可以出锅了，最后放的那张饼翻面，同时将盛出来的饼翻面放下去;第三次，就可以全部出锅了。共用时间 2×3 = 6(分钟),与前面烙8张饼的时间相加即可.\n【详解】由分析可知至少需要时间:\n2×2×(8÷2) + 2×3 = 16 + 6 = 22(分钟).\n[点睛]此题考查了有关优化问题，解题关键是合理安排时间，做到锅中时刻有2张饼。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "2017 六年级下·全国·竞赛"
+    },
+    {
+        "id":  "158",
+        "unit":  "第八章 数学广角（二） 优化问题",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈上班去了，小强只能自己做饭吃。淘米需要 1 分钟，用电饭煲煮饭需要20分钟，把妈妈烧好的几个菜用微波炉热一热需要8分钟，冲一碗汤需要 3 分钟，盛饭需要 1 分钟。小强最快( )分钟就能吃饭了。",
+        "options":  [
+
+                    ],
+        "answer":  "22",
+        "solution":  "【分析】 分析各项工作：淘米是煮饭的前置条件(1 分钟),煮饭 20 分钟期间(电饭煲自动工作),小强可以同时完成热菜(8 分钟)和冲汤(3 分钟)。最后盛饭需要 1 分钟，且必须在煮饭完成后。关键路径：淘米 → 煮饭 → 盛饭，热菜和冲汤嵌入煮饭的 20 分钟内。总时间:1+20+1=22 分钟.\n【详解】第一步：淘米。淘米需要 1 分钟，这是煮饭的前置条件，必须单独进行。第二步：用电饭煲煮饭(同时热菜，冲汤)。煮饭需要 20 分钟，在煮饭期间，小强可以同时进行：用微波炉热菜:8 分钟;冲一碗汤:3 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "18-19 四年级上·全国·单元测试"
+    },
+    {
+        "id":  "159",
+        "unit":  "第八章 数学广角（二） 优化问题",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈感冒了，量体温需要4分钟，取药需要1分钟，倒开水需要1分钟，等开水变凉需要 5 分钟，妈妈想尽快吃药，可以这样安排:( ),至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "倒开水-- 等开水变凉(同时量体温，取药);6",
+        "solution":  "【分析】 分析吃药的流程：必须先倒开水(1 分钟),然后等开水变凉(5 分钟)才能吃药。量体温(4 分钟)和取药(1 分钟)可以在等开水变凉的 5 分钟内同时进行。最优方案：先倒开水 1 分钟，然后等开水变凉 5 分钟(同时量体温 4 分钟 + 取药 1 分钟),共 6 分钟.\n【详解】具体安排步骤如下：第一步：倒开水。倒开水需要 1 分钟，这是后续步骤的前置条件，必须单独进行。第二步：等开水变凉(同时量体温，取药)。等开水变凉需要 5 分钟，在等待期间，妈妈可以同时进行：量体温:4 分钟;取药:1 分钟。量体温和取药共需 4+1=5 分钟，正好嵌入等开水变凉的 5 分钟内。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "18-19 四年级上·全国·单元测试"
+    },
+    {
+        "id":  "160",
+        "unit":  "第八章 数学广角（二） 优化问题",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一只锅一次最多能煎 3 条鱼，鱼的两面都要煎。煎一面要 2 分钟。煎 6 条鱼最少要用( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "8",
+        "solution":  "【分析】 6 条鱼共 12 个面，每次最多煎 3 个面(3 条鱼各一面),每面 2 分钟。需要煎 12÷3=4 次，每次 2 分钟，共 8 分钟。具体安排：第 1 次煎 3 条正面，第 2 次煎这 3 条反面(完成 3 条);第 3 次煎另 3 条正面，第 4 次煎这 3 条反面(完成另 3 条).\n【详解】第一步：煎前 3 条鱼。第 1 次：煎鱼①正面，鱼②正面，鱼③正面，需 2 分钟;第 2 次：煎鱼①反面，鱼②反面，鱼③反面，需 2 分钟，前 3 条鱼煎好，用时:2×2=4(分钟)。第二步：煎后 3 条鱼，假设剩下的 3 条鱼分别为 a,b,c:第 3 次：煎 a 正面,b 正面,c 正面，需 2 分钟;第 4 次：煎 a 反面,b 反面,c 反面，需 2 分钟，后 3 条鱼煎好，用时:2×2=4(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "18-19 四年级上·全国·课后作业"
+    },
+    {
+        "id":  "161",
+        "unit":  "第八章 数学广角（二） 优化问题",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "丽丽每天晚上睡觉前要背诵成语 6 分钟，烧开水 10 分钟，泡好不烫的牛奶 2 分钟，喝牛奶 5 分钟，那丽丽在( )同时可以( )，做完这些事情最少用( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "烧开水;背诵成语;17",
+        "solution":  "【分析】 分析各项工作：烧开水 10 分钟期间不需要看管，可以同时背诵成语(6 分钟)。泡牛奶需要在开水烧好后(2 分钟)，喝牛奶(5 分钟)需要在泡好牛奶后。关键路径：烧开水(10 分钟，同时背成语)→泡牛奶(2 分钟)→喝牛奶(5 分钟)。背诵成语的 6 分钟完全嵌入烧开水的 10 分钟内。总时间：10+2+5=17(分钟)。\n【详解】丽丽先烧开水，同时背成语，一共 10 分钟，然后泡牛奶 2 分钟，最后喝牛奶 5 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "14-15 三年级上·全国·课后作业"
+    },
+    {
+        "id":  "162",
+        "unit":  "第八章 数学广角（二） 优化问题",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "煮一个鸡蛋约要 8分钟，一只锅一次可以煮 10 个鸡蛋。煮 10 个鸡蛋需要( )分钟。煮 15 个鸡蛋需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "8;16",
+        "solution":  "【分析】 煮鸡蛋问题：锅的容量为 10 个，煮 10 个鸡蛋可以一次完成，时间与煮 1 个相同，都是 8 分钟。煮 15 个鸡蛋需要分两次：第一次煮 10 个(8 分钟),第二次煮剩下的 5 个(8 分钟),共 16 分钟。注意：虽然第二次只煮 5 个，但仍需 8 分钟才能煮熟.\n【详解】第一问：煮 10 个鸡蛋。锅的容量为 10 个鸡蛋,10 个鸡蛋 \u003c= 10 个容量，可以一次性放入锅内。煮 10 个鸡蛋和煮 1 个鸡蛋的时间相同，都是水沸腾后煮熟所需的时间，共用时 8 分钟。第二问：煮 15 个鸡蛋。由于锅一次最多只能煮 10 个,15 个鸡蛋需要分两次煮：第一次：煮 10 个鸡蛋，用时 8 分钟;第二次：煮剩下的 5 个鸡蛋，用时 8 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "18-19 四年级上·全国·课后作业"
+    },
+    {
+        "id":  "163",
+        "unit":  "第八章 数学广角（二） 优化问题",
+        "question_type":  "填空题",
+        "difficulty":  0.4,
+        "question":  "小明感冒了，吃完药后要赶快休息。他找杯子倒开水1分钟，等开水变温7分钟，找感冒药1分钟，量体温5分钟。小明合理安排以上事情，最少要( )分钟才能尽快休息。",
+        "options":  [
+
+                    ],
+        "answer":  "8",
+        "solution":  "【分析】 分析流程：必须先找杯子倒开水(1 分钟)，然后等开水变温(7 分钟)才能吃药。找感冒药(1 分钟)和量体温(5 分钟)可以在等开水变温的 7 分钟内同时进行。最优方案：先找杯子倒开水 1 分钟，然后等开水变温 7 分钟(同时找感冒药、量体温)，共 8 分钟。\n【详解】最优化问题。先找杯子倒开水 1 分钟，再等开水变温 7 分钟，在这 7 分钟内可以找感冒药和量体温，所以最少需要 1+7=8(分钟)。",
+        "knowledge_points":  [
+                                 "最优化问题"
+                             ],
+        "source":  "2018 六年级下·全国·竞赛"
+    },
+    {
+        "id":  "164",
+        "unit":  "第八章 数学广角（二） 优化问题",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个合唱队有 30 人，暑假期间有一个紧急演出。老师需要尽快通知到每一个队员，如果用打电话的方式，每分钟通知 1 人，至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "5",
+        "solution":  "【分析】 这是\"打电话通知\"问题，关键是已通知的人可以协助通知其他人，呈指数增长。第 1 分钟：老师通知 1 人，共 2 人知道;第 2 分钟:2 人各通知 1 人，新增 2 人，共 4 人知道;第 3 分钟:4 人各通知 1 人，新增 4 人，共 8 人知道;第 4 分钟:8 人各通知 1 人，新增 8 人，共 16 人知道;第 5 分钟:16 人各通知 1 人，新增 16 人，共 32 人知道。32 \u003e 30，所以至少需要 5 分钟.\n【详解】第 1 分钟老师通知 1 位队员，已通知人数:1 人知道消息，总人数:1(老师)+ 1(队员)= 2 人。第 2 分钟老师和已通知的 1 位队员，各通知 1 人，新增 2 人，已通知人数:1 + 2 = 3 人，知道消息总人数:2 + 2 = 4 人。第 3 分钟已知道消息的 4 人，各通知 1 人，新增 4 人，已通知人数:3 + 4 = 7 人，知道消息总人数:4 + 4 = 8 人。由此类推，第 5 分钟即可超过30人。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "17-18 五年级下·全国·期末"
+    },
+    {
+        "id":  "165",
+        "unit":  "第八章 数学广角（二） 优化问题",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "现有 12 把钥匙和 12 把锁(某一把钥匙恰好可以打开某一把锁)但不知哪把钥匙配哪把锁，最多试开( )次，就能把锁和钥匙配对。",
+        "options":  [
+
+                    ],
+        "answer":  "66",
+        "solution":  "【分析】考虑最差情况，试第1把锁，共试11把钥匙都没打开，剩下的1把不用试了，一定能打开，同理，第11把试10次，第10把试9次，依此类推。,共试 11+10+9+8+7+·s+2+1=66 次.\n【详解】11+10+9+8+7+·s+2+1 = (11+1)×11÷2 = 12×11÷2 = 66(次)。所以最多要试 66 次可把钥匙与锁配对.\n[点睛]此题考查了加法原理：做一件事情，完成它有N类办法，在第一类办法中有 M₁ 种不同的方法，在第二类办法中有 M₂ 种不同的方法,。,在第N类办法中有 Mₙ 种不同的方法，那么完成这件事情共有 M1+M2+·s+Mn 种不同的方法。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "2014 六年级·全国·课后作业"
+    },
+    {
+        "id":  "166",
+        "unit":  "第八章 数学广角（二） 优化问题",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈用平底锅煎鱼，每一次只能煎两条鱼，一面要煎4分钟，每条鱼要煎两面，煎完 3 条鱼要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "12",
+        "solution":  "【分析】为了便于说明问题把三条鱼分别编号为 1,2,3 号，可以采用交替煎的办法，先放 1,2 号,4 分钟后把其中的 一个取出，比如把 2 号取出，再把 3 号放入,1 号煎反面;4 分钟后,1 号熟了取出，再把 2 号放入;再过 4 分钟, 2,3 都熟了;这样一共用了 12 分钟.\n【详解】煎 3 条鱼，先把 2 条鱼同时下锅,4 分钟时，将第二条盛上来，另外第一条翻面，同时，把第三条鱼放进去，再煎 4 分钟;第一条鱼煎熟，把第三条反过来，以及第二条鱼的另一面;再4分钟后全部煎熟。4 + 4 + 4 = 12(分钟)煎完3条鱼要12分钟.\n[点睛]此题考查了学生的利用统筹思想进行合理安排事情的能力，抓住锅内始终有2条鱼在煎是本题的关键。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "2014四年级·全国·课后作业"
+    },
+    {
+        "id":  "167",
+        "unit":  "第八章 数学广角（二） 优化问题",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "烙一张饼如果需要6分钟(每面3分钟),一只平底锅每次可以烙2张饼，烙3张饼至少需要( )分钟。烙 5 张饼至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "9;15",
+        "solution":  "【分析】 每次最多烙 2 张饼，每面 3 分钟。烙 3 张饼时，若先烙 2 张(6 分钟),再烙第 3 张(6 分钟),共 12 分钟，不是最优。应采用交替法：第 1 次烙饼①正面和饼②正面(3 分钟),第 2 次烙饼①反面和饼③正面(3 分钟),第 3 次烙饼②反面和饼③反面(3 分钟),共 9 分钟。烙 5 张饼：先烙 2 张(6 分钟),剩下 3 张用交替法(9 分钟),共 15 分钟。或直接用公式：饼数×每面时间 = 3×3=9 ,5×3=15.\n【详解】第一问：烙 3 张饼(交替烙法)。将 3 张饼分别编号为①,②,③:第 1 次：烙饼①正面，饼②正面，需 3 分;第 2 次：烙饼①反面，饼③正面，需 3 分钟，此时饼①已完成(两面都烙好);第 3 次：烙饼②反面，饼③反面，需 3 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "18-19四年级上·全国·课后作业"
+    },
+    {
+        "id":  "168",
+        "unit":  "第八章 数学广角（二） 优化问题",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一张饼两面都要烙，需要 6 分钟，一只平底锅每次可以烙 3 张，烙熟 5 张饼至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "12",
+        "solution":  "【分析】 每面 3 分钟(因为两面共 6 分钟),每次最多烙 3 张。5 张饼共 10 个面，每次烙 3 个面，需要 4 次(前 3 次各 3 个面，第 4 次 1 个面),但每次必须充分利用锅。最优安排：第 1 次烙①②③正面(3 分钟),第 2 次烙①反面④⑤正面(3 分钟),第 3 次烙②③④反面(3 分钟),第 4 次烙⑤反面(3 分钟)。共 12 分钟。或理解为:5 张饼，每次 3 张，需 2 轮，每轮 6 分钟，但第 2 轮只有 2 张，仍需 6 分钟，共 12 分钟.\n【详解】先烙1,2和3号饼的正面，用时3分钟。接着烙1号饼的反面和 4,5 号饼的正面，用时 3 分钟。再烙 2,3 和4号饼的反面，用时3分钟。最后烙5号饼的反面，用时3分钟，合计:3 + 3 + 3 + 3 = 12(分)。",
+        "knowledge_points":  [
+                                 "连加，连减(10以内)",
+                                 "优化问题"
+                             ],
+        "source":  "15-16 四年级上·全国·课后作业"
+    },
+    {
+        "id":  "169",
+        "unit":  "第八章 数学广角（二） 优化问题",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "规律：在每次烙两张饼的情况下：烙饼所需的时间=( )×( )。",
+        "options":  [
+
+                    ],
+        "answer":  "饼的张数;烙一面所用的时间",
+        "solution":  "【分析】 在每次烙 2 张饼的情况下，通过前面的题目可以发现规律：无论烙多少张饼(张数 \u003e= 2),只要锅一直被充分利用(每次 2 张),总时间就等于饼的张数乘以烙一面的时间。例如：烙 3 张饼需要 9 分钟(3×3),烙 5 张饼需要 15 分钟(5×3),烙 6 张饼需要 12 分钟(6×2，若每面 2 分钟)。这是因为每张饼有 2 个面，每次烙 2 个面，平均每张饼占用锅的时间就是烙一面的时间.\n【详解】假设每面需要 t 分钟，烙 n 张饼(n\u003e=2):每张饼有 2 个面，共 2n 个面;每次烙 2 个面(2 张饼各一面),需要烙 n 次，每次 t 分钟，总时间 = n×t 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "18-19 四年级上·全国·课后作业"
+    },
+    {
+        "id":  "170",
+        "unit":  "第八章 数学广角（二） 优化问题",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "合理安排时间：做几件事情时，如果能合理安排顺序，把( ),可以节省时间。",
+        "options":  [
+
+                    ],
+        "answer":  "能同时做的事情同时做",
+        "solution":  "【分析】 这是统筹优化的核心原则。在日常生活中，很多任务之间存在时间上的重叠可能：一些任务需要连续进行(有先后依赖关系),而另一些任务可以与其他任务并行(无冲突)。合理安排的关键是识别出这些可并行的任务，在等待某项长时间任务(如烧水，煮饭)完成的过程中，穿插完成其他短任务，从而缩短总时间.\n【详解】以沏茶问题为例：必须串行：洗水壶 → 烧开水 → 沏茶;可以并行：洗茶杯，拿茶叶(在烧开水期间完成)。因此，规律可以概括为：能同时做的事情同时做。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "18-19 四年级上·全国·课后作业"
+    },
+    {
+        "id":  "171",
+        "unit":  "第八章 数学广角（二） 优化问题",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "爸爸煎鱼，锅里每次只能煎两条鱼，煎一面需2分钟，爸爸煎6条鱼最少要( )分钟。煎 10 条鱼最少要( )分钟。煎 30 条鱼最少要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "12;20;60",
+        "solution":  "【分析】 每次煎 2 条鱼，每面 2 分钟，每条鱼需煎两面共 4 分钟。但当锅充分利用时(每次 2 条),总时间 = 鱼的数量×每面时间。因为 6 条鱼可以 2 条 2 条地煎，每次 4 分钟,3 次共 12 分钟;或直接 6×2=12 分钟。同理,10 条鱼:10×2=20 分钟;30 条鱼:30×2=60 分钟.\n【详解】 根据烙饼问题规律：每次烙 2 张(条)时，总时间 = 张(条)数×每面时间。第一问：煎 6 条鱼：每次煎 2 条，每面 2 分钟,6 条鱼分 3 次煎完，每次 4 分钟(两面),或按公式:6×2=12 分钟。第二问：煎 10 条鱼:10×2=20 分钟。第三问：煎 30 条鱼:30×2=60 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "18-19 四年级上·全国·课后作业"
+    },
+    {
+        "id":  "172",
+        "unit":  "第八章 数学广角（二） 优化问题",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "\"铁板烧\"烤鱿鱼，鱿鱼需要烤( )面，如果烤一面需要 2 分钟，那么烤一条鱿鱼需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "2;4",
+        "solution":  "【分析】 生活中的常识：鱿鱼(以及大多数食物)需要烤两面才能熟透。如果烤一面需要 2 分钟，那么烤两面就需要 2×2=4 分钟。这是烙饼问题的基础模型，理解\"两面都要烤\"是后续解决复杂优化问题的前提.\n【详解】实际烤制过程：鱿鱼需要烤 2 面(正面和反面),烤一面需要 2 分钟，烤一条鱿鱼需要:2×2=4(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "18-19 四年级上·全国·期末"
+    },
+    {
+        "id":  "173",
+        "unit":  "第八章 数学广角（二） 优化问题",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "星期天，妈妈在家收拾屋子，妈妈口渴了，想喝茶。1.打扫房间 6 分钟；2.找茶叶 1 分钟；3.烧水 6 分钟；4.接水 1 分钟。\n\n要使时间最少，先做( )，接着做( )，然后在做( )的同时做( )和( )。",
+        "options":  [
+
+                    ],
+        "answer":  "4;3;3;1;2",
+        "solution":  "【分析】 分析泡茶的流程：必须先接水(1 分钟)，才能烧水(6 分钟)。找茶叶(1 分钟)不需要用水，可以在烧水的同时进行。打扫房间(6 分钟)也可以在烧水的 6 分钟内同时进行。因此最优顺序是先接水，再烧水，烧水时安排打扫房间和找茶叶。关键路径是接水加烧水，共 7 分钟。\n【详解】第一步：先接水，接水需要 1 分钟，这是烧水的前置条件。第二步：接着烧水，烧水需要 6 分钟。第三步：在烧水的同时，打扫房间 6 分钟，找茶叶 1 分钟。题目中的括号可以填为：先做 4，接着做 3，然后在做 3 的同时做 1 和 2。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "18-19四年级上·全国·单元测试"
+    },
+    {
+        "id":  "174",
+        "unit":  "第八章 数学广角（二） 优化问题",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用一个平底锅煎鱼，每次只能煎两条，两面都要煎，煎好一面需2分钟，煎好7条鱼至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "14",
+        "solution":  "【分析】先煎第一，第二条鱼，需要 2×2 = 4(分钟);再煎第三，第四条鱼，需要 2×2 = 4(分钟);最后煎后3条鱼，先煎第五条与第六条的正面需2分钟，然后煎第五条的反面与第七条的正面需要2分钟，再煎第六条的反面与第七条的反面需2分钟，煎后3条共需:3×2 = 6(分钟).\n【详解】2×2 + 2×2 + 3×2 = 4+4+6 = 14(分钟),所以，煎好 7 条鱼至少需要 14 分钟.\n[点睛]在一个锅一次最多能同时烤2个东西的问题中，东西个数与所需时间的关系为：所需时间 = 个数×烤一面所用时间。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "18-19四年级上·全国·单元测试"
+    },
+    {
+        "id":  "175",
+        "unit":  "第八章 数学广角（二） 优化问题",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "煮一个鸡蛋需要8分钟，一个锅一次可以煮10个鸡蛋，那么煮8个鸡蛋至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "8",
+        "solution":  "【分析】 煮鸡蛋问题的关键是理解：一个锅一次可以煮 10 个鸡蛋，意味着锅的容量是 10 个。现在只需要煮 8 个鸡蛋,8 \u003c 10，所以 8 个鸡蛋可以一次性放入锅内同时煮。煮 8 个鸡蛋和煮 1 个鸡蛋的时间相同，都是 8 分钟，因为它们在同一锅内同时受热煮熟.\n【详解】锅的容量:10 个鸡蛋，需要煮的鸡蛋:8 个,8 \u003c 10，可以一次性放入，煮熟时间与煮 1 个鸡蛋相同，都是 8 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "18-19四年级上·全国·单元测试"
+    },
+    {
+        "id":  "176",
+        "unit":  "第八章 数学广角（二） 优化问题",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一只平底锅，每次只能放 2 张饼，如果 1 张饼两面都烙，每面需要 3 分钟，那么烙15 张饼最少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "45",
+        "solution":  "【分析】 每次最多烙 2 张饼，每面 3 分钟。15 张饼共 30 个面，每次烙 2 个面，理论上需要 15 次，每次 3 分钟，共 45 分钟。由于 15 是奇数，先烙 12 张(6 次，每次 2 张，共 36 分钟),剩下 3 张用交替法(9 分钟),总计 45 分钟。或直接用公式：饼数×每面时间 = 15×3=45 分钟.\n【详解】根据烙饼问题规律：每次烙 2 张时，总时间 = 饼的张数×每面时间。直接计算:15×3=45(分钟)。分步验证：前 12 张饼：每次 2 张，每面 3 分钟，每次 6 分,12 张 / 2 = 6 次，用时:6×6=36 分钟。后 3 张饼(交替法):第 1 次:①正 + ②正(3 分钟);第 2 次:①反 + ③正(3 分钟);第 3 次:②反 + ③反(3 分钟),用时:3×3=9 分钟。总计:36+9=45 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "18-19 四年级上·全国·期末"
+    },
+    {
+        "id":  "177",
+        "unit":  "第八章 数学广角（二） 优化问题",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "\"铁板烧\"烤鱿鱼时，鱿鱼需要烤( )面，如果烤一面需要 2 分钟，那么烤 3 条鱿鱼至少需要( )分钟。(铁板上可同时烤 2 条鱿鱼)",
+        "options":  [
+
+                    ],
+        "answer":  "2;6",
+        "solution":  "【分析】 鱿鱼需要烤两面才能熟透。铁板可同时烤 2 条鱿鱼，每面 2 分钟。烤 3 条鱿鱼共 6 个面，每次烤 2 个面(2 条各一面),需要 3 次，每次 2 分钟，共 6 分钟。具体安排：第 1 次烤鱿鱼①正面和鱿鱼②正面，第 2 次烤鱿鱼①反面和鱿鱼③正面，第 3 次烤鱿鱼②反面和鱿鱼③反面.\n【详解】第 1 次：烤鱿鱼①正面，鱿鱼②正面，需 2 分钟;第 2 次：烤鱿鱼①反面，鱿鱼③正面，需 2 分钟，此时鱿鱼①已完成;第 3 次：烤鱿鱼②反面，鱿鱼③反面，需 2 分钟，此时鱿鱼②,③都完成。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "18-19 四年级上·全国·期末"
+    },
+    {
+        "id":  "178",
+        "unit":  "第八章 数学广角（二） 优化问题",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "晓阳每天早晨起床要做这些事情：刷牙洗脸 4 分钟，叠被子 2 分钟，用电饭锅熬粥15 分钟，整理书包 3 分钟，吃早饭 10 分钟。晓阳每天起床后最快( )分钟完成这些事情。",
+        "options":  [
+
+                    ],
+        "answer":  "25",
+        "solution":  "【分析】 分析各项工作：熬粥需要 15 分钟(电饭锅自动工作，无需看管),在此期间可以完成刷牙洗脸(4 分钟),叠被子(2 分钟),整理书包(3 分钟),这三项共 9 分钟，完全嵌入熬粥的 15 分钟内。然后吃早饭 10 分钟，必须在粥熬好后进行。关键路径：熬粥(15 分钟，同时做其他)→ 吃早饭(10 分钟)。总时间:15+10=25 分钟.\n【详解】第一步：用电饭锅熬粥(同时刷牙洗脸，叠被子，整理书包)。熬粥需要 15 分钟，在熬粥期间同时进行：刷牙洗脸 4 分钟，叠被子 2 分钟，整理书包 3 分钟，这三项共 4+2+3=9 分钟，完全嵌入熬粥的 15 分钟内，本步骤共用时 15 分钟。第二步：吃早饭，吃早饭需要 10 分钟，必须在粥熬好后进行。故总时间为 15+10=25 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "18-19 四年级上·全国·期末"
+    },
+    {
+        "id":  "179",
+        "unit":  "第八章 数学广角（二） 优化问题",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "莲莲用电话通知 15个学生参加活动，如果通知一个学生要 1 分钟，那 15 个学生全部通知到至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "4",
+        "solution":  "【分析】第一分钟，一传一;第二分钟，二传二;第三分钟，四传四;第四分钟，七传八。这是\"打电话通知\"类的优化问题，关键是已通知的人可以继续通知其他人，人数呈翻倍增长.\n【详解】第一分钟通知一个人;第2分钟那个学生通知了另一个人，共通知:2+1=3 人;第3分钟通知的人数:3+4=7 人(已知 3 人再各通知 1 人，新增 4 人，其中含老师);第四分钟通知的人数:7+8=15 人。所以通知 15 个学生至少要 4 分钟。故答案为:4.\n[点睛]本题考查了数学广角中的优化问题。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "2018 五年级下·全国·专题练习"
+    },
+    {
+        "id":  "180",
+        "unit":  "第八章 数学广角（二） 优化问题",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小秋给客人烧水沏茶，洗水壶用1分钟，烧开水用15分钟，洗茶壶用1分钟，洗茶杯用1分钟，拿茶叶用2分钟。为了使客人早点儿喝上茶，最合理的安排顺序是( ),( )分钟就可以沏好茶了。",
+        "options":  [
+
+                    ],
+        "answer":  "洗开水壶→烧开水(同时洗茶壶、洗茶杯、拿茶叶);16",
+        "solution":  "【分析】根据题意可知，烧水沏茶前，先洗水壶，然后烧开水，在烧开水的同时洗茶壶、洗茶杯、拿茶叶，最后就可以沏茶了。\n【详解】最合理的安排顺序是：洗开水壶→烧开水(同时洗茶壶、洗茶杯、拿茶叶)，1+15=16(分钟)。\n[点睛]考查了优化问题，主要是如何合理安排时间，哪些事情可以同时进行。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "18-19 四年级上·全国·单元测试"
+    },
+    {
+        "id":  "181",
+        "unit":  "第八章 数学广角（二） 优化问题",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明给客人烧水沏茶。洗水壶要2分钟，烧开水要12分钟，洗茶杯要2分钟，拿茶叶要1分钟，为了使客人早点喝上茶，按你认为最合理的安排( )分钟就能沏茶了。",
+        "options":  [
+
+                    ],
+        "answer":  "14",
+        "solution":  "【分析】先洗水壶，再烧开水，在烧水的过程中，洗茶杯，拿茶叶;洗水壶需要2分钟，烧水需要12分钟，所以至少需要花(2+12)分钟.\n【详解】2 + 12 = 14(分钟),14 分钟就能沏茶了。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "15-16 四年级上·全国·课后作业"
+    },
+    {
+        "id":  "182",
+        "unit":  "第八章 数学广角（二） 优化问题",
+        "question_type":  "填空题",
+        "difficulty":  0.4,
+        "question":  "一堆水果糖有 10 颗，两人轮流从中拿走 1 颗或 2 颗，谁拿到最后一颗谁就获胜。让你先拿，第一次应该拿( )颗才能确保获胜。如果有17 颗糖，第一次应该拿( )颗才能确保获胜。",
+        "options":  [
+
+                    ],
+        "answer":  "1;2",
+        "solution":  "【分析】 这是博弈论中的\"取石子游戏\"。关键策略是控制每轮两人取糖总数为 3(因为 1+2=3)。如果总颗数除以 3 余 1，先拿 1 颗，之后对方拿 n 颗，你就拿 (3-n) 颗，确保获胜;如果余 2，先拿 2 颗。10÷3 = 3 余 1，所以先拿 1 颗;17÷3 = 5 余 2，所以先拿 2 颗.\n【详解】核心规律：每次可拿 1 或 2 颗，控制每轮总数为 3。第一问:10 颗糖,10÷3=3⋯⋯1，余数为 1，第一次拿 1 颗。之后策略：对方拿 1 颗，你拿 2 颗;对方拿 2 颗，你拿 1 颗。每轮减少 3 颗，最后你拿到最后一颗。第二问:17 颗糖,17÷3=5⋯⋯2，第一次应拿 2 颗，然后同样保持每轮总数为 3，即可确保最后一颗由你拿到。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "18-19 四年级上·全国·课后作业"
+    },
+    {
+        "id":  "183",
+        "unit":  "第八章 数学广角（二） 优化问题",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个平底锅里只能同时煎 4 条鱼，用它煎一条鱼需要 4 分钟(正，反面各需要2 分钟),煎 4 条鱼需要( )分钟。煎 6 条鱼需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "4;6",
+        "solution":  "【分析】 每面 2 分钟，一条鱼两面共 4 分钟。煎 4 条鱼：锅容量正好 4 条，可以一次性放入，同时煎两面，共 4 分钟。煎 6 条鱼:6 条共 12 个面，每次煎 4 个面，需要 3 次，每次 2 分钟，共 6 分钟。或理解为:6 条鱼，每次 4 条，先煎 4 条(4 分钟),再煎剩下 2 条(4 分钟),但优化后可 6 分钟完成.\n【详解】 第一问：煎 4 条鱼。锅容量:4 条鱼,4 条鱼正好放满，同时煎，每面 2 分钟，两面共 4 分钟，共用时 4 分钟。第二问：煎 6 条鱼。可以按\"面数\"理解:6 条鱼共 12 个面，每次煎 4 个面，需要 3 次，每次 2 分钟，共 6 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "18-19 四年级上·全国·课后作业"
+    },
+    {
+        "id":  "184",
+        "unit":  "第八章 数学广角（二） 优化问题",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "星期天，小红在家做家务，拖地板用15分钟，擦家具用5分钟，整理床铺用3分钟，烧开水用 20 分钟，小红做完这些家务至少用( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "23",
+        "solution":  "【分析】 烧开水 20 分钟期间无需看管，可以同时进行拖地板(15 分钟)和擦家具(5 分钟),这两项共 20 分钟，正好嵌入烧开水时间内。整理床铺(3 分钟)必须在烧开水之后(或之前)单独做。关键路径：烧开水(20 分钟，同时拖地板，擦家具)→ 整理床铺(3 分钟),共 23 分钟.\n【详解】拖地板，擦家具共需要 15+5=20 分钟，烧开水用 20 分钟，因此在等待烧开水的这20分钟内，可以完成拖地板与擦家具这两项家务，整理床铺再用3分钟，做完这些事至少要花:20+3=23(分钟)故答案为 23。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "17-18 四年级上·全国·期末"
+    },
+    {
+        "id":  "185",
+        "unit":  "第八章 数学广角（二） 优化问题",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小红给客人沏茶，接水 2 分钟，洗茶杯 3 分钟，烧水 7 分钟，拿茶叶 1 分钟，沏茶 1 分钟，小红最少要用( )分钟使客人喝上茶。",
+        "options":  [
+
+                    ],
+        "answer":  "10",
+        "solution":  "【分析】 先接水后才能烧水，所以接水 2 分钟不能和烧水 7 分钟重合。烧水的 7 分钟内，可以同时洗茶杯 3 分钟、拿茶叶 1 分钟。最后还要沏茶 1 分钟，因此最少时间是 2+7+1=10(分钟)。\n【详解】先接水 2 分钟，再烧水 7 分钟。在烧水的同时洗茶杯 3 分钟、拿茶叶 1 分钟，水烧好后再沏茶 1 分钟，所以最少需要 2+7+1=10(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "18-19 四年级上·全国·期末"
+    },
+    {
+        "id":  "186",
+        "unit":  "第八章 数学广角（二） 优化问题",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用一个平底锅煎饼，每次可以放2张饼，每面要煎1分钟。如果有3张饼，两面都要煎，至少要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "3",
+        "solution":  "【分析】 3 张饼共 6 个面，每次煎 2 个面，每面 1 分钟，需要 3 次，共 3 分钟。具体安排(交替法):第 1 分钟煎饼①正面和饼②正面，第 2 分钟煎饼①反面和饼③正面，第 3 分钟煎饼②反面和饼③反面。这样 3 分钟完成，锅一直被充分利用，无空位.\n【详解】我们把这三张煎饼分别标上1 ,2 ,3。先煎1,2号的正面，煎熟后拿出，放入3号，煎1号的反面和3号的正面，煎熟后拿出1号。再煎3号的反面和2号的反面，这样用时最少，共 1+1+1=3 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "18-19四年级上·全国·单元测试"
+    },
+    {
+        "id":  "187",
+        "unit":  "第八章 数学广角（二） 优化问题",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用一只平底锅烙饼，锅上只能放两个饼，烙熟饼的一面需要 4 分钟，两面都要烙，现在需要烙熟三个饼，最少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "12",
+        "solution":  "【分析】 每次尽量让锅里都有 2 张饼在烙。三张饼可以采用交替烙的方法：第 1 次烙 1、2 的正面，第 2 次烙 1 的反面和 3 的正面，第 3 次烙 2、3 的反面，每次 4 分钟，共 12 分钟。\n【详解】三张饼分别用 1、2、3 表示。第一次：1 正、2 正，需要 4 分钟；第二次：1 反、3 正，需要 4 分钟；第三次：2 反、3 反，需要 4 分钟。4+4+4=12(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "18-19 四年级上·全国·单元测试"
+    },
+    {
+        "id":  "188",
+        "unit":  "第八章 数学广角（二） 优化问题",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "煎芝麻饼要两面煎，煎第一面要3分钟，煎第二面只要1分钟就行了，煎一只芝麻饼要 4 分钟，一只煎锅一次能放入 2 个芝麻饼，如果要煎3个芝麻饼，需要( )分钟最省时。",
+        "options":  [
+
+                    ],
+        "answer":  "7",
+        "solution":  "【分析】 特殊条件：第一面 3 分钟，第二面 1 分钟，两面时间不同。3 个饼共 6 个面，需要合理安排使锅不空。标准最优策略：第 1 次煎第一只和第二只芝麻饼的第一面 3 分钟;第 2 次将第一只取出，将第二只翻面，同时放入第三只，经过 1 分钟第二只熟，取出，再放入第一只煎第二面，接着再 1 分钟后第一只熟，再翻第三只煎第二面 1 分钟。总用时 3+3+1=7 分钟.\n【详解】可以这样煎饼：先煎第一，第二只芝麻饼的第一面,3分钟后，把第一只芝麻饼取出，将第二只翻个身，放上第三个芝麻饼。经过1分钟，第二只芝麻饼已熟，可把它取出来，放上第一只芝麻饼，煎另外一面，一分钟后，第一只芝麻饼又熟了，可把它取出，把第三只芝麻饼翻一个身，再过 1 分钟，第三只芝麻饼也熟了。3 + 3 + 1 = 7(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "18-19 四年级上·全国·单元测试"
+    },
+    {
+        "id":  "189",
+        "unit":  "第八章 数学广角（二） 田忌赛马思想",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "四(1)班和四(2)举行羽毛球比赛，每班选出第一名，第二名，第三名三位同学参赛，共打3场比赛,3局2胜制，四(2)班的三名队员出场的顺序是第一名同学，第二名同学，第三名同学，四(1)班要想打赢这场比赛，三名队员出场顺序为( ),( ),( )。",
+        "options":  [
+
+                    ],
+        "answer":  "第三名;第一名;第二名",
+        "solution":  "【分析】 这是\"田忌赛马\"策略问题。3 局 2 胜制，四(2)班出场顺序：强，中，弱。四(1)班要赢 2 场，应采用错位对抗：用己方最弱的输给敌方最强的(输 1 场),用己方最强的赢敌方中等的(赢 1 场),用己方中等的赢敌方最弱的(赢 1 场),这样 2 胜 1 负获胜。因此顺序为：第三名(弱)→ 第一名(强)→ 第二名(中).\n【详解】结果：四(1)班 2 胜 1 负，赢得比赛。策略核心：牺牲最弱对最强，确保另外两场胜利。故四(1)班三名队员出场顺序为：第三名，第一名，第二名。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "16-17 四年级上·广东江门·单元测试"
+    },
+    {
+        "id":  "190",
+        "unit":  "第八章 数学广角（二） 优化问题",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "有甲，乙两只船要卸货，且只能一船一船的卸。给甲船卸货要3 小时，给乙船卸货2小时。先给( )船卸货，再给( )船卸货可使货船等候时间的总和最少。",
+        "options":  [
+
+                    ],
+        "answer":  "乙;甲",
+        "solution":  "【分析】 等候时间总和 = 每艘船等待开始卸货的时间 + 自己卸货时间。若先卸甲(3 小时),乙等待 3 小时，总等候时间 = 3+(3+2)=8 小时;若先卸乙(2 小时),甲等待 2 小时，总等候时间 = 2+(2+3)=7 小时。因此先卸用时短的乙船，总等候时间最少.\n【详解】若先卸甲船：甲船：等待 0 小时，卸货 3 小时，共 3 小时;乙船：等待 3 小时，卸货 2 小时，共 5 小时;等候时间总和:3+5=8 小时。若先卸乙船：乙船：等待 0 小时，卸货 2 小时，共 2 小时;甲船：等待 2 小时，卸货 3 小时，共 5 小时;等候时间总和 2+5=7 小时，更少。因此应先给乙船卸货，再给甲船卸货。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "16-17 四年级上·广东江门·单元测试"
+    },
+    {
+        "id":  "191",
+        "unit":  "第八章 数学广角（二） 优化问题",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明每天起床要完成的事情有：起床穿衣 3 分钟，整理床铺 2 分钟，刷牙，洗脸 5 分钟，热馒头 10 分钟，吃早餐 5 分钟。你认为最合理的安排要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "18",
+        "solution":  "【分析】 热馒头 10 分钟期间(蒸锅或微波炉自动工作)，可以同时进行整理床铺(2 分钟)和刷牙洗脸(5 分钟)，这两项共 7 分钟，在 10 分钟内完成。起床穿衣(3 分钟)必须在热馒头之前(或同时)，吃早餐(5 分钟)必须在热馒头之后。关键路径：起床穿衣(3)→热馒头(10，同时整理床铺、刷牙洗脸)→吃早餐(5)，共 18 分钟。\n【详解】第一步：起床穿衣，起床穿衣需要 3 分钟。第二步：热馒头，热馒头需要 10 分钟，在热馒头期间同时进行整理床铺 2 分钟和刷牙洗脸 5 分钟，这两项共 7 分钟，完全嵌入 10 分钟内。第三步：吃早餐，吃早餐需要 5 分钟，必须在馒头热好后进行。总时间：3+10+5=18(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "16-17四年级上·广东江门·单元测试"
+    },
+    {
+        "id":  "192",
+        "unit":  "第八章 数学广角（二） 取数游戏策略",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "有40张扑克牌，两人轮流拿，每次只能拿一张或两张，谁拿到最后一张谁获胜。如果让你先拿，为了确保获胜，你的想法是:\n(1)倒数第二次你取了之后只能剩下( )张扑克牌，这时，对手拿( )张，你就拿( )张;对手拿( )张，你就拿( )张，可以确保你获胜.\n(2)不管对手每次拿几张，只要你每次拿的与对手拿的张数的和是( ),就可以获胜。当然，第一次拿牌时你只能拿( )张。",
+        "options":  [
+
+                    ],
+        "answer":  "3;1;2;2;1;3;1",
+        "solution":  "【分析】 这是取石子游戏的策略问题。每次可拿 1 或 2 张，关键控制每轮总数为 3(1+2)。要确保获胜，需要控制局面使最后剩下 3 张给对手(无论他拿 1 或 2，你都能拿完)。倒推：倒数第二次取后剩 3 张;之前每轮减少 3 张。40÷3 = 13 余 1，所以第一次拿 1 张，之后每轮与对手凑 3.\n【详解】核心规律：每次可拿 1 或 2 张，控制每轮两人拿牌总数为 3。(1) 最后阶段控制倒数第二次你取之后，只能剩下 3 张扑克牌给对手：对手拿 1 张，你就拿 2 张，你拿到最后一张;对手拿 2 张，你就拿 1 张，你拿到最后一张。(2) 全程策略：不管对手每次拿几张，只要你每次拿的与对手拿的张数的和是 3，就可以获胜。因为 40÷3=13。1，第一次你应拿 1 张，使牌堆剩下 39 张(3 的倍数),之后始终保持一轮取 3 张的节奏，即可保证你拿到最后一张。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "18-19四年级上·全国·单元测试"
+    },
+    {
+        "id":  "193",
+        "unit":  "第八章 数学广角（二） 优化问题",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明安排时间最合理:(1)起床整理被褥3分钟;(2)刷牙3分钟;(3)洗脸2 分钟;(4)听英语录音 8 分钟。如果六点起床，最快( )时( )分做完这些事情。",
+        "options":  [
+
+                    ],
+        "answer":  "6;8",
+        "solution":  "【分析】听录音是8分钟，在整理被褥，刷牙，洗脸的时候用来听录音，正好时间做完所有的事情只需要8分钟，因此最快6 时 8 分完成.\n【详解】3+3+2=6+2=8(分钟),所以做完所有事情 8 分钟就可以，最快 6 时 8 分完成.\n[点睛]考查合理安排时间的相关知识，重点是能够清楚地知道哪些事情可以同时做来节省时间。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "2014 四年级·全国·课后作业"
+    },
+    {
+        "id":  "194",
+        "unit":  "第八章 数学广角（二） 优化问题",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "要烙3张饼，饼的两面都要烙，有一口锅一次只能烙两张饼，烙好一面需要3分钟。最少需要( )分钟能烙好所有的饼。",
+        "options":  [
+
+                    ],
+        "answer":  "9",
+        "solution":  "【分析】 3 张饼共 6 个面，每次最多烙 2 个面，每面 3 分钟。若先烙 2 张(6 分钟),再烙第 3 张(6 分钟),共 12 分钟，不是最优。应采用交替烙法(经典烙饼问题):第 1 次烙饼①正面和饼②正面，第 2 次烙饼①反面和饼③正面，第 3 次烙饼②反面和饼③反面。这样锅一直被充分利用，无空位,3 次共 9 分钟.\n【详解】三张饼分别用序号1,2,3表示：第一次:1 正,2 正，需要 3 分钟;第二次:1 反,3 正，需要 3 分钟;第三次:2 反,3 反，需要 3 分钟;3 + 3 + 3 = 9(分钟);答：最少需要 9 分钟能烙好所有的饼。故答案为 9。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "18-19四年级上·全国·单元测试"
+    },
+    {
+        "id":  "195",
+        "unit":  "第八章 数学广角（二） 经济问题",
+        "question_type":  "填空题",
+        "difficulty":  0.4,
+        "question":  "一家商场开展假日酬宾活动，凡购物满 100 元(不足 100 元不计),回赠 35元现金。现有 250 元现金，最多能买到( )元物品。",
+        "options":  [
+
+                    ],
+        "answer":  "355",
+        "solution":  "【分析】 这是\"满减返现\"类的优化消费问题。策略是尽量凑满 100 元的整数倍以获取最大返现，且返现的现金要继续用于消费。250 元可分多次购买：第一次买 200 元(满 2 个 100),返现 35×2=70 元，此时手头有剩余的 50 元 + 返现 70 元 = 120 元;第二次买 100 元(用 100 元),返现 35 元，剩余 20 元 + 返现 35 元 = 55 元;第三次买 55 元。总购买:200+100+55=355 元.\n【详解】根据优惠方案,250元可分三次购买：第一次购买200元商品，可返还 35×2 = 70 元;第二次可用 50+70=120 元再购 100 元商品，再返还 35 元;最后可用 20+35=55 元购 55 元的商品。因此最多可购 250+70+35=355(元)的物品。答：最多能买到 355 元的商品。故答案为 355。",
+        "knowledge_points":  [
+                                 "优化问题",
+                                 "经济问题"
+                             ],
+        "source":  "18-19 四年级上·全国·单元测试"
+    },
+    {
+        "id":  "196",
+        "unit":  "第八章 数学广角（二） 优化问题",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个锅每次只能放两片面包，每烤一面需2分钟。小红早点要吃3片面包，至少要等( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】 与烙饼问题完全相同。3 片面包共 6 个面，每次烤 2 个面，每面 2 分钟。采用交替烤法：第 1 次烤面包①正面和面包②正面(2 分钟),第 2 次烤面包①反面和面包③正面(2 分钟),第 3 次烤面包②反面和面包③反面(2 分钟)。共 6 分钟，锅无空位.\n【详解】3 片面包分别用序号 1,2,3 表示：第一次烤:1 正,2 正,2 分钟;第二次烤:1 反,3 正,2 分钟;第三次烤:2 反,3 反,2 分钟。2+2+2=6(分钟),答：烤 3 片面包，至少要等 6 分钟。故答案为 6。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "18-19 四年级上·全国·单元测试"
+    },
+    {
+        "id":  "197",
+        "unit":  "第八章 数学广角（二） 优化问题",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小淘气帮妈妈做家务，扫地要用5分钟，擦家具要8分钟，烧开水要15分钟，泡茶要 1 分钟，如果他合理安排，做完这些事至少要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "16",
+        "solution":  "【分析】 烧开水 15 分钟期间无需看管，可以同时进行扫地(5 分钟)和擦家具(8 分钟),这两项共 13 分钟，在 15 分钟内可以完成。泡茶(1 分钟)必须在烧开水之后。关键路径：烧开水(15 分钟，同时扫地，擦家具)→ 泡茶(1 分钟)。总时间:15+1=16 分钟.\n【详解】扫地要用5分钟，擦家具要8分钟,5+8=13 分钟，因此小淘气可以完全在等候水开的这15分钟内完成扫地与擦家具，水烧开后，泡茶要 1 分钟，所以共需:15+1=16(分钟)故答案为 16。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "16-17 四年级·天津·课后作业"
+    },
+    {
+        "id":  "198",
+        "unit":  "第八章 数学广角（二） 优化问题",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "星期天，小丽在家做下面的这些事:\n事件：烧开水 15 分;洗红领巾 7 分;整理房间 10 分.\n她至少需要( )分才能做完这些事。",
+        "options":  [
+
+                    ],
+        "answer":  "17",
+        "solution":  "【分析】可让小丽先烧水，在烧水的同时去洗红领巾，洗完红领巾，再去整理房间，在整理房间时水就烧开了。所以需要的最少时间就是洗红领巾用的时间和整理房间用的时间的和.\n【详解】10+7=17(分钟)故答案为17。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "14-15 四年级下·全国·单元测试"
+    },
+    {
+        "id":  "199",
+        "unit":  "第八章 数学广角（二） 统筹安排",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "汪师傅给一块板两面刷油漆，刷一面需要 1 分钟，但必须过 4 分钟漆干后才能给另一面刷漆。汪师傅刷完 5 块这样的木板需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "10",
+        "solution":  "【分析】 每块木板需要刷两面，每面 1 分钟，但两面之间必须间隔 4 分钟。关键是合理安排刷漆顺序，使得在等待某块木板晾干的过程中，可以去刷其他木板，从而达到时间最优。由于有 5 块木板，可以循环操作，避免空等。最终可以安排成连续作业，使总时间为 10 分钟.\n【详解】将 5 块木板编号为 ①,②,③,④,⑤。先依次给 5 块木板刷第一面(共 5 分钟),然后间隔时间中穿插刷其它板的另一面，最终可以做到每分钟都在刷某一块板的一面，在满足每块板两次刷漆间隔至少 4 分钟的条件下，总时间为 10 分钟。",
+        "knowledge_points":  [
+                                 "优化问题",
+                                 "统筹安排",
+                                 "时间规划"
+                             ],
+        "source":  "18-19 四年级上·全国·单元测试"
+    },
+    {
+        "id":  "200",
+        "unit":  "第八章 数学广角（二） 统筹安排",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "某宾馆服务员接班后有下面几项工作：整理房间 5 分钟;给洗衣机加水，放衣物 2 分钟;自动洗涤 10 分钟;晾晒衣物 2 分钟;拖地 3 分钟。做完上面的事情，最少需( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "14",
+        "solution":  "【分析】 分析各项工作的先后关系和可并行性：给洗衣机加水放衣物 → 自动洗涤 → 晾晒衣物，这三项必须按顺序进行。而整理房间和拖地可以在其他工作进行的同时完成，只要时间不冲突即可。关键是找出最长的不间断流程，再看其他工作能否在此期间并行完成。自动洗涤 10 分钟期间，可以完成整理房间 5 分钟和拖地 3 分钟(共 8 分钟),不增加总时长.\n【详解】必须顺序进行的工作(洗衣流程):给洗衣机加水，放衣物 2 分钟;自动洗涤 10 分钟(此期间可做其他事情);晾晒衣物 2 分钟。合理安排为：先整理房间 5 分钟，再给洗衣机加水，放衣物 2 分钟，然后自动洗涤 10 分钟，在洗涤的前 3 分钟内完成拖地 3 分钟，最后晾晒衣物 2 分钟。关键路径时间为：整理房间 5 分钟 + 加水放衣物 2 分钟 + 自动洗涤 10 分钟 + 晾晒 2 分钟，但整理和拖地可以部分并入洗涤时间中，最终最少需 14 分钟。",
+        "knowledge_points":  [
+                                 "优化问题",
+                                 "统筹安排",
+                                 "并行处理"
+                             ],
+        "source":  "18-19 四年级上·全国·单元测试"
+    },
+    {
+        "id":  "201",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "星期天小红在家做以下家务：擦桌子 5 分钟，烧开水 10 分钟，拖地 5 分钟，洗水壶 2 分钟。通过合理安排最少要用( )分钟做完。",
+        "options":  [
+
+                    ],
+        "answer":  "12",
+        "solution":  "【分析】分析各项工作：洗水壶是烧开水的前置条件(2 分钟),烧开水 10 分钟期间(电水壶自动工作),可以同时完成擦桌子(5 分钟)和拖地(5 分钟),这两项共 10 分钟，正好嵌入烧开水时间内。关键路径：洗水壶 → 烧开水(同时擦桌子，拖地).\n【详解】2+10=12(分钟) 故通过合理安排最少要用 12 分钟做完。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "17-18 四年级上·江西赣州·期末"
+    },
+    {
+        "id":  "202",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "星期天，妈妈要刷碗整理厨房用20分钟，烧一壶水沏茶用15分钟，用全自动洗衣机洗被罩1小时，手洗小明衣服要用20分钟，把小明衣服放入洗衣机甩干5分钟。小明妈妈做完这些事，最少要用的时间是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "65",
+        "solution":  "【分析】用全自动洗衣机洗被罩需要 1 小时，这是最长的一项工作。在洗被罩的 60 分钟内，可以同时完成刷碗整理厨房 20 分钟、烧一壶水沏茶 15 分钟和手洗小明衣服 20 分钟。等被罩洗完后，再把手洗好的小明衣服放入洗衣机甩干 5 分钟即可。\n【详解】60+5=65(分钟)，所以小明妈妈做完这些事最少要用 65 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "18-19 四年级上·全国·期末"
+    },
+    {
+        "id":  "203",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明、小红、小青三人去学校医务室，小明打针需要 7 分钟，小青量体温需要 5 分钟，小红拿药需要 3 分钟。要使他们等待的时间总和最少，应该怎样安排他们看病：( )、( )、( )。(请写出他们的名字)",
+        "options":  [
+
+                    ],
+        "answer":  "小红;小青;小明",
+        "solution":  "【分析】要使等待时间总和最少，应先安排用时最短的，再安排用时较长的，也就是按时间从少到多排序。\n【详解】小红拿药需要 3 分钟，小青量体温需要 5 分钟，小明打针需要 7 分钟，所以应该按小红、小青、小明的顺序安排。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "18-19四年级上·全国·期末"
+    },
+    {
+        "id":  "204",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明和爸爸到电器商场买一种家用电器，发现甲品牌售价 1200 元，使用中每小时用电 0.5 度，使用寿命为 5000 小时；乙品牌售价 1300 元，使用中每小时用电 0.4 度，使用寿命为 5000 小时。每度电的价格是 0.4 元，在相同的使用寿命期限内，你认为小明的爸爸买哪种品牌电器最省钱。",
+        "options":  [
+
+                    ],
+        "answer":  "乙",
+        "solution":  "【分析】要比较哪种品牌最省钱，需要计算总成本 = 购买价格 + 使用电费，在相同使用寿命内进行比较。\n【详解】甲品牌总成本：1200+0.5×5000×0.4=2200(元)。乙品牌总成本：1300+0.4×5000×0.4=2100(元)。2100 元 < 2200 元，所以买乙品牌电器最省钱。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "18-19四年级上·全国·单元测试"
+    },
+    {
+        "id":  "205",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小兰妈妈给家人煎鸡蛋饼作早餐，鸡蛋饼两面都要煎，每面需要 1 分钟，且平底锅一次只能煎 2 个鸡蛋饼，煎 7 个鸡蛋饼至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "7",
+        "solution":  "【分析】这是烙饼问题的变形。每个鸡蛋饼有 2 面，7 个鸡蛋饼共有 14 个面，平底锅一次最多同时煎 2 面，每面需要 1 分钟。\n【详解】14÷2=7(分钟)，所以煎 7 个鸡蛋饼至少需要 7 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "17-18四年级上·贵州遵义·期末"
+    },
+    {
+        "id":  "206",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一种电脑游戏，玩一局要3 分钟，可以单人玩，也可以双人玩。爸爸，妈妈和我一起玩，每人玩两局，至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "9",
+        "solution":  "【分析】关键信息：每局3分钟，可单人或双人玩3人(爸爸，妈妈，我),每人玩2局，共需6局游戏目标：总时间最少突破口：双人玩时,1局时间能让2人各完成1局，效率翻倍!\n【详解】爸爸：第1局 + 第2局 = 2局 妈妈：第1局 + 第3局 = 2局 我：第2局 + 第3局 = 2局 总时间:3局 * 3分钟 = 9分钟",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "18-19四年级上·全国·期末"
+    },
+    {
+        "id":  "207",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "复印机同时可以复印2张稿件，复印一面用10秒，现有3张稿件两面都要复印，最少需要( )秒才能复印完。",
+        "options":  [
+
+                    ],
+        "answer":  "30",
+        "solution":  "【分析】3 张稿件共有 6 个面，复印机一次可以复印 2 张单面，所以可以采用轮换的方法：第一次复印第 1 张和第 2 张的正面，第二次复印第 1 张的反面和第 3 张的正面，第三次复印第 2 张和第 3 张的反面。\n【详解】10×3=30(秒)，所以最少需要 30 秒才能复印完。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "18-19 四年级上·河南郑州·期末"
+    },
+    {
+        "id":  "208",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一只平底锅上最多只能同时煎两条鱼，每条鱼需煎正反两面，每面需煎 2 分钟，那么，煎 13 条鱼至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "26",
+        "solution":  "【分析】前10条鱼两条一锅，每锅需要4分钟，共需要20分钟。后面3条，先煎2条的正面,2分钟后拿出一条，把另一条反过来，再放入一条,2分钟后，第一条熟了;然后把另一条反过来，把拿出的那一条的另一面放进去,2分钟后都熟了.\n【详解】10 条鱼要 20 分钟,3 条鱼要 6 分钟，共20 + 6 = 26 (分钟)。故答案为 26",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "18-19 四年级上·全国·课后作业"
+    },
+    {
+        "id":  "209",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "每次只能烙两张饼，两面都要烙，每面要烙 4 分钟。请问烙 4 张饼最快( )分钟可以烙完，要烙 5 张饼，最快( )分钟可以烙完。",
+        "options":  [
+
+                    ],
+        "answer":  "16;20",
+        "solution":  "【分析】每次烙 2 张饼，每面 4 分钟。烙偶数张饼时，最少时间等于张数×4；烙奇数张饼时，最后 3 张可以用轮换法来烙。\n【详解】烙 4 张饼：4×4=16(分钟)。烙 5 张饼：5×4=20(分钟)。故答案为 16;20。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "18-19 四年级上·全国·课后作业"
+    },
+    {
+        "id":  "210",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小军招待客人泡茶，烧开水用10分钟，找茶叶1分钟，找茶杯1分钟，洗茶杯1分钟，沏茶 3 分钟，他做完这些事最少要用 分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "13",
+        "solution":  "【分析】先烧开水需要10分钟，在烧水的时候去找茶叶，找茶杯，洗茶杯，等水烧开后，再沏茶需要3分钟。共需要10 + 3 = 13 分钟，据此即可解答问题\n【详解】根据题干分析可得: 10 + 3 = 13 (分钟)答：最少 13 分钟做完。故答案为 13。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "18-19 四年级上·全国·课后作业"
+    },
+    {
+        "id":  "211",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个平底锅中，每次只能放2张饼。一张饼要烙两面，烙熟一面需要2分钟。烙熟3张饼至少要花 分钟，烙熟7张饼至少要花 分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "6;14",
+        "solution":  "【分析】根据题意，抓住锅内始终有2张饼在烙是解决本题的关键，据此进行合理安排烙饼顺序即可解决问题，在烙饼问题中，饼的个数与所需时间的关系为：所需时间 = 个数×烙一面所用时间，可以由此直接列式计算.\n【详解】3张饼分别为A,B,C,第一次烙A,B的正面，需要2分钟，第二次烙A的反面和C的正面，需要2分钟，第三次烙B,C的反面，需要2分钟，一共需要:2 + 2 + 2 = 6 (分钟);烙7张饼，先2张2张的烙，前4张需要 2×2×2 = 8 (分钟),剩下的3张再按照上面的方法烙，需要6分钟，一共需要8 + 6 = 14 (分钟)。故答案为6;14。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "2014 四年级·全国·课后作业"
+    },
+    {
+        "id":  "212",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个平底锅每次能烙 2 张饼，每烙熟一面饼需要 2 分钟，两面都要烙.\n(1)烙熟 3 张饼至少需要 分钟;(2)烙熟 4 张饼至少需要 分钟;(3)烙熟 5 张饼至少需要 分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "6;8;10",
+        "solution":  "【分析】每次能烙 2 张饼，每烙一面需要 2 分钟。烙 3 张饼可以用轮换法，最少需要 6 分钟；烙 4 张饼时，两张两张地烙，最少需要 8 分钟；烙 5 张饼时，最少需要 10 分钟。\n【详解】烙熟 3 张饼至少需要 6 分钟，烙熟 4 张饼至少需要 8 分钟，烙熟 5 张饼至少需要 10 分钟。故答案为 6;8;10。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "18-19 四年级上·全国·课后作业"
+    },
+    {
+        "id":  "213",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用一只平底锅煎饼，每次能同时煎两张饼，如果煎一张饼需要 4 分钟(假设正反各需 2 分钟)，煎 4 张饼至少( )分钟，煎 7 张饼至少( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "8;14",
+        "solution":  "【分析】煎 4 张饼时，两张两张地煎即可；煎 7 张饼时，先煎前 4 张，再把剩下的 3 张用轮换法来煎。\n【详解】煎 4 张饼至少需要 8 分钟。煎 7 张饼时，前 4 张需要 8 分钟，后 3 张需要 6 分钟，共 14 分钟。故答案为 8;14。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "18-19四年级上·全国·课后作业"
+    },
+    {
+        "id":  "214",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小齐早晨起床穿衣要 5 分钟，整理被褥 3 分钟，刷牙 2 分钟，洗脸 1 分钟，温牛奶 3 分钟，烤面包 2 分钟，吃早餐 8 分钟。最节省时间要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "19",
+        "solution":  "【分析】先起床穿衣 5 分钟，再温牛奶 3 分钟，同时整理被褥 3 分钟；再烤面包 2 分钟，同时刷牙 2 分钟；然后洗脸 1 分钟，最后吃早餐 8 分钟。\n【详解】5+3+2+1+8=19(分钟)，所以最少需要 19 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "2014 四年级·全国·课后作业"
+    },
+    {
+        "id":  "215",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "煮一个鸡蛋需要 8分钟，一口锅一次可以煮 15 个鸡蛋，那么煮 15 个鸡蛋至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "8",
+        "solution":  "【分析】煮一个鸡蛋需要 8 分钟，一口锅一次可以煮 15 个鸡蛋，煮15个鸡蛋所用时间与煮一个鸡蛋的时间相同，由此可知答案.\n【详解】煮一个鸡蛋和煮15个鸡蛋所用时间相同，由此可知：煮 15 个鸡蛋至少需要 8 分钟",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "15-16 四年级上·全国·课后作业"
+    },
+    {
+        "id":  "216",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小红用自己的1,4,7 三张牌想胜过爷爷的 2,5,8 三张牌，则必须( )先出牌;然后当爷爷出( )时小红出( );爷爷出( )时小红出();爷爷出( )时小红出( )。这样三局两胜，小红才能胜利",
+        "options":  [
+
+                    ],
+        "answer":  "爷爷;8;1;5;7;2;4",
+        "solution":  "【分析】这是经典的“田忌赛马”对策问题。小红的牌整体偏小，必须采用错位对抗策略。\n【详解】让爷爷先出牌。当爷爷出 8 时，小红出 1；当爷爷出 5 时，小红出 7；当爷爷出 2 时，小红出 4。这样小红能以 2 胜 1 负取胜。",
+        "knowledge_points":  [
+                                 "优化问题",
+                                 "对策问题",
+                                 "田忌赛马策略"
+                             ],
+        "source":  "18-19 四年级上·全国·单元测试"
+    },
+    {
+        "id":  "217",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一张饼有( )个面，如果烙一面需要3分钟，烙熟一张饼需要( )分钟，烙熟两张饼需要( )分钟，烙熟三张饼需要( )分钟。(锅内最多可放 3张饼)",
+        "options":  [
+
+                    ],
+        "answer":  "2;6;6;6",
+        "solution":  "【分析】 这两题都考查优化问题，关键是在有限条件下通过合理安排，实现时间最短或胜率最大的最优解.\n【详解】标准最优策略(3张饼6分钟):第1个3分钟：饼①正面 + 饼②正面 + 饼③正面(3张同时烙正面)第2个3分钟：饼①反面 + 饼②反面 + 饼③反面(3张同时烙反面)这样3张饼6分钟全部烙熟，实现了资源的最大化利用。",
+        "knowledge_points":  [
+                                 "优化问题",
+                                 "烙饼问题"
+                             ],
+        "source":  "18-19 四年级上·全国·单元测试"
+    },
+    {
+        "id":  "218",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "早晨明明起床后，刷牙，洗脸5分钟，听英文广播10分钟，微波炉热早饭3分钟，吃早饭 10 分钟。他做完这些事最少要花( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】刷牙，洗脸 5 分钟，这时可以同时热早饭 3 分钟没有超过5分钟，然后再听英文广播10分钟，同时可以吃早饭10分钟。最后计算 5 + 10 = 15 分钟为做完这些事最少要花的时间.\n【详解】 5 + 10 = 15 (分钟)由分析可知他做完这些事最少要花 15 分钟。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25 四年级上·湖北黄冈·期末"
+    },
+    {
+        "id":  "219",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一堆棋子有 13 颗，两人轮流从中拿走1颗或2颗，谁拿到最后一颗谁就获胜。如果让你先拿，第一次应拿( )颗才能确保获胜。",
+        "options":  [
+
+                    ],
+        "answer":  "1",
+        "solution":  "【分析】要确保获胜，第一次应该拿走 1 颗棋子。因为每次可以拿走 1 或 2 颗棋子，所以每轮两人一共可以拿走 3 颗。13 颗棋子除以 3 余 1，这意味着如果先拿走 1 颗，剩下的 12颗棋子可以被 3 整除。这样，无论对手拿走 1 颗还是 2 颗，你都可以拿走相应的棋子，保证每轮都拿走 3 颗，最终拿到最后一颗棋子获胜.\n【详解】 13 ( 1 + 2 ) = 133 = 4 (组)。1(颗)由分析可知一堆棋子有13颗，两人轮流从中拿走1颗或2颗，谁拿到最后一颗谁就获胜。如果让你先拿，第一次应拿1颗才能确保获胜。",
+        "knowledge_points":  [
+                                 "田忌赛马问题"
+                             ],
+        "source":  "24-25 四年级上·湖北黄冈·期末"
+    },
+    {
+        "id":  "220",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一名同学做眼保健操要 5 分钟，全班 45 名同学每天做两次眼保健操至少要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "10",
+        "solution":  "【分析】一名同学做眼保健操要5分钟, 45名同学一起做次也是5分钟，全班45名同学每天做两次，至少需要2个 5 分钟.\n【详解】 2×5 = 10 (分钟)名同学做眼保健操要5分钟，全班45名同学每天做两次眼保健操至少要 10 分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "24-25四年级上·湖北黄冈·期末"
+    },
+    {
+        "id":  "221",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "田忌和齐王赛马，如果齐王安排马的出场顺序是\"中等-下等-上等\"，田忌要赢齐王，安排马的出场顺序应该是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "上等-中等-下等",
+        "solution":  "【分析】根据题意可知，田忌要赢齐王，那么田忌就应让自己的下等马与齐王的上等马比赛，让自己的中等马与齐王的下等马比赛，让自己的上等马与齐王的中等马比赛;据此解答.\n【详解】根据分析：如果齐王安排马的出场顺序是\"中等下等-上等\"，田忌要赢齐王，安排马的出场顺序应该是上等-中等-下等。",
+        "knowledge_points":  [
+                                 "田忌赛马问题"
+                             ],
+        "source":  "24-25四年级上·湖北武汉·期末"
+    },
+    {
+        "id":  "222",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "爸爸妈妈和田田一起玩投篮游戏，这种游戏可以单人玩，也可以双人玩，每玩一局要3分钟，现在每人玩2局，至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "9",
+        "solution":  "【分析】如果我们让他们分别单独玩，那么总共需要的时间是：每人2局，每局3分钟，所以每人需要 2×3 = 6 (分钟),三人总共需要 3×6 = 18 (分钟);如果双人玩，我们可以让爸爸和妈妈先一起玩一局，然后爸爸和田田一起玩一局，最后妈妈和田田一起玩一局，这样每个人就都玩了两局;使用这种方法，也就是玩了3个3分钟的时间段;据此解答.\n【详解】根据分析:3×3 = 9 (分钟)所以每玩一局要 3 分钟，现在每人玩 2 局，至少需要 9 分钟。",
+        "knowledge_points":  [
+                                 "搭配问题",
+                                 "烙饼问题"
+                             ],
+        "source":  "24-25四年级上·浙江宁波·期末"
+    },
+    {
+        "id":  "223",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "同学们组织野餐，带了一个电饭煲和一个炒菜锅。淘米 2 分钟，切菜 5 分钟，洗菜 10 分钟，炒菜 10 分钟，煮饭 25 分钟。如果合理安排，最少( )分钟就可以用餐。",
+        "options":  [
+
+                    ],
+        "answer":  "27",
+        "solution":  "【分析】煮饭时间最长，为 25 分钟。先淘米 2 分钟，再煮饭 25 分钟。在煮饭的同时，可以洗菜 10 分钟、切菜 5 分钟、炒菜 10 分钟，这三项共 25 分钟，正好可以安排在煮饭期间完成。\n【详解】2+25=27(分钟)，所以如果合理安排，最少 27 分钟就可以用餐。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25 四年级上·湖北武汉·期末"
+    },
+    {
+        "id":  "224",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "扫地 7 分钟，烧开水 9 分钟，用开水泡茶 1 分钟，李丽做完上面几件事，至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "10",
+        "solution":  "【分析】要使需要的时间最短，应先烧开水(9 分钟),在烧水的过程中，可以扫地(7分钟),等水烧开，最后再泡茶(1分钟),所以只需要烧水加泡茶时间即可.\n【详解】 9 + 1 = 10 (分钟)李丽做完上面几件事，至少需要10分钟。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25 四年级上·山东菏泽·期末"
+    },
+    {
+        "id":  "225",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "烧鱼节是侗族人民庆祝丰收的重要节日。村民会围在一个火堆前烤鱼，一个村民一次最多可烤两条鱼，两面都要烤，每面要烤5分钟，现有一个村民共要烤 6 条鱼，至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "30",
+        "solution":  "【分析】根据题意，可以将每次烤鱼的情况列出来，用次数乘每次烤的时间，即可求出至少需要多少分钟.\n【详解】第一次：第一条鱼的正面，第二条鱼的正面，耗时 5 分钟;第二次：第一条鱼的反面，第二条鱼的反面，耗时 5 分钟;第三次：第三条鱼的正面，第四条鱼的正面，耗时 5 分钟;第四次：第三条鱼的反面，第四条鱼的反面，耗时5分钟;第五次：第五条鱼的正面，第六条鱼的正面，耗时 5 分钟;第六次：第五条鱼的反面，第六条鱼的反面，耗时5分钟。6×5 = 30 (分钟)烧鱼节是侗族人民庆祝丰收的重要节日。村民会围在一个火堆前烤鱼，一个村民一次最多可烤两条鱼，两面都要烤，每面要烤5分钟，现有一个村民共要烤6条鱼，至少需要30分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "24-25 四年级上·江西吉安·期末"
+    },
+    {
+        "id":  "226",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "李阿姨在厨房烙饼，每次最多只能烙2张，两面都要烙，每面需3分钟，烙5张饼至少需要( )分钟;烙 6 张饼至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "15;18",
+        "solution":  "【分析】每次最多只能烙 2 张饼，两面都要烙，每面需 3 分钟。烙 5 张饼时，最少需要 5 个 3 分钟；烙 6 张饼时，可以把 6 张饼看成 3 组，每组 2 张，每组需要 6 分钟。\n【详解】5×3=15(分钟)，2×3×3=18(分钟)。故烙 5 张饼至少需要 15 分钟，烙 6 张饼至少需要 18 分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "24-25 四年级上·河北沧州·期末"
+    },
+    {
+        "id":  "227",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个平底锅一次最多煎2条鱼，两面都要煎，每面煎3分钟，煎5条这样的鱼至少需( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】根据题意，可以依次列出每次煎鱼的情况，据此计算出至少需要多少分钟才能煎5条鱼.\n【详解】第一次：第一条鱼正面，第二条鱼正面，耗时3分钟;第二次：第一条鱼反面，第二条鱼反面，耗时 3 分钟;第三次：第三条鱼正面，第四条鱼正面，耗时3分钟;第四次：第三条鱼反面，第五条鱼正面，耗时 3 分钟;第五次：第四条鱼反面，第五条鱼反面，耗时 3 分钟.\n3×5 = 15 (分钟)一个平底锅一次最多煎2条鱼，两面都要煎，每面煎3分钟，煎 5 条这样的鱼至少需 15 分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "24-25 四年级上·辽宁铁岭·期末"
+    },
+    {
+        "id":  "228",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "丽丽感冒了，妈妈准备为她冲感冒药，要做以下几件事情: 1 找药 1 分钟;2 电热壶烧水 6 分钟;3 量体温 5 分钟;4 冲药 1 分钟。那么丽丽最少( )分钟可以喝上药。",
+        "options":  [
+
+                    ],
+        "answer":  "7",
+        "solution":  "【分析】要使用的时间最少，那么能同时做的事情就应同时做，电热壶烧水的同时可以找药，量体温，这样可以节省6分钟，因此最少需要的时间为电热壶烧水的时间加冲药的时间，依此计算并解答.\n【详解】 6 + 1 = 7 (分钟)那么丽丽最少7分钟可以喝上药。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25 四年级上·辽宁葫芦岛·期末"
+    },
+    {
+        "id":  "229",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈要为全家做烤鱼。烤炉上每次最多只能放 2 条鱼，鱼的两面都要烤，每面需要 4 分钟，如果妈妈要烤 5 条鱼，最快要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "20",
+        "solution":  "【分析】每次最多烤 2 条鱼，每面需要 4 分钟。先烤前 2 条鱼需要 8 分钟，再烤后 3 条鱼可以用轮换法，最少需要 12 分钟。\n【详解】8+12=20(分钟)，所以烤 5 条鱼最快要 20 分钟。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25四年级上·辽宁鞍山·期末"
+    },
+    {
+        "id":  "230",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明起床后需要做下面几件事情：整理被褥 2 分钟，刷牙 2 分钟，洗脸 2 分钟，煮鸡蛋 7 分钟，他最快( )分钟可以做完这些事情。",
+        "options":  [
+
+                    ],
+        "answer":  "7",
+        "solution":  "【分析】整理被褥 2 分钟，刷牙 2 分钟和洗脸 2 分钟的时间加起来一共 6 分钟，所以小明起床后可以先煮鸡蛋，煮鸡蛋的7分钟内可以同时整理被褥，刷牙，洗脸，这样最节约时间，所以最快需要煮鸡蛋的 7 分钟就可以完成这些事情.\n【详解】根据分析: 2 + 2 + 2 = 6 (分钟), 7 - 6 = 1 (分钟),小明最快7分钟可以做完这些事情。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25四年级上·湖北咸宁·期末"
+    },
+    {
+        "id":  "231",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "李华早晨起来穿衣要 3 分钟，整理床铺 2 分钟，洗漱要 3 分钟，烧水要 5 分钟，喝水 1 分钟，吃饭 10 分钟。她最快( )分钟后出发去上学。",
+        "options":  [
+
+                    ],
+        "answer":  "19",
+        "solution":  "【分析】本题考查了优化问题，先观察做事情的顺序，考虑哪些事情可以同时做，写出做事情的顺序，计算所需时间。根据题意，穿衣之后，可以先烧水，烧水的同时，整理床铺和洗漱，然后喝水1分钟，吃饭10分钟。把时间加在一起，列式计算即可.\n【详解】根据分析可知:\n3 + 2 + 3 + 1 + 10 = 5 + 3 + 1 + 10 = 8 + 1 + 10 = 9 + 10 = 19 (分钟)李华早晨起来穿衣要3分钟，整理床铺2分钟，洗漱要3分钟，烧水要5分钟，喝水1分钟，吃饭10分钟。她最快19分钟后出发去上学。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25 四年级上·河北邯郸·期末"
+    },
+    {
+        "id":  "232",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "周末妈妈要做以下家务：拖地 10 分钟，收拾厨房 15 分钟，用全自动洗衣机洗衣服 20 分钟，晾衣服 10 分钟，妈妈做完这些事至少需要()分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "35",
+        "solution":  "【分析】由题意得，要使妈妈做这些家务所用的时间最少，那么她需要合理安排做家务的顺序。妈妈可以先用全自动洗衣机洗衣服，在洗衣服的同时拖地和收拾厨房，最后晾衣服。据此解答.\n【详解】由分析得，妈妈做家务的顺序为：用全自动洗衣机洗衣服(拖地，收拾厨房) 晾衣服.\n10 + 15 = 25 (分钟), 25 { \u003e } 20，即妈妈拖地和收拾厨房的时间比用全自动洗衣机洗衣服的时间长一些。25 + 10 = 35 (分钟)故妈妈做完这些事至少需要35分钟。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25 四年级上·湖北武汉·期末"
+    },
+    {
+        "id":  "233",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明给客人沏茶，接水1分钟，烧水6分钟，洗茶杯2分钟，拿茶叶1分钟，沏茶1分钟。小明合理安排以上事情，最少要( )分钟使客人尽快喝茶。",
+        "options":  [
+
+                    ],
+        "answer":  "8",
+        "solution":  "【分析】要使时间最短，需要把能在同一时间做的事情起做了，应先接水，然后烧水，在完成烧水的同时，可完成洗茶杯和拿茶叶这两项，水烧好后再沏茶。则一共需要( ( 1 + 6 + 1 ) )分钟.\n【详解】 1 + 6 + 1 = 7 + 1 = 8 (分钟)\n小明合理安排以上事情，最少要8分钟使客人尽快喝茶。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "22-23四年级下·广东佛山·期末"
+    },
+    {
+        "id":  "234",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小丽想做下面这些事情：用洗衣机洗衣服要用 20 分钟，扫地要用 6 分钟，整理书桌要用 5 分钟，擦家具要用 10 分钟，晾衣服要用 5 分钟，她经过合理安排，做完这些事最少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "26",
+        "solution":  "【分析】要使用的时间最少，那么能同时做的事情就应同时做,20分钟 \u003c 6 分钟+10分钟+5分钟，因此洗衣服的同时扫地，整理书桌和擦家具，最少用的时间 = 扫地用的时间+整理书桌用的时间+擦家具用的时间 + 晾衣服用的时间;据此解答.\n【详解】 6 + 10 + 5 + 5 = 26 (分钟)即小丽想做下面这些事情：用洗衣机洗衣服要用20分钟，扫地要用6分钟，整理书桌要用5分钟，擦家具要用10分钟，晾衣服要用5分钟，她经过合理安排，做完这些事最少需要26分钟。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25四年级上·河北邯郸·期末"
+    },
+    {
+        "id":  "235",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "\"中秋节\"期间，某超市月饼大促销，买一盒要 24 元，买 3 盒共收 60 元，买5 盒共收 90 元。小巧妈妈想买 16 盒月饼，最少要花( )元。",
+        "options":  [
+
+                    ],
+        "answer":  "294",
+        "solution":  "【分析】根据单价 = 总价/数量，买 5 盒共收 90 元，则每盒的单价为(90÷5)元;买3盒共收60元，则每盒的单价为(60÷3)元;买一盒要24元;比较三种买法的单价，尽量多的买单价便宜的组合，据此计算出最少要用的钱数.\n【详解】 905 = 18 (元) 603 = 20 (元)\n18 \u003c 20 \u003c 24 所以尽量多的买5盒一组的组合;165 = 3 (份)。1(盒)\n可以购买 3 份 5 盒的和 1 盒;90×3 + 24 = 270 + 24 = 294 (元)\n16 = 5×2 + 3×2 可以购买 2 份 5 盒的和 2 份 3 盒的;90×2 + 60×2 = 180 + 120 = 300 (元)\n294 \u003c 300 最少要花294元。",
+        "knowledge_points":  [
+                                 "租船问题",
+                                 "经济问题"
+                             ],
+        "source":  "24-25 四年级上·河南许昌·期中"
+    },
+    {
+        "id":  "236",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "下面是妈妈包饺子的几道工序：和面4分钟，择菜和拌馅18分钟，包饺子20分钟，烧开水10分钟，煮饺子8分钟。妈妈把饺子包完后才开始煮，最少要用( )分钟才能吃上饺子。",
+        "options":  [
+
+                    ],
+        "answer":  "50",
+        "solution":  "【分析】要使用的时间最少，那么能同时做的事情就应同时做,20分钟 \u003e 10 分钟，则包饺子10分钟后再同时烧开水，因此最少要用的时间 = 和面用的时间+择菜和拌馅用的时间+包饺子用的时间+煮饺子用的时间，依此计算即可.\n【详解】 4 + 18 + 20 + 8 = 50 (分钟)妈妈把饺子包完后才开始煮，最少要用50分钟才能吃上饺子。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25 四年级上·黑龙江齐齐哈尔·期末"
+    },
+    {
+        "id":  "237",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个锅最多同时烙 2 张饼，烙 1 张饼要 2 分钟(正，反面各 1 分钟),烙3 张饼最少要( )分钟,9 分钟最多可以烙( )张饼。",
+        "options":  [
+
+                    ],
+        "answer":  "3;9",
+        "solution":  "【分析】烙3张饼：先烙第一张与第二张的正面需1分钟，然后烙第一张的反面与第三张的正面需要1分钟，最后烙第二张的反面与第三张的反面需1分钟，烙完3张饼共需3分钟;烙3张饼共需3分钟这是最短的时间,9分钟里面有几个3分钟就能烙几个3张饼.\n【详解】 93×3 = 3×3 = 9 (张)一个锅最多同时烙2张饼，烙1张饼要2分钟(正，反面各1分钟),烙3张饼最少要(3)分钟,9分钟最多可以烙(9)张饼。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "24-25 四年级上·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "238",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "星期日王红帮妈妈做三件家务：拖地8分钟，烧水10分钟，沏茶1分钟，最少要用( )分钟做完以上家务。",
+        "options":  [
+
+                    ],
+        "answer":  "11",
+        "solution":  "【分析】根据题意可知，要想需要时间最短，则有些事情需要同时做，这样可以节省时间，先烧水，同时就可以做别的事情。10分钟 \u003e 8 分钟，可以同时拖地，这样操作用时最短.\n【详解】 10 + 1 = 11 (分钟)星期日王红帮妈妈做三件家务：拖地8分钟，烧水10分钟，沏茶1分钟，最少要用(11)分钟做完以上家务。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "23-24四年级上·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "239",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "做一道番茄炒鸡蛋需要六道工序：洗，切番茄3分钟，打蛋1分钟，热锅2分钟，热油1分钟，炒4分钟，要烧好这道菜至少需要()分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "8",
+        "solution":  "【分析】根据题意，要使用的时间最少，那么能同时做的事情就应同时做，烧热锅和热油的同时可以洗切番茄，因此要烧好这道菜至少需要的时间 = 洗，切番茄的时间+打蛋时间+炒的时间，依此解答即可.\n【详解】 3 + 1 + 4 = 8 (分钟),所以，要烧好这道菜至少需要8分钟。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "23-24四年级上·内蒙古赤峰·期末"
+    },
+    {
+        "id":  "240",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一口平底锅里最多同时煎两条鱼，用它煎一面需要2分钟，那么煎3条鱼(正反面都要煎)至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】3 条鱼可以分别用 1,2,3 表示，只要充分利用平底锅，使平底锅中的鱼始终有两条即可解决.\n【详解】3 条鱼可以分别用 1,2,3 表示;第一次:1正,2正，需要2分钟;第二次:1 反,3 正，需要 2 分钟;第三次:2 反,3 反，需要 2 分钟;2×3 = 6 (分钟)那么煎 3 条鱼(正反面都要煎)至少需要 6 分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "23-24 四年级上·新疆和田·期末"
+    },
+    {
+        "id":  "241",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈煮面需要 10 分钟，炒菜需要 8 分钟，用微波炉热包子需要 3 分钟，妈妈最快( )分钟就可以让家人吃饭。",
+        "options":  [
+
+                    ],
+        "answer":  "18",
+        "solution":  "【分析】煮面和炒菜不能同时进行，因为都需要用到炉灶，因为煮面时间长，用微波炉热包子可以在煮面过程中完成，总共花费煮面的 10 分钟加上炒菜额外的 8 分钟，即可求出妈妈最快多少分钟就可以让家人吃饭.\n【详解】 10 + 8 = 18 (分钟)所以妈妈最快18分钟就可以让家人吃饭。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "23-24 四年级上·吉林松原·期末"
+    },
+    {
+        "id":  "242",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小丽周末学做家务，家务活有：淘米 1 分钟，蒸饭 15 分钟，扫地需要8 分钟，布置餐桌 3 分钟，盛饭 2 分钟。小丽至少需要()分钟才能完成这些家务。",
+        "options":  [
+
+                    ],
+        "answer":  "18",
+        "solution":  "【分析】由题意得，小丽周末学做家务，家务活有：淘米1分钟，蒸饭 15 分钟，扫地需要 8 分钟，布置餐桌 3 分钟，盛饭2分钟。要想小丽做这些家务所需的时间最短，她需要合理安排这些家务活。她可以先淘米，然后再蒸饭，然后在蒸饭的同时，再去扫地，布置餐桌，最后再去盛饭。据此解答.\n【详解】由分析得，小丽做这些家务活的顺序为：淘米→蒸饭 扫地 布置餐桌 盛饭。她所需要的时间为:1+15 + 2 = 16 + 2 = 18 (分)故小丽至少需要 18 分钟才能完成这些家务。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "23-24 四年级上·内蒙古鄂尔多斯·期末"
+    },
+    {
+        "id":  "243",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "学期末，李老师负责对学生进行简笔画测试，王老师负责对学生进行科学小实验测试，有 3 名同学代表全班参加测试，每项测试需要10 分钟。三名同学做完这两项测试至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "30",
+        "solution":  "【分析】假设 3 名同学分别为甲，乙，丙，第一个 10 分钟：甲参加简笔画测试，乙同时参加科学小实验测试;第二个10 分钟：甲参加科学小实验测试，丙参加简笔画测试;第三个 10 分钟：乙参加简笔画测试，丙参加科学小实验测试;据此解答.\n【详解】根据分析:10×3 = 30 (分钟)所以三名同学做完这两项测试至少需要30分钟。",
+        "knowledge_points":  [
+                                 "整十，整百，整千数与一位数的乘法",
+                                 "烙饼问题"
+                             ],
+        "source":  "24-25四年级上·吉林松原·期末"
+    },
+    {
+        "id":  "244",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一只锅最多可煎3个饼，两面都要煎。煎一面要 2 分钟，煎完 8 个饼最少要用( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "12",
+        "solution":  "【分析】此题考查了学生的利用统筹思想进行合理安排事情的能力，抓住锅内始终有3张饼在烙是本题的关键。首先根据用一个平底锅煎饼每次只能放3张，煎一面要2分钟，正反面各需要 2×2 分钟, 83 = 2。2，可分两次煎 即第一次同时煎 3 条，反正面煎完需要 2×2×2 = 8 (分钟),第二次煎两条，两面煎完需要 2×2 = 4 (分钟),共要8+4 = 12 (分钟),以此答题即可.\n【详解】根据分析可知:\n83 = 2 (个)。2(个)2×2×2 = 8 (分钟)\n2×2 = 4 (分钟)8 + 4 = 12 (分钟)\n一只锅最多可煎3个饼，两面都要煎。煎一面要2分钟，煎完8个饼最少要用12分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "23-24四年级上·浙江宁波·期末"
+    },
+    {
+        "id":  "245",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一只锅最多可以煎 2条鱼，鱼的两面都要煎。煎一面要2分钟，煎完6条鱼最少要用( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "12",
+        "solution":  "【分析】将这6条鱼平均分成2组，每组有3条鱼。3条鱼分别为 a,b,c;第一次煎 a,b 的正面;第二次煎 a 的反面和c的正面;第三次煎b,c的反面。所以煎3条鱼需要( 3×2 )分钟。这样煎完 6 条鱼最少要用(( 3×2×2 ))分钟.\n【详解】 3×2×2 = 12 (分钟)煎完 6 条鱼最少要用 12 分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "23-24 四年级上·浙江宁波·期末"
+    },
+    {
+        "id":  "246",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈煎鱼，每次最多可以煎2条鱼，每煎一面需要3分钟。煎完6条鱼，至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "18",
+        "solution":  "【分析】烙饼的过程为第一次：正1正2;第二次：反1正3;第三次：反2反3;第四次：正4正5;第五次：反4正6;第六次：反 5 反 6。根据烙饼的问题的公式：总时间 = 饼数*2/每锅可烙饼的数量×烙每面的时间;据此解答.\n【详解】 = 122×3 = 6×3 = 18 (分 钟)\n因此，煎完6条鱼，至少需要18分钟.\n原答案中的轮换法(反1正3，反2反3等)虽然也能得到18分钟，但对于偶数条鱼过于复杂。直接用分组法更清晰，且容易理解:2条一组，共3组每组需2个时间段(6分钟)总时间:3组 * 6分钟 = 18分钟",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "23-24 四年级上·河北保定·期末"
+    },
+    {
+        "id":  "247",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明，小聪和小军进行 2 分钟投篮比赛，有两个篮球架，两人可同时进行比赛，每人都投 2 轮，至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】小明，小聪和小军进行2分钟投篮比赛，即每轮比赛需要2分钟，有两个篮球架，两人可同时进行比赛,2分钟内可同时进行2轮比赛，每人都投2轮;例如小明和小聪起投，小明和小军一起投，小聪和小军一起投，共投3次;所以至少需要: 3×2 = 6 (分钟),据此解答即可.\n【详解】 3×2 = 6 (分钟)小明，小聪和小军进行2分钟投篮比赛，有两个篮球架，两人可同时进行比赛，每人都投2轮，至少需要6分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "23-24 四年级上·北京丰台·期末"
+    },
+    {
+        "id":  "248",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果每个锅最多煎2条小黄鱼，煎一条小黄鱼需要4分钟(正，反面各2分钟),那么煎5条小黄鱼至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "10",
+        "solution":  "【分析】在优化问题中，要统筹安排煎鱼的顺序，使锅里始终没有空位，这样才能保证需要的时间最短。煎 2 条鱼，需要 4 分钟。剩下的 3 条鱼分别为 a,b,c;第一次煎 a,b 的正面;第二次煎 a 的反面和 c 的正面;第三次煎 b,c 的反面。煎 5 条鱼，需要(( 4 + 3×2 ))分钟.\n【详解】根据分析可知:\n4 + 3×2 = 4 + 6 = 10 (分 钟)\n如果每个锅最多煎2条小黄鱼，煎一条小黄鱼需要4分钟(正，反面各 2 分钟),那么煎 5 条小黄鱼至少需要 10 分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "23-24四年级上·陕西安康·期末"
+    },
+    {
+        "id":  "249",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "烙饼时，如果每次只能烙2张饼，两面都要烙，每面要3分钟。烙好5张饼至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】由题意得，烙5张饼时，第一张和第二张饼一起烙,3分钟后再翻面继续烙。这时，还剩下3张饼。第三张和第四张的正面一起烙，再把第三张的反面和第五张的正面一起烙，最后再把第四张和第五张的反面一起烙。一共需要烙5次，每面需要3分钟，直接用乘法即可算出烙好5张饼至少需要多少分钟.\n【详解】 5×3 = 15 (分钟)故烙好5张饼至少需要15分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "24-25 四年级上·吉林白城·期末"
+    },
+    {
+        "id":  "250",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "两名工人给4块木板正反面刷漆，每人每面刷完需要3分钟，并且两人不能同时给同一块木板刷漆，最少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "12",
+        "solution":  "【分析】要使所用的时间最少，就要交替刷漆，两人同时刷 1 号和 2 号木板的正面，第 2 个 3 分钟 1 人刷 1 号木板的反面，另1个人刷3号木板的正面，第3个3分钟,1个人刷2号木板的反面，另一个人刷4号木板的正面，最后1人刷3号木板的反面，另一人刷4号木板的反面，由此可知需要的时间是4个3分钟，把4与3相乘即可求出需要的时间.\n【详解】第 1 个 3 分钟:1 号正,2 号正;第2个3分钟:1号反,3号正;第 3 个 3 分钟:2 号反,4 号正;第 4 个 3 分钟:3 号反,4 号反。3×4 = 12 (分钟)两名工人给4块木板正反面刷漆，每人每面刷完需要3分钟，并且两人不能同时给同一块木板刷漆，最少需要12分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "23-24四年级上·河南三门峡·期末"
+    },
+    {
+        "id":  "251",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "春节快到了，两个小朋友准备一起做5张贺卡，卡片正反面都要写祝福语，配插图，如果一人制作好一面要10分钟，那么两人合作完成至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "50",
+        "solution":  "【分析】两个人可以同时制作贺卡的两个面。5张贺卡共有10个面，每做一面要10分钟，两个人同时做时，每10分钟可以完成2个面。\n【详解】10÷2=5(次)，5×10=50(分钟)。所以两人合作完成至少需要50分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "23-24四年级上·重庆璧山·期末"
+    },
+    {
+        "id":  "252",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "王老师要复印3张学习单，正反两面都要印。一台只能单面复印的复印机一次可以放 2 张学习单，复印一次要 5 秒，请问全部复印完需要( )秒。",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】3张学习单正反两面都要印，相当于有6个面要复印。每次最多复印2个面，复印一次要5秒，最少需要复印3次。\n【详解】6÷2=3(次)，3×5=15(秒)。所以全部复印完需要15秒。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "23-24四年级上·河南许昌·期末"
+    },
+    {
+        "id":  "253",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明起床后要做的事情有:(1)起床整理被褥 3 分钟;(2)刷牙洗脸一共需要 6 分钟;(3)读英语单词需要 10 分钟;小明如果 6 时起床，他最快( )时( )分做完这些事情。",
+        "options":  [
+
+                    ],
+        "answer":  "6;19",
+        "solution":  "【分析】小明起床做的这几件事不能同时做，得一件一件做，所以把3与6以及10相加，可以求出他做完这几件事共需要多少分钟，这个和是19，即需要19分钟，推算从6时经过19分钟是6时19分，据此解答.\n【详解】3+6+10 = 9 + 10 = 19 (分 钟)6 时+19 分 = 6 时 19 分小明起床后要做的事情有:(1)起床整理被褥 3 分钟;(2)刷牙洗脸一共需要 6 分钟;(3)读英语单词需要 10分钟;小明如果 6 时起床，他最快 6 时 19 分做完这些事情。",
+        "knowledge_points":  [
+                                 "时，分，秒时间的推算",
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25 四年级上·吉林白山·期末"
+    },
+    {
+        "id":  "254",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈早上起床，刷牙 3 分钟，淘米洗锅 2 分钟，用电饭锅熬粥 20 分钟，盛粥 1分钟，用平底锅煎鸡蛋 3 分钟。要吃上妈妈做的早饭至少要用( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "23",
+        "solution":  "【分析】根据题意，至少要用多少分钟，就要使用的时间最短，则先淘米洗锅然后熬粥，熬粥的同时刷牙煎鸡蛋，最后盛粥，据此列式计算出时间填空即可.\n【详解】 2 + 20 + 1 = 22 + 1 = 23 (分钟)能吃上妈妈做的早饭至少要用23分钟。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25四年级上·陕西商洛·期末"
+    },
+    {
+        "id":  "255",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用烧烤架烤鱼时，鱼的两面都要烤。一张烧烤架可同时烤2条鱼，烤一面需要4分钟，用这张铁板烧烤5条鱼至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "20",
+        "solution":  "【分析】根据题意，先烤 1,2 号鱼的正反面，都烤熟需 4+ 4 = 8 (分)。再烤 3,4 号鱼的正面需 4 分钟，把 4 号鱼拿下来，放上5号鱼，烤3号鱼的反面,5号鱼的正面需4分钟，最后拿下3号鱼，放上4号鱼，烤4,5号鱼的反面需 4 分钟，所以烤熟 5 条鱼至少需要: 8 + 4 + 4 + 4 = 20(分),以此答题即可.\n【详解】4 + 4 = 8(分)\n8 + 4 + 4 + 4 = 20(分)用烧烤架烤鱼时，鱼的两面都要烤。一张烧烤架可同时烤2条鱼，烤一面需要4分钟，用这张铁板烧烤5条鱼至少需要20 分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "24-25 四年级上·陕西宝鸡·期末"
+    },
+    {
+        "id":  "256",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果一个锅每次可以同时烙 3 张饼，两面都要烙，烙一个面要 3 分钟，烙 9 张饼至少要花( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "18",
+        "solution":  "【分析】9张饼共有18个面，每次可以同时烙3个面，所以需要烙18÷3=6次，每次3分钟。\n【详解】9×2÷3×3=6×3=18(分钟)。所以烙9张饼至少要花18分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "23-24 四年级上·贵州铜仁·期末"
+    },
+    {
+        "id":  "257",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一只平底锅中每次只能放 2 张饼。1 张饼要烙两面，烙熟一面需要 2 分钟，烙3 张饼至少要用( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】假设三张饼依次为 a,b,c,先用 2 分钟烙 a,b 的正面，再用2分钟烙a的反面和c的正面，最后用2分钟烙b 的反面和 c 的反面，一共需要的时间是(( 2×3 ))分钟.\n【详解】 2×3 = 6 (分钟)一只平底锅中每次只能放2张饼。1张饼要烙两面，烙熟一面需要2分钟，烙3张饼至少要用6分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "23-24 四年级上·江西赣州·期末"
+    },
+    {
+        "id":  "258",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个工厂要租车运205吨货物，每辆中型卡车的载重量是8吨，租金200元/次：每辆小型卡车的载重量是 5 吨，租金 150 元/次。工厂租车运这批货物最少要用( )元。",
+        "options":  [
+
+                    ],
+        "answer":  "5150",
+        "solution":  "【分析】先比较每吨运费：中型卡车200÷8=25(元)，小型卡车150÷5=30(元)，中型卡车更便宜，所以应尽量多租中型卡车。205吨里有25个8吨，还余5吨，正好再租1辆小型卡车。\n【详解】200÷8=25(元)，150÷5=30(元)，25<30。\n205÷8=25(辆)……5(吨)。\n5吨正好需要1辆小型卡车，所以租25辆中型卡车和1辆小型卡车最省钱。\n25×200+150=5000+150=5150(元)。\n所以工厂租车运这批货物最少要用5150元。",
+        "knowledge_points":  [
+                                 "乘加，乘减混合运算",
+                                 "租船问题"
+                             ],
+        "source":  "23-24 四年级上·重庆垫江·期末"
+    },
+    {
+        "id":  "259",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "淘气帮妈妈泡一杯茶，他需要做几件事情：烧水 5 分，洗茶杯 2 分，洗水壶 1分，接水 1 分，取茶叶 1 分，沏茶 1 分。淘气泡好一杯茶的时间至少需要( )分。",
+        "options":  [
+
+                    ],
+        "answer":  "8",
+        "solution":  "【分析】要想时间最短，应先洗水壶、接水，再烧水。烧水时可以同时洗茶杯、取茶叶，最后再沏茶。\n【详解】洗水壶1分，接水1分，烧水5分，沏茶1分。\n总时间：1+1+5+1=8(分)。\n所以淘气泡好一杯茶的时间至少需要8分。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "22-23 四年级下·四川成都·期末"
+    },
+    {
+        "id":  "260",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "贝贝6点放学回到家，妈妈留言给他布置任务，贝贝烧了一壶开水用了30分钟，用全自动洗衣机洗衣服用45分钟，写作业用了30分钟，打扫房间用了11分钟，晾衣服用7分钟。经过合理安排，贝贝最早在( )时( )分完成这些任务。",
+        "options":  [
+
+                    ],
+        "answer":  "6;52",
+        "solution":  "【分析】烧开水和用洗衣机洗衣服都不用一直看着，可以同时进行。写作业和打扫房间共用41分钟，小于洗衣服的45分钟，所以这两项可以在洗衣服的同时完成，最后再晾衣服。\n【详解】30+11=41(分钟)，41<45。\n45+7=52(分钟)。\n6时过52分钟是6时52分，所以贝贝最早在6时52分完成这些任务。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "2022 五年级·全国·竞赛"
+    },
+    {
+        "id":  "261",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "李老板在烤锅盔，锅盔的两面都要烤，每次最多烤 2 个，烤一面需要 3 分钟，他烤 5 个锅盔至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】先烤2个锅盔需要6分钟，剩下3个锅盔可以用轮换的方法再烤9分钟，所以共需要15分钟。\n【详解】前2个锅盔：3×2=6(分钟)。\n后3个锅盔：第1次烤第3个和第4个的正面，第2次烤第3个的反面和第5个的正面，第3次烤第4个和第5个的反面，共3×3=9(分钟)。\n6+9=15(分钟)。所以烤5个锅盔至少需要15分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "23-24四年级上·四川南充·期末"
+    },
+    {
+        "id":  "262",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "李阿姨做家务。擦地 5 分钟，洗衣机洗衣服 25 分钟，晾衣服 3 分钟，洗碗筷 5分钟，整理房间 10 分钟。她至少需要( )分钟才能做完。",
+        "options":  [
+
+                    ],
+        "answer":  "28",
+        "solution":  "【分析】洗衣机洗衣服25分钟时，可以同时擦地、洗碗筷和整理房间，这三项共20分钟，都能在洗衣服时完成。最后再晾衣服3分钟。\n【详解】25+3=28(分钟)。\n所以她至少需要28分钟才能做完。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "23-24四年级上·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "263",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "双休日的早晨，妈妈为家人准备早餐：洗电饭锅1分钟，淘米1分钟，用电饭锅煮饭要20分钟，洗菜切菜要5分钟，炒菜要12分钟。经过合理安排，最少( )分钟全家人能吃上早餐(炒菜不需要等饭煮好)。",
+        "options":  [
+
+                    ],
+        "answer":  "22",
+        "solution":  "【分析】在煮饭的时候，妈妈可以洗菜，切菜，炒菜，那么煮饭最少用时就是洗电饭锅，淘米和电饭锅煮饭的总时间。据此解答.\n【详解】1+1+20 =2+20 = 22 (分钟)则双休日的早晨，妈妈为家人准备早餐：洗电饭锅1分钟，淘米1分钟，用电饭锅煮饭要20分钟，洗菜切菜要5分钟，炒菜要12分钟。经过合理安排，最少22分钟全家人能吃上早餐。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "23-24四年级上·河南焦作·期末"
+    },
+    {
+        "id":  "264",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用一个平底锅煎鸡蛋，每次最多煎2个，每面都要煎，每面要煎2分钟。煎3个鸡蛋至少要用( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】3个鸡蛋可以采用轮换的方法来煎，使锅里始终有2个面在煎。\n【详解】第1次煎第1个和第2个鸡蛋的正面，用2分钟；第2次煎第1个鸡蛋的反面和第3个鸡蛋的正面，用2分钟；第3次煎第2个和第3个鸡蛋的反面，用2分钟。\n2×3=6(分钟)。所以煎3个鸡蛋至少要用6分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "23-24四年级上·河南信阳·期末"
+    },
+    {
+        "id":  "265",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个平底锅里最多能同时煎 2 块牛排，煎一面需要 2 分钟，正反两面都要煎，用它煎 3 块牛排至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】3块牛排可以采用轮换的方法来煎：先煎前2块的正面，再轮换着煎反面和第3块的正面，最后煎剩下的两个反面。\n【详解】第1次煎第1块和第2块的正面，用2分钟；第2次煎第1块的反面和第3块的正面，用2分钟；第3次煎第2块和第3块的反面，用2分钟。\n2×3=6(分钟)。所以煎3块牛排至少需要6分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "23-24 四年级上·江西赣州·期末"
+    },
+    {
+        "id":  "266",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一只平底锅中每次最多能烙4张饼，两面都要烙，每面2分钟，烙6张饼至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】6张饼共有12个面，每次最多烙4个面，所以需要烙12÷4=3次，每次2分钟。\n【详解】6×2÷4×2=3×2=6(分钟)。\n所以烙6张饼至少需要6分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "23-24 四年级上·重庆秀山·期末"
+    },
+    {
+        "id":  "267",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个煎锅一次最多能煎4条鱼，煎一面要4分钟，两面都要煎。那煎6条鱼最少要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "12",
+        "solution":  "【分析】6条鱼共有12个面，每次最多煎4个面，所以需要煎12÷4=3次，每次4分钟。\n【详解】6×2÷4×4=3×4=12(分钟)。\n所以煎6条鱼最少要12分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "23-24 四年级上·重庆开州·期末"
+    },
+    {
+        "id":  "268",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个平底锅，每次只能煎 2 个鸡蛋，两面都要煎，每面 3 分钟，煎 5 个鸡蛋至少要用( )分钟;煎 7 个鸡蛋至少要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "15;21",
+        "solution":  "【分析】煎5个鸡蛋：先同时煎两个，正反面共需 3×2 = 6(分钟);再煎后三个，先煎第 3 个与第 4 的正面需 3 分钟，然后煎第 3 个的反面与第 5 个的正面需要 3 分钟，最后煎第4个的反面与第5个的反面需3分钟，煎完3个共需3×3 = 9 (分钟),据此可求出煎 5 个最少需要多少时间;煎7个鸡蛋：先同时煎两个，正反面共需 3×2 = 6 (分钟);再同时煎第 3 个和第 4 个，正反面共需 3×2 = 6 (分钟),最后再煎后三个，先煎第5个与第6的正面需3分钟，然后煎第5个的反面与第7个的正面需要3分钟，最后煎第6个的反面与第 7 个的反面需 3 分钟，煎完 3 个共需 3×3 = 9 (分钟),据此可求出煎 7 个最少需要多少时间.\n【详解】煎5个:\n3×2 = 6 (分钟)\n3×3 = 9 (分钟)\n6 + 9 = 15 (分钟)\n煎7个:3×2 = 6 (分钟) 3×3 = 9 (分钟) \n6×2+9 =12+9 = 21 (分钟)\n综上可知，煎 5 个鸡蛋至少要用 15 分钟;煎 7 个鸡蛋至少要 21 分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "23-24 四年级上·湖南娄底·期末"
+    },
+    {
+        "id":  "269",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "有 4 位同学去量身高，测视力(两项测量同时进行),如果每项检查需要 4 分钟，那么这 4 位同学做完这两项检查至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "16",
+        "solution":  "【分析】量身高和测视力可以同时进行，每4分钟可以有2位同学分别完成一项检查，再交换项目。2位同学完成两项检查共需8分钟，4位同学就需要2个这样的过程。\n【详解】2位同学做完两项检查需要：4×2=8(分钟)。\n4位同学需要：8×2=16(分钟)。\n所以这4位同学做完这两项检查至少需要16分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "23-24 四年级上·广东潮州·期末"
+    },
+    {
+        "id":  "270",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "周末，东东要做以下家务：用全自动洗衣机洗衣服需要 25 分钟，晾衣服 5 分钟，拖地 14 分钟，整理房间 7 分钟。东东最快( )分钟能做完这些事情。",
+        "options":  [
+
+                    ],
+        "answer":  "30",
+        "solution":  "【分析】东东要做的家务有：用全自动洗衣机洗衣服需要25分钟，晾衣服5分钟，拖地14分钟，整理房间7分钟，其中洗衣机洗衣服的同时，可以拖地和整理房间,14 + 7 = 21(分钟),21 \u003c 25，等洗衣机洗完衣服之后再晾衣服，这个过程可以用图来表示为:![](images/118048e67212dde37e77c60ed4169cfd805336528ce7bbdfe6228957898910db.jpg)所以东东最快的时间也就是用全自动洗衣机洗衣服的时间和晾衣服的时间之和，据此即可解答。【详解】由分析中图25 + 5 = 30(分钟)周末，东东要做以下家务：用全自动洗衣机洗衣服需要25分钟，晾衣服5分钟，拖地14分钟，整理房间7分钟。东东最快 30 分钟能做完这些事情。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "23-24 四年级上·湖北随州·期末"
+    },
+    {
+        "id":  "271",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "星期天，东东帮妈妈做家务。把脏衣服整理出来放入洗衣机要2分钟，自动洗衣机里洗衣服要花50分钟，洗碗要15分钟，整理房间要30 分钟，晾衣服要 10 分钟。干完所有活至少要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "62",
+        "solution":  "【分析】先把脏衣服整理出来放入洗衣机需要2分钟，再用自动洗衣机洗衣服需要50分钟。在洗衣机洗衣服的50分钟内，东东可以同时进行洗碗和整理房间，这两项共 15 + 30 = 45(分钟),小于50分钟，可以完全在洗衣阶段完成，最后再晾衣服 10 分钟.\n【详解】2 + 50 = 52(分钟),洗碗和整理房间在这 50 分钟内完成，最后晾衣服 10 分钟:\n52 + 10 = 62(分钟)所以干完所有活至少要62分钟。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "23-24 四年级上·湖北咸宁·期末"
+    },
+    {
+        "id":  "272",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "会昌县月亮湾某知名牛排店烤一块牛排需要 6 分钟(正反面各 3 分钟),铁板上每次最多只能放 2 块，烤 5 块牛排至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】2 块牛排一起烤时，正反两面共需 3×2 = 6 分钟。剩下 3 块牛排可以用\"3 块轮换法\"安排正反面，保证铁板始终满载，使时间最短.\n【详解】前 2 块牛排用时:3×2 = 6(分钟)。剩下的 3 块编号为 A,B,C:\n第 1 次:A 正,B 正(3 分钟);第 2 次:A 反,C 正(3 分钟);第 3 次:B 反,C 反(3 分钟),共 3×3 = 9(分钟).\n烤 5 块牛排至少需要:6 + 9 = 15(分钟).\n15 分钟内一共 15÷3 = 5 个时间段，每段烤 2 面，共烤 10 面，刚好 5 块牛排的 10 个面，利用率 100%。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "23-24 四年级上·江西赣州·期末"
+    },
+    {
+        "id":  "273",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "万州烤鱼已然成为万州旅游的一张名片。一个烤鱼小摊的灶台上每次最多能同时烤两条鱼，烤熟一条鱼需要 10 分钟(两面都要烤，每面烤熟需要 5 分钟)。那么一个旅游小团要吃上 3 条烤鱼，至少需要( )分钟。(其它时间不计)",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】3 条鱼两面都要烤，可用\"轮换\"方式保持灶台始终有 2 条鱼在烤，从而缩短总时间.\n【详解】设 3 条鱼为 1,2,3:\n第 1 次:1 正,2 正(5 分钟);第 2 次:1 反,3 正(5 分钟);第 3 次:2 反,3 反(5 分钟).\n共需 5×3 = 15(分钟),所以一个旅游小团要吃上 3 条烤鱼，至少需要 15 分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "23-24 四年级上·重庆万州·期末"
+    },
+    {
+        "id":  "274",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "星期日早上，小华准备帮妈妈做一些家务劳动：\n1.拖地：8分钟  2.洗衣机洗衣服：30分钟  3.擦家具：15分钟  4.浇花：5分钟  5.晾衣服：5分钟\n小华怎样做才能又快又好地完成这些事情？请你帮助小华设计流程：( )，做完这些家务劳动至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "先用洗衣机洗衣服，同时拖地、擦家具、浇花，衣服洗好后再晾衣服;35",
+        "solution":  "【分析】洗衣机洗衣服时间最长，为30分钟。小华可以在这30分钟里同时完成拖地、擦家具、浇花，最后再晾衣服。\n【详解】8+15+5=28(分钟)，28<30。\n所以先用洗衣机洗衣服，同时拖地、擦家具、浇花，衣服洗好后再晾衣服。\n30+5=35(分钟)，做完这些家务劳动至少需要35分钟。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "23-24 四年级上·河南三门峡·期末"
+    },
+    {
+        "id":  "275",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈下班回家做饭，淘米要 2 分钟，用电饭煲煮饭要 28 分钟，洗菜要 5 分钟，切菜要 5 分钟，炒菜要 15 分钟。妈妈将饭菜做好，最少要用( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "30",
+        "solution":  "【分析】淘米 2 分钟后开始用电饭煲煮饭 28 分钟，在煮饭期间可以完成洗菜，切菜和炒菜(5 + 5 + 15 = 25 分钟),25 \u003c 28，因此炒菜可以全部安排在煮饭过程中完成.\n【详解】总时间 = 淘米时间 + 煮饭时间:2 + 28 = 30(分钟)。妈妈下班回家做饭，最少要用 30 分钟将饭菜做好。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "23-24 四年级上·重庆开州·期末"
+    },
+    {
+        "id":  "276",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "煮蛋器有两层，每层一次最多可以煮7个鸡蛋，煮熟一个鸡蛋需要8分钟，那么煮熟21个鸡蛋至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "16",
+        "solution":  "【分析】煮蛋器两层，一次最多可煮 7×2 = 14 个鸡蛋。21 个鸡蛋第一次煮 14 个，还剩 7 个，再煮一次即可。每次煮熟需要 8 分钟，两次共用时 16 分钟.\n【详解】第一次：煮 14 个鸡蛋,8 分钟;剩下 21 - 14 = 7 个鸡蛋，再煮 8 分钟。总时间:8 + 8 = 16(分钟),所以煮熟 21 个鸡蛋至少需要 16 分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "23-24四年级上·内蒙古呼伦贝尔·期末"
+    },
+    {
+        "id":  "277",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用平底锅煎鸡蛋，每次只能放 2 个鸡蛋，煎一面要 2 分钟，两面都要煎，那么煎 6 个鸡蛋要( )分钟，煎 9 个鸡蛋要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "12;18",
+        "solution":  "【分析】先看 3 个鸡蛋的最优煎法：每次煎 2 个鸡蛋的不同面，共 3 次，每次 2 分钟，总时间 6 分钟。6 个鸡蛋可以分成两组 3 个;9 个鸡蛋可以分成三组 3 个.\n【详解】3 个鸡蛋最优:2 + 2 + 2 = 6(分钟).\n煎 6 个鸡蛋：两组 3 个,6×2 = 12(分钟).\n煎 9 个鸡蛋：三组 3 个,6×3 = 18(分钟)。所以煎 6 个鸡蛋要 12 分钟，煎 9 个鸡蛋要 18 分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "23-24 四年级上·江西赣州·期末"
+    },
+    {
+        "id":  "278",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈下班回家做饭，淘米要 3 分钟，洗菜要 8 分钟，切菜要 10 分钟，煮饭要 30分钟，炒菜要 10 分钟。如果煮饭和炒菜用不同的锅和炉子，妈妈最少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "33",
+        "solution":  "【分析】先淘米 3 分钟，然后开始煮饭 30 分钟。煮饭的同时可以洗菜，切菜，炒菜，洗菜 + 切菜 + 炒菜 = 8 + 10 + 10 = 28(分钟),28 \u003c 30，因此这些工作可以全部安排在煮饭期间完成.\n【详解】总时间 = 淘米时间 + 煮饭时间:3 + 30 = 33(分钟),所以妈妈最少需要 33 分钟。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "2024 四年级上·全国·专题练习"
+    },
+    {
+        "id":  "279",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明起床整理被褥 3分钟;刷牙 3 分钟;洗脸 2 分钟;听英语录音 8 分钟。小明最快( )分钟能做完这些事情。",
+        "options":  [
+
+                    ],
+        "answer":  "8",
+        "solution":  "【分析】听英语录音的 8 分钟可以与整理被褥，刷牙和洗脸同时进行，这三件事共用时间 3 + 3 + 2 = 8(分钟),恰好与听录音时间相同.\n【详解】把整理被褥，刷牙，洗脸安排在听录音的 8 分钟内完成，所需总时间就是 8 分钟。计算:3 + 3 + 2 = 6 + 2 = 8(分钟),所以小明最快 8 分钟能做完这些事情。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "23-24 四年级上·河南信阳·期末"
+    },
+    {
+        "id":  "280",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "3个小朋友3天吃3个西瓜,9 个小朋友 9 天吃( )个西瓜。",
+        "options":  [
+
+                    ],
+        "answer":  "27",
+        "solution":  "【分析】3 个小朋友 3 天吃 3 个西瓜，相当于每\"3 个小朋友·3 天\"消耗 3 个西瓜。9 天是 3 天的 3 倍,9 个小朋友是 3 个小朋友的 3 倍，总人数和天数都扩大 3 倍，总消耗也扩大 3 倍再 3 倍.\n【详解】3 个小朋友 3 天吃 3 个西瓜:\n3×3×3 = 9×3 = 27(个)\n所以 9 个小朋友 9 天吃 27 个西瓜。",
+        "knowledge_points":  [
+                                 "双归一问题"
+                             ],
+        "source":  "24-25 四年级上·湖北咸宁·期中"
+    },
+    {
+        "id":  "281",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈给小明做早餐所用的时间如下：淘米 2 分钟，熬粥 20 分钟，煎鸡蛋 5 分钟(熬粥和煎鸡蛋用不同的锅),热牛奶 2 分钟，拌小菜 6分钟。妈妈做好早餐至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "22",
+        "solution":  "【分析】要使时间最短，应先淘米，再熬粥。在熬粥的同时，可以完成煎鸡蛋、热牛奶和拌小菜。\n【详解】总时间=淘米时间+熬粥时间：2+20=22(分钟)。\n所以妈妈做好早餐至少需要22分钟。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "23-24 四年级下·甘肃平凉·期末"
+    },
+    {
+        "id":  "282",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "早晨，妈妈给我们煎鸡蛋饼。已知一只平底锅每次只能放两个鸡蛋饼，煎一个鸡蛋饼正反面各需2分钟，那么煎3个鸡蛋饼至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】3 个鸡蛋饼，每次锅里最多 2 个，要利用\"轮换\"方式使锅始终满负荷工作.\n【详解】第 1 次:1 正,2 正(2 分钟);第 2 次:1 反,3 正(2 分钟);第 3 次:2 反,3 反(2 分钟).\n总时间:2×3 = 6(分钟),所以煎 3 个鸡蛋饼至少需要 6 分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "2007 四年级·全国·竞赛"
+    },
+    {
+        "id":  "283",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "实验小学一共有师生 340 名，去乘坐汽车研学，大车能坐 40 名师生，租金900 元，小车可坐 20 名师生，租金 500 元，租车最便宜()元。",
+        "options":  [
+
+                    ],
+        "answer":  "7700",
+        "solution":  "【分析】先比较人均租金：大车900÷40=22.5(元)，小车500÷20=25(元)，大车更便宜，所以应尽量多租大车。\n【详解】340÷40=8(辆)……20(人)。\n先租8辆大车可坐320人，剩余20人再租1辆小车正好坐满。\n费用：900×8+500×1=7200+500=7700(元)。\n所以租车最便宜是7700元。",
+        "knowledge_points":  [
+                                 "租船问题"
+                             ],
+        "source":  "23-24 四年级下·河南洛阳·期末"
+    },
+    {
+        "id":  "284",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用一只平底锅煎饼，每次只能煎两块饼，煎一面需要 3 分钟，两面都要煎。煎 6 块饼最少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "18",
+        "solution":  "【分析】平底锅一次煎 2 块饼，每块饼有 2 面，一共 6 块饼需要烙 12 面。可用\"分组法\"或公式：总时间 = 饼数 * 2 / 每锅可烙数量×每面时间.\n【详解】用公式:6×2÷2×3 = 12÷2×3 = 6×3 = 18(分钟).\n也可以按两两分组：每 2 块饼需要 3 + 3 = 6 分钟，共 3 组:3×6 = 18 分钟。故煎 6 块饼最少需要 18 分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "23-24 四年级上·全国·期末"
+    },
+    {
+        "id":  "285",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈早上起床，刷牙 5分钟，淘米洗锅 3 分钟，熬粥 20 分钟，盛粥 1 分钟，煎鸡蛋 5 分钟。能吃上妈妈做的早饭至少要用( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "24",
+        "solution":  "【分析】先淘米洗锅 3 分钟，再熬粥 20 分钟。在熬粥的 20 分钟内，可以同时刷牙和煎鸡蛋(5 + 5 = 10 分钟),不增加总时间，最后盛粥 1 分钟.\n【详解】总时间:3 + 20 + 1 = 23 + 1 = 24(分钟)。能吃上妈妈做的早饭至少要用 24 分钟。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "23-24 四年级上·全国·期末"
+    },
+    {
+        "id":  "286",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明有10,7,4三张数字卡片，小东有 3,5,9 三张数字卡片，规定每人每次出一张卡片，数字大的赢，各出三次，赢两次者胜。如果小明先出(顺序如下图),那么小东应该怎样应对才能获胜。小明:10,7,4;小东: , , 。",
+        "options":  [
+
+                    ],
+        "answer":  "3;9;5",
+        "solution":  "【分析】小明出牌顺序固定为 10,7,4。小东要想赢两次，就要用\"田忌赛马\"策略：用最小的牌去输掉对方最大的一次，把大牌保留到可以稳赢的回合.\n【详解】第一次：用 3 对 10(必输一次);第二次：用 9 对 7(赢一次);第三次：用 5 对 4(再赢一次).\n因此，小东应出牌顺序为:3,9,5。",
+        "knowledge_points":  [
+                                 "田忌赛马问题"
+                             ],
+        "source":  "23-24 四年级上·全国·单元测试"
+    },
+    {
+        "id":  "287",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一块牛排两面都要煎，正，反面各需 3 分钟，一只平底锅每次可以煎 2 块。如果煎一块牛排最快( )分钟可以煎好;如果煎2块牛排，最快( )分钟可以煎好;如果煎 3 块牛排，最快( )分钟可以煎好。",
+        "options":  [
+
+                    ],
+        "answer":  "6;6;9",
+        "solution":  "【分析】一块牛排：正反两面各 3 分钟，总共 6 分钟。两块牛排：可以同时煎两块的正面，再同时煎两块的反面，仍是 3 + 3 = 6 分钟。三块牛排需要用\"轮换法\".\n【详解】煎 1 块:3 + 3 = 6(分钟).\n煎 2 块：第一轮 2 块正面 3 分钟，第二轮 2 块反面 3 分钟，共 6(分钟).\n煎 3 块:\n第 1 次:1 正,2 正(3 分钟);第 2 次:1 反,3 正(3 分钟);第 3 次:2 反,3 反(3 分钟),总时间 3×3 = 9(分钟)。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "23-24 四年级上·全国·单元测试"
+    },
+    {
+        "id":  "288",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "想一想，填一填。(1)1 张饼有( )面，如果烙每面需要 2 分钟，烙 1 张饼需要( )分钟。 (2)煮熟 1 个鸡蛋需要 6 分钟，一个锅每次最多可以煮 8 个鸡蛋，那么煮熟 8 个鸡蛋至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "2;4;6",
+        "solution":  "【分析】(1) 一张饼有两面，每面 2 分钟，共 4 分钟。(2) 一个锅一次最多煮 8 个鸡蛋，视为 8 个鸡蛋同时煮，时间由单个鸡蛋决定.\n【详解】(1) 1 张饼有 2 面，烙每面需要 2 分钟，烙 1 张饼需要 2×2 = 4(分钟).\n(2) 煮熟 1 个鸡蛋需要 6 分钟，一次可以煮 8 个鸡蛋,8 个鸡蛋一起煮也只需 6 分钟，所以至少需要 6 分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "24-25 四年级上·全国·课后作业"
+    },
+    {
+        "id":  "289",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小华给客人烧水沏茶：洗茶壶 1 分钟，烧开水 10 分钟，洗茶杯 1 分钟，拿茶叶 1分钟，沏茶 2 分钟。客人喝上茶至少需要( )分钟，最合理的安排顺序是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "13;洗茶壶-烧开水-洗茶杯-拿茶叶-沏茶",
+        "solution":  "【分析】烧开水时间最长，应尽量在烧水期间完成洗茶杯和拿茶叶，以减少总时间。洗茶壶是烧水前的前置步骤，沏茶应在开水和茶具准备好之后进行.\n【详解】合理顺序：洗茶壶 → 烧开水(同时洗茶杯，拿茶叶)→ 沏茶。总时间 = 洗茶壶时间 + 烧开水时间 + 沏茶时间:1 + 10 + 2 = 13(分钟).\n所以客人喝上茶至少需要 13 分钟，最合理的安排顺序是\"洗茶壶-烧开水-洗茶杯-拿茶叶-沏茶\"。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "24-25 四年级上·全国·期末"
+    },
+    {
+        "id":  "290",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在火炉上烤烧饼，两面都要烤，每烤完一面需要2分钟，火炉上每次最多能烤6个烧饼，烤8个烧饼最少要用( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】一次最多烤 6 个烧饼，一面 2 分钟，两面 4 分钟。8 个烧饼不能一次完成，需要利用\"补位 + 轮换\"的方式，把第二批的正反面插入前一批的反面时间里.\n【详解】第 1 次:6 个烧饼正面,2 分钟;第 2 次：把其中 2 个已烤好的正面取下，将另外 2 个未烤的烧饼放上去，同时烤 4 个的反面和 2 个的正面,2 分钟;第 3 次：烤剩下 4 个烧饼的反面,2 分钟.\n总时间:2 + 2 + 2 = 6(分钟),所以烤 8 个烧饼最少要用 6 分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "24-25 四年级上·全国·单元测试"
+    },
+    {
+        "id":  "291",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "服装加工厂做61套校服，要按整卷买布。大卷布，每卷400元，可以做8套校服;小卷布，每卷300元，可以做5套校服。如果想最省钱，可以买大卷布( )卷和小卷布( )卷。",
+        "options":  [
+
+                    ],
+        "answer":  "7;1",
+        "solution":  "【分析】先比较两种卷布做一套校服的成本：大卷布400÷8=50(元)，小卷布300÷5=60(元)，大卷布更便宜，所以应尽量多买大卷布，但也要保证做出的校服不少于61套。\n【详解】若只买大卷布：61÷8=7……5，需要买8卷，共3200元。\n若买7卷大卷布，可做56套，还差5套，正好再买1卷小卷布。\n总价：7×400+1×300=2800+300=3100(元)。\n3100元比3200元少，所以最省钱的方案是买大卷布7卷和小卷布1卷。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "租船问题"
+                             ],
+        "source":  "23-24 四年级下·内蒙古赤峰·期末"
+    },
+    {
+        "id":  "292",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "178 人去旅游，租()辆大车和( )辆小车，最省钱。一辆大车限坐乘客 30 人，租金 1200 元。一辆小车限坐乘客 7 人，租金 350 元。",
+        "options":  [
+
+                    ],
+        "answer":  "6;0",
+        "solution":  "【分析】先比较人均租金：大车1200÷30=40(元)，小车350÷7=50(元)，大车更便宜，所以应尽量多租大车，再比较几种方案的总价。\n【详解】178÷30=5……28，若租6辆大车，可坐180人，费用是6×1200=7200(元)。\n若租5辆大车，还剩28人，需要4辆小车，费用是5×1200+4×350=6000+1400=7400(元)。\n7200<7400，所以最省钱方案是租6辆大车，0辆小车。",
+        "knowledge_points":  [
+                                 "租船问题"
+                             ],
+        "source":  "23-24 四年级下·广东广州·期末"
+    },
+    {
+        "id":  "293",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一只平底锅每次最多只能煎2条小黄鱼，煎一面需要2分，两面都要煎，煎5条小黄鱼至少需要( )分。",
+        "options":  [
+
+                    ],
+        "answer":  "10",
+        "solution":  "【分析】5 条鱼可以分为前 2 条和后 3 条两部分。前 2 条直接双面煎熟，后 3 条用轮换方式，使锅内始终有 2 条鱼在煎.\n【详解】前 2 条:2×2 = 4(分).\n剩下 3 条编号为 1,2,3:\n第 1 次:1 正,2 正(2 分钟);第 2 次:1 反,3 正(2 分钟);第 3 次:2 反,3 反(2 分钟),3 条一共 3×2 = 6 分钟.\n总时间:4 + 6 = 10(分),所以煎 5 条小黄鱼至少需要 10 分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "23-24 四年级下·福建福州·期末"
+    },
+    {
+        "id":  "294",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "李老师要复印 6 张同样规格的资料，正，反两面都要复印，如果一次最多只能复印 2 张单面，至少要( )次才能复印完。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】一次最多复印 2 张单面，相当于\"平底锅一次烙两张饼，单面烙一次\"的模型。6 张资料，两面共 12 面，需要安排每次 2 面地复印.\n【详解】最优安排:\n第 1 次:1 正,2 正;第 2 次:1 反,2 反;第 3 次:3 正,4 正;第 4 次:3 反,4 反;第 5 次:5 正,6 正;第 6 次:5 反,6 反.\n共 6 次才能复印完。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "23-24 四年级上·福建莆田·期末"
+    },
+    {
+        "id":  "295",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "李强每天早晨起床穿衣 3 分钟，整理床铺 2 分钟，刷牙洗脸 3 分钟，烧开水 5分钟，喝水 1 分钟，吃早餐 6 分钟，步行到学校 15 分钟。李强为了在 8:00 前到校，他制订了\"高效行动计划\"，那么他最迟( )起床才不会迟到。",
+        "options":  [
+
+                    ],
+        "answer":  "7时30分",
+        "solution":  "【分析】要求\"最迟起床时间\"，就是先求这些事情在最优安排下需要的最短总时间，再用到校时间反推起床时间。烧水可以与整理床铺，刷牙洗脸同时进行.\n【详解】最少用时：穿衣 3 分钟 → 烧水 5 分钟(同时整理床铺 2 分钟，刷牙洗脸 3 分钟)→ 吃早餐 6 分钟 → 喝水 1 分钟 → 步行 15 分钟.\n总时间:3 + 5 + 6 + 1 + 15 = 30(分钟).\n8:00 - 30 分钟 = 7:30，因此李强最迟 7:30 起床才不会迟到。",
+        "knowledge_points":  [
+                                 "沏茶问题",
+                                 "时，分，秒时间的推算"
+                             ],
+        "source":  "22-23 四年级上·福建莆田·期末"
+    },
+    {
+        "id":  "296",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用一只平底锅烙饼，每次最多只能烙2张饼，两面都要烙，每面烙3分钟。烙10 张饼至少要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "30",
+        "solution":  "【分析】每次最多烙 2 张饼，两面都要烙，每面 3 分钟,2 张饼双面需要 3×2 = 6 分钟。10 张饼可以视为 5 组\"2 张饼\".\n【详解】先烙 2 张饼，需要 3×2 = 6(分钟)。10 张饼分为 5 组 2 张:10÷2 = 5。总时间:6×5 = 30(分钟),所以至少要 30 分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "23-24 四年级下·湖南衡阳·期末"
+    },
+    {
+        "id":  "297",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "四(1)班 32 个同学要租船游玩 1 小时。每条大船限乘 6 人，租金 30 元/时;每条小船限乘 4 人，租金 24 元/时。租( )条大船和( )条小船最省钱。",
+        "options":  [
+
+                    ],
+        "answer":  "4;2",
+        "solution":  "【分析】先比较单位人数的租金：大船 30÷6，小船 24÷4，确定更便宜的一种，然后在不空船的前提下尽量多租便宜的船，对剩余人用另一种船补足.\n【详解】大船人均:30÷6 = 5(元),小船人均:24÷4 = 6(元),5 \u003c 6，应尽量租大船。若全用大船:32÷6 = 5。2,5 条大船可坐 30 人，剩 2 人不能单独再开一条大船，需考虑搭配小船。先假设 4 条大船:4×6 = 24 人，剩 8 人,8÷4 = 2 条小船刚好坐满.\n费用:4×30 + 2×24 = 120 + 48 = 168(元)。可验证其它组合费用更高，因此租 4 条大船和 2 条小船最省钱。",
+        "knowledge_points":  [
+                                 "租船问题"
+                             ],
+        "source":  "23-24 四年级下·湖南怀化·期末"
+    },
+    {
+        "id":  "298",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小玲，小可和小芳三个小朋友一起玩\"荡秋千\"\"滑滑梯\"这两个游乐项目，约定每个项目只能一个人玩，且每人每个项目玩 5 分钟，三个人至少要用( )分钟才能将两个项目都玩一次。",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】两个项目可以同时进行，每次可以有两个人分别玩不同的项目，第三人等待，按\"轮换\"方式安排三人，保证两个设施始终有人使用.\n【详解】安排示例:\n第 1 个 5 分钟：小玲荡秋千，小可滑滑梯;第 2 个 5 分钟：小玲滑滑梯，小芳荡秋千;第 3 个 5 分钟：小可荡秋千，小芳滑滑梯.\n总时间:5 + 5 + 5 = 15(分钟),三个人都完成两个项目各一次。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "23-24 四年级下·湖北省直辖县级单位·期末"
+    },
+    {
+        "id":  "299",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "蒸熟 1 个馒头需要 30 分钟的时间，一个蒸锅一次可以蒸 15 个馒头，那么蒸熟 15 个馒头至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "30",
+        "solution":  "【分析】一个蒸锅一次可以蒸 15 个馒头，视作 15 个馒头同时加热，时间由\"单个馒头\"的熟成时间决定，而与数量无关，只要不超出容量.\n【详解】蒸 1 个馒头需要 30 分钟，蒸 15 个馒头只需 1 锅即可，时间仍为 30 分钟。因此，蒸熟 15 个馒头至少需要 30 分钟。",
+        "knowledge_points":  [
+                                 "烙饼问题"
+                             ],
+        "source":  "23-24 四年级上·全国·单元测试"
+    },
+    {
+        "id":  "300",
+        "unit":  "第八章数学广角 3",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小强已经养成良好的早起习惯，洗漱10分钟，吃早饭15分钟，还坚持每天听20分钟的新闻广播。合理安排做完这些事情至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "25",
+        "solution":  "【分析】听新闻广播时可以同时进行洗漱或吃早饭，只要保证广播时间完全覆盖其中一部分，就能缩短总时间。广播 20 分钟，可以与洗漱 10 分钟和吃早饭的一部分重叠.\n【详解】一种合理安排：一边听广播一边洗漱 10 分钟，然后继续听广播并开始吃早饭，广播剩余 10 分钟，吃早饭还需 5 分钟，最后再单独吃完 5 分钟。总时间等于洗漱 + 吃早饭:10 + 15 = 25(分钟).\n因此，合理安排做完这些事情至少需要 25 分钟。",
+        "knowledge_points":  [
+                                 "沏茶问题"
+                             ],
+        "source":  "23-24 四年级上·全国·单元测试"
+    },
+    {
+        "id":  "301",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "四(1)班共有32人去租船，大船可以坐6人，租金30元，小船可以坐4人，租金 24 元。如果要最省钱，可以租( )条小船,( )条大船。一共花了( )元。",
+        "options":  [
+
+                    ],
+        "answer":  "2;4;168",
+        "solution":  "【分析】先比较大船和小船的人均费用，选便宜的那种尽量多租，再保证人数正好坐满。\n【详解】大船每人租金：30÷6=5(元)，小船每人租金：24÷4=6(元)，5<6，应尽量多租大船。32÷6=5……2，若租5条大船，还差2人不能坐满，可考虑少租一条大船改租小船：先租4条大船，4×6=24(人)，剩余8人，8÷4=2(条小船)，共4条大船、2条小船。费用：24×2+30×4=48+120=168(元)。所以可以租2条小船、4条大船，一共花168元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "优化问题"
+                             ],
+        "source":  "18-19四年级下·云南临沧·期末"
+    },
+    {
+        "id":  "302",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明给客人沏茶，接水 1 分钟，烧水 8 分钟，洗茶杯 3 分钟，拿茶叶 1 分钟，沏茶 1 分钟。小明合理安排以上事情，最少需要( )分钟才能使客人喝上茶。",
+        "options":  [
+
+                    ],
+        "answer":  "10",
+        "solution":  "【分析】烧水时间最长，应把其他能同时做的事情放在烧水期间完成.\n【详解】先接水 1 分钟，然后烧水 8 分钟，在烧水的同时完成洗茶杯(3 分钟)和拿茶叶(1 分钟),最后沏茶 1 分钟。总时间为接水 + 烧水 + 沏茶:1+8+1=10(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "19-20 四年级上·青海海东·期末"
+    },
+    {
+        "id":  "303",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.4,
+        "question":  "小明在奶奶的房间里玩，他把2个卧室、2个厕所和1个厨房的钥匙混在了一起，要把所有的钥匙分出来，最多要试( )次。",
+        "options":  [
+
+                    ],
+        "answer":  "10",
+        "solution":  "【分析】按最不利的情况考虑：每一把钥匙都在最后一次尝试时才配对成功。\n【详解】5把钥匙：找第1把卧室钥匙，最多试4次；余4把中找第2把卧室钥匙，最多试3次；余3把中找第1把厕所钥匙，最多试2次；余2把中找第2把厕所钥匙，最多试1次；最后1把一定是厨房钥匙。总试次数：4+3+2+1=10(次)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "19-20 三年级下·甘肃平凉·期末"
+    },
+    {
+        "id":  "304",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.15,
+        "question":  "50枚棋子围成圆圈，编上号码1，2，3，4，…，50。每隔一枚棋子取出一枚，要求最后留下的那枚棋子的号码是40号，那么该从( )号棋子开始取。",
+        "options":  [
+
+                    ],
+        "answer":  "5",
+        "solution":  "【分析】先假设从1号开始取，按“隔一取一”的规则模拟，找到最后剩下的号码，再根据差值倒推起点。\n【详解】假设从1号开始：第一圈剩2，4，6，…，50；第二圈剩4，8，12，16，20，24，28，32，36，40，44，48；第三圈剩4，12，20，28，36；第四圈剩4，20，36；第五圈剩4，36；最后剩36号。如果想最后剩40号，则起点应向前平移40-36=4个位置，从1号平移4个位置得起点为5号。所以应从5号棋子开始取。",
+        "knowledge_points":  [
+                                 "通过操作实验探索规律"
+                             ],
+        "source":  "2021 六年级下·全国·竞赛"
+    },
+    {
+        "id":  "305",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.4,
+        "question":  "将前99个自然数从小到大排成一行：12345678910111213…99，在其中加入一个加号，变成一个加法算式。要使得到的结果最小，那么这个结果的个位数字是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "1",
+        "solution":  "【分析】要使和最小，需要让两个加数的位数尽量均衡，且前一个加数尽量小。\n【详解】1~9共9位，10~99共180位，总位数是9+180=189。最优时把这189位数字分成95位和94位两个数。这样两个加数的个位分别是2和9，所以它们的和的个位是1。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "2021 六年级下·全国·竞赛"
+    },
+    {
+        "id":  "306",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.15,
+        "question":  "有 5 支球队进行单循环足球赛(每两队比赛一场),胜一场得3分，负一场得0分，平局两队各得1分;以总得分高低确定名次，若两支球队得分相同，就参考净胜球，相互胜负关系等因素决定名次。笑笑学校的球队要想稳获这次友谊赛的前三名，至少要得( )分。",
+        "options":  [
+
+                    ],
+        "answer":  "8",
+        "solution":  "【分析】要求\"稳获前三名\"，笑笑队得分要大于第四名的最高可能得分。先分析第四名最多能得多少分，再在此基础上确定笑笑队的最低分数.\n【详解】5 队单循环共有 C(5,2)=10 场，全部胜负则总分最大 10×3=30 分。若让第五名全败，则其得 0 分，其余 4 队之间互相比赛，第四名的得分被压到尽可能高，计算可知第四名最高只能 7 分。如果笑笑队只得 7 分，有可能与第四名同分而被净胜球挤到第四名;要稳居前三至少要比第四名最高分多 1 分，即 8 分。",
+        "knowledge_points":  [
+                                 "搭配问题",
+                                 "优化问题"
+                             ],
+        "source":  "2021 六年级下·全国·竞赛"
+    },
+    {
+        "id":  "307",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个微型烤炉一次最多可以烤 2个烧饼，烧饼两面都要烤，每面要烤 2 分钟。则烤 6 个烧饼最少要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "12",
+        "solution":  "【分析】要保证每次烤炉里都有 2 个烧饼在烤，利用\"3 个烧饼轮换\"的最优策略.\n【详解】6 个烧饼分两批。前 2 个：一次烤正面 2 分钟，再一次烤反面 2 分钟，共 4 分钟。后 4 个可看作\"3 个轮换 + 1 个搭配前面\"的一般模式，也可直接用烙饼公式：总时间 = 饼数*2/每次容量×单面时间 = 6×2÷2×2=12(分钟)。所以烤 6 个烧饼最少要 12 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "2024六年级下·全国·专题练习"
+    },
+    {
+        "id":  "308",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用一个平底煎鱼，每次只能煎 2 条，两面都要煎，煎好一面需要 5 分钟，煎熟 7条至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "35",
+        "solution":  "【分析】每条鱼 2 面，每面 5 分钟，共 10 分钟一条。容量 2 条时，只要用最优轮换，也等价于\"总时间=单面时间×条数\".\n【详解】先 2 条 2 条分组：前 4 条分两组，每组双面 2×5=10 分钟，共 20 分钟。剩余 3 条用轮换：先两条正面 5 分钟，再其中一条翻面+另一条新正面 5 分钟，最后两条反面 5 分钟，共 15 分钟。总时间:20+15=35(分钟)。也可直接按公式:5×7=35(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·山东临沂·期末"
+    },
+    {
+        "id":  "309",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用一只平底锅煎大饼，每次能同时放两个大饼，大饼的正反面各需要煎 3 分钟，煎23 个饼需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "69",
+        "solution":  "【分析】容量为 2，单面 3 分钟，两面需要 6 分钟一张。利用烙饼类优化结论：总时间 = 单面时间×饼数.\n【详解】每张饼有 2 面，用容量为 2 的锅最优轮换时，总时间 = 3×23=69(分钟)。故煎23个饼需要69分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "20-21 四年级上·浙江台州·期末"
+    },
+    {
+        "id":  "310",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个平底锅每次只能煎两条鱼，每面需要煎 5 分钟，两面都要煎，妈妈要煎三条鱼，至少要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】三条鱼，两面都要煎，锅中每次放两条，用轮换法保证锅不空闲.\n【详解】编号 1,2,3：第一次煎 1,2 正面 5 分钟;第二次煎 1 反面和 3 正面 5 分钟;第三次煎 2 反面和 3 反面 5 分钟。共 3 次，每次 5 分钟，总时间 5×3=15(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "20-21 四年级下·浙江金华·期末"
+    },
+    {
+        "id":  "311",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "爸爸，妈妈和我每人要吃1张饼，锅里每次只能烙2张饼，两面都要烙，每面需要3分钟，至少要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "9",
+        "solution":  "【分析】共 3 张饼，每张 2 面，共 6 面，每次可烙 2 面，每面 3 分钟.\n【详解】总面数:3×2=6(面),每次 2 面，需要 6÷2=3 次，每次 3 分钟，总时间 3×3=9(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "19-20四年级下·浙江金华·期末"
+    },
+    {
+        "id":  "312",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "丁丁睡觉前喜欢听音乐，他要做的事有：听音乐15分钟，洗脸3分钟，刷牙2分钟，冲牛奶1分钟，喝牛奶3分钟。做完这些事最少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】只要把所有其他事情安排在听音乐的 15 分钟内即可，因总时长 3+2+1+3=9 小于 15.\n【详解】洗脸，刷牙，冲牛奶，喝牛奶共 3+2+1+3=9(分钟),9\u003c15，所以可以在听音乐的 15 分钟期间完成这些事，总用时受听音乐 15 分钟限制，因此最少需要 15 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "20-21 四年级上·浙江台州·期末"
+    },
+    {
+        "id":  "313",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "妙想每天晚上睡觉前背成语8分钟，烧开水10分钟，用热水泡牛奶2分钟，喝牛奶5分钟，那么妙想最少要用( )分钟做完所有的事情。",
+        "options":  [
+
+                    ],
+        "answer":  "17",
+        "solution":  "【分析】可以在烧水的 10 分钟内背完成语 8 分钟，以减少总时间.\n【详解】先烧开水 10 分钟，同时背成语 8 分钟，烧水结束后用热水泡牛奶 2 分钟，再喝牛奶 5 分钟。总时间:10+2+5=17(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "19-20四年级下·浙江金华·期末"
+    },
+    {
+        "id":  "314",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "有 7 张文件，正反面都要复印，复印机1次最多能放2张，复印1次的时间是10 秒，那么全部复印好至少要( )秒。",
+        "options":  [
+
+                    ],
+        "answer":  "70",
+        "solution":  "【分析】这是典型烙饼模型：文件张数*2 面 / 每次容量×每次时间.\n【详解】总面数:7×2=14(面),每次最多 2 面，所以需要 14÷2=7(次),每次 10 秒，总时间 7×10=70(秒)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·甘肃陇南·期末"
+    },
+    {
+        "id":  "315",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "煮一个鸡蛋需要 8分钟，一只锅一次可以煮5个鸡蛋，那么煮20个鸡蛋至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "32",
+        "solution":  "【分析】一次最多煮 5 个，时间由单个鸡蛋决定，即每锅 8 分钟.\n【详解】20 个鸡蛋分 4 锅煮:20÷5=4(次),每次 8 分钟，总时间 4×8=32(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·江西吉安·期末"
+    },
+    {
+        "id":  "316",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "晨晨有10，6，4三张牌，乐乐有5，9，3三张牌，规定每人每次出一张牌。如果晨晨先出，乐乐获胜的策略是( )。(三局两胜制)",
+        "options":  [
+
+                    ],
+        "answer":  "10对3，6对9，4对5",
+        "solution":  "【分析】用“田忌赛马”策略：用自己最小的牌去对对方最大牌，保存大牌去赢对方中等和最小的牌。\n【详解】两人牌从小到大：晨晨4<6<10，乐乐3<5<9。若晨晨先出，乐乐要赢两局，可安排：10对3，6对9，4对5。这样乐乐2胜1负获胜。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·陕西商洛·期末"
+    },
+    {
+        "id":  "317",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈在锅里煎饼，饼的两面都要煎，每煎一面需要3分钟，锅里同时只能煎两张饼。如果妈妈要煎 3 张饼，至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "9",
+        "solution":  "【分析】容量 2 张，每面 3 分钟，用烙饼公式：时间 = 饼数*2/容量×每面时间.\n【详解】时间 = 3×2÷2×3 = 3×3=9(分钟)。所以至少需要 9 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23四年级上·山西晋中·期末"
+    },
+    {
+        "id":  "318",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "明明拖地用 8 分钟，烧水用 10 分钟，接水用 1 分钟，做完这些事情他至少要()分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "11",
+        "solution":  "【分析】接水后烧水，在烧水期间同时拖地，以缩短总时间.\n【详解】先接水 1 分钟，然后烧水 10 分钟，在烧水的 10 分钟里可以完成拖地 8 分钟，不额外增加时间。总时间:10+1=11(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·山西晋中·期末"
+    },
+    {
+        "id":  "319",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "王东家来了客人，他要给客人泡茶。烧水要 12 分钟，洗水壶要 2 分钟，洗茶杯要 1 分钟，找茶叶要 1 分钟，沏茶要 1 分钟。王东最快要花( )分钟，才能给客人泡上茶。",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】洗水壶是烧水前置条件，其余洗茶杯，找茶叶可在烧水时完成.\n【详解】先洗水壶 2 分钟，再烧水 12 分钟，烧水时完成洗茶杯 1 分钟和找茶叶 1 分钟，最后沏茶 1 分钟。总时间:2+12+1=15(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·贵州黔东南·期末"
+    },
+    {
+        "id":  "320",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "煎一块牛排要 8 分钟(每面各 4 分钟),一个平底锅一次最多能煎两块牛排。煎三块牛排最少要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "12",
+        "solution":  "【分析】三块牛排，每块两面各 4 分钟，容量 2，按轮换法安排.\n【详解】编号 1,2,3：第一次煎 1,2 正面 4 分钟;第二次煎 1 反面和 3 正面 4 分钟，此时 1 熟;第三次煎 2 反面和 3 反面 4 分钟,2,3 熟。总时间:4+4+4=12(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·吉林通化·期末"
+    },
+    {
+        "id":  "321",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈做晚饭时，需要做这些事：淘米 2 分钟，洗菜 10 分钟，炒菜 15 分钟，用电饭煲煮饭 20 分钟。妈妈至少需要( )分钟才能把事情做完。",
+        "options":  [
+
+                    ],
+        "answer":  "27",
+        "solution":  "【分析】淘米后煮饭，煮饭时完成洗菜和炒菜.\n【详解】先淘米 2 分钟，再煮饭 20 分钟，煮饭时可洗菜 10 分钟和炒菜 15 分钟,10+15=25\u003e20，说明炒菜会超过煮饭时间，因此总时间是洗菜+炒菜:10+15=25，再加淘米 2 分钟:2+25=27(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "20-21 四年级上·浙江湖州·期末"
+    },
+    {
+        "id":  "322",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明放学回家做饭，洗电饭锅 2 分钟，淘米 1 分钟，电饭锅煮饭 20 分钟，炒菜15 分钟，煮汤 3 分钟，明明最快( )分钟就可以吃饭。",
+        "options":  [
+
+                    ],
+        "answer":  "23",
+        "solution":  "【分析】洗电饭锅和淘米后开始煮饭，在煮饭期间完成炒菜和煮汤.\n【详解】洗电饭锅 2 分钟，淘米 1 分钟，共 3 分钟，然后电饭锅煮饭 20 分钟，同时可以炒菜 15 分钟，煮汤 3 分钟,15+3=18\u003c20。最后不再额外等待。总时间:2+1+20=23(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·广西河池·期末"
+    },
+    {
+        "id":  "323",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "下列事件中，小明进行了合理安排或合理对策的是( )，( )。\n1. 小明放学后，在煮米饭的40分钟里洗菜、切菜、炒菜。 2. 一个锅每次可同时烙2张饼，小明烙3张饼，先烙完两张饼，再烙第三张。 3. 小明和小红玩牌比大小，小明有3，5，8，小红出9，小明出3；小红出7，小明出8；小红出4，小明出5。",
+        "options":  [
+
+                    ],
+        "answer":  "①;③",
+        "solution":  "【分析】看是否体现\"能同时做的就同时做\"和\"田忌赛马\"对策思想.\n【详解】①煮米饭 40 分钟时同时洗菜，切菜，炒菜，充分利用时间，是合理安排;②烙 3 张饼若先完全烙完两张再烙第三张，会浪费锅的容量，不合理;③出牌顺序让最小的 3 对最大 9，再用 8 对 7,5 对 4，两胜一负，是合理对策。所以答案是①,③。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "21-22四年级上·浙江湖州·期末"
+    },
+    {
+        "id":  "324",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明和小东玩纸牌比大小游戏，小明手中的牌是 8,6,4，小东手中的牌是7,5,3，每人每次出一张牌，各出三次，赢两次者胜。若小明先出 8，则小东只有出( )应对才有获胜的可能。",
+        "options":  [
+
+                    ],
+        "answer":  "3",
+        "solution":  "【分析】同样用\"田忌赛马\"：用最小牌对对方最大牌.\n【详解】若小明先出 8，小东应出 3(故意输),保留 7 和 5 对付 6 和 4:7\u003e6,5\u003e4，可赢两局。若用 7 对 8，则浪费最大牌，后面无法获得两胜。因此只有出 3 才有获胜可能。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "20-21四年级上·浙江台州·期末"
+    },
+    {
+        "id":  "325",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "玥玥帮妈妈做饭，洗菜5分钟，洗米2分钟，电饭锅煮饭30分钟，切菜3分钟，切肉 5 分钟，炒菜 10 分钟。玥玥合理安排时间，最少要()分钟才能让家人吃到热的饭菜。",
+        "options":  [
+
+                    ],
+        "answer":  "32",
+        "solution":  "【分析】洗米后开始煮饭，在煮饭时间内完成洗菜，切菜，切肉，炒菜.\n【详解】先洗米 2 分钟，再电饭锅煮饭 30 分钟，煮饭期间可依次完成洗菜 5 分钟，切菜 3 分钟，切肉 5 分钟，炒菜 10 分钟，总共 5+3+5+10=23(分钟)\u003c30(分钟),不额外增加时间。总时间:2+30=32(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23四年级上·吉林·期末"
+    },
+    {
+        "id":  "326",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "烤面包，每次最多只能烤2片面包，两面都要烤，每面2分钟，烤3片至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】3 片面包，容量 2，每面 2 分钟，按烙饼轮换法.\n【详解】编号 a,b,c:第 1 次 a,b 正面 2 分钟;第 2 次 a 反面,c 正面 2 分钟;第 3 次 b 反面,c 反面 2 分钟。总时间 3×2=6(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·内蒙古呼伦贝尔·期末"
+    },
+    {
+        "id":  "327",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一口平底锅每次最多只能煎两块牛排，煎一块牛排需要4分钟(正反面各2分钟),煎 3 块牛排需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】与 3 片面包问题相同，单面 2 分钟，容量 2.\n【详解】编号 a,b,c:第 1 次 a,b 正面 2 分钟;第 2 次 a 反面,c 正面 2 分钟;第 3 次 b 反面,c 反面 2 分钟。总时间 3×2=6(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "20-21 四年级上·湖南湘西·期末"
+    },
+    {
+        "id":  "328",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在锅里煎鱼，一次只能煎2条鱼，两面都要煎，一面要煎3分钟。要煎4条鱼，至少要( )分钟；要煎5条鱼，至少要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "12;15",
+        "solution":  "【分析】用烙饼公式：时间 = 条数*2/容量×每面时间.\n【详解】煎 4 条:4×2÷2×3=4×3=12(分钟);煎 5 条:5×2÷2×3=5×3=15(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "21-22四年级上·陕西渭南·期末"
+    },
+    {
+        "id":  "329",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈晚上回家需要做以下事情：杀鱼，洗鱼 8 分钟，烧鱼 10 分钟，淘米 2 分钟，用电饭锅煮饭 20 分钟。妈妈做完这些事情至少需要()分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "22",
+        "solution":  "【分析】用电饭锅煮饭时可同时杀鱼，洗鱼，烧鱼，只要总时长不超过 20 分钟.\n【详解】杀鱼洗鱼 8 分钟，烧鱼 10 分钟，共 18 分钟，小于煮饭 20 分钟，可安排在煮饭期间。先淘米 2 分钟，再煮饭 20 分钟，同时完成杀鱼洗鱼和烧鱼。总时间:2+20=22(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·湖南湘潭·期末"
+    },
+    {
+        "id":  "330",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "四(1)班和四(2)班举行100米赛跑，采取三局两胜制。两个班各选出3位选手。(1)班的3位选手的成绩分别是A:17秒,B:16 秒,C:20 秒。(2)班的 3 位选手的成绩分别是甲:15秒，乙:18秒，丙:23秒。如果(1)班的选手比赛顺序是 A→B→C ,那么(2)班选手以( )→( )→( )的比赛顺序才能获胜。",
+        "options":  [
+
+                    ],
+        "answer":  "甲;丙;乙",
+        "solution":  "【分析】100米赛跑用时越短成绩越好。要想两胜一负，可以采用“田忌赛马”策略。\n【详解】四(1)班顺序是A(17秒)→B(16秒)→C(20秒)。四(2)班若按甲(15秒)→丙(23秒)→乙(18秒)出场，则甲对A获胜，丙对B失败，乙对C获胜，正好两胜一负。所以四(2)班应按甲→丙→乙的顺序出场。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "21-22 四年级上·陕西渭南·期末"
+    },
+    {
+        "id":  "331",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "聪聪和明明进行扑克牌点数大小比赛，三局两胜制。聪聪的牌是 3,5,7，明明的牌是 4,6,8。明明先出 8，聪聪若想要赢，必须出( )。",
+        "options":  [
+
+                    ],
+        "answer":  "3",
+        "solution":  "【分析】同样用\"以小博大\"的田忌赛马策略：用最小的牌去输掉对方最大牌.\n【详解】明明出 8 时，聪聪若想整体获胜，应出 3(故意输这一局),保留 5,7 对付 6 和 4:7\u003e6,5\u003e4，两局都赢。若聪聪用 7 或 5 对 8，则后面无法保证两胜。因此必须出 3。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "20-21 四年级上·浙江嘉兴·期末"
+    },
+    {
+        "id":  "332",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "下列事件中，小明进行了合理安排或合理对策的是( ),( ).\n1 小明放学后，在煮米饭的 40 分钟里洗菜，切菜，炒菜。 2 一个锅每次可同时烙 2 张饼，小明烙 3 张饼，先烙完两张饼，再烙第三张。 3 小明和小红玩牌比大小，小明有 3,5,8，小红有 9,7,4。小明请小红先出牌，小红出 9，小明出 3;小红出7，小明出 8;小红出 4，小明出 5。",
+        "options":  [
+
+                    ],
+        "answer":  "①;③",
+        "solution":  "【分析】与第 23 题类似，考察统筹安排和田忌赛马策略.\n【详解】①利用煮米饭 40 分钟的长时间同时完成洗菜，切菜，炒菜，时间安排合理;②先完全烙完两张再烙第三张，锅有空闲，不是最优;③出牌时用 3 对 9,8 对 7,5 对 4，两胜一负，是合理对策。所以是①,③。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "20-21 四年级上·浙江湖州·期末"
+    },
+    {
+        "id":  "333",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个锅里最多可以同时烙2张饼。每张饼的正反面都要烙，烙一面要2分钟，烙 5 张饼至少要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "10",
+        "solution":  "【分析】5 张饼，两面都烙，容量 2，使用\"2+3 分批轮换\"的最优策略.\n【详解】先烙前 2 张饼，双面共需 2×2=4(分钟)。剩下 3 张 a,b,c:第一次烙 a,b 正面 2 分钟;第二次烙 a 反面和 c 正面 2 分钟;第三次烙 b,c 反面 2 分钟，共 3×2=6 分钟。总时间:4+6=10(分钟)。也可用公式:5×2÷2×2=5×2=10(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "21-22 四年级上·浙江台州·期末"
+    },
+    {
+        "id":  "334",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一只平底锅上最多只能同时煎两条鱼，用它来煎一条鱼需要 6 分钟(正反两面各 3 分钟)。那么，煎熟 5 条鱼至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】一次最多煎2条鱼，正反都要煎，要使时间最少，要保证每次锅中尽量有2条鱼，不让锅空闲。可以给 5 条鱼编号 1,2,3,4,5，列举最优煎法.\n【详解】第一次:1 正,2 正;第二次:1 反,2 反;第三次:3 正,4 正;第四次:3 反,5 正;第五次:4 反,5 反。共煎 5 次，每次 3 分钟，共 5×3=15(分钟),所以煎熟 5 条鱼至少需要 15 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "21-22 四年级上·贵州遵义·期末"
+    },
+    {
+        "id":  "335",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "煮稀饭约 30 分钟，炒菜约 10 分钟，洗菜约 15 分钟，如果由你完成，最少要用( )分钟才能做完这些事情。",
+        "options":  [
+
+                    ],
+        "answer":  "30",
+        "solution":  "【分析】要使时间最少，应在煮稀饭时同时完成洗菜和炒菜.\n【详解】洗菜和炒菜共需 15+10=25(分钟),25\u003c30，可以安排在煮稀饭的 30 分钟内完成。总用时由煮稀饭决定，为 30 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "18-19 四年级上·湖南永州·期末"
+    },
+    {
+        "id":  "336",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用一只平底锅烙饼，每次能同时烙两张饼，每面烙2分钟，如果要烙9张饼，两面都要烙，那么最少要用( )分钟才能烙好。",
+        "options":  [
+
+                    ],
+        "answer":  "18",
+        "solution":  "【分析】两面都要烙，每面 2 分钟，一次最多烙 2 张，可先批量烙前 6 张，再用轮换法烙剩余 3 张.\n【详解】前 6 张饼：每次 2 张饼，两面共需 2×2=4(分钟)一批,3 批共 4×3=12(分钟)。剩余 3 张 a,b,c:第 1 次烙 a,b 正面 2 分钟;第 2 次烙 a 反面,c 正面 2 分钟;第 3 次烙 b,c 反面 2 分钟，共 3×2=6(分钟)。总时间:12+6=18(分钟)。也可用公式:9×2÷2×2=18。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "21-22 四年级上·贵州黔东南·期末"
+    },
+    {
+        "id":  "337",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小红在烤炉上烤面包，每次最多只能烤2片，两面都要烤，每面1分钟，烤好3 片面包需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "3",
+        "solution":  "【分析】这是烙饼问题：每片 2 面，每面 1 分钟，一次 2 片.\n【详解】用公式：总时间 = 片数*2 / 每次容量×每面时间 = 3×2÷2×1 = 3×1=3(分钟)。所以烤好 3 片面包需要 3 分钟。",
+        "knowledge_points":  [
+                                 "表内乘除混合运算",
+                                 "优化问题"
+                             ],
+        "source":  "20-21四年级上·陕西渭南·期末"
+    },
+    {
+        "id":  "338",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "体育课上，四(1)班和四(2)班各选出 3 名学生参加跳远比赛，三局两胜。四(1)班 3 名学生的跳远成绩是：王伟 140 厘米，李建126 厘米，张波 134 厘米，四(2)班 3 名学生的跳远成绩是：丁杰124厘米，李乐138厘米，秦凯130厘米。比赛中每名学生都正常发挥，四(1)班3名学生的比赛顺序是王伟 张波 李建。四(2)班的 3 名学生按( )→( )→( )的比赛顺序才能战胜四(1)班。",
+        "options":  [
+
+                    ],
+        "answer":  "丁杰;李乐;秦凯",
+        "solution":  "【分析】要战胜对方，采用田忌赛马策略：本方第三名对对方第一名，本方第一名对对方第二名，本方第二名对对方第三名.\n【详解】四(1)班顺序：王伟140，张波134，李建126。四(2)班成绩：丁杰124，李乐138，秦凯130。安排顺序为丁杰→李乐→秦凯，与四(1)班对位:\n第1局：丁杰124 对 王伟140(输);第2局：李乐138 对 张波134(赢);第3局：秦凯130 对 李建126(赢).\n三局两胜，四(2)班获胜。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "21-22四年级上·陕西渭南·期末"
+    },
+    {
+        "id":  "339",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈准备一顿午饭需要做下面的几件事情: 1 洗菜5分钟, 2 择菜7分钟, 3淘米 2 分钟, 4 用电饭锅蒸米饭 20 分钟, 5 切菜 2 分钟,6 炒菜 5 分钟。妈妈准备一顿午饭至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "22",
+        "solution":  "【分析】淘米后蒸米饭，在蒸米饭时同时完成洗菜，择菜，切菜，炒菜.\n【详解】洗菜，择菜，切菜，炒菜共需 5+7+2+5=19(分钟)\u003c蒸饭 20 分钟，可完全嵌入蒸饭过程。先淘米 2 分钟，再蒸饭 20 分钟，最后盛饭即可。总时间:2+20=22(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "21-22四年级上·陕西渭南·期末"
+    },
+    {
+        "id":  "340",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小芳星期天帮妈妈炒蛋，要做七项工作，敲蛋 1 分钟，搅蛋 3 分钟，切葱 2 分钟，洗锅2分钟，烧热锅2分钟，烧热油4分钟，炒蛋4分钟，完成这些工作最少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "12",
+        "solution":  "【分析】要减少时间，需把能同时完成的工作重叠安排，如烧锅，烧油时进行敲蛋，搅蛋，切葱.\n【详解】合理安排为：先洗锅 2 分钟，再烧热锅(2 分钟)和烧热油(4 分钟),在烧锅和烧油的 4 分钟里完成敲蛋 1 分钟，搅蛋 3 分钟，切葱 2 分钟(共 6 分钟，部分重叠),最后炒蛋 4 分钟。总体可以压缩到约 12 分钟，题解给出总时间 2+2+4+4=12(分钟),说明洗锅与后续加热，准备工作交叉安排。",
+        "knowledge_points":  [
+                                 "优化问题",
+                                 "时，分，秒有关的计算"
+                             ],
+        "source":  "20-21四年级上·陕西西安·期末"
+    },
+    {
+        "id":  "341",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在学校体检中，有四位同学要去量身高和验视力，每项检查都要 3 分钟，他们至少要用( )分钟能做完这些检查。",
+        "options":  [
+
+                    ],
+        "answer":  "12",
+        "solution":  "【分析】两项检查可同时进行，每次安排两位同学分别做不同项目，轮换交叉，使两台设备始终在工作.\n【详解】设四人 A,B,C,D:\n第1个3分钟:A量身高,B验视力;第2个3分钟:B量身高,C验视力;第3个3分钟:C量身高,D验视力;第4个3分钟:D量身高,A验视力。总时间 3×4=12(分钟),四人都完成两项检查。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "21-22四年级上·辽宁抚顺·期末"
+    },
+    {
+        "id":  "342",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "小芳在家里烙饼，锅里每次最多只能烙2张饼，每张饼两面都要烙，每面要3分钟，小芳烙 5 张这样的饼至少要用( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】典型烙饼问题，用公式：时间 = 饼数*2/每次容量×每面时间.\n【详解】烙 5 张饼:5×2÷2×3 = 5×3=15(分钟)。也可理解为 5 张饼共 10 面，每次 2 面，共 5 次，每次 3 分钟，总 15 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "21-22四年级上·福建福州·期末"
+    },
+    {
+        "id":  "343",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小红感冒了，吃完药后要赶快休息。小红找杯子倒开水1分钟，等开水变温6分钟，找感冒药1分钟，量体温4分钟，至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "7",
+        "solution":  "【分析】等开水变温 6 分钟时可同时进行找药和量体温，尽量重叠时间.\n【详解】先找杯子倒开水 1 分钟，然后等水变温 6 分钟，在这 6 分钟中完成找感冒药(1 分钟)和量体温(4 分钟),不增加额外时间。总时间:1+6=7(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "21-22四年级上·安徽芜湖·期末"
+    },
+    {
+        "id":  "344",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈每天早上起床后要做以下事情：刷牙(3 分钟),洗脸(2 分钟),煮鸡蛋(8分钟),烤面包(2分钟),吃早餐(6分钟)。妈妈至少用( )分钟才能完成这些事情。",
+        "options":  [
+
+                    ],
+        "answer":  "14",
+        "solution":  "【分析】煮鸡蛋时间最长，可在此期间安排刷牙，洗脸，烤面包.\n【详解】在煮鸡蛋的 8 分钟中可以完成刷牙 3 分钟，洗脸 2 分钟，烤面包 2 分钟，共 7 分钟，均可嵌入。最后吃早餐 6 分钟。总时间：煮鸡蛋 8 分钟 + 吃早餐 6 分钟 = 14 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "19-20 四年级上·天津东丽·期末"
+    },
+    {
+        "id":  "345",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "3名同学排队打水，只有一个水龙头，甲同学需要3分钟，乙同学需要2分钟，丙同学需要5分钟。最后一人打完水时，三人所用时间总和最少是( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "17",
+        "solution":  "【分析】要使三人所用时间总和最少，应按用时从小到大排队：先用时短的，后用时长的。\n【详解】最优顺序是乙、甲、丙。乙用2分钟完成，甲用2+3=5(分钟)完成，丙用2+3+5=10(分钟)完成。三人所用时间总和是2+5+10=17(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "20-21四年级上·宁夏石嘴山·期末"
+    },
+    {
+        "id":  "346",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "三位同学复印材料，甲要7分钟，乙要13分钟，丙要5分钟，若要三人等待的总时间最短，三人复印的先后顺序是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "丙;甲;乙",
+        "solution":  "【分析】同样让用时最短的先复印，按时间从小到大排序可使等待总时间最少。\n【详解】用时分别是：丙5分钟，甲7分钟，乙13分钟，5<7<13，所以按丙→甲→乙的顺序复印最优。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "21-22 四年级上·天津·期末"
+    },
+    {
+        "id":  "347",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "星期日笑笑做家务：用洗衣机洗衣服要 30 分钟，扫地要 5 分钟，拖地板要 15 分钟，擦家具要 6 分钟，晾衣服要 4 分钟。经过合理安排，她做完这些事至少要花( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "34",
+        "solution":  "【分析】洗衣机工作时可以同时扫地，拖地板和擦家具，最后晾衣服.\n【详解】扫地，拖地板，擦家具共需 5+15+6=26(分钟)\u003c洗衣 30 分钟，可全部安排在洗衣机运行期间。先启动洗衣机同时做这三项，洗完后再晾衣服 4 分钟。总时间:30+4=34(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "21-22 四年级上·海南省直辖县级单位·期末"
+    },
+    {
+        "id":  "348",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "有礼貌的小红给李叔叔沏茶，接水要1分钟，洗茶杯要1分钟，烧开水要5分钟，沏茶要1分钟。她至少需要( )分钟才能沏好茶。",
+        "options":  [
+
+                    ],
+        "answer":  "7",
+        "solution":  "【分析】要缩短时间，应在烧水时同时洗茶杯.\n【详解】先接水 1 分钟，然后烧开水 5 分钟，在这 5 分钟里可以洗茶杯 1 分钟，最后沏茶 1 分钟。总用时:1+5+1=7(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "21-22 四年级上·甘肃陇南·期末"
+    },
+    {
+        "id":  "349",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用平底锅煎一条鱼需要 6 分钟(正反面各 3 分钟),已知锅里可同时煎两条鱼，那么煎 5 条鱼至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】先用轮换法煎前 3 条，再煎后 2 条，或直接用烙饼结论：时间 = 单面时间×条数.\n【详解】按分析：先 3 条按轮换策略煎需 3 次，每次 3 分钟，共 9 分钟;再 2 条鱼一起煎两面需 2×3=6 分钟，总时间 9+6=15 分钟。也等价于 3 分钟*5 条 =15 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "21-22 四年级上·甘肃陇南·期末"
+    },
+    {
+        "id":  "350",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "三位同学要去验视力，量体重，每项检查都要4分钟，他们三人做完这些检查至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "12",
+        "solution":  "【分析】两项检查可并行进行，安排两位同学分别做不同项目，第三人等待，轮换三次.\n【详解】设三人 1,2,3:\n第1个4分钟:1 测视力,2 量体重;第2个4分钟:1 量体重,3 测视力;第3个4分钟:2 测视力,3 量体重。三人都完成两项检查，共需 3×4=12 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·福建福州·期末"
+    },
+    {
+        "id":  "351",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "丹丹周六早上这样安排：起床穿衣要 3 分钟，刷牙洗脸要 3 分钟，整理房间要5分钟，听音乐要10分钟，吃早点要6分钟。丹丹做完这些事至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "17",
+        "solution":  "【分析】听音乐 10 分钟的过程可以与刷牙洗脸、整理房间、吃早点部分重叠。\n【详解】先起床穿衣 3 分钟。听音乐 10 分钟时，同时完成刷牙洗脸 3 分钟和整理房间 5 分钟，共 8 分钟，还可以再吃早点 2 分钟，吃早点还剩 4 分钟。总时间为起床 3 分钟 + 听音乐 10 分钟 + 吃早点剩余 4 分钟 = 17(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·甘肃庆阳·期末"
+    },
+    {
+        "id":  "352",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一只平底锅上只能烤两条鱼，用它烤一条鱼需要4分钟(正，反面各需2分钟),那么烤 7 条鱼至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "14",
+        "solution":  "【分析】当锅里始终有 2 条鱼时，最少时间 = 单面时间×条数.\n【详解】每条鱼 2 面，单面 2 分钟。用 2 条容量的锅最优轮换时，总时间=2×7÷2×2=7×2=14(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·广东汕头·期末"
+    },
+    {
+        "id":  "353",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用一只平底锅煎饼，每次只能同时煎两张饼，如果煎一张饼需要2分钟(正反面各需 1 分钟),那么煎 3 张饼至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "3",
+        "solution":  "【分析】每张饼正、反面各需要 1 分钟，3 张饼用轮换法安排最省时。\n【详解】编号 1、2、3：第一次煎 1、2 的正面 1 分钟；第二次煎 1 的反面和 3 的正面 1 分钟；第三次煎 2、3 的反面 1 分钟。共需 3 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "21-22四年级上·新疆省直辖县级单位·期末"
+    },
+    {
+        "id":  "354",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果锅里每次最多可以放 2 条鱼，每条鱼需要煎两面，每煎一面需要 3 分钟，要煎 5 条鱼最少需要用( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】标准烙饼公式：时间 = 条数*2/容量×单面时间.\n【详解】5 条鱼，两面都煎，每面 3 分钟，一次 2 条:5×2÷2×3=5×3=15(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "21-22 四年级上·新疆喀什·期末"
+    },
+    {
+        "id":  "355",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一只平底锅每次最多只能烙 2 张饼，两面都要烙，每面 3 分钟，烙 9 张饼至少需要( )分钟,15 分钟最多能烙( )张饼。",
+        "options":  [
+
+                    ],
+        "answer":  "27;5",
+        "solution":  "【分析】烙 9 张饼用公式求总时间，用给定时间反推最多张数.\n【详解】烙 9 张饼：时间=3×9=27(分钟)。15 分钟内最多烙的张数：每张饼\"占用\"单面时间 3 分钟，按最优轮换 15÷3=5(张)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "20-21 四年级上·福建莆田·期末"
+    },
+    {
+        "id":  "356",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈用平底锅煎鸡蛋，每锅可以煎两个蛋，每个面煎 1 分钟，两面都要煎。煎熟 9 个蛋需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "9",
+        "solution":  "【分析】每蛋 2 面，每面 1 分钟，容量 2，用公式：时间 = 个数×单面时间.\n【详解】在最优轮换下，锅中始终有 2 个蛋，总时间=1×9=9(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·江西宜春·期末"
+    },
+    {
+        "id":  "357",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "明明每天晚上睡觉前都要背诵英语单词 8 分钟，刷牙 3 分钟，热牛奶 10 分钟，喝牛奶 2 分钟，明明做完这些事最少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "13",
+        "solution":  "【分析】热牛奶时间最长，可以在热牛奶的同时背单词和刷牙，以减少总时长。\n【详解】先热牛奶 10 分钟，在热牛奶的同时背单词 8 分钟、刷牙 3 分钟，这两件事共 11 分钟，所以热牛奶结束后还差 1 分钟完成刷牙。然后再喝牛奶 2 分钟。最少需要 10 + 1 + 2 = 13(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·新疆克孜勒苏·期末"
+    },
+    {
+        "id":  "358",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "假期的一天，小红想帮妈妈做家务：用洗衣机洗衣服(20分钟),扫地(6分钟),擦家具(10分钟),晾衣服(5分钟)。经过合理安排，她做完这些事情至少要用( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "25",
+        "solution":  "【分析】洗衣机工作时可以同时扫地和擦家具，最后晾衣服.\n【详解】扫地和擦家具共需 6+10=16(分钟)\u003c洗衣服 20 分钟，可全部安排在洗衣机工作期间。最后晾衣服 5 分钟。总时间:20+5=25(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "19-20 四年级上·全国·单元测试"
+    },
+    {
+        "id":  "359",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小红每天起床后要做完下面这些事情：收拾房间4分钟，听广播10分钟，洗漱6分钟，吃早餐12分钟。她每天做完这些事情，最少要()分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "22",
+        "solution":  "【分析】听广播 10 分钟过程中可同时收拾房间和洗漱.\n【详解】收拾房间 4 分钟和洗漱 6 分钟共 10 分钟，可全部安排在听广播的 10 分钟里，最后吃早餐 12 分钟。总时间:10+12=22(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23四年级上·陕西宝鸡·期末"
+    },
+    {
+        "id":  "360",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "煎一条鱼要6分钟(正反两面各 3 分钟),一个平底锅一次只能同时煎 2 条鱼。妈妈要煎 3 条鱼，最少要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "9",
+        "solution":  "【分析】容量 2，单面 3 分钟,3 条鱼用轮换法最优.\n【详解】第 1 次:1,2 正面,3 分钟;第 2 次:1 反面,3 正面,3 分钟，此时 1 熟;第 3 次:2 反面,3 反面,3 分钟。总时间:3+3+3=9(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·陕西宝鸡·期末"
+    },
+    {
+        "id":  "361",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个平底锅每次只能烙两张饼，两面都要烙，每面要烙 2 分钟，3 张饼最快( )分钟烙完。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】典型的 3 张饼轮换法，始终保持锅里有 2 张饼在烙。\n【详解】编号 1、2、3：第一次烙 1、2 的正面，2 分钟；第二次烙 1 的反面和 3 的正面，2 分钟；第三次烙 2、3 的反面，2 分钟。总时间 2+2+2=6(分钟)，为最快时间。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·广东佛山·期末"
+    },
+    {
+        "id":  "362",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明帮妈妈做家务，需要做：用洗衣机洗衣服(20分钟),扫地(10分钟),整理书桌(10分钟),晾衣服(5分钟)。如果小明的时间安排合理，做完这些家务最少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "25",
+        "solution":  "【分析】洗衣机洗衣时同时完成扫地和整理书桌，最后晾衣服.\n【详解】扫地 10 分钟，整理书桌 10 分钟共 20 分钟，可全部安排在洗衣机 20 分钟工作中，最后晾衣服 5 分钟。总时间:20+5=25(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·吉林·期末"
+    },
+    {
+        "id":  "363",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈做午饭的过程及时间：淘米(2 分钟),用电饭煲蒸米饭(30 分钟),择菜(3 分钟),洗菜(5 分钟),炒热菜(8 分钟),拌凉菜(5 分钟),盛米饭(1 分钟)。妈妈从开始做饭到吃饭至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "33",
+        "solution":  "【分析】蒸米饭时可以同时完成择菜，洗菜，炒热菜，拌凉菜.\n【详解】择菜，洗菜，炒热菜，拌凉菜共 3+5+8+5=21(分钟)\u003c蒸饭 30 分钟，可完全嵌入其间。总时间 = 淘米 2 分钟 + 蒸饭 30 分钟 + 盛饭 1 分钟 = 33 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·福建厦门·期末"
+    },
+    {
+        "id":  "364",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用一只平底锅烙饼，每次只能烙2张饼，两面都要烙，烙好一面要3分钟，烙好6张饼最少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "18",
+        "solution":  "【分析】6 张饼，两面都烙，容量 2，用烙饼总时间公式.\n【详解】总时间 = 饼数*2/每次容量×每面时间 = 6×2÷2×3=6×3=18(分钟)。也可以按轮换分组：每 2 张需要 3×2=6 分钟，共 3 组,6×3=18 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "21-22四年级上·河南三门峡·期末"
+    },
+    {
+        "id":  "365",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小馋猫烤鱼，鱼的两面都要烤，每烤一面需要烤 5 分钟，小猫每次最多只能烤两条鱼，现在需要烤三条鱼，至少要用( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】3 条鱼，两面各烤 5 分钟，容量是 2，用轮换法最省时。\n【详解】设鱼为甲、乙、丙：第一轮烤甲、乙的第一面，5 分钟；第二轮烤甲的第二面和丙的第一面，5 分钟；第三轮烤乙、丙的第二面，5 分钟。3 条鱼都烤熟了，共需 5×3=15(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "21-22四年级上·河南郑州·期末"
+    },
+    {
+        "id":  "366",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个锅每次最多只能烙2张饼，两面都要烙，每面3分钟。如果要烙4张饼，至少要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "12",
+        "solution":  "【分析】烙 4 张饼，容量 2，两面各 3 分钟，用总时间公式即可.\n【详解】总时间=4×2÷2×3=4×3=12(分钟),即至少要 12 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·新疆克孜勒苏·期末"
+    },
+    {
+        "id":  "367",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一只平底锅每次只能同时煎两个荷包蛋，每面煎 2 分钟，两面都要煎，妈妈这样煎 3 个荷包蛋至少要用( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】3 个荷包蛋，两面都煎，每面 2 分钟，容量 2，用轮换法.\n【详解】总时间 = 个数×单面时间 = 3×2=6(分钟)。也可具体排程为：第 1 次 1,2 正面,2 分钟;第 2 次 1 反面,3 正面,2 分钟;第 3 次 2,3 反面,2 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·江西抚州·期末"
+    },
+    {
+        "id":  "368",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一年级游园乐考，两位老师对三位同学进行测试，每位同学要完成书包整理和系鞋带，每项测试都需要用 3 分钟，两位老师完成这次测试至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "9",
+        "solution":  "【分析】两位老师对应两项任务，可同时测试两名同学，轮换 3 轮即可完成.\n【详解】设三位同学为 1,2,3:\n第1个3分钟:1 整理书包,2 系鞋带;第2个3分钟:2 整理书包,3 系鞋带;第3个3分钟:3 整理书包,1 系鞋带。三轮共 3×3=9(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·浙江湖州·期末"
+    },
+    {
+        "id":  "369",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈用平底锅煎蛋，每次能同时煎两个鸡蛋，两面都要煎，每煎一面需3分钟。丫丫家有5个人，妈妈要煎5个荷包蛋，最少要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】5 个蛋，两面都煎，每面 3 分钟，容量 2，用\"2+3\"分批轮换.\n【详解】先煎 2 个蛋，两面共需 3×2=6(分钟);剩 3 个蛋，用轮换策略:3×3=9(分钟)。总时间:6+9=15(分钟)。也可以用公式:5×2÷2×3=5×3=15(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·陕西西安·期末"
+    },
+    {
+        "id":  "370",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一口平底锅每次能烙 4 张饼，烙好一面需要 3 分钟，两面都要烙，要烙好 10 张饼至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】把问题看成\"烙饼问题\"，总时间 = 饼数*2 / 每锅可烙的张数×烙一面的时间.\n【详解】10×2÷4×3 = 20÷4×3 = 5×3 = 15(分钟),所以至少需要 15 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23四年级上·贵州遵义·期末"
+    },
+    {
+        "id":  "371",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈下班回家做饭的过程及时间：淘米(2 分钟)，蒸米饭(20 分钟)，炒菜(10 分钟)，盛饭菜(2 分钟)，妈妈做这顿饭至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "24",
+        "solution":  "【分析】蒸米饭时可以同时炒菜，蒸米饭 20 分钟大于炒菜 10 分钟，因此这两项可以重叠进行。\n【详解】总时间 = 淘米 2 分钟 + 蒸米饭 20 分钟 + 盛饭菜 2 分钟 = 2 + 20 + 2 = 24(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·甘肃庆阳·期末"
+    },
+    {
+        "id":  "372",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小强骑在马背上赶马过河，共有甲，乙，丙，丁4匹马，甲马过河需要2分钟，乙马过河需要3分钟，丙马过河需要5分钟，丁马过河需要6 分钟，每次只赶两匹马过河，要把 4 匹马都赶到对岸去，最少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "17",
+        "solution":  "【分析】两匹马一起过河，其时间由较慢的那匹决定;应让时间短的多次往返，使总时间最少.\n【详解】先骑甲赶乙过河，时间 3 分钟，再骑甲返回 2 分钟，共 5 分钟;再骑丙赶丁过河，时间 6 分钟，再骑乙返回 3 分钟，共 9 分钟;最后再骑甲赶乙一起过河 3 分钟。总时间:5 + 9 + 3 = 17(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·贵州铜仁·期末"
+    },
+    {
+        "id":  "373",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一位客人到早餐店吃早餐，告诉老板说加5个鸡蛋。早餐店为了让客人吃到新鲜的鸡蛋，需要现给客人煎蛋。一口平底锅可同时煎2个蛋，每个鸡蛋两面都要煎，煎好每一面需要2分钟，客人最快()分钟可以吃到鸡蛋。",
+        "options":  [
+
+                    ],
+        "answer":  "10",
+        "solution":  "【分析】先把 5 个鸡蛋分成\"先完全煎熟 2 个 + 剩下 3 个用轮换法煎熟\".\n【详解】前 2 个鸡蛋：两面都煎，需要 2×2 = 4(分钟);剩下 3 个，按 3 个鸡蛋轮换策略，需要 3×2 = 6(分钟)。总时间 4 + 6 = 10(分钟),客人最快 10 分钟可以吃到鸡蛋。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·贵州铜仁·期末"
+    },
+    {
+        "id":  "374",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小琴每天晚上睡觉前要背诵英语单词6分钟，烧开水9分钟，用烧好的开水烫牛奶3分钟，整理书包2分钟。如果小琴合理安排，最少用( )分钟就可以做完全部事情。",
+        "options":  [
+
+                    ],
+        "answer":  "12",
+        "solution":  "【分析】烧开水时间最长，可在烧水时同时背单词和整理书包.\n【详解】先烧开水 9 分钟，同时完成背诵英语单词 6 分钟和整理书包 2 分钟;再用开水烫牛奶 3 分钟。总时间:9 + 3 = 12(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·安徽芜湖·期末"
+    },
+    {
+        "id":  "375",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明在火炉上烤鱼，鱼的两面都要烤，每次最多烤2条，烤一面需要3分钟，则烤 9 条鱼至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "27",
+        "solution":  "【分析】把烤鱼看作烙饼问题，只要始终保持火炉中有 2 条鱼在烤即可达到最优.\n【详解】最少时间 = 烤一面所需时间×条数 = 3×9 = 27(分钟),所以至少需要 27 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·江西抚州·期末"
+    },
+    {
+        "id":  "376",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一张饼有 2 面，烙熟一面需要花 2 分钟，两面都要烙，每次最多烙两张，烙熟3 张饼至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】仍是烙饼问题，锅内最多 2 张，合理轮换，使锅不空闲.\n【详解】最少时间 = 烙一面时间×张数 = 2×3 = 6(分钟),所以烙熟 3 张饼至少需要 6 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·江西赣州·期末"
+    },
+    {
+        "id":  "377",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "欢欢和乐乐进行扑克牌比大小游戏，三局两胜制。欢欢的牌是 8,6,3，乐乐的牌是 10,7,5，乐乐先出牌 10，欢欢若想赢，必须出()。",
+        "options":  [
+
+                    ],
+        "answer":  "3",
+        "solution":  "【分析】按\"田忌赛马\"思想，先用自己最小的牌去对对手最大牌，故意输掉一局，保住后面两局胜利.\n【详解】乐乐先出 10 时，欢欢应出 3，剩下 8 和 6 对乐乐的 7 和 5，可以赢两局，所以欢欢必须出 3。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23四年级上·浙江宁波·期末"
+    },
+    {
+        "id":  "378",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "李阿姨要复印 5 张资料，正反面都要复印，复印机一次可以放 2 张，复印一次要 6 秒，最少要复印( )秒。",
+        "options":  [
+
+                    ],
+        "answer":  "30",
+        "solution":  "【分析】把复印看作烙饼问题，始终保证复印机一次放满 2 张，时间最短.\n【详解】总时间 = 复印一面时间×张数 = 6×5 = 30(秒),所以至少要复印 30 秒。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "21-22 四年级上·浙江宁波·期末"
+    },
+    {
+        "id":  "379",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小文准备冲一杯橙汁，洗杯子和汤勺各用1分钟，烧开水4分钟，取出橙汁粉冲泡 2 分钟，至少需要( )分钟，小文才能喝到橙汁。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】烧开水时间最长，可以在烧水的 4 分钟里洗杯子和汤勺.\n【详解】先烧水 4 分钟，同时完成洗杯子和汤勺;再取橙汁粉冲泡 2 分钟。总时间:4 + 2 = 6(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·广东汕头·期末"
+    },
+    {
+        "id":  "380",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小思周末在家做家务，收拾脏衣服放进洗衣机 1 分钟，洗衣机快洗 15 分钟，晒衣服 5 分钟，洗米 1 分钟，煮饭 12 分钟。她做完这些家务至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "21",
+        "solution":  "【分析】洗衣机洗衣 15 分钟时，可以同时洗米和煮饭，只要不超过 15 分钟即可.\n【详解】洗米和煮饭共 1 + 12 = 13(分钟)\u003c15 分钟，可嵌入洗衣过程。总时间 = 洗衣 15 分钟 + 收拾衣服 1 分钟 + 晒衣服 5 分钟 = 15 + 1 + 5 = 21(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·广西柳州·期末"
+    },
+    {
+        "id":  "381",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "张叔叔每天早晨需要完成下列几件事才能出门上班：起床 3 分钟，洗漱 5 分钟，电饭煲煮粥 15 分钟，整理公文包 3 分钟，喝粥 5 分钟。他完成这些事至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "23",
+        "solution":  "【分析】煮粥 15 分钟时，可以同时洗漱和整理公文包，煮好粥后再喝粥。\n【详解】先起床 3 分钟，再开始煮粥 15 分钟。在煮粥的同时可以完成洗漱 5 分钟和整理公文包 3 分钟。粥煮好后再喝粥 5 分钟。总时间是 3 + 15 + 5 = 23(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "21-22 四年级上·浙江宁波·期末"
+    },
+    {
+        "id":  "382",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "平底锅中每次只能煎两条鱼，每条鱼煎两面，一面要煎3分钟。煎4条鱼需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "12",
+        "solution":  "【分析】按烙饼公式：总时间 = 条数*2 / 每次容量×单面时间.\n【详解】4×2÷2×3 = 8÷2×3 = 4×3 = 12(分钟),所以需要 12 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23四年级上·广西河池·期末"
+    },
+    {
+        "id":  "383",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "炸一根油条需要 5分钟，一只锅头里一次只能炸 2 根。炸 5 根油条需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】前 4 根可以两锅两根一起炸，最后剩 1 根再炸一次.\n【详解】前 4 根:2 锅需要 2 轮，共 10 分钟;剩 1 根再炸一轮 5 分钟。总时间:10 + 5 = 15(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·广东汕头·期末"
+    },
+    {
+        "id":  "384",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一只平底锅每次最多能煎 2 个饼，每面需要煎 3 分钟，煎 7 个饼最少需要()分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "21",
+        "solution":  "【分析】把 7 个饼分为\"先煎熟 4 个 + 再用轮换法煎 3 个\".\n【详解】前 4 个饼：两面都煎,2 个一组，需要 2×3×2 = 12(分钟);剩 3 个饼用三张轮换法，需 3×3 = 9(分钟)。总时间:12 + 9 = 21(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "21-22 四年级上·浙江绍兴·期末"
+    },
+    {
+        "id":  "385",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "李老师要复印 7 张获奖证书，正反面都要复印，如果复印机一次最多放2张证书，她最少要复印( )次。",
+        "options":  [
+
+                    ],
+        "answer":  "7",
+        "solution":  "【分析】可先按 4 张 + 3 张分步:4 张两面都印，再对 3 张使用\"3 张轮换\"的复印策略.\n【详解】前 4 张：一共 4×2 = 8 个面，每次印 2 个面，需要 8÷2 = 4(次);剩 3 张按 3 张轮换，要 3(次)。总次数:4 + 3 = 7(次)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·浙江宁波·期末"
+    },
+    {
+        "id":  "386",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "本学期我们学习了《数学广角--优化》,你也能用优化的策略计算出\"497+498+498+498+499\"的结果吗。我算的结果是( ),我是这样算的: 。",
+        "options":  [
+
+                    ],
+        "answer":  "2490",
+        "solution":  "【分析】通过调整加数，把表达式化成若干个相同的数相加，转化为乘法.\n【详解】497 + 499 = 498 + 498，所以 497+498+498+498+499 = 498+498+498+498+498 = 498×5 = 2490。",
+        "knowledge_points":  [
+                                 "两，三位数的加减混合运算",
+                                 "优化问题"
+                             ],
+        "source":  "20-21 四年级上·广西河池·期末"
+    },
+    {
+        "id":  "387",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "张杰做饭，淘米3分钟，洗菜10分钟，切菜8分钟，炒菜15分钟，电饭锅煮饭 30 分钟。至少需要( )分钟才能把饭菜做好。",
+        "options":  [
+
+                    ],
+        "answer":  "36",
+        "solution":  "【分析】煮饭 30 分钟时可以同时洗菜，切菜，炒菜，但这三项共 10+8+15 = 33 分钟，比 30 分钟长，需要分两段重叠.\n【详解】洗菜，切菜，炒菜共 33 分钟，煮饭 30 分钟，重叠后超出 3 分钟;再加淘米 3 分钟。总时间:30 + 3 + 3 = 36(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23四年级上·云南曲靖·期末"
+    },
+    {
+        "id":  "388",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "恬恬周末在家烙饼，锅里每次最多能烙两张饼，烙一张饼需要2分钟(正反面各需1分钟)。恬恬想给爸爸，妈妈，爷爷，奶奶和自己各烙一张饼，至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "5",
+        "solution":  "【分析】把烙饼问题转化为\"单面时间×张数\"，只要锅里尽量放满两张饼即可.\n【详解】一面 1 分钟,5 张饼共 5 个\"面时间\"，最少时间 = 1×5 = 5(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23四年级上·云南曲靖·期末"
+    },
+    {
+        "id":  "389",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "有 5 张文稿，正反面都要复印，复印一次最多只能同时放 2 张文稿，复印一次的时间是 20 秒，那么全部复印好至少需要( )秒。",
+        "options":  [
+
+                    ],
+        "answer":  "100",
+        "solution":  "【分析】与烙饼问题完全同构，始终保证复印机中有 2 张文稿即可最优.\n【详解】每张文稿两面，总面数 5×2 = 10，每次印 2 个面，需要 10÷2 = 5(次),每次 20 秒，共 5×20 = 100(秒)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "20-21 四年级上·浙江杭州·期末"
+    },
+    {
+        "id":  "390",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明给客人沏茶，接水 1 分钟，烧水 6 分钟，拿茶叶 1 分钟，沏茶 1 分钟。小明合理安排以上事情，最少要( )分钟使客人尽快喝茶。",
+        "options":  [
+
+                    ],
+        "answer":  "8",
+        "solution":  "【分析】烧水时间最长，可在烧水时完成拿茶叶这件事.\n【详解】先接水 1 分钟，再烧水 6 分钟(同时拿茶叶 1 分钟),最后沏茶 1 分钟。总时间:1 + 6 + 1 = 8(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "20-21 四年级上·浙江宁波·期末"
+    },
+    {
+        "id":  "391",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明饿了，但是妈妈洗米要2分钟，用电饭锅煮饭要20分钟，洗菜要3分钟，炒菜要 15 分钟，那么小明至少要等( )分钟才能吃上饭。",
+        "options":  [
+
+                    ],
+        "answer":  "22",
+        "solution":  "【分析】煮饭时可以同时洗菜和炒菜,3+15 = 18 分钟\u003c20 分钟，可以完全嵌入.\n【详解】总时间 = 洗米 2 分钟 + 煮饭 20 分钟 = 22(分钟),所以小明至少要等 22 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "21-22 四年级上·浙江温州·期末"
+    },
+    {
+        "id":  "392",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小王要复印 5 张图片，正反面都要复印，如果复印机里一次最多放2张图片，每印一次要 10 秒钟，复印 5 张图片最少需要花( )秒。",
+        "options":  [
+
+                    ],
+        "answer":  "50",
+        "solution":  "【分析】仍然是烙饼模型，保证每次都放满 2 张图片即可.\n【详解】最少时间 = 印一面时间×张数 = 10×5 = 50(秒)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "20-21 四年级上·浙江杭州·期末"
+    },
+    {
+        "id":  "393",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明帮妈妈蒸鸡蛋羹，前后花掉了 23 分钟。他是这样安排的:1 分钟打蛋,2 分钟切葱花,2 分钟搅蛋,2 分钟洗锅,6 分钟把水烧开,10 分钟蒸熟。如果你是小明，做完这些事情最少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "18",
+        "solution":  "【分析】洗锅和烧水必须按顺序，但烧水时可以同时打蛋，切葱花，搅蛋.\n【详解】先洗锅 2 分钟，再烧水 6 分钟，同时完成 1 分钟打蛋,2 分钟切葱花和 2 分钟搅蛋;最后蒸熟 10 分钟。总时间:2 + 6 + 10 = 18(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "20-21四年级上·浙江杭州·期末"
+    },
+    {
+        "id":  "394",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "理发店同时来了3位客人，如果只有1名理发师，按他们所要理的发型，甲需要 20 分钟，乙需要 10 分钟，丙需要 15 分钟，要使他们的等候时间总和最少，他们的顺序应该是( );( );( )，最少是( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "乙;丙;甲;80",
+        "solution":  "【分析】等候时间总和最少时，应按用时从小到大理发，也就是短的先理，长的后理。\n【详解】排序为乙(10 分钟)、丙(15 分钟)、甲(20 分钟)。乙理发时，3 人都在等待，等待时间是 10×3；丙理发时，2 人在等待，等待时间是 15×2；甲理发时，1 人在等待，等待时间是 20。总等待时间：10×3 + 15×2 + 20 = 30 + 30 + 20 = 80(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "20-21四年级上·湖南永州·期末"
+    },
+    {
+        "id":  "395",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈给客人泡茶，接水 1 分钟，烧水要 6 分钟，洗茶杯要 2 分，拿茶叶要 1 分钟，泡茶1分钟，妈妈合理安排以上事情，做完最少需要()分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "8",
+        "solution":  "【分析】烧水 6 分钟时，可以同时洗茶杯和拿茶叶.\n【详解】先接水 1 分钟，再烧水 6 分钟(内同时完成洗茶杯 2 分钟和拿茶叶 1 分钟),最后泡茶 1 分钟。总时间:1 + 6 + 1 = 8(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·云南楚雄·期末"
+    },
+    {
+        "id":  "396",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "聪聪，明明和亮亮三位同学去验视力，测肺活量这两项检查，每项检查都要 2分钟，他们最短要( )分钟才能全部做完这些检查。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】合理安排让两项检查同时有人做，三轮交叉完成.\n【详解】第 1 轮：聪聪验视力，明明测肺活量(2 分钟);第 2 轮：明明验视力，亮亮测肺活量(2 分钟);第 3 轮：亮亮验视力，聪聪测肺活量(2 分钟)。共 3 轮，总时间 3×2 = 6(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23四年级上·浙江绍兴·期末"
+    },
+    {
+        "id":  "397",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "文印员阿姨要复印材料，一次能同时印2张，两面都要印，每面要3秒，复印 3 张最少需要( )秒，复印 8 张最少需要( )秒。",
+        "options":  [
+
+                    ],
+        "answer":  "9;24",
+        "solution":  "【分析】仍用烙饼思路，保证每次印两张材料，使总时间最少.\n【详解】3 张材料：总面数 3×2 = 6，每次 2 面，需要 3 次，每次 3 秒，共 3×3 = 9(秒);8 张材料：总面数 16，每次 2 面，需要 8 次，每次 3 秒，共 8×3 = 24(秒)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23四年级上·河北石家庄·期末"
+    },
+    {
+        "id":  "398",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小勤做家务，准备洗衣1分钟，用洗衣机洗完衣服要35分钟，晾晒衣服3分钟，拖地15分钟，整理房间8分钟，擦窗户5分钟。做完这些家务最少要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "39",
+        "solution":  "【分析】洗衣机工作 35 分钟时，可以同时拖地，整理房间和擦窗户.\n【详解】拖地，整理房间，擦窗户共 15+8+5 = 28(分钟)\u003c35 分钟，可完全嵌入洗衣时间。总时间：洗衣 35 分钟 + 准备洗衣 1 分钟 + 晾晒衣服 3 分钟 = 35 + 1 + 3 = 39(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·广东佛山·期末"
+    },
+    {
+        "id":  "399",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "乐乐，明明两人玩比大小的游戏，采取三局两胜制。明明的三张牌分别是5,7,9，乐乐的三张牌分别是 4,6,8，如果乐乐想获胜，应该让( )(填\"乐乐\"或\"明明\")先出牌，并且乐乐要用( )对战明明的 9。",
+        "options":  [
+
+                    ],
+        "answer":  "明明;4",
+        "solution":  "【分析】要用\"田忌赛马\"策略，先让对方亮出最大牌，再用自己最小的牌去对.\n【详解】乐乐想获胜，应让明明先出牌。当明明出 9 时，乐乐用自己最小的 4 去对，让这一局输掉;再用 8 对 7,6 对 5，赢两局，所以答案是\"明明;4\"。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "21-22 四年级上·浙江杭州·期末"
+    },
+    {
+        "id":  "400",
+        "unit":  "第八章数学广角 4",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用一只平底锅烙饼，每次只能放2张饼，烙一面需要2分钟，烙三张饼至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】3 张饼用“3 张轮换法”，保证每次锅里有 2 张饼。\n【详解】第一次烙 a、b 的正面 2 分钟；第二次烙 a 的反面和 c 的正面 2 分钟；第三次烙 b、c 的反面 2 分钟，共 2+2+2 = 6(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "21-22四年级上·浙江绍兴·期末"
+    },
+    {
+        "id":  "401",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小华给客人沏茶，接水1分钟，烧水7分钟，洗茶杯2分钟，拿茶叶1分钟，沏茶 1 分钟，小华合理安排以上事情，最少要( )分钟使客人尽快喝茶。",
+        "options":  [
+
+                    ],
+        "answer":  "9",
+        "solution":  "【分析】烧水时间最长，应在烧水过程中完成洗茶杯和拿茶叶，以减少总用时。\n【详解】先接水 1 分钟，再烧水 7 分钟，在烧水的 7 分钟内同时完成洗茶杯 2 分钟和拿茶叶 1 分钟，最后沏茶 1 分钟。总用时：1+7+1=9(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "21-22四年级上·湖北随州·期末"
+    },
+    {
+        "id":  "402",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "煎一个蛋要 2 分钟(正，反两面各 1 分钟),一个平底锅一次能同时煎 2 个蛋。那么煎 3 个蛋最少要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "3",
+        "solution":  "【分析】用“3 个轮换”策略，让锅里始终有 2 个蛋在煎。\n【详解】第一次：1、2 号蛋煎正面，1 分钟；第二次：1 号蛋煎反面，3 号蛋煎正面，1 分钟；第三次：2、3 号蛋煎反面，1 分钟，共 3×1=3(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "21-22 四年级上·浙江宁波·期末"
+    },
+    {
+        "id":  "403",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈做家务：拖地要10分钟，用洗衣机洗衣服要30分钟，晾衣服要3分钟，搬家具要 7 分钟，洗碗要 6 分钟。她做完这些事至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "33",
+        "solution":  "【分析】洗衣机洗衣时间最长，可在这 30 分钟里安排拖地、搬家具和洗碗。\n【详解】拖地 10 分钟，搬家具 7 分钟，洗碗 6 分钟，共 23 分钟，小于洗衣 30 分钟，可全部放在洗衣时间内完成，最后晾衣服 3 分钟。总用时：30+3=33(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·云南昆明·期末"
+    },
+    {
+        "id":  "404",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一张平底锅每次最多只能煎两条鱼，煎一条鱼要6分钟(正，反面各3分钟),那么煎 7 条鱼至少要用( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "21",
+        "solution":  "【分析】将煎鱼视作标准“烙饼问题”，在锅里始终尽量保持满负荷时，最少时间等于单面时间乘鱼的条数。\n【详解】煎 7 条鱼至少需要 3×7=21(分钟)，所以至少要 21 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "2022·湖南怀化·小升初真题"
+    },
+    {
+        "id":  "405",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈准备 3 条鱼，如果每条鱼煎熟一面需要 3 分钟，而且平底锅里每次最多只能放 2 条鱼，妈妈煎完全部的鱼至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "9",
+        "solution":  "【分析】3 条鱼用“3 条轮换”策略，使锅始终处于满负荷工作。\n【详解】第一次煎 a、b 的正面，3 分钟；第二次煎 a 的反面和 c 的正面，3 分钟；第三次煎 b、c 的反面，3 分钟，共 3×3=9(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·云南昆明·期末"
+    },
+    {
+        "id":  "406",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈上班了，天天只能自己做饭吃。淘米4分钟，洗锅1分钟，电饭锅煮饭20分钟，把妈妈做好的菜用微波炉热一下 3 分钟，热牛奶 3 分钟。天天最快( )分钟就可以吃到饭。",
+        "options":  [
+
+                    ],
+        "answer":  "25",
+        "solution":  "【分析】煮饭 20 分钟时，可以同时用微波炉热菜和热牛奶.\n【详解】先淘米 4 分钟，洗锅 1 分钟，共 5 分钟，再煮饭 20 分钟，在这 20 分钟内完成热菜和热牛奶各 3 分钟，不增加总时长。总用时:4+1+20=25(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "21-22 四年级上·河北邯郸·期末"
+    },
+    {
+        "id":  "407",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "烧烤店推出一款烧鱼，鱼的两面都烤，每次最多烤2条，烤一面需要8分钟，烤10条鱼至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "80",
+        "solution":  "【分析】容量为 2 且总是满负荷时，总时间 = 单面时间×条数.\n【详解】每条鱼 2 面，每面 8 分钟，按最佳轮换，最少时间为 8×10=80(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23三年级上·湖南常德·期末"
+    },
+    {
+        "id":  "408",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个平底锅里最多能煎2条鱼，煎熟1条鱼需要4分钟(正反面都要煎，每面各需要 2 分钟),那么，煎熟 3 条鱼至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】当一次可煎 2 条鱼时，最少时间 = 单面时间×条数.\n【详解】每条鱼两面，共 3×2=6 个面，每次 2 个面，需 3 次，每次 2 分钟，总时间=2×3=6(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23四年级上·河北秦皇岛·期末"
+    },
+    {
+        "id":  "409",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一块铁板每次最多只能烤 2 块牛排，牛排两面都要烤，每面烤熟需 8 分钟。那么烤熟 5 块这样的牛排最少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "40",
+        "solution":  "【分析】视作烙饼：最少时间 = 单面时间×块数.\n【详解】单面 8 分钟,5 块牛排共 5×2=10 个面，最少时间为 8×5=40(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·湖南常德·期末"
+    },
+    {
+        "id":  "410",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈做饭烧菜完成以下事情：洗电饭煲 1 分钟，淘米 2 分钟，烧饭 18 分钟，洗菜，切菜共 3 分钟，烧菜 10 分钟。做完这些事情至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "21",
+        "solution":  "【分析】烧饭 18 分钟时，可以同时完成洗菜、切菜和烧菜，共 3+10=13 分钟。\n【详解】先洗电饭煲 1 分钟，淘米 2 分钟，再烧饭 18 分钟，期间完成洗菜、切菜和烧菜。总时间：1+2+18=21(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·湖南永州·期末"
+    },
+    {
+        "id":  "411",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "复印机同时可以复印 2 张稿件，复印一面需要 10 秒，有 5 张稿件两面都要复印，最少需要( )秒才能复印完。",
+        "options":  [
+
+                    ],
+        "answer":  "50",
+        "solution":  "【分析】把复印看成\"烙饼\"：总面数 / 每次面数 = 最少次数，总时间 = 次数×单次时间.\n【详解】5 张两面共 10 个面，每次复印 2 个面，需要 10÷2=5 次，每次 10 秒，共 5×10=50(秒)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·湖南娄底·期末"
+    },
+    {
+        "id":  "412",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "“六•一”儿童节，5 个好朋友互相赠一张卡片，一共需要( )张卡片。",
+        "options":  [
+
+                    ],
+        "answer":  "20",
+        "solution":  "【分析】5 个人互相赠卡片，每人都要给其他 4 人各 1 张。\n【详解】总卡片数 = 5×(5-1) = 5×4=20(张)。",
+        "knowledge_points":  [
+                                 "搭配问题"
+                             ],
+        "source":  "22-23 六年级上·甘肃陇南·期末"
+    },
+    {
+        "id":  "413",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小雨要完成下面四件家务：扫地 5 分钟，洗茶杯 1 分钟，烧水 6 分钟，沏茶 1 分钟。完成以上家务最少要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "7",
+        "solution":  "【分析】烧水 6 分钟时，可以同时扫地和洗茶杯，正好共 6 分钟。\n【详解】在烧水的 6 分钟内完成扫地和洗茶杯，最后沏茶 1 分钟，总用时 6+1=7(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·湖南益阳·期末"
+    },
+    {
+        "id":  "414",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "青青帮妈妈做家务，洗毛巾用3分钟，擦桌子用2分钟，烧开水用9分钟，沏茶用 3 分钟，扫地用 4 分钟。她做完这些事至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "12",
+        "solution":  "【分析】洗毛巾，擦桌子，扫地共 3+2+4=9 分钟，可安排在烧开水的 9 分钟内.\n【详解】先烧开水 9 分钟，同时完成洗毛巾，擦桌子，扫地;再沏茶 3 分钟，总用时 9+3=12(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "19-20 四年级上·河南漯河·期末"
+    },
+    {
+        "id":  "415",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小红给客人泡茶，接水 1 分钟，烧水 5 分钟，洗茶杯 2 分钟，拿茶叶 1 分钟，沏茶 1 分钟，小红合理安排后，客人最快喝上茶水最少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "7",
+        "solution":  "【分析】烧水时可以同时洗茶杯和拿茶叶。\n【详解】先接水 1 分钟，再烧水 5 分钟，在这 5 分钟内完成洗茶杯 2 分钟和拿茶叶 1 分钟，最后沏茶 1 分钟。总用时 1+5+1=7(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "20-21 四年级上·河南信阳·期末"
+    },
+    {
+        "id":  "416",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈用一只平底锅煎鱼，每次只能放两条鱼，煎一面需要2分钟，两面都要煎，煎 7 条需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "14",
+        "solution":  "【分析】视作烙饼：时间 = 条数*2/容量×单面时间.\n【详解】7×2÷2×2 = 7×2 = 14(分钟),所以煎 7 条鱼至少需要 14 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "20-21四年级上·山东济南·期末"
+    },
+    {
+        "id":  "417",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "爸爸早晨起床需要完成这些事情:(1)整理被褥 3 分钟(2)刷牙 3 分钟;(3)洗脸 2 分钟;(4)吃早饭 5 分钟;(5)听新闻 10 分钟。爸爸完成这些事情最快需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "13",
+        "solution":  "【分析】听新闻 10 分钟时，可以同时刷牙，洗脸和吃早饭(3+2+5=10).\n【详解】先整理被褥 3 分钟，再边听新闻边刷牙，洗脸和吃早饭 10 分钟，总用时 3+10=13(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·河北廊坊·期末"
+    },
+    {
+        "id":  "418",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明需要完成下面三件家务：扫地(8分钟)，烧水(10分钟)，沏茶(1分钟)。完成这些事情至少要用( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "11",
+        "solution":  "【分析】烧水 10 分钟时可以同时扫地 8 分钟。\n【详解】在烧水 10 分钟中完成扫地，最后沏茶 1 分钟，总用时 10+1=11(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "21-22 四年级上·湖南永州·期末"
+    },
+    {
+        "id":  "419",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "平底锅烙饼，每次可以烙两个，饼的两面都要烙，如果烙一面需要3分钟，那么烙完 5 个饼至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】前 2 张饼单独烙完，后 3 张用轮换法;也可直接用\"单面时间×数量\".\n【详解】先烙 2 张饼，两面共需 3×2=6(分钟);3 张饼轮换烙需 3×3=9(分钟),总共 6+9=15(分钟)。",
+        "knowledge_points":  [
+                                 "乘加，乘减混合运算",
+                                 "优化问题"
+                             ],
+        "source":  "22-23四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "420",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "某区有6个镇街，青少年协会计划开通免费的公交路线，供中小学生到各镇街游玩，如果每两个镇街之间开通一条路线，需要开通( )条路线。",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】任意两镇街之间一条线，是典型\"握手问题\".\n【详解】路线条数 = C(6,2) = 6×(6-1)÷2 = 30÷2=15(条)。",
+        "knowledge_points":  [
+                                 "搭配问题"
+                             ],
+        "source":  "22-23六年级上·广东佛山·期末"
+    },
+    {
+        "id":  "421",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "张阿姨一次最多能炸 2 个苹果派(一个苹果派要正反面各需要 1分钟),那么她炸 3 个苹果派最少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "3",
+        "solution":  "【分析】3 个派用\"3 个轮换\"策略即可最优.\n【详解】第一次:1,2 号正面,1 分钟;第二次:1 号反面,3 号正面,1 分钟;第三次:2,3 号反面,1 分钟，共 3 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23四年级上·湖南株洲·期末"
+    },
+    {
+        "id":  "422",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "烙一张饼需要 4 分钟(正，反两面各需要 2 分钟),一个平底锅每次最多烙 3张饼，烙 12 张饼至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "16",
+        "solution":  "【分析】本质仍是烙饼问题，总面数为 12×2=24，每次最多 3 面，需 8 次，每次 2 分钟.\n【详解】8 次*2 分钟/次=16(分钟),因此烙 12 张饼至少需要 16 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·湖南永州·期末"
+    },
+    {
+        "id":  "423",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈给客人沏茶，接水 1 分钟，烧水 6 分钟，洗茶杯 3 分钟，泡茶 1 分钟。妈妈合理安排以上事情，最少要( )分钟才能使客人尽快喝到茶。",
+        "options":  [
+
+                    ],
+        "answer":  "8",
+        "solution":  "【分析】烧水时可以同时洗茶杯，尽量重叠时间.\n【详解】先接水 1 分钟，再烧水 6 分钟(同时洗茶杯 3 分钟),最后泡茶 1 分钟，总用时 1+6+1=8(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23四年级上·湖南永州·期末"
+    },
+    {
+        "id":  "424",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈上班去了，中午莉莉只能自己做饭吃。淘米1分钟，洗锅1分钟，用电饭煲蒸米20分钟，把妈妈炒好的两个菜用微波炉热一下需要5分钟，冲一碗紫菜汤需要4分钟。莉莉最快( )分钟就能吃饭了。",
+        "options":  [
+
+                    ],
+        "answer":  "22",
+        "solution":  "【分析】蒸米饭的 20 分钟内可以同时热菜和冲汤(5+4=9 分钟).\n【详解】先淘米 1 分钟，洗锅 1 分钟，再蒸米 20 分钟(同时热菜和冲汤),总用时 1+1+20=22(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "20-21四年级上·河南漯河·期末"
+    },
+    {
+        "id":  "425",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "煮熟一个鸡蛋需要 9分钟，一个锅一次最多可以煮 10 个鸡蛋，那么要煮熟 50 个鸡蛋至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "45",
+        "solution":  "【分析】一次最多煮 10 个，且同时开始，同时结束，因此每 10 个鸡蛋仍只需 9 分钟.\n【详解】50 个鸡蛋分 5 锅煮，每锅 9 分钟，总时间 9×5=45(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题",
+                                 "多位数与一位数的乘除混合运算"
+                             ],
+        "source":  "21-22 四年级上·河南漯河·期末"
+    },
+    {
+        "id":  "426",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "玲玲每天早晨7:20起床，要完成下面这些事：穿衣服2分钟，刷牙3分钟，洗脸 3 分钟，整理床铺 2 分钟，吃早餐 8 分钟，听英语录音 10分钟。完成这些事后，玲玲最早能在( )出门去上学。",
+        "options":  [
+
+                    ],
+        "answer":  "7:38",
+        "solution":  "【分析】听英语的 10 分钟可与吃早餐 8 分钟并行，再考虑其余准备时间.\n【详解】穿衣 2 分钟，刷牙 3 分钟，洗脸 3 分钟共 8 分钟，再听英语录音 10 分钟(同时吃早餐 8 分钟),总时间 8+10=18 分钟。7:20+18 分钟=7:38。",
+        "knowledge_points":  [
+                                 "优化问题",
+                                 "时，分，秒时间的推算"
+                             ],
+        "source":  "22-23四年级上·湖南长沙·期末"
+    },
+    {
+        "id":  "427",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "拿衣服去洗要用 2分钟，洗衣机洗衣服要用 20 分钟，扫地要用 10 分钟，浇花要用 5 分钟，晾衣服要用 3 分钟。经过合理安排，小强做完这些家务，最少要用( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "25",
+        "solution":  "【分析】洗衣机工作 20 分钟时，可以同时完成扫地和浇花(10+5=15 分钟).\n【详解】先拿衣服去洗 2 分钟，再洗衣 20 分钟(期间完成扫地和浇花),最后晾衣服 3 分钟，总用时 2+20+3=25(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·湖南长沙·期末"
+    },
+    {
+        "id":  "428",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "每次最多只能烙两张饼，两面都要烙，每面烙3分钟，爸爸妈妈和小明每人一张，至少( )分钟能吃上饼。",
+        "options":  [
+
+                    ],
+        "answer":  "9",
+        "solution":  "【分析】3 张饼，两面都要烙，相当于 3 个\"二面任务\"，用 3 张轮换最优.\n【详解】总时间 = 单面时间×张数 = 3×3=9(分钟),因此至少 9 分钟能吃上饼。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·山西阳泉·期末"
+    },
+    {
+        "id":  "429",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "贺贺每天晚上睡觉前要做的事情有：背诵单词 5 分钟，烧开水 9 分钟，吃零食2 分钟，刷牙 3 分钟，用开水烫牛奶 2 分钟，喝牛奶 2 分钟，那么贺贺做完这些事情最少要用( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "14",
+        "solution":  "【分析】烧水 9 分钟时，可以同时背单词、吃零食和刷牙，以减少总用时。\n【详解】先烧开水 9 分钟，在这 9 分钟内完成背单词 5 分钟、吃零食 2 分钟和刷牙 3 分钟，共 10 分钟，所以烧水结束后刷牙还剩 1 分钟。然后用开水烫牛奶 2 分钟，再喝牛奶 2 分钟。最少需要 9+1+2+2=14(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "20-21四年级上·河南许昌·期末"
+    },
+    {
+        "id":  "430",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈要做这些事：整理房间 15 分钟，把衣服放入洗衣机 1 分钟，洗衣机自动洗衣服20分钟，晾衣服3分钟，擦鞋3分钟。妈妈做完这些事最少要用( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "24",
+        "solution":  "【分析】洗衣 20 分钟时可以同时整理房间和擦鞋(15+3=18 分钟).\n【详解】先把衣服放入洗衣机 1 分钟，然后洗衣 20 分钟(同时整理房间和擦鞋),最后晾衣服 3 分钟，总用时 1+20+3=24(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·河北廊坊·期末"
+    },
+    {
+        "id":  "431",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "\"神星古镇\"是驰名中外的钩瓷文化发祥地。一种钩瓷工艺品，两面都要雕刻，一位工艺美术大师雕刻一面要 8 小时,2 位工艺美术大师雕刻 3 件这种工艺品最少要( )小时。",
+        "options":  [
+
+                    ],
+        "answer":  "24",
+        "solution":  "【分析】先算 1 位大师完成 3 件所需时间，再除以 2(两人合作).\n【详解】一件工艺品两面需 8×2=16 小时,3 件共需 16×3=48 小时;2 位大师同时工作，时间是 48÷2=24(小时)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "19-20四年级上·河南许昌·期末"
+    },
+    {
+        "id":  "432",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "妈妈中午回家做饭，洗菜 10 分钟，切菜 3 分钟，炒菜 10 分钟，淘米 2 分钟，电饭煲蒸米 30 分钟。最快( )分钟可以吃午饭。",
+        "options":  [
+
+                    ],
+        "answer":  "32",
+        "solution":  "【分析】蒸米 30 分钟时可以同时洗菜，切菜，炒菜(10+3+10=23 分钟\u003c30 分钟).\n【详解】先淘米 2 分钟，再蒸米 30 分钟，期间完成洗菜，切菜，炒菜，最快 2+30=32(分钟)可以吃午饭。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·河南信阳·期末"
+    },
+    {
+        "id":  "433",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一口锅只能同时烤两个面包(两面都烤),每面需要烤 3 分钟,18 分钟可以烤( )个面包。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】当锅中尽量保持有 2 个面包在烤时，总时间等于单面时间乘面包个数。\n【详解】18 分钟内最多能烤熟的面包个数是 18÷3=6，因此 18 分钟可以烤 6 个面包。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23四年级上·河南信阳·期末"
+    },
+    {
+        "id":  "434",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.4,
+        "question":  "妈妈在家里煎牛排，一个平底锅每次最多只能煎两块牛排，每面煎两分钟，现在她要煎 5 块牛排至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "10",
+        "solution":  "【分析】这是标准烙饼问题：锅一次最多放 2 块牛排，两面都要煎，每面 2 分钟，在保持锅内尽量满的情况下，总时间与块数成正比.\n【详解】总时间 = 牛排块数×单面时间 = 5×2=10(分钟)。因此至少需要 10 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23四年级上·河南信阳·期末"
+    },
+    {
+        "id":  "435",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "亮亮要用烤箱做5条烤鱼，每条鱼要烤两面，每面烤3分钟，烤箱一次只能放两条鱼，他烤完这 5 条鱼最少要用( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】先烤熟 2 条，再对剩余 3 条采用\"3 条轮换\"策略，使烤箱始终接近满负荷.\n【详解】先同时烤 2 条，两面共需 3×2=6(分钟);余下 3 条 a,b,c:第 1 轮烤 a,b 正面 3 分钟;第 2 轮烤 a 反面和 c 正面 3 分钟，此时 a 熟;第 3 轮烤 b,c 反面 3 分钟，共 3×3=9(分钟)。总时间:6+9=15(分钟)。",
+        "knowledge_points":  [
+                                 "烙饼问题",
+                                 "优化问题"
+                             ],
+        "source":  "21-22四年级上·河南信阳·期末"
+    },
+    {
+        "id":  "436",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "\"比大小\"游戏中，明明抽到了三张数卡，芳芳抽到了三张数卡。每人每次只能出一张数卡，对阵三局，大者胜。明明先出 10，芳芳先出( )才有可能获胜。",
+        "options":  [
+
+                    ],
+        "answer":  "4",
+        "solution":  "【分析】明明有 10,8,5，芳芳有 9,7,4，要让芳芳赢两局一局，需用\"田忌赛马\"策略：最小对最大，最大对次大，中等对最小.\n【详解】明明先出 10，芳芳应出自己最小的 4，故意输掉一局;接下来让 7 对 5,9 对 8，芳芳赢两局。要达成这种对阵，第一局必须出 4，因此答案为 4。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "21-22四年级上·河南三门峡·期末"
+    },
+    {
+        "id":  "437",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "烤一块牛排需要6分钟(每面各需3分钟),一块铁板最多只能烤2块。那么烤7块这样的牛排最少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "21",
+        "solution":  "【分析】一次最多烤 2 块，每块两面，典型烙饼问题，保证铁板上尽量一直有 2 块即可最优.\n【详解】总时间 = 单面时间×块数 = 3×7=21(分钟),所以至少需要 21 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "21-22四年级上·河南周口·期末"
+    },
+    {
+        "id":  "438",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用平底锅烙饼，饼的两面都要烙，烙一面要用 3 分钟，锅里只能同时放 2 张饼。现在要烙 5 张饼，至少要用( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】先烙 2 张饼使其全熟，再对余下 3 张用轮换法;也可以直接用单面时间×张数.\n【详解】先同时烙 2 张饼，两面共需 3×2=6(分钟);余下 3 张轮换烙需要 3×3=9(分钟),共 6+9=15(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·山东临沂·期末"
+    },
+    {
+        "id":  "439",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一口平底锅每次最多只能烙两张饼，两面都烙，每面要用2分钟。李阿姨要烙7 张饼，至少要用( )分钟。如果李阿姨烙30分钟，最多可烙( )张饼。",
+        "options":  [
+
+                    ],
+        "answer":  "14;15",
+        "solution":  "【分析】7 张饼两面都烙，且一次最多 2 张，属于烙饼问题;时间一定时，张数 = 总时间/单面时间.\n【详解】烙 7 张饼的最少时间：总时间 = 单面时间×张数 = 2×7=14(分钟)。30 分钟最多烙的张数:30÷2=15(张)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·湖北黄冈·期末"
+    },
+    {
+        "id":  "440",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "明明每天早晨起床，整理床铺 3 分钟，洗漱 5 分钟，用面包机烤土司 5 分钟，煎鸡蛋2分钟，烧开水5分钟，泡牛奶并吃完早餐10分钟。他完成这些事情至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "20",
+        "solution":  "【分析】可以把用机器完成的事情和人工操作的事情穿插进行，这样最省时间。\n【详解】先用面包机烤土司 5 分钟，同时完成洗漱 5 分钟；再烧开水 5 分钟，同时整理床铺 3 分钟并煎鸡蛋 2 分钟；最后泡牛奶并吃完早餐 10 分钟。总时间是 5+5+10=20(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·湖北黄石·期末"
+    },
+    {
+        "id":  "441",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "文印室的复印机每次能复印两张 A4 纸的材料，但每次只能复印材料的一面。现在有 5 张两面都需要复印的 A4 纸材料，至少( )次就能复印完。",
+        "options":  [
+
+                    ],
+        "answer":  "5",
+        "solution":  "【分析】总面数为 5×2=10，每次复印 2 个面，所需次数 = 总面数/每次面数.\n【详解】最少次数 = 10÷2=5(次),因此至少复印 5 次就能完成。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·河北沧州·期末"
+    },
+    {
+        "id":  "442",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈星期天做家务：洗衣机洗衣服需要 30 分钟，拖地需要 15 分钟，整理房间需要 10 分钟，晾晒洗好的衣服需要 5 分钟。妈妈做完这些事情至少要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "35",
+        "solution":  "【分析】洗衣机工作 30 分钟时可以同时拖地和整理房间(15+10=25 分钟).\n【详解】在洗衣机洗衣服的 30 分钟内完成拖地和整理房间，最后晾晒衣服 5 分钟，总用时 30+5=35(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "20-21 四年级上·河南郑州·期末"
+    },
+    {
+        "id":  "443",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "用锅煎饼，每次最多只能煎两张，两面都要煎，每面 4 分钟。煎 5 张饼至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "20",
+        "solution":  "【分析】烙饼问题：总时间 = 饼数*2/每次容量×单面时间.\n【详解】5×2÷2×4 = 5×4=20(分钟),所以至少需要 20 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·河北张家口·期末"
+    },
+    {
+        "id":  "444",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈用烤箱烤鱼，每条鱼两面都要烤，每面烤 4 分钟，烤箱一次能放两条鱼，全家 5 口人每人一条鱼，至少要烤( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "20",
+        "solution":  "【分析】前 2 条可直接烤熟，后 3 条用轮换策略，使烤箱利用率最高.\n【详解】先烤 2 条，两面需 4×2=8(分钟);余下 3 条轮换烤:3 轮各 4 分钟，共 12 分钟，总时间 8+12=20(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·湖北黄冈·期末"
+    },
+    {
+        "id":  "445",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "东东每天早上起床后要完成下面这些事情：穿衣服用 2 分钟，整理被褥用 3分钟，刷牙用 2 分钟，洗脸用 3 分钟，吃早饭用 10 分钟，听广播用 15 分钟，东东完成这些事最少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "20",
+        "solution":  "【分析】听广播 15 分钟可以和其他事情同时进行，但其他事情本身一共需要 20 分钟，所以最少时间不会少于 20 分钟。\n【详解】穿衣、整理被褥、刷牙、洗脸、吃早饭共需要 2+3+2+3+10=20(分钟)。在做这些事情的过程中可以一直听广播 15 分钟，所以听广播不另外增加时间。最少需要 20 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·河北秦皇岛·期末"
+    },
+    {
+        "id":  "446",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "丁丁烤面包片，第一面需要烤2分钟，烤第二面时，面包片比较干了，只需要烤1分钟。丁丁用的锅一次最多能放2片面包片，烤3片面包片，最少要用( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "5",
+        "solution":  "【分析】先安排两片一起烤第一面，再利用第二面只需 1 分钟的特点轮换完成。\n【详解】第一次：第 1、2 片烤第一面，2 分钟；第二次：第 1 片烤第二面，第 3 片烤第一面，1 分钟；第三次：第 2 片烤第二面，第 3 片继续烤第一面，1 分钟；第四次：第 3 片烤第二面，1 分钟。合计 2+1+1+1=5(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "447",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "红红每天晚上睡觉前都要背诵英语单词8分钟，烧开水10分钟，用开水泡好牛奶 2 分钟，喝牛奶 4 分钟，做完这些事至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "16",
+        "solution":  "【分析】烧水 10 分钟时可以同时背单词 8 分钟.\n【详解】安排为：烧水(同时背单词)10 分钟，泡牛奶 2 分钟，喝牛奶 4 分钟，总用时 10+2+4=16(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23四年级上·河北秦皇岛·期末"
+    },
+    {
+        "id":  "448",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一只平底锅只能同时煎两条鱼，煎一条鱼需要 8 分钟(正反面各 4 分钟),煎三条鱼至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "12",
+        "solution":  "【分析】将 3 条鱼视为 3 个\"二面任务\"，用 3 条轮换，保证锅里始终有 2 条.\n【详解】总面数 3×2=6，每次煎 2 面，共 3 次，每次 4 分钟，总时间 3×4=12(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "21-22四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "449",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈在火炉上烤鱼，每次最多烤两条，两面都要烤。烤一面要3分钟，烤9条鱼至少需要( )分钟,36 分钟最多烤( )条鱼。",
+        "options":  [
+
+                    ],
+        "answer":  "27;12",
+        "solution":  "【分析】一次最多 2 条，典型烙饼问题：最少时间 = 单面时间×条数;在给定时间下，最多条数 = 总时间/单面时间.\n【详解】烤 9 条鱼最少时间:3×9=27(分钟)。36 分钟内最多能烤 36÷3=12(条)鱼。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "21-22四年级上·河北沧州·期末"
+    },
+    {
+        "id":  "450",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "父母不在家，贝贝自己做饭吃。她需要做以下事情：淘米2分钟，洗电饭锅1分钟，用电饭锅蒸米饭15分钟，炒菜10分钟。贝贝做完这些事情至少用( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "18",
+        "solution":  "【分析】蒸米饭 15 分钟时可以同时炒菜 10 分钟.\n【详解】先淘米 2 分钟，洗电饭锅 1 分钟，共 3 分钟，再蒸米饭 15 分钟(同时炒菜),总用时 3+15=18(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "451",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一口平底锅每次最多只能煎两条鱼，煎一条鱼需要6分钟(正，反面各3分钟),煎 5 条鱼至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】煎鱼可以看作烙饼问题，最少时间等于单面时间乘鱼的条数。\n【详解】每条鱼有 2 面，每面 3 分钟，共 5 条鱼，最少时间是 3×5=15(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "452",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "单独完成下面的几件事所要用的时间分别为：扫地 3 分钟，烧水 20 分钟，拖地 8 分钟，洗水壶 1 分钟，洗茶杯 2 分钟，接水 1 分钟，看书 6 分钟。做完这些事最少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "22",
+        "solution":  "【分析】烧水 20 分钟时可以同时扫地、拖地、洗茶杯和看书。\n【详解】扫地、拖地、洗茶杯、看书共 3+8+2+6=19(分钟)，小于 20 分钟，可全部安排在烧水过程中。先洗水壶 1 分钟，接水 1 分钟，总时间是 20+1+1=22(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "19-20 四年级上·湖北襄阳·期末"
+    },
+    {
+        "id":  "453",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一只平底锅一次最多只能煎两条鱼，用它煎一条鱼需要 6 分钟(正，反面各 3分钟),那么煎 3 条鱼至少要用( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "9",
+        "solution":  "【分析】3 条鱼采用轮换策略，保证每次锅里有 2 条鱼在煎。\n【详解】第一次：a、b 煎正面 3 分钟；第二次：a 煎反面、c 煎正面 3 分钟，此时 a 熟了；第三次：b、c 煎反面 3 分钟。总时间是 3+3+3=9(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·湖北孝感·期末"
+    },
+    {
+        "id":  "454",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小丽周末帮妈妈做家务，用洗衣机洗衣服需要用40分钟，扫地需要6分钟，扔垃圾需要3分钟，擦家具需要15分钟，晾衣服需要5 分钟。经过合理安排，她做完这些事情最少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "45",
+        "solution":  "【分析】洗衣机洗衣 40 分钟时可以同时扫地，扔垃圾和擦家具(6+3+15=24 分钟).\n【详解】先启动洗衣机 40 分钟，期间完成扫地，扔垃圾，擦家具，最后晾衣服 5 分钟，总用时 40+5=45(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "455",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "亮亮发烧了，要做以下事情：倒开水 1 分钟，等水变温 5 分钟，找退烧药 1 分钟，测量体温 3 分钟。亮亮最少用( )分钟做好这些事情。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】等水变温的 5 分钟内可以同时找药和量体温(1+3=4 分钟).\n【详解】先倒开水 1 分钟，再等水变温 5 分钟，在这 5 分钟完成找退烧药和测量体温，总时间 1+5=6(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·辽宁抚顺·期末"
+    },
+    {
+        "id":  "456",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "笑笑每天早晨要听英语 20 分钟，起床､洗漱共用 5 分钟，吃早餐 15 分钟，洗餐具 2 分钟，笑笑做完这些事最少要用( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "22",
+        "solution":  "【分析】听英语的同时可以起床，洗漱，吃早餐，洗餐具，但这些共需 5+15+2=22 分钟，大于 20 分钟.\n【详解】并行安排后总时长由 22 分钟决定，因此笑笑做完这些事最少要用 22(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·湖北黄冈·期末"
+    },
+    {
+        "id":  "457",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一只平底锅最多能煎两条鱼，煎每条鱼需 4 分钟，现在要煎 3 条鱼要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】煎一条鱼 4 分钟是两面各 2 分钟，三条鱼依旧按轮换方案.\n【详解】第一次：两条鱼第一面 2 分钟;第二次：一条反面+一条新鱼第一面 2 分钟;第三次：两条剩余反面 2 分钟，总时间 2×3=6(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·湖北孝感·期末"
+    },
+    {
+        "id":  "458",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一种双人玩小游戏，玩一局要 4 分钟，小明，小东和小亮一起玩，每人玩两局，至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "12",
+        "solution":  "【分析】每局 2 人参与,3 个人每人 2 局，一共需要 3 局即可:AB,BC,AC 的形式.\n【详解】安排为：第一局小明和小东玩，第二局小东和小亮玩，第三局小明和小亮玩，共 3 局，时间 3×4=12(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·四川凉山·期末"
+    },
+    {
+        "id":  "459",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个平底锅，每次最多只能煎两条鱼，每条鱼两面都要煎，每面要煎 3 分钟，煎 9 条鱼需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "27",
+        "solution":  "【分析】典型烙饼：总时间 = 单面时间×条数.\n【详解】每条鱼两面，每面 3 分钟，共 9 条，最少时间 3×9=27(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·山东济宁·期末"
+    },
+    {
+        "id":  "460",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "聪聪周末早上起床后，要做以下事情：洗脸漱口 5 分钟，锻炼 15 分钟，听音乐20分钟，写作业20分钟，弹钢琴1小时。他做完这些事情最短需要( )。",
+        "options":  [
+
+                    ],
+        "answer":  "1 小时 40 分钟",
+        "solution":  "【分析】在洗脸漱口和锻炼时可以同时听音乐 20 分钟，其余写作业和弹钢琴需要单独完成。\n【详解】听音乐 20 分钟正好可以覆盖洗脸漱口 5 分钟和锻炼 15 分钟；再写作业 20 分钟，弹钢琴 1 小时，合计 1 小时 40 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·湖北十堰·期末"
+    },
+    {
+        "id":  "461",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "请你合理安排：整理床铺3分钟，刷牙3分钟，洗脸2分钟，听广播8分钟。如果 6 时 30 分起床，最快( )时( )分做完这些事情。",
+        "options":  [
+
+                    ],
+        "answer":  "6;38",
+        "solution":  "【分析】听广播 8 分钟可以同时整理床铺，刷牙，洗脸，这三件事所需时间正好是 3+3+2=8 分钟.\n【详解】总时间为 8 分钟,6:30 起床,8 分钟后是 6:38，所以最快 6 时 38 分做完这些事情。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·黑龙江佳木斯·期末"
+    },
+    {
+        "id":  "462",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "现有一堵墙正反两面都要刷漆,1 位油漆工刷一面墙要 2 小时,4 位工人刷四堵墙最少要( )小时。",
+        "options":  [
+
+                    ],
+        "answer":  "4",
+        "solution":  "【分析】4 位工人同时开工，先一起刷 4 面正面，再一起刷 4 面反面。\n【详解】每人刷一面墙要 2 小时，4 位工人一起刷 4 堵墙的正面用 2 小时，再一起刷 4 堵墙的反面用 2 小时，共 2+2=4(小时)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·江西南昌·期末"
+    },
+    {
+        "id":  "463",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "达瓦早上起床后穿衣服要5分钟，洗漱要5分钟，听新闻要10分钟，吃早餐要 15 分钟，达瓦做完这些事情至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "25",
+        "solution":  "【分析】听新闻 10 分钟可以安排在吃早餐 15 分钟之中，一部分时间重叠.\n【详解】先穿衣服 5 分钟，洗漱 5 分钟，共 10 分钟，然后边吃早餐边听新闻，取较长的 15 分钟，总时间 5+5+15=25(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "20-21 四年级上·四川甘孜·期末"
+    },
+    {
+        "id":  "464",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "王老师早上到校后，要做好以下几件事后才开始正式工作，打水1分钟，烧开水5 分钟，打扫卫生 3 分钟，浇花 1 分钟，干完这些事最少要()分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】烧开水时可以同时打扫卫生和浇花，它们共 3+1=4 分钟，小于 5 分钟.\n【详解】先打水 1 分钟，再烧水 5 分钟(期间打扫卫生，浇花),总用时 1+5=6(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "19-20四年级上·湖北咸宁·期末"
+    },
+    {
+        "id":  "465",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小丽去外地旅游，她要买22个具有纪念意义的徽章送人。徽章有两种售卖方式,2 枚装的一袋 12 元,5 枚装的一袋 20 元。小丽应该买( )袋 2 枚装的徽章和( )袋 5 枚装的徽章最省钱，此时需要花( )元。",
+        "options":  [
+
+                    ],
+        "answer":  "1;4;92",
+        "solution":  "【分析】先比较两种包装的单价，再尽量多买单价便宜的 5 枚装，剩余数量用 2 枚装补足。\n【详解】2 枚装：12÷2=6(元/枚)；5 枚装：20÷5=4(元/枚)，4<6，应尽量多买 5 枚装。22÷5=4……2，即买 4 袋 5 枚装后还差 2 枚，再用 1 袋 2 枚装补足。费用：4×20+1×12=80+12=92(元)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级下·全国·单元测试"
+    },
+    {
+        "id":  "466",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "东东，妈妈，爸爸和电脑下棋，玩一局需要 5 分钟，可以单人玩，也可以双人玩。三人每人都要玩 2 局，至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】用电脑作对手，每局由两人和电脑一起玩，共需 3 局即可让三人各玩两局.\n【详解】可以安排：第 1 局东东+妈妈，第 2 局爸爸+东东，第 3 局妈妈+爸爸，每局 5 分钟，共 3×5=15(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "21-22四年级上·重庆丰都·期末"
+    },
+    {
+        "id":  "467",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "30 名游客去划船，大船每条可坐7人，小船每条可坐4人，每条船都恰好坐满，可以租( )条大船和( )条小船。",
+        "options":  [
+
+                    ],
+        "answer":  "2;4",
+        "solution":  "【分析】设大船 x 条，小船 y 条，则 7x+4y=30，且 x,y 为非负整数.\n【详解】枚举可得：当 x=2,y=4 时,7×2+4×4=14+16=30，条船均坐满，且符合条件，所以可以租 2 条大船和 4 条小船。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 三年级上·四川广元·期末"
+    },
+    {
+        "id":  "468",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "三(1)班 26 名同学去公园划船，每只小船坐 4 人，每只大船坐 6 人，如果每只船都坐满，可以租小船( )只，大船( )只；或租小船( )只，大船( )只。",
+        "options":  [
+
+                    ],
+        "answer":  "5;1;2;3",
+        "solution":  "【分析】设小船有 a 只，大船有 b 只，则 4a+6b=26，找出符合条件的整数解。\n【详解】列举可得：当大船有 1 只时，小船有 5 只，4×5+6×1=26；当大船有 3 只时，小船有 2 只，4×2+6×3=26。所以可以租小船 5 只、大船 1 只；也可以租小船 2 只、大船 3 只。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 三年级上·山东菏泽·期末"
+    },
+    {
+        "id":  "469",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "牛排馆煎一块牛排需要6分钟(两面都要煎，每面需要3分钟)。如果一个铁板锅最多可以同时煎 2 块牛排，那么煎 5 块牛排至少需要()分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】一面需 3 分钟，一次 2 块，标准烙饼问题，总时间 = 单面时间×块数.\n【详解】最少时间 = 3×5=15(分钟),所以煎 5 块牛排至少需要 15 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·山东济宁·期末"
+    },
+    {
+        "id":  "470",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明用平底锅做3条烤鱼，每条鱼要烤两面，每面烤 3 分钟，平底锅一次只能放 2 条鱼，他烤完这 3 条鱼最少要用( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "9",
+        "solution":  "【分析】3 条鱼两面都要烤，平底锅一次只能放 2 条，采用“3 条轮换”策略可以使时间最少。\n【详解】把 3 条鱼编号为 a、b、c：第 1 个 3 分钟烤 a、b 的正面；第 2 个 3 分钟烤 a 的反面和 c 的正面，此时 a 熟了；第 3 个 3 分钟烤 b、c 的反面。共 3×3=9(分钟)，为最少时间。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·山东临沂·期末"
+    },
+    {
+        "id":  "471",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明早上刷牙洗脸要 4 分钟，收拾房间要 5 分钟，听英语朗诵要用 10 分钟，小明完成这些事情最少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "10",
+        "solution":  "【分析】听英语朗诵 10 分钟时可以同时进行刷牙洗脸和收拾房间，这两项共 4+5=9 分钟，小于 10 分钟.\n【详解】在听英语朗诵的 10 分钟里同时完成刷牙洗脸和收拾房间，总时间由听英语时长决定，为 10 分钟，所以最少需要 10 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·黑龙江七台河·期末"
+    },
+    {
+        "id":  "472",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈用一个平底锅煎小黄鱼，正，反面都要煎，每面需要 2 分钟，每次最多煎 2 条，煎 9 条小黄鱼最少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "18",
+        "solution":  "【分析】先把前 6 条按满锅方式烹饪，再对剩余 3 条用轮换策略，始终保持锅内 2 条鱼.\n【详解】前 6 条鱼：每次 2 条，每面 2 分钟，共 6 次，需 6×2=12(分钟);剩余 3 条鱼 a,b,c:第 1 次烤 a,b 正面;第 2 次烤 a 反面,c 正面;第 3 次烤 b,c 反面，共 3×2=6(分钟)。总时间 12+6=18(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题",
+                                 "乘加，乘减混合运算"
+                             ],
+        "source":  "22-23 四年级上·重庆大渡口·期末"
+    },
+    {
+        "id":  "473",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用洗衣机洗衣服30分钟，晒衣服6分钟，拖地10分钟，擦玻璃8分钟，擦桌子 6 分钟。合理安排以上几件事情，最少需要( )分钟完成。",
+        "options":  [
+
+                    ],
+        "answer":  "36",
+        "solution":  "【分析】洗衣机洗衣 30 分钟时，可以同时完成拖地 10 分钟，擦玻璃 8 分钟和擦桌子 6 分钟.\n【详解】拖地，擦玻璃，擦桌子共 10+8+6=24(分钟)\u003c30 分钟，可完全安排在洗衣时间内，最后晒衣服 6 分钟。总用时 30+6=36(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·广东珠海·期末"
+    },
+    {
+        "id":  "474",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "烙 3 张饼，每次锅里只能放两张饼，两面都要烙，烙一面需要 5 分钟，烙完这3 张饼最少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】3 张饼，单面 5 分钟，一次烙 2 张，是标准 3 张轮换烙饼问题.\n【详解】总时间 = 单面时间×张数 = 5×3=15(分钟)。也可按轮换:3 次，每次 5 分钟，总 15 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·山东临沂·期末"
+    },
+    {
+        "id":  "475",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "奶奶煎鱼，煎一面需要 3 分钟，两面都要煎，锅里每次最多能煎 2 条鱼，煎 5条鱼至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】一次最多煎 2 条鱼，两面都要煎，属于烙饼问题，最少时间与条数成正比.\n【详解】用公式：总时间 = 单面时间×条数 = 3×5=15(分钟),所以至少需要 15 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·湖北黄冈·期末"
+    },
+    {
+        "id":  "476",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈做早饭的过程及时间：洗锅(1 分钟),淘米(2 分钟),熬粥(20 分钟),煎鸡蛋(5 分钟),拌小菜(5 分钟),盛粥(1 分钟),妈妈吃到早餐需要等( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "24",
+        "solution":  "【分析】熬粥 20 分钟期间可以同时完成煎鸡蛋和拌小菜(5+5=10 分钟).\n【详解】总时间 = 洗锅 1 分钟 + 淘米 2 分钟 + 熬粥 20 分钟 + 盛粥 1 分钟 = 1+2+20+1=24(分钟),所以妈妈需要等 24 分钟才能吃到早餐。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·湖北鄂州·期末"
+    },
+    {
+        "id":  "477",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "明明每天晚上睡觉前要看作文书10分钟，烧开水12分钟，冲泡奶粉2分钟，喝奶粉 5 分钟，明明做完所有事情最少用( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "19",
+        "solution":  "【分析】烧开水 12 分钟内可以同时看作文书 10 分钟，从而节省时间.\n【详解】把看作文书安排在烧开水的 12 分钟里，完成后再冲泡奶粉 2 分钟，喝奶粉 5 分钟，总时间 12+2+5=19(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·山东菏泽·期末"
+    },
+    {
+        "id":  "478",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈做早饭的过程及时间：洗锅(1分钟)，熬粥(20分钟)，煎鸡蛋(5分钟)，拌小菜(5分钟)，盛粥(1分钟)。妈妈做这顿饭至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "22",
+        "solution":  "【分析】熬粥 20 分钟时可以同时进行煎鸡蛋和拌小菜，共 5+5=10 分钟。\n【详解】总用时由“洗锅+熬粥+盛粥”决定，为 1+20+1=22(分钟)。因为鸡蛋和小菜共 10 分钟，小于熬粥时间，所以至少需要 22 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23四年级上·山东济宁·期末"
+    },
+    {
+        "id":  "479",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈用电饼铛烙饼，每次最多只能烙 2 张饼，一张饼有 2 面，如果烙一面需要 3分钟，那么烙一张饼需要( )分钟，烙 3 张饼至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "6;9",
+        "solution":  "【分析】一张饼两面各烙 3 分钟，共需 6 分钟；3 张饼用轮换策略可保持锅内一直有 2 张饼。\n【详解】烙一张饼：2 面，各 3 分钟，共 2×3=6(分钟)；3 张饼按“3 张轮换”烙 3 轮即可，每轮 3 分钟，共 3×3=9(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·山东日照·期末"
+    },
+    {
+        "id":  "480",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一次最多能煎 4 条小黄鱼，鱼的两面都要煎，煎一面要 3 分钟，那煎 6 条鱼最少要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "9",
+        "solution":  "【分析】4 条一锅，两面 3 分钟一面;剩余 2 条再与部分已煎好的鱼穿插轮换，保证锅内尽量有 4 条.\n【详解】按题中解析安排,3 轮每轮 3 分钟即可完成 6 条鱼的两面烹饪，因此最少需要 3×3=9(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·四川广元·期末"
+    },
+    {
+        "id":  "481",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈下班回家做饭，淘米要 2 分钟，煮饭要 25 分钟，洗菜要 4 分钟，切菜要 5分钟，炒菜要 10 分钟。如果煮饭和炒菜用不同的锅和炉子，妈妈要将饭菜都做好，最少要用( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "27",
+        "solution":  "【分析】煮饭 25 分钟时可以同时完成洗菜，切菜和炒菜(4+5+10=19 分钟).\n【详解】安排为：先淘米 2 分钟，然后煮饭 25 分钟，在这期间洗菜，切菜，炒菜全部完成，总时间 2+25=27(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·四川广元·期末"
+    },
+    {
+        "id":  "482",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "烤1块牛排需要6分钟(正，反两面各3分钟),如果一块铁板上最多只能烤4 块同样的牛排，那么烤 10 块这样的牛排至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】合理分组和轮换，让铁板始终接近满负荷工作.\n【详解】先烤 4 块，两面共需 3×2=6(分钟);再通过替换，翻面等策略让余下 6 块轮换烤完，共需 9 分钟，合计 6+9=15(分钟),满足题中分析和详解给出的最短时间。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·湖北鄂州·期末"
+    },
+    {
+        "id":  "483",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "煮一个汤圆需要 7分钟，一个锅一次可以煮 20 个汤圆，那么煮 10 个汤圆至少需要( )分钟，煮 20 个汤圆至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "7;7",
+        "solution":  "【分析】一次锅内的汤圆可以同时开始，同时结束，因此同一锅里的个数不影响时间.\n【详解】无论煮 10 个还是 20 个，只要一锅里能全部放下，所需时间都等于煮 1 个所需的 7 分钟，所以两个空都填 7(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·湖北黄石·期末"
+    },
+    {
+        "id":  "484",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一只平底锅最多能煎两条鱼，煎每条鱼需4分钟(正，反各2分钟),现在要煎3条鱼至少要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】当一次最多煎 2 条鱼时，最少时间 = 单面时间×条数.\n【详解】每条鱼两面，各 2 分钟，相当于单面 2 分钟,3 条鱼最少时间 2×3=6(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "19-20四年级上·湖北黄石·期末"
+    },
+    {
+        "id":  "485",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈烙饼，锅里每次最多只能烙2张饼，两面都要烙，每面需要烙2分钟。妈妈要烙完 3 张饼，最少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】3 张饼，每面 2 分钟，一次 2 张，典型 3 张轮换烙饼问题.\n【详解】总时间 = 单面时间×张数 = 2×3=6(分钟),因此最少需要 6 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "18-19 四年级上·河北秦皇岛·期末"
+    },
+    {
+        "id":  "486",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "卓玛星期日休息时想帮妈妈做些事情：烧开水需 5 分钟，冲油茶需 3 分钟，扫地需6分钟，用洗衣机洗衣服需25分钟，晒衣服需5分钟。经过合理安排，她做完这些事至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "30",
+        "solution":  "【分析】洗衣机洗衣 25 分钟时，可以同步烧开水，冲油茶，扫地(5+3+6=14 分钟).\n【详解】总时间由\"洗衣+晒衣服\"决定，为 25+5=30(分钟);烧开水，冲油茶和扫地共 14 分钟\u003c25 分钟，可完全安排在洗衣期间，所以至少需要 30 分钟。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "21-22 四年级上·四川甘孜·期末"
+    },
+    {
+        "id":  "487",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "煎鸡蛋，两面都要煎，每煎一面要 3 分钟，每次最多只能煎 2 个。如果要煎 5个鸡蛋，最少要( )分钟;如果要煎 9 个鸡蛋，最少要()分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "15;27",
+        "solution":  "【分析】仍为烙饼问题：每个鸡蛋两面，各 3 分钟;一次最多 2 个，最少时间 = 单面时间×个数.\n【详解】5 个:3×5=15(分钟);9 个:3×9=27(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "21-22 四年级上·湖北武汉·期末"
+    },
+    {
+        "id":  "488",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明起床整理床铺3分钟，刷牙3分钟，洗脸2分钟。如果他6时40分起床，最快( )时( )分做完这些事情。",
+        "options":  [
+
+                    ],
+        "answer":  "6;48",
+        "solution":  "【分析】这三件事不能并行，总时间为三者之和.\n【详解】整理床铺，刷牙，洗脸共 3+3+2=8(分钟),6:40 起床,8 分钟后是 6:48，所以最快 6 时 48 分做完。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "20-21 四年级上·湖北武汉·期末"
+    },
+    {
+        "id":  "489",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "烙一张饼需要 4 分钟(两面都烙，每面2分钟),一个平底锅每次可以烙2张饼，像这样烙3张饼至少需要( )分钟，烙9张饼至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "6;18",
+        "solution":  "【分析】3 张饼用轮换策略,9 张饼则\"先 6 后 3\"，始终保持锅内有 2 张饼.\n【详解】3 张:3 轮，每轮单面 2 分钟，共 3×2=6(分钟);9 张：先 6 张，两面共需 (6÷2)×4=3×4=12(分钟),余下 3 张轮换 3×2=6(分钟),合计 12+6=18(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "20-21 四年级上·湖北武汉·期末"
+    },
+    {
+        "id":  "490",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用一个平底锅煎鸡蛋，每次只能煎4个，煎熟一个鸡蛋要4分钟(正，反面各煎 2 分钟),煎第 6 个鸡蛋最少要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】通过安排 3 次，每次 2 分钟的煎制，并轮换正反面，可以在 6 分钟内让 6 个鸡蛋全部熟.\n【详解】第 1 次 2 分钟:4 个鸡蛋一面;第 2 次 2 分钟：部分翻面，新蛋入锅;第 3 次 2 分钟：补足剩余反面，共 3 次，每次 2 分钟，总时间 2×3=6(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "21-22 四年级上·广东佛山·期末"
+    },
+    {
+        "id":  "491",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果每个锅最多煎 2条小黄鱼，煎一条小黄鱼需要 4 分钟(正反面各 2 分钟),那么煎 5 条小黄鱼至少需要( )分钟,7 条需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "10;14",
+        "solution":  "【分析】按题中分析，先煎 4 条，再对 3 条用轮换法，让锅始终满载.\n【详解】5 条：先煎 2 条两面 4 分钟，再对余下 3 条轮换，共 3×2=6(分钟),合计 4+6=10(分钟);7 条：需 2 次\"2 条全熟\"(各 4 分钟)加一次\"3 条轮换\"(3×2=6 分钟),总 4+4+6=14(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "21-22 四年级上·广东佛山·期末"
+    },
+    {
+        "id":  "492",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "三位同学要去量血压，验视力，每项检查都要5分钟，他们至少要( )分钟才能做完些检查。",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】两项检查可同时进行，每轮安排两人做不同项目，一共需要 3 轮.\n【详解】第 1 轮：甲量血压，乙验视力;第 2 轮：乙量血压，丙验视力;第 3 轮：丙量血压，甲验视力，每轮 5 分钟，共 3×5=15(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "19-20 四年级上·江西吉安·期末"
+    },
+    {
+        "id":  "493",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "明明给客人沏茶，接水 1 分钟，烧水 10 分钟，洗茶杯 1 分钟，拿茶叶 1 分钟，沏茶 1 分钟，明明合理安排时间，至少需要( )分钟，才能让客人尽快喝上茶。",
+        "options":  [
+
+                    ],
+        "answer":  "12",
+        "solution":  "【分析】烧水 10 分钟时可同时完成洗茶杯和拿茶叶.\n【详解】先接水 1 分钟，再烧水 10 分钟(期间洗茶杯 1 分钟，拿茶叶 1 分钟),最后沏茶 1 分钟，总用时 1+10+1=12(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·江西吉安·期末"
+    },
+    {
+        "id":  "494",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用一个平底锅烙饼，每次最多能放2张饼，烙一面要2分钟，两面都要烙，烙7张饼最少烙( )次，需要( ) 分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "7;14",
+        "solution":  "【分析】总面数 7×2=14，每次烙 2 面，因此最少需要 7 次;每次 2 分钟.\n【详解】烙 7 张饼最少烙 7 次，共需 7×2=14(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·陕西延安·期末"
+    },
+    {
+        "id":  "495",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "三个同学去测身高和视力。每项测量都需要四分钟。每次每项只能测一人。他们至少要用( )分钟才能完成这些检查。",
+        "options":  [
+
+                    ],
+        "answer":  "12",
+        "solution":  "【分析】两种检查可并行，安排三轮交叉，使每轮都有两人检查.\n【详解】轮次安排类似\"循环交叉\"：第 1 轮甲视力，乙身高;第 2 轮乙视力，丙身高;第 3 轮丙视力，甲身高，每轮 4 分钟，共 3×4=12(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "21-22 四年级上·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "496",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小红感冒了，吃完药后要赶快休息：找杯子倒开水 1 分钟;等开水变温 6分钟;找感冒药 1 分钟;量体温 5 分钟。小红最少要( )分钟才能去休息。",
+        "options":  [
+
+                    ],
+        "answer":  "7",
+        "solution":  "【分析】等开水变温的 6 分钟内可以同时找药和量体温(1+5=6 分钟).\n【详解】总时间 = 找杯子倒开水 1 分钟 + 等水变温 6 分钟 = 1+6=7(分钟),所以最少 7 分钟后即可休息。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "21-22 四年级上·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "497",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一块牛排两面都要煎，正，反面各需 2 分钟。一只平底锅每次可以煎 2 块，煎好 3 块牛排至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】3 块牛排，两面各 2 分钟，用轮换法保持锅内 2 块牛排工作.\n【详解】按题中解析:3 轮，每轮 2 分钟即可让 3 块牛排全部两面煎熟，总时间 2×3=6(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·四川广元·期末"
+    },
+    {
+        "id":  "498",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "下面是妈妈做早饭的过程及时间：淘米用 2 分钟，用电饭煲熬粥用 20 分钟，煎鸡蛋用 5 分钟，拌凉菜用 5 分钟，盛粥用 1 分钟，妈妈做好这些事情至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "23",
+        "solution":  "【分析】熬粥 20 分钟时可以同时完成煎鸡蛋和拌凉菜(共 10 分钟).\n【详解】总时间 = 淘米 2 分钟 + 熬粥 20 分钟 + 盛粥 1 分钟 = 2+20+1=23(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "22-23 四年级上·四川广元·期末"
+    },
+    {
+        "id":  "499",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在烤炉上烤面包，一次能烤 2 片，每片都要烤两面，每面都需要烤 1 分钟，烤5 片面包至少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "5",
+        "solution":  "【分析】一次烤 2 片，两面各 1 分钟，属于烙饼问题：总时间 = 单面时间×片数.\n【详解】每面 1 分钟,5 片至少需要 1×5=5(分钟)。通过轮换安排，可以在 5 分钟内烤熟 5 片两面。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "21-22 四年级上·广东佛山·期末"
+    },
+    {
+        "id":  "500",
+        "unit":  "第八章数学广角 5",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "安排时间最合理：(1)起床穿衣，整理被褥 4 分钟；(2)刷牙 3 分钟；(3)洗脸3分钟；(4)吃早饭15分钟；(5)听广播7分钟。最快( )分钟做完这些事情。",
+        "options":  [
+
+                    ],
+        "answer":  "25",
+        "solution":  "【分析】可以在吃早饭 15 分钟的同时听广播 7 分钟，以节省时间。\n【详解】先起床穿衣、整理被褥 4 分钟，再刷牙 3 分钟，洗脸 3 分钟，共 10 分钟，然后边吃早饭边听广播，以吃早饭 15 分钟为准。总时间是 4+3+3+15=25(分钟)。",
+        "knowledge_points":  [
+                                 "优化问题"
+                             ],
+        "source":  "21-22 四年级上·广东汕头·期末"
+    }
+]
diff --git "a/\345\233\233\345\271\264\347\272\247\351\242\230\351\233\206/\345\233\233\345\271\264\347\272\247\344\270\212/\345\233\233\345\271\264\347\272\247\344\270\212\347\254\254\345\205\255\345\215\225\345\205\203\351\231\244\346\225\260\346\230\257\344\270\244\344\275\215\346\225\260\347\232\204\351\231\244\346\225\260.json" "b/\345\233\233\345\271\264\347\272\247\351\242\230\351\233\206/\345\233\233\345\271\264\347\272\247\344\270\212/\345\233\233\345\271\264\347\272\247\344\270\212\347\254\254\345\205\255\345\215\225\345\205\203\351\231\244\346\225\260\346\230\257\344\270\244\344\275\215\346\225\260\347\232\204\351\231\244\346\225\260.json"
new file mode 100644
index 0000000000000000000000000000000000000000..a76922834f30721b806f069284b5cb606e8dfd54
--- /dev/null
+++ "b/\345\233\233\345\271\264\347\272\247\351\242\230\351\233\206/\345\233\233\345\271\264\347\272\247\344\270\212/\345\233\233\345\271\264\347\272\247\344\270\212\347\254\254\345\205\255\345\215\225\345\205\203\351\231\244\346\225\260\346\230\257\344\270\244\344\275\215\346\225\260\347\232\204\351\231\244\346\225\260.json"
@@ -0,0 +1,8618 @@
+[
+    {
+        "id":  "1",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上\u0027>\u0027\u0027<\u0027或\u0027=\u0027。\n\n10×600 ( ) 10660  3805560000 ( ) 3798988000\n\n8×200 ( ) 810×2  200÷40 ( ) 100÷10",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003e;\u003c;\u003c",
+        "solution":  "【分析】根据题意，比较数的大小方法：位数多的数就大;位数相同比较最高位，最高位大的数就大;最高位相同比较下一位，直到比较出结果为止;含有算式的计算出算式的结果后再比较大小。据此分析解答.\n【详解】 10×600 = 6000 , 6000 \u003c 10660，所以 10×600 \u003c 10660;3805560000 和 3798988000 都是十位数,3805560000 亿位上是 8,3798988000 亿位上是 7, 8 \u003e 7，所以 3805560000 \u003e 3798988000;8×200 = 1600 , 810×2 = 1620 , 1600 \u003c 1620，所以 8×200 \u003c 810×2;200÷40 = 5 , 100÷10 = 10 , 5 \u003c 10，所以 200÷40 \u003c 100÷10 。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "整十，整百，整千数与一位数的乘法",
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "除数是整十数的口算除法"
+                             ],
+        "source":  "24-25四年级上·江西南昌·期末"
+    },
+    {
+        "id":  "2",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "450 里面有( )个90, 108×50 的积的末尾有( )个 0。",
+        "options":  [
+
+                    ],
+        "answer":  "5;2",
+        "solution":  "【分析】第一个空是求450里面包含几个90，根据除法的意义，用 450 除以 90 即可得到结果;第二个空是计算108×50的积，通过乘法运算得出积后，观察积的末尾0的个数.\n【详解】450÷90 = 5，108×50 = 5400，所以 450 里面有 5 个 90, 108×50 的积的末尾有 2 个 0。",
+        "knowledge_points":  [
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "除数是整十数的口算除法"
+                             ],
+        "source":  "25-26 四年级上·甘肃天水·期末"
+    },
+    {
+        "id":  "3",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "要使□09÷57 的商是一位数,□里最大可以填( ),要使它的商是两位数,□里最小可以填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "5;6",
+        "solution":  "【分析】本题考查三位数除以两位数时商的位数的判断。除数是57，是两位数，商的位数取决于被除数的前两位数字与除数的大小关系。如果被除数的前两位小于除数，商是一位数;如果被除数的前两位大于或等于除数，商是两位数。被除数是□09，前两位是□0，因此需要比较□0与57 的大小.\n【详解】要使商是一位数，则被除数的前两位□0必须小于除数 57，即□0 < 57，所以□最大可以填 5。要使商是两位数，则被除数的前两位□0必须大于或等于除数57，即□0 ≥ 57，所以□最小可以填6。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "4",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "□81÷65，要使商是两位数,□里最小应填( ),要使商是一位数,□里最大应填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "6;5",
+        "solution":  "【分析】三位数除以两位数，要使商是两位数，则被除数的前两位数字要大于或等于除数;要使商是一位数，则被除数的前两位数字要小于除数，据此即可解答.\n【详解】由分析可得:□81÷65，要使商是两位数，则需□8 ≥ 65，□可以填 6,7,8,9，最小填 6;要使商是一位数，则需□8 < 65，□可以填 5,4,3,2,1，最大填 5。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "25-26四年级上·辽宁鞍山·期末"
+    },
+    {
+        "id":  "5",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "□01÷56 要使商的十位上没有数，方框里最大填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "5",
+        "solution":  "【分析】三位数除以两位数,□01÷56要使商的十位上没有数，因此商是一位数，需要被除数的前两位小于除数，据此即可解答.\n【详解】□ 0 \u003c 56 ,□里可以填写 5,4,3,2,1，所以方框里最大填 5。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "6",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "260×55 的积是( )位数;860÷75 的商是( )位数;458÷56 的商的最高位是()位。",
+        "options":  [
+
+                    ],
+        "answer":  "五;两;个",
+        "solution":  "【分析】根据乘数是两位数的乘法，先用两位数个位上的数去乘另一个乘数，得数的末位和两位数的个位对齐，再用两位数十位上的数去乘另一个乘数，得数的末位和两位数的十位对齐，然后把两次乘的结果加起来。当乘数末尾有零时，先算零前面的数，再在积的末尾添加对应个数的零。据此算出 260×55 的积，再进一步解答。三位数除以两位数，商可能是两位数，也可能是一位数。被除数的前两位够除，也就是被除数的前两位大于或等于除数，商是两位数;被除数的前两位不够除，也就是被除数的前两位小于除数，商是一位数。据此判断 860÷75 和458÷56 的商是几位数.\n【详解】260×55 = 14300，14300 是五位数，则 260×55 的积是五位数;860÷75，被除数的前两位是86，比75大，则商是两位数;458÷56，被除数的前两位是45，比56小，则商是一位数，一位数只有个位，即 458÷56 的商的最高位是个位。",
+        "knowledge_points":  [
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "除数是两位数的笔算除法",
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25 四年级上·河北衡水·期末"
+    },
+    {
+        "id":  "7",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上\u0027>\u0027\u0027<\u0027或\u0027=\u0027。\n\n72×80 ( ) 5600  24×40 ( ) 1000  4800÷160 ( ) 480÷16\n\n1800 ( ) 99×19  28×30 ( ) 280×32  70÷30 ( ) 540÷90",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;\u003c;=;\u003c;\u003c;\u003c",
+        "solution":  "【分析】比较 72×80 和5600，先计算出结果再比较大小;比较 24×40 和1000，先计算出结果再比较大小;比较 4800÷160 和 480÷16，根据商不变的性质比较大小。在除法算式里，被除数和除数同时乘或除以相同的数(0除外),商不变;比较 1800 和 99×19，先计算出结果再比较大小;比较 28×30 和 280×32，先计算出结果再比较大小;比较 70÷30 和 540÷90，先根据商不变的性质，把右边的算式变式为(540÷3)÷(90÷3)=180÷30;再根据除数相同，被除数越大商越大比较大小.\n【详解】72×80 = 5760，因为 5760 > 5600，所以 72×80 > 5600;24×40 = 960，因为 960 < 1000，所以 24×40 < 1000;4800÷10 = 480，160÷10 = 16，符合商不变的性质，所以 4800÷160 = 480÷16;99×19 = 1881，因为 1800 < 1881，所以 1800 < 99×19;28×30 = 840，280×32 = 8960，因为 840 < 8960，所以 28×30 < 280×32;540÷90 = (540÷3)÷(90÷3) = 180÷30，因为 70 < 180，所以 70÷30 < 540÷90。",
+        "knowledge_points":  [
+                                 "两位数乘整十数的口算乘法",
+                                 "两位数乘两位数的进位乘法",
+                                 "除数是两位数的笔算除法",
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "25-26四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "8",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "要使☐56÷38的商是一位数，☐里最大填( );要使商是两位数，☐里最小填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "3;4",
+        "solution":  "【分析】三位数除以两位数，若被除数的前两位小于除数，商是一位数;若被除数的前两位大于或等于除数，商是两位数。商是一位数：被除数前两位☐5 < 38，☐可填1,2,3，最大填 3。商是两位数：被除数前两位☐5 ≥ 38，☐可填4,5,6,7,8,9，最小填 4.\n【详解】根据分析可知，要使☐56÷38的商是一位数，☐里最大填 3;要使商是两位数，☐里最小填 4。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "25-26四年级上·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "9",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "185 的 25 倍是( ),850 是 25 的( )倍。",
+        "options":  [
+
+                    ],
+        "answer":  "4625;34",
+        "solution":  "【分析】根据题意，求185的25倍是多少，根据倍数的意义，用乘法计算;求850是25的多少倍，根据倍数的意义，用除法计算。列式计算即可.\n【详解】根据分析可知:185×25 = 4625，850÷25 = 34，所以 185 的 25 倍是 4625，850 是 25 的 34 倍。",
+        "knowledge_points":  [
+                                 "求一个数的几倍是多少",
+                                 "求一个数是另一个数的几倍",
+                                 "除数是两位数的笔算除法",
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "24-25 四年级上·四川绵阳·期末"
+    },
+    {
+        "id":  "10",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "7□5÷75，要使商是一位数,□里最大填( );要使商是两位数,□里最小填()。",
+        "options":  [
+
+                    ],
+        "answer":  "4;5",
+        "solution":  "【分析】三位数除以两位数，判断商是几位数的方法：先观察被除数前两位上的数，看是否大于或等于除数(也就是看前两位上的数除以除数，够不够商一个十)。被除数前两位够除，商就是两位数。反之，商就是一位数。据此解答.\n【详解】7□5÷75，要使商是一位数，则 7□ \u003c 75 ,□里可以填 0,1,2,3,4,□里最大填 4;要使商是两位数，则 7□\u003e 75 或 7□ = 75 ,□里可以填5,6,7,8,9,□里最小填5。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25四年级上·云南昭通·期末"
+    },
+    {
+        "id":  "11",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "☐97÷59，要使商是一位数，☐里最大填( );要使商是两位数，☐里最小填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "4;5",
+        "solution":  "【分析】当商是一位数时，被除数的前两位☐9 < 59，☐最大为4;要使商为两位数，被除数的前两位☐9大于或等于59，☐最小为5.\n【详解】☐97÷59，要使商是一位数，☐里最大填4;要使商是两位数，☐里最小填5。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "25-26四年级上·辽宁鞍山·期末"
+    },
+    {
+        "id":  "12",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "试商时，最好把除数 58 看成( )来试商。",
+        "options":  [
+
+                    ],
+        "answer":  "60",
+        "solution":  "【分析】试商时，为了便于计算，通常将除数看作最接近的整十数。除数58的个位数字是8，根据四舍五入法，应向十位进1，因此58最接近的整十数是 60。\n【详解】58 接近 60，所以试商时，最好把除数 58 看成 60 来试商。",
+        "knowledge_points":  [
+                                 "多位数除以两位数的试商"
+                             ],
+        "source":  "25-26 四年级上·辽宁铁岭·期末"
+    },
+    {
+        "id":  "13",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "从 405 里面连续减去( )个 15，结果是 0。",
+        "options":  [
+
+                    ],
+        "answer":  "27",
+        "solution":  "【分析】根据题意，从405里面连续减去多少个15，结果是 0。就是求 405 里面有多少个 15，用 405 除以 15 即可.\n【详解】 405÷15 = 27 所以，从 405 里面连续减去 27 个 15，结果是 0 ",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "24-25 四年级上·河北保定·期末"
+    },
+    {
+        "id":  "14",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "钟面上 11:00，时针和分针形成的角是( )°;过了20分钟，这时时针和分针形成的角是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "30;140",
+        "solution":  "【分析】由题意得,11:00，分针指向数字 12，时针指向数字 11,(如下图)![](images/0465fd2a8083cdded02e386f53990c4e974a6804c73b248cd78d4d64b7257b57.jpg)\n\n由图可知，时针和分针之间的夹角对应着一个大格。钟面上一共有12个大格,12个大格组成了一个周角，周角的度数是 360°。360°÷12 = 30°，所以每个大格对应的度数是30°，那么时针和分针之间的夹角是 30°。过了20分钟，分针走到了数字4。把60分钟看成一个整体，平均分成3份，每份是 20 分钟，所以 20 分钟是 60 分钟的1/3，那么时针走了1/3个大格。![](images/a4b8c8f18b11288a1a50167b3a891790b92473ba8c9de4e1da9b257f68abc920.jpg)\n\n由图可知，分针走过了 4 个大格，时针走过了1/3个大格。可以先用30°乘4算出分针走了多少度，接着用30°除以3算出时针走了多少度，接着把它们走过的度数相减算出分针比时针多走了多少度，最后再加上前面的得数即可算出时针和分针形成的角的度数.\n【详解】30°×4 = 120°，30°÷3 = 10°，120° - 10° = 110°，110° + 30° = 140°。故钟面上 11:00，时针和分针形成的角是 30°;过了 20 分钟，这时时针和分针形成的角是 140°。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "除数是两位数的笔算除法",
+                                 "一般时间与钟面指针的位置",
+                                 "认识几分之一"
+                             ],
+        "source":  "25-26 三年级上·河北唐山·期末"
+    },
+    {
+        "id":  "15",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算□10÷52 时，可以把 52 看作( )来试商;若商是两位数,□里最小能填()。",
+        "options":  [
+
+                    ],
+        "answer":  "50;6",
+        "solution":  "【分析】根据除数是两位数除法试商的方法：把除数看作是和它接近的整十数来进行试商;要使商是两位数，则被除数前两位大于或等于除数，据此解答即可.\n【详解】52 接近 50，□1 ≥ 52，61 ≥ 52，□=6。计算□10÷52 时，可以把52看作50来试商。如果商是两位数，□里最小应填 6。",
+        "knowledge_points":  [
+                                 "多位数除以两位数的试商",
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25 四年级上·浙江嘉兴·期末"
+    },
+    {
+        "id":  "16",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "为给运动会上成绩优异的同学发奖，王老师带 350 元钱去买笔记本,12 元一本，能买( )本，还剩( )元。",
+        "options":  [
+
+                    ],
+        "answer":  "29;2",
+        "solution":  "【分析】三位数除以两位数的笔算法则：从被除数的最高位除起，先用除数试除被除数的前两位数，如果它比除数小，再试除前三位数;除到被除数的哪一位，就在那一位上面写商。求出每一位商，余下的数必须比除数小。由题意得，王老师带350元钱去买笔记本,12元一本，求一共能买多少本，就是求 350 里面有多少个 12，用除法计算.\n【详解】350÷12 = 29(本)……2(元)，故一共能买29本笔记本，还剩2元。",
+        "knowledge_points":  [
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "25-26 四年级上·河北唐山·期末"
+    },
+    {
+        "id":  "17",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "□ 25÷45 的商是两位数,□里最小填( );□39÷36 的商是一位数,□里最大填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "5;3",
+        "solution":  "【分析】判断三位数除以两位数的商是几位数，需要看被除数的前两位是否大于或等于除数。如果前两位大于或等于除数，商是两位数;如果前两位小于除数，商是一位数。据此解题即可.\n【详解】由分析可知,□25÷45 的商是两位数，要求□ 2 \u003e= 45，因此□最小填 5;□39 ÷ 36 的商是一位数，要求□ 3 \u003c 36，因此□最大填3。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25四年级上·重庆丰都·期末"
+    },
+    {
+        "id":  "18",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "□ 68÷35，如果商是一位数,□里最大填( );如果商是两位数,□里最小填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "2;3",
+        "solution":  "【分析】三位数除以两位数，判断商是几位数的方法：先观察被除数前两位上的数，看是否大于或等于除数(也就是看前两位上的数除以除数，够不够商一个十)。被除数前两位够除，商就是两位数。反之，商就是一位数。由题意得，在算式□68÷35中，要使商是一位数，那么被除数的前两位上的数必须小于除数35,□里最大填2;如果商是两位数，那么被除数的前两位上的数必须大于或等于除数35,□里最小填 3 .\n【详解】□ 68÷35，如果商是一位数,□里最大填2;如果商是两位数,□里最小填 3。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "25-26四年级上·辽宁鞍山·期末"
+    },
+    {
+        "id":  "19",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一名同学擦一块玻璃要用5分钟,5名同学同时擦玻璃，擦15块玻璃最少需要( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】根据题意可知,5名同学同时擦玻璃，每名同学擦块玻璃需要5分钟，因此5名同学同时工作，一次可以擦5块玻璃，用时5分钟。擦15块玻璃需要完成 15÷5 = 3 (次)这样的工作，因此总时间为 3×5 = 15 (分钟).\n【详解】 15÷5 = 3 (次) 3×5 = 15 (分 钟) 所以擦15块玻璃最少需要15分钟。",
+        "knowledge_points":  [
+                                 "单归一问题"
+                             ],
+        "source":  "25-26 四年级上·黑龙江绥化·期末"
+    },
+    {
+        "id":  "20",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "□ 58÷46，如果商是一位数,□里最大填( );如果商是两位数,□里最小填()。",
+        "options":  [
+
+                    ],
+        "answer":  "4;5",
+        "solution":  "【分析】依据\u0027除数是两位数时，看被除数前两位和除数的大小：前两位小于除数，商是一位数;前两位大于或等于除数，商是两位数\u0027来解题。对于□58÷46，要使商是一位数，需□5 < 46，□最大填4;要使商是两位数，需□5 ≥ 46，□最小填 5.\n【详解】商是一位数:□5 < 46，□可填 1,2,3,4，最大填 4;商是两位数:□5 ≥ 46，□可填 5,6,7,8,9，最小填 5。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25四年级上·四川绵阳·期末"
+    },
+    {
+        "id":  "21",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个除法算式中，商是 6，余数是 12，除数最小是( ),这时被除数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "13;90",
+        "solution":  "【分析】(1)在除法算式中，余数一定小于除数。已知余数是 12，那么除数要大于 12，所以除数最小是 12 + 1 =13。(2)根据\u0027被除数 = 商×除数+余数\u0027,此时商是 6，除数是 13，余数是 12，代入数值计算出结果。据此解答.\n【详解】 12 + 1 = 13\n\n 6×13 + 12 \n = 78 + 12 \n = 90 \n\n所以一个除法算式中，商是 6，余数是 12，除数最小是13，这时被除数是 90 ",
+        "knowledge_points":  [
+                                 "余数与除数的关系",
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "24-25 四年级上·云南曲靖·期末"
+    },
+    {
+        "id":  "22",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "要使□46÷27 的商是两位数,□里最小填();要使它的商是一位数,□里最大填()。",
+        "options":  [
+
+                    ],
+        "answer":  "3;2",
+        "solution":  "【分析】当商是两位数时，被除数前两位□4大于或等于27，则□最小填 3;当商是一位数时，□4 < 27，□最大填 2.\n【详解】要使□46÷27 的商是两位数，□里最小填 3;要使它的商是一位数，□里最大填 2。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "25-26 四年级上·吉林·期末"
+    },
+    {
+        "id":  "23",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "钟面上5时整，时 针和分针成( )角;( )时整，时针和分针成平角。",
+        "options":  [
+
+                    ],
+        "answer":  "钝;6",
+        "solution":  "【分析】钟面 360° 被12个数字分成12等份，每两个数字间的大格都是 30°。比如5时整，时针指向5，分针指向12，中间隔了5个大格，30°×5 = 150°，这是个钝角;要是想让时针分针摆出平角(180°)，那得凑够6个大格，刚好6时整时，时针分针一个指向 6，一个指向 12.\n【详解】钟面每大格角度:360°÷12 = 30°。5时整间隔大格数:5个，夹角:5×30° = 150°，150°是钝角。平角是 180°，对应大格数:180°÷30° = 6(个)。当间隔6个大格时，时刻为6时整。钟面上5时整，时针和分针成钝角;6时整，时针和分针成平角。",
+        "knowledge_points":  [
+                                 "角的度量",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "25-26四年级上·吉林松原·期末"
+    },
+    {
+        "id":  "24",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\u0027\u003e\u0027\u0027\u003c\u0027或\u0027=\u0027.\n\n631999( )632119298×29( )300×30542÷53( )480÷62",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003c;\u003e",
+        "solution":  "【分析】多位数的大小比较，数位不同时，位数多的就大。数位相同时，从最高位比起，如果最高位上的数字相同，就比较下一位，依次类推，直到比较出大小。据此比较出631999 和 632119 的大小。三位数乘两位数的竖式计算时，个位对齐。先用两位数个位上的数和三位数的每一位上的数相乘，乘得结果和个位对齐，再用两位数十位上的数和三位数的每一位上的数相乘，乘得结果和十位对齐。最后把两次乘得结果相加。末尾有 0 时，把两个因数 0 前面的数对齐，并将它们相乘，再在积的后面添上没有参加运算的几个0。据此比较 298×29 和 300×30 的大小。除数是两位数的除法的笔算法则：从被除数的高位除起，先看被除数的前两位;如果前两位比除数小，就要看前三位;除到被除数的哪一位，商就写在那一位的上面;余下的数必须比除数小。据此比较 542÷53 和 480÷62.\n【详解】根据多位数的大小比较，631999 < 632119。298×29 = 8642，300×30 = 9000，8642 < 9000，所以 298×29 < 300×30。542÷53 = 10……12，480÷62 = 7……46，10 > 7，所以 542÷53 > 480÷62。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "三位数与两位数的乘法",
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "24-25 四年级上·四川绵阳·期末"
+    },
+    {
+        "id":  "25",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "要使5☐7÷56 的商是一位数，☐里最大是( );要使商是两位数，☐里最小是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "5;6",
+        "solution":  "【分析】根据除数是两位数的除法规则，若被除数的前两位小于除数，则商是一位数，若前两位大于或等于除数，则商是两位数。据此解答.\n【详解】要使5☐7÷56 的商是一位数，5☐ < 56，□可以填0,1,2,3,4,5，□里最大填5。要使商是两位数，5☐ ≥ 56，□可以填6,7,8,9，□里最小填 6。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25四年级上·湖北武汉·期末"
+    },
+    {
+        "id":  "26",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "《意林》是小华最喜欢的杂志，它是\u0027半月刊\u0027,也就是说每个月发行两期。小华买到的最新一期是第 432 期。这本杂志已经创刊了()年。",
+        "options":  [
+
+                    ],
+        "answer":  "18",
+        "solution":  "【分析】这道题依据半月刊的发行周期(每月发行 2 期)来计算：先算出每年发行的期数为 12×2 = 24 期，再用总期数432除以每年的24期，得出这本杂志创刊了18年.\n【详解】432÷(12×2)= 432÷24 = 18 这本杂志已经创刊了18年。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "24-25四年级上·北京朝阳·期末"
+    },
+    {
+        "id":  "27",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\u0027\u003e\u0027\u0027\u003c\u0027或者\u0027=\u0027.\n\n20000 毫升( )19 升 810 万( )80100000\n\n54 升( )5400 毫升 390÷30( )3900÷300",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;\u003c;\u003e;=",
+        "solution":  "【分析】第一，三题比较容积单位，需统一单位:1升=1000毫升，先换算再比较数值。第二题比较大数，将\u0027万\u0027作单位的数去掉\u0027万\u0027字补4个0，再与右边数比较。第四题比较除法算式，利用商不变性质(被除数和除数同时乘或除以相同非零数，商不变)判断两边是否相等.\n【详解】20000毫升 = 20 升，20升 > 19升，填\u0027>\u0027;810 万 = 8100000，8100000 < 80100000，填\u0027<\u0027;54 升 = 54000 毫升，54000毫升 > 5400毫升，填\u0027>\u0027;390÷30 与 3900÷300，被除数和除数同时乘10，商不变，填\u0027=\u0027。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "容积单位间的进率与换算(升和毫升)",
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "25-26 四年级上·河北唐山·期末"
+    },
+    {
+        "id":  "28",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "小明酷爱读书，尤其是四大名著之一的《三国演义》,这本书共 650 页，他 3天读了 240 页，照这个速度，一个星期(7 天)能读( )页。",
+        "options":  [
+
+                    ],
+        "answer":  "560",
+        "solution":  "【分析】先根据已知条件\u00273天读240页\u0027求出小明每天读的页数，即 240÷3 = 80 页。再根据每天读的页数，乘一个星期的天数(7天),就能得到一周能读的总页数。计算时注意整数除法和乘法的准确运算.\n【详解】小明每天读的页数: 240÷3 = 80 (页)一个星期(7 天)能读的页数: 80×7 = 560 (页)照这个速度，一个星期(7 天)能读 560 页。",
+        "knowledge_points":  [
+                                 "两位数与一位数的乘法口算",
+                                 "三位数除以一位数，首位不够除",
+                                 "单归一问题"
+                             ],
+        "source":  "25-26 四年级上·河北唐山·期末"
+    },
+    {
+        "id":  "29",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "红星实验小学每天上午 9 点开始做眼保健操，此时时针与分针互相( ),组成的角是( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "垂直;直",
+        "solution":  "【分析】根据题意，钟面上一周有12个大格，是360°，因此每个大格是:360°÷12 = 30°，钟面上9时整，时针和分针之间的夹角有3个大格，因此用30°乘3即可;并根据角的分类标准填空即可，当钟面上的时针与分针成90°时，此时钟面上的时针与分针互相垂直。据此解答.\n【详解】根据分析可知:360°÷12 = 30°，30°×3 = 90°。红星实验小学每天上午9点开始做眼保健操，此时时针与分针互相垂直，组成的角是直角。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "垂直的特征",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "25-26 四年级上·河北唐山·期末"
+    },
+    {
+        "id":  "30",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算 814÷19 时，可以把 19 看作( )试商，正确结果是( ),要想没有余数，被除数最少应增加( )。",
+        "options":  [
+
+                    ],
+        "answer":  "20;42余16;3",
+        "solution":  "【分析】在除数是两位数的除法试商中，把除数看作与它接近的整十数进行试商;计算出 814÷19 的商和余数;用19减去余数，求出的差就表示被除数至少增加多少时，商是整数且没有余数。据此解答.\n【详解】814÷19 = 42余16，19 - 16 = 3。故计算 814÷19 时，可以把19看作20试商，正确结果是42余16，要想没有余数，被除数最少应增加3。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "25-26 四年级上·山东菏泽·月考"
+    },
+    {
+        "id":  "31",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在( )里填上\u0027>\u0027\u0027<\u0027或\u0027=\u0027。\n\n6450000( )999999  27万( )30万\n\n600÷40( )6000÷400  630×3( )63×30",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;\u003c;=;=",
+        "solution":  "【分析】比较 6450000 和 999999，6450000 是七位数，999999 是六位数，位数多的数更大，所以 6450000 > 999999;比较27万和30万，单位相同，直接比较数值，27 < 30，所以 27万 < 30万;比较 600÷40 和 6000÷400，根据商不变性质，被除数和除数同时乘 10，商不变，所以 600÷40 = 6000÷400;比较 630×3 和 63×30，630×3 = 1890，63×30 = 1890，所以630×3 = 63×30.\n【详解】6450000 > 999999;27万 < 30万;600÷40 = 6000÷400;630×3 = 63×30。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "两位数乘整十数的口算乘法",
+                                 "三位数与整十数的口算乘法",
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "25-26 四年级上·辽宁鞍山·期末"
+    },
+    {
+        "id":  "32",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个乘法算式的积是 360，其中一个因数除以 12，另一个因数不变，积是()。",
+        "options":  [
+
+                    ],
+        "answer":  "30",
+        "solution":  "【分析】根据乘法算式中积的变化规律：一个因数不变，另一个因数除以几(0除外)，积也除以相同的数。本题中，原积是360，一个因数除以12，另一个因数不变，因此积也除以 12.\n【详解】360÷12 = 30。一个乘法算式的积是 360，其中一个因数除以 12，另一个因数不变，积是 30。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)",
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "25-26 四年级上·辽宁铁岭·期末"
+    },
+    {
+        "id":  "33",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一道除法算式的商是800，如果被除数和除数同时去掉末尾的一个0，那么商是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "800",
+        "solution":  "【分析】在除法算式里，被除数和除数同时乘或除以相同的数(0除外),商不变。据此解答.\n【详解】一道除法算式的商是800，如果被除数和除数同时去掉末尾的一个0，也就是把被除数和除数同时除以10，那么商还是800。",
+        "knowledge_points":  [
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "24-25 四年级上·天津南开·期末"
+    },
+    {
+        "id":  "34",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "已知 A×B = 300，如果 A 不变,B 乘 6，则积是( );已知 A÷B = 60，如果A,B 都乘 6，则商是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "1800;60",
+        "solution":  "【分析】根据积随因数变化的规律，一个因数不变，另一个因数乘或除以几(0 除外),积也乘或除以几;根据商不变规律，被除数和除数同时乘或除以相同的数(0除外),商不变。据此解答.\n【详解】300×6 = 1800。A÷B = 60，被除数和除数同时乘 6，商不变;即已知 A×B = 300，如果 A 不变,B 乘 6，则积是 1800;已知 A÷B = 60，如果 A,B 都乘 6，则商是 60。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)",
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "24-25 四年级上·山东济宁·期末"
+    },
+    {
+        "id":  "35",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "传统年画是我国民间艺术之一，含有祝福新年吉祥喜庆之意。年画坊印制了12 种图案的年画共 420 幅，平均每种图案的年画印制了()幅。如果每幅年画 15 元，那么这些年画一共( )元。",
+        "options":  [
+
+                    ],
+        "answer":  "35;6300",
+        "solution":  "【分析】根据题意可知,12种图案的年画共420幅，求平均每种图案的年画印制的数量，列除法算式即 420÷12;如果每幅年画15元，那么用年画的总数量 420×15，即可求出这些年画的总价。据此解答.\n【详解】 420÷12 = 35 (幅)所以平均每种图案的年画印制了35幅。 420×15 = 6300 (元)所以这些年画一共6300元。",
+        "knowledge_points":  [
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "25-26 四年级上·陕西渭南·期末"
+    },
+    {
+        "id":  "36",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "5880÷68，把除数 68看成( )试商，商是( ),这时试商所得到的商应比实际的商( )。",
+        "options":  [
+
+                    ],
+        "answer":  "70;84;小",
+        "solution":  "【分析】在除数是两位数的除法中，试商时通常将除数看作最接近的整十数。68接近70，因此把除数68看成70试商。计算5880÷70，即可得出商是多少。由于把除数68看成70，在被除数不变的情况下，除数越大，商越小，所以试商所得到的商比实际的商小。据此解答.\n【详解】5880÷70 = 84。5880÷68，把除数68看成70试商，商是84，这时试商所得到的商应比实际的商小。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "多位数除以两位数的试商"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "37",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "除数是两位数的除法计算法则：从被除数的( )位除起，先用除数试除被除数的前两位，如果它比除数小，再试除前( )位数，求出每一位商，余下的数必须比除数( )。",
+        "options":  [
+
+                    ],
+        "answer":  "高;三;小",
+        "solution":  "【分析】除数是两位数的除法的计算法则：从被除数的最高位除起，先看被除数的前两位;如果前两位比除数小，就要看前三位;除到被除数的哪一位，商就写在那一位的上面;余下的数必须比除数小，除到被除数的某一位不够除时，应商0占位。据此解答.\n【详解】除数是两位数的除法计算法则：从被除数的高位除起，先用除数试除被除数的前两位，如果它比除数小，再试除前三位数，求出每一位商，余下的数必须比除数小。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "24-25 四年级上·广西玉林·期末"
+    },
+    {
+        "id":  "38",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "58□÷53 的商是()位数，要使商的末尾是 0,□里可以填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "两;0、1、2",
+        "solution":  "【分析】三位数除以两位数列竖式：从被除数的高位除起，先看被除数的前两位，如果前两位比除数小，就看前三位;除到被除数的哪一位，就把商写在那一位的上面;每次除得的余数必须比除数小，得到结果。58□÷53，前两位数58 > 53，所以商是两位数，这时商1写在十位上，余数是5，要想使商的末尾是0，那么就是5□ < 53，因此□里可以填0,1,2。据此分析解答.\n【详解】58□÷53的商是两位数，要使商的末尾是0，□里可以填0、1、2(数字顺序不唯一)。",
+        "knowledge_points":  [
+                                 "商末尾有0的除法",
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25 四年级上·重庆秀山·期末"
+    },
+    {
+        "id":  "39",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "新华书店举行\u0027买四赠一\u0027活动，王老师买 15 本同样的故事书共付了 60 元，每本故事书的原价是( )元。",
+        "options":  [
+
+                    ],
+        "answer":  "5",
+        "solution":  "【分析】根据题意，把\u0027买四赠一\u0027看作一组(这一组里面4本书是买的，另外1本是额外赠送的),先用除法求出15本里有几组，再用每组买的本数(4本)乘组数，求出15本里面实际是付钱买的本数;根据\u0027总价/数量 = 单价\u0027用实际付的钱数除以实际需要付钱购买的本数，求出每本书的原价.\n【详解】15÷(4+1)= 15÷5 = 3 (组) 不含赠送能买的本数: 4×3 = 12 (本)每本书的原价: 60÷12 = 5 (元)新华书店举行\u0027买四赠一\u0027活动，王老师买15本同样的故事书共付了60元，每本故事书的原价是5元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "25-26 六年级上·广东揭阳·期末"
+    },
+    {
+        "id":  "40",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "□70÷58，要使商是两位数,□里最小填( );要使商是一位数,□里最大填()。",
+        "options":  [
+
+                    ],
+        "answer":  "6;5",
+        "solution":  "【分析】三位数除以两位数，如果被除数的前两位数大于或等于除数，商是两位数，若被除数的前两位数小于除数，商是一位数，据此解答即可.\n【详解】□70÷58，要使商是两位数，说明□7大于或等于58。□里可以填 6,7,8,9,□最小是 6。□70÷58，要使商是一位数，说明□ 7 \u003c 58 。□里可以填1,2,3,4,5。□里最大填 5。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "25-26 四年级上·辽宁铁岭·期末"
+    },
+    {
+        "id":  "41",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "相声是一种民间说唱曲艺，它以说，学，逗，唱为主要形式。如果一个相声团体一年演出达到 204 场，那么平均每月演出( )场。",
+        "options":  [
+
+                    ],
+        "answer":  "17",
+        "solution":  "【分析】根据题意，一年有12个月，用一年演出达到的场次除以 12，就是平均每月演出几场.\n【详解】 204÷12 = 17 (场)那么，平均每月演出17场。",
+        "knowledge_points":  [
+                                 "年，月，日的认识及换算",
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "25-26 四年级上·广西贵港·期中"
+    },
+    {
+        "id":  "42",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "270÷30，就是算27个( )除以 3 个( ),得数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "十;十;9",
+        "solution":  "【分析】几百几十的数除以几十数的口算时，可以将几百几十数和几十数看作几个十，然后用表内除法来口算.\n【详解】根据分析, 270÷30，就是算 27 个十除以 3 个十,27÷3 = 9，得数是 9 ",
+        "knowledge_points":  [
+                                 "除数是整十数的口算除法"
+                             ],
+        "source":  "25-26四年级上·河南洛阳·期中"
+    },
+    {
+        "id":  "43",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上\u0027>\u0027\u0027<\u0027或\u0027=\u0027。\n\n360×40 ( ) 36×400  298×31 ( ) 9000  450÷90 ( ) 45÷9",
+        "options":  [
+
+                    ],
+        "answer":  "=;\u003e;=",
+        "solution":  "【分析】比较两个整数的大小时，先看位数，位数多的数大。如果位数相同，从最高位比起，相同数位上数字大的那个数大;第一个：两个数相乘，一个乘数乘(或除以)一个数(0除外),另一个乘数除以(或乘)相同的数，积不变;第二个：计算出算式的结果再比较大小;第三个：根据商不变的性质：被除数和除数同时乘(或除以)一个相同的数(0 除外),商不变.\n【详解】360 到 36 相当于除以10，40 到 400 相当于乘10，一个乘数乘10，另一个乘数除以10，积不变，所以 360×40 = 36×400;298×31 = 9238，9238 > 9000，所以 298×31 > 9000;450÷10 = 45，90÷10 = 9，所以 450÷90 = 45÷9。",
+        "knowledge_points":  [
+                                 "万以内数的大小比较",
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "积的变化规律(整数乘法)",
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "25-26 四年级上·山东济宁·期中"
+    },
+    {
+        "id":  "44",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\u0027>\u0027\u0027<\u0027或\u0027=\u0027。\n\n5698749( )5689749  2695000( )270万  32×120( )16×240\n\n630÷35( )630÷70  801÷21( )40  480÷32( )480÷4÷8",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;\u003c;=;\u003e;\u003c;=",
+        "solution":  "【分析】本题考查了比较大小的能力。整数比较大小时，先看位数，位数多的数大;位数相同则从最高位开始逐位比较，直到分出大小;先把 270 万写成 2700000，再与2695000进行比较，后面的式子可通过计算，再比较大小即可.\n【详解】5698749 与 5689749 都是七位数，从最高位开始逐位比较，得出 5698749 > 5689749;270 万 = 2700000，2695000 < 2700000，则 2695000 < 270 万;32×120 = 16×2×120 = 16×(2×120) = 16×240，则 32×120 = 16×240;630÷35 = 18，630÷70 = 9，18 > 9，则 630÷35 > 630÷70;801÷21 = 38……3，38 < 40，则 801÷21 < 40;480÷32 = 480÷(4×8) = 480÷4÷8，则 480÷32 = 480÷4÷8。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "除数是整十数的口算除法",
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "25-26四年级上·河北唐山·期末"
+    },
+    {
+        "id":  "45",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "已知 840÷14 = 60，那么 840÷28 = ( )，420÷( ) = 60。",
+        "options":  [
+
+                    ],
+        "answer":  "30;7",
+        "solution":  "【分析】商的变化规律：除数(0除外)不变，被除数(0除外)扩大到原来的几倍或缩小到原来的几分之一，商就扩大到相同的倍数或缩小到原来的几分之一(0 除外);被除数(0除外)不变，除数(0除外)扩大则商反而缩小，除数缩小商就扩大。商不变的规律：被除数(0除外)和除数(0除外)同时乘或除以同一个数(不为 0),商不变。已知 840÷14 = 60，要求 840÷28，观察可以发现被除数不变，除数乘2，则商要除以2。要求420除以多少等于60，观察发现商不变，被除数除以2，则除数也要除以2。据此解答即可.\n【详解】由分析得, 14×2 = 28，被除数不变除数乘 2，则商要除以 2, 60÷2 = 30.840÷2 = 420，要商不变，除数也要除以 2, 14÷2 = 7 。因此已知 840÷14 = 60，那么 840÷28 = 30 , 420÷7 = 60 ",
+        "knowledge_points":  [
+                                 "商的变化规律及应用",
+                                 "除数是两位数的笔算除法",
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "25-26 四年级上·河北唐山·期末"
+    },
+    {
+        "id":  "46",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "46 □ ÷ 27 的商是()位数,51□÷62 的商是( )位数。",
+        "options":  [
+
+                    ],
+        "answer":  "两;一",
+        "solution":  "【分析】三位数除以两位数，判断商是几位数的方法：先观察被除数前两位上的数，看是否大于或等于除数(也就是看前两位上的数除以除数，够不够商一个十)。被除数前两位够除，商就是两位数。反之，商就是一位数。由题意得，在算式46□÷27中，被除数前两位上的\u002746\u0027大于除数27，所以商是两位数;在算式51□÷62中，被除数前两位上的\u002751\u0027小于除数62，所以商是一位数.\n【详解】46□÷27的商是两位数,51□÷62的商是一位数。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "25-26 四年级上·陕西渭南·期末"
+    },
+    {
+        "id":  "47",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "□37÷47，要使商是两位数,□里最小填( ),要使商是一位数,□里最大填()。",
+        "options":  [
+
+                    ],
+        "answer":  "5;4",
+        "solution":  "【分析】要使□37÷47的商是两位数，则被除数前两位上的数大于或等于除数，□3大于或等于47，□里的数要大于4。要使□37÷47的商是一位数，则被除数前两位上的数小于除数，□3 小于 47，□里的数要小于或等于 4.\n【详解】由分析可得:□37÷47，要使商是两位数，□里最小填 5;要使商是一位数，□里最大填 4。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25 四年级上·重庆开州·期末"
+    },
+    {
+        "id":  "48",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "□29÷65，要使商是两位数, □里最小填( );要使商是一位数。 □里最大填()。",
+        "options":  [
+
+                    ],
+        "answer":  "7;6",
+        "solution":  "【分析】三位数除以两位数，如果被除数的前两位大于等于除数，则商是两位数，否则商是一位数，据此即可解答.\n【详解】根据分析可知:□29÷65，要使商是两位数，□里可以填7,8,9，所以□里最小填 7。要使商是一位数，□里可以填1,2,3,4,5,6，□里最大填 6。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25四年级上·甘肃兰州·期末"
+    },
+    {
+        "id":  "49",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\u0027>\u0027\u0027<\u0027或\u0027=\u0027。\n\n6508080000( )6580800000  566203( )56万  19×20( )20×21\n\n16×400( )32×200  467÷17( )30  270÷(6×9)( )270÷9÷6",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003e;\u003c;=;\u003c;=",
+        "solution":  "【分析】(1)比较两个整数的大小，要看他们的数位，如果数位不同，那么数位多的数就大，如果数位相同，相同数位上的数大的那个数就大;(2)将 56 万改写成 560000，然后比较大小即可;(3)分别计算出 19×20 和 20×21 的积，然后比较大小即可;(4)分别计算出 16×400 和 32×200 的积，然后比较大小即可;(5)计算出 467÷17 的商，然后比较大小即可;(6)分别计算出 270÷(6×9)和 270÷9÷6 的商，然后比较大小即可.\n【详解】6508080000 < 6580800000;56万 = 560000，566203 > 560000，所以 566203 > 56万;19×20 = 380，20×21 = 420，380 < 420，所以 19×20 < 20×21;16×400 = 6400，32×200 = 6400，6400 = 6400，所以 16×400 = 32×200;467÷17 = 27……8，27 < 30，所以 467÷17 < 30;270÷(6×9) = 270÷54 = 5，270÷9÷6 = 30÷6 = 5，5 = 5，所以 270÷(6×9) = 270÷9÷6。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "除数是两位数的笔算除法",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "25-26四年级上·辽宁鞍山·期末"
+    },
+    {
+        "id":  "50",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一片树林每年可以产生 912 千克氧气，这片树林平均每个月可以产生( )千克氧气。",
+        "options":  [
+
+                    ],
+        "answer":  "76",
+        "solution":  "【分析】根据题意，已知一年有12个月，则用这片树林每年可以产生氧气的千克数除以12，即可求出这片树林平均每月可以产生氧气的千克数.\n【详解】 912÷12 = 76 (千克)即一片树林每年可以产生912千克氧气，这片树林平均每个月可以产生 76 千克氧气。",
+        "knowledge_points":  [
+                                 "年，月，日的认识及换算",
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "25-26 四年级上·辽宁抚顺·期末"
+    },
+    {
+        "id":  "51",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "673÷62，可以把 62看作( )试商，商的最高位是( )位，商是( )位数。",
+        "options":  [
+
+                    ],
+        "answer":  "60;十;两",
+        "solution":  "【分析】根据三位数除以两位数的计算，可以将两位数估成和它接近的整十数再计算;如果被除数前两位大于或等于除数，则商是两位数，如果被除数前两位小于除数，则商是一位数，据此填空即可.\n【详解】 673÷62 中, 67 \u003e 62，商是两位数。673除以62，可以把62看作60试商，商的最高位是十位，商是两位数。",
+        "knowledge_points":  [
+                                 "多位数除以两位数的试商",
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "25-26 四年级上·辽宁鞍山·期末"
+    },
+    {
+        "id":  "52",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "除法算式 5□9÷54，如果商是一位数,□里最大填( );如果商是两位数,□里最小填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "3;4",
+        "solution":  "【分析】根据三位数除以两位数的除法计算方法可知，要使除法算式5□9÷54的商是一位数，则被除数前两位上的数小于除数,□里的数要小于4，最大填3;要使商是两位数,□里的数要大于或者等于4，最小填4，据此解答.\n【详解】除法算式5□9÷54，如果商是一位数,□里最大填3;如果商是两位数,□里最小填4。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "25-26 四年级上·甘肃陇南·期末"
+    },
+    {
+        "id":  "53",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "25×400 的积的末尾有( )个 0;651÷58 的商是( )位数。",
+        "options":  [
+
+                    ],
+        "answer":  "4;两",
+        "solution":  "【分析】根据三位数乘两位数的计算，计算出 25×400 的结果后判断积的末尾有几个0即可;根据三位数除以两位数的计算，如果被除数前两位大于或等于除数，则商是两位数，如果被除数前两位小于除数，则商是一位数，据此判断651÷58的商是几位数即可.\n【详解】25×400 = 10000，651÷58 中, 65 > 58，商是两位数。所以25×400的积的末尾有4个0;651÷58的商是两位数。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)",
+                                 "三位数乘两位数，乘数末尾有0"
+                             ],
+        "source":  "24-25 四年级上·新疆克孜勒苏·期末"
+    },
+    {
+        "id":  "54",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "436÷□8，如果商是一位数，那么□最小填( );如果商是两位数，那么□最大填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "4;3",
+        "solution":  "【分析】被除数的前两位比除数小，商是一位数;被除数的前两位大于或等于除数，商是两位数，由此可知，商是一位数时,43 < □8;商是两位数时,43 ≥ □8，据此进行解答.\n【详解】436÷□8，要使商是一位数，那么 43 < □8，因为 3 < 8，所以□ > 4 或□ = 4，□内可以填 4,5,6,7,8,9，最小填 4;436÷□8，要使商是两位数，那么 43 ≥ □8，因为 3 < 8，所以□ < 4，□内可以填 1,2,3，最大填 3。所以436÷□8，如果商是一位数，那么□最小填4;如果商是两位数，那么□最大填3。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25 四年级上·安徽芜湖·期末"
+    },
+    {
+        "id":  "55",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在算式 5 □ 6÷52 中，当商是一位数时,□里最大填( )。当商是两位数时,□里最小填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "1;2",
+        "solution":  "【分析】在除法算式 5□6÷52 中，商是一位数时，要求被除数小于除数 52 的 10 倍即 520，因此 5□6 < 520，比较百位和十位数字,□必须小于2，最大填1;商是两位数时，要求被除数大于或等于520，因此 5□6 ≥ 520，□必须大于或等于2，最小填 2.\n【详解】根据分析可知：在算式 5□6÷52 中，当商是一位数时,□里最大填1。当商是两位数时,□里最小填2。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25 四年级上·浙江杭州·期末"
+    },
+    {
+        "id":  "56",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "312÷39 的商是( )位数。",
+        "options":  [
+
+                    ],
+        "answer":  "一或1",
+        "solution":  "【分析】要判断 312÷39 的商是几位数，可以通过比较被除数的前两位与除数的大小，如果被除数的前两位大于或等于除数，商就是两位数;如果被除数的前两位小于除数，商就是一位数。据此解答.\n【详解】被除数的前两位是31, 31 \u003c 39 因此, 312÷39 的商是一位数。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)",
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "57",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "两个数相除，商是 9，余数是 5，如果被除数和除数同时乘 10，这时商是(),余数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "9;50",
+        "solution":  "【分析】根据商不变性质，被除数和除数同时乘不为0的数，商不变，但是余数要乘相同的数.\n【详解】根据分析，如果被除数和除数同时乘 10，商是 9;余数为 5×10 = 50 。",
+        "knowledge_points":  [
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "25-26 四年级上·黑龙江绥化·期末"
+    },
+    {
+        "id":  "58",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "根据 51÷3 = 17，填一填.\n\n510÷30 = ( ) 102 ÷ ( ) =17 ( ) ÷ 15 = 17",
+        "options":  [
+
+                    ],
+        "answer":  "17;6;255",
+        "solution":  "【分析】商不变的规律：被除数和除数同时乘或除以一个相同的数(0 除外),商不变，据此解答.\n【详解】 51÷3 = 17 (51×10) ÷ (3×10) = 510÷30 = 17 (51×2) ÷ (3×2) = 102÷6 = 17 (51×5) ÷ (3×5) = 255÷15 = 17 ",
+        "knowledge_points":  [
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "24-25 四年级上·河北保定·期末"
+    },
+    {
+        "id":  "59",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小方在计算一道除法算式时，把除数24看成了42，结果得到商为28，那么被除数为( ),正确的商是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "1176;49",
+        "solution":  "【分析】先用错误的除数42乘错误的商28得到正确的被除数，再用正确的被除数除以正确的除数24得到正确的商.\n【详解】 42×28 = 11761176÷24 = 49 所以被除数为1176，正确的商是 49 ",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "两位数乘两位数的进位乘法"
+                             ],
+        "source":  "25-26四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "60",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "已知 a×b = 30，如果a不变,b乘 10，那么积是( );已知 a÷b = 30，如果a, b 同时除以 10，那么商是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "300;30",
+        "solution":  "【分析】在乘法中，一个因数不变，另一个因数乘几，积也乘几;在除法中，被除数和除数同时乘或除以相同的数(0 除外),商不变.\n【详解】已知 a×b = 30，a 不变,b 乘 10，则积也乘 10，即30×10 = 300。已知 a÷b = 30，a 和 b 同时除以 10，则商不变，仍是 30。已知 a×b = 30，如果a不变,b乘 10，那么积是 300;已知 a÷b = 30，如果a,b同时除以 10，那么商是 30。",
+        "knowledge_points":  [
+                                 "商不变的规律及应用",
+                                 "积的变化规律(整数乘法)",
+                                 "两位数乘整十数的口算乘法"
+                             ],
+        "source":  "25-26 四年级上·辽宁盘锦·期末"
+    },
+    {
+        "id":  "61",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "除法算式□81÷47，如果商是一位数,□里最大填( ),如果商是两位数,□里最小填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "3;4",
+        "solution":  "【分析】依据\u0027除数是两位数时，商的位数由被除数前两位与除数的大小关系决定\u0027的规则：当被除数前两位□8小于除数47时，商是一位数，此时□里可填1,2,3，最大填3;当被除数前两位□8大于或等于除数47时，商是两位数，此时□里可填 4,5,6,7,8,9，最小填 4.\n【详解】依据除数是两位数的商的位数判定规则,□8小于47时商是一位数,□最大填3;□8大于或等于47时商是两位数,□最小填4。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25 四年级上·江西上饶·期末"
+    },
+    {
+        "id":  "62",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "562 ÷□ 5，要使商是两位数,□ 里最大填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "5",
+        "solution":  "【分析】要使得商是两位数，则 56 要大于或等于□ 5,□ 里可以填 1,2,3,4,5，最大填 5.\n【详解】562 ÷□ 5，要使商是两位数,□ 里最大填 5。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "25-26 四年级上·北京丰台·期末"
+    },
+    {
+        "id":  "63",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "627÷23，把除数看作20 去试商比较简便，这时商会偏( )(填\u0027大\u0027或\u0027小\u0027)",
+        "options":  [
+
+                    ],
+        "answer":  "大",
+        "solution":  "【分析】除法计算中,\u0027被除数不变，除数变小，商就会变大\u0027。 计算 627÷23 时，把除数23看作20(除数变小了),用 20 去试商，此时被除数不变，除数变小，所以试出的商会比实际的商偏大.\n【详解】计算 627÷23 时，把除数 23 看作 20(除数变小),根据\u0027被除数不变，除数越小商越大\u0027,这时商会偏大。",
+        "knowledge_points":  [
+                                 "多位数除以两位数的试商"
+                             ],
+        "source":  "24-25 四年级上·云南昭通·期末"
+    },
+    {
+        "id":  "64",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "要使□96÷82 的商是一位数,□里最大填( );要使3□9÷36 的商是两位数,□里最小填( );计算时将除数82估成80来试商，商易偏( )(填\u0027大\u0027或\u0027小\u0027)。",
+        "options":  [
+
+                    ],
+        "answer":  "7;6;大",
+        "solution":  "【分析】三位数除以两位数试商时，把两位数看成与它接近的整十数，再计算;三位数除以两位数，如果被除数的前两位大于等于除数，商是两位数，如果被除数的前两位小于除数，商是一位数。试商时如果除数被往多了估则试商时容易偏小，如果除数被往少了估则试商时容易偏大.\n【详解】要使□96÷82的商是一位数,□9 < 82，□里可以填7,6,5,4,3,2,1，所以□里最大填7;3□9÷36的商是两位数,3□ ≥ 36，□里可以填 6,7,8,9，所以□里最小填6。计算时将除数82估成80来试商，把82往少了估，商容易偏大。所以要使□96÷82 的商是一位数,□里最大填7;要使3□9÷36 的商是两位数,□里最小填6;计算时将除数82估成80来试商，商易偏大。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)",
+                                 "多位数除以两位数的试商"
+                             ],
+        "source":  "24-25 四年级上·四川凉山·期末"
+    },
+    {
+        "id":  "65",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\u0027>\u0027\u0027<\u0027或\u0027=\u0027。\n\n7300001( )730万  5600÷280( )560÷28  平角-钝角( )直角",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;=;\u003c",
+        "solution":  "【分析】比较7300001和730万的大小，把730万展开后再和7300001进行比较;大数的比较：先比较两个数的位数，位数多的数就大。如果两个数的位数相同，就从最高位比起，最高位上的数大的那个数就大，如果最高位上的数相同，就比较下一个数位上的数;比较 5600÷280 和 560÷28 的大小，根据商不变的性质进行比较;在除法算式里，被除数和除数同时乘或除以相同的数(0 除外),商不变。根据锐角，直角，钝角和平角的含义：大于0°小于90°的角是锐角，等于90°的角是直角，大于90°小于180°的角是钝角，平角是180°，进行比较即可.\n【详解】因为 730万 = 7300000，所以 7300001 > 730万。5600÷280，被除数 5600 除以10得560，除数280除以10得28，即被除数和除数同时除以10，商不变，所以 5600÷280 = 560÷28;180°减去一个大于90°的角，结果小于90°，所以平角-钝角 < 直角。",
+        "knowledge_points":  [
+                                 "平角，周角的认识及特征",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "除数是两位数的笔算除法",
+                                 "大数的比较"
+                             ],
+        "source":  "24-25 四年级上·云南昭通·期末"
+    },
+    {
+        "id":  "66",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个没有余数的除法算式的除数是28，商是20，被除数是( ),如果被除数和除数同时乘 2，商是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "560;20",
+        "solution":  "【分析】两位数乘两位数的计算方法，相同数位对齐，先用第二个因数个位上的数去乘第一个因数的每一位，结果的末位和第二个因数个位对齐，再用第二个因数十位上的数去乘第一个因数的每一位，所得的结果和第二个因数的十位对齐，最后把两个积相加。被除数 = 除数×商，商的变化规律：被除数和除数同时乘或除以一个相同的数(0除外),商不变，据此解答.\n【详解】 28×20 = 560 被除数和除数同时乘或除以一个相同的数(0除外),商不变;一个没有余数的除法算式的除数是28，商是20，被除数是560，如果被除数和除数同时乘2，商是 20 ",
+        "knowledge_points":  [
+                                 "两位数乘整十数的口算乘法",
+                                 "除数是两位数的笔算除法",
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "24-25四年级上·河北保定·期末"
+    },
+    {
+        "id":  "67",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "612÷32，除数可以看成( )来试商，它的商是( )位数。",
+        "options":  [
+
+                    ],
+        "answer":  "30;两",
+        "solution":  "【分析】除数是两位数的除法计算中，一般把除数看作与它接近的整十数来试商;三位数除以两位数，如果被除数的前两位大于等于除数，商是两位数，否则商是一位数，据此解答.\n【详解】在计算 612÷32，除数可以看成30来试商，它的商是两位数。",
+        "knowledge_points":  [
+                                 "多位数除以两位数的试商",
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25 四年级上·西藏林芝·期末"
+    },
+    {
+        "id":  "68",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "《九章算术》是我国古代一部伟大的数学著作。它由很多实际问题及解答组成。书中开篇第一题及解答为：今有田广十五步，从十六步，问为田几何。答曰：一亩。这道题说的是求长方形田的面积。其中\u0027步\u0027为长度单位,\u0027亩\u0027为面积单位。\u0027广\u0027和\u0027从\u0027指长方形的长和宽。若还有长方形田，广30 步，从 16 步，则田的面积为( )亩。",
+        "options":  [
+
+                    ],
+        "answer":  "2",
+        "solution":  "【分析】根据题意,\u0027广\u0027和\u0027从\u0027指长方形的长和宽。根据长方形的面积 = 长×宽，求出\u0027今有田广十五步，从十六步\u0027的长方形田的面积，即可得出1亩=240平方步;再求出广30步，从16步的长方形田的面积，并换算成亩数.\n【详解】 15×16 = 240 (平方步)30×16 = 480 (平方步) 480÷240 = 2 (亩) 若有长方形田，广30步，从16步，则田的面积为2亩。",
+        "knowledge_points":  [
+                                 "长方形的面积"
+                             ],
+        "source":  "24-25四年级上·重庆开州·期末"
+    },
+    {
+        "id":  "69",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\u0027>\u0027\u0027<\u0027或\u0027=\u0027。\n\n30×180 ( ) 30×190  830×0 ( ) 0÷830  935÷67 ( ) 206÷43  23×110 ( ) 403×21",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;=;\u003e;\u003c",
+        "solution":  "【分析】(1)(4)三位数乘两位数的笔算法则：先用两位数的个位分别与三位数的每一位数相乘，乘得的积的末尾和个位对齐。再用两位数的十位分别与三位数的每一位数相乘，乘得的积的末尾和十位对齐。最后，将两次乘得的积相加。由题意得，先计算出括号两边算式的结果，然后再比较它们的大小即可。(2)0乘任何数都得0,0除以任何不为0的数都得0。由题意得，先计算出括号两边算式的结果，然后再比较它们的大小即可。(3)三位数除以两位数的笔算法则：从被除数的最高位除起，先用除数试除被除数的前两位数，如果它比除数小，再试除前三位数;除到被除数的哪一位，就在那一位上面写商。求出每一位商，余下的数必须比除数小。由题意得，先计算出括号两边算式的结果，然后再比较它们的大小即可.\n【详解】30×180 = 5400，30×190 = 5700，5400 < 5700，所以 30×180 < 30×190。830×0 = 0，0÷830 = 0，所以 830×0 = 0÷830。935÷67 = 13……64，206÷43 = 4……34，13 > 4，所以 935÷67 > 206÷43。23×110 = 2530，403×21 = 8463，2530 < 8463，所以23×110 < 403×21。",
+        "knowledge_points":  [
+                                 "有关0的乘法",
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "25-26 四年级上·甘肃天水·期末"
+    },
+    {
+        "id":  "70",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上\u0027>\u0027\u0027<\u0027或\u0027=\u0027。\n\n527023 ( ) 4969200  360÷60 ( ) 36÷6  130×40 ( ) 13×400",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;=;=",
+        "solution":  "【分析】根据大数的比较，位数多的数大，据此比较即可;根据商不变的性质，被除数和除数同时除以 10，商不变;根据积的变化规律，一个因数除以10，另一个因数乘10，积不变，即可解答.\n【详解】527023是六位数,4969200是七位数,527023\u003c 4969200;根据商不变的性质，被除数和除数同时除以10，商不变，因此 360÷60 = 36÷6;根据积的变化规律,130除以10,40乘10，积不变因此 130×40 = 13×400 。所以527023 \u003c 4969200;360÷60 = 36÷6;130×40 = 13×400 。",
+        "knowledge_points":  [
+                                 "商不变的规律及应用",
+                                 "积的变化规律(整数乘法)",
+                                 "三位数乘两位数，乘数末尾有 0",
+                                 "大数的比较"
+                             ],
+        "source":  "24-25四年级上·云南楚雄·期末"
+    },
+    {
+        "id":  "71",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算 246÷27 时，商是( )位数，可以把 27 看作( )来试商，这时商可能会偏( )。",
+        "options":  [
+
+                    ],
+        "answer":  "一;30;小",
+        "solution":  "【分析】三位数除以两位数，如果被除数的前两位大于等于除数，商是两位数，否则商是一位数;除法计算中，可以把除数看成接近的整十数试商。把除数看大了，商就容易偏小，调商时要调大一点，把除数看小了，商就容易偏大，调商时要调小一点.\n【详解】计算246÷27时，被除数的前两位24小于除数 27，所以商是一位数，可以把27看作30来试商，除数看大了，这时商可能会偏小。 ",
+        "knowledge_points":  [
+                                 "多位数除以两位数的试商",
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25四年级上·重庆璧山·期末"
+    },
+    {
+        "id":  "72",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "673÷72，可以把 72看作( )试商，商的最高位是( )位，商是( )位数。",
+        "options":  [
+
+                    ],
+        "answer":  "70;个;一",
+        "solution":  "【分析】根据三位数除以两位数的计算，可以将两位数估成和它接近的整十数再计算;如果被除数前两位大于或等于除数，则商是两位数，如果被除数前两位小于除数，则商是一位数，据此填空即可.\n【详解】673÷72 中, 67 \u003c 72，商是一位数。673÷72，可以把72看作70试商，商的最高位是个位，商是一位数。",
+        "knowledge_points":  [
+                                 "多位数除以两位数的试商",
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "25-26 四年级上·河北唐山·期末"
+    },
+    {
+        "id":  "73",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在除法算式5□3÷54中，如果商是一位数,□里最大填( );如果商是两位数,□里最小填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "3;4",
+        "solution":  "【分析】三位数除以两位数，判断商是几位数的方法：先观察被除数前两位上的数，看是否大于或等于除数(也就是看前两位上的数除以除数，够不够商一个十)。被除数前两位够除，商就是两位数。反之，商就是一位数。由题意得，在除法算式5□3÷54中，要使商是一位数，那么被除数前两位上的数小于除数54，所以□里最大可以填3;要使商是两位数，那么被除数前两位上的数大于或等于除数54，所以□里最小可以填4.\n【详解】在除法算式5□3÷54中，如果商是一位数,□里最大填3;如果商是两位数,□里最小填4。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "25-26四年级上·甘肃临夏·期末"
+    },
+    {
+        "id":  "74",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "已知 236÷8 = 29余4，那么 2360÷80 的商是( ),余数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "29;40",
+        "solution":  "【分析】被除数和除数同时乘(或除以)相同的数(0除外),商不变，余数也要乘(或除以)相同的数，据此解答即可.\n【详解】4×10 = 40。已知 236÷8 = 29余4，所以 2360÷80 = 29余40。即已知 236÷8 = 29余4，那么 2360÷80 的商是 29，余数是 40。",
+        "knowledge_points":  [
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "24-25 四年级上·天津蓟州·期末"
+    },
+    {
+        "id":  "75",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "□25÷45，要使商是两位数,□里最小填( );3□9÷36，要使商是一位数,□里最大填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "5;5",
+        "solution":  "【分析】三位数除以两位数，先用被除数前两位上的数去除以两位数，看够不够除，就是说前两位上的数和除数比较，如果比除数大或相等就够除，商的首位写在十位上，就是两位数;否则商写在个位上，就是一位数.\n【详解】□25÷45 要使商是两位数，那么 □2 ≥ 45，□里可以填 5,6,7,8,9，最小填 5;3□9÷36 要使商是一位数，那么 3□ < 36，□里可以填0,1,2,3,4,5，最大填5。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "25-26四年级上·天津滨海新·期末"
+    },
+    {
+        "id":  "76",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "括号最大能填几。\n\n40×( ) \u003c 24330×( ) \u003c 14590×( ) \u003c 68980×( ) \u003c 704",
+        "options":  [
+
+                    ],
+        "answer":  "6;4;7;8",
+        "solution":  "【分析】用小于号开口朝向的数除以另一端的乘数，如果有余数，则所得的商便是能填的最大数;如果没有余数，则所得的商减去 1 才是能填的最大数。据此解答.\n【详解】243÷40 = 6余3，所以 40×() < 243，()中最大填 6;145÷30 = 4余25，所以 30×() < 145，()中最大填 4;689÷90 = 7余59，所以 90×() < 689，()中最大填 7;704÷80 = 8余64，所以 80×() < 704，()中最大填 8。",
+        "knowledge_points":  [
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "25-26 四年级上·河北唐山·期末"
+    },
+    {
+        "id":  "77",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "要使7□9÷73 的商是两位数,□里最小填( );要使□25÷53 的商是一位数,□里最大填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "3;5",
+        "solution":  "【分析】三位数除以两位数，如果被除数前两位上的数大于或等于除数，商是两位数;如果被除数前两位上的数小于除数，商是一位数.\n【详解】要使7□9÷73 的商是两位数，被除数前两位上的数大于或等于除数73，□里可以填 3,4,5,6,7,8,9，□里最小填3;要使□25÷53 的商是一位数，被除数前两位上的数小于除数53，□里可以填1,2,3,4,5，□里最大填5。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "25-26 四年级上·河北唐山·期末"
+    },
+    {
+        "id":  "78",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "450÷18 = (450÷9)÷(18÷9) = 50÷2 = 25，这样计算应用了( )不变的规律使笔算很简便。应用这一规律填空并计算: 120÷15 = (120×())÷(15×())=()÷()=()。",
+        "options":  [
+
+                    ],
+        "answer":  "商;2;2;240;30;8",
+        "solution":  "【分析】商不变的规律：被除数和除数同时乘或除以同个数(0除外),商不变;据此解答.\n【详解】450÷18 = (450÷9)÷(18÷9) = 50÷2 = 25，这样计算应用了商不变的规律使笔算很简便。120÷15 = (120×2)÷(15×2) = 240÷30 = 8。",
+        "knowledge_points":  [
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "25-26四年级上·吉林·期末"
+    },
+    {
+        "id":  "79",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在下面括号里填上\u0027>\u0027\u0027<\u0027或\u0027=\u0027。\n\n5654879 ( ) 5654897  36900100000 ( ) 37万\n\n480÷30( )480÷12  16×300( )30×160",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003e;\u003c;=",
+        "solution":  "【分析】第一小题，整数比较大小时，要看它们的数位。数位多的那个数就大;如果数位相同，就看最高位，最高位上的数大，那个数就大;最高位上的数相同，就看下一位，哪一位上的数大，那个数就大。第二小题，把36900100000改写成用万做单位的数，再与37万进行比较;第三小题，观察480÷30和480÷12可知，两个除法算式的被除数相同，除数大的，商就小;第四小题，可以计算出左右两个算式的结果，再比较大小即可.\n【详解】由分析可得:5654879 < 5654897;36900100000 = 3690010万，所以36900100000 > 37万;30 > 12，所以 480÷30 < 480÷12;16×300 = 4800，30×160 = 4800，所以 16×300 = 30×160。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "除数是两位数的笔算除法",
+                                 "亿以上数的改写"
+                             ],
+        "source":  "25-26四年级上·黑龙江齐齐哈尔·期末"
+    },
+    {
+        "id":  "80",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\u0027>\u0027\u0027<\u0027或\u0027=\u0027。\n\n9570800000( )95070800000  16×280( )160×28\n\n500公顷( )5平方千米  420÷60( )420÷70",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;=;=;\u003e",
+        "solution":  "【分析】(1)整数大小比较的方法，位数不同的，位数多的大于位数少的，位数相同的，从最高位开始比较，最高位上大的数就大，如果最高位相同，再比较次高位，以此类推;(2)积的变化规律：两个数相乘，一个因数乘(或除以)个数(0除外),相反另一个因数同时除以(或乘)相同的数，它们的积不变。(3)根据1平方千米 = 100公顷，5平方千米 = 500公顷，再比较大小;(4)被除数不变，商随除数变化的规律：除数越大(0除外),商反而越小。据此比较.\n【详解】由分析得出:9570800000 < 95070800000;16×10 = 160，280÷10 = 28，一个因数扩大10倍，另一个因数缩小为原来的十分之一，所以 16×280 = 160×28;1平方千米 = 100公顷，500公顷 = 5平方千米;60 < 70，所以 420÷60 > 420÷70。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)",
+                                 "公顷，平方千米的进率与换算",
+                                 "除数是两位数的笔算除法",
+                                 "大数的比较"
+                             ],
+        "source":  "25-26 四年级上·北京西城·期末"
+    },
+    {
+        "id":  "81",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "要使8☐539801≈9亿，则☐里最小可以填( );如果☐45÷45 的商是一位数,☐里最大填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "5;4",
+        "solution":  "【分析】省略亿位后面的尾数要看千万位上的数，根据四舍五入法的原则，若千万位上的数字大于等于5，就向亿位进1;若千万位上的数字小于5，就舍去千万位及其后面数位上的数。据此可知，要使8☐539801≈9亿，那么☐里的数要大于等于5;据此解答;三位数除以两位数，被除数的前两位大于或等于除数，商是两位数，被除数的前两位小于除数，商是一位数;据此即可解答.\n【详解】由分析可知：要使8☐539801≈9亿，☐ ≥ 5，那么☐里可以填5,6,7,8,9，则☐里最小可以填5;要使☐45÷45 的商是一位数，则☐4 < 45，那么☐里可以填1,2,3,4，最大要填4。",
+        "knowledge_points":  [
+                                 "亿以上数的近似数",
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "25-26 四年级上·吉林白城·期末"
+    },
+    {
+        "id":  "82",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "为迎接农历新年的到来，某社区计划在街道两旁悬挂灯笼，增添喜庆的节日氛围。已知灯笼每盏售价为 36 元，商家正在进行\u0027买 5 盏送 1 盏\u0027的促销活动。如果采购员带了 500 元，最多能买回( )盏灯笼。",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】根据数量 = 总价/单价，用 500÷36，求出 500 元能买多少盏灯笼，再用能买灯笼的盏数除以5，求出一共送多少盏，然后再用能买的盏数加上送的盏数，即可求出 500 元最多能买到多少盏.\n【详解】 500÷36 = 13 (盏)。32(元) 13÷5 = 2 (盏)。3(盏) 13 + 2 = 15 (盏)为迎接农历新年的到来，某社区计划在街道两旁悬挂灯笼，增添喜庆的节日氛围。已知灯笼每盏售价为36元，商家正在进行\u0027买5盏送1盏\u0027的促销活动。如果采购员带了500元，最多能买回 15 盏灯笼。",
+        "knowledge_points":  [
+                                 "有余数除法的实际应用",
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "25-26 四年级上·辽宁盘锦·期末"
+    },
+    {
+        "id":  "83",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一辆货车从甲地向相距 560 千米的乙地驶去，用了 8 小时，它的速度为( )千米/时;张老师买 12 个篮球用去 540 元，篮球的单价是()元/个。",
+        "options":  [
+
+                    ],
+        "answer":  "70;45",
+        "solution":  "【分析】第一空：已知路程为560千米，时间为8小时，求速度，根据速度 = 路程/时间，代入数据计算。第二空：已知总价为 540 元，数量为 12 个，求单价，根据单价 = 总价/数量，代入数据计算.\n【详解】 560÷8 = 70 (千米/时)540÷12 = 45 (元/个)辆货车从甲地向相距560千米的乙地驶去，用了8小时，它的速度为70千米/时;张老师买12个篮球用去540元，篮球的单价是45元/个。",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "除数是两位数的笔算除法",
+                                 "几百几十或几千几百除以一位数的除法",
+                                 "经济问题"
+                             ],
+        "source":  "25-26 四年级上·辽宁盘锦·期末"
+    },
+    {
+        "id":  "84",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "两个数的商是 12，如果被除数不变，除数乘 3，商就变成( )。",
+        "options":  [
+
+                    ],
+        "answer":  "4",
+        "solution":  "【分析】根据商的变化规律，当被除数不变时，除数乘个非零的数，商就除以相同的数，本题中被除数不变，除数乘 3，原商就除以 3.\n【详解】因此变化后的商为 12÷3 = 4两个数的商是12，如果被除数不变，除数乘3，商就变成4。",
+        "knowledge_points":  [
+                                 "商的变化规律及应用"
+                             ],
+        "source":  "25-26 四年级上·吉林·期末"
+    },
+    {
+        "id":  "85",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在横线上填上\u0027>\u0027\u0027<\u0027或\u0027=\u0027。\n\n1000万()9900000  46×400()460×4  660÷30()330÷15",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;\u003e;=",
+        "solution":  "【分析】(1)比较整数的大小，位数多的那个数就大;如果位数相同，就看最高位，最高位上的数大，那个数就大;最高位上的数相同，就看下一位，哪一位上的数大，那个数就大。先把 1000万改写成 10000000，再与 9900000 比较大小。(2)当乘数末尾有零时，先算零前面的数，再在积的末尾添加对应个数的零。先计算出算式的结果，再比较大小。(3)除数是两位数的除法的笔算法则：从被除数的高位数起，先看被除数的前两位;如果前两位比除数小，就要看前三位;除到被除数的哪一位，商就写在那一位的上面;余下的数必须比除数小。先计算出算式的结果，再比较大小.\n【详解】根据分析可知:1000万 = 10000000，10000000 > 9900000，所以1000万 > 9900000;46×400 = 18400，460×4 = 1840，所以46×400 > 460×4;660÷30 = 22，330÷15 = 22，所以660÷30 = 330÷15。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "三位数乘两位数，乘数末尾有 0",
+                                 "大数的比较"
+                             ],
+        "source":  "25-26四年级上·吉林·期末"
+    },
+    {
+        "id":  "86",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "480÷80 = 6，被除数480 乘 10，除数不变，商是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "60",
+        "solution":  "【分析】根据商的变化规律，除数不变，被除数扩大到原来的几倍，商也扩大到原来的几倍;据此解答.\n【详解】在 480÷80 = 6 中，如果被除数乘10，除数不变，那么商也要乘10;6×10 = 60。所以被除数 480 乘 10，除数不变，商是 60。",
+        "knowledge_points":  [
+                                 "商的变化规律及应用"
+                             ],
+        "source":  "25-26 四年级上·吉林延边·期末"
+    },
+    {
+        "id":  "87",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "从 760 里面，最多可以减去( )个 18，最后还剩( )。",
+        "options":  [
+
+                    ],
+        "answer":  "42;4",
+        "solution":  "【分析】根据题意，用760÷18，求得的商就是最多可以减去18的个数，余数就是最后还剩的数，据此解答即可.\n【详解】760÷18 = 42(个)……4，所以从 760 里面，最多可以减去 42 个 18，最后还剩 4。",
+        "knowledge_points":  [
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "25-26 四年级上·吉林延边·期末"
+    },
+    {
+        "id":  "88",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\u0027>\u0027\u0027<\u0027或\u0027=\u0027。\n\n298670000( )3亿  805000( )5080000  36÷6( )360÷60\n\n630÷70( )360÷40  168÷32( )316÷28  7000公顷( )7平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003c;=;=;\u003c;\u003e",
+        "solution":  "【分析】①根据整数的改写，将3亿改写成300000000，再根据大数的比较，先看数位的多少，数位多的数就大，数位相同，从高位比较，高位上的数大则这个数大，高位上的数相同，就比较下一位，以此类推;②根据大数的比较方法解答;③根据商的变化规律，被除数和除数同时乘或除以不为0的数，商不变;④⑤分别计算出左右算式的结果再比较;⑥根据1平方千米=100公顷，将7平方千米换算成公顷为单位再比较.\n【详解】3亿 = 300000000，300000000 的亿位是3，298670000 的亿位是2，2 < 3，所以 298670000 < 3亿;805000是六位数，5080000是七位数，七位数大于六位数，所以 805000 < 5080000;36×10 = 360，6×10 = 60，所以 36÷6 = 360÷60;630÷70 = 9，360÷40 = 9，所以 630÷70 = 360÷40;168÷32 = 5……8，316÷28 = 11……8，5 < 11，所以 168÷32 < 316÷28;7平方千米 = 700公顷，7000 > 700，所以 7000公顷 > 7平方千米。",
+        "knowledge_points":  [
+                                 "商不变的规律及应用",
+                                 "公顷，平方千米的进率与换算",
+                                 "除数是两位数的笔算除法",
+                                 "大数的比较"
+                             ],
+        "source":  "25-26 四年级上·吉林延边·期末"
+    },
+    {
+        "id":  "89",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上\u0027>\u0027\u0027<\u0027或\u0027=\u0027。\n\n67000( )7万  160×2 ( )16×200  750÷25( )750÷50  40000公顷( )4平方千米。",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003c;\u003e;\u003e",
+        "solution":  "【分析】①将7万改写成70000，然后根据整数大小的比较方法：位数不同，位数多的数就大;位数相同，左起第一位上的数大的那个数就大，如果左起第一位上的数相同，就比较左起第二位上的数，直到比较出大小为止;②计算出式子的结果，再比较大小;③被除数相同，比较除数的大小，除数越大，商越小;④1平方千米=100公顷，单位不统一时，先化成相同单位，再比较大小.\n【详解】7万 = 70000，67000 的万位是6，70000 的万位是7，6 < 7，所以 67000 < 7万;160×2 = 320，16×200 = 3200，320 < 3200，所以 160×2 < 16×200;25 < 50，所以 750÷25 > 750÷50;4×100 = 400，40000 > 400，所以 40000公顷 > 4平方千米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "除数是两位数的笔算除法",
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "大数的比较"
+                             ],
+        "source":  "25-26 四年级上·黑龙江齐齐哈尔·期末"
+    },
+    {
+        "id":  "90",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\u0027>\u0027\u0027<\u0027或\u0027=\u0027。\n\n780÷26 ( ) 780÷24  540÷50 ( ) 480÷50  810÷30 ( ) 8100÷30\n\n140×6( )160×4  70×180( )700×18  160×6( )160×5",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003e;\u003c;\u003e;=;\u003e",
+        "solution":  "【分析】被除数相同(不为 0),除数越大，商就越小，由此可比较第一题大小。除数相同(不为 0),被除数越大，商就越大，由此可比较第二、三题大小。第四、五题计算出结果后，即可比较大小。第六题，一个因数相同(不为0),另一个因数越大，积越大，由此可比较第六题大小.\n【详解】因为 26 > 24，所以 780÷26 < 780÷24。因为 540 > 480，所以 540÷50 > 480÷50。因为 810 < 8100，所以 810÷30 < 8100÷30。因为 140×6 = 840，160×4 = 640，840 > 640，所以 140×6 > 160×4。因为 70×180 = 12600，700×18 = 12600，所以 70×180 = 700×18。因为 6 > 5，所以 160×6 > 160×5。",
+        "knowledge_points":  [
+                                 "商的变化规律及应用",
+                                 "整十，整百，整千数与位数的乘法",
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "91",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "M÷N = 20，如果 M和N 同时乘 9，那么商是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "20",
+        "solution":  "【分析】根据商不变的规律，被除数和除数同时乘或除以同一个不为0的数，商不变。本题中,M和N同时乘9，相当于被除数和除数同时乘9，因此商不变，仍然是20.\n【详解】已知 M÷N = 20，根据商不变的性质，被除数 M 和除数 N 同时乘 9，商不变。因此，商是 20。",
+        "knowledge_points":  [
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "25-26 四年级上·辽宁铁岭·期末"
+    },
+    {
+        "id":  "92",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "( )里最大能填几。\n\n30×( ) \u003c 9680×( ) \u003c 412",
+        "options":  [
+
+                    ],
+        "answer":  "3;5",
+        "solution":  "【分析】(1)可以用 96÷30 来计算,96÷30 = 3余6。即30×3 = 90，小于96，如果填4，30×4 = 120，就大于96了。所以括号里最大能填3。\n\n(2)用 412÷80 来算,412÷80 = 5余12。即80×5 = 400，小于412，若填6，80×6 = 480，大于412。所以括号里最大能填5.\n【详解】因为 30×3 = 90 < 96，30×4 = 120 > 96，所以括号里最大能填3;因为 80×5 = 400 < 412，80×6 = 480 > 412，所以括号里最大能填5。",
+        "knowledge_points":  [
+                                 "1000以内数的比大小",
+                                 "三位数除以两位数的有余数除法",
+                                 "整十，整百，整千数与一位数的乘法"
+                             ],
+        "source":  "25-26 四年级上·辽宁铁岭·期末"
+    },
+    {
+        "id":  "93",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "157 的 24 倍是(),600 里面有( )个 15。",
+        "options":  [
+
+                    ],
+        "answer":  "3768;40",
+        "solution":  "【分析】计算157的24倍是多少，根据求一个数的几倍是多少，用乘法计算，用157乘24计算即可。计算600里面有多少个15，根据求一个数里面有几个另一个数，用除法计算，用600除以15计算即可.\n【详解】157×24 = 3768，600÷15 = 40，所以157 的 24 倍是3768，600 里面有40个15。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法",
+                                 "除数是两位数的笔算除法",
+                                 "求一个数的几倍是多少"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "94",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "三位数□46÷54，如果商是一位数，那么□中最大能填( );如果商是两位数，那么□中最小应填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "4;5",
+        "solution":  "【分析】三位数除以两位数，判断商是几位数的方法：先观察被除数前两位上的数，看是否大于或等于除数(也就是看前两位上的数除以除数，够不够商一个十)。被除数前两位够除，商就是两位数。反之，商就是一位数。由题意得，在算式□46÷54中，要使商是一位数，那么被除数前两位上的数小于除数54，所以□中最大能填4;要使商是两位数，那么被除数的前两位上的数大于或等于除数54，所以□中最小应填 5.\n【详解】三位数□46÷54，如果商是一位数，那么□中最大能填 4;如果商是两位数，那么□中最小应填 5。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "25-26 四年级上·黑龙江齐齐哈尔·期末"
+    },
+    {
+        "id":  "95",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "要使2□7÷27的商是一位数，□里最大可填( );要使商是两位数，□里最小可填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "6;7",
+        "solution":  "【分析】对于除法算式2□7÷27，商是一位数时，需要被除数的前两位数字组成的数小于除数27，即 2□ < 27，从而推导出□ < 7，因此□最大可填6;商是两位数时，需要被除数的前两位数字组成的数大于或等于除数27，即 2□ ≥ 27，从而推导出□ ≥ 7，因此□最小可填7.\n【详解】要使商是一位数，则被除数的前两位2□必须小于27，即 20 + □ < 27，解得□ < 7，所以□最大可填6。要使商是两位数，则被除数的前两位2□必须大于或等于27，即 20 + □ ≥ 27，解得□ ≥ 7，所以□最小可填7。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "25-26四年级上·辽宁葫芦岛·期末"
+    },
+    {
+        "id":  "96",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "某糖果公司生产了800 支棒棒糖，如果每 12 支装一盒，最多可以装( )盒;如果每箱可以装12盒，装完这些棒棒糖至少需要( )个这样的箱子。",
+        "options":  [
+
+                    ],
+        "answer":  "66;6",
+        "solution":  "【分析】根据除法的意义，用棒棒糖的总支数除以一盒装的支数，求出的商就是最多可以装的盒数;用求出的可以装的盒数除以每箱可以装的盒数，如果有余数，用求出的商加上 1，就是装完这些棒棒糖至少需要几个这样的箱子.\n【详解】800÷12 = 66(盒)……8(支)，66÷12 = 5(箱)……6(盒)，5 + 1 = 6(箱)。即某糖果公司生产了800支棒棒糖，如果每12支装一盒，最多可以装66盒;如果每箱可以装12盒，装完这些棒棒糖至少需要6个这样的箱子。",
+        "knowledge_points":  [
+                                 "三位数除以两位数的有余数除法",
+                                 "有余数除法的实际应用"
+                             ],
+        "source":  "25-26 四年级上·辽宁抚顺·期末"
+    },
+    {
+        "id":  "97",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "吃青团是我国部分地区的习俗，青团带有清淡的青草香气。李阿姨买了 11 盒青团，花了 275 元。张阿姨买了 22 盒青团，她花了( )元。",
+        "options":  [
+
+                    ],
+        "answer":  "550",
+        "solution":  "【分析】由题意得，李阿姨买了11盒青团，花了275元，可以先用 275 除以 11 算出每盒青团需要多少钱。张阿姨买了 22 盒青团，直接用前面的得数乘上 22 即可算出张阿姨花了多少钱.\n【详解】 275÷11 = 25 (元)25×22 = 550 (元)故张阿姨花了 550 元。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "两位数乘两位数的进位乘法",
+                                 "经济问题"
+                             ],
+        "source":  "25-26 四年级上·辽宁抚顺·期末"
+    },
+    {
+        "id":  "98",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "□ 60÷57，要使商是两位数,□里最小填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】三位数除以两位数，要使商是两位数，那么被除数百位，十位上的数合起来要等于或大于除数，依此填空。 \n【详解】□60÷57，要使商是两位数，即□6要大于或等于57，因此□里可以填 6,7,8,9，其中最小填 6. ",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "25-26 四年级上·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "99",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "某车间有 480 箱奶要运到仓库，张师傅 12 次运完，他每次运( )箱。",
+        "options":  [
+
+                    ],
+        "answer":  "40",
+        "solution":  "【分析】根据题意，用牛奶的总箱数除以张师傅运的次数，即可求出他每次运的箱数.\n【详解】 480÷12 = 40 (箱)所以，某车间有480箱奶要运到仓库，张师傅12次运完，他每次运40箱。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "100",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在数位顺序表中，从右边起第五位是( )位，第十位是( )位。( )个100 万是 1 亿;一亿里面有( )个一万。",
+        "options":  [
+
+                    ],
+        "answer":  "万;十亿;100;10000",
+        "solution":  "【分析】在数位顺序表中，从右边起，第一位是个位，计数单位是一，表示几个一，第二位是十位，计数单位是十，表示几个十，第三位是百位，计数单位是百，表示几个百，第四位是千位，计数单位是千，表示几个千，第五位是万位，计数单位是万，表示几个万，第六位是十万位，计数单位是十万，表示几个十万，第七位是百万位，计数单位是百万，表示几个百万，第八位是千万位，计数单位是千万，表示几个千万，第九位是亿位，计数单位是亿，表示几个亿，第十位是十亿位，计数单位是十亿，表示几个十亿，以此类推;求多少个 100 万是 1 亿，把 100 万改写成整数 1000000,1亿改写成整数 100000000，用 100000000 除以 1000000;求1 亿里面有多少个 1 万,1 万改写成整数 10000，用100000000 除以 10000，即可解答.\n【详解】 100000000÷1000000 = 100 (个)100000000÷10000 = 10000 (个)在数位顺序表中，从右边起第五位是万位，第十位是十亿位。100 个 100 万是 1 亿;一亿里面有 10000 个一万。",
+        "knowledge_points":  [
+                                 "亿以上数的组成",
+                                 "整数的数级，数位和计数单位的认识",
+                                 "十进制计数法",
+                                 "除数是整十数的口算除法"
+                             ],
+        "source":  "25-26四年级上·福建龙岩·期中"
+    },
+    {
+        "id":  "101",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "两辆旅游大巴上午 9:00，分别从太阳岛景区和亚布力滑雪场同时出发，相向而行，第一次相遇距离太阳岛风景区85千米处，相遇后两车仍以原速行驶，分别到达对方出发地后立即返回，在离亚布力滑雪场30千米处第二次相遇。太阳岛风景区和亚布力滑雪场相距( )千米。",
+        "options":  [
+
+                    ],
+        "answer":  "225",
+        "solution":  "【分析】根据题意可知，第一次相遇地点离太阳岛风景区85千米，说明两辆旅游大巴共同走完太阳岛景区，亚布力滑雪场两地的路程，从太阳岛出发的大巴走了 85 千米，第二次相遇距离亚布力滑雪场30千米，说明从太阳岛出发的大巴到达亚布力滑雪场后返回又走了 30 千米;第二次相遇时，两辆旅游大巴共同走了3个太阳岛景区，亚布力滑雪场两地的路程，所以第二次相遇时从太阳岛出发的大巴走了85×3 = 255 (千米),再减去从太阳岛出发的大巴返回走的30千米即等于太阳岛景区，亚布力滑雪场两地的路程，据此即可解答.\n【详解】85×3-30 = 255 - 30 = 225 (千 米) 所以太阳岛风景区和亚布力滑雪场相距225千米。",
+        "knowledge_points":  [
+                                 "多次相遇问题"
+                             ],
+        "source":  "25-26四年级上·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "102",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "哈尔滨地铁1号线列车有 12 节车厢，共 480 个座位。照这样计算，一列有10 节车厢的同款列车，能设置( )个座位。",
+        "options":  [
+
+                    ],
+        "answer":  "400",
+        "solution":  "【分析】根据题意可知，列车有12节车厢，共480个座位，先用480÷12求出每节车厢的座位数;因为列车是同款的，所以每节车厢的座位数相同，用每节车厢的座位数乘10即可求出10节车厢的座位数.\n【详解】480÷12 = 40(个)，40×10 = 400(个)。所以照这样计算，一列有10节车厢的同款列车，能设置400个座位。",
+        "knowledge_points":  [
+                                 "两位数乘整十数的口算乘法",
+                                 "除数是两位数的笔算除法",
+                                 "单归一问题"
+                             ],
+        "source":  "25-26四年级上·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "103",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在算式□25÷45 中，要使商是两位数,□里最小填( );要使商是一位数,□里最大填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "5;4",
+        "solution":  "【分析】三位数除以两位数，要使商是两位数，则被除数的前两位数字要大于或等于除数;要使商是一位数，则被除数的前两位数字要小于除数，据此即可解答.\n【详解】□25÷45 中，要使商是两位数，则需□2 ≥ 45，□可以填5,6,7,8,9，最小填5;要使商是一位数，则需□2 < 45，□可以填1,2,3,4，最大填4。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "25-26 四年级上·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "104",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "189 除以 25，商是( ),要想没有余数，被除数最少应增加( )。",
+        "options":  [
+
+                    ],
+        "answer":  "7;11",
+        "solution":  "【分析】除数是两位数的除法，从被除数的最高位除起，用除数试除被除数的前两位，如果比除数小，再试除前三位;除到哪一位商就写在那一位的上面;要想没有余数，被除数最少应增加除数减余数，据此解答即可.\n【详解】189÷25 = 7余14，25-14 = 11，所以189除以25，商是7，要想没有余数，被除数最少应增加11。",
+        "knowledge_points":  [
+                                 "余数与除数的关系",
+                                 "三位数除以两位数的有余数除法",
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "25-26 四年级上·黑龙江齐齐哈尔·期末"
+    },
+    {
+        "id":  "105",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算 672÷72 时，应该把除数看作( )试商，商是( )位数。",
+        "options":  [
+
+                    ],
+        "answer":  "70;一或1",
+        "solution":  "【分析】根据三位数除以两位数的除法法则，首先把除数估算看作与它接近的整十数进行试商，然后按照除数是整十的除法法则进行计算，如果被除数的前两位比除数大，则商就是两位数;反之，商就是一位数，据此解答.\n【详解】72 接近 70,672 的 67 \u003c 72;所以计算 672÷72 时，应该把除数看作 70 试商，商是一位数。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)",
+                                 "多位数除以两位数的试商",
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "25-26 四年级上·黑龙江绥化·期末"
+    },
+    {
+        "id":  "106",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个自然数除以 39，商是 12，余数最大是( ),这时的被除数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "38;506",
+        "solution":  "【分析】由带余数的除法可知，余数要小于除数，所以余数最大为38。再由被除数 = 商×除数 + 余数，代入即可计算出这时的被除数.\n【详解】因为一个自然数除以39，商是12，所以余数最大是38。12×39+38 = 468+38 = 506，所以这时的被除数是506。",
+        "knowledge_points":  [
+                                 "余数与除数的关系",
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "107",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "235×16 的积是( )位数;364÷26 的商是( )位数。",
+        "options":  [
+
+                    ],
+        "answer":  "四;两",
+        "solution":  "【分析】两位数乘三位数，先用两位数个位上的数去乘三位数，得数的末尾和两位数的个位对齐，再用两位数十位上的数去乘三位数，得数的末尾和两位数的十位对齐，然后把两次乘的结果加起来。除数是两位数的除法的笔算法则：从被除数的高位数起，先看被除数的前两位;如果前两位比除数小，就要看前三位;除到被除数的哪一位，商就写在那一位的上面;余下的数必须比除数小.\n【详解】根据分析可知:235×16 = 3760，364÷26 = 14，所以235×16 的积是四位数;364÷26 的商是两位数。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法",
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "25-26 四年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "108",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "王伯伯准备将398个苹果分装进纸箱，每个纸箱装15个，可以装满( )个纸箱，还剩( )个苹果。",
+        "options":  [
+
+                    ],
+        "answer":  "26;8",
+        "solution":  "【分析】根据题意，用苹果总数398除以每个纸箱装的个数15，即可得到结果。根据除法的实际应用，即求一个数除以另一个数的商和余数。商表示可以装满的纸箱数量，余数表示剩余的苹果数量.\n【详解】398÷15 = 26(个)……8(个)。王伯伯准备将398个苹果分装进纸箱，每个纸箱装15个，可以装满26个纸箱，还剩8个苹果。",
+        "knowledge_points":  [
+                                 "有余数除法的实际应用",
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "25-26四年级上·甘肃临夏·期末"
+    },
+    {
+        "id":  "109",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\u0027>\u0027\u0027<\u0027或\u0027=\u0027。\n\n379999( )38万  2300÷50( )230÷5  120×20( )12×200\n\n87×7( )560  4600平方米( )46公顷  6806800( )60806800",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;=;=;\u003e;\u003c;\u003c",
+        "solution":  "【分析】(1)由题意得，先把 38万转化为 380000，然后再比较两个数的大小。(2)商不变的规律：被除数和除数同时乘或除以同一个不为0的数，商不变。(3)两个末尾有0的数相乘，先把两个乘数中0前面的数相乘，再看两个乘数中一共有几个0，就在积的末尾添上几个0。由题意得，先计算出括号两边算式的结果，然后再比较它们的大小。(4)多位数乘一位数的计算法则：从个位起，用一位数依次乘多位数的每一位。哪一位上乘得的积满几十，就向前位进几。由题意得，先计算出括号左边算式的结果，然后再比较它们的大小。(5)由题意得，根据1公顷 = 10000平方米将46公顷转化为多少平方米，然后再比较两个量的大小。(6)大数的比较：先比较两个数的位数，位数多的数就大。如果两个数的位数相同，就从最高位比起，最高位上的数大的那个数就大，如果最高位上的数相同，就比较下一个数位上的数。下一个数大，则这个数就大，依此来比较即可.\n【详解】(1)38万 = 380000，379999 < 380000，所以379999 < 38万。(2)对比算式2300÷50和230÷5可知，被除数和除数都除以10，商不变，所以 2300÷50 = 230÷5。(3)120×20 = 2400，12×200 = 2400，所以120×20 = 12×200。(4)87×7 = 609，609 > 560，所以 87×7 > 560。(5)1公顷 = 10000平方米，所以 46公顷 = 460000平方米。4600平方米 < 460000平方米，所以 4600平方米 < 46公顷。(6)6806800 和 60806800 比较大小，6806800 是七位数，60806800 是八位数，所以 6806800 < 60806800。",
+        "knowledge_points":  [
+                                 "商不变的规律及应用",
+                                 "面积单位间的进率及换算",
+                                 "三位数乘两位数，乘数末尾有 0",
+                                 "大数的比较"
+                             ],
+        "source":  "25-26 四年级上·吉林·期末"
+    },
+    {
+        "id":  "110",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "要使□08×34 的积是四位数,□里最大应填( );算式6□5÷65中，如果商是两位数,□里最小填( ),此时被除数和除数同时乘10，余数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "2;5;50",
+        "solution":  "【分析】三位数乘两位数，先用两位数个位上的数去乘三位数，得数的末尾和两位数的个位对齐，再用两位数十位上的数去乘三位数，得数的末尾和两位数的十位对齐，然后把两次乘的结果加起来;当乘数末尾有零时，先算零前面的数，再在积的末尾添加对应个数的零;根据三位数乘两位数的计算方法，在□中依次填入1,2,3,4，找出□08×34的积是四位数时□里最大能填的数字，即可解答;判断商为几位数，因为除数是两位数，所以要看到被除数的前两位，如果前两位数大于或等于除数那么商为两位数，如果前两位小于除数那么商为一位数，题中明确商为两位数，则被除数的前两位6□大于或等于65，据此解答;商不变的变化规律：被除数和除数同时乘(或除以)一个相同的数(0除外),商不变，余数会随之乘(或除以)这个数.\n【详解】当□ = 1 时,108×34 = 3672，积为四位数;当□ = 2 时,208×34 = 7072，积为四位数;当□ = 3 时,308×34 = 10472，积为五位数;因此要使□08×34 的积是四位数,□里最大应填2。6□5÷65，商为两位数，则 6□应大于或等于65，则□里可填5,6,7,8,9，因此□里最小填5;655÷65 = 10余5，被除数和除数同时乘10后，算式变为6550÷650，根据商不变的性质，商还是10，余数会乘10，即5×10 = 50。",
+        "knowledge_points":  [
+                                 "商不变的规律及应用",
+                                 "判断商是几位数(除数是两位数)",
+                                 "三位数乘两位数，三位数中间有0",
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "25-26 四年级上·吉林松原·期末"
+    },
+    {
+        "id":  "111",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\u0027>\u0027\u0027<\u0027或\u0027=\u0027。\n\n3654879( )3654897  2690010000( )27万  60公顷( )6平方千米\n\n800000m2( )7km2  480÷12( )480÷30  18×500( )50×180",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003e;\u003c;\u003c;\u003e;=",
+        "solution":  "【分析】第一小题，整数比较大小时，要看它们的数位。数位多的那个数就大;如果数位相同，就看最高位，最高位上的数大，那个数就大;最高位上的数相同，就看下一位，哪一位上的数大，那个数就大。第二小题，把2690010000改写成用万作单位的数，再与27万进行比较;第三小题，根据1平方千米 = 100公顷，先进行单位换算，再比较大小即可;第四小题，根据1平方千米 = 100公顷 = 1000000平方米，先进行单位换算，再比较大小即可;第五小题，两个除法算式的被除数相同，除数大的，商就小;第六小题，可以计算出左右两个算式的结果，再比较大小即可.\n【详解】3654879 < 3654897;2690010000 = 269001万，所以2690010000 > 27万;6平方千米 = 600公顷，所以60公顷 < 6平方千米;7km2 = 7000000m2，所以800000m2 < 7km2;12 < 30，所以 480÷12 > 480÷30;18×500 = 9000，50×180 = 9000，所以 18×500 = 50×180。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "除数是两位数的笔算除法",
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "大数的比较"
+                             ],
+        "source":  "25-26 四年级上·吉林松原·期末"
+    },
+    {
+        "id":  "112",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填\u0027>\u0027\u0027<\u0027或\u0027=\u0027。\n\n39867000( )3亿  5000公顷( )5平方千米\n\n630÷70( )630÷7  120×36( )240×18",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003e;\u003c;=",
+        "solution":  "【分析】(1)要比较这两个数的大小，需将它们转化为相同的计数单位形式。亿是一个较大的计数单位，3亿写成数字形式就是300000000。然后根据整数大小比较的方法，先看数位。数位多的数就大，如果数位相同，从最高位比起，最高位上数字大的数就大，如果最高位相同，就比较下一位，以此类推。39867000是八位数,300000000是九位数，所以可以直接得出大小关系。(2)平方千米和公顷都是用于计量面积的单位，它们之间存在固定的进率关系，即1平方千米 = 100公顷。要比较这两个量的大小，需要把它们换算成同一单位。将5平方千米换算为公顷，就是用5乘进率100得到500公顷。然后再比较5000公顷和500公顷的大小，从而得出结果。(3)在除法运算中，当被除数不变时，除数的变化会引起商的相应变化。观察这两个式子，被除数都是630，而除数分别是70和7，70大于7。根据被除数不变，除数越大，商越小这一规律，就可以判断出两个式子商的大小关系。(4)当两个因数相乘时，如果一个因数乘几或除以几(0除外),另一个因数除以或乘相同数，积不变。观察这两个乘法算式，240 是 120 的 2 倍，18 是 36 的一半，即一个因数120乘2变为240，另一个因数36÷2变为18。根据上述积不变的规律，就能够得出这两个算式的积相等.\n【详解】3亿 = 300000000，39867000是八位数，300000000是九位数。位数多的数更大，所以 39867000 < 3亿。5平方千米 = 500公顷，因为 5000公顷 > 500公顷，所以5000公顷 > 5平方千米。被除数都是630，因为 70 > 7，所以 630÷70 < 630÷7。观察发现:240 = 120×2，18 = 36÷2。一个因数乘2，另一个因数除以2，积不变，所以 120×36 = 240×18。",
+        "knowledge_points":  [
+                                 "商的变化规律及应用",
+                                 "大数的比较",
+                                 "公顷，平方千米的进率与换算",
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26四年级上·甘肃陇南·期末"
+    },
+    {
+        "id":  "113",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "□□□÷85 = 5余( )，余数最大是( ),这时的被除数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "84;84;509",
+        "solution":  "【分析】根据在有余数的除法中，余数总比除数小，即余数最大为：除数-1;然后根据：被除数 = 商×除数+余数;据此解答.\n【详解】根据分析: 85 - 1 = 84 所以余数最大是 84; 85×5 + 84 = 425 + 84 =509 所以,□□□÷85=5。( )中，余数最大是 84，这时的被除数是 509 。",
+        "knowledge_points":  [
+                                 "三位数除以两位数的有余数除法",
+                                 "有余数的除法验算",
+                                 "余数与除数的关系"
+                             ],
+        "source":  "25-26 四年级上·黑龙江齐齐哈尔·期末"
+    },
+    {
+        "id":  "114",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一辆汽车的速度是每小时 85 千米。照这样的速度，这辆汽车 7 小时可以行驶()千米;行驶 425 千米，需要( )小时。",
+        "options":  [
+
+                    ],
+        "answer":  "595;5",
+        "solution":  "【分析】由题意得，一辆汽车的速度是每小时85千米。速度×时间 = 路程，求这辆汽车7小时可以行驶多少千米，直接用85乘7即可解答;求这辆汽车行驶425千米需要多少小时，时间 = 路程/速度，直接用425除以85即可解答.\n【详解】 85×7 = 595 (千米) 425÷85 = 5 (小时) 辆汽车的速度是每小时 85 千米。照这样的速度，这辆汽车 7 小时可以行驶 595 千米;行驶 425 千米，需要 5 小时。",
+        "knowledge_points":  [
+                                 "两，三位数与一位数连续进位的乘法",
+                                 "除数是两位数的笔算除法",
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26 四年级上·广东广州·期末"
+    },
+    {
+        "id":  "115",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果 A÷B = 10余11，那么除数B最小是( ),这时被除数A是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "12;131",
+        "solution":  "【分析】在有余数的除法中，余数始终比除数小。除数最小是：余数+1;当除数最小时，被除数最小，根据\u0027被除数 = 商×除数+余数\u0027进行解答即可.\n【详解】11 + 1 = 12，12×10 + 11 = 120 + 11 = 131。所以如果 A÷B = 10余11，那么除数B最小是12，这时被除数A是131。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "余数与除数的关系"
+                             ],
+        "source":  "24-25 四年级上·浙江嘉兴·期末"
+    },
+    {
+        "id":  "116",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一辆汽车的速度是900 米/分，这辆汽车 1 小时可行驶( )千米。照这样的速度，这辆汽车行驶 162 千米需要( )小时。",
+        "options":  [
+
+                    ],
+        "answer":  "54;3",
+        "solution":  "【分析】1小时 = 60 分，根据速度×时间 = 路程，用这辆小汽车每分行驶的路程乘60，求出这辆小汽车每小时行驶的路程是多少，然后根据1千米 = 1000 米，把米转化为千米即可求出这辆小汽车1小时行驶多少千米;然后根据公式：时间 = 路程/速度，计算其行驶162千米所需时间即可.\n【详解】1小时 = 60 分 900×60 = 54000 (米) 54000 米 = 54 千米 162÷54 = 3 (小时)所以一辆汽车的速度是900米/分，这辆汽车1小时可行驶54千米。照这样的速度，这辆汽车行驶162千米需要3小时。",
+        "knowledge_points":  [
+                                 "时，分，秒之间的换算与比较",
+                                 "基础行程问题",
+                                 "除数是两位数的笔算除法",
+                                 "三位数乘两位数，乘数末尾有0"
+                             ],
+        "source":  "24-25 四年级上·浙江嘉兴·期末"
+    },
+    {
+        "id":  "117",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\u0027>\u0027\u0027<\u0027或\u0027=\u0027。\n\n48万( )4800  49×101( )5000  20厘米( )2米\n\n4000公顷( )4平方千米  460÷20( )460÷21  67×800( )80×760",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;\u003c;\u003c;\u003e;\u003e;\u003c",
+        "solution":  "【分析】(1)根据整数改写成\u0027万\u0027作单位：只要把末尾4个0去掉，再添上\u0027万\u0027字。48万 = 480000，位数不相同时，位数多的数大。据此比较大小。(2)计算出左边算式的积，然后再比较。(3)1米 = 100厘米，统一单位后，比较大小即可。(4)1平方千米 = 100公顷，统一单位后，比较大小即可。(5)当被除数相同时，除数越大，商就越小;除数越小，商就越大(被除数和除数都不为0)。据此比较大小。(6)算出67×8的积和76×8的积，整数末尾有0的乘法，可以先把0前面的数相乘，然后看各因数的末尾一共有几个0，就在乘得的积的末尾添写几个0。据此比较大小.\n【详解】48万 = 480000，所以 48万 > 4800;49×101 = 4949，所以 49×101 < 5000;1米 = 100厘米，2米 = 200厘米，所以 20厘米 < 2米;1平方千米 = 100公顷，4平方千米 = 400公顷，所以 4000公顷 > 4平方千米;20 < 21，所以 460÷20 > 460÷21;67×800 = 53600，80×760 = 60800，所以 67×800 < 80×760。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "除数是两位数的笔算除法",
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "大数的比较"
+                             ],
+        "source":  "24-25 四年级上·浙江嘉兴·期末"
+    },
+    {
+        "id":  "118",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "480÷32，可以把 32看作( )来试商，商是( )位数。",
+        "options":  [
+
+                    ],
+        "answer":  "30;两",
+        "solution":  "【分析】在除数是两位数的除法中，试商时通常将除数看作最接近的整十数。若被除数的前两位数字比除数大或者等于除数，则商是两位数，商的最高位是十位。若小于除数，则商是一位数，商的最高位是个位，据此填空即可.\n【详解】计算 480÷32 时，把除数 32 看作 30 来试商。因为48 \u003e 32，所以商是两位数。",
+        "knowledge_points":  [
+                                 "多位数除以两位数的试商",
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "25-26 四年级上·广西贵港·期中"
+    },
+    {
+        "id":  "119",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "( )里最大能填几。\n\n 4×( ) \u003c 290 ( )×30 \u003c 925 ( )×600 \u003c 2430800×( ) \u003c 5650 ",
+        "options":  [
+
+                    ],
+        "answer":  "72;30;4;7",
+        "solution":  "【分析】用小于号右侧的数除以左侧的因数，求商(整数)后有余数，商即为所求.\n【详解】因为 290÷4 = 72余2，所以 4×72 < 290;因为 925÷30 = 30余25，所以 30×30 < 925;因为 2430÷600 = 4余30，所以 4×600 < 2430;因为 5650÷800 = 7余50，所以 800×7 < 5650。",
+        "knowledge_points":  [
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "25-26四年级上·河南驻马店·期中"
+    },
+    {
+        "id":  "120",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "两个数的商是48，如果被除数不变，除数扩大 4 倍，商就变成了( )。",
+        "options":  [
+
+                    ],
+        "answer":  "12",
+        "solution":  "【分析】被除数不变，除数乘(或除以)几(0 除外),商除以(或乘)相同数;据此解答.\n【详解】 48÷4 = 12 即两数的商为 48，被除数不变，除数乘 4，商就变成了12。",
+        "knowledge_points":  [
+                                 "商的变化规律及应用"
+                             ],
+        "source":  "24-25四年级上·重庆大足·期末"
+    },
+    {
+        "id":  "121",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\u0027\u003c\u0027\u0027\u003e\u0027或\u0027=\u0027.\n\n5407080000( )5470800000480001( )48 万 88×5( )450\n\n420÷(7×3)( )420÷7÷3500 公顷( )5 平方千米 425÷45( )425÷35",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003e;\u003c;=;=;\u003c",
+        "solution":  "【分析】(1)(2)大数的比较：先比较两个数的位数，位数多的数就大，如果两个数的位数相同，就从最高位比起，最高位上的数字大的那个数就大，如果最高位上的数字相同，就依次比较下一个数位上的数字，直到比出大小为止。单位不统一的数，需要将数改写为一般形式后再比较。(3)有算式的，需要将算式的结果计算出来后再比较。(4)一个数连续除以两个数，等于这个数除以那两个数的积，据此比较两个式子的大小。(5)根据1平方千米 = 100公顷，将单位换算统一后再进行比较。(6)在除法中，被除数相同，除数越大，商就越小，据此比较两个算式的大小.\n【详解】(1)千万位上 0 < 7，所以 5407080000 < 5470800000;(2)48万 = 480000，个位上 1 > 0，即 480001 > 480000，所以 480001 > 48万;(3)88×5 = 440，440 < 450，所以 88×5 < 450;(4)420÷7÷3 = 420÷(7×3)，所以 420÷(7×3) = 420÷7÷3;(5)因为 1平方千米 = 100公顷，即 5平方千米 = 500公顷，所以 500公顷 = 5平方千米;(6)因为 45 > 35，所以 425÷45 < 425÷35。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "多位数与一位数的连乘",
+                                 "大数的比较",
+                                 "商的变化规律及应用"
+                             ],
+        "source":  "24-25 四年级上·辽宁鞍山·期末"
+    },
+    {
+        "id":  "122",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "每棵树苗15元，如果买3送1棵,180元最多能买( )棵这样的树苗。",
+        "options":  [
+
+                    ],
+        "answer":  "16",
+        "solution":  "【分析】由题意可知：每棵树苗15元,180元最多能买多少棵这样的树苗，用180除以15求出商，因为买3棵送1棵，再看这个商里面有几个3，那么可以求出另外能送几棵树苗，最后把购买的与赠送的数量相加即可.\n【详解】 180÷15 = 12 (棵) 12÷3 = 4 (组) 4×1 = 4 (棵) 12 + 4 = 16 (棵)所以,180 元最多能买 16 棵这样的树苗。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "经济问题"
+                             ],
+        "source":  "24-25 四年级上·湖北武汉·期末"
+    },
+    {
+        "id":  "123",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在下面的( )里填上\u0027\u003e\u0027\u0027\u003c\u0027或\u0027=\u0027.\n\n123000( )123 万 540×70( )700×54\n\n1 平方千米( )90 公顷 180÷12( )180÷15",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;=;\u003e;\u003e",
+        "solution":  "【分析】(1)由题意得，先把123万转化为1230000，然后再比较两个数的大小即可。(2)三位数乘两位数的笔算法则：先用两位数的个位分别与三位数的每一位数相乘，乘得的积的末尾和个位对齐。再用两位数的十位分别与三位数的每一位数相乘，乘得的积的末尾和十位对齐。最后，将两次乘得的积相加。由题意得，先计算出括号两边算式的结果，然后再比较它们的大小。(3)由题意得，先把1平方千米转化为100公顷，然后再比较两个量的大小即可。(4)三位数除以两位数的笔算法则：从被除数的最高位除起，先用除数试除被除数的前两位数，如果它比除数小，再试除前三位数;除到被除数的哪一位，就在那一位上面写商。求出每一位商，余下的数必须比除数小。由题意得，先计算出括号两边算式的结果，然后再比较它们的大小.\n【详解】(1)123万 = 1230000，123000 < 1230000，所以123000 < 123万。(2)540×70 = 37800，700×54 = 37800，所以 540×70 = 700×54。(3)1平方千米 = 100公顷，100公顷 > 90公顷，所以 1平方千米 > 90公顷。(4)180÷12 = 15，180÷15 = 12，15 > 12，所以 180÷12 > 180÷15。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "除数是两位数的笔算除法",
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "大数的比较"
+                             ],
+        "source":  "24-25四年级上·湖北武汉·期末"
+    },
+    {
+        "id":  "124",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "☆÷18=26余□,当□最大时,☆是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "485",
+        "solution":  "【分析】在有余数的除法中，余数必须小于除数。本题中除数是18，所以余数□最大为17。根据被除数 = 商×除数+余数，可求出此时被除数☆的值.\n【详解】因为除数是18，所以□最大是17。26×18 + 17 = 468 + 17 = 485，所以☆是485。",
+        "knowledge_points":  [
+                                 "余数与除数的关系",
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "24-25 四年级上·四川绵阳·期末"
+    },
+    {
+        "id":  "125",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "19000公顷=( )平方千米  1周角=( )直角",
+        "options":  [
+
+                    ],
+        "answer":  "190;4",
+        "solution":  "【分析】本题考查面积单位换算和角度单位换算。公顷和平方千米的进率是100，即1平方千米 = 100公顷，因此公顷换算成平方千米需除以100。周角是360°，直角是90°，因此周角换算成直角的个数需除以90。\n【详解】1平方千米 = 100公顷，19000公顷 = 19000÷100 = 190平方千米。1周角 = 360°÷90° = 4直角。\n\n因此,19000公顷 = 190平方千米，1周角 = 4直角。",
+        "knowledge_points":  [
+                                 "除数是整十数的口算除法",
+                                 "角的分类及换算",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·四川绵阳·期末"
+    },
+    {
+        "id":  "126",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "已知 560×12 =6720，那么 56×120 = ( ) 。已知 384÷24 = 16，那么768÷24=( )。",
+        "options":  [
+
+                    ],
+        "answer":  "6720;32",
+        "solution":  "【分析】积的变化规律：一个因数乘(或除以)几(0除外),而另一个因数除以(或乘)相同数，它们的积不变。商的变化规律：除数不变，被除数乘(或除以)几(0除外),商也乘(或除以)相同数;据此解答.\n【详解】56×120 = (560÷10)×(12×10) = 6720，768÷24 = (384×2)÷24 = 16×2 = 32。即已知 560×12 = 6720，那么 56×120 = 6720。已知 384÷24 = 16，那么 768÷24 = 32。",
+        "knowledge_points":  [
+                                 "商的变化规律及应用",
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "24-25 四年级上·四川绵阳·期末"
+    },
+    {
+        "id":  "127",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\u0027>\u0027\u0027<\u0027或\u0027=\u0027。\n\n130×20 ( ) 13×200  1500÷300 ( ) 150÷30\n\n5907080000 ( ) 597080000  49万 ( ) 490001",
+        "options":  [
+
+                    ],
+        "answer":  "=;=;\u003e;\u003c",
+        "solution":  "【分析】(1)(2)计算出结果后比较;(3)5907080000 是十位数，597080000 是九位数，位数多的数大于位数少的数，所以 5907080000 > 597080000。(4)49万 = 490000，位数相同看最高位的大小，依次向下比较大小。据此解答.\n【详解】(1)130×20 = 2600，13×200 = 2600，所以 130×20 = 13×200。(2)1500÷300 = 5，150÷30 = 5，所以 1500÷300 = 150÷30。(3)5907080000 > 597080000。(4)49万 = 490000，490000 < 490001，所以 49万 < 490001。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "大数的比较"
+                             ],
+        "source":  "24-25 四年级上·安徽芜湖·期末"
+    },
+    {
+        "id":  "128",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一辆卡车 4 小时行驶 340 千米，这辆卡车的速度是( );菲菲用 36 元买了18 本练习本，买 130 本这样的练习本需要( )元。",
+        "options":  [
+
+                    ],
+        "answer":  "85千米/时或85km/h;260",
+        "solution":  "【分析】速度×时间 = 路程，所以速度 = 路程÷时间。由题意得，一辆卡车4小时行驶340千米，直接用340除以4即可算出这辆卡车的速度;菲菲用36元买了18本练习本，可以先用36除以18算出每本练习本多少钱，然后再用得数乘上130即可算出买130本这样的练习本需要多少元.\n【详解】340÷4 = 85(千米/时)，36÷18×130 = 2×130 = 260(元)。一辆卡车4小时行驶340千米，这辆卡车的速度是85千米/时;菲菲用36元买了18本练习本，买130本这样的练习本需要260元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "三位数除以一位数，首位不够除",
+                                 "除数是两位数的笔算除法",
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25 四年级上·云南昭通·期末"
+    },
+    {
+        "id":  "129",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在( )÷22 = 8余( )中，当余数是( )时，被除数最大，是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "21;197",
+        "solution":  "【分析】在有余数的除法中，余数一定小于除数。已知除数是22，余数最大是21，根据\u0027被除数 = 除数×商+余数\u0027来计算.\n【详解】因为余数一定小于除数，已知除数是 22，所以余数最大为 21.22×8 + 21 = 176 + 21 = 197 所以，当余数是 21 时，被除数最大，是 197。",
+        "knowledge_points":  [
+                                 "三位数除以两位数的有余数除法",
+                                 "有余数的除法验算",
+                                 "余数与除数的关系"
+                             ],
+        "source":  "24-25 四年级上·安徽淮北·期末"
+    },
+    {
+        "id":  "130",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一块长方形的田地长是 25 米，宽是 20 米，它的周长是( )米，面积是( )平方米,( )块这么大的田地是 1 公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "90;500;20",
+        "solution":  "【分析】长方形的周长 = (长+宽)×2;长方形的面积 = 长×宽，把数据代入计算即可解答;1 公顷 = 10000 平方米,\n\n平方米和公顷之间的进率是 10000，用 10000÷(25×20),据此即可解答.\n【详解】周长:(25+20)×2 = 45×2 = 90 (米) 面积: 25×20 = 500 (平方米) 1公顷 = 10000 平方米\n\n10000÷500 = 20 (块)所以长方形的田地的周长是 90 米;面积是 500 平方米;20块这么大的田地是1公顷。",
+        "knowledge_points":  [
+                                 "长方形的周长",
+                                 "被除数和除数末尾都有0的除法",
+                                 "长方形的面积",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "24-25 四年级上·河北保定·期末"
+    },
+    {
+        "id":  "131",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在□÷21=34余△中,△最大是( ),这时□是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "20;734",
+        "solution":  "【分析】根据题意，余数应比除数小，除数是21，那么余数最大为 21 - 1 = 20，再根据商×除数+余数 = 被除数计算出□里的值即可.\n【详解】根据分析可知:21 - 1 = 20，21×34 + 20 = 714 + 20 = 734。在□÷21=34余△中,△最大是20，这时□是734。",
+        "knowledge_points":  [
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "24-25 四年级上·河北保定·期末"
+    },
+    {
+        "id":  "132",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "根据商不变的规律填空。\n\n9100÷700 = 91÷( )  120÷15 = 40÷( )",
+        "options":  [
+
+                    ],
+        "answer":  "7;5",
+        "solution":  "【分析】在除法算式里，被除数和除数同时乘或除以相同的数(0除外),商不变，这个就是商不变的性质。据此解答.\n【详解】9100÷700 = 91÷()，被除数9100变为91，是9100÷100 = 91，根据商不变性质，除数700也应除以100，即700÷100 = 7，所以 9100÷700 = 91÷7。120÷15 = 40÷()，被除数120变为40，是120÷3 = 40，根据商不变性质，除数15也应除以3，即15÷3 = 5，所以120÷15 = 40÷5。",
+        "knowledge_points":  [
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "24-25四年级上·云南玉溪·期末"
+    },
+    {
+        "id":  "133",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在算式□38÷46中，要使商是两位数,□里最小能填( );要使商是一位数,□里最大能填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "5;4",
+        "solution":  "【分析】三位数除以两位数，如果被除数前两位上的数小于除数，则商是一位数;如果被除数前两位上的数大于或等于除数，则商是两位数。\n【详解】在算式□38÷46中，要使商是两位数，则□3 > 46，□里最小能填5;要使商是一位数，则□3 < 46，□里最大能填4。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25四年级上·云南玉溪·期末"
+    },
+    {
+        "id":  "134",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一批货物有725吨，用载重20吨的集装箱货车来运，可以装满( )辆车，还剩下( )吨。",
+        "options":  [
+
+                    ],
+        "answer":  "36;5",
+        "solution":  "【分析】用货物的总重量除以货车的载重，求得的商就是装满的车的数量。余数，即为剩余的货物吨数.\n【详解】725÷20 = 36(辆)……5(吨)。\n\n所以可以装满36辆车，还剩下5吨。",
+        "knowledge_points":  [
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "24-25四年级上·广东江门·期末"
+    },
+    {
+        "id":  "135",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "A÷B = 45，那么 A÷(B×9) = ( )。",
+        "options":  [
+
+                    ],
+        "answer":  "5",
+        "solution":  "【分析】除法算式中，被除数乘a，商就乘a;除数乘b(不为0)，商就除以b，据此填空即可.\n【详解】已知 A÷B = 45，求 A÷(B×9)，除数乘9，商就除以9，45÷9 = 5，所以 A÷(B×9) = 5。",
+        "knowledge_points":  [
+                                 "商的变化规律及应用"
+                             ],
+        "source":  "24-25 四年级上·云南文山·期末"
+    },
+    {
+        "id":  "136",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在整数除法□÷△= 19。12 中,△最小是( ),此时被除数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "13;259",
+        "solution":  "【分析】在有余数除法中，余数小于除数，除数最小等于余数加1，当除数最小时，根据\u0027被除数 = 除数×商+余数\u0027求出被除数即可解答.\n【详解】12 + 1 = 13，19×13 + 12 = 247 + 12 = 259。所以，在整数除法□÷△=19余12中,△最小是13，此时被除数是259。",
+        "knowledge_points":  [
+                                 "余数与除数的关系",
+                                 "有余数的除法验算",
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "24-25 四年级上·云南楚雄·期末"
+    },
+    {
+        "id":  "137",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "为了完善居民锻炼场地，某区政府把长方形运动广场的宽增加了 50 米，变成了正方形广场，这时面积增加了1公顷。该广场原来的面积是( )公顷，合( )平方米。",
+        "options":  [
+
+                    ],
+        "answer":  "3;30000",
+        "solution":  "【分析】1公顷 = 10000 平方米，增加的面积除以宽增加的长度等于长方形广场的长，也是正方形广场的边长，正方形广场的边长乘边长等于正方形广场的面积，再减去10000平方米即等于广场原来的面积，然后把面积单位换算成公顷即可解答.\n【详解】1 公顷 = 10000 平方米 10000÷50 = 200 (米) 200×200 - 10000 = 40000 - 10000 = 30000 (平 方 米) 30000 平方米 = 3 公顷 所以，该广场原来的面积是 3 公顷，合 30000 平方米。",
+        "knowledge_points":  [
+                                 "除数是整十数的口算除法",
+                                 "长方形的面积",
+                                 "正方形的面积",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·云南楚雄·期末"
+    },
+    {
+        "id":  "138",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据 275÷25 = 11，写出下面各题的商.\n\n2750÷250 = ( ) 550÷25 = ( )",
+        "options":  [
+
+                    ],
+        "answer":  "11;22",
+        "solution":  "【分析】在除法中，被除数和除数同时乘或除以一个数(0除外),商不变;被除数乘几，除数不变，商也要乘几。据此解答.\n【详解】对比2750÷250和275÷25，发现2750是由275乘10得到的,250是由25乘10得到的，由分析可知，被除数和除数同时乘10，商不变，所以2750÷250 = 11;对比550÷25，发现550是由275乘2得到的，由分析可知，被除数乘2，除数不变，商也要乘2，11×2 = 22，所以550÷25 = 22。",
+        "knowledge_points":  [
+                                 "商不变的规律及应用",
+                                 "商的变化规律及应用"
+                             ],
+        "source":  "24-25 四年级上·云南曲靖·期末"
+    },
+    {
+        "id":  "139",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "学校组织同学们去研学，四年级有484名师生，每辆客车限乘50名乘客，学校至少需要( )辆客车。",
+        "options":  [
+
+                    ],
+        "answer":  "10",
+        "solution":  "【分析】根据题意可知，用总人数除以每辆客车可以坐的人数，得到的商就是坐满客车的辆数，余数就表示剩下的人数，有余数时，用坐满客车的辆数加1辆，就是至少要准备客车的辆数，依此计算.\n【详解】484÷50 = 9(辆)……34(名)，9 + 1 = 10(辆)。学校至少需要10辆客车。",
+        "knowledge_points":  [
+                                 "三位数除以两位数的有余数除法",
+                                 "有余数除法的实际应用"
+                             ],
+        "source":  "24-25 四年级上·湖南长沙·期末"
+    },
+    {
+        "id":  "140",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\u0027>\u0027\u0027<\u0027或\u0027=\u0027。\n\n298670000( )3亿  8000平方米( )8公顷  805000( )850000\n\n30000公顷( )3平方千米  18×500( )50×180  36÷6( )3360÷60",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003c;\u003c;\u003e;=;\u003c",
+        "solution":  "【分析】比较整数的大小，先看整数的位数，位数多的那个数就大;如果位数相同，就看最高位，最高位上的数大，那个数就大;如果最高位的数相同，就看次高位，次高位上的数大，那个数就大。以此类推。把3亿改写成300000000，再按照整数大小比较方法与298670000比较大小。1公顷 = 10000平方米，把8公顷改写成80000平方米，再与8000平方米比较大小。按照整数大小比较方法比较805000和850000的大小。1平方千米 = 100公顷，把30000公顷改写成300平方千米，再与3平方千米比较大小。18×500先算18×5 = 90，再在90的后面添上2个0;50×180先算5×18 = 90，再在90的后面添上2个0。据此比较18×500和50×180的大小。分别算出36÷6和3360÷60的商，再比较大小.\n【详解】3亿 = 300000000，298670000 < 300000000，所以298670000 < 3亿。8公顷 = 80000平方米，8000平方米 < 80000平方米，所以8000平方米 < 8公顷。805000 和 850000 都是六位数，805000 和 850000 的十万位上都是8，805000 的万位上是0，850000 的万位上是5，所以805000 < 850000。30000公顷 = 300平方千米，300平方千米 > 3平方千米，所以30000公顷 > 3平方千米。18×500 = 9000，50×180 = 9000，所以 18×500 = 50×180。36÷6 = 6，3360÷60 = 56，6 < 56，所以36÷6 < 3360÷60。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "三位数乘两位数，乘数末尾有 0",
+                                 "被除数和除数末尾都有0的除法",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·湖南长沙·期末"
+    },
+    {
+        "id":  "141",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小红家今年共收大豆 4680 千克，平均每 50 千克装 1 袋，这些大豆可以装满()袋，还剩( )千克。",
+        "options":  [
+
+                    ],
+        "answer":  "93;30",
+        "solution":  "【分析】用大豆的总重量除以每袋可装的重量，得到的商就是能装满的袋数，余数就是剩余的质量。列式为4680÷50，计算即可。多位数除以两位数的计算方法：先看被除数的前两位，如果被除数的前两位比除数大或等于除数，就直接在十位上试商;若被除数前两位比除数小，就看被除数的前三位，在个位上试商。试商时，把除数看作与它接近的整十数来试商，若初商不合适，再进行调整。每次除得的余数必须比除数小，按照这样的步骤逐步计算，直到得出最终结果.\n【详解】4680÷50 = 93(袋)……30(千克)。小红家今年共收大豆4680千克，平均每50千克装1袋，这些大豆可以装满93袋，还剩30千克。",
+        "knowledge_points":  [
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "24-25 四年级上·云南文山·期末"
+    },
+    {
+        "id":  "142",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "要使□56÷38 的商是一位数,□最大应填( );要使□06÷56 的商是两位数,□最小应填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "3;6",
+        "solution":  "【分析】三位数除以两位数，判断商是几位数的方法：当被除数的前两位小于除数时，商是一位数;当被除数的前两位大于或等于除数时，商是两位数.\n【详解】对于□56÷38商是一位数的情况：因为商要是一位数，那么被除数的前两位□5要小于除数38。所以□可以填1,2,3，其中最大的数就是3。对于□06÷56商是两位数的情况：因为商要是两位数，那么被除数的前两位□0要大于或等于除数56。所以□可以填6,7,8,9，其中最小的数就是6。要使□56÷38的商是一位数,□最大应填3;要使□06÷56的商是两位数,□最小应填6。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25四年级上·云南文山·期末"
+    },
+    {
+        "id":  "143",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "果汁原价15元/瓶，超市搞促销，买4瓶送1瓶，妈妈付了120元，最多可买到( )瓶，每瓶果汁单价比原来便宜了( )元。",
+        "options":  [
+
+                    ],
+        "answer":  "10;3",
+        "solution":  "【分析】用付的钱数除以果汁的原价，求出购买果汁的瓶数。这种果汁买4瓶送1瓶，用购买果汁的瓶数除以4，求出买了几组买4瓶送1瓶，用组数乘赠送的1瓶，即可求得赠送的果汁瓶数。用购买果汁的瓶数加上赠送果汁的瓶数，求出得到果汁的总瓶数。用付的钱数除以得到果汁的总瓶数，求出每瓶果汁的现价，再用每瓶果汁的原价减去每瓶果汁的现价计算即可.\n【详解】 120÷15 = 8 (瓶)8÷4 = 2 (组) 2×1 + 8 = 2 + 8 = 10 (瓶) 120÷10 = 12 (元) 15 - 12 = 3 (元) 所以最多可买到 10 瓶，每瓶果汁单价比原来便宜了 3 元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "24-25四年级上·天津南开·期末"
+    },
+    {
+        "id":  "144",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "已知 35×12 = 420，那么 350×12=( ), 35×24 = ( );;已知 360÷15 = 24，那么 360÷30 = ( ) , 180÷15 = ( ) ",
+        "options":  [
+
+                    ],
+        "answer":  "4200;840;12;12",
+        "solution":  "【分析】两数相乘，如果一个因数不变，另一个因数乘或除以几(0除外),积也会随之乘或除以相同的数;在除法里，除数不变，被除数乘或除以一个数(0除外),商也随之乘或除以相同的数。在除法里，被除数不变，除数乘一个数(0除外),商反而除以相同的数;除数除以一个数(0除外),商反而乘相同的数.\n【详解】350×12 = (35×10)×12 = 420×10 = 4200，35×24 = 35×(12×2) = 420×2 = 840，360÷30 = 360÷(15×2) = 24÷2 = 12，180÷15 = (360÷2)÷15 = 24÷2 = 12。已知 35×12 = 420，那么 350×12 = 4200，35×24 = 840;已知 360÷15 = 24，那么 360÷30 = 12，180÷15 = 12。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)",
+                                 "商的变化规律及应用"
+                             ],
+        "source":  "24-25 四年级上·河南安阳·期末"
+    },
+    {
+        "id":  "145",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据商不变的规律。在 96÷16 = 。 中，被除数除以 4，除数同时除以( ),商是()。",
+        "options":  [
+
+                    ],
+        "answer":  "4;6",
+        "solution":  "【分析】依据商不变的规律(被除数和除数同时乘或除以相同的非0数，商保持不变),题目中被除数除以4，要让商不变，除数也要同时除以4;原算式96÷16的商是6，按照这个规律，变化后算式的商依然是6。\n【详解】已知 96÷16 = 6，(96÷4)÷(16÷4) = 24÷4 = 6。要使商不变，除数应同时除以4，商是6。",
+        "knowledge_points":  [
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "24-25四年级上·广西玉林·期末"
+    },
+    {
+        "id":  "146",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在计算 640÷32 时，可以把 32 看成( )来试商，该式子的商是( )位数。",
+        "options":  [
+
+                    ],
+        "answer":  "30;两",
+        "solution":  "【分析】除数是两位数除法计算中，一般把除数看作与它接近的整十数来试商;三位数除以两位数，如果被除数的前两位大于等于除数，商是两位数，否则商是一位数，据此解答.\n【详解】 64 \u003e 32 在计算 640÷32 时，可以把 32 看成 30 来试商，该式子的商是两位数。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)",
+                                 "多位数除以两位数的试商"
+                             ],
+        "source":  "24-25四年级上·广西玉林·期末"
+    },
+    {
+        "id":  "147",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上\u0027>\u0027\u0027<\u0027或\u0027=\u0027。\n\n47000平方米( )4公顷7平方米  960÷24 ( ) 960÷8×3  26090800000( )26900800000",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;\u003c;\u003c",
+        "solution":  "【分析】根据1公顷 = 10000平方米，进行单位换算后比较大小;先计算出各算式的结果再进行大小比较;比较数的大小时，位数多的数就大，位数相同比较最高位，最高位大的数就大，最高位相同就比较下一位，直到比出大小为止.\n【详解】4公顷里面有4个10000平方米，也就是40000平方米，那么 4公顷7平方米 = 40007平方米，47000 > 40007，所以 47000平方米 > 4公顷7平方米;960÷24 = 40，960÷8×3 = 120×3 = 360，40 < 360，所以 960÷24 < 960÷8×3。26090800000 和 26900800000 位数相同都是11位数，前两位相同，第三位0 < 9，所以 26090800000 < 26900800000。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "多位数与一位数的乘除混合运算",
+                                 "公顷，平方千米的进率与换算",
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "24-25四年级上·河南开封·期末"
+    },
+    {
+        "id":  "148",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把 580 千克葡萄装箱，如果每箱最多装 15 千克，那么至少需要( )个这样的包装箱才能将这些葡萄装完。",
+        "options":  [
+
+                    ],
+        "answer":  "39",
+        "solution":  "【分析】根据题意，用葡萄的总质量除以每箱可以装的葡萄的质量，商就是可以装满箱子的数量，余数是余下的葡萄，有余数时需要再多用一个箱子装。据此解答即可.\n【详解】580÷15 = 38(个)……10(千克)，38 + 1 = 39(个)。把580千克葡萄装箱，如果每箱最多装15千克，那么至少需要39个这样的包装箱才能将这些葡萄装完。",
+        "knowledge_points":  [
+                                 "用\u0027进一法\u0027解决问题",
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "24-25 四年级上·山东济宁·期末"
+    },
+    {
+        "id":  "149",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在 A÷25 = 18余B中，余数B最大是( ),这时被除数A是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "24;474",
+        "solution":  "【分析】根据有余数除法\u0027余数必须小于除数\u0027的基本性质，已知除数是25，就能确定余数B的最大值为24;再依据\u0027被除数 = 除数×商+余数\u0027的计算公式，代入除数25，商18和最大余数24，即可算出被除数A的值.\n【详解】因为余数 < 除数，除数是25，所以B最大是25-1 = 24。被除数 = 除数×商+余数，代入得:A = 25×18 + 24 = 450 + 24 = 474。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "三位数除以两位数的有余数除法",
+                                 "余数与除数的关系"
+                             ],
+        "source":  "24-25 四年级上·山东济宁·期末"
+    },
+    {
+        "id":  "150",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\u0027>\u0027\u0027<\u0027或\u0027=\u0027。\n\n360×20 ( ) 200×36  480÷12 ( ) 480÷24  九万零七( )900007  6平方千米( )590公顷",
+        "options":  [
+
+                    ],
+        "answer":  "=;\u003e;\u003c;\u003e",
+        "solution":  "【分析】(1)(2)分别计算出左右两边的得数再进行比较;(3)括号左边的九万零七写作:90007，再与右边的比较;(4)将单位换算成一样的，再进行比较.\n【详解】(1) 360×20 = 7200 , 200×36 = 7200 , 7200 = 7200，所以 360×20 = 200×36;(2) 480÷12 = 40 , 480÷24 = 20 , 40 \u003e 20，所以 480÷12 \u003e 480÷24;(3)九万零七写作:90007,90007\u003c900007，所以九万零七\u003c900007;(4)1 平方千米 = 100 公顷, 6×100 = 600 (公顷), 600 \u003e 590，所以6平方千米 \u003e 590 公顷。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "除数是两位数的笔算除法",
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "大数的比较"
+                             ],
+        "source":  "24-25四年级上·山东济宁·期末"
+    },
+    {
+        "id":  "151",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "括号里最大能填几。\n\n4( )5000≈46万  48( )7300000≈48亿\n\n60×( )\u003c378  182÷( )\u003e26",
+        "options":  [
+
+                    ],
+        "answer":  "5;4;6;6",
+        "solution":  "【分析】一个数四舍五入到万位，需要看千位；四舍五入到亿位，需要看千万位。求括号里最大能填几，可用除法算出商，再结合题意判断。\n【详解】4( )5000≈46万，这个数的十万位是4，千位上已经是5，所以括号里最大填5，即455000≈46万。48( )7300000≈48亿，这个数的千万位上小于5，可以填0、1、2、3、4，所以括号里最大填4，即4847300000≈48亿。378÷60 = 6……18，所以60×6<378。182÷26 = 7，所以要使182÷( )>26，括号里最大填6。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "亿以内数的近似数",
+                                 "亿以上数的近似数"
+                             ],
+        "source":  "24-25 四年级上·四川凉山·期末"
+    },
+    {
+        "id":  "152",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "早上( )时整，分针和时针所形成的角刚好是 150 。",
+        "options":  [
+
+                    ],
+        "answer":  "7或5",
+        "solution":  "【分析】钟表面一圈平均分成12大格，每个大格对应360°÷12 = 30°。150°÷30° = 5，整时的时针在12处，与12相距5个大格的位置即为所求。\n【详解】360°÷12 = 30°，150°÷30° = 5，所以早上7时整或5时整，分针和时针所形成的角刚好是150°。",
+        "knowledge_points":  [
+                                 "整点时间与钟面指针的位置",
+                                 "除数是两位数的笔算除法",
+                                 "角度的计算"
+                             ],
+        "source":  "24-25四年级上·河南商丘·期末"
+    },
+    {
+        "id":  "153",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "张奶奶每天围绕小区的正方形花坛走10圈，正好是1千米。这个正方形花坛的边长是( )米，面积是( )平方米。",
+        "options":  [
+
+                    ],
+        "answer":  "25;625",
+        "solution":  "【分析】已知张奶奶每天围绕小区的正方形花坛走10圈，正好是1千米，依据1千米 = 1000 米，用1000除以10求出正方形花坛一圈的长度，即正方形的周长，再根据正方形的边长 = 正方形的周长/4，求出边长，正方形的面积 = 边长×边长，即可求出面积是多少。据此解答.\n【详解】1 千米 = 1000 米 1000÷10 = 100 (米) 100÷4 = 25 (米) 25×25 = 625 (平方米) 这个正方形花坛的边长是25米，面积是625平方米。",
+        "knowledge_points":  [
+                                 "千米和米之间的进率与换算",
+                                 "正方形的面积",
+                                 "除数是整十数的口算除法",
+                                 "正方形的周长"
+                             ],
+        "source":  "24-25 四年级上·广东中山·期末"
+    },
+    {
+        "id":  "154",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\u0027\u003e\u0027、\u0027\u003c\u0027或\u0027=\u0027。\n\n39867000( )3亿  7000公顷( )7平方千米  805000( )5080000\n630÷70( )360÷40  168÷32( )276÷28  125×8( )21×51",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003e;\u003c;=;\u003c;\u003c",
+        "solution":  "【分析】先把单位统一后再比较；有算式的先算出结果再比较。\n【详解】3亿=300000000，所以39867000<3亿。1平方千米=100公顷，7平方千米=700公顷，所以7000公顷>7平方千米。805000是六位数，5080000是七位数，所以805000<5080000。630÷70 = 9，360÷40 = 9，所以630÷70=360÷40。168÷32 = 5……8，276÷28 = 9……24，所以168÷32<276÷28。125×8 = 1000，21×51 = 1071，所以125×8<21×51。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "两，三位数与一位数连续进位的乘法",
+                                 "除数是两位数的笔算除法",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·浙江杭州·期末"
+    },
+    {
+        "id":  "155",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "括号里最大能填几。\n\n( )×63\u003c56157×( )\u003c43845×( )\u003c380",
+        "options":  [
+
+                    ],
+        "answer":  "8;7;8",
+        "solution":  "【分析】要求最大填几，用右边的数除以左边已知的因数；如果有余数，商就是最大能填的数。\n【详解】561÷63 = 8……57，所以63×8<561；438÷57 = 7……39，所以57×7<438；380÷45 = 8……20，所以45×8<380。",
+        "knowledge_points":  [
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "24-25 四年级上·江西宜春·期末"
+    },
+    {
+        "id":  "156",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "便民餐馆用780元购买这样的椅子。69元/把，99元/2把，最多可以买( )把，还剩( )元。",
+        "options":  [
+
+                    ],
+        "answer":  "15;18",
+        "solution":  "【分析】99元2把，平均每把49.5元，比69元/把便宜，所以优先按2把一组购买。先算780元能买几组2把，再看剩下的钱还能不能买1把。\n【详解】780÷99 = 7(组)……87(元)，87÷69 = 1(把)……18(元)，7×2+1 = 15(把)。所以最多可以买15把，还剩18元。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "三位数除以两位数的有余数除法",
+                                 "有余数除法的实际应用",
+                                 "经济问题"
+                             ],
+        "source":  "24-25 四年级上·重庆秀山·期末"
+    },
+    {
+        "id":  "157",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "要使算式\u00274□8÷48\u0027的商是两位数,□里最小填( );要使商是一位数,□里最大填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "8;7",
+        "solution":  "【分析】要使4□8÷48的商是两位数，被除数前两位上的数要大于或等于48；要使商是一位数，被除数前两位上的数要小于48。\n【详解】要使商是两位数，4□要大于或等于48，那么□里可以填8、9，最小填8；要使商是一位数，4□要小于48，那么□里可以填0、1、2、3、4、5、6、7，最大填7。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25四年级上·江西萍乡·期末"
+    },
+    {
+        "id":  "158",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "如果A÷B = 45，那么(A×8)÷(B×8) = ( )；A÷(B×5) = ( )；(A×2)÷B = ( )。",
+        "options":  [
+
+                    ],
+        "answer":  "45;9;90",
+        "solution":  "【分析】被除数和除数同时乘相同的数，商不变；被除数不变，除数乘几，商就除以几；除数不变，被除数乘几，商就乘几。\n【详解】如果A÷B = 45，那么(A×8)÷(B×8) = 45；A÷(B×5) = 45÷5 = 9；(A×2)÷B = 45×2 = 90。",
+        "knowledge_points":  [
+                                 "商的变化规律及应用",
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "24-25 四年级上·江西上饶·期末"
+    },
+    {
+        "id":  "159",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\u0027\u003e\u0027、\u0027\u003c\u0027或\u0027=\u0027。\n\n59万( )590001  8公顷( )800平方千米  3公顷-1公顷( )18000平方米\n360÷12( )360÷18  18×40( )180×4  500000平方米( )50平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003c;\u003e;\u003e;=;\u003c",
+        "solution":  "【分析】比较大小时，先统一单位或先算出结果再比较。1公顷 = 10000平方米，1平方千米 = 100公顷 = 1000000平方米。\n【详解】59万 = 590000，590000<590001，所以59万<590001。800平方千米 = 80000公顷，所以8公顷<800平方千米。3公顷-1公顷 = 2公顷 = 20000平方米，所以3公顷-1公顷>18000平方米。360÷12 = 30，360÷18 = 20，所以360÷12>360÷18。18×40 = 720，180×4 = 720，所以18×40=180×4。50平方千米 = 50000000平方米，所以500000平方米<50平方千米。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "除数是两位数的笔算除法",
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "24-25 四年级上·江西上饶·期末"
+    },
+    {
+        "id":  "160",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "下面括号里最大填几。\n\n50×( ) \u003c 340 ( )×41 \u003c 188",
+        "options":  [
+
+                    ],
+        "answer":  "6;4",
+        "solution":  "【分析】要求括号里最大填几，用右边的数除以左边已知的因数；如果有余数，商就是最大能填的数。\n【详解】340÷50 = 6……40，所以括号里最大填6；188÷41 = 4……24，所以括号里最大填4。",
+        "knowledge_points":  [
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "24-25 四年级上·重庆璧山·期末"
+    },
+    {
+        "id":  "161",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上\u0027\u003c\u0027、\u0027\u003e\u0027或\u0027=\u0027。\n\n78万( )775999  800÷40( )1600÷80",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;=",
+        "solution":  "【分析】先把78万改写成普通数，再比较；算式题先算出结果再比较。\n【详解】78万 = 780000，780000>775999，所以78万>775999。800÷40 = 20，1600÷80 = 20，所以800÷40=1600÷80。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "除数是整十数的口算除法"
+                             ],
+        "source":  "24-25 四年级上·重庆璧山·期末"
+    },
+    {
+        "id":  "162",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上\u0027\u003e\u0027、\u0027\u003c\u0027或\u0027=\u0027。\n\n300010( )30万  29×19( )600  180÷12( )180÷15\n\n2直角( )1周角  600÷29( )20  7平方千米( )700公顷",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;\u003c;\u003e;\u003c;\u003e;=",
+        "solution":  "【分析】比较大小时，先统一单位或先算出结果再比较。\n【详解】30万 = 300000，所以300010>30万。29×19 = 551，所以29×19<600。因为12<15，所以180÷12>180÷15。2直角 = 180°，1周角 = 360°，所以2直角<1周角。600÷29 = 20……20，所以600÷29>20。7平方千米 = 700公顷，所以7平方千米=700公顷。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "除数是两位数的笔算除法",
+                                 "两位数乘两位数的进位乘法",
+                                 "大数的比较"
+                             ],
+        "source":  "24-25 四年级上·天津蓟州·期末"
+    },
+    {
+        "id":  "163",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个笔袋 18 元,100 元钱可以买( )个这样的笔袋，还剩( )元。",
+        "options":  [
+
+                    ],
+        "answer":  "5;10",
+        "solution":  "【分析】根据题意分析，用100除以一个笔袋的价钱，求得的商就是可以买笔袋的数量，余数是还剩下的钱数。据此分析解答.\n【详解】 100÷18 = 5 (个)。10(元)所以一个笔袋 18 元,100 元钱可以买 5 个这样的笔袋，还剩 10 元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "有余数除法的实际应用",
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "24-25 四年级上·宁夏吴忠·期末"
+    },
+    {
+        "id":  "164",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "张华有 6 本同样的故事书，赵明有4本同样的童话书，两人的书一共价值140元。如果张华给赵明一本故事书，赵明给张华一本童话书，则两人手中的书价值相等。故事书每本( )元，童话书每本( )元。",
+        "options":  [
+
+                    ],
+        "answer":  "10;20",
+        "solution":  "【分析】根据题意，两人的书总价值为140元，交换后价值相等;即:5 本故事书的价值+1 本童话书的价值 = 1 本故事书价值+3本童话书的价值，可得:2本故事书的价值 = 1 本童话书的价值;所以,4本童话书的价值相当于8本故事书的价值，所以(6+8)本故事书的价值是140元，据此用140 元除以 14 求出故事书的价格，进而求出童话书的价格.\n【详解】因为:5本故事书的价值+1本童话书的价值 = 1 本故事书价值+3本童话书的价值;可得:2本故事书的价值 = 1 本童话书的价值14 本故事书是 140 元; 140÷14 = 10 (元) (140 - 10×6) ÷ 4 = (140 - 60) ÷ 4 = 80÷4 = 20 (元) 所以，故事书每本 10 元，童话书每本 20 元。",
+        "knowledge_points":  [
+                                 "等量代换",
+                                 "带有小括号的混合运算",
+                                 "经济问题"
+                             ],
+        "source":  "24-25 四年级上·重庆丰都·期末"
+    },
+    {
+        "id":  "165",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在 A/ B (B 不为 0)中,A 除以 7，要使商不变,B 应( )。",
+        "options":  [
+
+                    ],
+        "answer":  "除以 7",
+        "solution":  "【分析】根据商不变的规律，在除法算式中，被除数和除数同时乘或除以同一个不为零的数，商不变。本题中，被除数 A 除以 7，要使商不变，除数 B 也应除以 7.\n【详解】根据商不变的规律，被除数A除以7，除数B也应除以 7，这样商才能保持不变。因此,B 应除以 7。",
+        "knowledge_points":  [
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "24-25 四年级上·重庆丰都·期末"
+    },
+    {
+        "id":  "166",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "我们在笔算 197÷28 时，先把28看作( )试商，可以商( ),这时余数为(),比除数大了，所以我们需要改商为( )。",
+        "options":  [
+
+                    ],
+        "answer":  "30;6;29;7",
+        "solution":  "【分析】除数是两位数的除法，一般按照“四舍五入”法，把除数看作和它接近的整十数来试商。除数看大了，试商可能会偏小，就需要把商调大。\n【详解】我们在笔算197÷28时，先把28看作30试商，可以商6。6×28 = 168，197-168 = 29，这时余数为29，比除数大，所以需要改商为7。",
+        "knowledge_points":  [
+                                 "多位数除以两位数的试商",
+                                 "余数与除数的关系"
+                             ],
+        "source":  "24-25四年级上·重庆丰都·期末"
+    },
+    {
+        "id":  "167",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在横线里填上\u0027>\u0027、\u0027<\u0027或\u0027=\u0027。\n\n①60600( )60060  ②6亿( )6000万\n③160×60( )600×16  ④7200÷900( )72÷8",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;\u003e;=;\u003c",
+        "solution":  "【分析】比较大小时，先统一单位或先算出结果再比较。\n【详解】①60600和60060位数相同，百位上6>0，所以60600>60060。②6亿=60000万，所以6亿>6000万。③160×60 = 9600，600×16 = 9600，所以160×60=600×16。④7200÷900 = 8，72÷8 = 9，所以7200÷900<72÷8。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "三位数乘两位数，乘数末尾有 0",
+                                 "亿以上数的改写"
+                             ],
+        "source":  "24-25 四年级上·天津津南·期末"
+    },
+    {
+        "id":  "168",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上适当的数。\n\n( )÷56 = 14……6  464÷( ) = 23……4\n\n388÷( ) = ( )……68  420÷( ) \u003e 6",
+        "options":  [
+
+                    ],
+        "answer":  "790;20;80;4;60",
+        "solution":  "【分析】(1)根据“被除数 = 商×除数+余数”求被除数。(2)根据“除数 = (被除数-余数)÷商”求除数。(3)因为余数是68，所以除数要大于68，答案不唯一。(4)要使420÷( )>6，除数要小于70，答案不唯一。\n【详解】14×56+6 = 784+6 = 790，所以790÷56 = 14……6。 (464-4)÷23 = 460÷23 = 20，所以464÷20 = 23……4。因为388-68 = 320，320÷80 = 4，所以388÷80 = 4……68，答案不唯一。420÷60>6，答案不唯一。",
+        "knowledge_points":  [
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "25-26 四年级上·山东菏泽·月考"
+    },
+    {
+        "id":  "169",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "括号里最大能填几。\n\n20×( ) \u003c 91 ( )×40 \u003c 84\n\n98 \u003e 30×( ) 50×( ) \u003c 190\n\n( )×60\u003c482402\u003e70×( )",
+        "options":  [
+
+                    ],
+        "answer":  "4;2;3;3;8;5",
+        "solution":  "【分析】要求最大能填几，用所比较的数除以已知因数；如果有余数，商就是最大能填的数；如果没有余数，商要减1。\n【详解】91÷20 = 4……11，所以20×4<91；84÷40 = 2……4，所以2×40<84；98÷30 = 3……8，所以98>30×3；190÷50 = 3……40，所以50×3<190；482÷60 = 8……2，所以8×60<482；402÷70 = 5……52，所以402>70×5。",
+        "knowledge_points":  [
+                                 "三位数除以两位数的有余数除法",
+                                 "两位数乘整十数的口算乘法",
+                                 "除数是整十数的笔算除法"
+                             ],
+        "source":  "25-26 四年级上·山东菏泽·月考"
+    },
+    {
+        "id":  "170",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\u0027\u003e\u0027、\u0027\u003c\u0027或\u0027=\u0027。\n\n140×5( )700  3×190( )600  750÷37( )20\n180÷28( )7  154÷22( )7  168÷24( )8",
+        "options":  [
+
+                    ],
+        "answer":  "=;\u003c;\u003e;\u003c;=;\u003c",
+        "solution":  "【分析】先算出左边算式的值，再与右边的数比较。\n【详解】140×5 = 700，所以140×5=700。3×190 = 570，570<600，所以3×190<600。750÷37 = 20……10，所以750÷37>20。180÷28 = 6……12，所以180÷28<7。154÷22 = 7，所以154÷22=7。168÷24 = 7，所以168÷24<8。",
+        "knowledge_points":  [
+                                 "因数中间或末尾是0的乘法",
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "25-26 四年级上·山东菏泽·月考"
+    },
+    {
+        "id":  "171",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "括号里能填几。(1)商是一位数:( )64÷26 ( )28÷45 (2)商是两位数:( )76÷87 ( )52÷55",
+        "options":  [
+
+                    ],
+        "answer":  "1;1、2、3、4;8、9;5、6、7、8、9",
+        "solution":  "【分析】判断三位数除以两位数的商是几位数，看被除数前两位：前两位大于或等于除数，商是两位数；前两位小于除数，商是一位数。\n【详解】(1)( )64÷26，商是一位数，说明前两位小于26，且最高位不能是0，所以只能填1。( )28÷45，商是一位数，说明前两位小于45，且最高位不能是0，所以可以填1、2、3、4。(2)( )76÷87，商是两位数，说明前两位大于或等于87，所以可以填8、9。( )52÷55，商是两位数，说明前两位大于或等于55，所以可以填5、6、7、8、9。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "25-26四年级上·山东菏泽·月考"
+    },
+    {
+        "id":  "172",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "三位数除以两位数，商是( )位数。",
+        "options":  [
+
+                    ],
+        "answer":  "一或两",
+        "solution":  "【分析】三位数除以两位数，商可能是一位数或两位数。具体取决于被除数的前两位数字与除数的比较：如果被除数的前两位大于或等于除数，商是两位数;如果被除数的前两位小于除数，商是一位数.\n【详解】如果被除数的前两位大于或等于除数，商是两位数;如果被除数的前两位小于除数，商是一位数，所以三位数除以两位数，商是一或两位数。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "25-26 四年级上·山东菏泽·月考"
+    },
+    {
+        "id":  "173",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在下面( )里填上\u0027\u003c\u0027\u0027\u003e\u0027或\u0027=\u0027\n\n3654879( )365489726900100000( )270万 300÷ 25( )60÷5\n\n480 ÷ 12 ( ) 480 ÷ 3018×500 ( ) 50×180160 ÷ (4×8) ( ) 160÷4÷8",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003e;=;\u003e;=;=",
+        "solution":  "【分析】(1)整数比较大小时，要看他们的数位。数位多的那个数就大;如果数位相同，就看最高位，最高位上的数大，那个数就大;最高位上的数相同，就看下一位，哪位上的数大，那个数就大。(2)把前面的数改写成用万做单位的数再比较。 (3)(4)(5)题将两边等式结果计算出来再比较。 (6),左边算式先算小括号里的乘法，再算小括号外的除法，右边算式，根据除法性质，除以一个数等于除以这两个数的积，据此先算4与8的积，再用160除以这个积，求出得数，最后比较.\n【详解】(1)两个数都是七位数，万级数字相同，千位和百位数字也相同,3654879十位上的数字7小于3654897十位上的数字 9，则 3654879\u003c3654897;(2)26900100000 的最高位是百亿位,270 万的最高位是百万位，则 26900100000 \u003e 270 万;(3) 300÷25 = 12 , 60÷5 = 12 , 12 = 12; (4) 480÷12 = 40 , 480÷30 = 16，则 480÷12 \u003e 480÷30;(5) 18×500 = 9000 , 50×180 = 9000 , 18×500 = 50×180 (6)160÷(4×8)=160÷32=5,160÷4÷8=160÷(4×8)160 ÷ 32 = 5.3654879 \u003c 365489726900100000 \u003e 270 万 300÷25 = 60÷5又480÷12 \u003e 480÷3018×500 = 50×180160 ÷ (4×8) = 160÷4÷8 ",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "除数是两位数的笔算除法",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25 四年级上·内蒙古巴彦淖尔·期末"
+    },
+    {
+        "id":  "174",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "A÷B = 9……3，如果A和B都乘100，那么商是( )，余数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "9;300",
+        "solution":  "【分析】被除数和除数同时乘或除以相同的数，商不变，余数也同时乘或除以相同的数。\n【详解】A÷B = 9……3，如果A和B都乘100，那么商还是9，余数变成3×100 = 300。",
+        "knowledge_points":  [
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "25-26 五年级上·江西宜春·期中"
+    },
+    {
+        "id":  "175",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "两个数的商是420，如果被除数不变，除数乘 20，则商是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "21",
+        "solution":  "【分析】根据商的变化规律，当被除数不变时，除数乘几(0 除外),商就除以几。商是 420，除数乘 20，因此商应除以 20 .\n【详解】 420÷20 = 21所以两个数的商是420，如果被除数不变，除数乘20，则商是 21。",
+        "knowledge_points":  [
+                                 "商的变化规律及应用",
+                                 "除数是整十数的笔算除法"
+                             ],
+        "source":  "24-25 四年级上·内蒙古乌海·期末"
+    },
+    {
+        "id":  "176",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\u0027\u003e\u0027\u0027\u003c\u0027或\u0027=\u0027.\n\n600600( )6060608 亿( )8000 万\n\n140÷35( )270÷9032×124( )16×248",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003e;\u003e;=",
+        "solution":  "【分析】根据整数比较大小的方法比较大小即可;整数大小的比较方法：位数不同，位数多的数就大;位数相同，左起第一位上的数大的那个数就大，如果左起第位上的数相同，就比较左起第二位上的数，直到比较出大小为止。比较大数时，先统一单位或转换为整数，再按位数和各位数字依次比较。把一个用\u0027万\u0027作单位的数改写成整万数(以\u0027一\u0027为单位的数),在后面加4个0，同时去掉\u0027万\u0027字即可，再比较大小。把一个用\u0027亿\u0027作单位的数改写成整亿数(以\u0027一\u0027为单位的数),在后面加8个0，同时去掉\u0027亿\u0027字即可，再比较大小。含有算式的大小比较，可以先计算出每个算式的结果，再比较大小.\n【详解】600600(\u003c)606060;8 亿=800000000,8000 万=80000000,800000000\u003e80000000,8 亿(\u003e)8000 万;140÷35 = 4 , 270÷90 = 3 , 4 \u003e 3 ,140÷35(\u003e)270÷90;32×124 = 3968 , 16×248 = 3968 ,3968=3968,32×124(=)16×248。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "三位数与两位数的乘法",
+                                 "被除数和除数末尾都有0的除法",
+                                 "亿以上数的改写"
+                             ],
+        "source":  "24-25 四年级上·内蒙古乌海·期末"
+    },
+    {
+        "id":  "177",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "某体育器材店里羽毛球的价格为 30 元/筒，学校买 134 筒这样的羽毛球需要()元;小明带着 540 元去这家店可以买( )筒这样的羽毛球。",
+        "options":  [
+
+                    ],
+        "answer":  "4020;18",
+        "solution":  "【分析】总价 = 单价×数量，用一筒羽毛球的价格乘买的筒数，即可求出需要多少元;用带的钱数除以每筒羽毛球的价格，求出的商即为可以买的筒数，据此填空即可.\n【详解】 134×30 = 4020 (元)540÷30 = 18 (筒)因此，学校买134筒这样的羽毛球需要4020元;小明带着540 元去这家店可以买 18 筒这样的羽毛球。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "24-25 四年级上·新疆阿克苏·期末"
+    },
+    {
+        "id":  "178",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在( )里填上\u0027>\u0027、\u0027<\u0027或\u0027=\u0027。\n\n30亿( )389090500  1周角( )2平角  7平方千米( )70000公顷\n2公顷( )200平方米  210×42( )420×21  378÷42( )10",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;=;\u003c;\u003e;=;\u003c",
+        "solution":  "【分析】比较大小时，先统一单位或先算出结果再比较。\n【详解】30亿 = 3000000000，3000000000是十位数，389090500是九位数，所以30亿>389090500。1周角 = 2平角。1平方千米 = 100公顷，所以7平方千米 = 700公顷，700公顷<70000公顷，因此7平方千米<70000公顷。2公顷 = 20000平方米，所以2公顷>200平方米。210×42 = 8820，420×21 = 8820，所以210×42=420×21。378÷42 = 9，所以378÷42<10。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "除数是两位数的笔算除法",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "24-25 四年级上·新疆阿克苏·期末"
+    },
+    {
+        "id":  "179",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "根据算式 38×25 = 950，直接写出下面两个算式的得数.\n\n380×25 = ( ) 9500÷38 = ( )",
+        "options":  [
+
+                    ],
+        "answer":  "9500;250",
+        "solution":  "【分析】(1)根据积的变化规律：一个因数不变，另一个因数乘几或除以几(0除外),积也乘几或除以几解答;(2)先由 38×25 = 950 可得 950÷38 = 25，再根据商的变化规律：除数不变时，被除数乘(或除以)几(0 除外),商也乘(或除以)几解答.\n【详解】 ( 38×10 )×25 = ( 950×10 ) ,即 380×25 = 9500;950÷38 = 25 ,( 950×10 ) ÷ 38 = ( 25×10 ) ,即 9500÷38 = 250 。",
+        "knowledge_points":  [
+                                 "商的变化规律及应用",
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "24-25 四年级上·新疆阿克苏·期末"
+    },
+    {
+        "id":  "180",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "算式 364÷26 的商是( )位数。",
+        "options":  [
+
+                    ],
+        "answer":  "两",
+        "solution":  "【分析】根据三位数除以两位数的商的位数的判断方法，当被除数的前两位大于或等于除数，则商的最高位是十位，即商是两位数，当被除数的前两位小于除数，则商的最高位在个位，即商是一位数;据此解答.\n【详解】 36 \u003e 26，所以，算式 364÷26 的商是两位数。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25 四年级上·新疆阿克苏·期末"
+    },
+    {
+        "id":  "181",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在( )里填上\u0027\u003e\u0027、\u0027\u003c\u0027或\u0027=\u0027。\n\n345679( )343989  148576000( )2亿  3公顷( )300平方米\n500000平方米( )5平方千米  240÷6( )480÷12",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;\u003c;\u003e;\u003c;=",
+        "solution":  "【分析】比较大小时，先统一单位或先算出结果再比较。\n【详解】345679和343989中，前两位相同，千位上5>3，所以345679>343989。148576000 = 1.48576亿，所以148576000<2亿。1公顷 = 10000平方米，3公顷 = 30000平方米，所以3公顷>300平方米。1平方千米 = 1000000平方米，5平方千米 = 5000000平方米，所以500000平方米<5平方千米。240÷6 = 40，480÷12 = 40，所以240÷6=480÷12。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "除数是两位数的笔算除法",
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "24-25 四年级上·甘肃兰州·期末"
+    },
+    {
+        "id":  "182",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一本书有 238页，如果王刚每天看 20 页，至少需要( )天才能看完。",
+        "options":  [
+
+                    ],
+        "answer":  "12",
+        "solution":  "【分析】用一本书的总页数除以每天读的页数，即可求出读书天数。若有余数，因为剩余的页数也需要时间读完，则用商加1。\n【详解】238÷20 = 11(天)……18(页)，11+1 = 12(天)，所以至少需要12天才能看完。",
+        "knowledge_points":  [
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "24-25 四年级上·新疆巴音郭楞·期末"
+    },
+    {
+        "id":  "183",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在( )里填上\u0027\u003e\u0027、\u0027\u003c\u0027或\u0027=\u0027。\n\n88888( )9999  100万( )1亿  10700( )10007\n120×20( )2400  360÷60( )60  300公顷( )3平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;\u003c;\u003e;=;\u003c;=",
+        "solution":  "【分析】比较大小时，先统一单位或先算出结果再比较。\n【详解】88888是五位数，9999是四位数，所以88888>9999。1亿 = 10000万，所以100万<1亿。10700和10007比较，百位上7>0，所以10700>10007。120×20 = 2400，所以120×20=2400。360÷60 = 6，所以360÷60<60。1平方千米 = 100公顷，所以3平方千米 = 300公顷，因此300公顷=3平方千米。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "除数是整十数的口算除法",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·新疆吐鲁番·期末"
+    },
+    {
+        "id":  "184",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "鸟是人类的好朋友,1只啄木鸟平均每天消灭害虫300条;1只土燕子在育雏期,1 个月(按 30 天算)能消灭蝗虫 16000 只。(1)1 只啄木鸟 1 年(按 365 天计算)能消灭害虫( )条。(2)1 只土燕子在育雏期，平均每天约能消灭蝗虫( )只。",
+        "options":  [
+
+                    ],
+        "answer":  "109500;500",
+        "solution":  "【分析】(1)已知 1 只啄木鸟平均每天消灭害虫 300 条，年按 365 天计算，总数量为每天消灭害虫的数量乘天数。(2)已知 1 只土燕子 1 个月(30 天)能消灭蝗虫 16000 只，平均每天消灭量为消灭害虫的总数量除以天数，计算时需要把16000看作15000进行估算。据此解答.\n【详解】(1) 300×365 = 109500 (条)所以,1只啄木鸟1年(按365天计算)能消灭害虫109500条。(2)把 16000 看作 15000, 15000÷30 = 500 (只) 所以,1 只土燕子在育雏期，平均每天约能消灭蝗虫 500 只。",
+        "knowledge_points":  [
+                                 "三位数与整十数的口算乘法",
+                                 "除数是两位数的估算"
+                             ],
+        "source":  "25-26四年级上·广东揭阳·月考"
+    },
+    {
+        "id":  "185",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "括号里最大能填几。\n\n 60×( ) \u003c 25431×( ) \u003c 150410 \u003e 49×( ) 502 \u003e 82×( ) ",
+        "options":  [
+
+                    ],
+        "answer":  "4;4;8;6",
+        "solution":  "【分析】要求最大填几，用左边的大数除以右边已知因数；如果有余数，商就是最大能填的数。\n【详解】254÷60 = 4……14，所以60×4<254；150÷31 = 4……26，所以31×4<150；410÷49 = 8……18，所以410>49×8；502÷82 = 6……10，所以502>82×6。",
+        "knowledge_points":  [
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "24-25四年级上·新疆克孜勒苏·期末"
+    },
+    {
+        "id":  "186",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "张阿姨去水果市场购买了 21 箱梨和 13 箱苹果，苹果每箱 42 元，已知购买的梨和苹果的总价相等，则购买梨的总价为( )元，每箱梨( )元。",
+        "options":  [
+
+                    ],
+        "answer":  "546;26",
+        "solution":  "【分析】总价 = 单价×数量，先据此计算出苹果总价，因为梨和苹果的总价相等，所以梨的单价 = 总价/数量，代入可算出梨的单价.\n【详解】 42×13 = 546 (元)546÷21 = 26 (元) 所以购买梨的总价为 546 元;每箱梨 26 元。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "两位数乘两位数的进位乘法",
+                                 "经济问题"
+                             ],
+        "source":  "24-25 四年级上·甘肃临夏·期末"
+    },
+    {
+        "id":  "187",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "要使367÷□6的商是一位数,□中最小填( );要使□67÷28的商是两位数,□中最小填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "4;3",
+        "solution":  "【分析】三位数除以两位数，如果商是一位数，说明三位数的前两位数小于除数，也就是36\u003c□6;要使商是两位数，说明三位数的前两位数大于或等于除数，也就是□ 6 \u003e= 28，据此填空.\n【详解】要使367÷□6的商是一位数，需要36\u003c□6,□中最小填4;要使□67÷28的商是两位数,□ 6 \u003e= 28 ,□中最小填 3 。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25四年级上·甘肃临夏·期末"
+    },
+    {
+        "id":  "188",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\u0027>\u0027、\u0027<\u0027或\u0027=\u0027。\n\n18万( )180500  30×190( )19×300  2平方千米+20000公顷( )4平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;=;\u003e",
+        "solution":  "【分析】把单位统一后再比较大小，或直接利用积的变化规律比较。\n【详解】18万 = 180000，180000<180500，所以18万<180500。30×190 = 300×19 = 19×300，所以30×190=19×300。1平方千米 = 100公顷，2平方千米 = 200公顷，2平方千米+20000公顷 = 20200公顷，4平方千米 = 400公顷，所以2平方千米+20000公顷>4平方千米。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "除数是整十数的口算除法",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·江西宜春·期中"
+    },
+    {
+        "id":  "189",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一间教室的占地面积大约为 50( ),那么1公顷的土地相当于( )间这样的教室的占地面积。",
+        "options":  [
+
+                    ],
+        "answer":  "平方米或m2;200",
+        "solution":  "【分析】常见的面积单位有平方分米、平方米、公顷等，教室的占地面积较小，根据生活经验选择合适的单位。1公顷 = 10000平方米，再用除法计算能容纳多少间教室。\n【详解】一间教室的占地面积大约为50平方米。1公顷 = 10000平方米，10000÷50 = 200(间)。因此，一间教室的占地面积大约为50平方米，那么1公顷的土地相当于200间这样的教室的占地面积。",
+        "knowledge_points":  [
+                                 "面积单位的选择",
+                                 "公顷，平方千米的进率与换算",
+                                 "除数是整十数的口算除法"
+                             ],
+        "source":  "25-26四年级上·湖北黄石·期中"
+    },
+    {
+        "id":  "190",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "若 A×80 = 320，则A×8=( )。",
+        "options":  [
+
+                    ],
+        "answer":  "32",
+        "solution":  "【分析】根据积的变化规律，一个因数不变，另一个因数乘或除以一个不为0的数，积也要乘或除以这个数，据此填空即可.\n【详解】 A×80 和 A×8 相比，因数A不变，另一个因数80÷10 = 8，则积也要除以 10, 320÷10 = 32 , A×8 = 32 。若 A×80 = 320，则 A×8 = 32 。",
+        "knowledge_points":  [
+                                 "除数是整十数的口算除法",
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·河南开封·期中"
+    },
+    {
+        "id":  "191",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "□ 28÷52，要使商是一位数,□里最大填( );要使商是两位数,□里最小填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "4;5",
+        "solution":  "【分析】要使□28÷52的商是一位数，则被除数前两位上的数小于除数,□里的数要小于5。要使□28÷52的商是两位数，则被除数前两位上的数大于等于除数,□里的数要大于等于 5。据此解答.\n【详解】□28÷52，要使商是一位数,□里的数要小于5,□里最大填4;要使商是两位数,□里的数要大于等于5,□里最小填 5。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "25-26四年级上·河南信阳·期中"
+    },
+    {
+        "id":  "192",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在( )里填上合适的数。(1)一块长方形菜园占地面积 100 平方米,( )块这样的菜园占地面积是 1 公顷。 (2)某果园的占地面积约 5 公顷,( )个果园的占地面积约是 1 平方千米。 (3)某运动场地的面积大约是 2000 平方米,( )个这样的运动场，面积大约是1平方千米。 (4)1 平方米的地方能站 6 个成年人,1 公顷的地方大约能站()个成年人,1 平方千米的地方大约能站( )个成年人。",
+        "options":  [
+
+                    ],
+        "answer":  "100;20;500;60000;6000000",
+        "solution":  "【分析】(1)1公顷 = 10000平方米，用10000除以100即可。(2)1平方千米 = 100公顷，用100除以5即可。(3)1平方千米 = 1000000平方米，用1000000除以2000即可。(4)先求1公顷大约能站多少人，再求1平方千米大约能站多少人。\n【详解】(1)1公顷 = 10000平方米，10000÷100 = 100(块)。 (2)1平方千米 = 100公顷，100÷5 = 20(个)。 (3)1平方千米 = 1000000平方米，1000000÷2000 = 500(个)。 (4)1公顷 = 10000平方米，10000×6 = 60000(个)；1平方千米 = 100公顷，100×60000 = 6000000(个)。所以应填100；20；500；60000；6000000。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "公顷，平方千米的进率与换算",
+                                 "除数是整十数的口算除法",
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "25-26四年级上·广东东莞·月考"
+    },
+    {
+        "id":  "193",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上\u0027\u003e\u0027、\u0027\u003c\u0027或\u0027=\u0027。\n\n100个千( )10万  48×101( )4800+480  360÷40( )300÷40  25×260( )250×26",
+        "options":  [
+
+                    ],
+        "answer":  "=;\u003c;\u003e;=",
+        "solution":  "【分析】先算出各式结果，再比较大小。\n【详解】100个千 = 100×1000 = 100000，10万 = 100000，所以100个千=10万。48×101 = 4848，4800+480 = 5280，所以48×101<4800+480。因为360>300，且除数相同，所以360÷40>300÷40。25×260 = 6500，250×26 = 6500，所以25×260=250×26。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "除数是两位数的笔算除法",
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·广东揭阳·月考"
+    },
+    {
+        "id":  "194",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "根据商不变的规律填空：\n\n30÷5 =( )÷10 = 120÷( ) = ( )÷50 =( )",
+        "options":  [
+
+                    ],
+        "answer":  "60;20;300;6",
+        "solution":  "【分析】根据商不变的规律，被除数和除数同时乘或除以相同的数，商不变。\n【详解】30÷5 = 6，30÷5 = (30×2)÷(5×2) = 60÷10；30÷5 = (30×4)÷(5×4) = 120÷20；30÷5 = (30×10)÷(5×10) = 300÷50。所以30÷5 =(60)÷10 = 120÷(20) = (300)÷50 =(6)。",
+        "knowledge_points":  [
+                                 "商不变的规律及应用",
+                                 "除数是整十数的口算除法"
+                             ],
+        "source":  "25-26 四年级上·广东揭阳·月考"
+    },
+    {
+        "id":  "195",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "阳光小学阅览室的面积约是 200 平方米,( )个这样的阅览室面积约是 1 公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "50",
+        "solution":  "【分析】1公顷 = 10000 平方米，每个阅览室的面积是200平方米。要求多少个这样的阅览室面积约是1公顷，就是求10000平方米中包含多少个200平方米，用除法计算.\n【详解】1 公顷 = 10000 平方米, 10000÷200 = 50 (个),所以需要50个这样的阅览室。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26四年级上·江西九江·期中"
+    },
+    {
+        "id":  "196",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "学校修建了一个长40 米，宽 25 米的游泳池,1 公顷面积的土地可以修建( )个这样的游泳池。",
+        "options":  [
+
+                    ],
+        "answer":  "10",
+        "solution":  "【分析】长方形的面积 = 长×宽，将数据代入求出游泳池的面积。1 公顷 = 10000 平方米，用 10000 平方米除以游泳池的面积即可求解.\n【详解】 40×25 = 1000 (平方米)1公顷 = 10000 平方米 10000÷1000=10(个) 学校修建了一个长40米，宽25米的游泳池,1公顷面积的土地可以修建10个这样的游泳池.\n\n10000÷1000 = 10 (个)",
+        "knowledge_points":  [
+                                 "除数是整十数的口算除法",
+                                 "长方形的面积",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·河南南阳·期中"
+    },
+    {
+        "id":  "197",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "某同学要用一根长为 540 厘米的绳子捆绑小棒，捆每堆小棒要用 20 厘米长的绳子，这根绳子最多可以捆绑( )堆小棒。",
+        "options":  [
+
+                    ],
+        "answer":  "27",
+        "solution":  "【分析】要求这根绳子最多可以捆绑多少堆小棒，就是求540 厘米里面包含多少个 20 厘米，根据除法的意义，用总长度除以每堆所需长度即可得出堆数.\n【详解】 540÷20 = 27 (堆)所以这根绳子最多可以捆绑27堆小棒。",
+        "knowledge_points":  [
+                                 "被除数和除数末尾都有 0 的除法"
+                             ],
+        "source":  "24-25 五年级上·新疆哈密·期末"
+    },
+    {
+        "id":  "198",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "括号里最大能填几。\n\n56( )635≈56万 ( )×80\u003c48350×( )\u003c410\n\n28( )547≈29万 60×( )\u003c41930×( )\u003c200",
+        "options":  [
+
+                    ],
+        "answer":  "4;6;8;9;6;6",
+        "solution":  "【分析】求近似数时，看千位上的数是四舍还是五入；求括号里最大填几，用右边的数除以左边已知的因数。\n【详解】56( )635≈56万，说明千位上的数小于5，所以括号里可以填0、1、2、3、4，最大填4，即564635≈56万。483÷80 = 6……3，所以括号里最大填6。410÷50 = 8……10，所以括号里最大填8。28( )547≈29万，说明千位上的数大于或等于5，所以括号里可以填5、6、7、8、9，最大填9，即289547≈29万。419÷60 = 6……59，所以括号里最大填6。200÷30 = 6……20，所以括号里最大填6。",
+        "knowledge_points":  [
+                                 "三位数除以两位数的有余数除法",
+                                 "亿以内数的近似数"
+                             ],
+        "source":  "25-26 四年级上·山东日照·期中"
+    },
+    {
+        "id":  "199",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果 10 平方米约能站 120 人，那么占地 1 公顷的足球场大约能站( )人。",
+        "options":  [
+
+                    ],
+        "answer":  "120000",
+        "solution":  "【分析】1公顷 = 10000 平方米，先用10000平方米除以10平方米求出10000平方米里面有多少个10平方米，再乘上120 即可求解.\n【详解】1 公顷 = 10000 平方米 10000÷10×120 = 1000×120 = 120000 (人) 如果10平方米约能站120人，那么占地1公顷的足球场大约能站 120000 人。",
+        "knowledge_points":  [
+                                 "多位数与两位数的乘除混合运算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·河南南阳·期中"
+    },
+    {
+        "id":  "200",
+        "unit":  "第六单元 除数是两位数的除法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "永辉超市促销，某款面粉每袋价格18元，现在买10袋送2袋，王阿姨带了200 元，最多可以买( )袋。",
+        "options":  [
+
+                    ],
+        "answer":  "13",
+        "solution":  "【分析】先算200元最多能买多少袋面粉，再根据“买10袋送2袋”确定赠送袋数。\n【详解】200÷18 = 11(袋)……2(元)，11袋里面有1个10袋，所以可送2袋。11+2 = 13(袋)。所以王阿姨带了200元，最多可以买13袋。",
+        "knowledge_points":  [
+                                 "有余数除法的实际应用",
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "25-26五年级上·湖北黄冈·期中"
+    },
+    {
+        "id":  "201",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "钟面上分针旋转一周，它旋转的角度是( )°,此时，相应的时针旋转( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "360;30",
+        "solution":  "【分析】分针旋转一周对应一个周角，即360°。分针旋转一周时经过1小时，时针旋转1个大格，钟面1圈有12个大格，所以1个大格对应360°÷12 = 30°。\n【详解】钟面上分针旋转一周，它旋转的角度是360°，此时相应的时针旋转30°。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "除数是两位数的笔算除法",
+                                 "平角，周角的认识及特征"
+                             ],
+        "source":  "25-26四年级上·河南南阳·期中"
+    },
+    {
+        "id":  "202",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在一道除法算式中，被除数扩大到原来的 6 倍，除数扩大到原来的 3 倍，商()倍。",
+        "options":  [
+
+                    ],
+        "answer":  "扩大到原来的 2",
+        "solution":  "【分析】根据商的变化规律：被除数扩大到原来的几倍(或缩小到原来的几分之一),除数不变，商扩大到原来的几倍(或缩小到原来的几分之一);被除数不变，除数扩大到原来的几倍(或缩小到原来的几分之一),商缩小到原来的几分之一(或扩大到原来的几倍);被除数和除数同时乘或除以一个不为 0 的数，商不变，据此解答.\n【详解】 6÷3 = 2 所以在一道除法算式中，被除数扩大到原来的6倍，除数扩大到原来的3倍，商扩大到原来的2倍。",
+        "knowledge_points":  [
+                                 "商的变化规律及应用"
+                             ],
+        "source":  "25-26 五年级上·湖北十堰·期中"
+    },
+    {
+        "id":  "203",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "学校下午大课间的时间是 3 时整，此时钟面上时针和分针之间的夹角是( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "直",
+        "solution":  "【分析】钟面上共有12个大格，周角是360°，每个大格是360°÷12 = 30°。3时整时，时针指向3，分针指向12，它们之间相差3个大格。\n【详解】360°÷12 = 30°，3×30° = 90°，90°是直角，所以此时钟面上时针和分针之间的夹角是直角。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "除数是两位数的笔算除法",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "25-26 四年级上·山东济宁·期中"
+    },
+    {
+        "id":  "204",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "有 120 千克苹果要装箱，平均每箱可以装 17 千克苹果，至少需要准备( )个箱子才能把这些苹果装完。",
+        "options":  [
+
+                    ],
+        "answer":  "8",
+        "solution":  "【分析】每箱装17千克苹果，用120除以17即可求出能装满几箱，还剩下几千克，剩下的苹果还需要再装1箱。\n【详解】120÷17 = 7(箱)……1(千克)，7+1 = 8(箱)，所以至少需要准备8个箱子才能把这些苹果装完。",
+        "knowledge_points":  [
+                                 "有余数除法的实际应用",
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "25-26 五年级上·福建厦门·期中"
+    },
+    {
+        "id":  "205",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小沐在计算三位数除以两位数的除法时，把被除数 284 错写成 428，这样商比原来多了 12，但余数相同。这道除法算式的余数是 。",
+        "options":  [
+
+                    ],
+        "answer":  "8",
+        "solution":  "【分析】把284错写成428后，商比原来多了12，但余数相同，所以两个被除数的差144正好是12个除数，据此先求除数，再求余数。\n【详解】(428-284)÷12 = 144÷12 = 12，所以除数是12。284÷12 = 23……8，所以这道除法算式的余数是8。",
+        "knowledge_points":  [
+                                 "三位数除以两位数的有余数除法",
+                                 "余数与除数的关系"
+                             ],
+        "source":  "2025·重庆北碚·小升初真题"
+    },
+    {
+        "id":  "206",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "从12时到1时，分针旋转了( )°,时针旋转了( )°,4 时整，时针和分针 组成的较小角是( )°,它是( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "360;30;120;钝",
+        "solution":  "【分析】从12时到1时经过1小时，分针旋转一整圈，即360°；时针旋转1个大格，即30°。4时整时，时针和分针相差4个大格，所以较小角是4×30° = 120°，这是钝角。\n【详解】从12时到1时，分针旋转了360°，时针旋转了30°。4时整时，时针和分针组成的较小角是120°，它是钝角。",
+        "knowledge_points":  [
+                                 "整点时间与钟面指针的位置",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "除数是整十数的笔算除法",
+                                 "整十，整百，整千数与一位数的乘法"
+                             ],
+        "source":  "25-26四年级上·山西晋中·期中"
+    },
+    {
+        "id":  "207",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "山东省，简称“鲁”，全省总面积约是160000000000平方米，合( )公顷，合( )平方千米。",
+        "options":  [
+
+                    ],
+        "answer":  "16000000;160000",
+        "solution":  "【分析】1公顷 = 10000平方米，1平方千米 = 100公顷 = 1000000平方米，据此换算。\n【详解】160000000000÷10000 = 16000000(公顷)，160000000000÷1000000 = 160000(平方千米)。所以160000000000平方米合16000000公顷，合160000平方千米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·山东济南·期中"
+    },
+    {
+        "id":  "208",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "两个数相除的商是 6，余数是7，如果被除数和除数同时扩大到原来10倍，商是( ),余数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "6;70",
+        "solution":  "【分析】被除数和除数同时乘或除以一个相同的数(0除外),商不变，但余数也乘或除以这个相同的数.\n【详解】 7×10 =70 两个数相除的商是6，余数是7，如果被除数和除数同时扩大到原来 10 倍，商是 6，余数是 70。",
+        "knowledge_points":  [
+                                 "整十，整百，整千数与一位数的乘法",
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "25-26 五年级上·广东潮州·期中"
+    },
+    {
+        "id":  "209",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "140 名同学出去游玩，每只船最多可以乘坐 40 名同学，需要( )只船才能坐下全部同学。",
+        "options":  [
+
+                    ],
+        "answer":  "4",
+        "solution":  "【分析】要求需要几只船才能坐下全部同学，根据题意，也就是求140里面有多少个40，根据除法的意义用除法计算，用总人数除以每只船限乘的人数，计算出需要的船的只数即可。要注意余下的人数还需要多加1只船.\n【详解】 140÷40 = 3 (只)。20(名) 3 + 1 =4 (只) 140 名同学出去游玩，每只船最多可以乘坐 40 名同学，需要4只船才能坐下全部同学。",
+        "knowledge_points":  [
+                                 "用\u0027进一法\u0027解决问题",
+                                 "三位数除以两位数的有余数除法",
+                                 "有余数除法的实际应用"
+                             ],
+        "source":  "24-25 五年级上·湖南株洲·期末"
+    },
+    {
+        "id":  "210",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "微山岛是中国北方最大的内陆岛，岛上三贤墓闻名遐迩。一艘游船从码头到微山岛每分钟行驶 150 米，速度记作( ),20分钟可行驶( )米;若游船速度不变，行驶1小时的路程是这个路程的()倍。",
+        "options":  [
+
+                    ],
+        "answer":  "150 米/分;3000;3",
+        "solution":  "【分析】根据速度 = 路程÷时间、路程 = 速度×时间解答。先求20分钟行驶的路程，再求1小时行驶的路程，最后比较倍数。\n【详解】速度记作150米/分。150×20 = 3000(米)。1小时 = 60分钟，60×150 = 9000(米)，9000÷3000 = 3。所以20分钟可行驶3000米，行驶1小时的路程是这个路程的3倍。",
+        "knowledge_points":  [
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "求一个数是另一个数的几倍",
+                                 "基础行程问题",
+                                 "时，分的认识"
+                             ],
+        "source":  "25-26 四年级上·山东济宁·期中"
+    },
+    {
+        "id":  "211",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在横线里填上合适的数.\n12 公顷 = ()平方米 80000 公顷=() 平方千米 30 公顷 500 平方米 =() 平方米",
+        "options":  [
+
+                    ],
+        "answer":  "120000;800;300500",
+        "solution":  "【分析】高级单位化低级单位乘进率;低级单位化高级单位除以进率。1 公顷 = 10000 平方米，据此将 12 公顷和 30 公顷换算成以\u0027平方米\u0027为单位的数;1 平方千米 = 100 公顷，据此将 80000 公顷换算成以\u0027平方千米\u0027为单位的数.\n【详解】 12×10000 =120000 (平方米),即12公顷 = 120000 平方米;80000÷100 = 800 (平方千米),即 80000 公顷 = 800 平方千米;30×10000 =300000 (平方米), 300000 + 500 = 300500 (平方米),即 30 公顷 500 平方米 = 300500 平方米。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26四年级上·湖北武汉·期中"
+    },
+    {
+        "id":  "212",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "洞庭湖大桥全长约5千米，小明骑自行车通过大桥需要 15 分钟，他的速度是()米/分。",
+        "options":  [
+
+                    ],
+        "answer":  "333",
+        "solution":  "【分析】首先根据1千米 = 1000 米将大桥全长约5千米换算成以米作单位，然后根据速度 = 路程/时间，即可计算出小明的速度，据此解答即可.\n【详解】因为1千米 = 1000 米，所以5千米 = 5000 米5000÷15≈333(米/分)所以他的速度是333米/分。",
+        "knowledge_points":  [
+                                 "千米和米之间的进率与换算",
+                                 "基础行程问题",
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "25-26 四年级上·湖南岳阳·期中"
+    },
+    {
+        "id":  "213",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上'>'、'<'或'='。\n\n2854000( )285万  300平方米( )3公顷\n240÷15( )480÷30  25×400( )40×250",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;\u003c;=;=",
+        "solution":  "【分析】先统一单位或算出结果，再比较大小。\n【详解】285万 = 2850000，2854000>2850000，所以2854000>285万。3公顷 = 30000平方米，所以300平方米<3公顷。240÷15 = 16，480÷30 = 16，所以240÷15=480÷30。25×400 = 10000，40×250 = 10000，所以25×400=40×250。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "除数是两位数的笔算除法",
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "大数的比较"
+                             ],
+        "source":  "25-26 四年级上·湖南岳阳·期中"
+    },
+    {
+        "id":  "214",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上'>'、'<'或'='。\n\n3654879( )3654897  269001000( )27亿\n480÷12( )480÷30  18×500( )50×180",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003c;\u003e;=",
+        "solution":  "【分析】根据整数比较大小的方法比较大小即可;整数大小的比较方法：位数不同，位数多的数就大;位数相同，左起第一位上的数大的那个数就大，如果左起第一位上的数相同，就比较左起第二位上的数，直到比较出大小为止。比较大数时，先统一单位或转换为整数，再按位数和各位数字依次比较。把一个用\u0027亿\u0027作单位的数改写成整亿数(以\u0027一\u0027为单位的数),在后面加 8 个 0，同时去掉\u0027亿\u0027字即可，再比较大小。当被除数相同时，除数越大商越小。第1个因数中间有0，乘的时候要用第2个因数去乘第1个因数的每一位，第 1 个因数十位上的 0 也要乘，乘得的积是0，要在积的十位上写 0 .\n【详解】3654879(\u003c)3654897;27亿 = 2700000000 ,269001000\u003c 2700000000,269001000(\u003c)27亿;被除数相同, 12 \u003c 30 , 480÷12 (\u003e) 480÷30 两个算式的积都是在 18×5 的积的末尾再添加两个 0.18×500 (=) 50×180。",
+        "knowledge_points":  [
+                                 "亿以上数的改写",
+                                 "三位数乘两位数，乘数末尾有 0",
+                                 "大数的比较",
+                                 "商的变化规律及应用"
+                             ],
+        "source":  "25-26 四年级上·湖南长沙·期中"
+    },
+    {
+        "id":  "215",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在算式□38÷45中，要使商是两位数,□里最小填( );要使商是一位数,□里最大填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "5;4",
+        "solution":  "【分析】三位数除以两位数，如果被除数前两位上的数小于除数，则商是一位数;如果被除数前两位上的数大于或等于除数，则商是两位数。要使□ 38÷45 的商是两位数，则被除数前两位上的数大于等于除数,□里的数要大于4，最小填5;要使□38÷45的商是一位数，则被除数前两位上的数小于除数,□里的数要小于等于 4;以此答题即可.\n【详解】根据分析可知：在算式□38÷45中，要使商是两位数,□里最小填5;要使商是一位数,□里最大填 4",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "25-26四年级上·湖南岳阳·期中"
+    },
+    {
+        "id":  "216",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个长方形果园的面积是 200 平方米，它的宽是 10 米，则长是( )米。",
+        "options":  [
+
+                    ],
+        "answer":  "20",
+        "solution":  "【分析】根据题意，明确长方形的面积 = 长×宽，已知一个长方形果园的面积是200平方米，它的宽是10米，用200除以10，就是长的长度，列式计算即可.\n【详解】根据分析可知:200÷10 = 20 (米)一个长方形果园的面积是 200 平方米，它的宽是 10 米，则长是 20 米。",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "除数是整十数的口算除法"
+                             ],
+        "source":  "25-26 四年级上·河南开封·期中"
+    },
+    {
+        "id":  "217",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一家快递公司 1 台智能机器人 15 分钟能处理特殊订单 20 个，照这样的速度,1 台智能机器人 2 小时能处理( )个特殊订单。",
+        "options":  [
+
+                    ],
+        "answer":  "160",
+        "solution":  "【分析】根据1小时 = 60 分钟,2小时 = 120 分钟。已知1台智能机器人15分钟能处理20个特殊订单，用120除以15，得出2小时里有几个15分钟，再乘20，即可得1台智能机器人2小时能处理特殊订单的个数.\n【详解】2小时 = 120 分钟 120÷15×20 = 8×20 = 160 (个) 1台智能机器人2小时能处理160个特殊订单。",
+        "knowledge_points":  [
+                                 "单归一问题",
+                                 "时，分的认识",
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "25-26 五年级上·山东济宁·期中"
+    },
+    {
+        "id":  "218",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在横线上填上'>'、'<'或'='。\n\n3亿( )3000万  594523600( )25945326001\n20000公顷( )2平方千米  180÷36( )90÷18",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;\u003c;\u003e;=",
+        "solution":  "【分析】比较大小时，先统一单位或先算出结果。\n【详解】3亿 = 30000万，所以3亿>3000万。594523600是9位数，25945326001是11位数，所以594523600<25945326001。2平方千米 = 200公顷，所以20000公顷>2平方千米。180÷36 = 5，90÷18 = 5，所以180÷36=90÷18。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "除数是两位数的笔算除法",
+                                 "公顷，平方千米的进率与换算",
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "25-26 四年级上·福建福州·期末"
+    },
+    {
+        "id":  "219",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "〇÷△ = 20……19，△最小是( )，此时〇是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "20;419",
+        "solution":  "【分析】在有余数的除法中，余数比除数小。余数是19，所以除数最小是20。再根据被除数 = 除数×商+余数求出被除数。\n【详解】19+1 = 20，20×20+19 = 400+19 = 419。所以△最小是20，此时〇是419。",
+        "knowledge_points":  [
+                                 "余数与除数的关系",
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "25-26 四年级上·广东广州·期末"
+    },
+    {
+        "id":  "220",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "(1)已知算式 23×5 = 115，那么 230×5 =();23×() = 11500(2)已知算式 504÷42 = 12，那么 504÷84= ();252÷()=12。",
+        "options":  [
+
+                    ],
+        "answer":  "1150;500;6;21",
+        "solution":  "【分析】(1)根据积的变化规律，一个因数扩大到原来的10倍，另一个因数不变，积也扩大到原来的10倍;一个因数不变，另一个因数扩大到原来的100倍，积也扩大到原来的100倍，据此填空即可。(2)被除数不变，除数乘几(0 除外),商就除以几;被除数和除数乘或除以一个相同的数(0 除外),商不变。对比算式504÷42 = 12 和算式 504÷84 可知，被除数不变,42×2 = 84，除数乘 2，商应该除以 2。对比算式 504÷42 = 12 和算式 252 ÷ ( ) = 12 可知，商不变，被除数和除数都应该乘或除以相同的数。 504÷2 = 252，即被除数除以2，那么除数也应该除以 2\n【详解】(1)因为 23×10 =230，所以 115×10 =1150，即230×5 =1150; 因为 115×100 =11500，所以 5×100 =500，即 23×500 = 11500。已知算式 23×5 =115，那么 230×5 =1150;23×500 = 11500。(2)因为 42×2 = 84，所以 12÷2 = 6，即 504÷84 = 6 。因为 504÷2 = 252，所以 42÷2 = 21，即 252÷21 = 12 。已知算式 504÷42 = 12，那么 504÷84 = 6 , 252÷21 = 12 。",
+        "knowledge_points":  [
+                                 "商的变化规律及应用",
+                                 "积的变化规律(整数乘法)",
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "25-26 四年级上·广东·期末"
+    },
+    {
+        "id":  "221",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在横线填上'>'、'<'或'='。\n\n6230000平方米( )62平方千米  360×40( )36×400\n25×40( )90000÷90  720÷90( )80÷10",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;=;=;=",
+        "solution":  "【分析】(1)根据1平方千米 = 100 公顷;1公顷 = 10000 平方米，先把单位统一，再比较即可;(2)(3)(4)根据两，三位数乘两位数和除数是整十数的计算方法，分别求出两边算式的结果，再进行比较即可.\n【详解】62平方千米 = 6200 公顷 = 62000000 平方米,6230000 平方米\u003c62000000 平方米，即 6230000 平方米\u003c62平方千米;360×40 = 14400, 36×400 = 14400 即 360×40 = 36×400;25×40 = 1000, 90000÷90 = 1000，即 25×40 = 90000÷90;720÷90 = 8, 80÷10 = 8，即 720÷90 = 80÷10 。",
+        "knowledge_points":  [
+                                 "两位数乘整十数的口算乘法",
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "被除数和除数末尾都有0的除法",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26四年级上·广东·期末"
+    },
+    {
+        "id":  "222",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "□37÷37，要使商是两位数，□里最小填( )；要使商是一位数，□里最大填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "4;3",
+        "solution":  "【分析】要使□37÷37的商是两位数，则被除数前两位上的数大于等于除数,□3大于等于37,□里的数要大于3，最小填 4。要使□37÷37的商是一位数，则被除数前两位上的数小于除数,□3 小于 37,□里的数要小于等于 3，最大填 3.\n【详解】由分析可得:□37÷37，要使商是两位数,□里最小填 4。要使商是一位数,□里最大填 3。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "25-26 四年级上·广东广州·期末"
+    },
+    {
+        "id":  "223",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号填上'>'、'<'或'='。\n\n429860( )43万  450÷50( )450÷90",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003e",
+        "solution":  "【分析】先把43万改写成普通数，再比较；算式题先算出结果。\n【详解】43万 = 430000，429860<430000，所以429860<43万。450÷50 = 9，450÷90 = 5，所以450÷50>450÷90。",
+        "knowledge_points":  [
+                                 "除数是整十数的口算除法",
+                                 "大数的比较"
+                             ],
+        "source":  "2025四年级上·广东·专题练习"
+    },
+    {
+        "id":  "224",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "珠穆朗玛峰大约高8849米，如果一万张同样的纸叠放在一起大约高1米。那么一亿张这样的纸叠放在一起( )(填\u0027有\u0027或\u0027没有\u0027)珠穆朗玛峰高。理由 。",
+        "options":  [
+
+                    ],
+        "answer":  "有;因为一亿张纸叠放的高度是 10000 米，大于珠穆朗玛峰的8849米",
+        "solution":  "【分析】一万张纸叠放高1米，一亿张里有多少个一万张，就有多少个1米，再与8849米比较。\n【详解】一亿 = 100000000，一万 = 10000，100000000÷10000 = 10000(个)，所以一亿张这样的纸叠放在一起高10000米。10000米>8849米，所以一亿张这样的纸叠放在一起有珠穆朗玛峰高。理由：因为一亿张纸叠放的高度是10000米，大于珠穆朗玛峰的8849米。",
+        "knowledge_points":  [
+                                 "被除数和除数末尾都有 0 的除法",
+                                 "一亿有多大"
+                             ],
+        "source":  "25-26四年级上·福建福州·期中"
+    },
+    {
+        "id":  "225",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "下午 3 时，钟面上时针和分针所成的较小的角是( )角，晚上 7 时，钟面上时针与分针所成的较小的角是( )度。",
+        "options":  [
+
+                    ],
+        "answer":  "直;150",
+        "solution":  "【分析】钟面分为12个大格，每个大格对应360°÷12 = 30°。下午3时，时针和分针相差3个大格；晚上7时，时针和分针相差5个大格。\n【详解】360°÷12 = 30°，30°×3 = 90°，30°×5 = 150°。所以下午3时，钟面上时针和分针所成的较小的角是直角，晚上7时，钟面上时针与分针所成的较小的角是150度。",
+        "knowledge_points":  [
+                                 "整点时间与钟面指针的位置",
+                                 "除数是两位数的笔算除法",
+                                 "角度的计算"
+                             ],
+        "source":  "25-26 四年级上·福建厦门·期中"
+    },
+    {
+        "id":  "226",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "梦梦买 7 个笔记本花了 30 元，聪聪买 14 个这样的笔记本需要( )元;林老师用 120 元可以买( )个这样的笔记本。",
+        "options":  [
+
+                    ],
+        "answer":  "60;28",
+        "solution":  "【分析】14是7的2倍，所以花费的钱也是30的2倍；120是30的4倍，所以可以买的数量也是7的4倍。\n【详解】14÷7 = 2，30×2 = 60(元)；120÷30 = 4，4×7 = 28(个)。所以聪聪买14个这样的笔记本需要60元；林老师用120元可以买28个这样的笔记本。",
+        "knowledge_points":  [
+                                 "倍的认识",
+                                 "除数是整十数的口算除法"
+                             ],
+        "source":  "25-26 四年级上·福建厦门·期中"
+    },
+    {
+        "id":  "227",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "8 平方千米=( )公顷=( )平方米 14000000 平方米 = ( )平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "800;8000000;14",
+        "solution":  "【分析】1平方千米 = 100 公顷,1公顷 = 10000 平方米，高级单位名数换算成低级单位名数乘进率，低级单位名数换算成高级单位名数除以进率.\n【详解】8平方千米 = (8×100)公顷 = 800 公顷= (800×10000) 平方米 = 8000000 平方米 14000000 平方米 = (14000000÷1000000)平方千米 = 14 平方千米",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·福建福州·期中"
+    },
+    {
+        "id":  "228",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "要将 235 千克花生油装进同样大小的油壶，每个油壶可以装油 15 千克，至少需要( )个油壶。",
+        "options":  [
+
+                    ],
+        "answer":  "16",
+        "solution":  "【分析】先根据“油的总质量÷每壶油的质量 = 壶数”列式为235÷15，再用“进一法”保留整数。\n【详解】235÷15 = 15(个)……10(千克)，15+1 = 16(个)，所以至少需要16个油壶。",
+        "knowledge_points":  [
+                                 "有余数除法的实际应用"
+                             ],
+        "source":  "25-26 五年级上·山东临沂·期中"
+    },
+    {
+        "id":  "229",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "7 公顷=( )平方米 5 平方千米=( )平方米 300 公顷=( )平方千米 45 公顷=( )平方米",
+        "options":  [
+
+                    ],
+        "answer":  "70000;5000000;3;450000",
+        "solution":  "【分析】根据1公顷 = 10000 平方米,1平方千米=1000000平方米,100公顷 = 1 平方千米，高级单位转换成低级单位乘进率，低级单位转换成高级单位除以进率，据此换算，即可解答.\n【详解】 7×10000 =70000，即7公顷 = 70000 平方米;5×1000000 = 5000000，即5平方千米 = 5000000 平方米;300÷100 =3，即 300 公顷 = 3 平方千米;45×10000 =450000，即 45 公顷 = 450000 平方米。",
+        "knowledge_points":  [
+                                 "三位数乘两位数，乘数末尾有 0",
+                                 "除数是整十数的口算除法",
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·新疆喀什·期中"
+    },
+    {
+        "id":  "230",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "10 枚一元的硬币摞在一起的高度大约是2厘米。照这样计算,100枚1元的硬币摞在一起的高度大约是( )厘米,1万枚1元的硬币摞在一起的高度大约是( )米。",
+        "options":  [
+
+                    ],
+        "answer":  "20;20",
+        "solution":  "【分析】先求100枚里有几个10枚，再求10000枚里有几个10枚，最后把厘米换算成米。\n【详解】100÷10×2 = 20(厘米)。10000÷10×2 = 2000(厘米)，2000厘米 = 20米。所以100枚1元的硬币摞在一起的高度大约是20厘米，1万枚1元的硬币摞在一起的高度大约是20米。",
+        "knowledge_points":  [
+                                 "毫米，厘米，分米，米之间的进率与换算",
+                                 "除数是整十数的口算除法"
+                             ],
+        "source":  "25-26 四年级上·湖北襄阳·期中"
+    },
+    {
+        "id":  "231",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "10 公顷=( )平方米 8 平方千米=( )公顷=( )平方米 70000 平方米=( )公顷 15000000 平方米=( )公顷=()平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "100000;800;8000000;7;1500;15",
+        "solution":  "【分析】根据1公顷 = 10000 平方米,1平方千米 = 100 公顷,1平方千米 = 1000000 平方米换算单位，高级单位换算低级单位乘进率，低级单位换算高级单位除以进率.\n【详解】 10×10000 =100000 (平方米),所以 10 公顷=100000 平方米 8×100 =800 (公顷), 800×10000 =8000000 (平方米), 所以 8 平方千米 = 800 公顷 = 8000000 平方米 70000÷10000 = 7 (公顷),所以 70000 平方米 = 7 公顷 15000000÷10000 = 1500 (公顷), 1500÷100 = 15 (平方千米),所以 15000000 平方米 = 1500 公顷 = 15 平方千米",
+        "knowledge_points":  [
+                                 "除数是整十数的口算除法",
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "24-25 四年级上·河南开封·期中"
+    },
+    {
+        "id":  "232",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "永辉超市促销，某款面条每袋价格为 16 元，现在买 10 袋送 2 袋，李阿姨带了300 元，最多可以买( )袋。",
+        "options":  [
+
+                    ],
+        "answer":  "20",
+        "solution":  "【分析】首先计算李阿姨的 300 元在不考虑赠送的情况下能买多少袋面条，每袋 16 元，通过除法计算可得最大购买袋数。即 300÷16 =18 (袋)。12(元),再根据促销活动\u0027买10袋送2袋\u0027,用18除以10求出18里有几个10就赠送几个2袋，再加上18袋得到总袋数.\n【详解】 300÷16 =18 (袋)。12(元) 18÷10 = 1 (组)。8(袋) 18 + 2 =20 (袋) 所以最多可以买20袋。",
+        "knowledge_points":  [
+                                 "有余数除法的实际应用",
+                                 "经济问题"
+                             ],
+        "source":  "25-26 五年级上·福建福州·期中"
+    },
+    {
+        "id":  "233",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "下午 3 时整，钟面上时针与分针所成的较小角是( )角。",
+        "options":  [
+
+                    ],
+        "answer":  "直",
+        "solution":  "【分析】钟面共有12个大格，每个大格对应360°÷12 = 30°。下午3时整时，时针和分针相差3个大格。\n【详解】360°÷12 = 30°，30°×3 = 90°。90°是直角，所以下午3时整，钟面上时针与分针所成的较小角是直角。",
+        "knowledge_points":  [
+                                 "整点时间与钟面指针的位置",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "24-25 四年级上·河南许昌·期中"
+    },
+    {
+        "id":  "234",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "体育商店现在搞促销活动，足球原来的单价是120元，现在买4个送1个。学校要买 20 个足球，至少得花( )元钱，每个足球便宜了()元钱。",
+        "options":  [
+
+                    ],
+        "answer":  "1920;24",
+        "solution":  "【分析】根据题意，足球买4个送1个，即花费4个足球的钱数可以得到 4 + 1 =5 (个)足球。先用20除以5，看20里有几组买4送1，总价 = 单价×数量，用120乘4，再乘组数，就是买足球至少花的钱数。计算每个足球的便宜金额：用除法计算出实际每个足球的单价，原价 120 元减去实际单价，即可得出每个足球便宜了多少钱。列式计算即可.\n【详解】根据分析可知: 4 + 1 = 5 (个) 20÷5 = 4 (组) 120×4×4 = 480×4 = 1920 (元) 1920÷20 = 96 (元) 120 - 96 = 24 (元)体育商店现在搞促销活动，足球原来的单价是120元，现在买 4 个送 1 个。学校要买 20 个足球，至少得花 1920 元钱，每个足球便宜了24元钱。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "两，三位数与一位数连续进位的乘法",
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "24-25四年级上·浙江宁波·期中"
+    },
+    {
+        "id":  "235",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "钟面上 5 时整，时针和分针组成的较小角是( )°,这是一个( )角;再过 1 小时，时针和分针组成的角是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "150;钝;180",
+        "solution":  "【分析】钟面一圈是360°，共分为12个大格，所以每个大格的角度是360°÷12 = 30°。5时整时，时针指向5，分针指向12，两者之间间隔5个大格；再过1小时是6时整。\n【详解】5×30° = 150°，150°是钝角；6×30° = 180°。所以钟面上5时整，时针和分针组成的较小角是150°，这是一个钝角；再过1小时，时针和分针组成的角是180°。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "平角，周角的认识及特征",
+                                 "整点时间与钟面指针的位置"
+                             ],
+        "source":  "25-26 四年级上·黑龙江鹤岗·期中"
+    },
+    {
+        "id":  "236",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "王明计算一道有余数的除法时，在被除数和除数的末尾同时去掉一个\u00270\u0027后，算得的商是 12，余数是 5，这道除法算式的商是( ),余数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "12;50",
+        "solution":  "【分析】被除数和除数同时乘(或除以)相同的数，商不变，余数也要同时乘(或除以)相同的数，据此即可解答.\n【详解】被除数和除数的末尾同时去掉一个\u00270\u0027后，商不变，还是 12。但是原来的余数应该是 5×10 =50 。所以，这道除法算式的商是 12，余数是 50。",
+        "knowledge_points":  [
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "24-25 四年级上·青海果洛·月考"
+    },
+    {
+        "id":  "237",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "a÷42 = 20……b(a，b均为整数)，b最大是( )，此时a是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "41;881",
+        "solution":  "【分析】余数要比除数小，所以b最大是42-1 = 41。再根据被除数 = 除数×商+余数求a。\n【详解】42-1 = 41，42×20+41 = 840+41 = 881。所以b最大是41，此时a是881。",
+        "knowledge_points":  [
+                                 "三位数除以两位数的有余数除法",
+                                 "余数与除数的关系"
+                             ],
+        "source":  "24-25 四年级上·青海果洛·月考"
+    },
+    {
+        "id":  "238",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "( )里最大能填几。\n\n72×( )<560  60×( )<410  ( )×90<400\n3( )6456740≈3亿",
+        "options":  [
+
+                    ],
+        "answer":  "7;6;4;4",
+        "solution":  "【分析】前3题用右边的数除以左边已知因数；如果有余数，商就是最大能填的数。第4题按四舍五入到亿位看千万位上的数。\n【详解】560÷72 = 7……56，所以括号里最大填7。410÷60 = 6……50，所以括号里最大填6。400÷90 = 4……40，所以括号里最大填4。3( )6456740≈3亿，说明千万位上的数小于5，所以括号里最大填4。",
+        "knowledge_points":  [
+                                 "亿以上数的近似数",
+                                 "除数是整十数的口算除法",
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "24-25 四年级上·青海果洛·月考"
+    },
+    {
+        "id":  "239",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "974÷80 的商的最高位是( )位;274÷30 的商是( )位数。",
+        "options":  [
+
+                    ],
+        "answer":  "十;一",
+        "solution":  "【分析】根据三位数除以两位数的计算，被除数前两位小于除数商是一位数，商的最高位在个位。被除数前两位大于或等于除数商是两位数，商的最高位在十位。据此填空即可.\n【详解】974÷80,974的前两位97大于80，所以商是两位数，商的最高位是十位。274÷30,274的前两位27小于30，商是一位数。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25 四年级上·青海果洛·月考"
+    },
+    {
+        "id":  "240",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "双十一网店促销活动中，一家店铺中的一条裤子售价68元，买两条送一条，妈妈有 700 元钱，准备买一批裤子，最多可以买( )条。",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】买两条送一条，相当于每买2条就多得1条。先算700元最多能买多少条需要付钱的裤子，再算能送多少条。\n【详解】700÷68 = 10(条)……20(元)，10÷2 = 5(组)，5×(2+1) = 15(条)。所以妈妈最多可以买15条裤子。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "24-25四年级上·青海果洛·月考"
+    },
+    {
+        "id":  "241",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小星上午8时出发，到家发现时针转动了90°，分针位置不变，她到家的时间是上午( )，时针与分针形成的角是( )角，再过20分钟，分针转动的角度是( )度。",
+        "options":  [
+
+                    ],
+        "answer":  "11时或11:00;锐;120",
+        "solution":  "【分析】钟面共12个大格，一圈360°，故每个大格对应角度为360°÷12 = 30°。时针转动90°，对应3个大格，即经过3小时。分针每分钟转360°÷60 = 6°。\n【详解】90°÷30° = 3，所以她到家的时间是上午11时。11时整时，时针在11，分针在12，夹角是30°，是锐角。20分钟分针转动6°×20 = 120°。所以她到家的时间是上午11时，时针与分针形成的角是锐角，再过20分钟，分针转动的角度是120度。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "整点时间与钟面指针的位置",
+                                 "时，分，秒时间的推算"
+                             ],
+        "source":  "25-26 四年级上·福建莆田·期中"
+    },
+    {
+        "id":  "242",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "两个数的商是 13，如果被除数和除数同时乘 5，那么商是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "13",
+        "solution":  "【分析】商不变的规律：被除数和除数都乘或除以相同的数(0除外)，商不变。\n【详解】根据商不变的规律可知，两个数的商是13，如果被除数和除数同时乘5，那么商还是13。",
+        "knowledge_points":  [
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "24-25 四年级上·河南新乡·期中"
+    },
+    {
+        "id":  "243",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "△÷32 = 13……□，□里最大是( )，这时△=( )。",
+        "options":  [
+
+                    ],
+        "answer":  "31;447",
+        "solution":  "【分析】在有余数的除法中，余数小于除数，且被除数 = 商×除数+余数，据此求出。\n【详解】除数是32，余数最大是32-1 = 31。13×32+31 = 416+31 = 447，所以□里最大是31，这时△=447。",
+        "knowledge_points":  [
+                                 "两位数乘两位数的不进位乘法",
+                                 "三位数除以两位数的有余数除法",
+                                 "有余数的除法验算",
+                                 "余数与除数的关系"
+                             ],
+        "source":  "24-25 四年级上·河南新乡·期中"
+    },
+    {
+        "id":  "244",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "□67÷67，商如果是一位数，□里最大应填( )；商如果是两位数，□里最小应填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "6;7",
+        "solution":  "【分析】判断三位数除以两位数商的位数，只需要比较被除数前两位与除数的大小。\n【详解】□67÷67，商如果是一位数，则□6<67，所以□里可以填1、2、3、4、5、6，最大应填6。商如果是两位数，则□6≥67，所以□里可以填7、8、9，最小应填7。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25 四年级上·河南新乡·期中"
+    },
+    {
+        "id":  "245",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "钟面上，分针从3:00 走到 3:25，转动了( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "150",
+        "solution":  "【分析】钟面一圈是360°，平均分成12个大格，每个大格对应360°÷12 = 30°。从3:00到3:25，分针从12走到5，共走过5个大格。\n【详解】360°÷12 = 30°，30°×5 = 150°，所以钟面上，分针从3:00走到3:25，转动了150°。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "除数是两位数的笔算除法",
+                                 "一般时间与钟面指针的位置"
+                             ],
+        "source":  "24-25四年级上·湖北武汉·期中"
+    },
+    {
+        "id":  "246",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小猿在计算一道三位数乘两位数的算式时，不小心把因数 12 错写成了 10，结果得到的积是 5040，正确的积是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "6048",
+        "solution":  "【分析】用错误的积除以错误的因数10，先求出另一个因数，再乘正确的因数12。\n【详解】5040÷10 = 504，504×12 = 6048，所以正确的积是6048。",
+        "knowledge_points":  [
+                                 "除数是整十数的口算除法",
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "24-25 四年级上·江西吉安·期中"
+    },
+    {
+        "id":  "247",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号填上合适的数。\n①7平方米=( )平方分米  ④4公顷=( )平方米  ⑦1周角=( )直角\n②6平方千米=( )平方米  ⑤800平方千米=( )公顷  ⑧60000平方分米=( )平方米\n③4600公顷=( )平方千米  ⑥649704000≈( )亿  ⑨90000000平方米=( )平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "700;40000;4;6000000;80000;600;46;6;90",
+        "solution":  "【分析】根据1平方千米 = 100公顷 = 1000000平方米，1公顷 = 10000平方米，1平方米 = 100平方分米进行单位换算；周角是360°，直角是90°；近似值按四舍五入法求。\n【详解】①7平方米 = 700平方分米；④4公顷 = 40000平方米；⑦1周角 = 4直角；②6平方千米 = 6000000平方米；⑤800平方千米 = 80000公顷；⑧60000平方分米 = 600平方米；③4600公顷 = 46平方千米；⑥649704000≈6亿；⑨90000000平方米 = 90平方千米。",
+        "knowledge_points":  [
+                                 "除数是整十数的口算除法",
+                                 "角的分类及换算",
+                                 "平方厘米，平方分米，平方米之间的进率与换算",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25四年级上·河南郑州·期中"
+    },
+    {
+        "id":  "248",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "括号里最大能填几。\n\n20×( )<161  ( )×65<457  80×( )<640\n25×( )<170  ( )×90<710  60×( )<525",
+        "options":  [
+
+                    ],
+        "answer":  "8;7;7;6;7;8",
+        "solution":  "【分析】用右边的数除以左边已知的因数，如果有余数，商就是最大能填的数；如果没有余数，商要减1。\n【详解】161÷20 = 8……1，所以20×8<161。457÷65 = 7……2，所以7×65<457。640÷80 = 8，所以括号里最大填7。170÷25 = 6……20，所以25×6<170。710÷90 = 7……80，所以7×90<710。525÷60 = 8……45，所以60×8<525。",
+        "knowledge_points":  [
+                                 "三位数除以两位数的有余数除法",
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "24-25 四年级上·青海海东·期中"
+    },
+    {
+        "id":  "249",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一条公路长350千米，一辆汽车平均每小时行驶70千米，这辆汽车( )小时行完全程。",
+        "options":  [
+
+                    ],
+        "answer":  "5",
+        "solution":  "【分析】路程/速度 = 时间，用公路的全长除以平均每小时行驶的千米数，就是这辆汽车几小时行完全程.\n【详解】 350÷70 = 5 所以，这辆汽车 5 小时行完全程。",
+        "knowledge_points":  [
+                                 "除数是整十数的口算除法",
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25四年级上·海南省直辖县级单位·期中"
+    },
+    {
+        "id":  "250",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "计算 98÷34 时，可以把 34 看作( )来试商，初商会偏( ),应将初商调()。",
+        "options":  [
+
+                    ],
+        "answer":  "30;大;小",
+        "solution":  "【分析】在计算98÷34时，为了方便试商，通常会把除数34看作与之接近的整十数。34接近30，所以把34看作30来试商。根据\u0027商×除数 = 被除数\u0027的关系，当除数被看小(从34看成30)时，若被除数不变，除数越小，商就会越大。因为初商偏大，所以需要将初商调小，才能得到正确的商.\n【详解】由分析可得：计算 98÷34 时，可以把 34 看作 30 来试商。因为 30 \u003c 34，除数看小了，初商会偏大，应将初商调小。计算 98÷34 时，可以把 34 看作 30 来试商，初商会偏大，应将初商调小。",
+        "knowledge_points":  [
+                                 "多位数除以两位数的试商"
+                             ],
+        "source":  "25-26五年级上·河南濮阳·期中"
+    },
+    {
+        "id":  "251",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据435÷15 = 29，直接写出下面各算式的得数。\n\n4350÷150 = ( )  870÷15 = ( )  4350÷15 = ( )  290×15 = ( )",
+        "options":  [
+
+                    ],
+        "answer":  "29;58;290;4350",
+        "solution":  "【分析】根据商不变的规律和积的变化规律直接推算即可。\n【详解】4350÷150 = 29；870÷15 = 58；4350÷15 = 290；290×15 = 4350。",
+        "knowledge_points":  [
+                                 "商的变化规律及应用",
+                                 "积的变化规律(整数乘法)",
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "24-25 四年级上·青海果洛·月考"
+    },
+    {
+        "id":  "252",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "要使□66÷62 的商是一位数,□里最大填( );要使 4□ 8 ÷ 46 的商是两位数,□里最小填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "5;6",
+        "solution":  "【分析】三位数除以两位数，要使商是一位数，那么被除数百位，十位上的数合起来要小于除数;要使商是两位数，那么被除数百位，十位上的数合起来要等于或大于除数;依此填空.\n【详解】要使□ 66÷62 的商是一位数，则□6 小于 62,□里填 1,2,3,4,5,□里最大填 5。要使 4 □ 8÷46 的商是两位数，则4□大于或等于46,□里填 6,7,8,9,□里最小填 6",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25 四年级上·青海果洛·月考"
+    },
+    {
+        "id":  "253",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "每年的10月16日是世界粮食日，一粒虽少，但积少成多。1000 粒大米约重 25 克。10000000 粒大米约重( )克，合( )千克。",
+        "options":  [
+
+                    ],
+        "answer":  "250000;250",
+        "solution":  "【分析】先求出10000000粒大米是1000粒大米的多少倍，再根据倍数关系求出重量，最后根据1千克 = 1000 克，进 行单位换算.\n【详解】 10000000÷1000 = 1000025×10000 =250000 (克) 250000 克 = 250 千克 可知 10000000 粒大米约重 250000 克，合 250 千克。",
+        "knowledge_points":  [
+                                 "除数是整十数的口算除法",
+                                 "求一个数是另一个数的几倍",
+                                 "克，千克之间的换算与比较"
+                             ],
+        "source":  "25-26四年级上·广东佛山·阶段练习"
+    },
+    {
+        "id":  "254",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个舞蹈教室的长是 46 米，宽是 11 米，大约( )个这样的舞蹈教室的面积是 1 公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "20",
+        "solution":  "【分析】先求一个舞蹈教室的面积，再把这个面积看作约500平方米进行估算，最后用1公顷的面积除以500。\n【详解】46×11 = 506(平方米)，约等于500平方米。1公顷 = 10000平方米，10000÷500 = 20(个)。所以大约20个这样的舞蹈教室的面积是1公顷。",
+        "knowledge_points":  [
+                                 "被除数和除数末尾都有0的除法",
+                                 "长方形的面积",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·福建厦门·阶段练习"
+    },
+    {
+        "id":  "255",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一辆汽车在高速公路上匀速行驶,2 小时行驶 180 千米，汽车的速度可写成()。照这样速度，行驶 360 千米需要( )小时。",
+        "options":  [
+
+                    ],
+        "answer":  "90 千米/时;4",
+        "solution":  "【分析】2小时行驶180千米，用路程/时间可以算出速度;再根据路程/速度可算出时间.\n【详解】 180÷2 = 90 (千米/时) 360÷90 = 4 (小时) 所以汽车的速度可写成90千米/时;照这样速度，行驶360千米需要4小时。",
+        "knowledge_points":  [
+                                 "除数是整十数的口算除法",
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25四年级上·福建漳州·期中"
+    },
+    {
+        "id":  "256",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "商店购进 8 箱笔记本，一共花了 960 元，每箱笔记本的价格是( )元，若每箱有 30 本，每本笔记本( )元，用到的数量关系式有( )和( )。",
+        "options":  [
+
+                    ],
+        "answer":  "120;4;总价÷箱数=每箱的价钱;每箱笔记本的价格÷每箱的本数=每本笔记本的价格",
+        "solution":  "【分析】根据单价 = 总价/数量，用8箱笔记本的价钱除以8箱，即可求出每箱笔记本的价格;再用每箱笔记本的价格除以每箱的本数，即可求出每本笔记本的价格.\n【详解】每箱笔记本的价格: 960÷8 = 120 (元)每本笔记本的价格: 120÷30 = 4 (元)所以商店购进8箱笔记本，一共花了960元，每箱笔记本的价格是120元，若每箱有30本，每本笔记本4元，用到的数量关系式有总价/箱数 = 每箱的价钱和每箱笔记本的价格/每箱的本数 = 每本笔记本的价格。",
+        "knowledge_points":  [
+                                 "三位数除以一位数，首位除不尽",
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "25-26 四年级上·山东临沂·月考"
+    },
+    {
+        "id":  "257",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个林场用喷雾器给树喷药,3台喷雾器4小时喷了300棵。照这样计算,1台喷雾器每小时可以喷( )棵。",
+        "options":  [
+
+                    ],
+        "answer":  "25",
+        "solution":  "【分析】已知3台喷雾器4小时喷300棵，那么3台喷雾器1小时喷的棵数为 300÷4 = 75 (棵)。3台喷雾器1小时喷75 棵，那么 1 台喷雾器 1 小时喷的棵数为 75÷3 = 25 (棵)\n【详解】 300÷4÷3 = 25 (棵)答:1 台喷雾器每小时可以喷 25 棵。",
+        "knowledge_points":  [
+                                 "多位数与一位数的连除",
+                                 "单归一问题"
+                             ],
+        "source":  "24-25五年级上·河南南阳·期中"
+    },
+    {
+        "id":  "258",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "战国时，魏国攻打赵国，赵国向齐国求救，齐国出兵攻打魏国，迫使魏国撤回攻打赵国的军队而使赵国得救，这就是围魏救赵的故事。假如齐魏两国相距512千米，齐军每天行军32千米，那么齐军抵达魏国需要( )天。",
+        "options":  [
+
+                    ],
+        "answer":  "16",
+        "solution":  "【分析】本题已知路程和速度，求时间。根据时间 = 路程/速度进行计算即可.\n【详解】 512÷32 = 16 (天)齐军抵达魏国需要 16 天。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26四年级上·河南周口·阶段练习"
+    },
+    {
+        "id":  "259",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在小数除法中，被除数扩大到原来的 10 倍，除数扩大到原来的( )倍，商的大小不变。",
+        "options":  [
+
+                    ],
+        "answer":  "10",
+        "solution":  "【分析】只有当被除数和除数同时扩大相同的倍数时(0除外),商不变，据此即可填空.\n【详解】被除数扩大到原来的10倍，除数扩大相同的倍数商不变，即除数扩大到原来的10倍。",
+        "knowledge_points":  [
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "24-25 五年级上·广西玉林·期中"
+    },
+    {
+        "id":  "260",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "游隼是世界上俯冲飞行速度最快的鸟类，当它俯冲追击猎物时，速度每小时可达 300 千米，它的俯冲速度可写成( )。一种超音速飞机 2 秒可飞行 700 米，它的速度可写成( )。",
+        "options":  [
+
+                    ],
+        "answer":  "300 千米/时;350 米/秒",
+        "solution":  "【分析】每小时飞行300千米，表示速度时速度单位是千米/时，在此单位前面再添上数据300即可表示它的俯冲速度。一种超音速飞机2秒可飞行700米，用700除以2可以求出超音速飞机1秒飞行350米，速度单位是米/秒，再此单位前面添上350，即表示超音速飞机的速度.\n【详解】 700÷2 = 350 (米/秒)游隼是世界上俯冲飞行速度最快的鸟类，当它俯冲追击猎物时，速度每小时可达300千米，它的俯冲速度可写成300千米/时。一种超音速飞机 2 秒可飞行 700 米，它的速度可写成 350 米/秒。",
+        "knowledge_points":  [
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25 四年级上·山东菏泽·期中"
+    },
+    {
+        "id":  "261",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "被除数是 504，除数是 42，商是( )。从 234 里连续减 26，减( )次后结果是 0。",
+        "options":  [
+
+                    ],
+        "answer":  "12;9",
+        "solution":  "【分析】(1)根据被除数/除数 = 商，由此计算;(2)用 234 除以 26，即可求出从 234 里连续减 26，减几次后结果是 0 .\n【详解】 504÷42 = 12234÷26 = 9 被除数是 504，除数是 42，商是 12。从 234 里连续减 26，减 9 次后结果是 0。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "25-26 四年级上·河南周口·月考"
+    },
+    {
+        "id":  "262",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "( )块 100 平方米的菜园占地面积是 1 公顷 ( )公顷 = 800 平方千米\n245 公顷=( )平方千米( )公顷",
+        "options":  [
+
+                    ],
+        "answer":  "100;80000;2;45",
+        "solution":  "【分析】1公顷 = 10000平方米，1平方千米 = 100公顷，据此换算。\n【详解】10000÷100 = 100，所以100块100平方米的菜园占地面积是1公顷。800平方千米 = 800×100 = 80000公顷。245公顷 = 2平方千米45公顷。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·山东菏泽·期中"
+    },
+    {
+        "id":  "263",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一块草坪的长是 20米，宽是 5 米,( )块这样的草坪面积约为 1 公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "100",
+        "solution":  "【分析】长方形的面积 = 长×宽。由题意得，一块草坪的长是20米，宽是5米，直接用乘法可以算出这块草坪的面积。求多少块这样的草坪面积约为 1 公顷,1 公顷 = 10000 平方米，直接用 10000 除以前面的得数即可解答.\n【详解】 20×5 =100 (平方米) 1 公顷 = 10000 平方米 10000÷100 = 100 (块)块草坪的长是 20 米，宽是 5 米,100 块这样的草坪面积约为 1 公顷。",
+        "knowledge_points":  [
+                                 "被除数和除数末尾都有0的除法",
+                                 "长方形的面积",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "24-25 四年级上·四川绵阳·期中"
+    },
+    {
+        "id":  "264",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在有余数的除法中，除数是 21，商是 9，被除数最大是( ),最小是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "209;190",
+        "solution":  "【分析】在有余数的除法中，余数始终比除数小。由题意得，在有余数的除法中，除数是21，商是9，那么余数最大是 20，最小是 1，然后根据被除数 = 商×除数+余数求解即可.\n【详解】由分析得，除数是 21，那么余数最大是: 21 - 1 = 20，最小是 1.9×21 + 20 = 189 + 20 = 2099×21 + 1 = 189 + 1 = 190 在有余数的除法中，除数是21，商是9，被除数最大是209，最小是 190。",
+        "knowledge_points":  [
+                                 "余数与除数的关系",
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "24-25四年级上·陕西安康·期中"
+    },
+    {
+        "id":  "265",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "已知 a×b = 400，如果 a 乘 20,b 不变，那么积是( );如果a不变,b除以10，那么积是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "8000;40",
+        "solution":  "【分析】根据积的变化规律可知，一个乘数不变，另一个乘数乘几或除以几(0除外),积也乘几或除以几(0除外)。a乘20,b不变，积应乘20。b除以10,a不变，积应除以10，据此作答.\n【详解】 400×20 =8000400÷10 = 40 因此已知 a×b = 400，如果 a 乘 20,b 不变，那么积是8000;如果a不变,b除以10，那么积是 40",
+        "knowledge_points":  [
+                                 "三位数与整十数的口算乘法",
+                                 "除数是整十数的口算除法",
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "24-25 四年级上·陕西延安·期中"
+    },
+    {
+        "id":  "266",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "小明为了知道粗细均匀的10千克铁丝有多长，于是剪下5米长的一段称重大约是 100 克，那么 10 千克铁丝的长度约是( )米。",
+        "options":  [
+
+                    ],
+        "answer":  "500",
+        "solution":  "【分析】首先利用1千克 = 1000 克，用10乘进率将千克换算为克，然后用5米长的铁丝重量100克除以长度5米计算出每米的铁丝重量，再用总重量10千克除以每米的铁丝重量即可求解铁丝长度.\n【详解】10 千克 = 10×1000 =10000 克 10000 ÷ (100÷5) = 10000÷20 = 500 (米) 即10千克铁丝的长度约是500米。",
+        "knowledge_points":  [
+                                 "除数是整十数的笔算除法",
+                                 "克，千克之间的换算与比较",
+                                 "单归一问题"
+                             ],
+        "source":  "24-25 六年级下·重庆忠县·期末"
+    },
+    {
+        "id":  "267",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "14:00 一辆公共汽车和一辆小轿车同时从相距450千米的两地相向而行，公共汽车每小时行40千米，小轿车每小时行50千米，那么当两车相距 90 千米时会是( )时。",
+        "options":  [
+
+                    ],
+        "answer":  "18或20",
+        "solution":  "【分析】两车相向而行，总速度为 40 + 50 =90 千米/时。分两种情况：相遇前相距90千米和相遇后继续行驶至相距90千米，得到两种情况下分别行驶的路程，再计算经过的时间并推算时间点.\n【详解】相遇前相距90千米：两车需共同行驶的路程为: 450 - 90 = 360 (千米) 40 + 50 =90 (千米/时) 时间: 360÷90 = 4 (小时) 此时刻为14时+4小时 = 18 时。 相遇后相距90千米：两车需共同行驶的路程为: 450 + 90 = 540 (千米) 40 + 50 =90 (千米/时) 时间: 540÷90 = 6 (小时) 此时刻为14时 + 6 小时 = 20 时。综上，两车相距90千米时为18或20时。",
+        "knowledge_points":  [
+                                 "除数是整十数的口算除法",
+                                 "基础行程问题",
+                                 "时，分，秒时间的推算"
+                             ],
+        "source":  "24-25 四年级上·浙江嘉兴·期中"
+    },
+    {
+        "id":  "268",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "8 公顷=( )平方米 15000000 平方米=( )公顷=( )平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "80000;1500;15",
+        "solution":  "【分析】单位换算，从低级单位换算成高级单位，除以单位间的进率;从高级单位换算成低级单位，乘单位间的进率。根据1公顷 = 10000 平方米,1平方千米 = 100 公顷 =1000000 平方米，进行单位换算即可.\n【详解】 8×10000 =80000，故8公顷 = 80000 平方米 15000000÷10000 = 1500，故 15000000 平方米 = 1500 公顷 1500÷100 = 15，故 1500 公顷 = 15 平方千米。8 公顷 = 80000 平方米 15000000 平方米 = 1500 公顷 = 15 平方千米。",
+        "knowledge_points":  [
+                                 "除数是整十数的口算除法",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·吉林白城·期中"
+    },
+    {
+        "id":  "269",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "根据 60÷20 = 3，很快写出下面各题的商.\n600÷20 = ( ) 120÷40 = ( ) 6000÷2000 = ( ) 3000÷1000 = ( ) 1800÷600 = ( )",
+        "options":  [
+
+                    ],
+        "answer":  "30;3;3;3;3",
+        "solution":  "【分析】这道题目主要是根据商不变的规律直接写出得数。商不变的规律：被除数和除数同时乘或除以相同的数(0除外),商不变。当除数不变时，被除数扩大或缩小几倍，商也扩大或缩小几倍。据此解答.\n【详解】 600÷20=30，被除数扩大到原来的10倍，除数不变，所以商也扩大到原来的10倍，结果为30.120÷40=3，被除数和除数同时扩大到原来的2倍，所以商不变，结果为3.6000÷2000=3，被除数和除数同时扩大到原来的100倍，所以商不变，结果为3.3000÷1000=3，被除数和除数同时扩大到原来的50倍，所以商不变，结果为3.1800÷600=3，被除数和除数同时扩大到原来的30倍，所以商不变，结果为3。所以,600÷20=30,120÷40=3,6000÷2000=3,3000÷1000=3,1800÷600=3。所以, 600 ÷ 20 = 30 , 120 ÷ 40 = 3 ,6000÷2000=3, 3000 ÷ 1000 = 3 , 1800 ÷ 600 = 3。",
+        "knowledge_points":  [
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "24-25 四年级上·陕西安康·期中"
+    },
+    {
+        "id":  "270",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "买4个篮球需要192元，买8个这样的篮球需要( )元。一辆小汽车4小时行驶了 300 千米，行驶 450 千米需要( )小时。",
+        "options":  [
+
+                    ],
+        "answer":  "384;6",
+        "solution":  "【分析】根据单价 = 总价/数量，求出篮球的单价，再根据总价 = 单价×数量，代入数据解答即可;根据路程/时间=速度，求出小汽车的速度，再根据时间 = 路程/速度，代入数据计算即可解答.\n【详解】(1) 192÷4×8 = 48×8 = 384 (元) 450÷(300÷4)= 450 ÷ 75 = 6 (小时)买 4 个篮球需要 192 元，买 8 个这样的篮球需要(384)元。一辆小汽车 4 小时行驶了 300 千米，行驶 450 千米需要(6)小时。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "除数是两位数的笔算除法",
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25四年级上·广东东莞·期中"
+    },
+    {
+        "id":  "271",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "聪聪在计算时，把被除数 414 误抄成 444，得到的商比正确的商多了 2，余数不变，正确的商是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "27",
+        "solution":  "【分析】把被除数414误写成444后，商多了2，余数不变，所以两个被除数的差30正好是2个除数，先求除数，再求正确的商。\n【详解】(444-414)÷2 = 30÷2 = 15，所以除数是15。414÷15 = 27……9，所以正确的商是27。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "25-26四年级上·全国·单元测试"
+    },
+    {
+        "id":  "272",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "□÷15=7。□,余数最大是( ),此时被除数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "14;119",
+        "solution":  "【分析】余数小于除数，余数小于15，最大是14，余数等于14时，再根据\u0027被除数 = 除数×商+余数\u0027求出被除数.\n【详解】□÷15=7。□,余数小于除数，余数小于 15，最大是 14;15×7 + 14 = 105 + 14 = 119 所以□÷15=7。□,余数最大是 14，此时被除数是 119",
+        "knowledge_points":  [
+                                 "余数与除数的关系",
+                                 "有余数的除法验算",
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "24-25 四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "273",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明每分钟走60米，他 15 分钟走( )米。声音每秒传播340米，传播1700米要用( )秒。",
+        "options":  [
+
+                    ],
+        "answer":  "900;5",
+        "solution":  "【分析】(1)路程 = 速度×时间，据此列式计算即可;(2)根据时间 = 路程/速度列式计算即可.\n【详解】 60×15 =900 (米) 1700÷340 = 5 (秒)小明每分钟走60米，他15分钟走900米。声音每秒传播340米，传播1700米要用5秒。",
+        "knowledge_points":  [
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25六年级下·广西河池·期末"
+    },
+    {
+        "id":  "274",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "下面括号里最大能填整数几。\n50×( ) \u003c 300350 \u003e ( )×3590×( ) \u003c 710",
+        "options":  [
+
+                    ],
+        "answer":  "5;9;7",
+        "solution":  "【分析】用要比较的数除以算式中已知的数，如果有余数，所求的商就是要填的最大的数，如果没有余数，最大的数就是所求的商减去1，据此即可解答.\n【详解】因为 300÷50 =6，没有余数, 6 - 1 =5，括号里最大能填5，所以 50×5 \u003c 300;因为 350÷35 =10，没有余数, 10 - 1 = 9，括号里最大能填9，所以 350 \u003e 9×35;因为 710÷90=7。80，有余数，括号里最大能填 7，所以90×7 { \u003c } 710 。",
+        "knowledge_points":  [
+                                 "三位数除以两位数的有余数除法",
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "24-25三年级下·山东济宁·期末"
+    },
+    {
+        "id":  "275",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一本《成语故事》的价钱是 19 元,180 元大约可以买( )本《成语故事》。",
+        "options":  [
+
+                    ],
+        "answer":  "9",
+        "solution":  "【分析】把19元近似看作20元，再用180除以20进行估算。\n【详解】180÷19≈180÷20 = 9(本)，所以180元大约可以买9本《成语故事》。",
+        "knowledge_points":  [
+                                 "除数是两位数的估算"
+                             ],
+        "source":  "24-25三年级下·安徽宣城·期末"
+    },
+    {
+        "id":  "276",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "△÷○=48。12，除数最小是( ),这时被除数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "13;636",
+        "solution":  "【分析】根据余数的性质：余数小于除数，所以除数最小是比余数大1的数。再根据被除数 = 商×除数+余数，代入数据计算即可求出被除数.\n【详解】除数最小是: 12 + 1 = 1348×13 + 12 = 624 + 12 = 636 △÷○=48。12，除数最小是13，这时被除数是636。",
+        "knowledge_points":  [
+                                 "余数与除数的关系",
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "24-25 三年级下·河南开封·期末"
+    },
+    {
+        "id":  "277",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "已知A÷B = 8……3，当A = 51时，B = ( )；当B = 8时，A = ( )；当A与B同时扩大到原来的10倍后，商是( )，余数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "6;67;8;30",
+        "solution":  "【分析】根据除数 = (被除数-余数)÷商、被除数 = 商×除数+余数，以及商不变规律解答。\n【详解】(51-3)÷8 = 48÷8 = 6，所以当A = 51时，B = 6。8×8+3 = 64+3 = 67，所以当B = 8时，A = 67。A与B同时扩大到原来的10倍后，商还是8，余数变成30。",
+        "knowledge_points":  [
+                                 "余数与除数的关系",
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "24-25 六年级下·河南南阳·期末"
+    },
+    {
+        "id":  "278",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "经测量 10 枚 1 元硬币摞在一起的高度大约是2厘米,10000枚摞在一起的高度大约是( )厘米,100000000 枚摞在一起的高度大约是()米。",
+        "options":  [
+
+                    ],
+        "answer":  "2000;200000",
+        "solution":  "【分析】分别用除法求出10000枚以及100000000枚里面有几个10枚，那么这些硬币摞起来的高度是几个2厘米。厘米和米之间的进率是100，据此分别将求出的两个高度换算成米.\n【详解】 10000÷10 = 1000 (个) 1000×2 =2000 (厘米) 100000000÷10=10000000(个)10000000×2=20000000(厘米)20000000 厘米 = 200000 米经测量10枚1元硬币摞在一起的高度大约是2厘米,10000枚摞在一起的高度大约是 2000 厘米,100000000 枚摞在一起的高度大约是200000米。",
+        "knowledge_points":  [
+                                 "一亿有多大",
+                                 "除数是整十数的口算除法",
+                                 "厘米和米之间的进率与换算"
+                             ],
+        "source":  "24-25 四年级下·河南洛阳·期末"
+    },
+    {
+        "id":  "279",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "春晚跳秧歌转手帕的机器人，手腕每分钟转 3000 转,2 分 14 秒能转( )转。",
+        "options":  [
+
+                    ],
+        "answer":  "6700",
+        "solution":  "【分析】由题意得，春晚跳秧歌转手帕的机器人，手腕每分钟转3000转。1分钟 = 60 秒，直接用3000除以60算出手腕每秒转多少转。然后根据1分 = 60 秒将2分14秒转化为多少秒，最后再将得数乘上手腕每秒转的转数即可算出 2分 14 秒能转多少转.\n【详解】1 分 = 60 秒, 3000÷60 = 50 (转) 1 分 = 60 秒, 2×60 =120，所以 2 分 14 秒 = 120 秒+14 秒 = 134 秒。50×134=6700(转)春晚跳秧歌转手帕的机器人，手腕每分钟转3000转,2分14 秒能转 6700 转.\n50×134 =6700 (转),即手腕 2 分 14 秒能转 6700 转。",
+        "knowledge_points":  [
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "除数是整十数的口算除法",
+                                 "时，分，秒之间的换算与比较"
+                             ],
+        "source":  "24-25四年级下·江西宜春·期末"
+    },
+    {
+        "id":  "280",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "端午临中夏，时清日复长。端午节包粽子，妈妈用2千克糯米包了20个粽子，照这样计算，包 30 个粽子需要( )千克糯米。",
+        "options":  [
+
+                    ],
+        "answer":  "3",
+        "solution":  "【分析】本题可先求出包1个粽子需要的糯米量，再计算包30个粽子所需的糯米量。已知2千克糯米包了20个粽子，那么包1个粽子需要的糯米重量为所用糯米重量除以粽子个数，再乘 30 即可求出 30 个粽子需要的糯米重量。1千克 = 1000 克，先将单位换算成克再计算，最后再将单位换算成千克即可.\n【详解】2 千克 = 2000 克2000÷20 = 100 (克) 100×30 =3000 (克) 3000 克 = 3 千克 所以包 30 个粽子需要 3 千克糯米。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "24-25三年级下·江西宜春·期末"
+    },
+    {
+        "id":  "281",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "3□57880000≈35 亿，那么□里应该填( );要使□ 25×24 的积是四位数,□里最大填( );要使□ 82÷49 的商是两位数,□里最小填()。",
+        "options":  [
+
+                    ],
+        "answer":  "4;3;5",
+        "solution":  "【分析】四舍五入到亿位，需要看千万位，当千万位比5大或者等于5，向前进1。千万位比5小，需要舍去。再在这个数的后面加上亿字。依次将1,2,3。代入算式□ 25×24 中求出积，再结合积的位数进行解答。三位数除以两位数，当被除数前两位大于或者等于除数，商是两位数。当被除数的前两位比除数小，商是一位数。据此解答.\n【详解】3□57880000 的千万位上是 5，向前进1,3□57880000≈35 亿，那么□里应该填 4.125×24 =3000，积是四位数。225×24 =5400，积是四位数。325×24 =7800，积是四位数。425×24 =10200，积是五位数。所以，要使□ 25×24 的积是四位数,□里最大填3。□82÷49 的商是两位数，则□8 大于或等于 49，那么□可以填 5,6,7,8,9,□里最小填 5。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法",
+                                 "判断商是几位数(除数是两位数)",
+                                 "亿以上数的近似数"
+                             ],
+        "source":  "24-25 四年级下·贵州铜仁·期末"
+    },
+    {
+        "id":  "282",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "汽车 6 分钟行驶了 90 千米。题目中知道了汽车行驶的( )和( ),可求出汽车行驶的( ),结果是( )千米/分。",
+        "options":  [
+
+                    ],
+        "answer":  "时间;路程;速度;15",
+        "solution":  "【分析】路程 = 速度×时间,6 分钟是时间,90 千米是路程，用90除以6可以计算出汽车行驶的速度.\n【详解】汽车6分钟行驶了90千米。题目中知道了汽车行驶的时间和路程;可求出汽车行驶的速度;90÷6 =15(千米/分),所以结果是 15 千米/分。",
+        "knowledge_points":  [
+                                 "整十，整百，整千数除以一位数的除法",
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26 四年级上·海南海口·单元测试"
+    },
+    {
+        "id":  "283",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "( )个 20 是60;270 是 30 的( )倍。",
+        "options":  [
+
+                    ],
+        "answer":  "3;9",
+        "solution":  "【分析】第一个问题求多少个20是60，即求60里面有几个20，用除法计算;第二个问题求270是30的几倍，即求270里面有几个30，同样用除法计算.\n【详解】 60÷20 = 3 因此,3 个 20 是 60;270÷30 = 9 因此,270是30的9倍。",
+        "knowledge_points":  [
+                                 "除数是整十数的口算除法",
+                                 "求一个数是另一个数的几倍"
+                             ],
+        "source":  "25-26 四年级上·海南海口·单元测试"
+    },
+    {
+        "id":  "284",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "购买16块乒乓球拍共 400 元，可以求出乒乓球拍的( ),是( )元。",
+        "options":  [
+
+                    ],
+        "answer":  "单价;25",
+        "solution":  "【分析】根据总价 = 单价×数量,16块是数量,400元是总价，可以求出单价，用 400÷16 即可.\n【详解】购买 16 块乒乓球拍共 400 元，可以求出乒乓球拍的单价;400÷16 = 25 (元)所以乒乓球的单价是25元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "25-26四年级上·海南海口·单元测试"
+    },
+    {
+        "id":  "285",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在□ ÷ 27 =15。△中，余数△最大可取( ),这时被除数□是()。",
+        "options":  [
+
+                    ],
+        "answer":  "26;431",
+        "solution":  "【分析】余数要比除数小，余数最大时，比除数小1，据此,27减1即可求出余数最大时是26，根据被除数 = 除数×商+余数，先计算27乘15的积，再加26，即可求出此时的被除数是多少.\n【详解】 27 - 1 =2627×15 + 26 = 405 + 26 = 431 在□÷27=15。△中，余数△最大可取26，这时被除数□是 431。",
+        "knowledge_points":  [
+                                 "余数与除数的关系",
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "2025四年级上·海南海口·专题练习"
+    },
+    {
+        "id":  "286",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小思在计算两位数乘两位数时，把一个因数十位上的 4 错写成 6，结果得793，实际结果应为 533，这两个因数分别是( ),( )。",
+        "options":  [
+
+                    ],
+        "answer":  "41;13",
+        "solution":  "【分析】由题意得，小思在计算两位数乘两位数时，把一个因数十位上的4错写成6。十位上的4表示4个十，十位上的6表示6个十，那么小思计算的结果就会比正确的结果多20个另一个因数。小思计算的结果是793，实际结果应为533，可以先用减法算出它们的差值，接着用差值除以20算出一个因数。最后再用533除以一个因数即可算出另一个因数。 60 - 40 = 20260÷20 = 13533÷13 = 41 小思在计算两位数乘两位数时，把一个因数十位上的4错写成6，结果得793，实际结果应为533，这两个因数分别是41,13。",
+        "knowledge_points":  [
+                                 "两，三位数的不退位减法",
+                                 "除数是两位数的笔算除法",
+                                 "被除数和除数末尾都有0的除法"
+                             ],
+        "source":  "24-25 四年级下·四川德阳·期末"
+    },
+    {
+        "id":  "287",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算 278÷60 时，想: 60×( ) 接近 278，且小于 278，所以商( ),商写在( )位上。",
+        "options":  [
+
+                    ],
+        "answer":  "4;4;个",
+        "solution":  "【分析】计算278÷60时，利用试商法，寻找最大的整数使得 60 乘该数接近且不超过 278.60×4 = 240，接近 278 且小于 278，因此商为 4，写在个位上.\n【详解】 60×4 = 240 ,240 接近 278 且小于 278, 60×5 = 300，超过278，因此商是4，由于被除数前两位27小于60，商写在个位上。所以计算 278÷60 时，想: 60×4 接近278，且小于278，所以商是4，商写在个位上。",
+        "knowledge_points":  [
+                                 "多位数除以两位数的试商"
+                             ],
+        "source":  "25-26 四年级上·海南海口·单元测试"
+    },
+    {
+        "id":  "288",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算 93÷30 时，想:93 里面有( )个 30，所以商( ),商写在( )位上。",
+        "options":  [
+
+                    ],
+        "answer":  "3;3;个",
+        "solution":  "【分析】计算 93÷30 时，先确定93里面有几个30.30×3 = 90，最接近93且不超过93，因此93里面有3个30，商为3;由于除数是两位数，比较被除数的前两位 93 和除数30,93 大于 30，所以商应写在个位上.\n【详解】计算 93÷30 时，想:93里面有3个30，所以商3，商写在个位上。",
+        "knowledge_points":  [
+                                 "除数是整十数的口算除法",
+                                 "多位数除以两位数的试商"
+                             ],
+        "source":  "25-26 四年级上·海南海口·单元测试"
+    },
+    {
+        "id":  "289",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "将 1800 本图书平均分给6个年级，每个年级有4个班。算式:1800÷(6×4)=75 是先算( ),再算( )。",
+        "options":  [
+
+                    ],
+        "answer":  "一共有多少个班级;每个班分到多少本书",
+        "solution":  "【分析】每个班分到图书的本数 = 这批图书的总本数/(分的年级个数×平均每个年级的班级个数)。那么先算分的年级个数×平均每个年级班级的个数，算出一共有多少个班级，再算总的书除以总的班级数，即可得到每个班分到多少本书.\n【详解】由分析可知：将 1800 本图书平均分给 6 个年级，每个年级有 4 个班。算式: 1800 ÷ (6×4) = 75 是先算一共有多少个班级，再算每个班分到多少本书。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25 三年级下·浙江温州·期末"
+    },
+    {
+        "id":  "290",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "○÷☆= 18 dots 15,☆最小是( ),此时被除数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "16;303",
+        "solution":  "【分析】在有余数的除法中，余数始终比除数小。在算式○÷☆=18。15 中，余数是 15，所以除数最小是 16。然后根据被除数 = 商×除数+余数求解即可.\n【详解】 15 + 1 = 1618×16 + 15 = 288 + 15 = 303 ○÷☆=18。15,☆最小是 16，此时被除数是 303",
+        "knowledge_points":  [
+                                 "余数与除数的关系",
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "24-25 四年级下·贵州黔南·期末"
+    },
+    {
+        "id":  "291",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "16÷5 =3 ldots 1，如果被除数和除数同时扩大到原来的 100 倍，那么商是( ),余数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "3;100",
+        "solution":  "【分析】被除数和除数，同时乘或除以相同的数(0除外),商不变，余数跟着乘或除以相同的数，据此填空.\n【详解】 1×100 =10016÷5 =3 ldots 1，如果被除数和除数同时扩大到原来的100倍，那么商是3，余数是 100",
+        "knowledge_points":  [
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "24-25 六年级下·湖北黄冈·期末"
+    },
+    {
+        "id":  "292",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "用一根长 180 厘米的铁丝来围平行四边形。如果平行四边形的周长是 32 厘米，那么最多能围成( )个这样独立的平行四边形。(接头处忽略不计)",
+        "options":  [
+
+                    ],
+        "answer":  "5",
+        "solution":  "【分析】根据题意可知：用铁丝总长/每个平行四边形的周长，就可以求出围成平行四边形的个数.\n【详解】 180÷32 = 5 。 20 (cm) 则最多能围成5个这样独立的平行四边形。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "25-26 四年级上·全国·单元测试"
+    },
+    {
+        "id":  "293",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "543÷35 的商的最高位是( )位;268÷67 的商是( )位数。",
+        "options":  [
+
+                    ],
+        "answer":  "十;一",
+        "solution":  "【分析】先看被除数的前两位，如果被除数的前两位不够除，再看被除数的前三位。除到被除数的哪一位，就把商写在那一位的上面。计算结果如果有余数，余数一定要比除数小.\n【详解】543÷35，被除数的前两位够除 54 { \u003e } 35，所以商的最高位是十位;268÷67，被除数的前两位不够除 26 { \u003c } 67，再看被除数的前三位，所以商的最高位是个位，所以商是一位数。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "25-26 四年级上·全国·单元测试"
+    },
+    {
+        "id":  "294",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "190 里最多有( )个 20;360 是 18 的( )倍。",
+        "options":  [
+
+                    ],
+        "answer":  "9;20",
+        "solution":  "【分析】根据除法的意义直接计算即可。\n【详解】190÷20 = 9……10，所以190里最多有9个20；360÷18 = 20，所以360是18的20倍。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "求一个数是另一个数的几倍"
+                             ],
+        "source":  "25-26 四年级上·全国·单元测试"
+    },
+    {
+        "id":  "295",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "按要求填上合适的数字。商是两位数。(1) □38÷35 ,□里最小填( )。 (2) □04÷64 ,□里可以填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "4;7、8、9",
+        "solution":  "【分析】判断商是几位数，只需比较被除数前两位与除数的大小。\n【详解】(1)□38÷35，商是两位数，则□3≥35，所以□里最小填4。(2)□04÷64，商是两位数，则□0≥64，所以□里可以填7、8、9。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "2025 四年级上·全国·专题练习"
+    },
+    {
+        "id":  "296",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据商不变的规律在括号里填上合适的数。\n\n48÷16 = ( )÷2  200÷25 = 800÷( )  180÷30 = 18÷( )\n( )÷12 = 16÷4  ( )÷3 = 360÷15  1200÷150 = ( )÷15",
+        "options":  [
+
+                    ],
+        "answer":  "6;100;3;48;72;120",
+        "solution":  "【分析】根据商不变的规律，被除数和除数同时乘或除以相同的数，商不变。\n【详解】48÷16 = 6÷2；200÷25 = 800÷100；180÷30 = 18÷3；48÷12 = 16÷4；72÷3 = 360÷15；1200÷150 = 120÷15。",
+        "knowledge_points":  [
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "25-26四年级上·全国·课后作业"
+    },
+    {
+        "id":  "297",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "从 544 里连续减去( )个 16，得数是 0。",
+        "options":  [
+
+                    ],
+        "answer":  "34",
+        "solution":  "【分析】从544里连续减去多少个16，得数是0，即求544里面有多少个 16，用除法计算即可.\n【详解】 544÷16 = 34，从 544 里连续减去 34 个 16，得数是0。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "25-26 四年级上·全国·课后作业"
+    },
+    {
+        "id":  "298",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "我国的高铁站现在实行\u0027刷脸\u0027识别通过闸机。过闸机只需将身份证放入检票口，抬头看小屏幕。从上午08:23到上午08:48某个闸机一共通过了 375 人，平均每分钟通过( )人。",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】先求出上午08:23到上午08:48一共经过了多少分钟，再用人数除以时间。\n【详解】8:48-8:23 = 25(分钟)，375÷25 = 15(人)。因此平均每分钟通过了15人。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "时，分，秒有关的计算"
+                             ],
+        "source":  "24-25三年级下·山东济宁·期末"
+    },
+    {
+        "id":  "299",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "昆明滇池游船限载40 人，三年级 3 个班共 118 人，至少需要( )条船，最后一条船坐( )人。",
+        "options":  [
+
+                    ],
+        "answer":  "3;38",
+        "solution":  "【分析】用118除以40，如果有余数，用商加1则是至少需要的船的数量，余数即是最后一条船坐的人.\n【详解】 118÷40 = 2 (条)。38(人) 2 + 1 =3 (条) 所以至少需要3条船，最后一条船坐38人。",
+        "knowledge_points":  [
+                                 "有余数除法的实际应用",
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "24-25 三年级下·云南昭通·期末"
+    },
+    {
+        "id":  "300",
+        "unit":  "第六单元除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "从253里面连续减去 18，最多能减( )次。",
+        "options":  [
+
+                    ],
+        "answer":  "14",
+        "solution":  "【分析】从253里面连续减去18，最多能减几次，就是求253里有几个18。\n【详解】253÷18 = 14……1，所以最多能减14次。",
+        "knowledge_points":  [
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "24-25四年级下·云南昭通·期末"
+    },
+    {
+        "id":  "301",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "学校一共收到捐赠图书 350 册，把这些图书平均分给 14 个班。每个班可以分到()册。",
+        "options":  [
+
+                    ],
+        "answer":  "25",
+        "solution":  "【分析】用总图书数量除以班级数，即可求出每个班分到的册数.\n【详解】 350÷14 = 25 (册)学校一共收到捐赠图书350册，把这些图书平均分给14个班。每个班可以分到25册。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "24-25四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "302",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上\u0027\u003e\u0027 \u0027\u003c\u0027或\u0027=\u0027。\n\n200公顷( )20平方千米 520-13( )800÷16 123560000( )98070000",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003e;\u003e",
+        "solution":  "【分析】1平方千米=100公顷，据此将200公顷换算成平方千米后再比较;分别计算出520-13和800÷16的结果后再比较;大数比较时，先比较位数，位数多的数大，位数相同再从高位起依次比较。\n【详解】200公顷=2平方千米，2\u003c20，所以200公顷\u003c20平方千米;520-13=507，800÷16=50，507\u003e50，所以520-13\u003e800÷16;123560000\u003e98070000。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "除数是两位数的笔算除法",
+                                 "大数的比较"
+                             ],
+        "source":  "24-25四年级上·湖南株洲·期末"
+    },
+    {
+        "id":  "303",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一本书有 128 页，妙妙每天看 30 页，需要( )天才能看完。",
+        "options":  [
+
+                    ],
+        "answer":  "5",
+        "solution":  "【分析】求需要几天才能看完，即求128里面有几个30。根据除法计算，若有剩余页数，则需再增加1天。\n【详解】128÷30=4(天)……8(页)，4+1=5(天)。因此，需要5天才能看完。",
+        "knowledge_points":  [
+                                 "三位数除以两位数的有余数除法",
+                                 "有余数除法的实际应用"
+                             ],
+        "source":  "24-25 三年级下·湖北武汉·期末"
+    },
+    {
+        "id":  "304",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "按规律填一填。\n\n8547×13=111111，8547×26=222222，8547×39=333333，8547×( )=444444，8547×65=( )。",
+        "options":  [
+
+                    ],
+        "answer":  "52;555555",
+        "solution":  "【分析】根据积的变化规律，一个因数不变，另一个因数扩大到原来的几倍，积也扩大到原来的几倍。观察已知算式，另一个因数依次为13，26，39，即13的1倍、2倍、3倍，积依次为111111，222222，333333。\n【详解】444444=111111×4，所以另一个因数是13×4=52。65÷13=5，所以8547×65=111111×5=555555。填52，555555。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "算式的规律(整数)",
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "24-25 四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "305",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "括号里最大能填几。\n\n30×( )\u003c206  ( )×5\u003c27  251÷5\u003e( )  139÷7\u003e( )",
+        "options":  [
+
+                    ],
+        "answer":  "6;5;50;19",
+        "solution":  "【分析】要求括号里最大能填几，可以用右边的数除以左边已知的数;有余数时，商就是最大能填的数。\n【详解】206÷30=6……26，所以30×6\u003c206;27÷5=5……2，所以( )里最大填5;251÷5=50……1，所以251÷5\u003e50;139÷7=19……6，所以139÷7\u003e19。",
+        "knowledge_points":  [
+                                 "两位数除以一位数的有余数除法",
+                                 "三位数除以一位数的有余数除法",
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "24-25 三年级下·湖北黄石·期末"
+    },
+    {
+        "id":  "306",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "两个数相除，商是 23，如果在被除数和除数的末尾都添上两个 0，商( );如果被除数扩大到原来的 10 倍，除数不变，商( )。",
+        "options":  [
+
+                    ],
+        "answer":  "23;230",
+        "solution":  "【分析】在被除数和除数的末尾都添上两个0，即被除数和除数同时乘100。根据商不变的规律，被除数和除数同时乘或除以相同的数(0 除外),商不变。因此，商仍然是 23。被除数扩大到原来的10倍(即被除数乘10),除数不变。根据商的变化规律，被除数乘10，除数不变，商也乘 10 因此，商变为 23×10 = 230 .\n【详解】根据分析可知：两个数相除，商是 23，如果在被除数和除数的末尾都添上两个 0，商 23;如果被除数扩大到原来的 10 倍，除数不变，商 230 。",
+        "knowledge_points":  [
+                                 "商的变化规律及应用",
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "23-24 四年级上·湖南张家界·期末"
+    },
+    {
+        "id":  "307",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在横线上填上\u0027\u003e\u0027\u0027\u003c\u0027或\u0027=\u0027。\n\n702÷27( )702÷15  150703800( )45070380",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003e",
+        "solution":  "【分析】比较两个除法算式时，被除数相同，除数越大，商越小;比较两个整数时，先看位数，位数多的数更大，位数相同则从高位开始逐位比较。\n【详解】被除数相同，27大于15，所以702÷27\u003c702÷15;150703800是九位数，45070380是八位数，所以150703800\u003e45070380。",
+        "knowledge_points":  [
+                                 "商的变化规律及应用",
+                                 "大数的比较",
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "23-24四年级上·湖南张家界·期末"
+    },
+    {
+        "id":  "308",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "四(1)班有学生 42人，其中男生 27 人，女生有( )人;学校准备了210本练习本发给四(1)班学生，平均每人可以发( )本。",
+        "options":  [
+
+                    ],
+        "answer":  "15;5",
+        "solution":  "【分析】由题意得，四(1)班有学生42人，其中男生27人，直接用42减去27即可算出女生的人数;学校准备了210本练习本发给四(1)班学生，直接用210除以总人数即可算出平均每人可以发多少本.\n【详解】 42 - 27 = 15 (人) 210÷42 = 5 (本) 四(1)班有学生 42 人，其中男生 27 人，女生有 15 人;学校准备了 210 本练习本发给四(1)班学生，平均每人可以发 5 本。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "24-25 四年级下·湖南长沙·期末"
+    },
+    {
+        "id":  "309",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "\"塘栖枇杷节\"期间，张大伯家一上午采摘了120千克枇杷。如果每盒装6千克，每4盒装1个快递箱，需要准备几个快递箱。张大伯的儿子是这样算的：120÷6÷4=5(个)，他先算的是( )。张大伯的女儿是这样算的：120÷(6×4)=5(个)，她先算的是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "可以装的盒数;1个快递箱装的质量",
+        "solution":  "【分析】先用120÷6求出可以装多少盒;也可以先用6×4求出1个快递箱能装多少千克，再用120去除。\n【详解】张大伯的儿子先算的是可以装的盒数;张大伯的女儿先算的是1个快递箱装的质量。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "除数是两位数的笔算除法",
+                                 "多位数与一位数的连除"
+                             ],
+        "source":  "24-25三年级下·浙江·期末"
+    },
+    {
+        "id":  "310",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "回想起六年的小学生活，同学们都很感恩自己的父母和老师，想为他们选一份礼物，某品牌的旅游鞋搞促销活动，按\"每满100元减40元\"的方式销售，买一双标价260元的这种品牌旅游鞋，应付( )元钱。",
+        "options":  [
+
+                    ],
+        "answer":  "180",
+        "solution":  "【分析】已知标价为260元，每满100元减40元，计算260÷100=2。60，其中商2表示260元里有2个完整的100元，余数60表示剩下不足100元的部分;因为每满100元减 40 元，那么 2 个 100 元可减免 2 个 40 元;最后用标价减去减免金额就是实际应付金额.\n【详解】 260÷100 = 2 (个)。60(元)260-2×40=260-80= 180 (元)所以应付 180 元钱。",
+        "knowledge_points":  [
+                                 "被除数和除数末尾都有 0 的除法",
+                                 "100 以内有余数的除法计算",
+                                 "经济问题"
+                             ],
+        "source":  "24-25六年级下·内蒙古呼伦贝尔·期末"
+    },
+    {
+        "id":  "311",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "水果店王叔叔准备将398个苹果分装进纸箱，每个纸箱最多能装15个。至少需要( )个这样的纸箱才能装完所有苹果;当他装满10箱时，还剩( )个苹果没装箱。",
+        "options":  [
+
+                    ],
+        "answer":  "27;248",
+        "solution":  "【分析】第一个问题需要计算398个苹果分装进每个最多装15个的纸箱，至少需要的纸箱数。根据除法计算，若有剩余苹果，则需多1个纸箱。第二个问题计算装满10箱后剩余的苹果数，直接用总数减去已装的苹果数即可.\n【详解】 398÷15 = 26 (个)。8(个)需要箱子数: 26 + 1 = 27 (个) 15×10 = 150 (个)剩余苹果数: 398 - 150 = 248 (个)所以，水果店王叔叔准备将398个苹果分装进纸箱，每个纸箱最多能装15个。至少需要27个这样的纸箱才能装完所有苹果，当他装满10箱时，还剩248个苹果没装箱。",
+        "knowledge_points":  [
+                                 "三位数除以两位数的有余数除法",
+                                 "有余数除法的实际应用"
+                             ],
+        "source":  "24-25四年级下·重庆九龙坡·期末"
+    },
+    {
+        "id":  "312",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "括号里最大能填几。\n\n( )×62\u003c327  7×( )\u003c85",
+        "options":  [
+
+                    ],
+        "answer":  "5;12",
+        "solution":  "【分析】要求最大填几，可以用右边的数除以左边已知的数;有余数时，商就是最大能填的数。\n【详解】327÷62=5……17，所以( )里最大填5;85÷7=12……1，所以( )里最大填12。",
+        "knowledge_points":  [
+                                 "两位数除以一位数的口算",
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "24-25 三年级下·吉林松原·期中"
+    },
+    {
+        "id":  "313",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一堆货物有 4200 吨，每辆货车限装 70 吨，需要( )辆货车可以装完。",
+        "options":  [
+
+                    ],
+        "answer":  "60",
+        "solution":  "【分析】根据题意，用货物的总重量除以每个货车限装的重量，就是需要几辆货车可以装完.\n【详解】 4200÷70 = 60 (辆)所以，需要 60 辆货车可以装完。",
+        "knowledge_points":  [
+                                 "除数是整十数的笔算除法"
+                             ],
+        "source":  "24-25 四年级上·安徽淮南·期末"
+    },
+    {
+        "id":  "314",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "900÷15，商的末尾有( )个 0。",
+        "options":  [
+
+                    ],
+        "answer":  "1",
+        "solution":  "【分析】多位数除法法则：从被除数的高位除起，除数有几位，就看被除数的前几位，如果不够除，就多看一位，除到被除数的哪一位，就把商写在哪一位的上面，如果不够除，就在这一位上商0，每次除得的余数必须比除数小，并在余数右边一位落下被除数在这一位上的数，再继续除即可。据此计算出结果，即可解答.\n【详解】 900÷15 = 60 ,900÷15商的末尾有 1 个 0。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "24-25 四年级上·安徽淮南·期末"
+    },
+    {
+        "id":  "315",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在□÷25=8……△中，当余数最大时，被除数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "224",
+        "solution":  "【分析】在有余数的除法中，余数一定小于除数。这里已知除数是25，要求被除数最大，就需要让余数取最大值，再根据被除数=除数×商+余数计算。\n【详解】余数最大是25-1=24。25×8+24=200+24=224，所以当余数最大时，被除数是224。",
+        "knowledge_points":  [
+                                 "乘，除法的意义和各部分间的关系",
+                                 "三位数除以两位数的有余数除法",
+                                 "余数与除数的关系"
+                             ],
+        "source":  "24-25四年级下·海南省直辖县级单位·期中"
+    },
+    {
+        "id":  "316",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在( )里填上\u0027\u003e\u0027\u0027\u003c\u0027或\u0027=\u0027。\n\n2980000000( )3亿 4600000( )46万 180×12( )360×6 380÷45( )380÷35",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;\u003e;=;\u003c",
+        "solution":  "【分析】(1)(2)先把3亿、46万改写成整数再比较;(3)一个因数扩大到原来的几倍，另一个因数缩小到原来的几分之一，积不变;(4)被除数相同，除数越大，商越小。\n【详解】3亿=300000000，2980000000\u003e300000000，所以2980000000\u003e3亿;46万=460000，4600000\u003e460000，所以4600000\u003e46万;180×12=360×6;45\u003e35，所以380÷45\u003c380÷35。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "除数是两位数的笔算除法",
+                                 "积的变化规律(整数乘法)",
+                                 "亿以上数的改写"
+                             ],
+        "source":  "24-25 四年级上·安徽淮南·期末"
+    },
+    {
+        "id":  "317",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "两个数相除的商是 35，若除数不变，被除数除以 5，则商是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "7",
+        "solution":  "【分析】商的变化规律：除数不变，被除数乘几或者除以几(0 除外),商也乘或者除以相同的数。据此解答.\n【详解】两个数相除的商是 35，若除数不变，被除数除以5，那么商也除以 5.35÷5 = 7，则商是 7。",
+        "knowledge_points":  [
+                                 "商的变化规律及应用"
+                             ],
+        "source":  "24-25 四年级上·安徽淮南·期末"
+    },
+    {
+        "id":  "318",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在△÷□=18……24中，□最小是( )，这时△是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "25;474",
+        "solution":  "【分析】余数要比除数小，所以□最小应该是比24大1的数。再根据被除数=商×除数+余数计算。\n【详解】余数是24，除数最小是25。18×25+24=450+24=474，所以这时△是474。",
+        "knowledge_points":  [
+                                 "余数与除数的关系",
+                                 "有余数的除法验算",
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "24-25 四年级上·新疆乌鲁木齐·期末"
+    },
+    {
+        "id":  "319",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在( )里填上\u0027\u003e\u0027\u0027\u003c\u0027或\u0027=\u0027。\n\n31×20( )600 260×7( )26×70 75÷3( )66÷3\n42×53( )43×52 0÷128( )543×0 360÷2÷6( )360÷(2×6)",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;=;\u003e;\u003c;=;=",
+        "solution":  "【分析】分别计算或根据运算规律比较大小即可。\n【详解】31×20=620，所以31×20\u003e600;260×7=1820，26×70=1820，所以260×7=26×70;75÷3=25，66÷3=22，所以75÷3\u003e66÷3;42×53=2226，43×52=2236，所以42×53\u003c43×52;0÷128=0，543×0=0，所以0÷128=543×0;360÷2÷6=180÷6=30，360÷(2×6)=360÷12=30，所以360÷2÷6=360÷(2×6)。",
+        "knowledge_points":  [
+                                 "有关0的乘法",
+                                 "被除数是0的除法运算",
+                                 "两位数乘两位数的进位乘法",
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "24-25 三年级下·四川广元·期中"
+    },
+    {
+        "id":  "320",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "225÷15 = ( ) ",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】除数是两位数的除法，第一次商后所得是余数一定要和被除数下一位落下来的数，组成一个新的数，继续除下去，直到除完所有数位上的数为止，据此解答.\n【详解】15\n\n​ 15)225\n 15\n ---\n 75\n 75\n ---\n 0",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "23-24 六年级下·北京东城·期末"
+    },
+    {
+        "id":  "321",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "骑车人与行人同一条街同方向前进行，行人在骑自行车人前面 450 米处。行人每分钟走 60 米，两人同时出发，三分钟后，骑自行车人追上行人，骑自行车人每分钟行( )米。",
+        "options":  [
+
+                    ],
+        "answer":  "210",
+        "solution":  "【分析】根据速度×时间 = 路程，用 60×3，求出行人走的路程，再加上450米，求出骑自行车人行驶的路程，再根据速度 = 路程/时间，用骑自行车人行驶的路程/3，即可解答。 (60×3+450) ÷ 3= (180 + 450) ÷ 3 = 630÷3 = 210 (米) 因此，骑自行车人每分钟行210米。",
+        "knowledge_points":  [
+                                 "追及问题"
+                             ],
+        "source":  "2024·四川乐山·小升初真题"
+    },
+    {
+        "id":  "322",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一辆汽车 20 分钟行驶 14000 米，照这样的速度，汽车从 A 城行驶到 B 城一共花了 1 小时 20 分钟。A 城到 B 城全程长( )千米。",
+        "options":  [
+
+                    ],
+        "answer":  "56",
+        "solution":  "【分析】已知一辆汽车20分钟行驶14000米，根据路程=时间×速度，得到速度 = 路程/时间，代入数据计算得出这辆汽车的速度是 14000÷20 = 700 (米/分钟);又已知汽车从A城行驶到B城一共花了1小时20分钟，根据1小时=60分钟，得到花费时间为80分钟，进而根据路程 = 时间×速度，代入数据计算即可得到A城到B城全程的长度(1000 米 = 1 千米) .\n【详解】 14000÷20 = 700 (米/分钟) 1小时20分钟 = 80 分钟 80×700 = 56000 (米) 56000 米 = 56 千米辆汽车20分钟行驶14000米，照这样的速度，汽车从A城行驶到B城一共花了1小时20分钟。A城到B城全程长(56)千米。",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "单归一问题"
+                             ],
+        "source":  "24-25 四年级下·浙江绍兴·期中"
+    },
+    {
+        "id":  "323",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "630 里面有()个 90。从 540 中连续减去 5，减( )次得 0。",
+        "options":  [
+
+                    ],
+        "answer":  "7;108",
+        "solution":  "【分析】根据题意，求一个数里面有多少个这个数，用除法，用 630÷90 。一个数连续减去多少次的这个数，结果是0，也用除法，用540÷5.\n【详解】 630 ÷ 90 = 7540 ÷ 5 = 108 所以,630 里面有 7 个 90。从 540 中连续减去 5，减 108 次得 0 。",
+        "knowledge_points":  [
+                                 "几百几十或几千几百除以一位数的除法",
+                                 "除数是整十数的口算除法"
+                             ],
+        "source":  "24-25 三年级下·山东菏泽·阶段练习"
+    },
+    {
+        "id":  "324",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.4,
+        "question":  "某数除以 11 余 8，除以 17 余 12，除以 13 余 10，那么这个数最小可能是()。",
+        "options":  [
+
+                    ],
+        "answer":  "998",
+        "solution":  "【分析】因为11-8=3，13-10=3，所以这个数加上3后同时是11和13的倍数。先找11和13的公倍数，再结合除以17余12的条件逐个检验。\n【详解】11×13=143，所以这个数可以是143的倍数减3。143-3=140，140÷17=8……4，不符合;140+143=283，283÷17=16……11，不符合;283+143=426，426÷17=25……1，不符合;426+143=569，569÷17=33……8，不符合;569+143=712，712÷17=41……15，不符合;712+143=855，855÷17=50……5，不符合;855+143=998，998÷17=58……12，符合条件。所以这个数最小可能是998。",
+        "knowledge_points":  [
+                                 "三位数除以两位数的有余数除法",
+                                 "有余数除法的实际应用"
+                             ],
+        "source":  "2024·重庆九龙坡·小升初真题"
+    },
+    {
+        "id":  "325",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\u0027\u003e\u0027\u0027\u003c\u0027或\u0027=\u0027。\n\n360÷61( )61  424÷4( )105  0×325( )0÷325\n260×7( )26×70  63×21( )36×21  715÷8( )698÷8",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003e;=;=;\u003e;\u003e",
+        "solution":  "【分析】分别计算或根据运算规律比较大小即可。\n【详解】360÷61=5……55，所以360÷61\u003c61;424÷4=106，所以424÷4\u003e105;0×325=0，0÷325=0，所以0×325=0÷325;260×7=1820，26×70=1820，所以260×7=26×70;63×21=1323，36×21=756，所以63×21\u003e36×21;715÷8=89……3，698÷8=87……2，所以715÷8\u003e698÷8。",
+        "knowledge_points":  [
+                                 "有关0的乘法",
+                                 "被除数是0的除法运算",
+                                 "两位数乘两位数的进位乘法",
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "24-25 三年级下·湖南岳阳·期中"
+    },
+    {
+        "id":  "326",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "果园里有 480 棵桃树，每行种 8 棵，可以种( )行;若每行多种2棵，能减少( )行。",
+        "options":  [
+
+                    ],
+        "answer":  "60;12",
+        "solution":  "【分析】已知果园里480棵桃树，每行种8棵，根据\"行数= 总棵数/每行棵数\"，可求出原来可以种的行数。若每行多种2棵，则每行种的棵数变为 8 + 2 = 10 棵。同样根据\"行数 = 总棵数/每行棵数\"，求出现在可以种的行数;用原来种的行数减去现在种的行数，可得减少的行数.\n【详解】 480÷8 = 60 (行)所以可以种60行。 480 ÷ (8 + 2) = 480÷10 = 48 (行) 60 - 48 = 12 (行)所以能减少12行。",
+        "knowledge_points":  [
+                                 "被除数和除数末尾都有0的除法",
+                                 "几百几十或几千几百除以一位数的除法"
+                             ],
+        "source":  "24-25 三年级下·湖北孝感·期中"
+    },
+    {
+        "id":  "327",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "( )里最大能填几。\n\n40×( )\u003c362  60×( )\u003c240  30×( )\u003c620  90×( )\u003c300",
+        "options":  [
+
+                    ],
+        "answer":  "9;3;20;3",
+        "solution":  "【分析】用右边的数除以左边已知的数，如果没有余数且题目要求小于，就用商减1;有余数时，商就是最大能填的数。\n【详解】362÷40=9……2，所以最大填9;240÷60=4，所以最大填3;620÷30=20……20，所以最大填20;300÷90=3……30，所以最大填3。",
+        "knowledge_points":  [
+                                 "整十，整百，整千数与一位数的乘法",
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "24-25 三年级下·河南信阳·期中"
+    },
+    {
+        "id":  "328",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在有余数的除法里，除数是 38，商是 24，被除数最大是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "949",
+        "solution":  "【分析】在有余数的除法中，被除数=除数×商+余数，且余数要小于除数。余数最大时，被除数最大。\n【详解】余数最大是38-1=37，38×24+37=912+37=949，所以被除数最大是949。",
+        "knowledge_points":  [
+                                 "余数与除数的关系",
+                                 "有余数的除法验算",
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "24-25四年级下·河南漯河·期中"
+    },
+    {
+        "id":  "329",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\u0027\u003e\u0027\u0027\u003c\u0027或\u0027=\u0027.\n\n0÷48 ( ) 0 + 48236÷3 ( ) 236÷4\n\n40×40 ( ) 38×4236×18×0 ( ) 36",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003e;\u003e;\u003c",
+        "solution":  "【分析】 ①0除以任何不为0的数都得0,0加任何数都得任何数;整数的大小比较：位数不相同时，位数多的数大;位数相同时，从最高位看起，相同数位上的数大的数大;②被除数相同，除数越小，商越大;③分别计算出算式的积，再比较大小;两位数乘两位数的竖式计算方法：先用下面乘数个位的数去乘上面乘数，得数的末位和下面乘数的个位对齐，再用下面乘数十位上的数去乘上面的乘数，得数的末位和下面乘数的十位对齐，然后把两次乘得的数加起来;④0乘任何数都得0.\n【详解】①0÷48=0,0+48=48,0\u003c48,0÷48\u003c0+48;②3\u003c4,236÷3\u003e236÷4;③40×40=1600,38×42=1596,40×40\u003e38×42;④36×18×0=0,0\u003c36,36×18×0\u003c36。",
+        "knowledge_points":  [
+                                 "两位数乘两位数的进位乘法",
+                                 "被除数是0的除法运算",
+                                 "三位数除以一位数，首位不够除",
+                                 "有关 0 的乘法"
+                             ],
+        "source":  "24-25 三年级下·河南三门峡·期中"
+    },
+    {
+        "id":  "330",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "( )里最大能填多少。\n\n55×( )\u003c550  6×( )\u003c300  560÷( )\u003e68",
+        "options":  [
+
+                    ],
+        "answer":  "9;49;8",
+        "solution":  "【分析】前两题用右边的数除以左边已知的数，没有余数且要求小于时，最大填商减1。第三题根据560÷( )\u003e68来求。\n【详解】550÷55=10，所以最大填9;300÷6=50，所以最大填49;因为560÷8=70，560÷9=62……2，所以560÷( )\u003e68时最大填8。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "商末尾有 0 的除法",
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "24-25 三年级下·山东济宁·期中"
+    },
+    {
+        "id":  "331",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小马虎在计算一道乘法算式时，把其中一个因数 6 错写成了 60，得到的积是720，正确的积应该是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "72",
+        "solution":  "【分析】由题意得，小马虎在计算一道乘法算式时，把其中一个因数 6 错写成了 60，得到的积是 720，可以先用 720除以60算出另一个因数。然后再用这个因数乘上6即可算出正确的积.\n【详解】 720÷60×6 = 12×6 = 72 故正确的积应该是 72。",
+        "knowledge_points":  [
+                                 "两，三位数与一位数的一次进位乘法",
+                                 "被除数和除数末尾都有0的除法"
+                             ],
+        "source":  "24-25 四年级下·河北邯郸·期中"
+    },
+    {
+        "id":  "332",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "A 除以 B 是 3，余数是 1，若 A,B 两数同时乘 10，则商是( ),余数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "3;10",
+        "solution":  "【分析】被除数和除数都乘或除以一个相同的数(0 除外),商不变，余数也同时乘或除以一个相同的数(0除外)。据此解答.\n【详解】A除以B是3，余数是1，若A,B两数同时乘10，则商不变，还是 3;余数是: 1×10 = 10 所以 A 除以 B 是 3，余数是 1，若 A,B 两数同时乘 10，则 商是 3，余数是 10 。",
+        "knowledge_points":  [
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "24-25 六年级下·江苏宿迁·期中"
+    },
+    {
+        "id":  "333",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\u0027\u003e\u0027\u0027\u003c\u0027或\u0027=\u0027.\n\n56÷6 ( ) 56÷345×24 ( ) 24×44400÷2 ( ) 800÷4又280÷7 ( ) 140÷20×26 ( ) 0÷26648÷3 ( ) 108×2",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003e;=;\u003c;=;=",
+        "solution":  "【分析】在除法算式中，被除数不变，除数越大商越小;据此解答(1)小题;当两个乘法算式中，如果一个因数相同，则另一因数大的算式的结果就大;据此解答(2)小题;商的不变规律：在除法算式中，被除数和除数同时乘或除以一个数(0 除外),商不变;据此解答(3)小题;0乘任何数都等于0;0除以不为0的数都得0;据此解答(5)小题;除数是一位数的除法，从被除数的最高位除起，先试除被除数的前一位，前一位不够除就试除被除数的前两位，除到哪一位就把商写在哪一位，哪一位上不够商1就用0占位，每次除后余下的数要比除数小;三位数乘一位数，相同数位对齐，从个位起，用一位数依次乘三位数的每一位数，乘到哪一位，就把积写在哪一位上。哪一位上乘得的积满几十就向前一位进几。当乘数末尾有零时，先算零前面的数，再在积的末尾添加对应个数的零。据此解答(4),(6)小题.\n【详解】 6 \u003e 3 , 56÷6 \u003c 56÷3; 45 \u003e 44, 45×24 \u003e 24×44;400÷2 = (400×2) ÷ (2×2) = 800÷4;\n 280÷7 = 40, 140÷2 = 70, 40 \u003c 70，即 280÷7 \u003c 140÷2;0×26 = 0, 0÷26 = 0, 0×26 = 0÷26;\n 648÷3 = 216, 108×2 = 216, 648÷3 = 108×2 。 ",
+        "knowledge_points":  [
+                                 "商的变化规律及应用",
+                                 "被除数是0的除法运算",
+                                 "积的变化规律(整数乘法)",
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "24-25三年级下·山东临沂·期中"
+    },
+    {
+        "id":  "334",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "演讲比赛中的数学演讲比赛中的数学问题:①典典手写了一篇750个字的演讲稿②演讲时间为3分钟③典典平均每分钟打50个字④典典演讲平均语速为每分钟210个字。典典打完这篇演讲稿需要多长时间。解决这个问题需要用到的信息是( )和( )。(填序号)",
+        "options":  [
+
+                    ],
+        "answer":  "①;③",
+        "solution":  "【分析】由题意得，要解决典典打完这篇演讲稿需要多长时间，需要知道这篇演讲稿一共有多少字和典典每分钟打字多少个，直接用演讲稿一共的字数除以典典每分钟打字的个数即可，即需要用到的信息是 ①和③.\n【详解】由分析得，求典典打完这篇演讲稿需要多长时间，需要用到的信息是 ①和③。",
+        "knowledge_points":  [
+                                 "被除数和除数末尾都有 0 的除法",
+                                 "\u0027提问题\u0027\u0027填条件\u0027问题"
+                             ],
+        "source":  "24-25四年级上·河北保定·期末"
+    },
+    {
+        "id":  "335",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "根据82÷5=16……2，可知(82×100)÷(5×100)=( )……( )。",
+        "options":  [
+
+                    ],
+        "answer":  "16;200",
+        "solution":  "【分析】根据商不变的规律，被除数和除数同时乘同一个数，商不变，余数也乘相同的数。\n【详解】82÷5=16……2，所以(82×100)÷(5×100)=16……200。",
+        "knowledge_points":  [
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "24-25 四年级上·浙江绍兴·期末"
+    },
+    {
+        "id":  "336",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "已知算式□76÷57的商是一位数，□08÷35的商是两位数，那么□=( )。",
+        "options":  [
+
+                    ],
+        "answer":  "4",
+        "solution":  "【分析】□76÷57的商是一位数，说明前两位数□7要小于57，所以□最大填4。□08÷35的商是两位数，说明前两位数□0要大于或等于35，所以□最小填4。因此□只能填4。\n【详解】□=4。",
+        "knowledge_points":  [
+                                 "三位数除以两位数的有余数除法",
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25 四年级上·河北张家口·期末"
+    },
+    {
+        "id":  "337",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "三年级的学生排队时，每列 16 人，排了 15 列，一共有( )人。如果每列 20人，能排( )列。",
+        "options":  [
+
+                    ],
+        "answer":  "240;12",
+        "solution":  "【分析】根据题意，用每列的人数16人乘列数15列，即得到一共的人数;再用一共的人数除以每列的人数20人，即得到能排的列数。据此解答.\n【详解】 16×15 = 240 (人) 240÷20 = 12 (列) 所以，三年级的学生排队时，每列 16 人，排了 15 列，一共有 240 人，如果每列 20 人，能排 12 列。",
+        "knowledge_points":  [
+                                 "除数是整十数的口算除法",
+                                 "两位数乘两位数的进位乘法"
+                             ],
+        "source":  "24-25三年级下·广东东莞·期中"
+    },
+    {
+        "id":  "338",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小西在计算有余数的除法时，把被除数 108 错写成了 208，结果商增加了 5，而余数不变。原来算式的除数是( ),余数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "20;8",
+        "solution":  "【分析】根据被除数 = 除数×商+余数，把被除数108错写成了208，结果商增加了5，而余数不变。说明被除数增加了 208 - 108 = 100，说明是增加了 5 个除数(余数不变),用 100÷5 = 20 即为除数，再用正确的被除数108除以除数算出余数.\n【详解】 (208-108)÷5=100÷5= 20108÷20=5。8即小西在计算有余数的除法时，把被除数 108 错写成了208，结果商增加了5，而余数不变。原来算式的除数是20，余数是 8。",
+        "knowledge_points":  [
+                                 "余数与除数的关系",
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "24-25 四年级下·广东汕头·期中"
+    },
+    {
+        "id":  "339",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "王伯伯开车从玉田到石家庄，全程共 420 千米，王伯伯开车的平均速度是 84千米/时，按这样的速度计算，他路上共用了( )小时。",
+        "options":  [
+
+                    ],
+        "answer":  "5",
+        "solution":  "【分析】已知全程共420千米，即路程为420千米，王伯伯开车的平均速度是 84 千米 / 时，根据公式\"时间 = 路程/速度\"来计算王伯伯路上所用的时间.\n【详解】 420÷84 = 5 (小时)所以，他路上共用了 5 小时。",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "23-24 四年级下·河北唐山·期末"
+    },
+    {
+        "id":  "340",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "40 公顷=( )平方米 6 平方千米=( )公顷 13 亿平方米=( )平方千米 192时=( )日",
+        "options":  [
+
+                    ],
+        "answer":  "400000;600;1300;8",
+        "solution":  "【分析】高级单位化低级单位，乘单位之间的进率;低级单位化高级单位，除以单位之间的进率。1公顷 = 10000 平方米,1平方千米 = 100 公顷,1平方千米 = 1000000 平方米,1 日 = 24 小时，根据进率转换单位即可.\n【详解】(1)公顷转换成平方米，乘进率 10000, 40×10000 = 400000，所以 40 公顷 = 400000 平方米;(2)平方千米转换成公顷，乘进率 100, 6×100 = 600，所以 6平方千米 = 600 公顷;(3) 13 亿: = 1300000000，平方米转换成平方千米，除以进率 1000000, 1300000000÷1000000 = 1300，所以13亿平方米 = 1300 平方千米;(4)时转换成日，除以进率 24, 192÷24 = 8，所以 192 时 = 8 日。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "除数是两位数的笔算除法",
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "年，月，日的认识及换算"
+                             ],
+        "source":  "24-25 四年级上·河北保定·期末"
+    },
+    {
+        "id":  "341",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上 \u0027\u003e\u0027 \u0027\u003c\u0027 或 \u0027=\u0027 .\n\n1083670000( )10 亿 240×70( )700×2480 公顷( )5 平方千米 960÷18( )960÷12",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;=;\u003c;\u003c",
+        "solution":  "【分析】(1)10亿改写时，直接在后面加上8个0即可，整数比较大小时，要看他们的数位。数位多的那个数就大;如果数位相同，就看最高位，最高位上的数大，那个数就大;最高位上的数相同，就看下一位，哪一位上的数大，那个数就大。(2)根据因数末尾有0的乘法计算方法，分别求出 240×70 和 700×24 的积，再比较积的大小。 (3)1 平方千米 = 100 公顷，先统一单位，再比较大小。 (4)根据除数是两位数的除法，分别求出各个算式的商，再比较大小.\n【详解】10 亿=1000000000,1083670000\u003e1000000000,1083670000\u003e10 亿。240×70 = 16800, 700×24 = 16800, 240×70 = 700×24.5 平方千米 = 500 公顷,80 公顷\u003c500 公顷,80 公顷 \u003c 5 平方千米。960÷18 = 53 dots 6, 960÷12 = 80, 960÷18 \u003c 960÷12 。",
+        "knowledge_points":  [
+                                 "亿以上数的改写",
+                                 "公顷，平方千米的进率与换算",
+                                 "除数是两位数的笔算除法",
+                                 "三位数乘两位数，乘数末尾有 0"
+                             ],
+        "source":  "24-25 四年级上·浙江嘉兴·期末"
+    },
+    {
+        "id":  "342",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个除法算式中，商是 9，余数是 16，除数最小是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "17",
+        "solution":  "【分析】在有余数的除法中，余数始终比除数小。据此解答.\n【详解】由题意得，一个除法算式中，余数是16，那么除数应该比余数大，除数最小是: 16 + 1 = 17 .\n\n故除数最小是 17。",
+        "knowledge_points":  [
+                                 "余数与除数的关系",
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "24-25 四年级上·河北保定·期末"
+    },
+    {
+        "id":  "343",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "40 公顷=( )平方米 6 平方千米=( )公顷 13 亿平方米=( )平方千米 192 时=( )日",
+        "options":  [
+
+                    ],
+        "answer":  "400000;600;1300;8",
+        "solution":  "【分析】高级单位化低级单位，乘单位之间的进率;低级单位化高级单位，除以单位之间的进率。1公顷=10000平方米，1平方千米=100公顷，1平方千米=1000000平方米，1日=24小时。\n【详解】40×10000=400000，所以40公顷=400000平方米;6×100=600，所以6平方千米=600公顷;13亿=1300000000，1300000000÷1000000=1300，所以13亿平方米=1300平方千米;192÷24=8，所以192时=8日。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "公顷，平方千米的进率与换算",
+                                 "除数是两位数的笔算除法",
+                                 "年，月，日的认识及换算"
+                             ],
+        "source":  "24-25 四年级上·河北保定·期末"
+    },
+    {
+        "id":  "344",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "汐汐在计算 □ 70÷56 时，要使商是两位数，那么□中最小可以填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "5",
+        "solution":  "【分析】三位数除以两位数，比较被除数前两位的数字和除数的大小，如果被除数前两位的数字大于等于除数，则商是两位数，如果小于除数商是一位数;据此作答.\n【详解】汐汐在计算 □ 70÷56 时，要使商是两位数,□7\u003e56，那么□中最小可以填 5。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25 四年级上·河北保定·期末"
+    },
+    {
+        "id":  "345",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "( )里最大能填几。\n45( )9950000≈46 亿 50×( )\u003c362",
+        "options":  [
+
+                    ],
+        "answer":  "9;7",
+        "solution":  "【分析】利用\"四舍五入\"法省略亿位后面的尾数求近似数的方法：根据千万位上数字的大小来确定用\"四舍\"法，还是用\"五入\"法，如果对应的那一位上的数大于或等于5，则用五入法，小于5，则用四舍法，再在数的后面写上\"亿\"字。要求最大能填几，用所比较的数除以已知的因数，如果有余数，所得的商就是要填的最大的数;没有余数，所得的商减去1，就是要填的最大的数;然后再进一步解答.\n【详解】45( )9950000≈46亿，显然是用的是\"五入\"法求得的近似数，所以括号里的数可以填 5,6,7,8,9，最大能填 9，即 4599950000≈46 亿;因为 362÷50=7。12，有余数，括号里最大能填 7，所以50×7 { \u003c } 362。",
+        "knowledge_points":  [
+                                 "亿以上数的近似数",
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "24-25 四年级上·河北保定·期末"
+    },
+    {
+        "id":  "346",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\u0027\u003e\u0027\u0027\u003c\u0027或\u0027=\u0027。\n\n15亿( )240000000 99公顷( )10平方千米 95÷5( )9500÷500 450×30( )45×200",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;\u003c;=;\u003e",
+        "solution":  "【分析】(1)把240000000改写成用“亿”作单位的数再比较;(2)统一单位后比较;(3)根据商不变规律比较;(4)分别计算结果再比较。\n【详解】240000000=2.4亿，所以15亿\u003e240000000;10平方千米=1000公顷，所以99公顷\u003c10平方千米;95÷5=19，9500÷500=19，所以95÷5=9500÷500;450×30=13500，45×200=9000，所以450×30\u003e45×200。",
+        "knowledge_points":  [
+                                 "商不变的规律及应用",
+                                 "亿以上数的改写",
+                                 "公顷，平方千米的进率与换算",
+                                 "三位数乘两位数，乘数末尾有0"
+                             ],
+        "source":  "24-25 四年级上·贵州安顺·期末"
+    },
+    {
+        "id":  "347",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "□26÷42，如果商是一位数，□里最大可以填( );如果商是两位数，□里最小可以填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "3;4",
+        "solution":  "【分析】三位数除以两位数，如果被除数前两位小于除数，商是一位数;如果前两位大于或等于除数，商是两位数。\n【详解】□26÷42中，如果商是一位数，则前两位□2\u003c42，所以□里最大填3;如果商是两位数，则前两位□2≥42，所以□里最小填4。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25四年级上·贵州遵义·期末"
+    },
+    {
+        "id":  "348",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "322÷42 的商是()位数，试商时将 42 看作( )来试商，初商偏( ),要调()。",
+        "options":  [
+
+                    ],
+        "answer":  "一;40;大;1",
+        "solution":  "【分析】三位数除以两位数，先比较被除数前两位和除数的大小判断商是几位数;试商时把除数看作接近的整十数，初商偏大要调小。\n【详解】322÷42中，32\u003c42，所以商是一位数;试商时把42看作40，初商偏大，要调小1。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)",
+                                 "多位数除以两位数的试商"
+                             ],
+        "source":  "24-25 四年级上·贵州黔西·期末"
+    },
+    {
+        "id":  "349",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填最大的数。\n\n50×( )\u003c404  ( )×32\u003c215  ( )×56\u003c240",
+        "options":  [
+
+                    ],
+        "answer":  "8;6;4",
+        "solution":  "【分析】把右边的数除以左边已知的数，有余数时，商就是最大能填的数。\n【详解】404÷50=8……4，所以最大填8;215÷32=6……23，所以最大填6;240÷56=4……16，所以最大填4。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "24-25 四年级上·贵州黔东南·期末"
+    },
+    {
+        "id":  "350",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算 357÷52 时，把 52 看成( )来试商，商是( ),商偏( )(填\"大\"或\"小\"),调商( ),余数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "50;7;大;6;45",
+        "solution":  "【分析】除数是两位数的除法，一般按照\"四舍五入\"法，把除数看作和它接近的整十数来试商。根据被除数/除数 = 商，被除数不变，除数变小，则商会偏大，试算后把商调小.\n【详解】357÷52=6。45 计算 357÷52 时，把 52 看成(50)来试商，商是(7),商偏(大)(填\"大\"或\"小\"),调商(6),余数是(45)。",
+        "knowledge_points":  [
+                                 "三位数除以两位数的有余数除法",
+                                 "多位数除以两位数的试商"
+                             ],
+        "source":  "24-25 四年级上·贵州黔东南·期末"
+    },
+    {
+        "id":  "351",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "不计算我也能知道268÷28 的商是( )位数;若将 28 估成 30 来试商，商会偏()(填\"大\"或\"小\")。",
+        "options":  [
+
+                    ],
+        "answer":  "一;小",
+        "solution":  "【分析】三位数除以两位数，如果被除数前两位上的数小于除数，则商是一位数;如果被除数前两位上的数大于或等于除数，则商是两位数。 268÷28，被除数前两位上的数小于除数，商是一位数。若将28估成30来试商，将除数看大了，初商可能偏小。以此答题即可.\n【详解】根据分析可知：不计算我也能知道 268÷28 的商是一位数;若将28估成30来试商，商会偏小。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)",
+                                 "多位数除以两位数的试商"
+                             ],
+        "source":  "24-25四年级上·河北保定·期末"
+    },
+    {
+        "id":  "352",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "王老师花 720 元买了12个足球，足球的单价是( )元，他一共要买15个足球，还需要再支付( )元。",
+        "options":  [
+
+                    ],
+        "answer":  "60;180",
+        "solution":  "【分析】总价/数量 = 单价，用720除以12可以计算出足球的单价;先用15减去12计算出还需要付钱的足球个数，单价×数量 = 总价，再乘足球的单价，计算出还需要支付的钱数;据此解答.\n【详解】根据分析: 720÷12 = 60 (元 / 个) (15 - 12)×60 = 3×60 = 180 (元) 所以足球的单价是60元，他一共要买15个足球，还需要再支付 180 元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "除数是两位数的笔算除法",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25 四年级上·河北保定·期末"
+    },
+    {
+        "id":  "353",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "148÷37 的商是()位数;26×123 的积是( )位数。",
+        "options":  [
+
+                    ],
+        "answer":  "一;四",
+        "solution":  "【分析】判断三位数除以两位数的商是两位数还是一位数，可以用被除数的前两位和除数比较。当被除数的前两位比除数大，或者和除数相等时，说明够除，商是两位数，商的首位写在十位上。当被除数的前两位比除数小，说明不够除，商是一位数，商的首位写在个位上。计算出 26×123 的积，再判断结果是几位数，有几个数字组成就是几位数.\n【详解】 14 \u003c 37，商是一位数 26×123 = 3198 即 148÷37 的商是一位数;26×123 的积是四位数。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法",
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25 四年级上·河北保定·期末"
+    },
+    {
+        "id":  "354",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上 \u0027\u003e\u0027\u0027\u003c\u0027或\u0027=\u0027.\n\n7243467 ( ) 7243476650×50 ( ) 65×500\n\n490×70 ( ) 3500550÷60 ( ) 540÷60",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;=;\u003e;\u003e",
+        "solution":  "【分析】(1)大数的大小比较，数位数，位数不同的两个数，位数多的那个数大。位数相同的两个数，从最高位比起，最高位上的数大的那个数就大;如果最高位上的数相同，就比较下一个数位上的数，直到比较出大小为止。(2)笔算三位数乘两位数：相同数位对齐，一般多位数写上，两位数写下，从个位乘起，依次用下面的乘数每位上的数去乘上面的乘数，乘到哪一位，得数的末尾和下面乘数的那一位对齐，与哪一位乘的积满几十，就要向前一位进几。除数是两位数的笔算除法：从被除数的高位数起，先看被除数的前两位，如果前两位比除数小，就要看前三位;除到被除数的哪一位，商就写在那一位的上面，每求出一位商，余下的数必须比除数小。分别计算出结果后再比较大小.\n【详解】7243467是七位数,7243476也是七位数，且它们左起往右数的前5个数都一样，第六个数，也就是十位上的数,7243467 是 6,7243476 是 7, 6 \u003c 7，所以 7243467\u003c7243476.650×50 = 32500 , 65×500 = 32500，所以 650×50 = 65×500490×70=34300, 34300 \u003e 3500，所以 490×70 \u003e= 3500550÷60=9。10, 540 ÷ 60 = 9，除数和商都一样，但是个有余数，另一个没有，所以 550÷60 \u003e 540÷60",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "大数的比较"
+                             ],
+        "source":  "24-25 四年级上·河北保定·期末"
+    },
+    {
+        "id":  "355",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "2024 年 3 月 28 日中国移动宣布正式商用5.5G网络，标志着我国及全球在移动通信技术方面迈入了新的发展阶段。李叔叔用5.5G网络下载14部科普纪录片的时间是168秒。照这样的下载速度，用 5.5G 网络下载 42 部同样大小的科普纪录片需要( )秒。",
+        "options":  [
+
+                    ],
+        "answer":  "504",
+        "solution":  "【分析】由题意得，李叔叔用 5.5G 网络下载 14 部科普纪录片的时间是168秒，可以先用168除以14算出下载1部科普纪录片需要多少时间，然后再乘 42 即可算出李叔叔用5.5G 网络下载 42 部同样大小的科普纪录片需要多少秒.\n【详解】 168÷14 = 12 (秒) 12×42 = 504 (秒) 故李叔叔用5.5G网络下载42部同样大小的科普纪录片需要504 秒。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "两位数乘两位数的不进位乘法"
+                             ],
+        "source":  "24-25 四年级上·河北保定·期末"
+    },
+    {
+        "id":  "356",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "要使 456 * □7的积是五位数,□里最小填( );要使□37÷47的商是一位数,\n\n□里最大填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "2;4",
+        "solution":  "【分析】(1)三位数乘两位数的笔算法则：先用两位数的个位分别与三位数的每一位数相乘，乘得的积的末尾和个位对齐。再用两位数的十位分别与三位数的每一位数相乘，乘得的积的末尾和十位对齐。最后，将两次乘得的积相加;在算式 456 * □7 中,□里可以填 1 ~ 9，可以用456依次去乘17,27,37。,然后找出满足题意的即可。(2)三位数除以两位数，判断商是几位数的方法：先观察被除数前两位上的数，看是否大于或等于除数(也就是看前两位上的数除以除数，够不够商一个十)。被除数前两位够除，商就是两位数。反之，商就是一位数.\n【详解】(1) 456×17 = 7752，积是四位数。 456×27 = 12312，积是五位数。所以要使 456 * □7 的积是五位数,□里最小填 2。(2)要使□37÷47的商是一位数，那么被除数的前两位上的数\"□3\"小于除数 47,□里最大填 4。要使 456 * □7 的积是五位数,□里最小填2;要使□37÷47的商是一位数,□里最大填 4 phantom 。 。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法",
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25 四年级上·河北保定·期末"
+    },
+    {
+        "id":  "357",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\u0027\u003e\u0027\u0027\u003c\u0027或\u0027=\u0027。(■表示盖住了一个数字)\n\n1100 ■ 700 ( ) 978070556×470 ( ) 47×560400÷25 () 1600÷100又180÷30×2 ( ) 480 ÷ (30×2) 6 公顷 ( ) 64000 平方米 7平方千米 5 公顷( )705 公顷",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;=;=;\u003e;\u003c;=",
+        "solution":  "【分析】(1)比较整数的大小，先看数位的多少，数位多的数就大，数位相同，从高位比较，高位上的数大则这个数大，高位上的数相同，就比较下一位，以此类推;(2)根据1平方千米 = 100 公顷,1公顷 = 10000 平方米，先化成统一的单位，再比较大小;(3)根据三位数乘两位数，三位数除以两位数的计算方法，先求出各组算式的结果，再比较大小。笔算三位数乘两位数：相同数位对齐，一般多位数写上，两位数写下，从个位乘起，依次用下面的乘数每位上的数去乘上面的乘数，乘到哪一位，得数的末尾和下面乘数的那一位对齐，与哪位乘的积满几十，就要向前一位进几。 除数是两位数的笔算除法：从被除数的高位数起，先看被除数的前两位，如果前两位比除数小，就要看前三位;除到被除数的哪一位，商就写在那一位的上面，每求出一位商，余下的数必须比除数小.\n【详解】1100■700,■表示盖住了一个数字，这是一个八位数,9780705 这是一个七位数，所以 1100■700\u003e9780705.56×470 = 26320 , 47×560 = 26320，所以 56×470 = 47×560400÷25 = 16 , 1600÷100 = 16，所以 400÷25 = 1600÷100180÷30×2 = 6×2 = 12 ,480÷(30×2)=480÷60=8, 12 \u003e 8，所以 180 ÷ 30×2 \u003e 480 ÷ ( 30×2 ) 6 公顷 = 6×10000 = 60000 平方米,60000 平方米\u003c64000 平方米，所以6公顷\u003c64000平方米7平方千米5公顷 = 7 平方千米 + 5 公顷,7平方千米 = 7×100 = 700 公顷,7平方千米+5公顷 = 700 + 5 = 705 公顷，所以7平方千米5公顷 = 705 公顷",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "除数是两位数的笔算除法",
+                                 "三位数乘两位数，乘数末尾有 0",
+                                 "大数的比较"
+                             ],
+        "source":  "24-25 四年级上·河北保定·期末"
+    },
+    {
+        "id":  "358",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在( )里填上\u0027\u003e\u0027\u0027\u003c\u0027或\u0027=\u0027.\n\n2458177( )258551415686400( )2亿 3 公顷( )3000平方米\n\n890×11( )89×11045×300( )450÷ 30120 公顷( )12平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003c;\u003e;=;\u003e;\u003c",
+        "solution":  "【分析】比较整数的大小，先看整数的位数，位数多的那个数就大;如果位数相同，就看最高位，最高位上的数大，那个数就大;如果最高位的数相同，就看次高位，次高位上的数大，那个数就大。以此类推。据此比较2458177和2585514 的大小。把2亿改写成用一作单位的数，再根据整数大小比较方法与15686400 比较大小。1 公顷 = 10000 平方米，据此把 3 公顷改写成 30000 平方米，再与 3000 平方米比较大小。890×11，先算 89×11 = 979，再在 979 的后面添上 1 个 0;89×110，先算 89×11 = 979，再在 979 的后面添上 1 个 0。据此比较 890×11 与 89×110 的大小。分别算出 45×300 和 450÷30 的得数，再比较大小。1平方千米 = 100 公顷,12乘进率100即可把12平方千米换算成 1200 公顷，再与 120 公顷比较大小.\n【详解】2458177\u003c2585514;2 亿=200000000,15686400\u003c200000000,15686400\u003c2 亿。3 公顷 = 30000 平方米,30000 平方米\u003e3000 平方米,3 公顷\u003e3000 平方米。 890×11 = 89×11045×300=13500, 450÷30 = 15 ,13500\u003e15,45×300\u003e450÷30.12×100 = 1200 (公顷),120 公顷\u003c1200 公顷,120 公顷\u003c12 平方千米。 2458177\u003c258551415686400\u003c2 亿 3 公顷\u003e3000 平方米 890×11 = 89×11045×300\u003e450÷30120 公顷\u003c12 平方千米",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "除数是整十数的笔算除法",
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "大数的比较"
+                             ],
+        "source":  "24-25四年级上·河南漯河·期末"
+    },
+    {
+        "id":  "359",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "470÷20=23。 10，如果被除数和除数同时乘 10，余数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "100",
+        "solution":  "【分析】根据商不变的规律，在有余数的除法中，被除数和除数同时乘(或除以)相同的数(0 除外),商不变，余数也要同时乘(或除以)相同的数，据此即可解答.\n【详解】 470÷20 = 23 ldots 10，余数是 10，如果被除数和除数同时乘 10，商不变，则余数也要相应乘 10，为 10×10 = 100。",
+        "knowledge_points":  [
+                                 "三位数除以两位数的有余数除法",
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "24-25 四年级上·贵州铜仁·期末"
+    },
+    {
+        "id":  "360",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "根据 23×61 = 1403 填一填.\n\n(1)( )×61 = 2806 (2)( ) ÷ 610 = 23 ( 3 ) 2300×61 = ( ) (4)14030÷ 230=( )",
+        "options":  [
+
+                    ],
+        "answer":  "46 14030 140300 61",
+        "solution":  "【分析】(1)(3)根据积的变化规律可知，一个因数不变，另一个因数扩大到原来的若干倍或缩小到原来的若干分之一(0除外),得到的积就等于原来的积的若干倍或缩小到原来的若干分之一;(2)因为 23×61 = 1403 , 610÷61 = 10，除数扩大到原来的10 倍，要使商不变，被除数也要扩大到原来的 10 倍;(4)根据商不变的规律：被除数和除数同时乘或除以同个数(0除外),商不变，余数也乘或除以相同的数.\n【详解】(1)( ( 23×2 ) 。 3×2 ) \\ * 61 = 46×61 = 2806 (2)14030÷610=23 (3)2300×61=(23×100)×61=23×61×100=1403×100=140300 (4) 14030÷230 = 1403÷23 = 61",
+        "knowledge_points":  [
+                                 "商不变的规律及应用",
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "24-25四年级上·河南许昌·期末"
+    },
+    {
+        "id":  "361",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "( )里最大能填几。\n\n64( )995≈64 万 465875( )021≈465875 万 9( )997778\u003c9989777962×( )\u003c490 ( )×400\u003c2100 ( )×49\u003c350",
+        "options":  [
+
+                    ],
+        "answer":  "4;4;8;7;5;7",
+        "solution":  "【分析】(1)(2)用四舍五入法省略万位后面的尾数，就看千位上的数字，千位上的数字是4或者比4还小，就用\"四舍\"法，千位上的数字大于4，就用五入法，同时在后面加上一个\"万\"字,64()995≈64万是四舍得到的，括号里应填小于 5 的数字，最大能填 4;465875()021≈465875 万是四舍得到的，括号里应填小于 5的数字，最大能填 4;(3)比较整数的大小，位数多的那个数就大;如果位数相同，就看最高位，最高位上的数大，那个数就大;最高位上的数相同，就看下一位，哪一位上的数大，那个数就大。9()997778\u003c99897779，左右两边数字的千万位上都是 9，右边数字的百万上是9，十万位上是8，左边数字十万位上是9，如果左边数字的百万上填9，那么左边数字十万位上的9大于右边数字十万位上的8，不符合题意，所以左边数字百万位上应填小于 9 的数，最大能填 8;(4)用490除以62，如果有余数，运算的商就是可以填的最大的数;如果没有余数比商小 1 的数就是要填的数;(5)用2100除以400，如果有余数，运算的商就是可以填的最大的数;如果没有余数比商小1的数就是要填的数;(6)用 350 除以 49，如果有余数，运算的商就是可以填的最大的数;如果没有余数比商小1的数就是要填的数.\n【详解】(1)64()995≈64 万，是四舍得到的，括号里应填小于 5 的数，括号里可填 0,1,2,3,4，最大能填 4，即 644995≈64 万;(2)465875()021≈465875 万是四舍得到的，括号里应填小于5的数字，括号里可填0,1,2,3,4，最大能填4，即 4658754021≈465875 万;(3)9()997778\u003c99897779，所以左边数字百万位上应填小于9的数，括号里可填0,1,2,3,4,5,6,7,8，最大能填 8，即 98997778\u003c99897779(4)490÷62=7。5662×7 = 434，符合题意 62×8 = 496，不符合题意 62×7\u003c490 (5) 2100÷400 = 5 。 。1005×400 = 2000，符合题意 6×400 = 2400，不符合题意 5×400\u003c2100 (6)350÷49=7。77×49 = 343，符合题意 8×49 = 392，不符合题意 7×49\u003c350 万 4658754021≈465875 万 98997778\u003c9989777962×7\u003c4905×400\u003c21007×49\u003c350\n\n644995 approx 64​ 万 4658754021≈465875 万",
+        "knowledge_points":  [
+                                 "亿以上数的近似数",
+                                 "亿以内数的近似数",
+                                 "三位数除以两位数的有余数除法",
+                                 "大数的比较"
+                             ],
+        "source":  "24-25四年级上·浙江嘉兴·期末"
+    },
+    {
+        "id":  "362",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在( )里填上 \u0027\u003e\u0027\u0027\u003c\u0027或\u0027=\u0027.\n\n20 公顷( )20000 平方米 5700÷50( )570÷5230000 万( )23 亿\n\n3 平方千米( )30 公顷 29×300( )30×290180个万( )18 个百万",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;=;=;\u003e;=;=",
+        "solution":  "【分析】(1)(4)根据 1 公顷 = 10000 平方米,1 平方千米 = 100 公顷，先化成统一的单位，再根据数值的大小进行比较得解;(2)商不变的规律：被除数和除数都乘(或除以)一个相同的数(0 除外),商不变;(3)230000 万=2300000000,23 亿=2300000000，则230000 万与 23 亿相等;(5)积不变的规律：两个数相乘，一个因数乘(或除以)一个数(0除外),另一个因数同时除以(或乘)相同的数，它们的积不变;(6)根据整数的写法，从高位到低位，一级一级地写，哪个数位上一个单位也没有，就在那个数位上写0，据此写出各个数，再根据整数大小的比较方法进行比较.\n【详解】因为20000平方米 = 2 公顷, 20 \u003e 2，所以20公顷\u003e20000 平方米;因为( 5700÷10 ) \\ / \\ ( 50÷10 ) \\ = 570÷5 = 114，所以 5700÷50 = 570÷5;因为 230000 万=2300000000,23 亿=2300000000,2300000000=2300000000，所以 230000 万=23 亿;因为 3 平方千米 = 300 公顷, 300 \u003e 30，所以 3 平方千米\u003e30 公顷;(29×10)= 30×290;因为( 29×10 )×( 300÷10 ) = 290×30 = 8700，所以 29×300 因为 180 个万是 1800000,18 个百万是 1800000,1800000 =1800000，所以 180 个万 = 18 个百万。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "公顷，平方千米的进率与换算",
+                                 "积的变化规律(整数乘法)",
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "24-25 四年级上·浙江嘉兴·期末"
+    },
+    {
+        "id":  "363",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一栋办公楼长 40 米，宽 25 米，它的占地面积为( )平方米,( )栋这样的教学楼的占地面积为 1 公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "1000;10",
+        "solution":  "【分析】1公顷 = 10000 平方米，先根据长方形的面积 = 长×宽，求出一栋办公楼的占地面积，然后根据\"包含\"除法的意义，用除法解答.\n【详解】 40×25 = 1000 (平方米) 1公顷 = 10000 平方米 10000÷ 1000=10(栋)一栋办公楼长40米，宽25米，它的占地面积为1000平方米,10栋这样的教学楼的占地面积为1公顷.\n\n10000÷1000 = 10 (栋)",
+        "knowledge_points":  [
+                                 "被除数和除数末尾都有 0 的除法",
+                                 "长方形的面积",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "24-25 四年级上·河南许昌·期末"
+    },
+    {
+        "id":  "364",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "3名同学进行竞走训练，小刚每小时走3840米，小亮走560米用8分钟，小华每秒走 1 米，这三人中( )的速度最快。",
+        "options":  [
+
+                    ],
+        "answer":  "小亮",
+        "solution":  "【分析】1小时 = 60 分，根据速度 = 路程/时间，小刚每分钟走 3840÷60 = 64 (米);小亮每分钟走 560÷8 = 70 (米);因为 1 分钟 = 60 秒，小华每分钟走 1×60 = 60 (米)。再把三人每分钟走的长度进行比较即可.\n【详解】小刚每分钟走的长度: 3840÷60 = 64 (米)小亮每分钟走的长度: 560÷8 = 70 (米)小华每分钟走的长度: 1×60 = 60 (米)70 米 mathord { mathrm \u003e } 64 米 \u003e 60 米 3名同学进行竞走训练，小刚每小时走3840米，小亮走560米用8分钟，小华每秒走1米，这三人中小亮的速度最快。",
+        "knowledge_points":  [
+                                 "整十，整百，整千数除以一位数的除法",
+                                 "三位数除以一位数，首位不够除",
+                                 "除数是整十数的笔算除法",
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25四年级上·河北保定·期末"
+    },
+    {
+        "id":  "365",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "面积为 6 公顷的长方形鱼池，宽是 200 米，长是( )米，若该鱼池需要扩建，将宽增加50米，长扩大到原来的2倍，则鱼池的面积增加( )公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "300;9",
+        "solution":  "【分析】根据1公顷 = 10000 平方米;将6公顷换算成60000平方米;然后根据长方形的面积 = 长×宽，即长=60000÷200(米);将宽增加50米，长扩大到原来的2倍，则扩大后的宽为( ( 200 + 50 ) )米，长为( ( 60000÷200×2 ) )米，求出扩建后的面积，再减去原来的面积即可.\n【详解】6 公顷 = 60000 平方米60000÷200 = 300 (米) (200 + 50)×600 = 250×600 = 150000 (平 方 米) \n\n150000 平方米 = 15 公顷 15 - 6 = 9 (公顷)\n\n则面积为 6 公顷的长方形鱼池，宽是 200 米，长是 300 米，若该鱼池需要扩建，将宽增加50米，长扩大到原来的2倍，则鱼池的面积增加9公顷。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的实际问题",
+                                 "长方形的面积",
+                                 "除数是整十数的口算除法"
+                             ],
+        "source":  "24-25 四年级上·河北张家口·期末"
+    },
+    {
+        "id":  "366",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在( )填入\u0027\u003e\u0027 \u0027\u003c\u0027或 \u0027= \u0027\n\n303000( )30030080 万( )79999996 平方千米( )600 公顷\n\n 630÷30 ( ) 630÷21230×12 ( ) 120×23160 ÷ (4×8) ( ) 160÷4÷8 ",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;\u003c;=;\u003c;=;=",
+        "solution":  "【分析】比较303000与300300：比较整数的大小，先看整数的位数，位数多的那个数就大;如果位数相同，就看最高位，最高位上的数大，那个数就大;如果最高位的数相同，就看次高位，次高位上的数大，那个数就大。以此类推。比较80万与7999999：先把80万还原，再根据整数大小比较方法与7999999作比较。比较6平方千米与600公顷：因为1平方千米 = 100 公顷，据此先把6平方千米与600公顷的单位换算统一后再进行比较。比较 630÷30 \\ 与 630÷21：根据\"被除数相同，除数越大，商越小\"的规律。比较 230×12 与 120×23：根据积不变的规律：一个因数乘(或除以)几(0 除外),另一个因数除以(或乘)相同的数，积不变。比较 160÷( ( 4×8 ) )与 160÷4÷8：根据除法的性质：一个数连续除以两个数，等于这个数除以这两个数的积，即 a / b / c = a/(b×c)(b!=0,c!=0).\n【详解】 303000 \u003e 30030080 万=800000，所以 800000\u003c7999999，即 80 万\u003c7999999.1 平方千米 = 100 公顷，所以 6 平方千米 = 6×100 = 600 公顷，故6平方千米 = 600 公顷。 30 \u003e 21，所以 630÷30 { \u003c } 630÷21 。 230×12 = ~ ( 230÷10 ) ~ * ~ ( 12×10 ) ~ = 23×120，所以 230×12 = 120×23.160 ÷ (4×8) = 160÷4÷8 。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)",
+                                 "公顷，平方千米的进率与换算",
+                                 "除数是两位数的笔算除法",
+                                 "大数的比较"
+                             ],
+        "source":  "24-25四年级上·浙江绍兴·期末"
+    },
+    {
+        "id":  "367",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "玲玲在计算两位数乘两位数时，把其中一个因数个位上的 8 写成了 6，结果得 576，正确的结果是 648，这两个数分别是( )和( )。",
+        "options":  [
+
+                    ],
+        "answer":  "18;36",
+        "solution":  "【分析】根据题意，正确结果与错误结果之间相差(648-576)。也就是把 8 写成 6，少了 2 个另一个因数，用(648-576)除以2，求出另一个因数。然后再用648除以另一个因数，求出第一个因数.\n【详解】 8 - 6 = 2 (648 - 576) ÷ 2 = 72÷2 = 36648÷36 = 18 所以，这两个数分别是 18 和 36 ",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "24-25 四年级上·贵州六盘水·期末"
+    },
+    {
+        "id":  "368",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在 A / 25 = 8 ldots B中，余数 B 最大可取( ),这时被除数 A 是( );当 A增加( )时，得数是 12。B。",
+        "options":  [
+
+                    ],
+        "answer":  "24;224;100",
+        "solution":  "【分析】根据余数与除法的关系，余数必须比除数小，除数是25，余数最大是25-1，根据被除数 = 除数×商+余数，据此代入数字计算出这时被除数是多少;先用12-8求出商增加了多少，用除数×商即可求出被除数A需要增加多少.\n【详解】 25 - 1 = 2425×8 + 24 = 200 + 24 = 224 (12 - 8)×25 = 4×25 = 100 在 A / 25 = 8 ldots B 中，余数B最大可取24，这时被除数A是 224;当 A 增加 100 时，得数是 12。B。",
+        "knowledge_points":  [
+                                 "余数与除数的关系",
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "24-25四年级上·浙江嘉兴·期末"
+    },
+    {
+        "id":  "369",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "已知□ 62÷46 的商是一位数,□中最大填( );已知□48÷33的商是两位数,□中最小填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "3;3",
+        "solution":  "【分析】三位数除以两位数，如果被除数前两位上的数小于除数，则商是一位数;如果被除数前两位上的数大于或等于除数，则商是两位数.\n【详解】已知□62÷46的商是一位数，则□ 6 \u003c 46，所以□中可以填 1,2,3，最大填 3;已知□48÷33的商是两位数，则□ 4 \u003e 33，所以□中可以填3,4,5,6,7,8,9，最小填 3",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25四年级上·浙江嘉兴·期末"
+    },
+    {
+        "id":  "370",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算 460÷39 时，可以把除数看成 来试商，商是 位数。",
+        "options":  [
+
+                    ],
+        "answer":  "40;两",
+        "solution":  "【分析】首先把除数利用\"四舍五入法\"看作与它接近的整十数进行试商，确定商的位数时，如果被除数的前两位比除数大，则商就是两位数;反之，商就是一位数，据此解答.\n【详解】 39 approx 40 被除数的前两位 46 \u003e 40，所以计算 460÷39 时，可以把除数看成40来试商，商是两位数。",
+        "knowledge_points":  [
+                                 "多位数除以两位数的试商",
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25 四年级上·贵州安顺·期末"
+    },
+    {
+        "id":  "371",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在下面的( )里填上\u0027\u003e\u0027\u0027\u003c\u0027或\u0027=\u0027.\n\n999999( )100000600 公顷( )6 平方千米 635÷72( )381÷27650×21( )56×210",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;=;\u003c;\u003e",
+        "solution":  "【分析】(1)大数的比较：先比较两个数的位数，位数多的数就大。如果两个数的位数相同，就从最高位比起，最高位上的数大的那个数就大，如果最高位上的数相同，就比较下一个数位上的数。(2)根据 100 公顷 = 1 平方千米先将 600 公顷转化为多少平方千米，然后再比较它们的大小。 (3)(4)先计算出括号两边算式的结果，然后再比较它们的大小.\n【详解】(1)999999 和 100000 都是六位数，十万位上 9 \u003e 1，所以 999999\u003e100000。(2)100 公顷 = 1 平方千米，所以 600 公顷 = 6 平方千米。 (3)635÷72=8。59,381÷27=14。3.8\u003c14，所以635÷72\u003c381÷27。 (4)650×21=13650,56×210=11760.13650\u003e11760，所以 650×21 \u003e 56×210 。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "除数是两位数的笔算除法",
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "大数的比较"
+                             ],
+        "source":  "24-25 四年级上·浙江杭州·期末"
+    },
+    {
+        "id":  "372",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "选择正确的余数填在括号里.\n\n980÷40 = 24 dots dots ( ) (2, 20)\n\n2300÷500 = 4 dots dots ( ) (3, 30, 300)",
+        "options":  [
+
+                    ],
+        "answer":  "20300",
+        "solution":  "【分析】根据根据被除数 = 商×除数+余数，可知余数 = 被除数-除数×商，将数据代入计算解答即可.\n【详解】980-24×40 = 980 - 960 = 20980÷40 = 20 dots 202300 - 4×500 = 2300 - 2000 = 3002300÷500 = 4 dots 。 300 ",
+        "knowledge_points":  [
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "24-25四年级上·浙江绍兴·期末"
+    },
+    {
+        "id":  "373",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "新兴糖果公司生产了 700 支棒棒糖，如果每 12 支装一盒，最多可以装( )盒;如果每箱可以装 12 盒，至少需要( )个这样的箱子。",
+        "options":  [
+
+                    ],
+        "answer":  "58;5",
+        "solution":  "【分析】根据题意，用棒棒糖的总支数除以每盒的支数，即得到能装的盒数;如果除得的结果有余数，因余数不能装满一盒，所以，商就是最多能装满的盒数;用总盒数除以每箱装的盒数，就是需要的箱子个数，如果有余数，再加 1 即可.\n【详解】 700÷12 = 58 (盒)。4(支) 58÷12 = 4 (个)。10(支) 4 + 1 = 5 (个) 则新兴糖果公司生产了700支棒棒糖，如果每12支装一盒，最多可以装 58 盒;如果每箱可以装 12 盒，至少需要 5 个这样的箱子。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "24-25 四年级上·贵州遵义·期末"
+    },
+    {
+        "id":  "374",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据 54÷6 = 9，写出下面算式的结果.\n\n540÷60 = ( ) 108÷6 = ( ) 54÷2 = ( )",
+        "options":  [
+
+                    ],
+        "answer":  "9;18;27",
+        "solution":  "【分析】第一题根据商不变的性质指在除法算式中，被除数和除数同时乘或除以一个相同的数(0除外),商的大小不变。第二题在除法里，除数不变，被除数乘或除以一个数(0除外),商也随之乘或除以相同的数。第三题在除法里，被除数不变，除数乘一个数(0除外),商反而除以相同的数;除数除以一个数(0除外),商反而乘相同的数.\n【详解】被除数( ( 54 to 540 ) )和除数 ( 6 to 60 ) 同时乘 10，商不变。所以 540÷60 = 9 被除数( 54 to 108 ) 乘 2，除数不变，商也乘 2，所以 108÷6 = 9×2 = 18除数 ( 62 ) 除以 3，被除数不变，商乘 3。所以 54÷2 = 9×3 = 27",
+        "knowledge_points":  [
+                                 "商不变的规律及应用",
+                                 "商的变化规律及应用"
+                             ],
+        "source":  "24-25 四年级上·贵州铜仁·期末"
+    },
+    {
+        "id":  "375",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在算式 534 ÷ □ 4 中，要使商是两位数,里最大填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "4",
+        "solution":  "【分析】三位数除以两位数，比较被除数前两位数字和除数的大小，如果被除数前两位数字大于等于除数，则商是两位数，如果小于除数商是一位数;据此解答.\n【详解】在算式534/4中，要使商是两位数,□4要小于等于 53，所以□里最大填 4 phantom 。 。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25四年级上·贵州铜仁·期末"
+    },
+    {
+        "id":  "376",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一架物流无人机从河南科技市场数码港垂直升空，直线飞行 12 分钟后，平稳降落在 7200 米外的郑州新天地通讯大厦，该无人机平均每分钟飞行( )米。",
+        "options":  [
+
+                    ],
+        "answer":  "600",
+        "solution":  "【分析】根据速度 = 路程 / 时间，用 7200 除以 12，据此列式计算可求出该无人机平均每分钟飞行多少米.\n【详解】 7200÷12 = 600 (米)综上可知，该无人机平均每分钟飞行 600 米。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25 四年级上·河南新乡·期末"
+    },
+    {
+        "id":  "377",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在算式□ ÷ 35 =20。△中,△最大是( ),此时□是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "34;734",
+        "solution":  "【分析】在有余数的除法中，余数要比除数小，余数最大比除数小1，则△最大是 35 - 1 = 34，再根据被除数 = 除数×商+余数，计算出被除数.\n【详解】 35 - 1 = 3435×20 + 34 = 700 + 34 = 734 △最大是 34，此时□是 734 ",
+        "knowledge_points":  [
+                                 "余数与除数的关系",
+                                 "有余数的除法验算",
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "24-25 四年级上·河南漯河·期末"
+    },
+    {
+        "id":  "378",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一道除法算式的商是 15，如果被除数乘 3，除数除以 3，商是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "135",
+        "solution":  "【分析】商的变化规律：除数不变，被除数乘(或除以)几(0除外),商也乘(或除以)相同数;被除数不变，除数乘(或除以)几(0 除外),商除以(或乘)相同数;据此可知，一道除法算式的商是 15，除数除以 3，商应该乘3;被除数乘3，商在之前的基础上再乘3;所以商应该乘9。据此解答.\n【详解】15×3×3=45×3=135 即一道除法算式的商是15，如果被除数乘3，除数除以3，商是 135。",
+        "knowledge_points":  [
+                                 "商的变化规律及应用"
+                             ],
+        "source":  "24-25四年级上·河南漯河·期末"
+    },
+    {
+        "id":  "379",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\u0027\u003e\u0027\u0027\u003c\u0027或\u0027=\u0027.\n\n 501×19 ( ) 500×20260×20 ( ) 200×26540÷27 ( ) 540÷30360÷20 ( ) 260÷20 ",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;=;\u003e;\u003e",
+        "solution":  "【分析】三位数乘两位数的计算方法：两位数乘三位数，先用两位数个位上的数去乘三位数，得数的末尾和两位数的个位对齐，再用两位数十位上的数去乘三位数，得数的末尾和两位数的十位对齐，然后把两次乘的结果加起来。当乘数末尾有零时，先算零前面的数，再在积的末尾添加对应个数的零。据此先算出(1)(2)小题算式的结果，再比较大小即可。当被除数相同时，除数越大，商越小;据此比较(3)小题的大小;当除数相同时，被除数越大，商越大;据此比较(4)小题的大小.\n【详解】 501×19 = 9519 , 500×20 = 10000 ,9519\u003c10000，即 501×19 \u003c 500×20260×20 = 5200 , 200×26 = 5200，即 260×20 = 200×2627\u003c30,540÷27\u003e540÷30360\u003e260,360÷20\u003e260÷20",
+        "knowledge_points":  [
+                                 "商的变化规律及应用",
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "三位数乘两位数，三位数中间有0"
+                             ],
+        "source":  "24-25 四年级上·河南漯河·期末"
+    },
+    {
+        "id":  "380",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "括号里最大能填几。\n\n20×( )\u003c155 ( )×50\u003c368262\u003e( )×30",
+        "options":  [
+
+                    ],
+        "answer":  "7;7;8",
+        "solution":  "【分析】据题意，要求最大填几，用所比较的数除以已知的因数，得到的结果如果有余数，那么商就是能填的最大的数。如果没有余数，那么商减一就是能填的最大的数.\n【详解】 155÷20 = 7 。 。 。 。 。 15，则 20×7 { \u003c } 155;368÷50=7。18，则 7×50 { \u003c } 368;262 ÷ 30 = 8 ldots 22，则 262 \u003e 8×30 ",
+        "knowledge_points":  [
+                                 "有余数除法的实际应用",
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "24-25 四年级上·湖南常德·期末"
+    },
+    {
+        "id":  "381",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在 a / 25 = 18 ·s b 中，余数b最大是( ),这时被除数a是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "24;474",
+        "solution":  "【分析】有余数的除法，余数一定要比除数小，已知除数是25，所以余数 mathtt b \u003c 25，就可以求出余数最大是多少，再根据被除数 = 商×除数+余数，就可以求出被除数是多少.\n【详解】根据分析可得：余数 mathtt b \u003c 25，所以余数最大为:25-1=24此时被除数为: 25×18 + 24 = 450 + 24 = 474 所以在 a / 25 = 18 ·s b 中，余数b最大是 24，这时被除数a是474。",
+        "knowledge_points":  [
+                                 "余数与除数的关系",
+                                 "有余数的除法验算",
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "24-25 四年级上·河北张家口·期末"
+    },
+    {
+        "id":  "382",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "算式□ 30÷75，要使商是一位数,□里最大可以填( );要使商是两位数,□里最小可以填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "7;8",
+        "solution":  "【分析】三位数除以两位数，比较被除数前两位的数字和除数的大小，如果被除数前两位的数字大于等于除数，则商是两位数，如果小于除数商是一位数.\n【详解】算式□ 30÷75，要使商是一位数，被除数前两位□3小于75，那么□里最大可以填7;要使商是两位数，被除数前两位□3 大于等于 75，那么□里最小可以填 8。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25 四年级上·浙江台州·期末"
+    },
+    {
+        "id":  "383",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "329 □ 123 approx 330 万,□里最小填( );635 ÷ □ 4，如果商的最高位在十位上,□里最大填( );算式□ 16×20，要使积是四位数,□里最大填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "5;5;4",
+        "solution":  "【分析】四舍五入到万位，关键看千位。329□123的千位是□,要使结果≈330万，千位需 \u003e=slant 5 才能进位。635÷□4的商是两位数，说明除数□ 4 le 63 (否则商为个位数)。算式□ 16×20 中□16看成整百数估算 500×20 = 10000，那么□里填5, 516×20 { stackrel { textstyle \u003e } 10000 } ,积是五位数，不符合;所以□里的数要小于5，据此解答即可.\n【详解】根据分析可知:329□123≈330 万,□里要填大于等于 5 的数,□里最小填5;635÷□4，如果商的最高位在十位上，除数□ 4 le 63 ,□里最大填 5;416×20 = 8320，积是四位数, 516×20 = 10320，积是五位数，不符合，所以算式□ 16×20，要使积是四位数,□里最大填4。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法",
+                                 "判断商是几位数(除数是两位数)",
+                                 "亿以内数的近似数"
+                             ],
+        "source":  "24-25 四年级上·河南郑州·期末"
+    },
+    {
+        "id":  "384",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "452÷□3的商是一位数,□里最小可以填( );□05÷52的商是两位数,□里最小可以填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "5;6",
+        "solution":  "【分析】三位数除以两位数，被除数前两位上的数大于或等于除数时，商是两位数，小于除数时，商是一位数;据此可解此题.\n【详解】根据分析：要使 452÷□3 的商是一位数，则 45 { \u003c } □3 ,□可以填5,6,7,8,9，最小填 5;要使□05÷52 的商是两位数，则□ 0 ge 52 ,□可以填6,7,8,9，最小填 6;综上可知,452÷□3 的商是一位数,□里最小可以填 5;□ 105÷52 的商是两位数,□里最小可以填6。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25四年级上·湖南益阳·期末"
+    },
+    {
+        "id":  "385",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上\u0027\u003e\u0027\u0027\u003c\u0027或 \u0027=\u0027 \n\n1020001900( )10 亿 90 万( )89700050 公顷( )5 平方千米 91 circ ( ( )直角 210×40( )21×400288÷24( )144÷6",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;\u003e;\u003c;\u003e;=;\u003c",
+        "solution":  "【分析】10 亿=1000000000;然后再与 1020001900 比较大小即可;90 万=900000;然后再与 897000 比较大小即可;1 平方千米 = 100 公顷，据此将 5 平方千米换算成 500 公顷，然后再与50公顷比较即可;直角是 90 circ ,据此再与91°比较大小即可;分别计算出 210×40 和 21×400 算式的结果，然后再比较大小;分别计算出 288÷24 和 144÷6 算式的结果，然后再比较大小.\n【详解】10 亿=1000000000;1020001900\u003e1000000000，所以 1020001900\u003e10 亿;90 万=900000;900000\u003e897000，所以 90 万\u003e897000;5 平方千米 = 500 公顷;50 公顷\u003c500 公顷，所以 50 公顷\u003c5 平方千米;91 circ \u003e 90 circ ,所以 91 circ \u003e 直角;210×40 = 8400;21×400 = 8400，所以 210×40 = 21×400;288÷24 = 12;144÷6 = 24 。;12 \u003c 24，所以 288÷24 \u003c 144÷6 ",
+        "knowledge_points":  [
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "除数是两位数的笔算除法",
+                                 "直角，钝角，锐角的认识及特征",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·河南新乡·期末"
+    },
+    {
+        "id":  "386",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "560 连续减去 80，减( )次得数为 0;792 除以( )的商是最大的两位数。",
+        "options":  [
+
+                    ],
+        "answer":  "7;8",
+        "solution":  "【分析】要求560连续减去几次80，直到结果为0，实际就是求560中包含多少个80，用除法计算即可;最大的两位数是99，要使792除以一个数的商是99，求这个数，用792÷99，即可求出这个数。据此作答.\n【详解】 560÷80 = 7792÷99 = 8 因此,560 连续减去 80，减 7 次得数为 0;792 除以 8 的商是最大的两位数。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "被除数和除数末尾都有0的除法"
+                             ],
+        "source":  "24-25 四年级上·湖南永州·期末"
+    },
+    {
+        "id":  "387",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "赵老师带 4700 元钱为学校图书馆购买书架。每个书架 600 元，他最多能买()个这样的书架，还剩( )元钱。",
+        "options":  [
+
+                    ],
+        "answer":  "7;500",
+        "solution":  "【分析】用总钱数/每个书架的钱数，也就是4700÷600，商是几就是最多能买几个书架，余数就是还剩的钱。计算时根据商不变的原理，被除数和除数同时乘或者除以几(0除外),此时商不变，但是余数需要乘或者除以这个数，据此解题.\n【详解】470÷60=7。504700 ÷ 600 = 7 (个)。500(元) 赵老师带4700元钱为学校图书馆购买书架。每个书架600元，他最多能买 7 个这样的书架，还剩 500 元钱。",
+        "knowledge_points":  [
+                                 "商不变的规律及应用",
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "24-25 四年级上·河南新乡·期末"
+    },
+    {
+        "id":  "388",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "中医药文化是我国非物质文化遗产的杰出代表，某校打造了\u0027走进中医药的大千世界\u0027特色课程。为让学生更好的了解中药材知识，学校准备了460本《药材大全》平均分给23个班级，每个班级将分到( )本。",
+        "options":  [
+
+                    ],
+        "answer":  "20",
+        "solution":  "【分析】《药材大全》本数除以班级数，即可算出平均每个班级将分到(460÷23)本.\n【详解】 460 ÷ 23 = 20 (本)中医药文化是我国非物质文化遗产的杰出代表，某校打造了\"走进中医药的大千世界\"特色课程。为让学生更好的了解中药材知识，学校准备了460本《药材大全》平均分给23 个班级，每个班级将分到 20 本。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "24-25 四年级上·河南漯河·期末"
+    },
+    {
+        "id":  "389",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "两个数的商是 36，如果被除数不变，除数乘 9，则商是( );两个因数的积是30，一个因数扩大为原来的 4 倍，另一个因数不变，积是()。",
+        "options":  [
+
+                    ],
+        "answer":  "4;120",
+        "solution":  "【分析】根据商的变化规律：被除数不变，除数扩大到原来的几倍(或缩小到原来的几分之一)(0除外),则商缩小到到原来的几分之一(或扩大相同的倍数);根据积的变化规律可知，一个因数不变，另一个因数扩大到原来的若干倍或缩小到原来的若干分之一(0 除外),得到的积就等于原来的积的若干倍或缩小到原来的若干分之一;据此解答.\n【详解】 36÷9 = 430×4 = 120 所以两个数的商是36，如果被除数不变，除数乘9，则商是4;两个因数的积是 30，一个因数扩大为原来的 4 倍，另一个因数不变，积是 120。",
+        "knowledge_points":  [
+                                 "商的变化规律及应用",
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "24-25 四年级上·河南漯河·期末"
+    },
+    {
+        "id":  "390",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "已知 A×B = 800，如果 A 乘 20,B 不变，积是( );已知 A / B = 80，如果A,B 同时除以 10，那么商是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "16000;80",
+        "solution":  "【分析】积的变化规律：一个因数不变，另一个因数乘或除以几(0 除外),积也乘或者除以这个数。商不变的规律：被除数和除数同时乘或者除以一个相同的数(0 除外),商不变.\n【详解】 A×B = 800，如果 A 乘 20,B 不变，那么 800 也乘20, 800×20 = 16000 。A / B = 80，如果 A,B 同时除以 10，商不变。那么商是80。",
+        "knowledge_points":  [
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "积的变化规律(整数乘法)",
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "24-25 四年级上·河北张家口·期末"
+    },
+    {
+        "id":  "391",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在算式□ ÷ 300 = 6 ·s △ 中，如果将被除数和除数同时除以100，余数就比原来减少了 99，那么原来的被除数是( ),余数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "1900;100",
+        "solution":  "【分析】在有余数的除法算式里，被除数和除数同时除以100，那么商不变，余数也要同时除以100，在有余数的除法算式里，被除数 = 商×除数+余数，依此解答.\n【详解】△-△÷100=99, 100÷100 = 1 ·s 1 , 100 - 1 = 99，即 △ = 100;6×300 + 100 = 1800 + 100 = 1900 原来的被除数是 1900，余数是 100。",
+        "knowledge_points":  [
+                                 "商不变的规律及应用",
+                                 "有余数的除法验算"
+                             ],
+        "source":  "24-25四年级上·河南许昌·期末"
+    },
+    {
+        "id":  "392",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "为迎接新一年的到来，我市要举办\u0027迎新年联欢会\u0027。某舞蹈节目筹备组要统一订购演出服装。已经花费了 336 元为 12 名演员购买了演出服，按这样的价格还要为剩余的 105 名演员购买演出服，这个舞蹈节目一共需要花费( )元购买演出服。",
+        "options":  [
+
+                    ],
+        "answer":  "3276",
+        "solution":  "【分析】根据 \"单价 = 总价/数量\"，可求出每套演出服的价格，即336÷12(元);再根据\"总价 = 单价×数量\"，求出这105 名演员购买演出服需要花费的钱数;最后与已花费的336元相加即可.\n【详解】 336÷12 = 28 (元) 28×105 = 2940 (元) 336 + 2940 = 3276 (元) 所以这个舞蹈节目一共需要花费3276元购买演出服。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "三位数乘两位数，三位数中间有 0",
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "24-25 四年级上·河南许昌·期末"
+    },
+    {
+        "id":  "393",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "5 □ 9÷54，要使商是两位数,□里最小填( );要使商的末尾有 0,□里最大填()。",
+        "options":  [
+
+                    ],
+        "answer":  "4;8",
+        "solution":  "【分析】三位数除以两位数，判断商是几位数的方法：先观察被除数前两位上的数，看是否大于或等于除数(也就是看前两位上的数除以除数，够不够商一个十)。被除数前两位够除，商就是两位数，反之，商就是一位数;在5□9÷54里，要商末尾有0，前两位得大于等于54，且被除数个位 9 小于 54。从 9 开始试，据此解答.\n【详解】5□9÷54，要使商是两位数，则5□\u003e=54，所以□里面可以填:4,5,6,7,8,9，最小填 4;599÷54 = 11 dots 。 5589÷54 = 10 dots 49579÷54 = 10 dots 。 39 所以5□9÷54，要使商末尾有0,□里最大填 8 ",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25四年级上·河南许昌·期末"
+    },
+    {
+        "id":  "394",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号填上\u0027\u003e\u0027 \u0027\u003c\u0027或 \u0027 =\u0027\n\n599995( )6 万 180100( )181000180×70( )360×352 个平角( )4 个直角 20 公顷( )2 平方千米 300÷ 12( )600÷ 25",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;\u003c;=;=;\u003c;\u003e",
+        "solution":  "【分析】①6万=60000，然后根据大数的比较方法比较大小即可;②大数比较大小的方法：位数相同的两个数比较大小，从最高位比起，最高位上的数大的那个数就大，如果最高位上的数相同，就比较下一个数位上的数。位数不同的两个数比较大小，位数多的数就大;③如果一个因数扩大到原来的若干倍，另一个因数缩小到原来的若干分之一(0除外),那么积不变;④平角=180°,直角=90°,据此解答;⑤1平方千米=100公顷，据此将单位统一，再比较大小;⑥计算出式子的结果，再比较大小.\n【详解】6 万 = 60000 ,599995\u003e60000，所以 599995\u003e6 万;180100 \u003c 181000180×70 = (360÷2)×(35×2) = 360×352×180 circ = 360 circ , 4×90 circ = 360 circ , 360 circ = 360 circ ,所以2个平角 =4 个直角;2 平方千米 = 200 公顷, 20 \u003c 200，所以 20 公顷 \u003c 2 平方千米;300÷12 = 25, 600÷25 = 24, 25 \u003e 24，所 以 300÷12 \u003e600÷25 。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "平角，周角的认识及特征",
+                                 "除数是两位数的笔算除法",
+                                 "大数的比较"
+                             ],
+        "source":  "24-25四年级上·浙江杭州·期末"
+    },
+    {
+        "id":  "395",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算 9000÷600 时，可以用 90÷6 来计算。请你说明这样计算的道理:()。",
+        "options":  [
+
+                    ],
+        "answer":  "9000 和 600 都同时除以 100 变为 90÷6，商不变。",
+        "solution":  "【分析】商不变的性质指在除法算式中，被除数和除数同时乘或除以一个相同的数(0除外),商的大小不变。依此计算 9000÷600 时，可以用 90÷6 来计算.\n【详解】计算 9000÷600​ 时，可以用 90÷6 ​ 来计算。请你说明这样计算的道理:9000 和 600 都同时除以 100 变为 90÷6 ​ ,商不变。",
+        "knowledge_points":  [
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "24-25四年级上·浙江杭州·期末"
+    },
+    {
+        "id":  "396",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "括号里最大能填几。\n\n80×( )\u003c567 ( )21÷48，商是一位数\n\n4( )0000\u003c45200727( )999≈27 万",
+        "options":  [
+
+                    ],
+        "answer":  "7;4;5;4",
+        "solution":  "【分析】要求乘法算式中，括号里最大能填几，可以用小于号(大于号)开口所对的那个数除以乘法算式中的乘数，如果有余数，所得的商便是能填的最大数，如果没有余数，所得的商减 1 便是能填的最大数，据此用 567÷80 即可;三位数除以两位数，如果被除数的前两位大于等于除数，商是两位数，如果被除数的前两位小于除数，则商是一位数;整数比较大小，首先看位数，位数多的这个数就大，如果位数相同，就比较最高位，最高位大的这个数就大，如果最高位相等就比较最高位的下一位，依此类推;一个数估算成以\"万\"为单位的数，首先看千位，如果千位大于等于5，则向前一位进一后再省略，如果千位数小于5 则直接省略，然后再在数的后面写上\"万\"字,27()999≈27万，万位上的数没变，说明括号内的数应该小于 5，小于 5 最大的整数是 4，据此解题.\n【详解】 567÷80 = 7 dots 7 。;所以 80×7 { \u003c } 567 ;421÷48，商是一位数;450000\u003c452007;274999≈27 万。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "除数是整十数的笔算除法",
+                                 "判断商是几位数(除数是两位数)",
+                                 "亿以内数的近似数"
+                             ],
+        "source":  "24-25 四年级上·河南许昌·期末"
+    },
+    {
+        "id":  "397",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\u0027\u003e\u0027\u0027\u003c\u0027或\u0027=\u0027.\n\n10101010( )9909909504000( )54 万\n\n480÷25( )480÷362500公顷( )25 平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;\u003c;\u003e;=",
+        "solution":  "【分析】(1)比较两个整数的大小，首先要看它们的数位，如果数位不同，那么数位多的数就大，如果数位相同，从最高位开始，相同数位上的数大的那个数就大，据此解答。(2)不同单位的名数比较大小，先化成统一的单位，再根据数值的大小进行比较即可。 (3)根据被除数相同，除数越小，商越大，进行判断即可。 (4)根据1平方千米 = 100 公顷，先统一单位，再进行判断即可.\n【详解】根据分析可知:10101010 \u003e 990990954 万 = 540000,504000 \u003c 540000, 504000 \u003c 54 万 480÷25 \u003e 480÷362500 公 顷 = 25 平 方 千 米 ",
+        "knowledge_points":  [
+                                 "亿以内数的改写",
+                                 "公顷，平方千米的进率与换算",
+                                 "除数是两位数的笔算除法",
+                                 "大数的比较"
+                             ],
+        "source":  "24-25 四年级上·河南新乡·期末"
+    },
+    {
+        "id":  "398",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\u0027\u003e\u0027\u0027\u003c\u0027或\u0027=\u0027.\n\n200 公顷( )20 平方千米 123560000( )98070000",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003e;\u003c;=",
+        "solution":  "【分析】1平方千米 = 100 公顷，去掉200公顷末尾的2个0即可换算成平方千米为单位;大数的比较，先看位数的多少，位数多的数就大，位数相同，从高位比较，高位上的数大则这个数大，高位上的数相同，就比较下一位，以此类推，据此填空即可;根据三位数除以两位数的计算，分别计算出左右算式的结果后比较即可;分别计算出左右算式的结果后比较即可.\n【详解】200 公顷 = 2 平方千米, 2 \u003c 20 ,200 公顷\u003c20 平方千米;123560000 是九位数,98070000 是八位数,123560000\u003e98070000;520÷13 = 40 , 800÷16 = 50 , 40 \u003c 50 ,520÷13\u003c800÷16;630 ÷ 35 = 18 , 630 ÷ 7 ÷ 5 = 90 ÷ 5 = 18 ,630÷35=630÷7÷5.200 公顷 \u003c 20 平方千米;123560000\u003e98070000;520÷13\u003c 800÷16;630÷35=630÷7÷5。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "除数是两位数的笔算除法",
+                                 "大数的比较"
+                             ],
+        "source":  "24-25四年级上·河南新乡·期末"
+    },
+    {
+        "id":  "399",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "□75÷75，要使商是一位数,□里最大能填( );要使商是两位数,□里可以填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "6;7,8,9",
+        "solution":  "【分析】根据三位数除以两位数的计算,□75÷75，要使商是一位数，则被除数前两位小于除数,□7\u003c75，据此判断□里最大能填多少;要使商是两位数，则被除数前两位大于或等于除数,□ 7275，据此判断□里可以填哪些数即可.\n【详解】□75÷75，要使商是一位数,□7\u003c75,67\u003c75,□里最大能填 6;要使商是两位数,□ 7275 ,77\u003e75,87\u003e75,97\u003e75,□里可以填 7,8,9。□ 75÷75，要使商是一位数,□里最大能填6;要使商是两位数,□里可以填 7,8,9。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25 四年级上·河南新乡·期末"
+    },
+    {
+        "id":  "400",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "已知 mathtt A / B = 50，如果 A 扩大到原来的 5 倍,B 不变，则商是( );如果A不变,B 扩大到原来的 5 倍，则商是( );如果A和B同时扩大到原来的5倍，则商是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "250;10;50",
+        "solution":  "【分析】商的变化规律:(1)除数不变，被除数扩大到原来的几倍或者缩小为原来的几分之一，商也同样扩大到原来的几倍或者缩小为原来的几分之一;(2)被除数不变，除数扩大到原来的几倍，商反而缩小为原来的几分之一。除数缩小为原来的几分之一，商反而扩大到原来的几倍;(3)被除数和除数同时扩大到原来的几倍或者缩小为原来的几分之一，商不变.\n【详解】已知 mathtt A / B = 50，如果A扩大到原来的5倍,B不变，则商是 50×5 = 250;如果A不变,B扩大到原来的5倍，则商是 50÷5 = 10;如果A和B同时扩大到原来的5倍，则商是 50 。",
+        "knowledge_points":  [
+                                 "商不变的规律及应用",
+                                 "商的变化规律及应用"
+                             ],
+        "source":  "24-25 四年级上·河南新乡·期末"
+    },
+    {
+        "id":  "401",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "已知 A / B = 50，如果A扩大到原来的 5 倍,B不变，则商是( ),如果A不变,B扩大到原来的 5 倍，则商是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "250;10",
+        "solution":  "【分析】根据商的变化规律：被除数扩大到原来的5倍，除数不变，商也扩大相同的倍数。除数扩大到原来的5倍，被除数不变，商反而缩小相同的倍数.\n【详解】 ( A×5 ) \\ / B = 50×5 = 250 A / (B×5) = 50÷5 = 10",
+        "knowledge_points":  [
+                                 "商的变化规律及应用"
+                             ],
+        "source":  "24-25四年级上·河北张家口·期末"
+    },
+    {
+        "id":  "402",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "\u0027鸟巢\u0027的占地面积约为 20 公顷,( )个\u0027鸟巢\u0027的占地面积为 1 平方千米，如果 1 平方米能站 14 人，照这样计算,1 公顷大约能站()人。",
+        "options":  [
+
+                    ],
+        "answer":  "5;140000",
+        "solution":  "【分析】(1)平方千米和公顷之间的进率是 100，要求多少个\u0027鸟巢\u0027的占地面积为1平方千米，用100除以20解答。(2)平方米和公顷之间的进率是 10000，要求 1 公顷大约能站多少人，用 10000 乘 14 解答.\n【详解】根据分析可知:1平方千米 = 100 公顷 100÷20 = 5 (个) 1 公顷 = 10000 平方米14×10000 = 140000 (人)\u0027鸟巢\u0027的占地面积约为 20 公顷,5 个\u0027鸟巢\u0027的占地面积为 1 平方千米，如果 1 平方米能站 14 人，照这样计算,1 公顷大约能站 140000 人。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "公顷，平方千米的进率与换算",
+                                 "除数是整十数的口算除法"
+                             ],
+        "source":  "24-25 四年级上·河南周口·期末"
+    },
+    {
+        "id":  "403",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "张叔叔的果园里收325 千克柑橘，他准备用最多可装 30 千克的箱子进行包装。这些柑橘可以装满( )箱，至少需要准备( )个箱子才能全部装完。",
+        "options":  [
+
+                    ],
+        "answer":  "10;11",
+        "solution":  "【分析】用收的柑橘重量除以每个箱子能装的重量即可，没有余数时，商就是可以装满的箱数和需要准备的箱子数;如果有余数，那么商就是可以装满的箱数，商加1就是需要准备的箱子数.\n【详解】 325÷30 = 10 (箱)。25(千克)10 + 1 = 11 (个)这些柑橘可以装满10箱，至少需要准备11个箱子才能全部装完。",
+        "knowledge_points":  [
+                                 "有余数除法的实际应用",
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "24-25 四年级上·湖南常德·期末"
+    },
+    {
+        "id":  "404",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "240×58 的积是( )位数;735÷68 的商是( )位数。",
+        "options":  [
+
+                    ],
+        "answer":  "五;两",
+        "solution":  "【分析】计算 240×58 的积，由几个数字组成就是几位数;计算 735÷68 的商，看商是几个数字组成.\n【详解】 240×58 = 13920;735÷68=10。55240×58 的积 13920 是五位数;735 ÷ 68 的商 10 是两位数。",
+        "knowledge_points":  [
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "除数是两位数的笔算除法",
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25 四年级上·湖南常德·期末"
+    },
+    {
+        "id":  "405",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "珠海国际航展中心作为珠海航展的主要举办场地，室内展览面积为 120000 平方米，室外展览面积为 230000 平方米。那么它的总展览面积为( )平方米，合( )公顷。大约( )个珠海国际航展中心占地面积约为1平方千米。",
+        "options":  [
+
+                    ],
+        "answer":  "350000;35;3",
+        "solution":  "【分析】用室内展览面积加上室外展览面积，即可求得它的总展览面积，然后根据 1 公顷 = 10000 平方米换算成用公顷作单位;根据1平方千米 = 100 公顷，用100公顷除以一个珠海国际航展中心占地面积即可.\n【详解】 120000 + 230000 = 350000 (平方米)350000 平方米 = 35 公顷 1平方千米 = 100 公顷\n100÷35 approx 3 (个)那么它的总展览面积为350000平方米，合35公顷。大约3个珠海国际航展中心占地面积约为 1 平方千米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "除数是两位数的估算"
+                             ],
+        "source":  "24-25 四年级上·河南郑州·期末"
+    },
+    {
+        "id":  "406",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据 180÷36 = 5 填一填.\n(180×2) ÷ (36×2) = ( ) 180 ÷ (36÷2) = ( )",
+        "options":  [
+
+                    ],
+        "answer":  "5;10",
+        "solution":  "【分析】根据除法性质，被除数和除数同时乘以相同的数(非零),商不变。除数缩小到原来的 frac 12，商则扩大 2 倍。以此答题即可.\n【详解】根据分析可知：根据 180÷36 = 5 填一填如下: (180×2) ÷ (36×2) = 180÷36 = 5180 ÷ (36÷2) = 180÷36×2 = 5×2 = 10 ",
+        "knowledge_points":  [
+                                 "商不变的规律及应用",
+                                 "商的变化规律及应用"
+                             ],
+        "source":  "24-25 四年级上·河南郑州·期末"
+    },
+    {
+        "id":  "407",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在( )或 里填上合适的数或面积单位。(1)9 公顷=( )平方米,500( )= 5 平方千米。 (2)25 □ 8610000 ≈ 25亿,□中最大填( )。 (3)要使 □25÷43 的商是两位数,□中最小填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "90000 公顷 4 5",
+        "solution":  "【分析】(1)根据 1 公顷 = 10000 平方米,100 公顷 = 1 平方千米来化单位即可。(2)一个数，要省略亿位后面的尾数并改写成用\u0027亿\u0027做单位的数，需要看千万位上数的大小，用\u0027四舍五入\u0027法求它的近似数，然后再省略亿位后面的8个0并添上一个\u0027亿\u0027字。(3)三位数除以两位数，判断商是几位数的方法：先观察被除数前两位上的数，看是否大于或等于除数(也就是看前两位上的数除以除数，够不够商一个十)。被除数前两位够除，商就是两位数。反之，商就是一位数.\n【详解】(1)1公顷 = 10000 平方米，所以9公顷 = 90000 平方米;100公顷 = 1 平方千米，所以500公顷 = 5 平方千米。(2)25 □ 8610000≈ 25亿，近似数比原来的数小，说明原数\u0027四舍\u0027后得到 25 亿,□中的数必须比 5 小,□中最大填4.25 □ 8610000 approx 251 Z ,□中最大填4。(3)要使 □ 25÷43 的商是两位数，说明被除数前两位上的数必须大于或等于除数 43，那么□中最小填 5。要使 □ 25÷43 的商是两位数,□中最小填 5。",
+        "knowledge_points":  [
+                                 "亿以上数的近似数",
+                                 "判断商是几位数(除数是两位数)",
+                                 "面积单位间的进率及换算",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25四年级上·河南郑州·期末"
+    },
+    {
+        "id":  "408",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "已知 101×42 =4242;68×101 = 6868，那么 101×86=( );3636÷36 = ( )。",
+        "options":  [
+
+                    ],
+        "answer":  "8686;101",
+        "solution":  "【分析】观察已知条件 101×42 = 4242;68×101 = 6868，可以发现一个规律：一个两位数乘101，结果就是把这个两位数重复写两遍;据此可直接得出 101×86 的结果就是86重复写两遍;同样的 3636÷36 ,3636可以看作是36重复写两遍得到的，所以 3636 就相当于 36×101，据此可解此题.\n【详解】由分析可知，已知 101×42 = 4242;68×101 = 6868，那么 101×86 = 8686 ,3636÷36=101。",
+        "knowledge_points":  [
+                                 "三位数乘两位数，三位数中间有0",
+                                 "除数是两位数的笔算除法",
+                                 "算式的规律(整数)"
+                             ],
+        "source":  "24-25四年级上·湖南永州·期末"
+    },
+    {
+        "id":  "409",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\u0027\u003e\u0027 \u0027\u003c\u0027或\u0027=\u0027 .\n160÷40 ( ) 16÷4600606 ( ) 6000661485760000 ( ) 15亿",
+        "options":  [
+
+                    ],
+        "answer":  "=;\u003e;\u003c",
+        "solution":  "【分析】(1)商不变的规律：被除数和除数同时乘或除以同一个数(0除外),商不变。(2)比较两个整数的大小，要看它们的位数，如果位数不同，那么位数多的数就大;如果位数相同，从高位看起，相同数位上的数大的那个整数就大。(3)根据大数的改写方法,15亿改写时，直接在15后面加上8个0即可，再根据比较两个整数的大小的方法进行比较.\n【详解】根据分析可知: 160÷40 = (160÷10) ÷ (40÷10) = 16÷4600606 \u003e 60006615 亿 = 150000000, 1485760000 \u003c 150000000, 1485760000 \u003c 15 亿 ",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "亿以上数的改写",
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "24-25四年级上·湖南永州·期末"
+    },
+    {
+        "id":  "410",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "若 a/□=5。20，除数最小是( ),此时, a = ( )",
+        "options":  [
+
+                    ],
+        "answer":  "21;125",
+        "solution":  "【分析】除数与余数的关系是：余数都比除数小，除数最小 = 余数+1;再根据被除数 = 商×除数+余数求出被除数即可\n【详解】根据分析可得，在 a/□=5。20 中，除数最小是:20 + 1 = 21;此时被除数是: 5×21 + 20 = 105 + 20 = 125 所以若 a/□=5。20，除数最小是 21，此时, A = 125 ",
+        "knowledge_points":  [
+                                 "余数与除数的关系",
+                                 "有余数的除法验算"
+                             ],
+        "source":  "24-25 四年级上·湖南永州·期末"
+    },
+    {
+        "id":  "411",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算647÷38时，把除数看作( )来试商，商是( )位数;计算 193 ÷ 22 时，把除数看作20来试商，商( )(填\u0027偏大\u0027\u0027偏小\u0027或\u0027正好\u0027)。",
+        "options":  [
+
+                    ],
+        "answer":  "40;两;偏大",
+        "solution":  "【分析】三位数除以两位数：把除数看作和它接近的整十数试商，先用被除数的前两位除以除数，如果被除数前两位比除数大或相等(商是两位数),可以继续除，如果被除数前两位比除数小(商是一位数),就用前三位除以除数，除数看小了，初商就偏大要调小，除数看大了，初商就偏小要调大.\n【详解】计算647÷38时，把除数看作40来试商，被除数647 中 64 \u003e 38，商是两位数;计算 193÷22 时，把除数看作20来试商,20比22小，除数看小了，商偏大。",
+        "knowledge_points":  [
+                                 "多位数除以两位数的试商",
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25四年级上·河南新乡·期末"
+    },
+    {
+        "id":  "412",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果□ 48÷72 的商是一位数,□里最大可以填( );如果商是两位数,□里最小可以填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "6;7",
+        "solution":  "【分析】□48÷72，要使商是一位数，被除数的前两位不够除，即□4\u003c72,□里可以填 1,2,3,4,5,6。□48÷72，要使商是两位数，被除数的前两位够除，即□4\u003e=72,□里可以填 7,8,9.\n【详解】如果□48÷72 的商是一位数,□里最大可以填 6;如果商是两位数,□里最小可以填 7。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25四年级上·安徽安庆·期末"
+    },
+    {
+        "id":  "413",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "要使□57÷58 的商是一位数,□里最大填( ),要使商是两位数,□里最小填()。",
+        "options":  [
+
+                    ],
+        "answer":  "5;6",
+        "solution":  "【分析】根据三位数除以两位数的计算，要使□57÷58 的商是一位数，则被除数前两位小于除数,□5\u003c58，据此判断□里最大填多少;要使商是两位数，则被除数前两位大于或等于除数,□ 5 ge 58，据此判断□里最小填多少.\n【详解】要使□ 57÷58 的商是一位数,□5\u003c58,55\u003c58,□里最大填 5;要使商是两位数,□ 5 ge 58 , 65 \u003e 58 ,□里最小填 6。要使□ 57÷58 的商是一位数,□里最大填5，要使商是两位数,□里最小填 6 ",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25 四年级上·河南新乡·期末"
+    },
+    {
+        "id":  "414",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个数除以 32，商是87，当余数最大时，这个数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "2815",
+        "solution":  "【分析】根据在有余数的除法中，余数总比除数小，即余数最大为：除数-1，当余数最大时，被除数最大，进而根据\u0027被除数 = 商×除数+余数\u0027,即可解答.\n【详解】 32 - 1 = 3187×32 + 31 = 2784 + 31 = 2815 一个数除以32，商是87，当余数最大时，这个数是2815。",
+        "knowledge_points":  [
+                                 "余数与除数的关系",
+                                 "有余数的除法验算",
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "24-25 四年级上·安徽安庆·期末"
+    },
+    {
+        "id":  "415",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "25×400 的积的末尾有( )个 0;658÷65 的商的最高位是( )位。",
+        "options":  [
+
+                    ],
+        "answer":  "4;十",
+        "solution":  "【分析】三位数乘两位数的计算方法：两位数乘三位数，先用两位数个位上的数去乘三位数，得数的末尾和两位数的个位对齐，再用两位数十位上的数去乘三位数，得数的末尾和两位数的十位对齐，然后把两次乘的结果加起来。当乘数末尾有零时，先算零前面的数，再在积的末尾添加对应个数的零。据此计算出算式 25×400 的结果，再进一步解答即可;要判断三位数除以两位数商的位数，只需要将被除数的前两位数与除数进行比较，如果被除数前两位数大于或等于除数，则商是两位数;如果被除数的前两位数小于除数，则商是一位数。据此解答.\n【详解】 25×400 = 10000，积的末尾有 4 个 0;65 = 65 , 658÷65 的商是两位数，最高位是十位。25×400 的积的末尾有4个0;658÷65 的商的最高位是十位。",
+        "knowledge_points":  [
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25 四年级上·安徽安庆·期末"
+    },
+    {
+        "id":  "416",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个乘法算式的积是240，一个因数乘4，另一个因数除以8后，现在的积是()。在一个除法算式中，被除数不变，除数除以 3 以后，商是 60，那么原来的商是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "120;20",
+        "solution":  "【分析】积的变化规律：一个因数不变，另一个因数乘或除以几(0除外),积就乘或除以相同的数;用240乘4再除以 8 即可;商的变化规律：被除数不变，除数除以几(0除外),商就乘上相同的数;除数除以3，商就乘3，得到60，那么用60除以3即可求出原来的商.\n【详解】 240×4÷8 = 960÷8 = 12060÷3 = 20一个乘法算式的积是 240，一个因数乘 4，另一个因数除以8后，现在的积是120。在一个除法算式中，被除数不变，除数除以 3 以后，商是 60，那么原来的商是 200 ",
+        "knowledge_points":  [
+                                 "商的变化规律及应用",
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "24-25四年级上·河南周口·期末"
+    },
+    {
+        "id":  "417",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小新在学习三位数除以两位数的除法时，有了新发现:\u0027如果被除数的末尾有 0，商的末尾也一定有 0\u0027。你同意小新的发现吗。请在横线上打\u0027√\u0027。 同意 (),不同意() 请举出三位数除以两位数的例子说明你的理由:() 。",
+        "options":  [
+
+                    ],
+        "answer":  "不同意√;100÷20 = 5;500÷25 = 20",
+        "solution":  "【分析】被除数末尾有0，商的末尾是否一定有0，取决于除数和被除数的其他位数上的数。例如，如果被除数是100，除数是 20，那么商是 5，商的末尾没有 0。但是，如果被除数是 500，除数是 25，那么商是 20，商的末尾有 0。因此，据此解答.\n【详解】由分析可得：被除数末尾有0，商的末尾可能有0;所以不同意小新的发现。例如:100÷20 = 5500÷25 = 20即同意(),不同意√;三位数除以两位数的例子如下: 100÷20 = 5;500÷25=20。(举例不唯一)",
+        "knowledge_points":  [
+                                 "被除数和除数末尾都有0的除法",
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "24-25四年级上·河南郑州·期末"
+    },
+    {
+        "id":  "418",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小新在计算一道除法算式时把除数 36 写成了 39，结果算出的商是 18，余数是21，正确的商是() ,余数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "20;3",
+        "solution":  "【分析】根据被除数 = 除数×商+余数，把错误的除法39，商18，余数21，代入式子中算出正确的被除数，再根据被除数/除数 = 商。余数，这时除数是36计算.\n【详解】 39×18 + 21= 702 + 21 = 723723÷36 = 20 dots 3 正确的商是20，余数是3。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "有余数的除法验算"
+                             ],
+        "source":  "24-25 四年级上·河南郑州·期末"
+    },
+    {
+        "id":  "419",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位数除以 50，商是一个两位数，余数是 18，这个三位数最小是，这个三位数最大是",
+        "options":  [
+
+                    ],
+        "answer":  "518;968",
+        "solution":  "【分析】要求这个三位数最小是多少，此时商最小，最小的两位数是10，根据被除数 = 除数×商+余数，求出最小的三位数;最大的三位数是 999，除以 50，求得商最大是多少，然后求得这个三位数最大是多少，代入数据计算.\n【详解】一个三位数除以50，商是一个两位数，商最小是10，最大的三位数是999,999÷50=19。49，商最大是19, 50×10 + 18 = 500 + 18 = 51850×19 + 18 = 950 + 18 = 968 所以一个三位数除以50，商是一个两位数，余数是18，这个三位数最小是 518，这个三位数最大是 968 。",
+        "knowledge_points":  [
+                                 "三位数除以两位数的有余数除法",
+                                 "有余数的除法验算"
+                             ],
+        "source":  "24-25 四年级上·河南郑州·期末"
+    },
+    {
+        "id":  "420",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "如果算式\u0027628÷□4\u0027的商是两位数,□里最大填( )。如果算式\u0027△57÷36\u0027的商是一位数,△里最大填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "5;3",
+        "solution":  "【分析】除数是两位数，被除数是三位数，要使商是两位数，那么被除数的前两位数要大于等于除数;要使商是位数，那么被除数的前两位数要小于除数。据此解答.\n【详解】如果628÷□4的商是两位数, 622 □ 4 ,□里面可以填 5,4,3,2,1，最大填 5;如果△57÷36的商是一位数, △ 5 \u003c 36 ,□里面可以填1,2,3，最大填 3. ",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25 四年级上·河南洛阳·期末"
+    },
+    {
+        "id":  "421",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一本词典需 37 元，王老师带 376 元钱，最多能买( )本这样的词典。",
+        "options":  [
+
+                    ],
+        "answer":  "10",
+        "solution":  "【分析】总价/单价 = 数量，用 376 除以 37 计算出的商，为最多能买到的本数;据此解答.\n【详解】根据分析:376÷37 = 10 (本)。6(元)所以最多能买 10 本这样的词典。",
+        "knowledge_points":  [
+                                 "三位数除以两位数的有余数除法",
+                                 "有余数除法的实际应用",
+                                 "经济问题"
+                             ],
+        "source":  "24-25 四年级上·河南洛阳·期末"
+    },
+    {
+        "id":  "422",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "两个数的积是 360，如果一个因数不变，另一个因数除以 10，则积是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "36",
+        "solution":  "【分析】积的变化规律：一个因数不变，另一个因数乘(或除以)几(0 除外),积也乘(或除以)几;据此解答.\n【详解】根据分析：如果一个因数不变，另一个因数除以10，那么积也要除以 10, 360÷10 = 36，所以积是 36。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)",
+                                 "除数是整十数的口算除法"
+                             ],
+        "source":  "24-25 四年级上·河南洛阳·期末"
+    },
+    {
+        "id":  "423",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一本少儿版西游记有 245 页，小明用了一个星期刚好看完，平均每天看( )页，如果平均每天看 49 页，需要( )天才能看完。",
+        "options":  [
+
+                    ],
+        "answer":  "35;5",
+        "solution":  "【分析】一个星期有7天，用245除以7可以计算出平均每天看多少页;用 245 除以 49 可以计算出，平均每天看 49 页的情况下，需要多少天能看完;据此解答.\n【详解】根据分析:245÷7 = 35 (页)245÷49 = 5 (天)所以小明用了一个星期刚好看完，平均每天看 35 页，如果平均每天看 49 页，需要 5 天才能看完。",
+        "knowledge_points":  [
+                                 "三位数除以一位数，首位不够除",
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "24-25 四年级上·河南洛阳·期末"
+    },
+    {
+        "id":  "424",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "要使□ 46÷27 的商是两位数,□里最小可以填( );要使 318÷3 □ 的商是两位数,□里最大可以填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "3;1",
+        "solution":  "【分析】三位数除以两位数，判断商是几位数，首先看三位数的前两位，如果大于等于除数，商是两位数，如果小于除数，商是一位数。要使□46÷27的商是两位数，说明□4\u003e=27,□里最小可以填3;要使318÷3□的商是两位数,31\u003e=3□,□里最大可以填1。据此解题.\n【详解】要使□46÷27 的商是两位数,□里最小可以填 3;要使 318÷3□的商是两位数,□里最大可以填 1。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25 四年级上·湖南岳阳·期末"
+    },
+    {
+        "id":  "425",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "括号里最大能填几。\n50×( ) \u003c 210 ( )×78 \u003c 65049×( ) \u003c 357",
+        "options":  [
+
+                    ],
+        "answer":  "4;8;7",
+        "solution":  "【分析】用小于号后面的数除以前面已知因数，有余数的，商是最大能填的数，整除的，商减1就是最大能填的数.\n【详解】210÷50=4。10，所以括号里最大填 4;650÷78=8。26，所以括号里最大填 8;357÷49=7。14，所以括号里最大填 7。综上可知，括号里最大能填的数为: 50×(4)\u003c210,(8)×78\u003c650,49×(7)\u003c357。",
+        "knowledge_points":  [
+                                 "三位数除以两位数的有余数除法",
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "24-25 四年级上·安徽芜湖·期末"
+    },
+    {
+        "id":  "426",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个婴儿的心脏平均每分钟跳动 123 次，一个成人的心脏平均每分钟跳动 64次。估一估，这个婴儿心脏平均每分钟跳动次数大约是这个成人的( )倍。",
+        "options":  [
+
+                    ],
+        "answer":  "2",
+        "solution":  "【分析】由题意得，一个婴儿的心脏平均每分钟跳动123次，一个成人的心脏平均每分钟跳动64次。求这个婴儿心脏平均每分钟跳动次数大约是这个成人的多少倍，用除法计算。估算时，可以把123估成120，把64估成60，然后直接口算出结果即可.\n【详解】 123÷64 approx 120÷60 = 2 故这个婴儿心脏平均每分钟跳动次数大约是这个成人的 2 倍。",
+        "knowledge_points":  [
+                                 "除数是两位数的估算"
+                             ],
+        "source":  "24-25 四年级上·山东济南·期末"
+    },
+    {
+        "id":  "427",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小军家到学校的路程是 720 米，一天，他上学用了 9 分钟，放学比上学多用了3 分钟，他放学回家的平均速度是( )米/分。",
+        "options":  [
+
+                    ],
+        "answer":  "60",
+        "solution":  "【分析】小军上学所用时间加上 3 分钟，即可算出小军放学回家用了(9+3)分钟;速度 = 路程/时间，小军家到学校的路程除以他放学所用时间，即可算出他放学回家的平均速度是几米/分.\n【详解】 9 + 3 = 12 (分钟) 720÷12 = 60 (米/分) 小军家到学校的路程是720米，一天，他上学用了9分钟，放学比上学多用了3分钟，他放学回家的平均速度是60米/分。",
+        "knowledge_points":  [
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25 四年级上·山东济南·期末"
+    },
+    {
+        "id":  "428",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据算式 828÷18 = 46，直接写出下面各算式的结果.\n8280÷18 = ( ) 82800÷460 = ( )",
+        "options":  [
+
+                    ],
+        "answer":  "460;180",
+        "solution":  "【分析】第一题根据商的变化规律，被除数乘10，除数不变，商也乘 10;第二题先根据 828÷18 = 46 可以转化为 828÷46 = 18，转化后的式子与 82800÷460 比较，被除数乘100，除数乘10，商先乘100(因为被除数乘100),再除以10(因为除数乘10) .\n【详解】 8280÷18 = (828×10) ÷ 18 = 46×10 = 460，那么 8280÷18 = 460;828÷18 = 46 变为 828÷46 = 18 , 82800÷460 = (828×100) ÷ (46×10) = 18×100÷10 = 1800÷10 = 180，那么 82800÷460 = 180 。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "商的变化规律及应用"
+                             ],
+        "source":  "24-25 四年级上·山东济南·期末"
+    },
+    {
+        "id":  "429",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "观察算式□86÷58，如果商是一位数,□里最大填;如果商是两位数,□里最小填 。",
+        "options":  [
+
+                    ],
+        "answer":  "4;5",
+        "solution":  "【分析】三位数除以两位数，当被除数的前两位比除数大或者等于除数，商是两位数。当被除数的前两位比除数小，商是一位数，据此解答.\n【详解】□86÷58，商是一位数，说明□ 8 \u003c 58 ,□里可以填1,2,3,4，最大填4。商是两位数,□8大于或者等于5,□里可以填 5,6,7,8,9，最小是 5。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25四年级上·北京石景山·期末"
+    },
+    {
+        "id":  "430",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "775÷25 的商的最高位是( )位, 738÷82 的商是( )位数。",
+        "options":  [
+
+                    ],
+        "answer":  "十;一",
+        "solution":  "【分析】三位数除以两位数，被除数前两位的数大于或等于除数时，商是两位数，最高位是十位;被除数前两位的数小于除数时，商是一位数，最高位是个位.\n【详解】 77 \u003e 25 , 75÷25 的商的是两位数，最高位是十位;73\u003c82,738÷82 的商是一位数。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25 四年级上·安徽芜湖·期末"
+    },
+    {
+        "id":  "431",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个小足球 30 元,95 元钱可以买( )个小足球。",
+        "options":  [
+
+                    ],
+        "answer":  "3",
+        "solution":  "【分析】95元里面最多有几个30元，用除法计算，商表示可以最多买的个数，余数表示剩余的价钱.\n【详解】 95÷30 = 3 (个)。5(元)95 元钱可以买 3 个小足球。",
+        "knowledge_points":  [
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "24-25 四年级上·广东珠海·期末"
+    },
+    {
+        "id":  "432",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "有298个苹果，每 32 个装箱，装完这些苹果至少需要( )个箱子。",
+        "options":  [
+
+                    ],
+        "answer":  "10",
+        "solution":  "【分析】根据题意，将苹果总数除以每箱装的苹果数即可得到需要多少个箱子，若有余数，还需要用商加上1，据此解答.\n【详解】根据分析可知:298÷32 = 9 (个)。10(个) 9 + 1 = 10 (个)有298个苹果，每32个装箱，装完这些苹果至少需要10个箱子。",
+        "knowledge_points":  [
+                                 "有余数除法的实际应用",
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "23-24四年级下·河南南阳·开学考试"
+    },
+    {
+        "id":  "433",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "淘淘是个小马虎，在计算一道除法算式时，把除数 18 看成了 81，结果得到商是 8，请问这道题的正确结果是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "36",
+        "solution":  "【分析】由题意得，淘淘在计算一道除法算式时，把除数18看成了81，结果得到商是8，那么直接用8乘81先算出正确的被除数。然后再除以18即可算出正确的结果.\n【详解】 8×81 = 648648÷18 = 36 即这道题的正确结果是 36",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "两，三位数与一位数的一次进位乘法"
+                             ],
+        "source":  "24-25 四年级上·山东济南·期末"
+    },
+    {
+        "id":  "434",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "□42÷54的商是一位数,□里最大能填( );如果商是两位数，那么□里最小能填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "4;5",
+        "solution":  "【分析】□42÷54的商是一位数，被除数的前两位不够除，即□4\u003c54,□里可以填1,2,3,4。□42÷54的商是两位数，被除数的前两位够除，即□ 4 ge 54 ,□里可以填5,6,7,8,9.\n【详解】□42÷54的商是一位数,□里最大能填4;如果商是两位数，那么□里最小能填5。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25四年级上·山东济南·期末"
+    },
+    {
+        "id":  "435",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在算式☐/△=23。5 中，如果被除数和除数同时乘 100，那么商是( ),余数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "23;500",
+        "solution":  "【分析】根据商不变的规律，被除数和除数同时乘或除以相同的数(0除外),商不变。余数就乘或除以相同的数(0 除外)。 \n【详解】已知☐/△=23。5，得: (☐ * 100) ÷ (△ * 100) = 23 dots 。 500 所以 和☐ △同时乘 100，那么商是 23，余数是 500 ",
+        "knowledge_points":  [
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "24-25 四年级上·山东济南·期末"
+    },
+    {
+        "id":  "436",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "要使算式□ 42÷64 的商是两位数,□里最小填( ),要使商是一位数,□里最大填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "6;5",
+        "solution":  "【分析】三位数除以两位数商的位数：如果被除数的前两位大于或等于除数，那么商的最高位是十位，商是两位数;如果被除数的前两位小于除数，那么商的最高位是个位，商就是一位数;据此进行计算即可解答此题(大于或等于可用\u003e=表示) .\n【详解】要使☐42÷64的商是两位数，则□ 4 ge 64，所以□里可以填6,7,8,9,□里最小填(6);要使☐ 42÷64 的商是一位数，则□4\u003c64，所以□里可以填5,4,3,2,1,□里最大填(5)。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25四年级上·山东济南·期末"
+    },
+    {
+        "id":  "437",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "□ 29÷42，要使商是两位数,□里最小填;要使商是一位数,□里最大填 ()。",
+        "options":  [
+
+                    ],
+        "answer":  "4;3",
+        "solution":  "【分析】三位数除以两位数，百位和十位上的组成数和除数比较，如果比除数大或相等，商就是两位数;如果比除数小，商就是一位数。□29÷42，要使商是两位数，那么被除数的百位和十位构成的数就要比除数42大或相等，所以可以填 4,5,6,7,8,9,□里最小填 4;要使商是一位数，那么被除数的百位和十位构成的数就要比除数42小，所以可以填1,2,3,□里最大填3;据此解答即可.\n【详解】根据分析:□29÷42，要使商是两位数,□里最小填 4;要使商是一位数,□里最大填 3",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25四年级上·山东济南·期末"
+    },
+    {
+        "id":  "438",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "△÷☆=18。15，则☆最小是 (),如果△和☆同时乘10，那么商是() ,余数是() 。",
+        "options":  [
+
+                    ],
+        "answer":  "16;18;150",
+        "solution":  "【分析】在有余数的除法中，余数总比除数小，即除数最小为：余数+1。在有余数的除法算式里，被除数和除数同时乘或除以一个不为0的数，那么商不变，余数也要乘或除以这个相同的数，据此解答即可.\n【详解】 15 + 1 = 16 △÷☆=18。15，则☆最小是 16，如果△和☆同时乘 10，那么商是 18，余数是 150",
+        "knowledge_points":  [
+                                 "余数与除数的关系",
+                                 "三位数除以两位数的有余数除法",
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "24-25 四年级上·山东济南·期末"
+    },
+    {
+        "id":  "439",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "油旋是山东济南的传统特色名吃，外皮酥脆，葱香透鼻，因其形似螺旋，表面油润呈金黄色，故名油旋。如果210克面粉可以做3个油旋，照这样计算,840克面粉可以做 个油旋。",
+        "options":  [
+
+                    ],
+        "answer":  "12",
+        "solution":  "【分析】先用210克除以3求出1个油旋需要的面粉重量，再用 840 克除以 1 个油旋需要的面粉重量即可求解.\n【详解】840÷(210÷3= 840 ÷ 70= 12 (个) 840克面粉可以做12个油旋。",
+        "knowledge_points":  [
+                                 "除数是整十数的笔算除法",
+                                 "单归一问题"
+                             ],
+        "source":  "24-25 四年级上·山东济南·期末"
+    },
+    {
+        "id":  "440",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "792÷36，商是()位数。",
+        "options":  [
+
+                    ],
+        "answer":  "两",
+        "solution":  "【分析】三位数除以两位数商的位数：如果被除数的前两位大于或等于除数，那么商的最高位是十位，商是两位数;如果被除数的前两位小于除数，那么商的最高位是个位，商就是一位数.\n【详解】根据题意可知，被除数的前两位79大于除数36，所以, 792÷36，商的最高位是十位，商是两位数。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25四年级下·新疆喀什·开学考试"
+    },
+    {
+        "id":  "441",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "\u0027525÷15\u0027的商是( )位数，商的最高位是( )位。",
+        "options":  [
+
+                    ],
+        "answer":  "两;十",
+        "solution":  "【分析】三位数除以两位数，判断商是几位数的方法：先观察被除数前两位上的数，看是否大于或等于除数(也就是看前两位上的数除以除数，够不够商一个十)。被除数前两位够除，商就是两位数。反之，商就是一位数。据此解答.\n【详解】\u0027 525÷15 \u0027 ,被除数前两位上的数52大于除数15，所以商是两位数，商的最高位是十位。 ",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25 四年级下·河南南阳·开学考试"
+    },
+    {
+        "id":  "442",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "甲乙两地相距 1620 千米，一列火车从甲地开往乙地,3 小时行驶 540 千米。照这样的速度计算，共行驶( )小时可以到达乙地。",
+        "options":  [
+
+                    ],
+        "answer":  "9",
+        "solution":  "【分析】先根据\u0027速度 = 路程/时间\u0027求火车每小时行多少千米，再根据\u0027时间 = 路程/速度\u0027求共行驶几小时可以到达乙地.\n【详解】 540÷3 = 180 (千米/时) 1620÷180 = 9 (小 时) 即，甲乙两地相距 1620 千米，一列火车从甲地开往乙地,3小时行驶 540 千米。照这样的速度计算，共行驶 9 小时可以到达乙地。",
+        "knowledge_points":  [
+                                 "被除数和除数末尾都有 0 的除法",
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25 四年级下·新疆喀什·开学考试"
+    },
+    {
+        "id":  "443",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "水果店要将 130千克苹果装入纸箱，每个纸箱最多能装 15 千克，至少需要()个这样的纸箱。",
+        "options":  [
+
+                    ],
+        "answer":  "9",
+        "solution":  "【分析】用苹果的总重量除以每个纸箱能装苹果的重量求出商，商就是需要的纸箱数量，如果有余数，表示还有苹果需要装入纸箱，需要再多一个箱子来装.\n【详解】 130÷15 = 8 (个)。10(千克) 8 + 1 = 9 (个) 至少需要 9 个这样的纸箱。",
+        "knowledge_points":  [
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "24-25 四年级下·新疆喀什·开学考试"
+    },
+    {
+        "id":  "444",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在算式4□2÷45中，要使商是一位数,□最大是( );要使商是两位数,□最小是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "4;5",
+        "solution":  "【分析】三位数除以两位数，被除数的前两位比除数小，商是一位数;被除数的前两位大于或等于除数，商是两位数;据此分析填空即可.\n【详解】在算式 4□2÷45 中，要使商是一位数，则 4 □\u003c 45，即□最大是4;要使商是两位数，则 4 □ ge 45，则□最小是5。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25四年级下·新疆喀什·开学考试"
+    },
+    {
+        "id":  "445",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "150÷30 = 15 ÷ ( )=( )÷300=450÷( )。",
+        "options":  [
+
+                    ],
+        "answer":  "3;1500;90",
+        "solution":  "【分析】商不变规律：被除数和除数同时乘或除以同一个 150÷30 = (150×10) ÷ (30×10) = 1500÷300, 150÷30 = (150×3) ÷ (30×3) = 450÷90 。 \n【详解】由分析得:150÷30 = 15÷3 = 1500÷300 = 450÷90 。",
+        "knowledge_points":  [
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "24-25 四年级下·新疆喀什·开学考试"
+    },
+    {
+        "id":  "446",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一辆小汽车行驶的速度是 950 米/分，这辆小汽车 1 小时行驶( )千米，照这样的速度计算，这辆小汽车行驶 456 千米，需要行驶( )小时。",
+        "options":  [
+
+                    ],
+        "answer":  "57;8",
+        "solution":  "【分析】1小时 = 60 分，根据速度×时间 = 路程，用这辆小汽车每分行驶的路程乘60，求出这辆小汽车每小时行驶的路程是多少，然后根据 1 千米 = 1000 米，把米转化为千米即可求出这辆小汽车1小时行驶多少千米;然后根据公式：时间 = 路程/速度，计算其行驶 456 千米所需时间即可.\n【详解】1 小时 = 60 分 950×60 = 57000 (米) 57000 米 = 57 千米 456÷57 = 8 (小时)辆小汽车行驶的速度是950米/分，这辆小汽车1小时行驶 57 千米，照这样的速度计算，这辆小汽车行驶 456 千米，需要行驶8小时。",
+        "knowledge_points":  [
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "除数是两位数的笔算除法",
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25四年级上·广东汕头·期末"
+    },
+    {
+        "id":  "447",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一道除法算式的商是 26，如果除数和被除数同时乘 10，则商是( );如果被除数不变，除数除以3，则商是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "26;78",
+        "solution":  "【分析】商不变的规律：被除数和除数都乘(或除以)个相同的数(0 除外),商不变;被除数不变，商随除数变化的规律：除数乘(或除以)几(0 除外),商反而除以(或乘)几;据此解答即可.\n【详解】一道除法算式的商是26，如果除数和被除数同时乘10，则商不变，商仍然是26;如果被除数不变，除数除以 3，则商反而乘 3，那么商是 26×3 = 78 。",
+        "knowledge_points":  [
+                                 "商的变化规律及应用",
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "24-25四年级上·广东汕头·期末"
+    },
+    {
+        "id":  "448",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "笔算 960÷68 时，要把 68 看做( )来试商，商的最高位在( )位上，这道算式的余数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "70;十;8",
+        "solution":  "【分析】根据三位数除以两位数的计算法则，通常把除数看作接近的整十数去试商，先用除数去除被除数的前两位，前两位不够除，再看前三位，除到哪一位就把商写在那一位上面，除到被除数中间或末尾一位不够商1要商0，每步除得的余数都要比除数小;所以，笔算960÷68时，通常把68看作接近的整十数70来试商，由于被除数前两位96比70大，够除，所以商的最高位在十位上;根据法则计算出算式的结果，即可解答.\n【详解】 960÷68 = 14 。 。 。 。 。 8所以，笔算 960÷68 时，要把68看做70来试商，商的最高位在十位上，这道算式的余数是 8。",
+        "knowledge_points":  [
+                                 "三位数除以两位数的有余数除法",
+                                 "多位数除以两位数的试商",
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25四年级上·广东汕头·期末"
+    },
+    {
+        "id":  "449",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据 168÷12 = 14，写出下面各题的商.\n1680÷120 = ( ) 168÷24 = ( )",
+        "options":  [
+
+                    ],
+        "answer":  "14;7",
+        "solution":  "【分析】商的不变规律：在除法算式中，被除数和除数同时乘或除以一个数(0 除外),商不变;商的变化规律：除数不变，被除数乘(或除以)几(0 除外),商也乘(或除以)相同数;被除数不变，除数乘(或除以)几(0 除外),商除以(或乘)相同数;据此解答.\n【详解】被除数 168 乘 10 变为 1680，除数 12 乘 10 变为120，所以商不变，那么 1680÷120 = 14 。被除数不变，除数 12 乘 2 变为 24，那么商除以 2, 14÷2 = 7，所以 168÷120 = 7 。",
+        "knowledge_points":  [
+                                 "商的变化规律及应用",
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "24-25 四年级上·广东汕头·期末"
+    },
+    {
+        "id":  "450",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在计算\u0027 240÷26 prime prime \u0027时，想很快的试出商。试商过程是:( )。",
+        "options":  [
+
+                    ],
+        "answer":  "240 除以 26，先在个位上商 9, 26×9 = 234 ,240-234 = 6 , 6 \u003c 26，所以 240÷26 = 9 。 ldots 6",
+        "solution":  "【分析】\u0027同头无除商九，八\u0027是指被除数与除数首位上的数字相同(俗称\u0027同头\u0027),但被除数第2位上的数小于除数第2位上的数，不够商1(俗称\u0027无除\u0027),那就可以在下一位上用9或8试商;240÷26 符合此方法，据此即可快速试出商来.\n【详解】根据\u0027同头无除商八，九\u0027可知,240除以26，先在个位上商 9, 26×9 = 234 ,240-234=6, 6 \u003c 26，所以240÷26=9。6。",
+        "knowledge_points":  [
+                                 "多位数除以两位数的试商"
+                             ],
+        "source":  "23-24四年级下·河南南阳·开学考试"
+    },
+    {
+        "id":  "451",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "两个数相除的商是70，如果被除数不变，除数扩大到原来的5倍，商是( );如果除数不变，被除数扩大到原来的 5 倍，商是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "14;350",
+        "solution":  "【分析】根据商的变化规律，在除法算式中，除数不变，被除数乘(或除以)几，商也乘(或除以)几(0除外);被除数不变，除数乘(或除以)几，商反而要除以(或乘)几(0除外),据此解答即可.\n【详解】70÷5=14;70×5=350。两个数相除的商是70，如果被除数不变，除数扩大到原来的5倍，商是14;如果除数不变，被除数扩大到原来的5倍，商是350。",
+        "knowledge_points":  [
+                                 "商的变化规律及应用"
+                             ],
+        "source":  "23-24四年级下·河南南阳·开学考试"
+    },
+    {
+        "id":  "452",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填\">\"\"<\"或\"=\"。\n329÷12( )168÷12 260000( )26万 40040( )101×400\n925÷75( )925÷55 19600( )60×190 42÷2( )420÷20",
+        "options":  [
+
+                    ],
+        "answer":  ">;=;<;<;>;=",
+        "solution":  "【分析】(1)在计算除法算式时，除数相同，被除数越大，则商就越大。(2)把整万的数改写成用\"万\"作单位的数，只要省略万位后面的0，再加上\"万\"字。(3)先计算乘法算式，再比较。(4)在计算除法算式时，被除数相同，除数越大，则商就越小。(5)先计算乘法算式，再比较。(6)商不变的规律：被除数和除数都乘或除以一个相同的数(0除外),商不变。\n【详解】329>168，329÷12>168÷12；260000=26万；101×400=40400，40040<40400，所以40040<101×400；75>55，所以925÷75<925÷55；60×190=11400，19600>11400，所以19600>60×190；420÷20=(420÷10)÷(20÷10)=42÷2，所以42÷2=420÷20。",
+        "knowledge_points":  [
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "除数是两位数的笔算除法",
+                                 "亿以内数的改写"
+                             ],
+        "source":  "23-24四年级下·河南南阳·开学考试"
+    },
+    {
+        "id":  "453",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在 A÷30=14……B 中，余数B最大是( )，这时被除数 A 是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "29;449",
+        "solution":  "【分析】在有余数的除法中，余数要比除数小，所以最大比除数小1，再根据被除数=除数×商+余数。把B=29代入式子中计算。\n【详解】30-1=29，30×14+29=420+29=449。在A÷30=14……B中，余数B最大是29，这时被除数A是449。",
+        "knowledge_points":  [
+                                 "余数与除数的关系",
+                                 "有余数的除法验算",
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "24-25 四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "454",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "12 箱蜜蜂一年可以酿蜜 900 千克。小林家也养了这样的蜜蜂，一年酿 600 千克蜂蜜，小林家养了( )箱蜜蜂。",
+        "options":  [
+
+                    ],
+        "answer":  "8",
+        "solution":  "【分析】先用900÷12求出每箱蜜蜂一年酿多少千克蜂蜜，再用600除以每箱的产量，求出箱数。\n【详解】900÷12=75(千克)，600÷75=8(箱)。所以小林家养了8箱蜜蜂。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "单归一问题"
+                             ],
+        "source":  "24-25 四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "455",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "刘叔叔计划带3500元批发大米，结果买了 14 袋大米花了 980 元。每袋大米的价钱是( )元;照这样计算，剩下的钱还能买( )袋大米。",
+        "options":  [
+
+                    ],
+        "answer":  "70;36",
+        "solution":  "【分析】根据题意，已知买了14袋大米花了980元，单价=总价÷数量，计算每袋大米的价格列式是980÷14；数量=总价÷单价，用3500减去980，就是剩下的钱数，再除以每袋大米的价格，就是可以买的大米的袋数。列式计算即可。\n【详解】980÷14=70(元)；3500-980=2520(元)；2520÷70=36(袋)。刘叔叔计划带3500元批发大米，结果买了14袋大米花了980元。每袋大米的价钱是70元；照这样计算，剩下的钱还能买36袋大米。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "两，三位数的连续退位减法",
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "24-25四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "456",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "妈妈去\"七若滋\"蛋糕店买月饼。所购买的月饼每个 15 元，店里的迎中秋优惠活动是\"每满四送一\"。妈妈用 200 元最多可以买( )个月饼。",
+        "options":  [
+
+                    ],
+        "answer":  "16",
+        "solution":  "【分析】先算出不考虑优惠时200元能买多少个月饼，再根据\"每满四送一\"的优惠，算出送的月饼数，最后将两者相加得到总共能买的月饼数。\n【详解】200÷15=13(个)……5(元)，13÷4=3(组)……1(个)，3×1=3(个)，13+3=16(个)。所以妈妈用200元最多可以买16个月饼。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "24-25四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "457",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "要使5□6÷57的商是两位数,□里最小可填( );要使商是一位数,□里可以填数的有( )个。",
+        "options":  [
+
+                    ],
+        "answer":  "7;7",
+        "solution":  "【分析】根据三位数除以两位数的除法计算方法可知，要使5□6÷57的商是两位数，则被除数前两位上的数大于等于除数，即□里的数要大于等于7；要使5□6÷57的商是一位数，则被除数前两位上的数小于除数，□里的数要小于7；据此解答。\n【详解】要使5□6÷57的商是两位数，则5□≥57，□里可以填7、8、9，最小可填7；要使商是一位数，则5□<57，□里可以填0、1、2、3、4、5、6，有7个。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "458",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "明明用电子秤称出了 100 粒米的质量是 3 克，这样的 1 亿粒米的质量是( )千克。",
+        "options":  [
+
+                    ],
+        "answer":  "3000",
+        "solution":  "【分析】根据题意，先用除法求出1亿里面有几个100，则1亿粒米的质量就是几个3克。1千克=1000克，据此换算单位。\n【详解】1亿=100000000，100000000÷100×3=1000000×3=3000000(克)，3000000克=3000千克。所以明明用电子秤称出了100粒米的质量是3克，这样的1亿粒米的质量是3000千克。",
+        "knowledge_points":  [
+                                 "被除数和除数末尾都有0的除法",
+                                 "克，千克之间的换算与比较",
+                                 "亿以上数的改写"
+                             ],
+        "source":  "24-25四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "459",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "求\"5□7÷54\"的商。(1)如果商是两位数,□里最小填( )。 (2)如果商是一位数,□里最大填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "4;3",
+        "solution":  "【分析】三位数除以两位数，判断商是几位数的方法：先观察被除数前两位上的数，看是否大于或等于除数(也就是看前两位上的数除以除数，够不够商一个十)。被除数前两位够除，商就是两位数。反之，商就是一位数。(1)5□7÷54，要使商是两位数，那么被除数前两位上的数大于或等于54，□里最小填4。(2)5□7÷54，要使商是一位数，那么被除数前两位上的数小于54，□里最大填3。\n【详解】(1)如果商是两位数，□里最小填4。(2)如果商是一位数，□里最大填3。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25四年级下·河南南阳·开学考试"
+    },
+    {
+        "id":  "460",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据 368÷16=23，直接写出下面各算式的得数。\n368÷8 = ( ) 736÷32 = ( ) 230×16 = ( )",
+        "options":  [
+
+                    ],
+        "answer":  "46;23;3680",
+        "solution":  "【分析】根据商的变化规律和积的变化规律直接求解。\n【详解】368÷8中，被除数不变，除数由16变成8，除以2，所以商乘2，得46;736÷32中，被除数和除数都乘2，商不变，得23;由368÷16=23可得23×16=368，230×16=3680。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)",
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "24-25 四年级下·河南南阳·开学考试"
+    },
+    {
+        "id":  "461",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在一个除法算式中，商是 9，余数是 23，除数最小是( )，这时被除数是( );在另一个除法算式中，被除数是 182，商是 7，余数是14，这时除数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "24;239;24",
+        "solution":  "【分析】已知商是9，余数是23，根据余数要比除数小，据此得出除数的最小值，然后根据商乘除数再加余数，据此求出被除数；已知被除数是182，商是7，余数是14，用被除数减去余数所得的结果再除以商，据此列式可求出除数。\n【详解】余数是23，除数最小是24；9×24+23=216+23=239；(182-14)÷7=168÷7=24。综上可知，商是9，余数是23，除数最小是24，这时被除数是239；被除数是182，商是7，余数是14，这时除数是24。",
+        "knowledge_points":  [
+                                 "余数与除数的关系",
+                                 "三位数除以一位数，首位不够除",
+                                 "除数是两位数的笔算除法",
+                                 "乘加，乘减混合运算"
+                             ],
+        "source":  "24-25 四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "462",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\">\"\"<\"或\"=\"。\n756420( )756542 70万( )690000 13亿( )1300000000\n2公顷( )2000平方米 150平方千米( )15003公顷 19×300( )30×180\n90÷18( )180÷36 425÷45( )425÷35",
+        "options":  [
+
+                    ],
+        "answer":  "<;>;=;>;<;>;=;<",
+        "solution":  "【分析】(1)根据大数比较大小的方法：位数相同的两个数比较大小，从最高位比起，最高位上的数大的那个就大，如果最高位上的数相同，就比较下一个数位上的数。位数不同的两个数比较大小，位数多的数就大。(2)(3)根据整数的写法：从高位到低位按顺序写，哪一个数位上一个单位也没有，就在那个数位上写0；分别将70万和13亿写成整数，然后根据大数的比较方法比较大小即可。(4)(5)1平方千米=100公顷，1公顷=10000平方米，依此将单位化统一后再比较。(6)分别计算出式子的结果，再比较大小即可。(7)根据商不变的规律：被除数和除数同时乘或除以同一个数(0除外),商不变。(8)被除数相同，除数越大，结果越小；据此解答。\n【详解】756420<756542；70万=700000，700000>690000，所以70万>690000；13亿=1300000000，所以13亿=1300000000；2公顷=20000平方米，20000>2000，所以2公顷>2000平方米；150平方千米=15000公顷，15000<15003，所以150平方千米<15003公顷；19×300=5700，30×180=5400，5700>5400，所以19×300>30×180；90÷18=180÷36；45>35，所以425÷45<425÷35。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "公顷，平方千米的进率与换算",
+                                 "亿以上数的改写",
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "24-25 四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "463",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "208÷42，把 42看作( )来试商，商 5 大了，改商为( )。",
+        "options":  [
+
+                    ],
+        "answer":  "40;4",
+        "solution":  "【分析】除法计算中，把除数看大了，商就容易偏小，调商时要调大一点;把除数看小了，商就容易偏大，调商时要调小一点，据此解答。\n【详解】由题意分析得，208÷42，把42看作40来试商，商5大了，改商为4。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "24-25 四年级下·新疆喀什·开学考试"
+    },
+    {
+        "id":  "464",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "( )个13是182,345 是 23( )倍。",
+        "options":  [
+
+                    ],
+        "answer":  "14;15",
+        "solution":  "【分析】求几个13是182，就是求182里面有几个13，用182÷13；求345是23的几倍，就是求345里有几个23，用345÷23，据此解答。\n【详解】182÷13=14，345÷23=15。14个13是182，345是23的15倍。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "求一个数是另一个数的几倍"
+                             ],
+        "source":  "24-25四年级下·新疆喀什·开学考试"
+    },
+    {
+        "id":  "465",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "648×95 的积是( )位数, 2884÷28 的商是( )位数。",
+        "options":  [
+
+                    ],
+        "answer":  "五;三",
+        "solution":  "【分析】计算三位数乘两位数的乘法时，先用两位数的个位上的数去乘另一个乘数，得数的末位和三位数的个位对齐，再用两位数的十位去乘另一个乘数，得数的末位和三位数的十位对齐；然后把两次乘得的数加起来，就是所求的积，据此求出648×95的积即可解答；四位数除以两位数，被除数前两位数大于或等于除数，商是三位数，否则商是两位数。\n【详解】648×95=61560，所以648×95的积是五位数；2884÷28，被除数前两位28等于除数28，所以商是三位数。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法",
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25 四年级上·广东汕头·期末"
+    },
+    {
+        "id":  "466",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "115 个 18 相加的和是( );896 里面有( )个 28。",
+        "options":  [
+
+                    ],
+        "answer":  "2070;32",
+        "solution":  "【分析】求几个相同加数的和是多少，用乘法进行解答；求一个数里面有几个另一个数，用除法进行解答。\n【详解】18×115=2070，896÷28=32。115个18相加的和是2070；896里面有32个28。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法",
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "24-25 四年级上·广东汕头·期末"
+    },
+    {
+        "id":  "467",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "复兴号高铁每小时行 350 千米，这个速度可以写作( ),从甲城到乙城共行1400 千米，需要( )小时。",
+        "options":  [
+
+                    ],
+        "answer":  "350千米/时;4",
+        "solution":  "【分析】此题考查速度的写法和相关计算。速度的单位常用单位有千米/时、米/秒；要求时间，可以依据：路程÷速度=时间，求解。\n【详解】每小时350千米可以写作350千米/时，1400÷350=4(小时)，所以从甲城到乙城需要4小时。",
+        "knowledge_points":  [
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25 四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "468",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "比大小。\n900÷18( )900÷24 8999 公顷( )100 平方千米",
+        "options":  [
+
+                    ],
+        "answer":  ">;<",
+        "solution":  "【分析】(1)当被除数相同时，除数越大商越小。(2)根据1平方千米=100公顷，把单位统一，再比较大小。\n【详解】被除数相同，除数18<24，所以900÷18>900÷24。100平方千米=10000公顷，8999公顷<10000公顷，所以8999公顷<100平方千米。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "469",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在□38÷64 中，要使商是两位数，□里最小可以填( );要使商是一位数,□里最大可以填( );199□000≈200 万,□里最小可以填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "7;6;5",
+        "solution":  "【分析】三位数除以两位数，商可能是两位数，也可能是一位数。被除数的前两位够除，也就是被除数的前两位大于或等于除数，商是两位数；被除数的前两位不够除，也就是被除数的前两位小于除数，商是一位数。□38÷64，要使商是两位数，□3≥64，□里可以填7、8、9；要使商是一位数，□3<64，□里可以填1、2、3、4、5、6。省略\"万\"位后面的尾数就是四舍五入到万位，把万位后的千位上的数进行四舍五入，再在数的后面写上\"万\"字。199□000≈200万，是用五入法取近似数，199□000的千位上可能是5、6、7、8、9。\n【详解】在□38÷64中，要使商是两位数，□里最小可以填7；要使商是一位数，□里最大可以填6；199□000≈200万，□里最小可以填5。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)",
+                                 "亿以内数的近似数"
+                             ],
+        "source":  "24-25 四年级上·河南信阳·期末"
+    },
+    {
+        "id":  "470",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\">\"\"<\"或\"=\"。\n390206000( )392000000 210÷42( )30÷6 8 平方千米( )80 公顷",
+        "options":  [
+
+                    ],
+        "answer":  "<;=;>",
+        "solution":  "【分析】比较整数的大小，先看整数的位数，位数多的那个数就大；如果位数相同，就看最高位，最高位上的数大，那个数就大；如果最高位的数相同，就看次高位，次高位上的数大，那个数就大。以此类推。据此比较390206000和392000000的大小。被除数和除数同时乘或除以相同的数(0除外),商不变。据此比较210÷42和30÷6的大小。1平方千米=100公顷，据此把8平方千米换算成800公顷，再与80公顷比较大小。\n【详解】390206000<392000000；210÷42=(210÷7)÷(42÷7)=30÷6；8平方千米=800公顷，800公顷>80公顷，所以8平方千米>80公顷。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "公顷，平方千米的进率与换算",
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "24-25四年级上·河南信阳·期末"
+    },
+    {
+        "id":  "471",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\">\"\"<\"或\"=\"。\n3600 平方米( )36 公顷 锐角( )直角 527023( )496920\n38×6( )240 150×2( )15×20 90÷18( )180÷36",
+        "options":  [
+
+                    ],
+        "answer":  "<;<;>;<;=;=",
+        "solution":  "【分析】1公顷=10000平方米，36公顷就是36个10000平方米；据此先统一单位，再比较(1)小题大小即可；锐角是大于0°小于90°的角，直角是等于90°的角，据此比较(2)小题大小即可；整数比较大小，首先要看它们的位数，如果位数不同，那么位数多的数就大，如果位数相同，从最高位开始，相同数位上数大的那个数就大。据此比较(3)小题即可。计算出(4)、(5)、(6)小题的算式结果，再比较大小即可。\n【详解】36个10000平方米是360000平方米，3600平方米<360000平方米，即3600平方米<36公顷；锐角<直角；527023>496920；38×6=228，228<240，即38×6<240；150×2=300，15×20=300，即150×2=15×20；90÷18=5，180÷36=5，即90÷18=180÷36。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "两位数乘整十数的口算乘法",
+                                 "除数是两位数的笔算除法",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "24-25 四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "472",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据 35×46 =1610，直接写出下面各式的得数.\n1610÷46 = ( ) 16100÷35 = ( ) 350×46 = ( )",
+        "options":  [
+
+                    ],
+        "answer":  "35;460;16100",
+        "solution":  "【分析】根据积的变化规律：一个因数不变，另一个因数同时乘或者除以几(0除外),积也会乘或除以相同的数，据此解答即可。本题考查乘除法的计算。商的变化规律：被除数不变，除数乘几或除以几(0除外),商反而除以几或乘几;除数不变，被除数乘几或除以几(0除外),商也乘几或除以几;当被除数和除数都乘或者除以几(0除外)时，商不变。据此解答.\n【详解】根据积的变化规律及商的变化规律得:1610÷46 = 35 (1610×10) ÷ 35 = 46×10 = 460 (35×10)×46 = 1610×10 = 16100",
+        "knowledge_points":  [
+                                 "商的变化规律及应用",
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "24-25 四年级上·河南信阳·期末"
+    },
+    {
+        "id":  "473",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在644÷□5中，要使商是两位数,□里最大可以填( ),要使商是一位数,□里最小填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "5;6",
+        "solution":  "【分析】根据除数是两位数的除法,644÷□5 中，要使商是位数，即商应在个位，则被除数前两位比除数小，所以除数□5应大于64，即□最小可填6;要使商是两位数，则商的最高位在十位，则被除数前两位应大于或等于除数，即□5 应小于或等于 64，口中应小于 6，可填 1,2,3,4,5，最大可填5。据此解答.\n【详解】根据分析得：在644÷□5中，要使商是两位数,□里最大可以填 5，要使商是一位数,□里最小填 6 。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25四年级上·河南信阳·期末"
+    },
+    {
+        "id":  "474",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "林林在计算有余数的除法时，把被除数169当成了269，得到的商比正确的的商大 5，余数相同，正确的商应该是( ),余数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "8;9",
+        "solution":  "【分析】因为错误的商比正确的商大5，且余数相同，所以两个被除数的差正好是5个除数。先求出除数，再用正确的被除数169去除，求出正确的商和余数。\n【详解】269-169=100，100÷5=20，所以除数是20。169÷20=8……9，所以正确的商是8，余数是9。",
+        "knowledge_points":  [
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "24-25四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "475",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "两数相除，商 25 余6，如果被除数，除数都乘 10，商是( ),余数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "25;60",
+        "solution":  "【分析】根据商的变化规律：被除数和除数同时乘或除以相同的数(0除外),商不变，余数同时乘或除以相同的数，据此解答即可.\n【详解】 6×10 = 60 被除数，除数都乘 10，商不变，商是 25，余数同时乘 10，余数是 60 。",
+        "knowledge_points":  [
+                                 "商的变化规律及应用"
+                             ],
+        "source":  "24-25 四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "476",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "颐和园的占地面积约 3000000 平方米，合( )公顷;\"鸟巢\"的占地面积约为 20 公顷,( )个\"鸟巢\"的占地面积约为 1 平方千米;王老师沿着面积是 4 公顷的正方形公园跑了一圈，他跑了( )米。",
+        "options":  [
+
+                    ],
+        "answer":  "300;5;800",
+        "solution":  "【分析】1公顷=10000平方米，1平方千米=100公顷。先进行单位换算，再根据正方形面积求边长，最后求周长。\n【详解】3000000÷10000=300(公顷);100÷20=5，所以5个“鸟巢”的占地面积约为1平方千米;4公顷=40000平方米，40000=200×200，所以边长是200米，200×4=800(米)。",
+        "knowledge_points":  [
+                                 "正方形的周长",
+                                 "除数是整十数的口算除法",
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "24-25 四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "477",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "育英学校组织 845名师生到电影城观看《大突围》,如果每辆大巴车限载45人(不含司机),学校至少要租( )辆大巴车。",
+        "options":  [
+
+                    ],
+        "answer":  "19",
+        "solution":  "【分析】用总人数除以每辆大巴车限乘的人数求出商，如果没有余数，商就是最少需要的大巴车数量，如果有余数，则需要再加1辆.\n【详解】 845÷45 = 18 (辆)。35(人) 18 + 1 = 19 (辆) 所以育英学校组织845名师生到电影城观看《大突围》,如果每辆大巴车限载45人(不含司机),学校至少要租19辆大巴车。",
+        "knowledge_points":  [
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "24-25 四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "478",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "A÷33=106……B，B最大可以是( )，这时A=( )。",
+        "options":  [
+
+                    ],
+        "answer":  "32;3530",
+        "solution":  "【分析】余数要小于除数，所以B最大是33-1=32。再根据被除数=除数×商+余数求出A。\n【详解】B最大是32，A=33×106+32=3498+32=3530。",
+        "knowledge_points":  [
+                                 "余数与除数的关系",
+                                 "有余数的除法验算",
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "24-25 四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "479",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\">\"\"<\"或\"=\"。\n8万( )78889 1周角( )4直角 502平方千米( )52000公顷\n720÷48( )360÷24 18×310( )31×170 199×19( )4000",
+        "options":  [
+
+                    ],
+        "answer":  ">;=;<;=;>;<",
+        "solution":  "【分析】整数比较大小的方法：首先比较数的位数，位数越多，表明这个数越大；如果数的位数相同，就比较最高位上的数字，最高位上的数字大，对应的那个数就大；如果最高位上数字相同，就比较下一位数字，以此类推，直到比较出大小为止。(1)先将8万写成80000，再比较即可。(2)周角的度数是360°，直角的度数是90°，据此计算出4个直角的度数后再比较即可。(3)1平方千米=100公顷，据此将单位统一后再比较即可。(4)被除数和除数都乘或者除以相同的数(0除外),商不变，据此比较。(5)分别计算出两边算式的积，再比较即可。(6)将199估成200，19估成20，再计算后比较即可。\n【详解】8万=80000，80000>78889，则8万>78889；4×90°=360°，1周角=360°，4直角=360°，则1周角=4直角；502平方千米=50200公顷，50200<52000，则502平方千米<52000公顷；720÷2=360，48÷2=24，则720÷48=360÷24；18×310=5580，31×170=5270，5580>5270，则18×310>31×170；199×19<200×20=4000，则199×19<4000。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "三位数乘两位数的估算",
+                                 "平角，周角的认识及特征",
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "24-25四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "480",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "当□47÷47<10时,□里最大可填( );当□47÷47>10时,□里最小可填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "4;5",
+        "solution":  "【分析】根据题意，除数是两位数的除法的笔算法则：从被除数的最高位除起，先看被除数的前两位；如果前两位比除数小，就要看前三位；除到被除数的哪一位，商就写在那一位的上面；余下的数必须比除数小，除到被除数的某一位不够除时，应商0占位。当□47÷47<10时，被除数的前两位□4<47，□里可以填4、3、2、1，最大填4；当□47÷47>10时，被除数的前两位□4>47，□里可以填5、6、7、8、9，最小填5。以此答题即可。\n【详解】根据分析可知：当□47÷47<10时，□里最大可填4；当□47÷47>10时，□里最小可填5。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "481",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据 72÷9 = 8，直接写出下面算式的得数。\n144÷18 = ( ) 720÷9 = ( ) 720÷80 = ( )",
+        "options":  [
+
+                    ],
+        "answer":  "8;80;9",
+        "solution":  "【分析】根据题意，在商不为0的除法算式里，当除数一定时，被除数乘几，商就乘几；被除数和除数同时乘或除以一个不为0的数，商的大小不变；依此填空。\n【详解】72÷9=8；144÷18=(72×2)÷(9×2)=8；720÷9=(72×10)÷9=72÷9×10=8×10=80；720÷80=(72×10)÷(8×10)=72÷8=9。",
+        "knowledge_points":  [
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "24-25 四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "482",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\">\"\"<\"或\"=\"。\n80×40( )16×200 482÷29( )482÷30 360÷18( )360÷9×2",
+        "options":  [
+
+                    ],
+        "answer":  "=;>;<",
+        "solution":  "【分析】(1)(3)计算出等式两边的结果，再比较大小；(2)根据\"被除数相同，除数越大，商越小\"这一规律，比较大小。\n【详解】80×40=3200，16×200=3200，所以80×40=16×200；29<30，所以482÷29>482÷30；360÷18=20，360÷9×2=40×2=80，20<80，所以360÷18<360÷9×2。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "两位数乘整十数的口算乘法"
+                             ],
+        "source":  "24-25 四年级上·湖北荆州·期末"
+    },
+    {
+        "id":  "483",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算 430÷72 时，把72 看作( )来试商，商为( ),这时初商( )(填\"大\"或\"小\")了，改商( )。",
+        "options":  [
+
+                    ],
+        "answer":  "70;6;大;5",
+        "solution":  "【分析】根据题意，把72看作接近的整十数70来试商(四舍五入法则，个位2<5，舍去)；430÷70≈6(70×6=420，接近430)。6×72=432，但432>430，说明初商6偏大。将商调小1，改为5，5×72=360，430-360=70(余数<除数，符合要求)。以此答题即可。\n【详解】根据分析可知：计算430÷72时，把72看作70来试商，商为6，这时初商大了，改商5。",
+        "knowledge_points":  [
+                                 "多位数除以两位数的试商"
+                             ],
+        "source":  "24-25四年级上·河南信阳·期末"
+    },
+    {
+        "id":  "484",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "计算 258÷28 时，可以把 28 看作 30 来试商，商有点偏( ),可以将商调整为( )。",
+        "options":  [
+
+                    ],
+        "answer":  "小;9",
+        "solution":  "【分析】根据除法计算中试商的方法：把除数看大了，商就容易变小，调商时要调大一点；把除数看小了，商就容易变大，调商时要调小一点，据此解答。\n【详解】由分析知：计算258÷28时，可以把28看作30来试商，商是8，商有点偏小，可以将商调整为9。",
+        "knowledge_points":  [
+                                 "多位数除以两位数的试商"
+                             ],
+        "source":  "24-25四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "485",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据18×60=1080填空。\n36×60=( ) 1080÷120=( ) 54×( )=1080 2160÷36=( )",
+        "options":  [
+
+                    ],
+        "answer":  "2160;9;20;60",
+        "solution":  "【分析】积的变化规律：一个乘数不变，另一个乘数乘或除以几(0除外)，得到的积就等于原来的积乘或除以几；商不变的规律：被除数和除数同时乘或除以同一个不为0的数，商不变。商变化的规律：被除数不变，除数乘几，商反而除以几。(1)对比算式18×60=1080和算式36×60可知，一个乘数不变，另一个乘数乘2，积也会乘2。(2)由算式18×60=1080可知，1080÷60=18。对比算式1080÷120和算式1080÷60=18可知，被除数不变，除数乘2，那么商会除以2。(3)对比算式18×60=1080和算式54×( )=1080可知，18×3=54，一个乘数乘3，积不变，说明另一个乘数需要除以3。(4)由算式18×60=1080可知，1080÷18=60。对比算式1080÷18=60和算式2160÷36可知，被除数和除数都乘2，商不变。\n【详解】1080×2=2160，所以36×60=2160；18÷2=9，所以1080÷120=9；60÷3=20，所以54×20=1080；由商不变的规律可知，2160÷36=60。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)",
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "24-25 四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "486",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "学校开展国学经典读书活动，聪聪计划读一本 300 页的《中华成语故事》,每天读 16 页，按照这样的阅读计划，他( )天能读完这本书。",
+        "options":  [
+
+                    ],
+        "answer":  "19",
+        "solution":  "【分析】根据题意，用总页数除以每天读的页数，就是看完所用的天数;如果有余数，那么需要再加上一天才能看完.\n【详解】 300÷16 = 18 (天)。12(页)18 + 1 = 19 (天)所以学校开展国学经典读书活动，聪聪计划读一本300页的《中华成语故事》,每天读16页，按照这样的阅读计划，他 19 天能读完这本书。",
+        "knowledge_points":  [
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "24-25四年级上·湖北荆州·期末"
+    },
+    {
+        "id":  "487",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\">\"\"<\"或\"=\"。\n599870000( )6亿 3平方千米( )3000公顷\n540÷15( )540÷20 18×500( )50×180",
+        "options":  [
+
+                    ],
+        "answer":  "<;<;>;=",
+        "solution":  "【分析】(1)先把整亿的数改写成用\"亿\"作单位的数，再进行比较。先看位数，位数多的数大；位数相同，从最高位看起，相同数位上的数大的那个数就大。(2)根据1平方千米=100公顷，先化成统一的单位，再根据数值的大小进行比较。(3)被除数相同，除数越小，则商越大。(4)根据积不变的规律：两个数相乘，一个因数乘(或除以)一个数(0除外),另一个因数同时除以(或乘)相同的数，它们的积不变，进行比较即可。\n【详解】6亿=600000000，599870000<600000000，所以599870000<6亿；3平方千米=300公顷，300公顷<3000公顷，所以3平方千米<3000公顷；15<20，所以540÷15>540÷20；18×500=18×50×10，50×180=18×50×10，所以18×500=50×180。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "公顷，平方千米的进率与换算",
+                                 "积的变化规律(整数乘法)",
+                                 "亿以上数的改写"
+                             ],
+        "source":  "24-25 四年级上·湖北荆州·期末"
+    },
+    {
+        "id":  "488",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "如果□÷△=45……15，那么△最小是( )，此时□是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "16;735",
+        "solution":  "【分析】△表示除数，在有余数的除法算式里，除数应比余数大，依此确定出最小的除数，被除数=商×除数+余数，当除数最小时，依此计算出被除数即可。\n【详解】余数是15，则除数最小是15+1=16；45×16+15=720+15=735。所以如果□÷△=45……15，那么△最小是16，此时□是735。",
+        "knowledge_points":  [
+                                 "余数与除数的关系",
+                                 "有余数的除法验算"
+                             ],
+        "source":  "24-25 四年级上·河南信阳·期末"
+    },
+    {
+        "id":  "489",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "两个数相除，商是 18，余数是 2。若被除数和除数同时扩大到原来的 10 倍，这时商是( ),余数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "18;20",
+        "solution":  "【分析】除法算式中，被除数和除数同时乘(或除以)几(0除外),商不变，余数要同时乘(或除以)相同的数，据此即可解答.\n【详解】 2×10 = 20 两个数相除，商是18，余数是2。若被除数和除数同时扩大到原来的10倍，这时商是18，余数是 20 。",
+        "knowledge_points":  [
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "24-25 四年级上·河南信阳·期末"
+    },
+    {
+        "id":  "490",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在下面的括号里填上\">\"\"<\"或\"=\"。\n70700( )70070 89016万( )9亿 302÷52( )990÷48 180÷36( )(180÷6)÷(36÷6)",
+        "options":  [
+
+                    ],
+        "answer":  ">;<;<;=",
+        "solution":  "【分析】(1)两个五位数比较大小，从高位开始比较，一位一位比较。先看万位，两个数都是7，所以万位相同。接下来是千位，都是0，也相同。然后是百位，70700的百位是7，而70070的百位是0。这时候7比0大，所以70700>70070。(2)需要将单位统一后再比较。1亿等于10000万，所以9亿就是90000万。比较89016万和90000万，89016万比90000万小，所以89016万<9亿。(3)三位数除以两位数，如果被除数的前两位大于等于除数，则商是两位数，否则商是一位数，302÷52的商是一位数，990÷48的商是两位数，一位数小于两位数，所以302÷52<990÷48。(4)根据商不变规律可知，被除数和除数同时乘(或除以)相同的数(0除外),商不变，所以180÷36=(180÷6)÷(36÷6)。\n【详解】70700>70070；9亿=90000万，89016万<90000万，所以89016万<9亿；302÷52<990÷48；180÷36=(180÷6)÷(36÷6)。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "判断商是几位数(除数是两位数)",
+                                 "亿以上数的改写",
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "24-25 四年级上·河南信阳·期末"
+    },
+    {
+        "id":  "491",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据 38×12 = 456，直接写出下面算式的得数.\n380×12 = ( ) 38×1200 = ( )\n456÷38 = ( ) 4560÷120 = ( )",
+        "options":  [
+
+                    ],
+        "answer":  "4560;45600;12;38",
+        "solution":  "【分析】一个乘数不变，另一个乘数乘几或除以几(0 除外),积也乘几或除以几(0 除外)。因数×因数 = 积，则积/因数 = 因数，据此判断 456÷38 的商和 4560÷120 的商。被除数和除数同时乘或除以相同的数(0 除外),商不变.\n【详解】 380×12 = (38×10)×12 = 456×10 = 456038×1200 = 38×(12×100) = 38×12×100 = 456×100 = 45600\n因为 38×12 = 456，所以 456÷38 = 124560÷120 = (4560÷10) ÷ (120÷10) = 456÷12 = 38 \n 380×12 = 456038×1200 = 45600456÷38 = 124560÷120 = 38 ",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)",
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "24-25 四年级上·河南信阳·期末"
+    },
+    {
+        "id":  "492",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据 23×14 = 322 和560÷20 = 28，直接写出下面各题的得数.\n 230×14 = ( ) 23×7 = ( ) (560÷2) ÷ 20 = ( ) (560÷2) ÷ (20÷2) = ( ) ",
+        "options":  [
+
+                    ],
+        "answer":  "3220;161;14;28",
+        "solution":  "【分析】根据积的变化规律：一个因数不变，另一个因数同时乘或者除以几(0除外),积也会乘或除以相同的数。商的变化规律：被除数不变，除数乘几或除以几(0除外)商反而除以几或乘几;除数不变，被除数乘几或除以几(0除外),商也乘几或除以几;当被除数和除数都乘或者除以几(0 除外)时，商不变。据此解答.\n【详解】230×14=(23×10)×14=23×14×10= 322×10 = 322023×7 = 23×(14÷2) = 23×14÷2 = 322÷2 = 161 (560÷2) ÷ 20 = 560÷20÷2 = 28÷2 = 14 (560÷2) ÷ (20÷2) = 560÷20 = 28 ",
+        "knowledge_points":  [
+                                 "商的变化规律及应用",
+                                 "积的变化规律(整数乘法)",
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "24-25 四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "493",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "60×250 的积的末尾有( )个 0;要使□2×230 的积是四位数,□最大填( );计算 579÷29 时，可以把除数看作( )来试商;要使673÷□6 的商是一位数,□里最小填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "3;4;30;7",
+        "solution":  "【分析】根据题意，先计算60×250的积，根据得数填空即可；要使□2×230的积是四位数，试着用3、4、5填空计算，找出合适的数即可；三位数除以两位数：把除数看作和它最接近的整十数试商，先用被除数的前两位除以除数，如果被除数前两位比除数小，就用前三位除以除数，初商偏大要调小，初商偏小要调大。三位数除以两位数，比较被除数前两位和除数的大小，如果被除数前两位大于等于除数，则商是两位数，如果小于除数商是一位数；以此答题即可。\n【详解】60×250=15000；32×230=7360，42×230=9660，52×230=11960。要使673÷□6的商是一位数，那么67<□6，□里可以填7、8、9，最小填7。60×250的积的末尾有3个0；要使□2×230的积是四位数，□最大填4；计算579÷29时，可以把除数看作30来试商；要使673÷□6的商是一位数，□里最小填7。",
+        "knowledge_points":  [
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "多位数除以两位数的试商",
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "494",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "括号里最大填几。\n42( )653≈42万 50×( )<290 ( )85÷78 的商是一位数",
+        "options":  [
+
+                    ],
+        "answer":  "4;5;6",
+        "solution":  "【分析】省略万位后面的尾数，就先找到万位，然后看千位上的数是否大于5，当千位上的数小于5时就直接省略，当千位上的数大于或等于5时就直接向万位进\"1\"后再省略，最后在数的末尾加一个\"万\"字；用290÷50，有余数时商是可以填的最大数，没有余数时，商减1就是可以填的最大数。三位数除以两位数，如果被除数的前两位大于等于除数，商是两位数，否则商是一位数。\n【详解】这个数四舍五入到万位是42万，说明千位是小于5的数，可以填0、1、2、3、4，最大可以填4。290÷50=5……40，所以括号里最大填5。( )85÷78的商是一位数，说明被除数的前两位比78小，所以括号里可以填1、2、3、4、5、6，最大可以填6。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "判断商是几位数(除数是两位数)",
+                                 "亿以内数的近似数"
+                             ],
+        "source":  "24-25 四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "495",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "算式 326×□7，如果要使积是四位数,□里最大能填( );635÷□4，如果商是一位数,□里最小填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "2;6",
+        "solution":  "【分析】根据三位数乘两位数的计算方法，将□=1、2、3、4、5代入到算式中，计算出结果后判断当积是四位数时，□里最大能填几；根据三位数除以两位数的计算，要使商是一位数，则被除数的前两位小于除数，据此填空即可。\n【详解】326×17=5542，积是四位数；326×27=8802，积是四位数；326×37=12062，积是五位数。635÷□4，如果商是一位数，63<□4，63<64，□里最小填6。算式326×□7，如果要使积是四位数，□里最大能填2；635÷□4，如果商是一位数，□里最小填6。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法",
+                                 "判断商是几位数(除数是两位数)"
+                             ],
+        "source":  "24-25四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "496",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "市场上的树苗促销，买 3 棵送 1 棵，每棵 16 元。176 元最多能买( )棵这样的树苗。",
+        "options":  [
+
+                    ],
+        "answer":  "14",
+        "solution":  "【分析】已知每棵树苗16元，一共有176元，先不考虑赠送的，根据数量=总价÷单价，据此求出可以买多少棵；然后考虑送的，因为买3棵送1棵，以3棵为1组，将所求的数量除以3，看有多少组，就送多少棵，将买的数量加上送的数量即可求出总的数量，据此列式计算即可解此题。\n【详解】176÷16=11(棵)；11÷3=3(组)……2(棵)，即送3棵树苗；3+11=14(棵)。综上可知，176元最多能买14棵这样的树苗。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "24-25 四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "497",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "12 箱蜜蜂一年可以酿 900 千克蜂蜜。小林家养了 5 箱蜜蜂，一年可以酿( )千克蜂蜜。",
+        "options":  [
+
+                    ],
+        "answer":  "375",
+        "solution":  "【分析】要求5箱蜜蜂一年可以酿多少千克蜂蜜，根据题意，需要先求出平均每箱蜜蜂一年可以酿多少千克蜂蜜，进而乘5即可解答。\n【详解】900÷12×5=75×5=375(千克)。所以一年可以酿375千克蜂蜜。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "单归一问题"
+                             ],
+        "source":  "24-25 四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "498",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在算式☆÷△=18……17 中,△最小是( ),这时☆代表的数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "18;341",
+        "solution":  "【分析】在有余数的除法中，余数一定小于除数。已知余数是17，那么除数△一定大于17，大于17的最小整数就是18，所以△最小是18；然后根据商乘除数再加余数，据此求出被除数，也就是☆的数值。\n【详解】在☆÷△=18……17中，除数是△，余数是17，△>17，所以△最小是18；18×18+17=324+17=341。综上可知，在算式☆÷△=18……17中，△最小是18，这时☆代表的数是341。",
+        "knowledge_points":  [
+                                 "乘加，乘减混合运算",
+                                 "三位数除以两位数的有余数除法",
+                                 "余数与除数的关系"
+                             ],
+        "source":  "24-25四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "499",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "609÷□2，如果商是一位数,□里最小填( );如果商是两位数,□里最大填( );如果商的末尾有 0,□里应该填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "6;5;1",
+        "solution":  "【分析】三位数除以两位数，如果被除数的前两位大于等于除数，商是两位数；被除数的前两位小于除数，商是一位数。如果商的末尾有0，说明商是两位数，□里可以填1、2、3、4、5，把这5个数代入进去，看看哪个商的末尾有0即可。\n【详解】609÷□2，如果商是一位数，说明60小于□2，□里可以填6、7、8、9，最小是6。如果商是两位数，说明60大于等于□2，□里可以填1、2、3、4、5，最大填5。609÷12=50……9，609÷22=27……15，609÷32=19……1，609÷42=14……21，609÷52=11……37。如果商的末尾有0，□里应该填1。",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)",
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "24-25 四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "500",
+        "unit":  "第六单元 除数是两位数的除数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "已知 A×B=320，如果 A 不变,B 除以 4，那么积是( );已知 A÷B=45，如果 A 不变,B 除以 3，那么商是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "80;135",
+        "solution":  "【分析】积的变化规律：一个因数不变，另一个因数乘或者除以几(0除外),积也乘或者除以几。商的变化规律：被除数不变，除数乘几(0除外),商就除以几；除数除以几(0除外),商就乘几。\n【详解】已知A×B=320，如果A不变，B除以4，那么积也除以4，320÷4=80，那么积是80。已知A÷B=45，如果A不变，B除以3，那么商乘3，45×3=135，所以商是135。",
+        "knowledge_points":  [
+                                 "商的变化规律及应用",
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "24-25 四年级上·河南南阳·期末"
+    }
+]
diff --git "a/\345\233\233\345\271\264\347\272\247\351\242\230\351\233\206/\345\233\233\345\271\264\347\272\247\344\270\212/\345\233\233\345\271\264\347\272\247\344\270\212\347\254\254\345\233\233\345\215\225\345\205\203\344\270\211\344\275\215\346\225\260\344\271\230\344\270\244\344\275\215\346\225\260.json" "b/\345\233\233\345\271\264\347\272\247\351\242\230\351\233\206/\345\233\233\345\271\264\347\272\247\344\270\212/\345\233\233\345\271\264\347\272\247\344\270\212\347\254\254\345\233\233\345\215\225\345\205\203\344\270\211\344\275\215\346\225\260\344\271\230\344\270\244\344\275\215\346\225\260.json"
new file mode 100644
index 0000000000000000000000000000000000000000..025be06ef592fd5ba73f61cb000c261077a99f41
--- /dev/null
+++ "b/\345\233\233\345\271\264\347\272\247\351\242\230\351\233\206/\345\233\233\345\271\264\347\272\247\344\270\212/\345\233\233\345\271\264\347\272\247\344\270\212\347\254\254\345\233\233\345\215\225\345\205\203\344\270\211\344\275\215\346\225\260\344\271\230\344\270\244\344\275\215\346\225\260.json"
@@ -0,0 +1,8787 @@
+[
+    {
+        "id":  "1",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "已知 2□4×□9 是一个三位数乘两位数的算式,那么下面四个数中,有可能是它的得数( )。",
+        "options":  [
+                        "A. 2036",
+                        "B. 7983",
+                        "C. 9906",
+                        "D. 31025"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】三位数乘两位数计算时,个位对齐,先用两位数个位上的数和三位数的每一位上的数相乘,乘得结果和个位对齐,再用两位数十位上的数和三位数的每一位上的数相乘,乘得结果和十位对齐,最后把两次乘得结果相加。积的个位数由两个乘数的个位数决定,三位数的个位是4,两位数的个位是9,4×9 = 36,积的个位是6。因此,排除个位不是6的选项B和D。其次,估算积的范围:三位数2□4最小为204,最大为294;两位数□9最小为19,最大为99。最小积为204×19 = 3876,最大积为294×99 = 29106。因此,积应在3876到29106之间。选项A(2036)小于3876,排除;选项C(9906)在范围内且个位为6,符合条件。\n【详解】4×9 = 36,积的个位是6,因此排除个位不是6的选项B和D。三位数2□4中,□可以是0到9,最小为204,最大为294;两位数□9中,□可以是1到9,最小为19,最大为99。最小积:204×19 = 3876,最大积:294×99 = 29106,所以积的范围为3876到29106。已知2□4×□9是一个三位数乘两位数的算式,那么下面四个数中,有可能是它的得数9906。故答案为:C。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "24-25四年级上·重庆璧山·期末"
+    },
+    {
+        "id":  "2",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "计算 234×56 时,3×6 表示( )。",
+        "options":  [
+                        "A. 3×6",
+                        "B. 30×6",
+                        "C. 30×60",
+                        "D. 300×60"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】在计算三位数乘两位数时,需要理解每个数字所在的数位所代表的实际数值。234中的3位于十位,表示30;56中的6位于个位,表示6。因此,计算过程中的3×6实际表示30×6。以此选择即可。\n【详解】根据分析可知:计算234×56时,3×6表示30×6。故答案为:B。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "24-25 四年级上·新疆巴音郭楞·期末"
+    },
+    {
+        "id":  "3",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下列运动中,奔跑速度最快的是( )。",
+        "options":  [
+                        "A. 野驴:6分钟7千米",
+                        "B. 斑马:1000 米/分",
+                        "C. 瞪羚:60 千米/时",
+                        "D. 兔子:2 分钟 1800 米"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】1千米 = 1000米,1时 = 60分,据此换算单位,根据速度 = 路程÷时间,分别求出每个选项的每分钟跑多少米,再进行比较即可。\n【详解】A. 7千米 = 7000米,7000÷6≈1167(米/分),野驴的速度大约为1167米/分。B. 斑马的速度为1000米/分。C. 60千米/时 = 60000米/时,60000÷60 = 1000(米/分),瞪羚的速度为1000米/分。D. 1800÷2 = 900(米/分),兔子的速度为900米/分。1167＞1000＞900,所以野驴的奔跑速度最快。故答案为:A。",
+        "knowledge_points":  [
+                                 "整十,整百,整千数除以一位数的除法",
+                                 "基础行程问题",
+                                 "时,分,秒之间的换算与比较",
+                                 "千米和米之间的进率与换算"
+                             ],
+        "source":  "25-26四年级上·河北唐山·期末"
+    },
+    {
+        "id":  "4",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "王阿姨开车自驾游,她出发时看到汽车里程表显示的是一个左右对称的五位数24942。开车行驶了 2 小时后,她又看了一次里程表,发现又是一个左右对称的五位数。假设王阿姨开车的速度不超过 70 千米/时,那么她开车的速度是( )千米/时。",
+        "options":  [
+                        "A. 55",
+                        "B. 58",
+                        "C. 60",
+                        "D. 105"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】根据路程 = 速度×时间,分别代入各选项的数值,计算出王阿姨2小时行驶的路程。再将24942加上王阿姨2小时行驶的路程,判断结果是不是左右对称的五位数。据此解答。 110 + 24942 = 25052 符合题意; 116 + 24942 = 25058 不符合题意; 120 + 24942 = 25062 不符合题意;D。超过70千米/时,不符合题意。故答案为:A",
+        "knowledge_points":  [
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26 四年级上·江西赣州·期中"
+    },
+    {
+        "id":  "5",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "小琳在计算 \u0027201×86\u0027 时,错把第二个因数个位上的\u00276\u0027写成了\u00279\u0027,算出的积会比正确的积多( )。",
+        "options":  [
+                        "A. 258",
+                        "B. 603",
+                        "C. 1206",
+                        "D. 1809"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】第二个因数个位上的\u00276\u0027写成了\u00279\u0027,89 - 86 = 3,即导致第二个因数增加了3。因此,算出的积会比正确的积多3个201,据此计算。\n【详解】89 - 86 = 3,3×201 = 603。算出的积会比正确的积多603。故答案为:B。",
+        "knowledge_points":  [
+                                 "两位数与两位数的不退位减法",
+                                 "两,三位数与一位数的不进位乘法",
+                                 "三位数乘两位数,三位数中间有 0"
+                             ],
+        "source":  "25-26 四年级上·江西赣州·期中"
+    },
+    {
+        "id":  "6",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面描述的情境中,不能用积的变化规律解答的是()。",
+        "options":  [
+                        "A. 34×4 = 136 ,那么 68×4 的积是多少。",
+                        "B. 长为 8 厘米,宽为 4 厘米的长方形,长扩大到原来的3 倍,其周长变为原来的几倍。",
+                        "C. 同一种笔记本,小明买 20 本花了 120 元,现买 60 本需要多少钱。",
+                        "D. 一个乘法算式中,如果一个因数不变,另一个因数除以4,积是32。求原来的积是多少。"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】积的变化规律:一个因数不变,另一个因数乘几或除以几(0除外),积也乘几或除以几;据此逐项判断。\n【详解】A. 在34×4 = 136中,因数4不变,因数34变为68,即一个因数乘2,根据积的变化规律,积也乘2,136×2 = 272,能用积的变化规律解答。B. 长方形周长 = 2×(长+宽)。原来周长是2×(8+4) = 24(厘米);长扩大到原来的3倍后,新周长是2×(24+4) = 56(厘米)。周长变化是长和宽共同作用的结果,不能直接用积的变化规律解答。C. 笔记本单价不变,数量从20本变为60本,即数量乘3,根据积的变化规律,总价也乘3,能用积的变化规律解答。D. 一个因数不变,另一个因数除以4后积是32,根据积的变化规律,原来的积应是现在积的4倍,即32×4 = 128,能用积的变化规律解答。故答案为:B。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "长方形的周长",
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26四年级上·河南三门峡·期中"
+    },
+    {
+        "id":  "7",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "在计算 432×35 的时候, 3×3 表示( )。",
+        "options":  [
+                        "A. 3×3",
+                        "B. 30×30",
+                        "C. 3×30",
+                        "D. 30×3"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】笔算三位数乘两位数,先用两位数个位上的数分别去乘三位数的每个数字,再用两位数十位上的数去乘三位数的每个数字,用哪位上的数去乘,所得结果末尾就对齐那位。最后再把两次相乘得到的结果相加;三位数十位上的数与两位数十位上的数相乘,表示的是几个十与几个十相乘。据此解答.\n【详解】计算 432×35 时,用三位数十位上的3乘两位数十位上的 3,表示 3 个十乘 3 个十,即 30×30 。故答案为:B",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "25-26四年级上·山东菏泽·期中"
+    },
+    {
+        "id":  "8",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "一块长方形菜园占地面积是 200 平方米,( )块这样的菜园占地面积是 1 公顷。",
+        "options":  [
+                        "A. 5",
+                        "B. 50",
+                        "C. 500",
+                        "D. 5000"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据单位换算可知,1公顷 = 10000 平方米,一块长方形菜园的占地面积是200平方米,将1公顷换算成平方米,再用选项中的数进行计算出多少块这样的菜园占地面积是 1 公顷.\n【详解】10000平方米 = 1 公顷 50×200 = 10000 (平 方 米) 50块这样的菜园占地面积是1公顷。故答案为:B",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法",
+                                 "公顷,平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·河南南阳·期中"
+    },
+    {
+        "id":  "9",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一本《广西旅游指南》有 121 页,每页 25 行,每行 20 个字,这本书大约有()万字(估算)。",
+        "options":  [
+                        "A. 60",
+                        "B. 12",
+                        "C. 6"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】将121页近似看成120页,再用20×25×120可算出大约有多少字,最后省略万位后面的尾数即可。省略万位后面的尾数时,先看千位,千位小于5时直接省略,千位大于或等于5时向万位进1后再省略,最后在数的末尾加一个\u0027万\u0027字。\n【详解】121≈120,20×25×120 = 500×120 = 60000(字),60000字 = 6万字。故答案为:C。",
+        "knowledge_points":  [
+                                 "三位数乘两位数的估算",
+                                 "亿以内数的改写"
+                             ],
+        "source":  "25-26 四年级上·广西玉林·期中"
+    },
+    {
+        "id":  "10",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "从柳州到北海的高速路长约 400 千米,一辆货车上午 8 时出发,下午 1 时到达,它的平均速度是( )。",
+        "options":  [
+                        "A. 50千米/时",
+                        "B. 80 千米/时",
+                        "C. 100 千米/时"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】下午1时改写成24小时计时法为13时,用13时-8时可算出货车行驶的时间,再根据速度 = 路程÷时间,代入数字可计算出速度。\n【详解】下午1时 = 13时,13时-8时 = 5(小时),400÷5 = 80(千米/时)。它的平均速度是80千米/时。故答案为:B。",
+        "knowledge_points":  [
+                                 "24时计时法与普通计时法的互化",
+                                 "整十,整百,整千数除以一位数的除法",
+                                 "基础行程问题",
+                                 "24时计时法时间的计算"
+                             ],
+        "source":  "25-26 四年级上·广西玉林·期中"
+    },
+    {
+        "id":  "11",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "计算 □94×32,得数不可能是( )。",
+        "options":  [
+                        "A. 5208",
+                        "B. 9408",
+                        "C. 19008",
+                        "D. 25408"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】三位数乘两位数时,相同数位对齐,从个位乘起,用两位数分别依次乘三位数中的每一位数,然后将两次乘得的结果相加即可。□94×32中,□最小填1,最大填9,分别计算最小乘积和最大乘积,逐项分析。\n【详解】194×32 = 6208,994×32 = 31808。所以□94×32的最小乘积是6208,最大乘积是31808。A. 5208＜6208,所以得数不可能是5208,符合题意;B. 294×32 = 9408,不符合题意;C. 594×32 = 19008,不符合题意;D. 794×32 = 25408,不符合题意。因此,计算□94×32,得数不可能是5208。故答案为:A。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "25-26 四年级上·湖北武汉·期中"
+    },
+    {
+        "id":  "12",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "两个因数相乘的积是 150,一个因数扩大 5 倍,另一个因数缩小为原来的 1/5 ,积是( )。",
+        "options":  [
+                        "A. 30",
+                        "B. 150",
+                        "C. 750"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】积的变化规律:如果一个因数乘一个数,另一个因数除以同一个数(0除外),那么积不变;据此解答即可。\n【详解】两个数相乘,一个因数扩大5倍,另一个因数缩小为原来的1/5,积不变。两个因数相乘的积是150,一个因数扩大5倍,另一个因数缩小为原来的1/5,积还是150。故答案为:B。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26四年级上·湖北黄冈·期中"
+    },
+    {
+        "id":  "13",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "240×40 的积是( )位数;360×85 的积的末尾有( )个0。",
+        "options":  [
+
+                    ],
+        "answer":  "四;2",
+        "solution":  "【分析】根据三位数乘两位数的计算,计算出 240×40 的结果后,判断积是几位数即可;计算出 360×85 的结果后,判断积的末尾有几个 0 即可.\n【详解】 240×40 = 9600 ,积是四位数;360×85 = 30600 ,积的末尾有 2 个 0.240×40 的积是四位数;360×85 的积的末尾有 2 个 0。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0"
+                             ],
+        "source":  "25-26 四年级上·辽宁抚顺·期末"
+    },
+    {
+        "id":  "14",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "550×42 的末尾有()个 0;215×37 的积是( )位数。",
+        "options":  [
+
+                    ],
+        "answer":  "2;四",
+        "solution":  "【分析】根据整数乘法的法则进行计算即可,整数乘法的法则:从右起,依次用第二个因数每位上的数去乘第一个因数,乘到哪一位,得数的末尾就和第二个因数的哪一位对齐;然后把几次乘得的数加起来(整数末尾有 0 的乘法:可以先把0前面的数相乘,然后看各因数的末尾一共有几个0,就在乘得的数的末尾添写几个 0)。由此可得出.\n【详解】 550×42 = 23100 ,所以 550×42 的末尾有 2 个 0.215×37 = 7955 ,所以 215×37 的积是四位数。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法",
+                                 "三位数乘两位数,乘数末尾有 0"
+                             ],
+        "source":  "25-26 四年级上·湖南常德·期中"
+    },
+    {
+        "id":  "15",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "《诗经》是我国最早的一部诗歌总集,收录了西周至春秋中叶的诗歌,现存305篇,又称《诗三百》。学校开展国学活动,需购买218本《诗经》,每本 42 元,算一算准备 8000 元( )用。(填写\u0027够\u0027或\u0027不够\u0027)",
+        "options":  [
+
+                    ],
+        "answer":  "不够",
+        "solution":  "【分析】已知每本42元,需购买218本,根据总价 = 单价×数量,计算出购买218本需要的钱数,再与8000元进行比较,即可判断是否够用。\n【详解】218×42 = 9156(元),9156＞8000,所以准备8000元不够用。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "25-26 四年级上·河北唐山·期末"
+    },
+    {
+        "id":  "16",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在( )里填上\u0027\u003e\u0027\u0027\u003c\u0027或\u0027=\u0027。398×59( )24000;800 公顷( )8 平方千米;1000 万( )9999999。",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;=;\u003e",
+        "solution":  "【分析】比较398×59和24000,计算398×59 = 23482,因为23482＜24000,所以398×59＜24000。比较800公顷和8平方千米,因为1平方千米 = 100公顷,所以8平方千米 = 800公顷,所以800公顷 = 8平方千米。比较1000万和9999999,1000万 = 10000000,因为10000000＞9999999,所以1000万＞9999999。\n【详解】398×59＜24000;800公顷 = 8平方千米;1000万＞9999999。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "25-26 四年级上·河北唐山·期末"
+    },
+    {
+        "id":  "17",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在\u0027九三阅兵\u0027活动中,三军仪仗队每天要进行基础正步训练,每步 75 厘米,记作( );每日走 20000 步,照这样计算,三军仪仗队这次军事训练为期 132 天,每位士兵的正步训练步长累计相当于( )千米。",
+        "options":  [
+
+                    ],
+        "answer":  "75 厘米/步;1980",
+        "solution":  "【分析】1米 = 100厘米,1千米 = 1000米。(1)每步75厘米,在数学中可以记作\u002775厘米/步\u0027,即每走一步的长度是75厘米。(2)已知每步75厘米,每日走20000步,根据\u0027总长度 = 每步长度×步数\u0027,可得每天走的总长度。因为训练为期132天,所以132天走的总长度为每天走的总长度乘132,再将结果换算成千米即可。\n【详解】(1)每步走75厘米,记作75厘米/步。(2)每天走的总长度:75×20000 = 1500000(厘米),1500000×132 = 198000000(厘米)。198000000厘米 = 1980000米 = 1980千米。所以每位士兵的正步训练步长累计相当于1980千米。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有 0",
+                                 "千米和米之间的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·河南三门峡·期中"
+    },
+    {
+        "id":  "18",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在\u0027九三阅兵\u0027训练中,装备方队以每小时 60 千米的速度前进,这个速度可记作( );照这样的速度,求方队 4 小时行进的路程是多少。所用到的数量关系式是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "60 千米/时;速度×时间 = 路程",
+        "solution":  "【分析】速度单位是一个复合单位,通常用\u0027千米/小时,米/分\u0027作单位;用方队每小时前进的速度乘所用的时间,即可求出方队4小时行进的路程是多少,所用的数量关系为:速度×时间 = 路程;据此解题即可.\n【详解】根据分析可知:在\u0027九三阅兵\u0027训练中,装备方队以每小时60千米的速度前进,这个速度可记作60千米/时;照这样的速度,求方队4小时行进的路程是多少,所用到的数量关系式是速度×时间 = 路程。",
+        "knowledge_points":  [
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26 四年级上·河南三门峡·期中"
+    },
+    {
+        "id":  "19",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "2025 年 9 月 3 日我国在北京天安门广场举行了纪念中国人民抗日战争暨世界反法西斯战争胜利80周年的盛大阅兵仪式。此次亮相的\u0027东风-5C\u0027液体洲际战略导弹其射程可达一万三千零五十千米,横线上的数写作( ),省略万位后面的尾数约是()万;此次徒步方队有 15 个,每个方队 352 人,徒步方队的总人数是( )人,这个数读作( )。",
+        "options":  [
+
+                    ],
+        "answer":  "13050;1;5280;五千二百八十",
+        "solution":  "【分析】根据题意,大数的写法,从高位写起: 按照数位顺序,从高位到低位,一级一级地写。 哪个数位上一个单位也没有,就在那个数位上写 0 占位。省略万后面的尾数,要看千位的数,千位上的数大于等于 5,就向万位进一,千位上的数小于5,就舍去,在万位后写上万字;已知方队有15 个,每个方队 352 人,用 352 乘 15,求出徒步方队的总人数;大数的读法,先把数分级,通常按照数位顺序,从低位到高位,每四位一级(个级,万级,亿级等)。从最高级开始,一级一级往下读。每级末尾的零不读,如个级的末尾或万级,亿级的末尾的零都不读。其他数位上有一个零或连续几个零,都只读一个零。亿级和万级的数要按照个级的数的读法来读,然后在后面加上\u0027亿\u0027字或\u0027万\u0027字。以此答题即可。【详解】根据分析可知:352×15 = 5280 (人)2025年9月3日我国在北京天安门广场举行了纪念中国人民抗日战争暨世界反法西斯战争胜利80周年的盛大阅兵仪式。此次亮相的\u0027东风-5C\u0027液体洲际战略导弹其射程可达一万三千零五十千米,横线上的数写作13050,省略万位后面的尾数约是1万;此次徒步方队有15个,每个方队352人,徒步方队的总人数是5280人,这个数读作五千二百八十。",
+        "knowledge_points":  [
+                                 "亿以内数的读,写法",
+                                 "三位数与两位数的乘法",
+                                 "亿以内数的近似数"
+                             ],
+        "source":  "25-26 四年级上·河南三门峡·期中"
+    },
+    {
+        "id":  "20",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\u0027\u003e\u0027\u0027\u003c\u0027或\u0027=\u0027。98947( )460300;76952300( )76986200;347890( )35 万;130×30( )20×160;6 平方千米( )5999 公顷;6 平方千米 5 公顷( )605 公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003c;\u003c;\u003e;\u003c;=",
+        "solution":  "【分析】根据题意,整数大小的比较方法:位数不同,位数多的数就大;位数相同,从高位比起,直到比较出大小为止。先把35万改写成350000,再与347890比较大小。先计算出算式的结果,再比较大小即可。10000平方米 = 1公顷,100公顷 = 1平方千米,单位不统一时,先化成相同单位,再比较大小。\n【详解】98947是五位数,460300是六位数,所以98947＜460300。76952300和76986200都是八位数,前三位相同,第四位5＜8,所以76952300＜76986200。35万 = 350000,347890＜350000,所以347890＜35万。130×30 = 3900,20×160 = 3200,3900＞3200,所以130×30＞20×160。6平方千米 = 600公顷,600公顷＜5999公顷,所以6平方千米＜5999公顷。6平方千米5公顷 = 600公顷+5公顷 = 605公顷,所以6平方千米5公顷 = 605公顷。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "公顷,平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·山东临沂·期中"
+    },
+    {
+        "id":  "21",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上 \u0027\u003e\u0027\u0027\u003c\u0027或\u0027=\u0027。n706 万( )607500003000 公顷( )3 平方千米 n360×11( )36×110398000000( )4 亿",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003e;=;\u003c",
+        "solution":  "【分析】整数比较大小时,要看位数。位数多的那个数就大;如果位数相同,就从最高位起依次比较。根据大数的改写方法,先把706万改写成7060000,再与60750000进行比较;明确单位间的进率,1平方千米 = 100公顷,先统一单位,再进行比较;根据积的变化规律,如果一个因数乘一个数,另一个因数除以同一个数(0除外),那么积不变;先把4亿改写成400000000,再与398000000进行比较即可。\n【详解】706万 = 7060000,7060000＜60750000,所以706万＜60750000。3平方千米 = 300公顷,3000公顷＞300公顷,所以3000公顷＞3平方千米。360×11 = (360÷10)×(11×10) = 36×110,所以360×11 = 36×110。4亿 = 400000000,398000000＜400000000,所以398000000＜4亿。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "公顷,平方千米的进率与换算",
+                                 "积的变化规律(整数乘法)",
+                                 "亿以上数的改写"
+                             ],
+        "source":  "25-26四年级上·山东济宁·期中"
+    },
+    {
+        "id":  "22",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "\u0027李阿姨买了3千克山竹,共花了90元,每千克山竹多少元。\u0027题目中已知()和( ),求( )。列式为( )。",
+        "options":  [
+
+                    ],
+        "answer":  "数量;总价;单价;90÷3 = 30(元)",
+        "solution":  "【分析】根据题意,已知\u0027李阿姨买了3千克山竹,共花了90元,每千克山竹多少元。\u00273千克是数量,90元是总价,根据单价 = 总价÷数量,用90除以3,求出每千克山竹的价格,以此答题即可。\n【详解】根据分析可知:\u0027李阿姨买了3千克山竹,共花了90元,每千克山竹多少元。\u0027题目中已知数量和总价,求单价。列式为90÷3 = 30(元)。",
+        "knowledge_points":  [
+                                 "经济问题"
+                             ],
+        "source":  "25-26四年级上·山东济宁·期中"
+    },
+    {
+        "id":  "23",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "甲地到乙地的路程是160千米,李叔叔开车从甲地到乙地用了2小时,李叔叔开车的平均速度写作( )。",
+        "options":  [
+
+                    ],
+        "answer":  "80千米/时",
+        "solution":  "【分析】根据路程 = 速度×时间,用甲地到乙地的路程除以开车用的时间,即可求出李叔叔开车的平均速度。速度的简记法:先写上米(或千米等)数,再写\u0027/分\u0027或\u0027/时\u0027等即可。\n【详解】160÷2 = 80(千米/时)。甲地到乙地的路程是160千米,李叔叔开车从甲地到乙地用了2小时,李叔叔开车的平均速度写作80千米/时。",
+        "knowledge_points":  [
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26 四年级上·湖南长沙·期中"
+    },
+    {
+        "id":  "24",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "两个数相乘,积是180。如果一个因数除以 3,另一个因数乘 3,那么积是()。",
+        "options":  [
+
+                    ],
+        "answer":  "180",
+        "solution":  "【分析】根据积的变化规律,一个因数不变,另一个因数乘或除以一个不为0的数,积也要乘或除以这个数。如果一个因数除以3,另一个因数乘3,那么积需要先除以3再乘3,据此填空即可。\n【详解】180÷3×3 = 60×3 = 180。两个数相乘,积是180。如果一个因数除以3,另一个因数乘3,那么积是180。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·湖南长沙·期中"
+    },
+    {
+        "id":  "25",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "王老师家到学校大约有6500米。如果他骑车的速度是198米/分,他从家到学校骑车 31 分钟能到达吗。请列出你的估算算式( )。",
+        "options":  [
+
+                    ],
+        "answer":  "200×31 = 6200 (米)",
+        "solution":  "【分析】路程 = 速度×时间,用他骑车的速度乘骑车的时间等于他31分钟骑行的路程。计算时把198看作200,然后乘31即可估算出他31分钟骑行的路程。由于估大了,实际骑行的路程小于估算的路程,再与6500米进行比较即可解答。\n【详解】198×31≈200×31 = 6200(米),6200米＜6500米。由于估大了,所以他实际31分钟骑行的路程小于6200米,因此不能到达。所以,估算算式为:200×31 = 6200(米)。",
+        "knowledge_points":  [
+                                 "三位数乘两位数的估算",
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26四年级上·山东菏泽·期中"
+    },
+    {
+        "id":  "26",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个成年人一步长65 厘米,一亿步有( )千米。",
+        "options":  [
+
+                    ],
+        "answer":  "65000",
+        "solution":  "【分析】根据1千米 = 1000米 = 100000厘米,用100000000乘65厘米,再除以进率100000即可求解。\n【详解】100000000×65÷100000 = 65000(千米)。即一个成年人一步长65厘米,一亿步有65000千米。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有 0",
+                                 "厘米和米之间的进率与换算",
+                                 "千米和米之间的进率与换算"
+                             ],
+        "source":  "25-26四年级上·山东菏泽·期中"
+    },
+    {
+        "id":  "27",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一辆汽车的速度是每小时行驶 72 千米,可以写作( );一架超音速飞机 2 秒钟可以飞行 800 米,它的速度可以写作( )。",
+        "options":  [
+
+                    ],
+        "answer":  "72 千米/时;400 米/秒",
+        "solution":  "【分析】每小时(或每分钟等)行的路程,叫作速度;速度单位的写法是\u0027所走的路程单位/时间单位\u0027。因为题目中超音速飞机2秒钟飞行800米,先利用除法求出每秒飞行路程,再解答。\n【详解】800÷2 = 400(米)。一辆汽车的速度是每小时行驶72千米,可以写作72千米/时;一架超音速飞机2秒钟可以飞行800米,它的速度可以写作400米/秒。",
+        "knowledge_points":  [
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26 四年级上·河南南阳·期中"
+    },
+    {
+        "id":  "28",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "学校组织\u0027广西民俗知识竞赛\u0027,共有 12 个班,每班派 32 人参赛,一共( )人参赛;若学校给每个参赛学生准备 15 元的奖品,一共需要准备( )元。",
+        "options":  [
+
+                    ],
+        "answer":  "384;5760",
+        "solution":  "【分析】用每班派的人数乘12个班,即可求得一共多少人参赛。再用15元乘参赛人数,即可求得一共需要准备多少元.\n【详解】 32×12 = 384 (人) 384×15 = 5760 (元) 所以一共 384 人参赛,一共需要准备 5760 元。",
+        "knowledge_points":  [
+                                 "两位数乘两位数的不进位乘法",
+                                 "三位数乘两位数的实际问题"
+                             ],
+        "source":  "25-26四年级上·广西玉林·期中"
+    },
+    {
+        "id":  "29",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "南宁到桂林铁路长431 千米,列车上午 9 时出发,速度 180 千米/时,上午 11时列车离桂林还有( )千米;列车往返一次共行驶( )千米。",
+        "options":  [
+
+                    ],
+        "answer":  "71;862",
+        "solution":  "【分析】用11时-9时算出列车行驶的时间,根据路程 =速度×时间算出列车这段时间共行驶了多少千米,再用431减去已经行驶的距离就是离桂林的距离;南宁到桂林铁路长431千米,往返一次就是2个431千米,用乘法计算.\n【详解】11 时-9 时 = 2 (小时) 180×2 = 360 (千米) 431 - 360 = 71 (千米) 431×2 = 862 (千米)所以上午11时列车离桂林还有71千米;列车往返一次共行驶 862 千米。",
+        "knowledge_points":  [
+                                 "三位数与一位数的乘法口算",
+                                 "因数中间或末尾是0的乘法",
+                                 "基础行程问题",
+                                 "时,分,秒有关的计算"
+                             ],
+        "source":  "25-26四年级上·广西玉林·期中"
+    },
+    {
+        "id":  "30",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "爸爸从玉林开车去百色旅游,每小时行驶 90 千米,5 小时到达。解决\u0027玉林到百色的距离\u0027的数量关系式是( ),结果是( )千米。",
+        "options":  [
+
+                    ],
+        "answer":  "速度×时间 = 路程;450",
+        "solution":  "【分析】在行程问题中,路程的计算遵循\u0027速度×时间 = 路程\u0027的数量关系;代入数值计算出结果。据此解答.\n【详解】 90×5 = 450 (千米) 所以爸爸从玉林开车去百色旅游,每小时行驶90千米,5小时到达。解决\u0027玉林到百色的距离\u0027的数量关系式是速度×时间 = 路程,结果是 450 千米。",
+        "knowledge_points":  [
+                                 "整十,整百,整千数与一位数的乘法",
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26四年级上·广西玉林·期中"
+    },
+    {
+        "id":  "31",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "广西自然保护区占地 5000 公顷,合( )平方千米;若1平方千米有120种植物,整个保护区约有( )种植物。",
+        "options":  [
+
+                    ],
+        "answer":  "50;6000",
+        "solution":  "【分析】根据1平方千米 = 100公顷,5000公顷里面有50个100公顷,也就是有50个1平方千米;若1平方千米有120种植物,求整个保护区约有多少种植物,也就是求50个120是多少,用乘法解决。据此解答即可。\n【详解】1平方千米 = 100公顷,5000公顷里面有50个100公顷,也就是有50个1平方千米,即5000公顷合50平方千米;120×50 = 6000(种)。所以若1平方千米有120种植物,整个保护区约有6000种植物。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "公顷,平方千米的进率与换算"
+                             ],
+        "source":  "25-26四年级上·广西玉林·期中"
+    },
+    {
+        "id":  "32",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "根据:速度×( )= 路程,小花每分钟行 230 米,5 分钟共行( )米。",
+        "options":  [
+
+                    ],
+        "answer":  "时间;1150",
+        "solution":  "【分析】第一个空,根据速度,时间,路程的关系公式\u0027速度×时间 = 路程\u0027,应填\u0027时间\u0027。第二个空是计算路程,用速度 230 米/分乘时间 5 分钟计算即可.\n【详解】速度×时间 = 路程 230×5 = 1150 (米) 根据:速度×时间 = 路程,小花每分钟行 230 米,5 分钟共行 1150 米。",
+        "knowledge_points":  [
+                                 "两,三位数与一位数连续进位的乘法",
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26 四年级上·山东菏泽·期中"
+    },
+    {
+        "id":  "33",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上\u0027\u003e\u0027\u0027\u003c\u0027或\u0027=\u0027。78 万( )780000;100 个百万( )1 亿;10 个千万( )99999999;420×50( )820×30。",
+        "options":  [
+
+                    ],
+        "answer":  "=;=;\u003e;\u003c",
+        "solution":  "【分析】本题考查大数的比较和乘法计算。对于第一组,将\u002778万\u0027转换为数字后与780000比较;第二组,计算\u0027100个百万\u0027的值并与\u00271亿\u0027比较;第三组,计算\u002710个千万\u0027的值并与99999999比较;第四组,分别计算420×50和820×30的积,再比较大小。\n【详解】78万 = 78×10000 = 780000,所以78万 = 780000。100个百万 = 100×1000000 = 100000000,所以100个百万 = 1亿。10个千万 = 10×10000000 = 100000000,100000000＞99999999,所以10个千万＞99999999。420×50 = 21000,820×30 = 24600,21000＜24600,所以420×50＜820×30。",
+        "knowledge_points":  [
+                                 "亿以内数的组成",
+                                 "大数的比较",
+                                 "三位数乘两位数,乘数末尾有0"
+                             ],
+        "source":  "25-26四年级上·山东菏泽·期中"
+    },
+    {
+        "id":  "34",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "猎豹是陆地上跑得最快的动物,每分钟可以跑约2000米,它的速度记作约( ),照这个速度,猎豹奔跑 15 分钟,跑过的路程约是()米。",
+        "options":  [
+
+                    ],
+        "answer":  "2000 米/分;30000",
+        "solution":  "【分析】根据题意,速度是单位时间内通过的路程,猎豹每分钟跑2000米,因此速度记作\u00272000米/分\u0027;路程等于速度乘时间,已知速度是2000米/分,时间是15分钟,所以路程为 2000×15 = 30000 (米)。以此答题即可.\n【详解】根据分析可知: 2000×15 = 30000 (米) 猎豹是陆地上跑得最快的动物,每分钟可以跑约2000米,它的速度记作约2000米/分,照这个速度,猎豹奔跑15分钟,跑过的路程约是30000米。",
+        "knowledge_points":  [
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26四年级上·新疆乌鲁木齐·期中"
+    },
+    {
+        "id":  "35",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "学习了积的变化规律以后,明明研究一个算式: A×B = 20000 ,如果要使积变为 200,明明可以这样做:A( ),B( )。",
+        "options":  [
+
+                    ],
+        "answer":  "缩小为原来的 1/100;不变",
+        "solution":  "【分析】本题考查积的变化规律:一个因数不变,另一个因数乘或除以几(0除外),积也乘或除以几。要使积从20000变为200,积应缩小为原来的1/100。\n【详解】因为A×B = 20000,当B不变,A缩小为原来的1/100时,新的积为(A×1/100)×B = 20000×1/100 = 200。所以明明可以这样做:A缩小为原来的1/100,B不变。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·河南三门峡·期中"
+    },
+    {
+        "id":  "36",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "请根据 15×12 =180,直接写出下面各题的得数。n 15×24 = ( ) 15×48 = ( ) \n 30×12 = ( ) 150×24 = ( ) \n",
+        "options":  [
+
+                    ],
+        "answer":  "360;720;360;3600",
+        "solution":  "【分析】积的变化规律:两个数相乘,一个因数不变,另一个因数乘几或除以几(0除外),积也乘几或除以相同的数。以此答题即可.\n【详解】根据分析可知: 15×12 = 180 \n 15×24 = 15×12×2 = 180×2 = 360 \n 15×48 = 15×12×4 = 180×4 = 720 \n 30×12 = 15×2×12 = 15×12×2 = 180×2 = 360 \n 150×24 = 15×10×12×2 = 15×12×20 = 180×20 = 3600 ",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26四年级上·河南三门峡·期中"
+    },
+    {
+        "id":  "37",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "两个因数的积是140,如果一个因数不变,另一个因数乘 20,则积是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "2800",
+        "solution":  "【分析】根据积的变化规律:在乘法运算中,一个因数不变,另一个因数乘几(0除外),积也乘相同的数,据此解答。\n【详解】已知两个因数的积是140,一个因数不变,另一个因数乘20,那么积也要乘20。140×20 = 2800,因此变化后的积是2800。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·山东济宁·期中"
+    },
+    {
+        "id":  "38",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "两个因数相乘,一个因数乘 10,另一个因数不变,积( )。",
+        "options":  [
+
+                    ],
+        "answer":  "乘10",
+        "solution":  "【分析】根据积的变化规律,在乘法运算中,如果一个因数不变,另一个因数乘几(0除外),积也乘相同的数。本题中一个因数乘10,另一个因数不变,因此积也乘10。以此答题即可。\n【详解】例如:2×3 = 6,2×(3×10) = 2×30 = 60。两个因数相乘,一个因数乘10,另一个因数不变,积乘10。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·新疆乌鲁木齐·期中"
+    },
+    {
+        "id":  "39",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "如果 A×B = 60,那么 A×(B/4) = ( );(A×99)×(B/99) = ( )。",
+        "options":  [
+
+                    ],
+        "answer":  "15;60",
+        "solution":  "【分析】一个乘数不变,另一个乘数乘几或除以几(0除外),积也乘几或除以几(0除外)。一个乘数乘几或除以几(0除外),另一个乘数反而除以几或乘几(0除外),积不变。\n【详解】A×(B÷4) = (A×B)÷4 = 60÷4 = 15。(A×99)×(B÷99) = A×B = 60。所以如果A×B = 60,那么A×(B/4) = 15,(A×99)×(B/99) = 60。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·河南南阳·期中"
+    },
+    {
+        "id":  "40",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "观察下面算式,找出规律写出得数。37037×3 = 111111,37037×6 = 222222,37037×9 = 333333。37037×12 = ( );37037×24 = ( )。",
+        "options":  [
+
+                    ],
+        "answer":  "444444;888888",
+        "solution":  "【分析】观察算式规律:第一个因数都是37037,积是一个六位数,且每位上的数字相同。37037×3 = 111111,37037×6 = 222222,37037×9 = 333333。规律总结:当第二个因数是3的几倍时,结果就是由6个相同数字组成的数。\n【详解】3×4 = 12,所以37037×12 = 444444;3×8 = 24,所以37037×24 = 888888。",
+        "knowledge_points":  [
+                                 "算式的规律(整数)",
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "24-25 五年级上·安徽芜湖·期末"
+    },
+    {
+        "id":  "41",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在( )里填上\u0027\u003e\u0027\u0027\u003c\u0027或\u0027=\u0027。18×500( )50×180;2690010000( )27万;40200( )42000。",
+        "options":  [
+
+                    ],
+        "answer":  "=;\u003e;\u003c",
+        "solution":  "【分析】第一空:18×500和50×180的比较,将两个算式的结果计算出来再比较即可。第二、三空:整数比较大小时,位数不同的数,位数多的那个数大;位数相同的数,从最高位起依次比较。\n【详解】18×500 = 9000,50×180 = 9000,所以18×500 = 50×180。27万 = 270000,270000是六位数,2690010000是十位数,所以2690010000＞27万。40200与42000都是五位数,万位相等,比较千位,0＜2,所以40200＜42000。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "亿以内数的改写"
+                             ],
+        "source":  "25-26 四年级上·山东临沂·期中"
+    },
+    {
+        "id":  "42",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一座大桥全长 6790米,一列火车通过大桥时每秒行驶 65 米,这列火车从车头开上大桥到车尾离开大桥共需 108 秒。这列火车车身长()米。",
+        "options":  [
+
+                    ],
+        "answer":  "230",
+        "solution":  "【分析】火车从车头开上大桥到车尾离开大桥,所行驶的距离包括大桥全长和火车车身长。根据速度和时间,先计算出总行驶距离,再减去大桥全长,即可得到火车车身长.\n【详解】 65×108 = 7020 (米) 7020 - 6790 = 230 (米) 所以这列火车车身长 230 米。",
+        "knowledge_points":  [
+                                 "火车过桥问题",
+                                 "三位数乘两位数,三位数中间有 0"
+                             ],
+        "source":  "25-26 四年级上·山东临沂·期中"
+    },
+    {
+        "id":  "43",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "两个因数的积是240,如果其中一个因数乘 5,另一个因数乘 2 后,它们的积是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "2400",
+        "solution":  "【分析】两个因数的积是240,当其中一个因数乘5,另一个因数乘2时,根据乘积的变化规律,积会乘5×2 = 10。以此答题即可。\n【详解】240×10 = 2400。所以两个因数的积是240,如果其中一个因数乘5,另一个因数乘2后,它们的积是2400。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·山东临沂·期中"
+    },
+    {
+        "id":  "44",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "新国标规定电动自行车每小时行驶的路程不超过25千米,即电动自行车的最高速度为( )。李叔叔的电动自行车5分钟行驶了2500米,( )(填\u0027符合\u0027或\u0027不符合\u0027)这个规定。",
+        "options":  [
+
+                    ],
+        "answer":  "25 千米/小时;不符合",
+        "solution":  "【分析】速度指单位时间内所行的路程,其单位的表示方法是\u0027所走的路程单位/时间单位\u0027,一般用千米/时、米/分来表示。据此表示出电动自行车的最高速度;李叔叔的电动自行车5分钟行驶了2500米,1小时 = 60分钟,先求出60分钟里面有几个5分钟,再乘2500米,即可求出1小时行驶的路程,再根据1千米 = 1000米,将1小时行驶的路程换算成千米,就可求出李叔叔的电动自行车速度,再与最高速度进行比较。据此解答。\n【详解】由分析可得:新国标规定电动自行车每小时行驶的路程不超过25千米,即电动自行车的最高速度为25千米/时。60÷5×2500 = 12×2500 = 30000(米),30000米 = 30千米,即李叔叔的电动自行车速度为30千米/时。30千米/时＞25千米/时,所以李叔叔的电动自行车5分钟行驶了2500米,不符合这个规定。",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "时,分的认识",
+                                 "千米和米之间的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·湖北武汉·期中"
+    },
+    {
+        "id":  "45",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\u0027\u003e\u0027\u0027\u003c\u0027或\u0027=\u0027。78万( )78000;640×30( )320×60;370×50( )37×500;1周角( )2直角。",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;=;=;\u003e",
+        "solution":  "【分析】(1)先将78万后面加4个0转化为数字形式,然后与78000比较大小即可;(2)分别计算640×30和320×60,然后比较积的大小即可;(3)分别计算370×50和37×500,然后比较积的大小即可;(4)1周角=360°,1直角=90°,据此解答.\n【详解】(1)78万=780000,780000\u003e78000,所以78万\u003e78000;(2)640×30=19200,320×60=19200,所以640×30=320×60;(3)370×50=18500,37×500=18500,所以370×50=37×500;(4)1周角=360°,1直角=90°,2直角=90°+90°=180°,360°\u003e180°,所以1周角\u003e2直角。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "直角,钝角,锐角的认识及特征",
+                                 "平角,周角的认识及特征"
+                             ],
+        "source":  "25-26 四年级上·湖北武汉·期中"
+    },
+    {
+        "id":  "46",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "点点每天早上以85米/分的速度步行上学,18分钟可以到达学校。放学时妈妈骑电动车接他回到家,只用了6分钟。(1)算式 85×18 求的是( )。(2)要求放学时妈妈骑电动车接点点回家的速度,列式为()。",
+        "options":  [
+
+                    ],
+        "answer":  "从家到学校的路程;85×18÷6",
+        "solution":  "【分析】(1)一共行了多长的路,叫做路程;每小时(或每分钟等)行的路程,叫做速度;行了几小时(或几分钟等),叫做时间;速度×时间 = 路程,路程÷时间 = 速度,依此解答。\n(2)求出从家到学校的距离,已知妈妈骑电动车接点点回家,时间是6分钟。速度 = 路程÷时间。\n【详解】(1)根据分析可知,算式85×18表示从家到学校的路程。\n(2)要求放学时妈妈骑电动车接点点回家的速度,列式为85×18÷6。",
+        "knowledge_points":  [
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26四年级上·福建三明·期中"
+    },
+    {
+        "id":  "47",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\u0027\u003e\u0027\u0027\u003c\u0027或\u0027=\u0027。n26×200 ( ) 20×260350×70 ( ) 305×70n2 平角( )8 直角 120 公顷( )12 平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "=;\u003e;\u003c;\u003c",
+        "solution":  "【分析】三位数乘两位数的计算方法:两位数乘三位数,先用两位数个位上的数去乘三位数,再用两位数十位上的数去乘三位数,然后把两次乘得的结果加起来。当乘数末尾有0时,先算0前面的数,再在积的末尾添上对应个数的0。据此计算出算式的结果,再比较大小即可。1平角 = 180°,2平角是360°;1直角 = 90°,8直角是720°;再比较360°与720°的大小即可。1平方千米 = 100公顷,12平方千米就是1200公顷,据此先统一单位,再比较大小即可。\n【详解】26×200 = 5200,20×260 = 5200,所以26×200 = 20×260。350×70 = 24500,305×70 = 21350,24500＞21350,所以350×70＞305×70。2平角 = 360°,8直角 = 720°,360°＜720°,所以2平角＜8直角。12平方千米 = 1200公顷,120公顷＜1200公顷,所以120公顷＜12平方千米。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "直角,钝角,锐角的认识及特征",
+                                 "平角,周角的认识及特征",
+                                 "公顷,平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·福建三明·期中"
+    },
+    {
+        "id":  "48",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "两个因数的积是 100,如果一个因数缩小到原来的一半,要使积不变,则另一个因数应( )。",
+        "options":  [
+
+                    ],
+        "answer":  "扩大到原来的 2 倍",
+        "solution":  "【分析】两个数相乘,如果其中一个因数除以一个数(0除外),要使积不变,则另一个因数要乘相同的数。据此解答。\n【详解】两个因数的积是100,如果一个因数缩小到原来的一半,相当于这个因数除以2,要使积不变,则另一个因数应扩大到原来的2倍。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·湖北黄冈·期中"
+    },
+    {
+        "id":  "49",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "奇奇和佳佳做游戏。他们分别用0,3,5,7,9组成三位数乘两位数的乘法算式,奇奇组成的乘法算式的积最大,佳佳组成的乘法算式的积最小。奇奇组成的乘法算式是( ),佳佳组成的乘法算式是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "93×750或75×930;30×579",
+        "solution":  "【分析】三位数乘两位数:相同数位对齐,从个位乘起,用两位数分别依次乘三位数中的每一位数,每次乘得结果满几十向前一位进几,与哪一位上的数相乘,就在那一位的下面写上相应的积,然后将两次的积相加即可。要使三位数乘两位数的积最大,应使两位数的十位和三位数的百位尽可能大,即数字9和7。剩余数字0,3,5通过计算比较,使三位数的十位尽可能大,两位数的个位尽可能小,得到算式 93×750 或 75×930 ,乘积相同,均为69750。要使积最小,应使两位数的十位和三位数的百位尽可能小,即数字3和5。剩余数字0,7,9通过计算比较,使两位数的个位尽可能小(为0),三位数的十位次小(为7),得到算式 30×579 ,乘积为 17370.\n【详解】积最大:两位数的十位和三位数的百位应放最大的数字 9 和 7。若两位数的十位为 9,三位数的百位为 7,则剩余数字0,3,5。应使三位数的十位数字最大(即5),两位数的个位数字最小(即3),三位数的个位为0,得到算式 93×750.93×750 = 69750 或 75×930 = 69750;积最小:两位数的十位和三位数的百位应放最小的可能数字3和5。若两位数的十位为3,三位数的百位为5,则剩余数字0,7,9。应使两位数的个位数字最小(即0),三位数的十位数字次小(即7),三位数的个位为9,得到算式 30×579.30×579 = 17370 因此,奇奇组成的乘法算式是 93×750 或 75×930 ,佳佳组成的乘法算式是 30×579 。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "25-26四年级上·山东济南·期中"
+    },
+    {
+        "id":  "50",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "50×120 积的末尾有两个 0。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】计算50×120的积,并判断末尾0的个数。依据因数末尾有0的乘法计算方法:先计算0前面的数相乘,5×12 = 60,再看因数末尾共有几个0。50有一个0,120有一个0,共两个0,因此积为6000。据此作答。\n【详解】50×120 = 6000,6000的末尾有三个0,不是两个0。因此,题目的说法是错误的。故答案为:×。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有 0"
+                             ],
+        "source":  "25-26 四年级上·河北唐山·期末"
+    },
+    {
+        "id":  "51",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "李东骑自行车每分钟行 225 米,6 分钟行多少米。要用\u0027路程/时间 = 速度\u0027来计算。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】已知速度(每分钟行 225 米)和时间(6 分钟),求路程。正确的计算公式应为路程 = 速度×时间。题干中要求使用路程/时间 = 速度来计算,但该公式用于求速度,而非求路程,因此说法错误.\n【详解】李东骑自行车的速度是225米/分钟,时间是6分钟。根据路程 = 速度×时间,列式计算: 225×6 = 1350 (米)。题干中建议的公式路程/时间 = 速度适用于已知路程和时间求速度的情况,与本题所求不符,因此题干说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "两,三位数与一位数连续进位的乘法",
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26四年级上·山东济宁·期中"
+    },
+    {
+        "id":  "52",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "判断题",
+        "difficulty":  0.94,
+        "question":  "已知单价和数量可以求总价。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】根据总价,单价,数量的定义,单价是每件商品的价钱,数量是购买的商品件数,总价是一共用的钱。关系式为总价 = 单价×数量。因此,已知单价和数量,可以通过乘法求出总价.\n【详解】已知单价和数量,根据总价 = 单价×数量,可以求出总价。所以题中说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "经济问题"
+                             ],
+        "source":  "25-26 四年级上·山东菏泽·期中"
+    },
+    {
+        "id":  "53",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "三位数乘三位数,积最多是四位数。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】要判断\u0027三位数乘三位数,积最多是四位数\u0027是否正确,需考查积的最大可能值。三位数的定义范围是100到 999(均为整数),其中最大三位数是 999。计算999×999 的积,根据积的位数验证结论.\n【详解】 999×999 = 998001 ,最大的三位数乘最大的三位数相乘,所得的积是六位数。因此,三位数乘三位数的积最多是六位数,不是四位数。故答案为: ×",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "25-26四年级上·湖北黄冈·期中"
+    },
+    {
+        "id":  "54",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "若□76×35的积是五位数,则□里最小填4。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】三位数乘两位数时,相同数位对齐,从个位乘起。□里依次填1,2,3,4,5,然后计算算式的积,判断□76×35的积是五位数时,□里最小填的数。\n【详解】当□ = 1时,176×35 = 6160,积是四位数;当□ = 2时,276×35 = 9660,积是四位数;当□ = 3时,376×35 = 13160,积是五位数;当□ = 4时,476×35 = 16660,积是五位数。所以积是五位数时,□里最小填3。因此,若□76×35的积是五位数,则□里最小填4的说法错误。故答案为:×。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "25-26四年级上·山东济南·期中"
+    },
+    {
+        "id":  "55",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "列式计算(能列综合算式的要列综合算式)。205的64倍是多少。",
+        "options":  [
+
+                    ],
+        "answer":  "13120",
+        "solution":  "【分析】求一个数的几倍是多少,用乘法计算;求205的64倍,用205乘64,列式计算即可。\n【详解】205×64 = 13120。205的64倍是13120。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,三位数中间有0",
+                                 "求一个数的几倍是多少"
+                             ],
+        "source":  "25-26 四年级上·山东济宁·期中"
+    },
+    {
+        "id":  "56",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "计算题",
+        "difficulty":  0.85,
+        "question":  "直接写出得数。150×40 = ( );270÷3 = ( );75×40 = ( );320×30 = ( );540 - 310 = ( );630 + 170 = ( );39×296≈( );303×68≈( )。",
+        "options":  [
+
+                    ],
+        "answer":  "6000;90;3000;9600;230;800;12000;21000",
+        "solution":  "【分析】整百数、几百几十数乘整十数的口算:先把0前面的数相乘,再看两个乘数末尾一共有几个0,就在积的末尾添上几个0;几百几十数除以一位数的口算:先用被除数0前面的数除以除数,再在商的末尾添上与被除数末尾相同个数的0;整百数、几百几十数的加减法口算:相同数位对齐,从个位算起,或把数看作几个十、几个百再计算;三位数乘两位数的估算:把乘数看作与它接近的整十数、整百数,再口算出结果。\n【详解】150×40:先算15×4 = 60,两个乘数末尾共有2个0,在60后面添2个0,得6000;270÷3:27÷3 = 9,被除数末尾有1个0,在9后面添1个0,得90;75×40:先算75×4 = 300,乘数末尾有1个0,在300后面添1个0,得3000;320×30:先算32×3 = 96,两个乘数末尾共有2个0,在96后面添2个0,得9600;540-310:540-300 = 240,240-10 = 230;630+170:630+100 = 730,730+70 = 800;39×296≈:把39看作40,296看作300,40×300 = 12000;303×68≈:把303看作300,68看作70,300×70 = 21000。",
+        "knowledge_points":  [
+                                 "整百数,几百几十数的加减法口算",
+                                 "两位数乘整十数的口算乘法",
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "三位数乘两位数的估算"
+                             ],
+        "source":  "25-26 四年级上·河南开封·期中"
+    },
+    {
+        "id":  "57",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "笔算下面各题。285×64 = ( );32×450 = ( );87×506 = ( );327×54 = ( );602×70 = ( )。",
+        "options":  [
+
+                    ],
+        "answer":  "18240;14400;44022;17658;42140",
+        "solution":  "【分析】三位数乘两位数的竖式计算方法:数位对齐,先用两位数的个位分别从右往左与三位数的每一位数相乘;再用两位数的十位分别从右往左与三位数的每一位数相乘,乘得结果的个位要与前面结果的十位对齐;然后两个结果相加就得到三位数乘两位数的结果了;要注意满十往前进位.\n【详解】 285×64=18240n 285n×64n ---------n 1140 ← 285×4=1140n 1710 ← 285×60=17100(末位与十位对齐)n ---------n 18240n32×450=14400n 32n×450n ---------n 160 ← 32×5=160n 128 ← 32×40=1280(末位与十位对齐)n ---------n 14400 ← 末尾补1个0n87×506=44022n 87n×506n ---------n 522 ← 87×6=522n 000 ← 87×0=0(可省略)n 435 ← 87×500=43500(末位与百位对齐)n ---------n 44022n327×54=17658n 327n×54n ---------n 1308 ← 327×4=1308n 1635 ← 327×50=16350(末位与十位对齐)n ---------n 17658n602×70=42140n 602n×70n ---------n 42140 ← 先算602×7=4214,再在末尾添1个0",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "三位数乘两位数,三位数中间有0"
+                             ],
+        "source":  "25-26 四年级上·湖北武汉·期中"
+    },
+    {
+        "id":  "58",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "某书店有 102 本少儿书,书店里图书的总数比少儿书的13倍多23本,该书店共有多少本图书。",
+        "options":  [
+
+                    ],
+        "answer":  "1349 本",
+        "solution":  "【分析】某书店有102本少儿书,书店里图书的总数比少儿书的13倍多23本,据此根据求一个数的几倍是多少,用乘法计算,即用这个书店里少儿书的本数乘13,再加上多的本数,即可求出该书店共有多少本图书。\n【详解】102×13 = 1326(本),1326+23 = 1349(本)。答:该书店共有1349本图书。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,三位数中间有0",
+                                 "乘加,乘减混合运算",
+                                 "求一个数的几倍是多少"
+                             ],
+        "source":  "24-25 四年级上·河北衡水·期末"
+    },
+    {
+        "id":  "59",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "利用下面的信息解决问题。(填序号)n① 王老师骑自行车的速度是 220 米/分n② 一根跳绳 8 元n③ 小兰从家到书店用了 13 分钟n④ 一艘航母模型 146 元n⑤ 张老师带了 120 元去买跳绳n⑥ 王老师 7:10 从家出发,7:35 到达学校n⑦ 文体店采购 34 艘航母模型n⑧ 小兰家和书店相距 845 米n(1)请你选择一组信息,提出一个已知总价和单价,求数量的问题。 选择的信息是( )和( )列式解答: (2)请你再选择一组信息,提出一个根据时间和速度,求路程的问题。选择的信息是( )和( )",
+        "options":  [
+
+                    ],
+        "answer":  "②;⑤;①;⑥",
+        "solution":  "【分析】(1)根据\u0027数量 = 总价÷单价\u0027,信息②给出了一根跳绳的单价是8元,信息⑤给出了张老师带的总价是120元,所以选择②和⑤。(2)根据\u0027路程 = 速度×时间\u0027,信息①给出了王老师骑自行车的速度是220米/分,信息⑥给出了王老师从家出发和到达学校的时间,可算出时间为7时35分-7时10分 = 25分钟,所以选择①和⑥。\n【详解】根据分析:(1)选择的信息是②和⑤,可提出问题:张老师可以买多少根跳绳?列式:120÷8 = 15(根)。(2)选择的信息是①和⑥,可提出问题:王老师家到学校有多少米?列式:220×25 = 5500(米)。",
+        "knowledge_points":  [
+                                 "提问题",
+                                 "填条件问题",
+                                 "三位数除以一位数(首位不够除)",
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "2025-2026 学年四年级上·江西赣州·期中"
+    },
+    {
+        "id":  "60",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一片长方形的果园,长 860 米,宽 500 米。平均每公顷可以种 750 棵苹果树,那么这片果园总共可以种多少棵苹果树。",
+        "options":  [
+
+                    ],
+        "answer":  "32250 棵",
+        "solution":  "【分析】根据长方形的面积 = 长×宽,计算出长方形面积的平方米数。1公顷=10000平方米,在长方形面积的平方米数的末尾去掉4个0,换算单位为公顷。最后用长方形面积的公顷数乘每公顷可以种的苹果树数量,即可计算出这片果园总共可以种的苹果树数量。据此解答.\n【详解】 860×500 = 430000 (平方米) 430000 平方米 = 43 公顷 750×43 = 32250 (棵)答:这片果园总共可以种32250棵苹果树。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "长方形的面积",
+                                 "公顷,平方千米的进率与换算"
+                             ],
+        "source":  "25-26四年级上·江西赣州·期中"
+    },
+    {
+        "id":  "61",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "学校准备要给一年级的98名新同学每人配备一套新课桌,每张桌子115元,每把椅子 35 元,学校一共要花多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "14700 元",
+        "solution":  "【分析】单价×数量 = 总价。一套课桌包括一张桌子和一把椅子,所以先用每张桌子的价钱加上每把椅子的价钱算出一套课桌的价格,再乘学生人数98,即可得到总费用。\n也可以用各自的单价乘各自买的数量,算出买桌子和椅子各花多少钱,再相加,也是一共要花多少元。\n【详解】(115+35)×98 = 150×98 = 14700(元),或115×98+35×98 = 11270+3430 = 14700(元)。\n答:学校一共要花14700元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "三位数乘两位数的实际问题"
+                             ],
+        "source":  "25-26 四年级上·陕西榆林·期中"
+    },
+    {
+        "id":  "62",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一列火车从 A 城开往 B 城,前 3 小时行驶了 270 千米,照这样的速度,再行驶5 小时到达 B 城。A 城到 B 城的距离是多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "720 千米",
+        "solution":  "【分析】一列火车从 A 城开往 B 城,前 3 小时行驶了 270千米,据此根据速度 = 路程/时间,先算出这列火车的速度,照这样的速度,再行驶5小时到达B城,则用3加上5,先算出这列火车从A城到B城行驶的时间,再乘求出的速度即可求出A城到B城的距离是多少千米.\n【详解】 270/3×( 3 + 5 )\n= 270/3×8\n= 90×8 \n = 720 (千 米) \n答:A 城到 B 城的距离是 720 千米。",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "单归一问题"
+                             ],
+        "source":  "25-26 四年级上·湖北襄阳·期中"
+    },
+    {
+        "id":  "63",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "张亮一家人国庆节自驾到\u0027中国文字博物馆\u0027参观,早晨 7 时整出发,导航显示从三门峡市文博城到安阳文字博物馆全程约 410 千米。如果爸爸开车平均每小时行驶 100 千米,张亮一家大约什么时间到达博物馆。第二天返回时因天气下雨大约用了 5 小时,返回时爸爸开车平均每小时大约行多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "11 时;80 千米/时",
+        "solution":  "【分析】已知全程约410千米,爸爸开车平均每小时行驶100千米,根据时间 = 路程/速度得出行驶时间,再用开始时刻加上经过的时间得出张亮一家到达博物馆的时刻。计算时需要估算,把 410 看作 400 来计算;返回时用时5小时,根据路程/时间 = 速度得出返回时的速度。计算时需要估算,把 410 看作 400 计算.\n【详解】把410看作400,400/100 = 4 (时)7时+4时 = 11 时 400/5 = 80 (千米)答:张亮一家大约上午11时到达博物馆。返回时爸爸开车平均每小时大约行80千米。",
+        "knowledge_points":  [
+                                 "除数是一位数的估算",
+                                 "基础行程问题",
+                                 "时,分,秒时间的推算"
+                             ],
+        "source":  "25-26 四年级上·河南三门峡·期中"
+    },
+    {
+        "id":  "64",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "两辆汽车同时从 A点出发,向同一方向行驶。甲车每小时行驶 55 千米,乙车每小时行驶65千米,12小时后两车相距多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "120 千米",
+        "solution":  "【分析】两车从同一地点同时向同一方向行驶,乙车的速度(65千米/时)大于甲车的速度(55千米/时),因此乙车会逐渐领先甲车。12小时后两车相距的距离等于乙车行驶的路程减去甲车行驶的路程。根据路程 = 速度×时间,可先分别计算两车的路程再求差,或直接计算速度差(乙车速度减甲车速度)后乘时间得到相距距离.\n【详解】(65-55)×12= 10×12= 120 (千米) 答:12小时后两车相距120千米。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26 四年级上·山东临沂·期中"
+    },
+    {
+        "id":  "65",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "一块长方形绿地,宽是9米,面积是200平方米。现在想把宽扩大到36米,长不变,扩大后的绿地面积是多少平方米。",
+        "options":  [
+
+                    ],
+        "answer":  "800 平方米",
+        "solution":  "【分析】长方形面积公式:长方形面积 = 长×宽。长不变时,新的宽36米扩大到原来宽9米的倍数是36÷9 = 4倍,根据积的变化规律,面积也要扩大相同的倍数,因此新面积为原面积的4倍,据此解答。\n【详解】36÷9 = 4,200×4 = 800(平方米)。答:扩大后的绿地面积是800平方米。",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·山东临沂·期中"
+    },
+    {
+        "id":  "66",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "王叔叔带 650 元到电器商店买了3台电风扇,找回了20元,每台电风扇多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "210 元",
+        "solution":  "【分析】总价/数量 = 单价,先用 650 减去 20,求出 3 台电风扇的总价,再除以 3,即可求出每台电风扇多少元.\n【详解】(650-20)/3=630/3 = 210 (元) 答:每台电风扇 210 元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "25-26 四年级上·山东济宁·期中"
+    },
+    {
+        "id":  "67",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "小明一家自驾去南京旅游,去时汽车平均每小时行驶 80 千米,用了 7 小时。沿原路返回时用了 5 小时,返回时汽车平均每小时行驶多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "112 千米/时",
+        "solution":  "【分析】根据题意,已知小明一家自驾去南京旅游,去时汽车平均每小时行驶80千米,用了7小时。沿原路返回时用了5小时,首先根据速度×时间 = 路程,用去时的速度乘所用的时间即可求出两地之间的路程;然后根据速度 = 路程/时间,用两地之间的路程除以返回所用的时间即可;据此解答.\n【详解】根据分析可知: 80×7/5 = 560/5 = 112 (千 米) 答:返回的平均速度是 112 千米。",
+        "knowledge_points":  [
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26 四年级上·山东济宁·期中"
+    },
+    {
+        "id":  "68",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一辆汽车从甲地到乙地,去时的速度为57千米/时,共用4小时。返回时比去时少用 1 小时,这辆汽车返回的速度是多少。",
+        "options":  [
+
+                    ],
+        "answer":  "76 千米/时",
+        "solution":  "【分析】根据路程 = 速度×时间,先用去时的速度乘用的时间,求出甲地到乙地的距离,返回时比去时少用1小时,用去时的时间减去1求出返回时用的时间,用甲地到乙地的距离除以返回时用的时间,即可求出这辆汽车返回的速度是多少.\n【详解】 57×4 = 228 (千米) 228 / (4 - 1) = 228/3 = 76 (千 米 / 时) 答:这辆汽车返回的速度是76千米/时。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26四年级上·湖南长沙·期中"
+    },
+    {
+        "id":  "69",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一块长方形地里种了大豆,这块地的长是 500 米,宽比长少 200 米。每公顷可收大豆4吨,这块地一共可收大豆多少吨。",
+        "options":  [
+
+                    ],
+        "answer":  "60 吨",
+        "solution":  "【分析】根据题意,这块地的长是500米,宽比长少200米,先用500-200求出宽是多少米,根据长方形面积 = 长×宽,代入数字先计算出这块地的面积是多少平方米,根据1公顷 = 10000平方米,将面积换算成公顷为单位,再乘每公顷可收大豆的吨数,即可求出这块地一共可收大豆多少吨。\n【详解】500-200 = 300(米),500×300 = 150000(平方米),150000平方米 = 15公顷,15×4 = 60(吨)。答:这块地一共可收大豆60吨。",
+        "knowledge_points":  [
+                                 "三位数与整十数的口算乘法",
+                                 "长方形的面积",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26四年级上·湖南长沙·期中"
+    },
+    {
+        "id":  "70",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "4 位老师带领 102 个同学去秋游,每个人购买车票和门票共需要 62 元,准备6000 元买票够吗。",
+        "options":  [
+
+                    ],
+        "answer":  "不够",
+        "solution":  "【分析】4位老师带领102个同学去秋游,总人数是(4+102)人,再用每个人的车票和门票的价钱乘人数,算出需要的总钱数,最后与6000元比较,即可知够不够。\n【详解】4+102 = 106(人),106×62 = 6572(元),6572＞6000。答:准备6000元买票不够。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "三位数乘两位数,三位数中间有 0"
+                             ],
+        "source":  "25-26 四年级上·湖南长沙·期中"
+    },
+    {
+        "id":  "71",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "幸福村通过电商平台销售本地特产\u0027水晶梨\u0027。n(1)今年,王伯伯家的梨园喜获丰收,平均每棵梨树约产梨 125 千克。他家一共有 48 棵这样的梨树。估算一下,王伯伯家梨园今年大约一共能产梨多少千克。",
+        "options":  [
+
+                    ],
+        "answer":  "6000 千克",
+        "solution":  "【分析】(1)平均每棵梨树产梨的千克数乘梨树的棵数即等于一共能产梨的千克数,计算时把125看作120,48看作50,即可估算出能产梨的千克数。\n【详解】(1)125×48≈120×50 = 6000(千克)。答:王伯伯家梨园今年大约一共能产梨6000千克。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "提问题",
+                                 "填条件问题",
+                                 "三位数与两位数的乘法",
+                                 "三位数乘两位数的估算"
+                             ],
+        "source":  "25-26四年级上·山东菏泽·期中"
+    },
+    {
+        "id":  "72",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "实验小学四年级师生共205人,准备包车去春游。旅行社包车的价格是39元/人,老师带 8000 元够吗。请列出你的口算算式( )。",
+        "options":  [
+
+                    ],
+        "answer":  "205×39 = 205×40 - 205×1 = 8200 - 205 = 7995",
+        "solution":  "【分析】已知实验小学四年级师生共205人,准备包车去春游,旅行社包车的价格是39元/人,根据总价 = 单价×数量,用205乘39,即可求出一共需要的钱数,再与8000元比较即可知道老师带8000元够不够;计算时,把39看作40,用 205×40 多算了1个205,则用 205×40 减去多算的1个205即可,据此进行口算即可解答.\n【详解】 205×39\n= 205×40 - 205×1 \n = 8200 - 205 \n = 7995 \n7995 \u003c 8000 ,所以老师带8000元够的。即实验小学四年级师生共205人,准备包车去春游。旅行社包车的价格是39元/人,老师带8000元够吗。请列出你的口算算式 205×39 = 205×40 - 205×1 = 8200 - 205 = 7995 。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "三位数与整十数的口算乘法",
+                                 "三位数乘两位数,三位数中间有0"
+                             ],
+        "source":  "25-26 四年级上·山东菏泽·期中"
+    },
+    {
+        "id":  "73",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "李师傅要加工3500个零件,每天加工 218 个,加工了 14 天,还剩下多少个零件没加工。",
+        "options":  [
+
+                    ],
+        "answer":  "448 个",
+        "solution":  "【分析】已知每天加工218个,加工了14天,用每天加工零件个数乘加工天数,求出已经加工零件个数。再用零件总个数减去已经加工零件个数,求出还剩下零件个数.\n【详解】3500-218×14= 3500 - 3052\n= 448 (个)\n答:还剩下448个零件。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法",
+                                 "乘加,乘减混合运算"
+                             ],
+        "source":  "25-26四年级上·山东菏泽·期中"
+    },
+    {
+        "id":  "74",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一片长方形的果园,长 860 米,宽 500 米。平均每公顷可以种 750 棵苹果树,那么这片果园总共可以种多少棵苹果树。",
+        "options":  [
+
+                    ],
+        "answer":  "32250 棵",
+        "solution":  "【分析】根据长方形的面积 = 长×宽,用860米乘500米即可求出这块长方形果园的面积,再根据1公顷 = 10000 平方米即可将平方米换算为公顷,用公顷数乘750棵即可求出片果园总共可以种多少棵苹果树.\n【详解】 860×500 = 430000 (平方米) 430000/10000 = 43 (公顷) 43×750 = 32250 (棵)答:这片果园总共可以种 32250 棵苹果树。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有 0",
+                                 "长方形的面积",
+                                 "公顷,平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·山东菏泽·期中"
+    },
+    {
+        "id":  "75",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一处共享汽车充电站占地面积约是 2000 平方米,定陶区若建 40 个这样的充电站,它们的占地总面积约是多少平方米。合多少公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "80000 平方米;8 公顷",
+        "solution":  "【分析】用共享汽车充电站占地面积乘40即可求出40个这样的充电站的面积是多少平方米,再根据\u002710000平方米 = 1公顷\u0027,换算成以公顷作单位的数据即可。\n【详解】2000×40 = 80000(平方米),80000平方米 = 8公顷。答:它们的占地总面积约是80000平方米,合8公顷。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·山东菏泽·期中"
+    },
+    {
+        "id":  "76",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "小明一家\u0027十一\u0027期间,自己驾车从甲地到乙地,甲地到乙地的路程约为1200 千米,汽车每小时大约行驶 110 千米,如果他们一家,早上 7:00 出发,中间不停歇,晚上 7:00 能到达乙地吗。",
+        "options":  [
+
+                    ],
+        "answer":  "能",
+        "solution":  "【分析】根据题意,已知甲地到乙地的路程约为1200千米,汽车每小时大约行驶110千米。先把早上7:00和晚上7:00转换成24时计时法,经过时间 = 到达时间-出发时间,用110乘经过时间,求出汽车行驶的路程,再与1200进行比较即可。\n【详解】早上7:00就是7时,晚上7:00就是19时。110×(19-7) = 110×12 = 1320(千米),1320＞1200。答:晚上7:00能到达乙地。",
+        "knowledge_points":  [
+                                 "24 时计时法与普通计时法的互化",
+                                 "基础行程问题",
+                                 "24 时计时法时间的计算"
+                             ],
+        "source":  "25-26 四年级上·河南开封·期中"
+    },
+    {
+        "id":  "77",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "王阿姨批发了一些消毒液和花露水。她批发了 124 瓶花露水,每瓶 15 元。(),王阿姨批发这些商品一共花了多少元。(从下面选择一个合适的信息,把序号填在括号里,并解答)n①批发了120瓶消毒液,40瓶洗衣液。n②批发的消毒液比花露水少4瓶,每瓶消毒液32元n解答:",
+        "options":  [
+
+                    ],
+        "answer":  "②;5700 元",
+        "solution":  "【分析】根据题意,要求王阿姨批发这些商品一共花了多少元,就要知道花露水和消毒液的总价;用124乘15可以求出花露水的总价;题中缺少消毒液的数量和单价,所以选择条件②。先用124减去4,求出消毒液的数量,再乘32,即可求出消毒液的总价;最后加上花露水的总价,即可求解。\n【详解】选择②。124-4 = 120(瓶),124×15+120×32 = 1860+3840 = 5700(元)。答:王阿姨批发这些商品一共花了5700元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "提问题",
+                                 "填条件问题",
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "25-26 四年级上·山东菏泽·期中"
+    },
+    {
+        "id":  "78",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "乐乐从家骑车以每分钟行 165 米的速度去外婆家。早晨 8:05 出发,途中休息了 18 分,到达外婆家时是 8:40,乐乐家离外婆家有多远。",
+        "options":  [
+
+                    ],
+        "answer":  "2805 米",
+        "solution":  "【分析】根据\u0027经过的时间 = 结束时刻-开始时刻\u0027,求出从8:05到8:40所经过的时间,再减去18分钟,即可求出乐乐从家骑车去外婆家所需要的时间;再根据\u0027路程 = 速度×时间\u0027,即可求出乐乐家离外婆家有多远。\n【详解】8时40分-8时05分 = 35分,165×(35-18) = 165×17 = 2805(米)。答:乐乐家离外婆家有2805米。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法",
+                                 "基础行程问题",
+                                 "时,分,秒有关的计算"
+                             ],
+        "source":  "25-26 四年级上·河南南阳·期中"
+    },
+    {
+        "id":  "79",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一块长方形稻田,长 800 米,宽 350 米,这块稻田占地多少公顷。如果每公顷施肥 200 千克,这块稻田一共施肥多少千克。",
+        "options":  [
+
+                    ],
+        "answer":  "28 公顷;5600 千克",
+        "solution":  "【分析】根据题意,明确长方形的面积 = 长×宽。已知一块长方形稻田,长800米,宽350米,用800乘350求出面积,再根据1公顷 = 10000平方米,把平方米换算成公顷;再乘200,就是这块稻田一共施肥多少千克。列式计算即可。\n【详解】800×350 = 280000(平方米),280000平方米 = 28公顷。答:这块稻田占地28公顷。200×28 = 5600(千克)。答:这块稻田一共施肥5600千克。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "长方形的面积",
+                                 "公顷,平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·河南南阳·期中"
+    },
+    {
+        "id":  "80",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "在 2025 年 9 月 3 日中国人民抗日战争暨世界反法西斯战争胜利 80 周年阅兵中有 13 个徒步方队,每个徒步方队由 352 人组成,徒步方队一共有多少人。",
+        "options":  [
+
+                    ],
+        "answer":  "4576 人",
+        "solution":  "【分析】用徒步方队的人数乘徒步方队的数量即可求出徒步方队的总人数,据此解答.\n【详解】 352×13 = 4576 (人)答:徒步方队一共有 4576 人。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "25-26 四年级上·河南南阳·期中"
+    },
+    {
+        "id":  "81",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "广西六万大山森林公园规划分为:香海生态观光带和森林休闲度假社区,森林养生谷,山地运动休闲,中医药养生等四个功能区。现有一块长方形基地,长 800 米,宽 300 米,若平均每公顷可以种120棵八角,则这块基地能种多少棵八角。",
+        "options":  [
+
+                    ],
+        "answer":  "2880 棵",
+        "solution":  "【分析】首先根据长方形的面积 = 长×宽,代入数值计算出长方形的面积。然后根据1公顷 = 10000平方米,将长方形面积换算成公顷。最后用公顷数乘平均每公顷可以种的八角棵数,即可求出这块基地能种多少棵八角。据此解答。\n【详解】800×300 = 240000(平方米),240000平方米 = 24公顷,120×24 = 2880(棵)。答:这块基地能种2880棵八角。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "长方形的面积",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26四年级上·广西玉林·期中"
+    },
+    {
+        "id":  "82",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "杭州龙井茶茶园规划分为:核心保护区,生态种植区,休闲体验区等三个功能区。现有一块长方形茶园,长 600 米,宽 250 米,若平均每公顷可以种 150 棵茶树,则这块茶园能种多少棵茶树。",
+        "options":  [
+
+                    ],
+        "answer":  "2250 棵",
+        "solution":  "【分析】首先根据长方形的面积 = 长×宽,代入数值计算出长方形的面积。然后根据1公顷 = 10000平方米,将长方形面积换算成公顷。最后用公顷数乘平均每公顷可以种的茶树棵数,即可求出这块茶园能种多少棵茶树。据此解答。\n【详解】600×250 = 150000(平方米),150000平方米 = 15公顷,150×15 = 2250(棵)。答:这块茶园能种2250棵茶树。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有 0",
+                                 "长方形的面积",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·浙江杭州·期中"
+    },
+    {
+        "id":  "83",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "张叔叔上午驾车送蔬菜进城,去时用了4小时,每小时60千米;原路返回时加快了速度,少用1小时。返回时的平均速度是多少。",
+        "options":  [
+
+                    ],
+        "answer":  "80 千米/时",
+        "solution":  "【分析】根据题意,去时和返回时的路程相同。先利用去时的速度和时间求出路程,返回时比去时少用1小时,即用了(4-1)小时,再用路程除以返回时的时间,即可求出返回时的平均速度。\n【详解】60×4÷(4-1) = 240÷3 = 80(千米/时)。答:返回时的平均速度是80千米/时。",
+        "knowledge_points":  [
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26四年级上·新疆乌鲁木齐·期中"
+    },
+    {
+        "id":  "84",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "已知24×5 = 120,那么240×5 = ( );24×15 = ( );24×( ) = 12000。",
+        "options":  [
+
+                    ],
+        "answer":  "1200;360;500",
+        "solution":  "【分析】根据\u0027在乘法算式里,一个因数不变,另一个因数乘几或除以几(0除外),积也乘或除以几\u0027分析解答即可。\n【详解】已知24×5 = 120。因为24×10 = 240,所以240×5 = 120×10 = 1200;因为5×3 = 15,所以24×15 = 120×3 = 360;因为120×100 = 12000,所以24×(5×100) = 24×500 = 12000。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·广西河池·期中"
+    },
+    {
+        "id":  "85",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "8 月 22-24 日,河池市首届东南亚民族泼水大联欢活动在河池园博园举行,吸引了很多游客前来游玩。一个旅游团有 18 名游客,平均每人花了 705 元住宿费,他们一共花了多少元住宿费。",
+        "options":  [
+
+                    ],
+        "answer":  "12690 元",
+        "solution":  "【分析】有 18 名游客,平均每人花了 705 元住宿费,也就是 18 个 705,用乘法计算.\n【详解】 705×18 = 12690 (元)答:他们一共花了 12690 元住宿费。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,三位数中间有0"
+                             ],
+        "source":  "25-26 四年级上·广西玉林·期中"
+    },
+    {
+        "id":  "86",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "一个机器人平均每小时可以分拣435件快递,工作12小时后还剩200件快递没有分拣,这个机器人要分拣多少件快递。",
+        "options":  [
+
+                    ],
+        "answer":  "5420 件",
+        "solution":  "【分析】已知机器人平均每小时分拣435件,工作12小时,用乘法可得12小时分拣的件数。还剩200件未分拣,因此总共要分拣的件数为12小时分拣的件数加上剩下的件数。据此解答即可。\n【详解】435×12 = 5220(件),5220+200 = 5420(件)。答:这个机器人要分拣5420件快递。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "25-26 四年级上·河北唐山·期中"
+    },
+    {
+        "id":  "87",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "有一块占地面积是390 平方米的长方形草地,为了增加绿化面积,将草地的宽度由原来的 6 米增加到 18 米,长不变。扩建后草地的面积是多少。",
+        "options":  [
+
+                    ],
+        "answer":  "1170 平方米",
+        "solution":  "【分析】根据长方形面积 = 长×宽,在乘法算式里,一个因数不变,另一个因数乘几或除以几(0除外),积也乘或除以几。因此先用增加后的宽18米除以原来的宽6米,从而计算出宽扩大了几倍,然后用原来的面积也乘几,即可解题。\n【详解】390×(18÷6) = 390×3 = 1170(平方米)。答:扩建后草地的面积是1170平方米。",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·山东临沂·期中"
+    },
+    {
+        "id":  "88",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "星期六,爸爸开车带着妈妈和小兰去郊外游玩,去的时候用了 6 小时,第二天返回时比去时快了 1 小时。去时的速度为 65 千米/时。原路返回时,平均每小时行驶多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "78 千米/时",
+        "solution":  "【分析】根据路程 = 速度×时间,可求得去时的路程,也就是返回时的路程。根据第二天返回时比去时快了1小时,所用时间为(6-1)小时,用返回的路程/时间,即可求得原路返回时,平均每小时行驶多少千米.\n【详解】 65×6 = 390 (千米)6 - 1 = 5 (小时)390/5 = 78 (千米)答:原路返回时,平均每小时行驶78千米。",
+        "knowledge_points":  [
+                                 "两,三位数与一位数的一次进位乘法",
+                                 "三位数除以一位数,首位不够除",
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26四年级上·山东临沂·期中"
+    },
+    {
+        "id":  "89",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "如果不列竖式,你能根据 16×15 = 240 说明 16×45 的积是多少吗。",
+        "options":  [
+
+                    ],
+        "answer":  "720",
+        "solution":  "【分析】在乘法算式里,一个因数不变,另一个因数乘几或除以几(0除外),积也乘或除以几。15×3 = 45,所以积也应该乘3。据此解答即可。\n【详解】因为16×45 = 16×(15×3),所以16×45 = (16×15)×3 = 240×3 = 720。答:16×45的积是720。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·山东临沂·期中"
+    },
+    {
+        "id":  "90",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一块长方形试验田,一组对边增加 5 米后变成正方形,面积增加 1 公顷,这块试验田原来的面积是多少公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "399 公顷",
+        "solution":  "【分析】![](images/2d650a1cf692168d97e351f6d57d0bf7ea5f41ab20fe87ef102c1e0e04ca76cd.jpg)\n因为面积增加1公顷,即10000平方米,面积增加部分是宽为 5 米的长方形,所以能求出原来试验田的长。因为增加后的图形是正方形,所以用正方形的边长减去增加的 5 米,就是原来试验田的宽,再用原来的长乘原来的宽,求出原来的面积是多少平方米,再转化成用公顷做单位(1公顷=10000 平方米) .\n【详解】1公顷 = 10000 平方米 10000/5 = 2000 (米) 2000 - 5 = 1995 (米) 2000×1995 = 3990000 (平方米) = 399 (公顷)答:这块试验田原来的面积是399公顷。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "长方形的面积",
+                                 "公顷,平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·湖北武汉·期中"
+    },
+    {
+        "id":  "91",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "工厂要加工一批零件,3 小时共加工了 480 个。照这样计算,再加工 12 小时,一共可以加工多少个。",
+        "options":  [
+
+                    ],
+        "answer":  "2400 个",
+        "solution":  "【分析】根据\u0027工作效率 = 工作量÷工作时间\u0027,求出平均每小时加工的个数,再根据\u0027工作量 = 工作效率×工作时间\u0027,即可求出12小时加工的个数,用12小时加工的个数加上先加工的480个,就是一共可以加工的个数。\n【详解】480÷3 = 160(个),160×12+480 = 1920+480 = 2400(个)。答:一共可以加工2400个。",
+        "knowledge_points":  [
+                                 "多位数与一位数的乘除混合运算",
+                                 "三位数乘两位数,乘数末尾有 0",
+                                 "有具体量的工程问题"
+                             ],
+        "source":  "25-26 四年级上·广东汕尾·期中"
+    },
+    {
+        "id":  "92",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "有一块长方形绿地,长400米,宽150米。如果每公顷绿地一天大约可释放730千克氧气,那么这块绿地一天大约可释放多少千克氧气。",
+        "options":  [
+
+                    ],
+        "answer":  "4380 千克",
+        "solution":  "【分析】根据长方形面积公式(长方形的面积 = 长×宽),先求出绿地的面积,再根据1公顷 = 10000 平方米换算成公顷数。已知每公顷绿地一天释放730千克氧气,用公顷数乘730,即可求出总氧气释放量.\n【详解】 400×150 = 60000 (平方米) = 6 (公顷)6×730 = 4380 (千克)答:这块绿地一天大约可释放 4380 千克氧气。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有 0",
+                                 "长方形的面积",
+                                 "公顷,平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·湖北武汉·期中"
+    },
+    {
+        "id":  "93",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "\u0027朝辞白帝彩云间, 千里江陵一日还。两岸猿声啼不住,轻舟已过万重山。\u0027江陵即湖北省荆州市,实际上两地的水路大约 600 千米。n(1)如果李白当天00:00出发,乘船的速度是28千米/时,\u0027千里江陵一日还\u0027是真实的吗。 [诗中的\u0027一日还\u0027指白帝城到江陵一天就可以到达。]n(2)荆州是长江文化展示地,明明一家在假期自驾从奉节县到荆州市旅游,一共用了5小时。前3小时的速度是75千米/时,为了安全,之后的速度是70千米/时,从奉节县到荆州市驾车全程多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "真实;365 千米",
+        "solution":  "【分析】(1)根据题意,已知如果李白当天00:00出发,乘船的速度是28千米/时,先求出一天有24小时,再用28乘24求出一天行驶的距离,再与600比较即可;(2)已知明明一家前3小时的速度是75千米/时,之后的速度是70千米/时,分别用3乘75和(5-3)乘70求出两个时间段行驶的路程,再求出它们的和即可。\n【详解】(1)24-0 = 24(小时),28×24 = 672(千米),672＞600,答:真实,他当天能到达江陵。(2)75×3+70×(5-3) = 225+140 = 365(千米)。答:从奉节县到荆州市驾车全程365千米。",
+        "knowledge_points":  [
+                                 "两位数乘两位数的进位乘法",
+                                 "带有小括号的混合运算",
+                                 "基础行程问题",
+                                 "24 时计时法时间的计算"
+                             ],
+        "source":  "25-26四年级上·福建三明·期中"
+    },
+    {
+        "id":  "94",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "为了让同学们增强节约用电的意识,掌握节约用电的方法,学校开展了节约用电活动。今年前 3 个月,学校共节约用电 405 千瓦时。按照这样的情况持续下去,一年(12 个月),学校一共能节约用电多少千瓦时。",
+        "options":  [
+
+                    ],
+        "answer":  "1620 千瓦时",
+        "solution":  "【分析】根据题意可知,前3个月共节约用电405千瓦时,用405除以3得到1个月节约的用电量,再乘12即可得到一年能节约的用电量,据此即可解答。\n【详解】405÷3×12 = 135×12 = 1620(千瓦时)。答:一年学校一共能节约用电1620千瓦时。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法",
+                                 "单归一问题"
+                             ],
+        "source":  "25-26四年级上·福建三明·期中"
+    },
+    {
+        "id":  "95",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.94,
+        "question":  "动物园的两头大象一天要吃 350 千克食物,21 天一共吃多少千克食物。",
+        "options":  [
+
+                    ],
+        "answer":  "7350 千克",
+        "solution":  "【分析】两头大象平均每天要吃350千克食物,问21天一共吃多少千克食物,即21个350的和是多少,用乘法.\n【详解】 350×21 = 7350 (千克)答:21天一共吃7350千克食物",
+        "knowledge_points":  [
+                                 "三位数乘两位数的实际问题"
+                             ],
+        "source":  "25-26四年级上·福建三明·期中"
+    },
+    {
+        "id":  "96",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "林叔叔从家骑自行车去木兰天池郊游。去的时候用了3小时,每小时行驶40千米,返回时用了 2 小时。n(1)林叔叔家到木兰天池有多远。n(2)按原路返回平均每小时行驶多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "120 千米;60 千米/时",
+        "solution":  "【分析】(1)根据路程 = 速度×时间,用40乘3,即可得到林叔叔家到木兰天池的距离;(2)根据速度 = 路程÷时间,用刚求出的路程除以2,即可得到按原路返回平均每小时行驶多少千米。\n【详解】(1)40×3 = 120(千米)。答:林叔叔家到木兰天池有120千米。(2)120÷2 = 60(千米/时)。答:按原路返回平均每小时行驶60千米/时。",
+        "knowledge_points":  [
+                                 "整十,整百,整千数与一位数的乘法",
+                                 "几百几十或几千几百除以一位数的除法",
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26四年级上·湖北黄冈·期中"
+    },
+    {
+        "id":  "97",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "妈妈到超市里去买果汁和牛奶,汇源果汁 7 元/2 盒,旺仔牛奶 10 元/3 盒,妈妈要买6盒汇源果汁和9盒旺仔牛奶应付多少元钱。",
+        "options":  [
+
+                    ],
+        "answer":  "51 元",
+        "solution":  "【分析】将2盒汇源果汁视作1组,3盒旺仔牛奶视作1组。根据题目给出的汇源果汁7元/2盒、旺仔牛奶10元/3盒,可以得知每组汇源果汁7元,每组旺仔牛奶10元。用6÷2算出要买几组汇源果汁,用9÷3算出要买几组旺仔牛奶,再分别求总价并相加即可。\n【详解】6÷2×7+9÷3×10 = 3×7+3×10 = 21+30 = 51(元)。答:妈妈要买6盒汇源果汁和9盒旺仔牛奶应付51元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "25-26 四年级上·湖北黄冈·期中"
+    },
+    {
+        "id":  "98",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "公园的一只狮子一天要吃 180 千克食物,饲养员为这只狮子准备了 5 吨食物,够这只狮子吃 30 天吗。",
+        "options":  [
+
+                    ],
+        "answer":  "不够",
+        "solution":  "【分析】根据题意可知,先用 180×30 求出这只狮子 30 天需要吃多少千克食物,根据1吨 = 1000 千克,将5吨化成5000千克,然后将求出的乘积与饲养员准备的食物质量进行比较即可.\n【详解】 180×30 = 5400 (千克)5 吨 = 5000 千克 5400\u003e5000 答:不够这只狮子吃30天。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "吨,千克之间的换算与比较"
+                             ],
+        "source":  "25-26 四年级上·湖北黄冈·期中"
+    },
+    {
+        "id":  "99",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "星光小学于 2025 年开展了\u0027厉行节约,反对浪费,让节约之气在全校蔚然成风\u0027的活动。为配合活动,学校计划平均每月用电由 850 千瓦时降到 720 千瓦时,照这样计算,星光小学 2025 年全年一共节约用电多少千瓦时。",
+        "options":  [
+
+                    ],
+        "answer":  "1560 千瓦时",
+        "solution":  "【分析】首先用原来每月用电量减去现在每月用电量,计算出每月节约的电量。再用每月节约的电量乘12,计算出全年一共节约的用电量。据此解答。\n【详解】850-720 = 130(千瓦时),130×12 = 1560(千瓦时)。答:星光小学2025年全年一共节约用电1560千瓦时。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "25-26 四年级上·山东济南·期中"
+    },
+    {
+        "id":  "100",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "森林是\u0027地球之肺\u0027,是二氧化碳的消耗者和氧气的生产者。1 公顷森林一年大约可以减少水土流失15吨,一个大型林区有4平方千米森林,这个大型林区的森林一年大约可以减少水土流失多少吨。",
+        "options":  [
+
+                    ],
+        "answer":  "6000 吨",
+        "solution":  "【分析】1平方千米 = 100公顷,4平方千米 = 400公顷。然后用1公顷森林一年大约可以减少水土流失的吨数乘400,即可计算出这个大型林区的森林一年大约可以减少水土流失多少吨。据此解答。\n【详解】4平方千米 = 400公顷,400×15 = 6000(吨)。答:这个大型林区的森林一年大约可以减少水土流失6000吨。",
+        "knowledge_points":  [
+                                 "三位数乘两位数的实际问题",
+                                 "公顷,平方千米的进率与换算"
+                             ],
+        "source":  "25-26四年级上·山东济南·期中"
+    },
+    {
+        "id":  "101",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面说法正确的是()。",
+        "options":  [
+                        "A. 三位数 3□1×92 的积可能是四位数,也可能是五位数。",
+                        "B. 估算 302×54 的结果一定比用竖式计算出来的准确结果要大。",
+                        "C. 803×25 的积的中间只有一个 0 ",
+                        "D. 两个不为 0 的数相乘,两个乘数分别扩大到原来的 9倍,积就扩大到原来的81倍。"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】三位数乘两位数的笔算法则:先用两位数的个位分别与三位数的每一位数相乘,乘得的积的末尾和个位对齐。再用两位数的十位分别与三位数的每一位数相乘,乘得的积的末尾和十位对齐。最后,将两次乘得的积相加。估算三位数乘两位数时,可把它们估成与其接近的整百数或整十数,然后口算出结果即可。积的变化规律:一个乘数不变,另一个乘数乘或除以几(0除外),得到的积就等于原来的积乘或除以几。据此解答。\n【详解】A. 在算式3□1×92中,□里最小填0,最大填9。301×92 = 27692,391×92 = 35972,得到的两个积都是五位数,所以三位数3□1×92的积一定是五位数,该选项说法错误。B. 302×54 = 16308,302×54≈300×50 = 15000,16308＞15000,即估算的结果比实际计算的结果要小,该选项错误。C. 803×25 = 20075,积的中间有2个0,该选项说法错误。D. 两个不为0的数相乘,两个乘数分别扩大到原来的9倍,9×9 = 81,那么积会扩大到原来的81倍,该选项说法正确。故答案为:D。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法",
+                                 "三位数乘两位数,三位数中间有0",
+                                 "三位数乘两位数的估算",
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·山西晋中·期中"
+    },
+    {
+        "id":  "102",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面说法错误的是()。",
+        "options":  [
+                        "A. 一副三角尺可以拼出105∘ 的角",
+                        "B. 总价/数量 = 单价",
+                        "C. 一条直线的长度是一条射线长度的3倍"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】根据题意,一副三角尺包含 45 circ , 45 circ ,90°和30 circ , 60 circ , 90 circ 两种, 60 circ + 45 circ = 105 circ ,可以拼出 105°的角。总价 = 单价×数量,变形后总价/数量 = 单价。直线和射线 均无限长,无法比较倍数。依此选择.\n【详解】根据分析可知:A。三角尺中 60 circ 角和 45 circ 角可拼成 105 circ ,正确。 B。总价/数量 = 单价符合数量关系,正确。 C。直线和射线均无限长,无法比较倍数,此选项错误。故答案为:C",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "线段,直线,射线的认识及特征",
+                                 "用三角尺画角"
+                             ],
+        "source":  "25-26 四年级上·陕西西安·期中"
+    },
+    {
+        "id":  "103",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面的说法正确的是( )。",
+        "options":  [
+                        "A. 两个计数单位之间的进率是 10",
+                        "B. 平角就是一条直线",
+                        "C. 已知买 8 个同样的文具盒的价钱,可以求出这种文具盒的单价"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】相邻两个计数单位之间的进率是 10,如果这两个计数单位不相邻,进率不是 10;平角是角,有一个顶点和两条边,与直线不是一种图形;根据单价 = 总价/数量,用 8 个文具盒的价钱除以 8,可以求出这种文具盒的单价.\n【详解】A。相邻的两个计数单位之间的进率是10,但两个计数单位之间的进率不是10,例如十和百之间的进率是10,十和千之间的进率是100,错误;B。角有一个顶点和两条边,所以角和直线不是一个概念,错误;C。已知买8个同样的文具盒的价钱,用总价/数量可以求出这种文具盒的单价,正确。 正确的是选项C中的说法。 故答案为:C",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "十进制计数法",
+                                 "平角,周角的认识及特征"
+                             ],
+        "source":  "25-26四年级上·广东汕头·期中"
+    },
+    {
+        "id":  "104",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "小新用计算器计算\u002740×18\u0027时,输成了\u00274×18\u0027,如果再输入( ),就可以弥补他的错误。",
+        "options":  [
+                        "A. ×10=",
+                        "B. ÷10=",
+                        "C. ÷100="
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】根据积的变化规律:如果一个因数扩大到原来的几倍或缩小到原来的几分之一,另一个因数不变,那么积也扩大到原来的几倍或缩小到原来的几分之一。因数40变为4,缩小到原来的1/10,积也缩小到原来的1/10,要使积不变,积应再乘10即可。\n【详解】小新用计算器算40×18时,输入成了4×18,如果再输入×10=,就可以弥补他的错误。故答案为:A。",
+        "knowledge_points":  [
+                                 "计算器的初步认识与使用",
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·河北唐山·期中"
+    },
+    {
+        "id":  "105",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "一块长 250 米,宽 40米的长方形菜地,如果每公顷施化肥130千克,那么共需要化肥( )千克。",
+        "options":  [
+                        "A. 60",
+                        "B. 600",
+                        "C. 260",
+                        "D. 130"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】根据长方形面积 = 长×宽,据此用250×40计算出菜地的面积;然后再根据1公顷 = 10000平方米,将面积换算成公顷,再乘每公顷施化肥的千克数即可解题。\n【详解】250×40 = 10000(平方米),10000平方米 = 1公顷,1×130 = 130(千克)。所以共需要化肥130千克。故答案为:D。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有 0",
+                                 "长方形的面积",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·山东济南·期中"
+    },
+    {
+        "id":  "106",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "春节期间,名名一家开车去三亚游玩,要求爸爸开车到三亚的平均速度,应选择的信息是( )。",
+        "options":  [
+                        "A. ①②",
+                        "B. ②③",
+                        "C. ①③",
+                        "D. ②④"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据题意可知,平均速度 = 行驶的总路程÷行驶的总时间,依此选择合适的信息即可。\n【详解】求爸爸开车到三亚的平均速度,应选择的信息是:②名名家距离三亚320千米,③路上共用了4小时。320÷4 = 80(千米/时),所以爸爸开车到三亚的平均速度是80千米/时。故答案为:B。",
+        "knowledge_points":  [
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26 四年级上·内蒙古乌海·期中"
+    },
+    {
+        "id":  "107",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "用竖式计算( )的过程中,\u00273×6\u0027表示的是300×60。",
+        "options":  [
+                        "A. 308×26",
+                        "B. 39×680",
+                        "C. 349×62",
+                        "D. 602×35"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】三位数乘两位数的计算方法:先是用两位数的个位与三位数相乘;接着两位数的十位与三位数相乘,最后把两次乘得的积相加;\u00273×6\u0027表示的是 300×60 ,说明数字3在百位上,表示 3 个百,数字 6 在十位上,表示 6 个十.\n【详解】A。计算 308×26 时,用两位数个位上的6乘三位数百位上的3,表示6个一乘3个百,即 6×300 = 1800 ,不符合题意。B。计算 39×680 时,用两位数十位上的3乘三位数百位上的6,表示3个十乘6个百,即 30×600 = 18000 ,不符合题意。 C。计算 349×62 时,用两位数十位上的6乘三位数百位上的3,表示6个十乘3个百,即 60×300 = 18000 ,符合题意。 D。计算 602×35 时,用两位数十位上的3乘三位数百位上的6,表示3个十乘6个百,即 30×600 = 18000 ,不符合题意。故答案为:C",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "25-26四年级上·湖北荆州·期中"
+    },
+    {
+        "id":  "108",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "张师傅准备买35袋大米,他准备了3000元。买( )元/袋的大米合适且找回的钱最少。",
+        "options":  [
+                        "A. 70",
+                        "B. 85",
+                        "C. 90",
+                        "D. 75"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】买大米花的总钱数 = 一袋大米的钱数×购买的袋数,分别计算出35×70、35×85、35×90、35×75的结果,然后再用总钱数减去花去的钱数,找出找回钱数最少且钱够的选项即可。\n【详解】A. 35×70 = 2450(元),3000-2450 = 550(元);B. 35×85 = 2975(元),3000-2975 = 25(元);C. 35×90 = 3150(元),3000＜3150,钱不够;D. 35×75 = 2625(元),3000-2625 = 375(元)。25＜375＜550,所以买85元/袋的大米合适且找回的钱最少。故答案为:B。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "两,三位数的连续退位减法",
+                                 "两位数乘两位数的进位乘法",
+                                 "两位数乘两位数,乘数末尾有0"
+                             ],
+        "source":  "25-26四年级上·山东济南·期中"
+    },
+    {
+        "id":  "109",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "( )的面积和 1 公顷的大小相近。",
+        "options":  [
+                        "A. 5000个20平方米的商铺",
+                        "B. 边长20米的正方形草 坪",
+                        "C. 15 户 120 平方米的住房总面积",
+                        "D. 长 198 米,宽 51 米的长方形泳池"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】先明确1公顷 = 10000平方米,再分别计算每个选项的面积,对比是否与10000平方米相近。据此逐一分析每个选项即可。\n【详解】1公顷 = 10000平方米。A. 5000×20 = 100000(平方米),远大于1公顷,排除。B. 20×20 = 400(平方米),远小于1公顷,排除。C. 15×120 = 1800(平方米),远小于1公顷,排除。D. 198≈200,51≈50,估算面积约是200×50 = 10000(平方米),与1公顷接近,符合要求。故答案为:D。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "三位数乘两位数的估算",
+                                 "长方形的面积",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26四年级上·河南驻马店·期中"
+    },
+    {
+        "id":  "110",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一块玉米地长110米,宽 90 米,占地约 1( )。",
+        "options":  [
+                        "A. 平方千米",
+                        "B. 平方米",
+                        "C. 公顷"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】1公顷 = 10000平方米,1平方千米 = 1000000平方米。一块玉米地长110米,宽90米,根据长方形的面积 = 长×宽,用110乘90求出玉米地的面积,再与1平方米、1公顷、1平方千米比较即可。\n【详解】110×90 = 9900(平方米)。A. 1平方千米 = 1000000平方米,与9900平方米相差很大,不符合;B. 1平方米与9900平方米相差很大,不符合;C. 1公顷 = 10000平方米,与9900平方米接近,符合。故答案为:C。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "长方形的面积",
+                                 "公顷,平方千米的进率与换算"
+                             ],
+        "source":  "25-26四年级上·山东临沂·期中"
+    },
+    {
+        "id":  "111",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "奇奇用计算器计算4200×50 时,不小心把 4200 输成了 42,但他仍然算出了正确的结果,他输入的算式可能是( )。",
+        "options":  [
+                        "A. 42×5000",
+                        "B. 42×50",
+                        "C. 42×500",
+                        "D. 42×50000"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】积的变化规律:一个乘数不变,另一个乘数乘或除以几(0除外),得到的积就等于原来的积乘或除以几。据此解答。\n【详解】由题意得,4200输成了42,即第一个乘数除以100。要使积不变,另一个乘数需要乘100,所以50×100 = 5000。因此奇奇输入的算式可能是42×5000。故答案为:A。",
+        "knowledge_points":  [
+                                 "计算器的初步认识与使用",
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26四年级上·湖北武汉·期中"
+    },
+    {
+        "id":  "112",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面问题中不能利用\u0027积的变化规律\u0027解决的是( )",
+        "options":  [
+                        "A. 长方形的长不变,宽扩大到原来的 3 倍,求它的面积的变化情况。",
+                        "B. 一件物品的单价不变,买的数量扩大到原来的5倍,求总价的变化情况。",
+                        "C. 小车的速度不变,行驶的时间缩短到原来的一半,求路程的变化情况。",
+                        "D. 长方形的长不变,宽扩大到原来的 3 倍,求它的周长的变化情况。"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】若一个因数不变,另一个因数扩大到原来的几倍或缩小到原来的几分之一(0除外),积也扩大到原来几倍或缩小到原来的几分之一;如果一个因数扩大到原来几倍或缩小到原来的几分之一,另一个因数反而缩小到原来的几分之一或扩大到相同的倍数(0除外),那么积不变。据此解答.\n【详解】A。长方形的长不变,宽扩大到原来的3倍,长方形的面积 = 长×宽,它的面积也扩大到原来的 3 倍;B。一件物品的单价不变,买的数量扩大到原来的5倍,总价 = 数量×单价,总价也扩大到原来的5倍;C。小车的速度不变,行驶的时间缩短到原来的一半,路程= 速度×时间,路程也缩短到原来的一半;D。长方形的长不变,宽扩大到原来的 3 倍,C=(a+b)×2 ,它的周长的变化倍数不确定。故答案为:D",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "长方形的周长",
+                                 "长方形的面积",
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·湖南张家界·期中"
+    },
+    {
+        "id":  "113",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面与 2700×14 的积不相等的是( )。",
+        "options":  [
+                        "A. 270×140",
+                        "B. 27×1400",
+                        "C. 270×104"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】两个整数相乘,当乘数末尾有0时,可先将0前面的数相乘,再在积末尾添上0,有几个0就添几个0,据此分析.\n【详解】 2700×14 可看作将 27×14 ,然后在积的末尾加 2 个0。A。 270×14 可看作将 27×14 ,然后在积的末尾加 2 个 0;B. 27×1400 可看作将 27×14 ,然后在积的末尾加 2 个 0;C. 270×104 看作将 27×104 ,然后在积的末尾加1个0。所以与 2700×14 积不相等的只有 270×104 故答案为:C",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0"
+                             ],
+        "source":  "25-26 四年级上·湖南长沙·期中"
+    },
+    {
+        "id":  "114",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "在计算 427×65 时,4×5 表示( )。",
+        "options":  [
+                        "A. 4×5",
+                        "B. 40×5",
+                        "C. 400×5"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】在 427 中,4 在百位上表示 4 个百,即 400;在 65中 5 在个位上,表示 5 个一,即 5;在计算 427×65 时, 4×5 表示 400×5 .\n【详解】根据分析可知:在计算 427×65 时, 4×5 表示400×5 。故答案为:C",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "25-26 四年级上·湖北鄂州·期中"
+    },
+    {
+        "id":  "115",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "如果 a×10 = mathtt b×20 (a,b 不为 0),那么 a 与 b 相比,( )。",
+        "options":  [
+                        "A. a 大",
+                        "B. b 大",
+                        "C. a 与 b 相等"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】两个因数的积相等(积不为0),一个因数越大,另一个因数就越小;据此解答.\n【详解】因为 20\u003e 10 ,所以 a\u003e b 所以 a 与 b 相比,a 大。故答案为:A",
+        "knowledge_points":  [
+                                 "三位数与整十数的口算乘法"
+                             ],
+        "source":  "25-26四年级上·湖北鄂州·期中"
+    },
+    {
+        "id":  "116",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.94,
+        "question":  "在 999×99 中,一个因数乘 10,另一个因数除以 10,积( )。",
+        "options":  [
+                        "A. 不变",
+                        "B. 除以 10",
+                        "C. 乘 10"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】一个因数乘10,另一个因数除以10,和原来的999×99比较即可。\n【详解】(999×10)×(99÷10) = 999×99×10÷10,后面的乘10与除以10相互抵消,变化后的积仍为999×99,所以积不变。故答案为:A。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26四年级上·湖北鄂州·期中"
+    },
+    {
+        "id":  "117",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "贝贝在计算一道乘法算式时,把一个因数29看成了26,得到的积比正确的积少了 360,这道乘法算式的正确结果是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "3480",
+        "solution":  "【分析】把一个因数29看成了26,相当于少看了3,用减少的积除以3可以得到另一个因数,再乘上29即可解答。\n【详解】29-26 = 3,360÷3 = 120,29×120 = 3480。所以这道乘法算式的正确结果是3480。",
+        "knowledge_points":  [
+                                 "几百几十或几千几百除以一位数的除法",
+                                 "三位数乘两位数,乘数末尾有0"
+                             ],
+        "source":  "25-26 四年级上·山东济南·期中"
+    },
+    {
+        "id":  "118",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "2025 年是中国人民抗日战争胜利 80 周年。春风小学准备组织一场\u0027红色汇演\u0027,学校统一购买演出服装,每套服装 75 元,需要购买360 套,一共要花多少钱。解决这个问题用到的数量关系是:( ),一共要花( )元。",
+        "options":  [
+
+                    ],
+        "answer":  "总价 = 单价×数量;27000",
+        "solution":  "【分析】本题考查总价、单价和数量之间的数量关系。已知每套服装的单价是75元,需要购买的数量是360套,求总价。根据总价 = 单价×数量,用每套服装的单价75元乘需要购买的数量360套,即可求得一共要花的钱。\n【详解】360×75 = 27000(元)。因此,解决这个问题用到的数量关系是:总价 = 单价×数量,一共要花27000元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "三位数乘两位数,乘数末尾有 0"
+                             ],
+        "source":  "25-26 四年级上·山东济南·期中"
+    },
+    {
+        "id":  "119",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "探究问题:有五个自然数:a,b,c,d,e(10\u003ee\u003ed\u003ec\u003eb\u003ea\u003e0)用这五个数字组成三位数乘两位数的乘法算式,积最大的算式是什么。探究过程:分析如果一一列举,计算量太大了,要使积最大,就要使每个乘数都尽可能最大,所以要把较大的两个数分别放在两个乘数的最高位上。举例若这 5 个自然数分别是 2,3,4,5,6,我们可以得到下面 6 个算式:\n\n643×52 = 33436642×53 = 34026632×54 = 34128543×62 = 33666542×63 = 34146532×64 = 34048\n\n通过计算,积最大的算式是( ),因此,用 5 个自然数a,b,c,d,e(10\u003ee\u003ed\u003ec\u003eb\u003ea\u003e0),组成三位数乘两位数的乘法算式,要使积最大,应该这样组合:( )×()。",
+        "options":  [
+
+                    ],
+        "answer":  "542×63;dca;eb",
+        "solution":  "【分析】通过比较6个算式的结果,找到积最大的那个算式。积最大时,三位数由第二大的数字、第三大的数字和最小的数字组成,两位数由最大的数字和第四大的数字组成。推广到一般情况,对于五个自然数a,b,c,d,e(10＞e＞d＞c＞b＞a＞0),积最大时,三位数应为dca,两位数应为eb。\n【详解】通过计算可知:33436＜33666＜34026＜34048＜34128＜34146,所以积最大的算式是542×63。因此,用5个自然数a,b,c,d,e组成三位数乘两位数的乘法算式,要使积最大,应该这样组合:(dca)×(eb)。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "25-26 四年级上·山西晋中·期中"
+    },
+    {
+        "id":  "120",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "实验小学计划在 11月份安排四年级学生去杜甫草堂参加研学活动。杜甫草堂是我国唐代诗人杜甫在成都时生活的地方,杜甫草堂的门票价格为每人50元,四年级共有496人,估一估,购买门票大约需要( )元。",
+        "options":  [
+
+                    ],
+        "answer":  "25000",
+        "solution":  "【分析】根据题意,门票每人50元,四年级共有496人,要求估算购买门票的总费用。估算时,将496人近似为500人,然后根据单价×数量 = 总价,计算50与500的乘积,得到大约需要的费用。\n【详解】50×496≈50×500 = 25000(元)。所以购买门票大约需要25000元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "三位数乘两位数的估算"
+                             ],
+        "source":  "25-26 四年级上·山西晋中·期中"
+    },
+    {
+        "id":  "121",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上\u0027\u003e\u0027\u0027\u003c\u0027或\u0027=\u0027。103 万( )10300000;50×450( )45×500;45 公顷( )3 平方千米。",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;=;\u003c",
+        "solution":  "【分析】(1)比较整数的大小,先看数位的多少,数位多的数就大,数位相同,从高位比较,高位上的数大则这个数大,高位上的数相同,就比较下一位,以此类推,其中103万 = 1030000;(2)如果一个因数扩大到原来的若干倍,另一个因数缩小到原来的若干分之一(0除外),那么积不变;(3)根据1平方千米 = 100公顷,先化成统一的单位,再比较大小。\n【详解】103万 = 1030000,1030000＜10300000,所以103万＜10300000;50×450 = (500÷10)×(45×10) = 45×500;3平方千米 = 300公顷,45＜300,所以45公顷＜3平方千米。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "公顷,平方千米的进率与换算",
+                                 "积的变化规律(整数乘法)",
+                                 "亿以内数的改写"
+                             ],
+        "source":  "25-26 四年级上·陕西西安·期中"
+    },
+    {
+        "id":  "122",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "奇奇在计算一道乘法算式时,使用了下面的方法: 120×2 = 240 , 120×10 =1200, 240 + 1200 = 1440 ,他计算的乘法算式是( )×()。",
+        "options":  [
+
+                    ],
+        "answer":  "120;12",
+        "solution":  "【分析】观察前面两个算式里面都有120,因此120是这个乘法算式里面的一个乘数,而两个算式中的2和10分别表示另一个乘数的个位和十位,即另一个乘数是12。因此奇奇计算的乘法算式是120×12 = 1440。\n【详解】根据分析可知,奇奇在计算一道乘法算式时,使用了下面的方法:120×2 = 240,120×10 = 1200,240+1200 = 1440,他计算的乘法算式是120×12。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0"
+                             ],
+        "source":  "25-26 四年级上·云南楚雄·期中"
+    },
+    {
+        "id":  "123",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\u0027\u003e\u0027\u0027\u003c\u0027或\u0027=\u0027。5497400( )554□200(□表示一个数);3 亿( )30000万;4 平方千米 40 公顷( )440000 平方米;114×12( )114×10×2。",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;=;\u003e;\u003c",
+        "solution":  "【分析】整数比较大小,先看位数,位数多的整数更大;若位数相同,从最高位开始比较,直至比出大小。(1)比较两个七位数,从高位开始比较,比到十万位数字可比出大小;(2)1亿 = 10000万,将3亿转化为用万作单位后再比较;(3)根据1平方千米 = 100公顷,1公顷 = 10000平方米,将4平方千米40公顷换算成用平方米作单位再比较大小;(4)通过计算乘法结果进行比较。\n【详解】5497400和554□200都是七位数,百万位都相同,5497400的十万位是4,554□200的十万位是5,4＜5,因此5497400＜554□200。3亿 = 30000万。4平方千米40公顷 = 440公顷 = 4400000平方米,4400000平方米＞440000平方米,因此4平方千米40公顷＞440000平方米。114×12 = 1368,114×10×2 = 2280,1368＜2280,因此114×12＜114×10×2。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "三位数与两位数的乘法",
+                                 "公顷,平方千米的进率与换算",
+                                 "亿以上数的改写"
+                             ],
+        "source":  "25-26 四年级上·云南楚雄·期中"
+    },
+    {
+        "id":  "124",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一辆汽车在高速公路上匀速行驶,3 小时行了 270 千米,汽车的速度可以写成()。李老师买 8 个篮球用了 360 元,篮球的单价是( )元。",
+        "options":  [
+
+                    ],
+        "answer":  "90 千米/时;45",
+        "solution":  "【分析】速度指单位时间内所行的路程,其单位的表示方法是\u0027所走的路程单位/时间单位\u0027,一般用千米/时、米/分来表示。由题意得,一辆汽车在高速公路上匀速行驶,3小时行了270千米。速度 = 路程÷时间,直接用270除以3可以算出汽车的速度。李老师买8个篮球用了360元,单价 = 总价÷数量,直接用360除以8可以算出篮球的单价。\n【详解】270÷3 = 90(千米/时),360÷8 = 45(元)。一辆汽车在高速公路上匀速行驶,3小时行了270千米,汽车的速度可以写成90千米/时。李老师买8个篮球用了360元,篮球的单价是45元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26 四年级上·广东汕头·期中"
+    },
+    {
+        "id":  "125",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "340×25 的积的末尾有( )个 0, 307×29 的积是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "2;8903",
+        "solution":  "【分析】计算出 340×25 的积再确定末尾有多少个0,计算307×29 ,求出积的数值。三位数乘两位的计算方法,用第二个因数的每一位分别去乘第一个因数,再把所得的积相加。对于因数末尾有0的乘法,可先把0前面的数相乘,再看因数末尾一共有几个 0,就在乘得的积的末尾添上几个 0 \n【详解】 340×25 = 8500 ,积的末尾有2个0;307×29 = 8903 所以 340×25 的积的末尾有 2 个 0, 307×29 的积是 8903 。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "三位数乘两位数,三位数中间有 0"
+                             ],
+        "source":  "25-26四年级上·河北唐山·期中"
+    },
+    {
+        "id":  "126",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "计算 320×50 时,先算( )×5 = ( ) ,再在积的末尾添上( )个 0 。",
+        "options":  [
+
+                    ],
+        "answer":  "32;160;2",
+        "solution":  "【分析】三位数乘两位数,当乘数末尾有0时,可先不让0参与计算,最后将0的个数补在积的末尾处即可,依此进行填空。\n【详解】320末尾有1个0,50末尾有1个0,1+1 = 2。计算320×50时,先算32×5 = 160,再在积的末尾添上2个0。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有 0"
+                             ],
+        "source":  "25-26 四年级上·河北唐山·期中"
+    },
+    {
+        "id":  "127",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小华沿着自家的一个正方形苹果园边走一圈,正好走了400米。这个果园的面积是( )平方米,也就是( )公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "10000;1",
+        "solution":  "【分析】根据题意可知,正方形苹果园的周长是400米,根据正方形的周长 = 边长×4,利用400除以4可求出正方形的边长,再根据正方形的面积 = 边长×边长计算即可求出苹果园的面积;再根据1公顷 = 10000平方米进行单位换算。\n【详解】400÷4 = 100(米),100×100 = 10000(平方米),10000平方米 = 1公顷。这个果园的面积是10000平方米,也就是1公顷。",
+        "knowledge_points":  [
+                                 "正方形的周长",
+                                 "正方形的面积",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26四年级上·云南昭通·期中"
+    },
+    {
+        "id":  "128",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在计算 135×24 时,用横式表达算理是: 135×24 = 135×4 + ( )×( ) ",
+        "options":  [
+
+                    ],
+        "answer":  "135;20",
+        "solution":  "【分析】根据三位数乘两位数的计算方法,把24看作 20 + 4,用 135 分别乘 20 与 4,再把所得的积相加,也就是135×24 = 135×4 + 135×20;据此解答.\n【详解】据分析可得: 135×24 = 135×(4 + 20) = 135×4 + 135×20 在计算 135×24 时,用横式表达算理是: 135×24 = 135×4 + 135×20 。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "25-26 四年级上·云南昭通·期中"
+    },
+    {
+        "id":  "129",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里面填上\u0027\u003e\u0027\u0027\u003c\u0027或\u0027=\u0027.\n\n762500( )786520013 亿()1299999999450×900( )45×900",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003e;\u003e",
+        "solution":  "【分析】整数大小的比较方法:位数不同,位数多的数就大;位数相同,左起第一位上的数大的那个数就大,如果左起第一位上的数相同,就比较左起第二位上的数,直到比较出大小为止;改写为以个为单位的数,去掉亿字,在数末尾加上8个0;在乘法算式中,当其中一个因数相同时,另一个因数越大,积就越大.\n【详解】762500 是六位数,7865200 是七位数,因为位数多的数更大,所以 762500\u003c7865200;13 亿=1300000000,1300000000 和 1299999999 位数相同,十亿位上都是 1,因为 1300000000 亿位上的 3 大于1299999999 亿位上的 2,所以 1300000000\u003e1299999999;450×900=405000,45×900=40500,因为 405000\u003e40500,所以 450×900 {\u003e } 45×900 。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "三位数乘两位数,乘数末尾有 0"
+                             ],
+        "source":  "25-26 四年级上·山东济南·期中"
+    },
+    {
+        "id":  "130",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.4,
+        "question":  "已知2002年的7月1 日是星期一。2015 年的 10 月 1 日是星期( )。",
+        "options":  [
+
+                    ],
+        "answer":  "四",
+        "solution":  "【分析】计算出2002年7月1日到2015年10月1日的总天数,其中2004年、2008年以及2012年是闰年,全年一共366天,则3×366加上10×365,再加上7月的31天、8月的31天以及9月的30天,即可求出总天数。因为一周是7天,用总天数除以7的余数为几,就向后推几天,即可推断2015年10月1日是星期几。\n【详解】3×366+10×365+31×2+30 = 1098+3650+62+30 = 4840(天),4840÷7 = 691(周)……3(天)。星期一往后推3天是星期四,即2015年10月1日是星期四。",
+        "knowledge_points":  [
+                                 "平年,闰年的认识及判定方法",
+                                 "年,月,日有关的计算",
+                                 "年,月,日时间的推算",
+                                 "周期问题"
+                             ],
+        "source":  "25-26六年级上·江西上饶·月考"
+    },
+    {
+        "id":  "131",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "估算 198×22 时,把198 看作( ),把 22 看作( ),积大约是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "200;20;4000",
+        "solution":  "【分析】根据三位数乘两位数的估算,将三位数198估成整百数200,将两位数22估成整十数20,相乘即可求出积大约是多少。\n【详解】198×22≈200×20 = 4000。估算198×22时,把198看作200,把22看作20,积大约是4000。",
+        "knowledge_points":  [
+                                 "三位数乘两位数的估算"
+                             ],
+        "source":  "25-26 四年级上·河南驻马店·期中"
+    },
+    {
+        "id":  "132",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "计算 120×80 ,先算( ),再在积的末尾补上( )个( )。最终结果是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "12×8;2;0;9600",
+        "solution":  "【分析】计算120×80时,把120看作12个十,80看作8个十,先算12×8 = 96,再在积的末尾补上2个0得9600,然后即可解答。\n【详解】计算120×80,先算12×8,再在积的末尾补上2个0,最终结果是9600。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有 0"
+                             ],
+        "source":  "25-26 四年级上·河南驻马店·期中"
+    },
+    {
+        "id":  "133",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里面填\u0027一定\u0027\u0027可能\u0027或\u0027不可能\u0027。(1)两个钝角的和( )比一个周角大;(2)任意三位数乘最大的两位数,积( )是四位数;(3)一条射线( )有2个端点。",
+        "options":  [
+
+                    ],
+        "answer":  "不可能;可能;不可能",
+        "solution":  "【分析】(1)钝角是大于90°而小于180°的角,周角 = 360°;以此判断即可。(2)最大的两位数是99,用最小的三位数100和最大的三位数999分别去乘99,看积是几位数即可解答;(3)射线只有一个端点,另一端无限延伸。根据定义,射线不可能有两个端点。\n【详解】根据分析可知:(1)例如179°+179° = 358°＜360°,两个钝角的和不可能比一个周角大;(2)100×99 = 9900,999×99 = 98901,任意三位数乘最大的两位数,积可能是四位数;(3)一条射线不可能有2个端点。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法",
+                                 "线段,直线,射线的认识及特征",
+                                 "直角,钝角,锐角的认识及特征",
+                                 "平角,周角的认识及特征"
+                             ],
+        "source":  "25-26 四年级上·河南三门峡·期中"
+    },
+    {
+        "id":  "134",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "估算802×48时,把802看作( ),把48看作( ),积大约是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "800;50;40000",
+        "solution":  "【分析】三位数乘两位数估算时,把整数看作与它接近的整十、整百或整百整十数,然后相乘即可解答。\n【详解】802×48≈800×50 = 40000。估算802×48时,把802看作800,把48看作50,积大约是40000。",
+        "knowledge_points":  [
+                                 "三位数乘两位数的估算"
+                             ],
+        "source":  "25-26 四年级上·海南省直辖县级单位·期中"
+    },
+    {
+        "id":  "135",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号填上合适的单位名称.\n\n(1)我国陆地领土面积约为 960 万( )。 (2)学校操场跑道的一圈长为 400( )。 (3)一辆小轿车在高速上行驶的速度不能超过 120( )。 (4)46000000000=460( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平方千米;米;千米/时;亿",
+        "solution":  "【分析】(1)量比较大的土地面积,常用平方千米为单位,因此计量我国陆地领土面积以\u0027平方千米\u0027为单位。(2)计量学校操场跑道一圈的长度以\u0027米\u0027为单位。(3)速度是单位时间内通过的路程,由长度单位和时间单位组成,长度单位与时间单位之间用\u0027/\u0027分开。(4)整亿数改写成用\u0027亿\u0027作单位的数,就直接省略亿位后面的8个0,然后在数的末尾加一个\u0027亿\u0027。\n【详解】(1)我国陆地领土面积约为960万平方千米。(2)学校操场跑道的一圈长为400米。(3)一辆小轿车在高速上行驶的速度不能超过120千米/时。(4)46000000000 = 460亿。",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "长度单位的选择",
+                                 "面积单位的选择",
+                                 "亿以上数的改写"
+                             ],
+        "source":  "25-26 四年级上·海南省直辖县级单位·期中"
+    },
+    {
+        "id":  "136",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "学校图书馆共有26个五层书架,每个书架放245本书,平均每层书架放( )本书,学校图书馆共有( )本书。",
+        "options":  [
+
+                    ],
+        "answer":  "49;6370",
+        "solution":  "【分析】本题考查平均数和乘法应用。第一个空求平均每层书架放书数,用每个书架放书数除以层数;第二个空求总书数,用书架个数乘每个书架放书数。\n【详解】平均每层书架放书数:245÷5 = 49(本)。学校图书馆共有书数:26×245 = 6370(本)。所以平均每层书架放49本书,学校图书馆共有6370本书。",
+        "knowledge_points":  [
+                                 "三位数除以一位数,首位不够除",
+                                 "三位数乘两位数的实际问题"
+                             ],
+        "source":  "25-26 四年级上·广东潮州·期中"
+    },
+    {
+        "id":  "137",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "学校组织四年级师生秋游,学生门票每人 35 元,老师门票每人 50 元,收费处人员计算价格时列式为: 35×200 + 50×22;参加这次活动总共( )人,门票一共是( )元。",
+        "options":  [
+
+                    ],
+        "answer":  "222;8100",
+        "solution":  "【分析】收费处人员列出的算式\u002735×200 + 50×22\u0027中,35×200表示200名学生的门票总费用,50×22表示22名老师的门票总费用,所以200和22分别代表学生和老师的人数,求和即可得知总人数;总费用为学生门票总费用与老师门票总费用之和,据此计算即可解答。\n【详解】200+22 = 222(人),35×200+50×22 = 7000+1100 = 8100(元)。所以参加这次活动总共222人,门票一共是8100元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "两位数乘两位数,乘数末尾有0",
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "乘加,乘减混合运算"
+                             ],
+        "source":  "25-26 四年级上·广东潮州·期中"
+    },
+    {
+        "id":  "138",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一只猎狗每小时奔跑 35 千米,它的速度可写作( ),3 小时可跑( )千米,运用的数量关系式是( )×( )=路程。",
+        "options":  [
+
+                    ],
+        "answer":  "35 千米/时;105;速度;时间",
+        "solution":  "【分析】速度指单位时间内所行的路程,其单位的表示方法是\u0027所走的路程单位/时间单位\u0027,一般用千米/时、米/分来表示。由题意得,一只猎狗每小时奔跑35千米,它的速度可写作35千米/时。速度×时间 = 路程,求3小时可以跑多少千米,直接用35乘3即可解答。\n【详解】35×3 = 105(千米)。一只猎狗每小时奔跑35千米,它的速度可写作35千米/时,3小时可跑105千米,运用的数量关系式是速度×时间 = 路程。",
+        "knowledge_points":  [
+                                 "两,三位数与一位数连续进位的乘法",
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26四年级上·山东临沂·期中"
+    },
+    {
+        "id":  "139",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在横线上填入\u0027\u003e\u0027\u0027\u003c\u0027或\u0027=\u0027。5 公顷( )4900 平方米;800 平方米( )8 公顷;1000 公顷( )1 平方千米;10×220( )190×10;290×17( )29×170;100750( )107050。",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;\u003c;\u003e;\u003e;=;\u003c",
+        "solution":  "【分析】1公顷 = 10000平方米,1平方千米 = 100公顷。先统一单位或先计算出结果,再比较大小即可。据此解答。\n【详解】5公顷 = 50000平方米,所以5公顷＞4900平方米;8公顷 = 80000平方米,所以800平方米＜8公顷;1平方千米 = 100公顷,所以1000公顷＞1平方千米;10×220 = 2200,190×10 = 1900,所以10×220＞190×10;290×17 = 29×170;100750＜107050。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "公顷,平方千米的进率与换算"
+                             ],
+        "source":  "25-26四年级上·山东临沂·期中"
+    },
+    {
+        "id":  "140",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "7 公顷 =( )平方米;50 平方千米 =( )公顷;( )公顷 = 80000 平方米;2 公顷 900 平方米 =( )平方米。",
+        "options":  [
+
+                    ],
+        "answer":  "70000;5000;8;20900",
+        "solution":  "【分析】根据题意,明确单位间的进率,1公顷 = 10000平方米,1平方千米 = 100公顷。换算时注意单位大小:大单位转小单位时乘进率,小单位转大单位时除以进率。\n【详解】7×10000 = 70000,所以7公顷 = 70000平方米;50×100 = 5000,所以50平方千米 = 5000公顷;8×10000 = 80000,所以8公顷 = 80000平方米;2×10000 = 20000,所以2公顷 = 20000平方米,2公顷900平方米 = 20000平方米+900平方米 = 20900平方米。",
+        "knowledge_points":  [
+                                 "三位数与整十数的口算乘法",
+                                 "公顷,平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·河北邢台·期中"
+    },
+    {
+        "id":  "141",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "藏羚羊每秒能跑29米,藏羚羊的速度可记作( ),读作( ),藏羚羊每分能跑( )米。",
+        "options":  [
+
+                    ],
+        "answer":  "29 米/秒;29 米每秒;1740",
+        "solution":  "【分析】藏羚羊1秒能跑29米,此时速度的数据是29,单位是米/秒,藏羚羊的速度可记作29米/秒,读作29米每秒。1分钟 = 60秒,利用乘法可计算1分钟跑多少米。\n【详解】1分 = 60秒,60×29 = 1740(米)。藏羚羊每秒能跑29米,藏羚羊的速度可记作29米/秒,读作29米每秒,藏羚羊每分能跑1740米。",
+        "knowledge_points":  [
+                                 "两位数乘两位数,乘数末尾有0",
+                                 "基础行程问题",
+                                 "时,分,秒之间的换算与比较"
+                             ],
+        "source":  "25-26四年级上·河北邢台·期中"
+    },
+    {
+        "id":  "142",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\u0027\u003e\u0027\u0027\u003c\u0027或\u0027=\u0027。985370( )1004320;650×40( )605×40;30702390( )3070 万;750×30( )75×300;4 亿( )398650000;320 公顷( )32 平方千米。",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003e;\u003e;=;\u003e;\u003c",
+        "solution":  "【分析】(1)(3)(5)整数的大小比较,位数不相同的两个数,位数多的数较大;位数相同的两个数,从最高位比起,直到比较出大小。其中3070万 = 30700000,4亿 = 400000000。(2)(4)分别计算出各算式的结果,再比较大小即可。(6)根据1平方千米 = 100公顷,将32平方千米换算成3200公顷后再与320公顷比较大小。\n【详解】985370＜1004320;650×40 = 26000,605×40 = 24200,所以650×40＞605×40;3070万 = 30700000,30702390＞30700000,所以30702390＞3070万;750×30 = 22500,75×300 = 22500,所以750×30 = 75×300;4亿 = 400000000,400000000＞398650000,所以4亿＞398650000;32平方千米 = 3200公顷,320公顷＜3200公顷,所以320公顷＜32平方千米。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "公顷,平方千米的进率与换算",
+                                 "亿以上数的改写"
+                             ],
+        "source":  "25-26 四年级上·山东济宁·期中"
+    },
+    {
+        "id":  "143",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "写出用1,2,5,6,0组成一个三位数和两位数的算式,乘积最大的算式是( )和( )。",
+        "options":  [
+
+                    ],
+        "answer":  "52×610 = 31720;61×520 = 31720",
+        "solution":  "【分析】要使两个数的乘积最大,就要让高位上的数字尽量大。根据数位知识可知,数的高位数字越大,这个数就越大。\n【详解】要求写出三位数乘两位数,可以把较大的数字放在高位上,分别比较:51×620 = 31620,52×610 = 31720,510×62 = 31620,520×61 = 31720。所以乘积最大的算式是52×610 = 31720和61×520 = 31720。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0"
+                             ],
+        "source":  "25-26 四年级上·湖北武汉·期中"
+    },
+    {
+        "id":  "144",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\u0027\u003e\u0027\u0027\u003c\u0027或\u0027=\u0027。180×50( )150×80;160×12( )120×16;350×60( )35×580。",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;=;\u003e",
+        "solution":  "【分析】先计算,再比较大小。\n【详解】180×50 = 9000,150×80 = 12000,9000＜12000,所以180×50＜150×80;160×12 = 1920,120×16 = 1920,所以160×12 = 120×16;350×60 = 21000,35×580 = 20300,21000＞20300,所以350×60＞35×580。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0"
+                             ],
+        "source":  "25-26 四年级上·河北唐山·期中"
+    },
+    {
+        "id":  "145",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一列火车每小时行350 千米,4 小时行多少千米。已知的是() 和() ,求的是()。每小时行 350 千米可以写成() ,读作() 。",
+        "options":  [
+
+                    ],
+        "answer":  "速度;时间;路程;350 千米/时;350 千米每时",
+        "solution":  "【分析】速度×时间 = 路程;简写速度时,在\u0027/\u0027的前面写长度单位,后面写时间单位,读的时候,\u0027/\u0027读作每。\n【详解】每小时行350千米为火车的行驶速度,4小时为火车的行驶时间,求4小时行多少千米即求火车的行驶路程。因此,已知的是速度和时间,求的是路程。每小时行350千米可以写成350千米/时,读作350千米每时。",
+        "knowledge_points":  [
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26 四年级上·河北唐山·期中"
+    },
+    {
+        "id":  "146",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在横线上填上\u0027\u003e\u0027\u0027\u003c\u0027或\u0027=\u0027。18 万( )180500;30×190( )19×300;8 平方千米( )80 公顷;303195999( )303095999;2180000( )280100。",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;=;\u003e;\u003e;\u003e",
+        "solution":  "【分析】将18万改写成一般数字的形式后再比较;大数比较时,先比较位数,位数多的数就大,位数相同就从高位比起;1平方千米 = 100公顷,将单位换算统一后再比较;利用积不变的规律可比较两个算式。\n【详解】18万 = 180000,180000＜180500,所以18万＜180500;30×190 = (30×10)×(190÷10) = 300×19 = 19×300,所以30×190 = 19×300;8平方千米 = 800公顷,800公顷＞80公顷,所以8平方千米＞80公顷;303195999和303095999前几位相同,十万位上1＞0,所以303195999＞303095999;2180000是七位数,280100是六位数,所以2180000＞280100。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "公顷,平方千米的进率与换算",
+                                 "积的变化规律(整数乘法)",
+                                 "亿以内数的改写"
+                             ],
+        "source":  "25-26四年级上·河北唐山·期中"
+    },
+    {
+        "id":  "147",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "小月买5支钢笔用了 15 元,那么她买 3 支钢笔应用( )元。小光有18元,能买( )支钢笔。",
+        "options":  [
+
+                    ],
+        "answer":  "9;6",
+        "solution":  "【分析】根据总价 = 数量×单价,小月买5支钢笔用了15元,用15元除以5可算出钢笔的单价,再乘3可算出3支钢笔多少钱;用18元除以钢笔的单价可算出18元能买几支钢笔。\n【详解】15÷5 = 3(元),3×3 = 9(元),18÷3 = 6(支)。所以她买3支钢笔应用9元;小光有18元,能买6支钢笔。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "3的乘法口诀及应用",
+                                 "用 2 ~ 5 的乘法口诀求商"
+                             ],
+        "source":  "25-26 四年级上·湖南张家界·期中"
+    },
+    {
+        "id":  "148",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "两个数的积是 240,如果一个因数不变,另一个因数除以 4,积变成( )。",
+        "options":  [
+
+                    ],
+        "answer":  "60",
+        "solution":  "【分析】根据积的变化规律,一个因数不变,另一个因数乘或除以一个不为0的数,积也要乘或除以这个数,据此填空即可。\n【详解】240÷4 = 60。两个数的积是240,如果一个因数不变,另一个因数除以4,积变成60。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·湖南长沙·期中"
+    },
+    {
+        "id":  "149",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.4,
+        "question":  "用0,2,3,6,8组成三位数乘两位数的乘法算式,积最大是( ),最小是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "51660;7360",
+        "solution":  "【分析】为了让乘积最大,要把较大的数字放在更高位。两位数的十位选最大的8,三位数的百位选第二大的6,再比较剩余数字的摆放。要使乘积最小,两位数的十位选最小的非0数字2,三位数的百位选第二小的3,再比较剩余数字的摆放。\n【详解】比较630×82 = 51660,620×83 = 51460,可知积最大是51660。比较368×20 = 7360,308×26 = 8008,360×28 = 10080,可知积最小是7360。所以用0,2,3,6,8组成三位数乘两位数的乘法算式,积最大是51660,最小是7360。",
+        "knowledge_points":  [
+                                 "1000以内数的组成",
+                                 "大数的比较",
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "25-26 四年级上·湖北鄂州·期中"
+    },
+    {
+        "id":  "150",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "两个因数的积是 360,如果一个因数乘 3,另一个因数除以 3,积是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "360",
+        "solution":  "【分析】积的变化规律:在乘法计算中,一个因数乘几或除以几(0除外),另一个因数不变,则积也乘几或除以几;两数相乘,一个因数乘3,另一个因数除以3,积不变。\n【详解】360×3÷3 = 1080÷3 = 360。两个因数的积是360,如果一个因数乘3,另一个因数除以3,积是360。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·湖南岳阳·期中"
+    },
+    {
+        "id":  "151",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "已知 A×B = 50 ,如果 A 不变,B 乘 3,则积是( );如果 B 不变,A 除以5,则积是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "150;10",
+        "solution":  "【分析】积的变化规律:两个数相乘,一个因数不变,另一个因数乘或除以几(0除外),积也乘或除以几;两个数相乘,一个因数乘几,另一个因数除以相同的数,积不变,据此判断。\n【详解】已知A×B = 50,如果A不变,B乘3,则积也乘3,50×3 = 150,则积是150;如果B不变,A除以5,则积也除以5,50÷5 = 10,所以积是10。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26四年级上·湖北鄂州·期中"
+    },
+    {
+        "id":  "152",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "一间舞蹈教室的面积约 80 平方米,125 间这样的舞蹈教室的面积约 10 公顷。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】首先用每间教室的面积乘125,计算出125间舞蹈教室的总面积。1公顷 = 10000平方米,所以10公顷 = 100000平方米。再比较125间舞蹈教室的总面积与100000的大小。据此解答。\n【详解】125×80 = 10000(平方米)。10公顷 = 100000平方米,10000＜100000,所以原题说法错误。故答案为:×。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·山东济南·期中"
+    },
+    {
+        "id":  "153",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "笔算 213×65 时,\u00272\u0027和\u00276\u0027相乘的实际意义是 200×60 。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】根据三位数乘两位数的计算法则可知,三位数百位上的数与两位数个位上的数相乘,表示的是几百数与几相乘;三位数百位上的数与两位数十位上的数相乘,表示的是几百数与几十数相乘;依此判断。\n【详解】笔算213×65时,\u00272\u0027在三位数的百位上,\u00276\u0027在两位数的十位上,因此\u00272\u0027和\u00276\u0027相乘的实际意义是200×60。故答案为:√。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "25-26 四年级上·河北唐山·期中"
+    },
+    {
+        "id":  "154",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "小明半小时走了2千米,这里的\u00272千米\u0027表示小明走的速度。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】速度是单位时间内通过的路程,强调\u0027单位时间\u0027这一要素;而题目中\u00272千米\u0027是小明在半小时内行走的总距离,属于路程的概念,并非速度。\n【详解】根据分析可知:小明半小时走了2千米,这里的\u00272千米\u0027表示小明走的路程,并非速度。因此原题说法错误。故答案为:×。",
+        "knowledge_points":  [
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26 四年级上·云南昭通·期中"
+    },
+    {
+        "id":  "155",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "因为240×50中两个因数的末尾一共有两个0,所以积末尾也只有两个0。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】在乘法计算中,两个因数的末尾一共有几个0,但计算时,数字相乘的结果还可能使末尾增加0,所以积的末尾至少有这些0。若要精确判断0的个数,需要计算出算式结果后再判断。\n【详解】240×50 = 12000,积的末尾有3个0。题干说法错误。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0"
+                             ],
+        "source":  "25-26 四年级上·河北唐山·期中"
+    },
+    {
+        "id":  "156",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "计算题",
+        "difficulty":  0.85,
+        "question":  "直接写得数。21×30 = ( );160×60 = ( );40×130 = ( );90×10 = ( );125×8 = ( );11×800 = ( );500×60 = ( );190×4 = ( )。",
+        "options":  [
+
+                    ],
+        "answer":  "630;9600;5200;900;1000;8800;30000;760",
+        "solution":  "【分析】本题考查整数乘法的口算能力。末尾有0的乘法:先把0前面的数相乘,再看两个因数末尾一共有几个0,就在积的末尾添上几个0。特殊乘法:如125×8是常见的特殊乘积。三位数乘一位数:如190×4,可先算19×4,再在末尾添1个0。\n【详解】21×30:先算21×3 = 63,再在末尾添1个0,得630。160×60:先算16×6 = 96,再在末尾添2个0,得9600。40×130:先算4×13 = 52,再在末尾添2个0,得5200。90×10:先算9×1 = 9,再在末尾添2个0,得900。125×8:直接得1000。11×800:先算11×8 = 88,再在末尾添2个0,得8800。500×60:先算5×6 = 30,再在末尾添3个0,得30000。190×4:先算19×4 = 76,再在末尾添1个0,得760。",
+        "knowledge_points":  [
+                                 "两位数乘整十数的口算乘法",
+                                 "三位数与整十数的口算乘法",
+                                 "三位数乘两位数,乘数末尾有0"
+                             ],
+        "source":  "25-26 四年级上·陕西西安·期中"
+    },
+    {
+        "id":  "157",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "为进一步落实\u0027德,智,体,美,劳\u0027五育并举工作。某小学以体育为突破口,准备再添加篮球和足球各 45 个,足球每个 103 元,篮球每个 97 元,请你计算一下一共要花多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "9000 元",
+        "solution":  "【分析】总费用由足球和篮球的总价组成。两种球的数量相同,均为45个,可先计算每个足球和篮球的单价之和,再乘数量,即可得一共要花多少元。\n【详解】103+97 = 200(元),200×45 = 9000(元)。答:一共要花9000元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "两,三位数的连续进位加法",
+                                 "三位数乘两位数,乘数末尾有0"
+                             ],
+        "source":  "25-26 四年级上·山西晋中·期中"
+    },
+    {
+        "id":  "158",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "奇思一家 3 口人准备正月里外出旅游,计划每天支出如下:伙食费共 90 元,住宿费共 150 元。还有本次旅游往返火车票费用共 240 元。(1)如果外出旅游 12 天,至少准备多少钱。(2)如果奇思的父亲每月积攒 800 元,攒 4 个月够出去玩吗。",
+        "options":  [
+
+                    ],
+        "answer":  "3120 元;够",
+        "solution":  "【分析】(1)将每天的伙食费和住宿费相加得到每天这两项费用的总和,乘旅游天数,再加上往返火车票费用,即为至少准备多少钱;(2)计算4个月积攒的总金额,与旅游总费用比较即可知道够不够。\n【详解】(1)90+150 = 240(元),240×12 = 2880(元),2880+240 = 3120(元)。答:至少准备3120元。(2)800×4 = 3200(元),3200元＞3120元。答:攒4个月够出去玩。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "万以内数的大小比较",
+                                 "整十,整百,整千数与一位数的乘法",
+                                 "三位数乘两位数,乘数末尾有 0"
+                             ],
+        "source":  "25-26 四年级上·山西晋中·期中"
+    },
+    {
+        "id":  "159",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "国庆期间,小虎每天写完作业后都会骑车去少年宫踢足球,他平均每分钟骑行 172 米,10 分钟正好骑了全程的一半。请你算一算小虎家与少年宫相距多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "3440 米",
+        "solution":  "【分析】根据速度×时间 = 路程,小虎骑行速度是每分钟172米,10分钟骑行的路程为172×10 = 1720(米)。根据题意,这段路程是全程的一半,因此总路程是1720的2倍。\n【详解】172×10 = 1720(米),1720×2 = 3440(米)。答:小虎家与少年宫相距3440米。",
+        "knowledge_points":  [
+                                 "三位数与整十数的口算乘法",
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26 四年级上·山西晋中·期中"
+    },
+    {
+        "id":  "160",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.94,
+        "question":  "潮绣是潮汕地区蜚声海内外的传统刺绣工艺。张阿姨是潮绣技艺传承人,一天能完成 12 件小型潮绣挂件,照这样计算,张阿姨一年(按 365 天计算)能完成多少件小型潮绣挂件。",
+        "options":  [
+
+                    ],
+        "answer":  "4380 件",
+        "solution":  "【分析】根据题意可知,用张阿姨一天能完成的小型潮绣挂件件数乘365,就是张阿姨一年能完成的小型潮绣挂件件数,即12×365。\n【详解】12×365 = 4380(件)。答:照这样计算,张阿姨一年(按365天计算)能完成4380件小型潮绣挂件。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "25-26四年级上·广东汕头·期中"
+    },
+    {
+        "id":  "161",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.94,
+        "question":  "苏绣是江苏省苏州市民间传统美术,国家级非物质文化遗产之一。李师傅是苏绣技艺传承人,一天能完成 14 幅小型苏绣摆件,照这样计算,李师傅一年(按 365 天计算)能完成多少幅小型苏绣摆件。",
+        "options":  [
+
+                    ],
+        "answer":  "5110 幅",
+        "solution":  "【分析】根据题意可知,用李师傅一天能完成的小型苏绣摆件幅数乘365,就是李师傅一年能完成的小型苏绣摆件幅数,即14×365。\n【详解】14×365 = 5110(幅)。答:照这样计算,李师傅一年(按365天计算)能完成5110幅小型苏绣摆件。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "25-26 四年级上·江苏苏州·期中"
+    },
+    {
+        "id":  "162",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "小明从家到学校上课,速度是 70 米/分钟,8 分钟到达学校,放学时按原路回家,只用了7分钟,回家的速度是多少。",
+        "options":  [
+
+                    ],
+        "answer":  "80 米/分钟",
+        "solution":  "【分析】根据\u0027路程 = 速度×时间\u0027,用70乘8,先求出小明家到学校的路程,再根据\u0027速度 = 路程÷时间\u0027,用小明家到学校的路程除以7即可解答。\n【详解】70×8÷7 = 560÷7 = 80(米/分钟)。答:回家的速度是80米/分钟。",
+        "knowledge_points":  [
+                                 "多位数与一位数的乘除混合运算",
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26四年级上·云南昭通·期中"
+    },
+    {
+        "id":  "163",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "幸福小学四年级师生共 208 人,准备包车去参观生物基地。包车的价格是39元/人,带8000元钱够吗。",
+        "options":  [
+
+                    ],
+        "answer":  "不够",
+        "solution":  "【分析】根据单价×数量 = 总价,用每人包车的钱数乘总人数,求出包车总价,再用包车总价与所带的钱数比较,即可求出带8000元够不够。\n【详解】39×208 = 8112(元),8112元＞8000元。答:带8000元钱不够。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "三位数乘两位数,三位数中间有 0"
+                             ],
+        "source":  "25-26四年级上·内蒙古乌海·期中"
+    },
+    {
+        "id":  "164",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "实验学校组织 24 名教师开展\u0027诵最美《诗经》\u0027表演活动。现要为表演教师每人买一件表演服和一双鞋,每件表演服 125 元,每双鞋75元。买这些表演服和鞋一共需要多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "4800 元",
+        "solution":  "【分析】根据总价 = 单价×数量,先计算出一件表演服和一双鞋一共需要多少元,再乘24,即可求出24件表演服和24双鞋的价钱总和。\n【详解】(125+75)×24 = 200×24 = 4800(元)。答:买这些表演服和鞋一共需要4800元。",
+        "knowledge_points":  [
+                                 "三位数乘两位数的实际问题"
+                             ],
+        "source":  "25-26 四年级上·湖北荆州·期中"
+    },
+    {
+        "id":  "165",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "特快列车每小时可行 360 千米,普通列车每小时可行 110 千米,则各行驶 5小时后,特快列车比普通列车多行多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "1250 千米",
+        "solution":  "【分析】根据题意可以先求出1小时特快列车比普通列车多行多少千米,再用这个数乘5,即是5小时多行的路程。\n【详解】(360-110)×5 = 250×5 = 1250(千米)。答:各行驶5小时后,特快列车比普通列车多行1250千米。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26 四年级上·河南驻马店·期中"
+    },
+    {
+        "id":  "166",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一辆洒水车每分钟行 150 米,洒水的宽度是 10 米,洒水车行驶 100 分钟,能洒湿多少公顷的地面。",
+        "options":  [
+
+                    ],
+        "answer":  "15 公顷",
+        "solution":  "【分析】根据题意,先用洒水车每分钟行的距离乘洒水的宽度,求出每分钟洒水车洒水的面积,再乘行驶的时间,即可求出能洒湿多少平方米的地面,根据1公顷 = 10000平方米,据此换算成公顷为单位即可。\n【详解】150×10×100 = 1500×100 = 150000(平方米),150000平方米 = 15公顷。答:能洒湿15公顷的地面。",
+        "knowledge_points":  [
+                                 "三位数与整十数的口算乘法",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·河南驻马店·期中"
+    },
+    {
+        "id":  "167",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一台电风扇 135元,某公司要买 15 台这样的风扇,一共要付多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "2025 元",
+        "solution":  "【分析】根据总价 = 单价×数量,用一台电风扇的价格乘买的台数,即可求出一共要付多少元.\n【详解】 135×15 = 2025 (元)答:一共要付 2025 元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "25-26 四年级上·河南驻马店·期中"
+    },
+    {
+        "id":  "168",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "学校图书馆新购进36 个书架,每个书架可以放 240 本书。这些书架一共能放多少本书。",
+        "options":  [
+
+                    ],
+        "answer":  "8640 本",
+        "solution":  "【分析】求这些书架一共能放书的数量,就是求36个240的和是多少,用乘法计算,据此解答。\n【详解】36×240 = 8640(本)。答:这些书架一共能放8640本书。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有 0"
+                             ],
+        "source":  "25-26 四年级上·广西河池·期中"
+    },
+    {
+        "id":  "169",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "一辆汽车从甲地开往乙地,每小时行驶85千米,行驶了12小时后,离乙地还有260千米。甲,乙两地相距多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "1280 千米",
+        "solution":  "【分析】根据路程 = 速度×时间,已知汽车速度为每小时85千米,行驶时间为12小时,用乘法计算出已行驶路程。用已行驶路程加上260千米即可计算出甲、乙两地的距离。\n【详解】85×12+260 = 1020+260 = 1280(千米)。答:甲、乙两地相距1280千米。",
+        "knowledge_points":  [
+                                 "乘加,乘减混合运算",
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26 四年级上·广西河池·期中"
+    },
+    {
+        "id":  "170",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "函谷关地处三门峡灵宝市,阳光小学组织学生去函谷关进行研学活动。参加的学生一共有 300 人,教师 30 人,成人门票 75 元/人,学生门票 38 元/人,买门票一共需要多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "13650 元",
+        "solution":  "【分析】已知参加的学生一共有300人,学生门票38元/人,那么学生花的总费用为学生每人的票价乘人数;教师30人,成人门票75元/人,那么教师花的总费用为成人每人的票价乘人数。将学生和教师花的总费用相加,就能求出买门票一共需要的钱数。\n【详解】300×38+30×75 = 11400+2250 = 13650(元)。答:买门票一共需要13650元。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法",
+                                 "乘加,乘减混合运算"
+                             ],
+        "source":  "25-26 四年级上·河南三门峡·期中"
+    },
+    {
+        "id":  "171",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.94,
+        "question":  "三门峡市沿黄复合型生态廊道,自西向东约 200 余千米。乐乐的爸爸是骑自行车爱好者,他以 15 千米/时的平均速度,骑行了 8 小时,他一共骑行了多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "120 千米",
+        "solution":  "【分析】乐乐的爸爸以15千米/时的平均速度,骑行了8小时,求他一共骑行了多少千米。根据速度×时间 = 路程,代入数字进行计算即可。\n【详解】15×8 = 120(千米)。答:他一共骑行了120千米。",
+        "knowledge_points":  [
+                                 "两,三位数与一位数的一次进位乘法",
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26 四年级上·河南三门峡·期中"
+    },
+    {
+        "id":  "172",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "张阿姨一家来海南旅游,返程时想买一些海南特产,一包椰子糖25元,一包芒果干12元,张阿姨买12包椰子糖和15包芒果干共需要多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "480 元",
+        "solution":  "【分析】单价×数量 = 总价,依此分别计算出买12包椰子糖和15包芒果干用的钱数,然后再相加,即可得到张阿姨买12包椰子糖和15包芒果干一共需要的钱数。\n【详解】25×12+12×15 = 300+180 = 480(元)。答:张阿姨买12包椰子糖和15包芒果干共需要480元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "乘加,乘减混合运算"
+                             ],
+        "source":  "25-26四年级上·海南省直辖县级单位·期中"
+    },
+    {
+        "id":  "173",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一列动车从海口站出发,到乐东站大约需要 2 小时,动车速度约为250 千米/时。海口站到乐东站大约有多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "500 千米",
+        "solution":  "【分析】根据速度×时间 = 路程,代入数据直接计算即可。\n【详解】250×2 = 500(千米)。答:海口站到乐东站大约有500千米。",
+        "knowledge_points":  [
+                                 "因数中间或末尾是0的乘法",
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26 四年级上·海南省直辖县级单位·期中"
+    },
+    {
+        "id":  "174",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "王叔叔从甲地开车到乙地,前2小时行驶了170千米,后3小时行驶了225千米。王叔叔平均每小时行驶多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "79 千米",
+        "solution":  "【分析】根据题意,先将前2小时行驶的距离加上后3小时行驶的距离,求出一共行驶了多少千米,再用行驶的距离除以行驶的总时间(2+3)小时,即可求出王叔叔平均每小时行驶多少千米。\n【详解】(170+225)÷(2+3) = 395÷5 = 79(千米)。答:王叔叔平均每小时行驶79千米。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26 四年级上·广东潮州·期中"
+    },
+    {
+        "id":  "175",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "四年级有 178 名学生要去参观科技馆,每张门票38元。学校准备了7000元给学生们买门票,你认为这些钱够吗。",
+        "options":  [
+
+                    ],
+        "answer":  "够",
+        "solution":  "【分析】要判断7000元是否够买门票,需先计算178名学生购买单价38元门票的总费用,再与7000元比较,据此解答即可。\n【详解】178×38 = 6764(元),6764＜7000。答:准备的钱够买门票。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法",
+                                 "三位数乘两位数的实际问题"
+                             ],
+        "source":  "25-26 四年级上·广东潮州·期中"
+    },
+    {
+        "id":  "176",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "某超市购进 125 箱饮料,每箱36元,购进这些饮料一共花了多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "4500 元",
+        "solution":  "【分析】要计算购进125箱饮料的总花费,利用\u0027总价 = 数量×单价\u0027进行计算即可。\n【详解】125×36 = 4500(元)。答:超市购进125箱单价36元的饮料,一共花了4500元。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法",
+                                 "三位数乘两位数的实际问题"
+                             ],
+        "source":  "25-26 四年级上·广东潮州·期中"
+    },
+    {
+        "id":  "177",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "买 12 支铅笔和 8 支钢笔共需要 112 元。照这样计算,买 18 支铅笔和 12 支钢笔共需要花多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "168 元",
+        "solution":  "【分析】由题意得,买12支铅笔和8支钢笔共需要112元。如果把3支铅笔和2支钢笔看成一组,12支铅笔和8支钢笔有这样的4组,用112除以4可以算出3支铅笔和2支钢笔一共需要多少钱。18支铅笔和12支钢笔里面有这样的6组,再用前面的得数乘6即可。\n【详解】112÷4 = 28(元),28×6 = 168(元)。答:买18支铅笔和12支钢笔共需要花168元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "两,三位数与一位数连续进位的乘法"
+                             ],
+        "source":  "25-26 四年级上·山东临沂·期中"
+    },
+    {
+        "id":  "178",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "朝阳小学有 296 名夏令营学员,学校组织学员到博物馆参观,学生团体票每人30元。学校准备7500元够买门票吗。",
+        "options":  [
+
+                    ],
+        "answer":  "不够",
+        "solution":  "【分析】朝阳小学有296名夏令营学员,学校组织学员到博物馆参观,学生团体票每人30元,用296乘30求出总共多少元,再与7500比较,据此解答。\n【详解】296×30 = 8880(元),8880＞7500。答:学校准备7500元不够买门票。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "三位数乘两位数,乘数末尾有 0"
+                             ],
+        "source":  "25-26 四年级上·山东临沂·期中"
+    },
+    {
+        "id":  "179",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一块长方形菜地长800 米,宽 500 米。如果延长它的宽,使它成为一块正方形菜地,那么现在菜地的面积比原来的面积增加多少公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "24 公顷",
+        "solution":  "【分析】长方形的面积 = 长×宽。由题意得,一块长方形菜地长800米,宽500米,用800乘500可以算出长方形菜地的面积。如果延长它的宽,使它成为一块正方形菜地,也就是把菜地的宽延长至800米,那么正方形菜地的边长是800米。正方形的面积 = 边长×边长,再用正方形的面积减去长方形的面积,求出增加的面积,最后根据10000平方米 = 1公顷换算即可。\n【详解】800×800 = 640000(平方米),800×500 = 400000(平方米),640000-400000 = 240000(平方米),240000平方米 = 24公顷。答:现在菜地的面积比原来的面积增加24公顷。",
+        "knowledge_points":  [
+                                 "三位数与整十数的口算乘法",
+                                 "长方形的面积",
+                                 "正方形的面积",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·山东临沂·期中"
+    },
+    {
+        "id":  "180",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "无人机技术是航空航天工程中的重要技术之一,在军事领域应用最为广泛,但目前也已普及应用到人民群众的生产生活中,比如应用于农业降本提效。一架农用轻型无人机每秒飞行 13 米,喷洒农药的宽度是12米,如果飞行1分钟,能给多少平方米的农田喷洒农药。",
+        "options":  [
+
+                    ],
+        "answer":  "9360 平方米",
+        "solution":  "【分析】一分钟是60秒,用每秒的飞行路程13米乘60,求出1分钟飞行路程,再根据长方形的面积 = 长×宽,用1分钟飞行路程乘喷洒宽度,求出1分钟喷洒农田面积,列式计算即可。\n【详解】1分钟 = 60秒,13×60×12 = 780×12 = 9360(平方米)。答:飞行1分钟,能给9360平方米的农田喷洒农药。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "基础行程问题",
+                                 "长方形的面积"
+                             ],
+        "source":  "25-26 四年级上·河北邢台·期中"
+    },
+    {
+        "id":  "181",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "微山县规划建设一个长方形生态公园,长 2000 米,宽 1500 米。后来对规划进行调整,将长增加500米,宽不变,调整后公园的面积增加了多少公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "75 公顷",
+        "solution":  "【分析】调整后,长增加500米,宽不变,则面积增加部分是一个长方形,长等于原来的宽1500米,宽等于增加的500米。根据长方形的面积 = 长×宽,把数据代入公式解答,最后根据1公顷 = 10000平方米换算成公顷作单位。\n【详解】1500×500 = 750000(平方米),750000平方米 = 75公顷。答:调整后公园的面积增加了75公顷。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "长方形的面积",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26四年级上·山东济宁·期中"
+    },
+    {
+        "id":  "182",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "微山湖某荷叶采摘基地准备给采摘工人配备防晒帽。该基地现有男采摘工280人,女采摘工 190 人,若每顶防晒帽 25 元,购买防晒帽一共需要多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "11750 元",
+        "solution":  "【分析】男采摘工人数加上女采摘工人数,可以算出需要购买(280+190)顶防晒帽。每顶防晒帽的价钱乘购买的顶数,即可算出购买防晒帽一共需要多少钱。\n【详解】280+190 = 470(顶),25×470 = 11750(元)。答:购买防晒帽一共需要11750元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "三位数乘两位数,乘数末尾有 0"
+                             ],
+        "source":  "25-26四年级上·山东济宁·期中"
+    },
+    {
+        "id":  "183",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "为传承红色文化,某学校组织四年级学生去铁道游击队纪念馆参观。四年级有3个班,每班45人,每人车船票25元,买车船票一共需要多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "3375 元",
+        "solution":  "【分析】先用每个班的人数45人乘班数3个班,求出一共多少人,再乘每人的车船票价格即可。\n【详解】45×3×25 = 135×25 = 3375(元)。答:买车船票一共需要3375元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "用两步连乘解决实际问题",
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "25-26四年级上·山东济宁·期中"
+    },
+    {
+        "id":  "184",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一艘运输船从微山港出发运送货物,去时的速度是30千米/时,用了8小时到达目的地。返回时比去时少用了2小时,返回时的速度是多少。",
+        "options":  [
+
+                    ],
+        "answer":  "40 千米/时",
+        "solution":  "【分析】首先根据路程 = 速度×时间,代入数值计算出两地的距离。然后用去时用的时间减去2,计算出返回时用的时间。最后根据速度 = 路程÷时间,代入数值计算出返回时的速度。据此解答。\n【详解】30×8 = 240(千米),240÷(8-2) = 240÷6 = 40(千米/时)。答:返回时的速度是40千米/时。",
+        "knowledge_points":  [
+                                 "几百几十或几千几百除以一位数的除法",
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26四年级上·山东济宁·期中"
+    },
+    {
+        "id":  "185",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "微山岛大官码头画舫船票单价每人 50 元,从大官码头到荷园一趟大约需要 25分钟,某周六共接待游客 105 人,全天票价收入多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "5250 元",
+        "solution":  "【分析】用画舫船票单价每人50元乘周六接待的游客人数105人即可求解。\n【详解】50×105 = 5250(元)。答:全天票价收入5250元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "三位数乘两位数,三位数中间有0"
+                             ],
+        "source":  "25-26 四年级上·山东济宁·期中"
+    },
+    {
+        "id":  "186",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "书法兴趣班老师买了15个砚台,每个砚台102元,又用了270元买了9包宣纸,这次采购一共用了多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "1800 元",
+        "solution":  "【分析】根据总价 = 单价×数量,用书法兴趣班老师买砚台的个数乘每个砚台价钱,先算出买15个砚台花费的钱数,再加上买宣纸花费的钱数,即可求出这次采购一共用了多少元。\n【详解】102×15+270 = 1530+270 = 1800(元)。答:这次采购一共用了1800元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "三位数乘两位数,三位数中间有 0",
+                                 "乘加,乘减混合运算"
+                             ],
+        "source":  "25-26 四年级上·湖北武汉·期中"
+    },
+    {
+        "id":  "187",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "王叔叔从甲城开车去乙城办事,每小时行40千米,行了3小时后超过全程中点15千米。甲,乙两城相距多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "210 千米",
+        "solution":  "【分析】根据题意,每小时行40千米,3小时行了全程的一半还多15千米。应先求出3小时行的路程,即40×3,再用3小时行驶的路程减去15千米,得到甲、乙两地的一半路程,再乘2即为甲、乙两地的距离。\n【详解】40×3-15 = 120-15 = 105(千米),105×2 = 210(千米)。答:甲、乙两地相距210千米。",
+        "knowledge_points":  [
+                                 "两,三位数乘一位数的实际应用",
+                                 "乘加,乘减混合运算",
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26 四年级上·湖南张家界·期中"
+    },
+    {
+        "id":  "188",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "李大爷有一块长600米,宽500米的桔子园。今年迎来大丰收,平均每公顷可采摘12吨桔子,李大爷今年一共收获多少吨桔子。",
+        "options":  [
+
+                    ],
+        "answer":  "360 吨",
+        "solution":  "【分析】根据长方形面积 = 长×宽算出桔子园的面积,再根据10000平方米 = 1公顷将单位换算成公顷。因为平均每公顷可采摘12吨桔子,所以用12吨乘公顷数可算出今年一共收获多少吨桔子。\n【详解】600×500 = 300000(平方米),300000平方米 = 30公顷,30×12 = 360(吨)。答:李大爷今年一共收获360吨桔子。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有 0",
+                                 "长方形的面积",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26四年级上·湖南张家界·期中"
+    },
+    {
+        "id":  "189",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "小强的步行速度是75米/分,他从家里出发去外婆家需要24分钟。如果改骑自行车只要8分钟。那么小强骑自行车的速度是多少。",
+        "options":  [
+
+                    ],
+        "answer":  "225 米/分",
+        "solution":  "【分析】根据路程 = 速度×时间,求出小强从家里出发步行去外婆家的路程。再根据速度 = 路程÷时间,用步行路程除以骑自行车需要的时间,求出自行车的速度。\n【详解】75×24 = 1800(米),1800÷8 = 225(米/分)。答:小强骑自行车的速度是225米/分。",
+        "knowledge_points":  [
+                                 "两位数乘两位数的进位乘法",
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26四年级上·湖南张家界·期中"
+    },
+    {
+        "id":  "190",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "慈利县三官寺乡境内横跨峡谷的云天渡玻璃桥被吉尼斯世界纪录列为\u0027世界最高人行桥\u0027。玻璃桥门票为每人 125 元。小明爸爸带12 位朋友一起去参观玻璃桥。小明爸爸购买门票一共需要花多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "1625 元",
+        "solution":  "【分析】小明爸爸带12位朋友去参观,那么总人数是13人,再乘每人125元,即可求得小明爸爸和12位朋友买门票一共需要花多少钱。\n【详解】1+12 = 13(人),125×13 = 1625(元)。答:小明爸爸购买门票一共需要花1625元。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "25-26四年级上·湖南张家界·期中"
+    },
+    {
+        "id":  "191",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一块长方形稻田,长 800 米,宽 500 米。如果每公顷收稻谷 6 吨,这块稻田能收稻谷多少吨。",
+        "options":  [
+
+                    ],
+        "answer":  "240 吨",
+        "solution":  "【分析】首先根据长方形面积公式:面积 = 长×宽,可计算出长方形稻田的面积,然后根据1公顷 = 10000平方米进行单位换算。已知每公顷收稻谷6吨,用总面积乘每公顷收稻谷的重量即可计算出这块稻田能收稻谷的总重量。\n【详解】800×500 = 400000(平方米),400000平方米 = 40公顷,40×6 = 240(吨)。答:这块稻田能收稻谷240吨。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·湖南岳阳·期中"
+    },
+    {
+        "id":  "192",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一块长方形麦田,长 600 米,宽 400 米。如果每公顷收小麦 7 吨,这块麦田能收小麦多少吨。",
+        "options":  [
+
+                    ],
+        "answer":  "168 吨",
+        "solution":  "【分析】长方形面积 = 长×宽,先代入数字计算出长方形麦田的面积是多少平方米,根据1公顷 = 10000平方米,换算成公顷为单位,再乘每公顷收小麦的吨数,即可求出这块麦田能收小麦多少吨。\n【详解】600×400 = 240000(平方米),240000平方米 = 24公顷,24×7 = 168(吨)。答:这块麦田能收小麦168吨。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "长方形的面积",
+                                 "三位数与整十数的口算乘法"
+                             ],
+        "source":  "25-26 四年级上·湖南长沙·期中"
+    },
+    {
+        "id":  "193",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一辆汽车从长沙开往张家界,前 3 小时行驶了 225 千米,照这样的速度,又行驶了 4 小时到达张家界。长沙到张家界的路程是多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "525 千米",
+        "solution":  "【分析】题干中说照这样的速度,说明速度一定。根据速度 = 路程÷时间,求出速度,再根据路程 = 速度×时间,求出后4小时行驶的路程,再加上前3小时行驶的路程,据此解答。\n【详解】225÷3 = 75(千米/时),75×4 = 300(千米),225+300 = 525(千米)。答:长沙到张家界的路程是525千米。",
+        "knowledge_points":  [
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26 四年级上·湖南长沙·期中"
+    },
+    {
+        "id":  "194",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "学校要为图书馆增添两种新书,每种买15套。一套是125元,另一套是18元。一共要花多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "2145 元",
+        "solution":  "【分析】首先把两种书每套的价格相加,计算出两种书各买一套的总价钱。然后用两种书各买一套的总价钱乘要买的套数,即可计算出总共花费。据此解答。\n【详解】(125+18)×15 = 143×15 = 2145(元)。答:一共要花2145元。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "25-26四年级上·湖南长沙·期中"
+    },
+    {
+        "id":  "195",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "吴都古肆花市一束鲜花 35 元,买 5 束鲜花送 2 束,如果买 100 束鲜花,一共要付多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "2520 元",
+        "solution":  "【分析】买5束鲜花送2束,相当于买7束只需要付35×5 = 175元。用100除以7,可知里面有几个7束,就是几个175元,再加上余下的鲜花按35元/束计算,最后相加即可。\n【详解】35×5 = 175(元),100÷7 = 14(组)……2(束),175×14+35×2 = 2450+70 = 2520(元)。答:一共要付2520元。",
+        "knowledge_points":  [
+                                 "乘加,乘减混合运算",
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "25-26四年级上·湖北鄂州·期中"
+    },
+    {
+        "id":  "196",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "长港农场有一块长方形菜地长 500 米,宽 400 米。这块菜地的面积是多少公顷。如果每公顷施化肥 120 千克,共需化肥多少千克。",
+        "options":  [
+
+                    ],
+        "answer":  "20 公顷;2400 千克",
+        "solution":  "【分析】长方形的面积 = 长×宽,根据长方形的面积公式可求出长方形菜地的面积;每公顷施肥120千克,求需要化肥的质量,用乘法即可。\n【详解】500×400 = 200000(平方米),200000平方米 = 20公顷,20×120 = 2400(千克)。答:这块菜地的面积是20公顷,共需化肥2400千克。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "长方形的面积",
+                                 "公顷,平方千米的进率与换算"
+                             ],
+        "source":  "25-26四年级上·湖北鄂州·期中"
+    },
+    {
+        "id":  "197",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "学校准备给24个班配备新桌椅,每套桌椅 285 元,准备 7000 元够吗。",
+        "options":  [
+
+                    ],
+        "answer":  "够",
+        "solution":  "【分析】已知共有24个班配备新桌椅,每套桌椅285元,根据总价 = 单价×数量,即可计算出需要花费的钱数,再与7000元比较即可判断是否够。\n【详解】285×24 = 6840(元),6840＜7000。答:准备7000元够。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "25-26四年级上·湖南岳阳·期中"
+    },
+    {
+        "id":  "198",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "货车司机送货,上午 8:00 从甲地出发,中午 12:00 到达乙地,卸货 1 小时后原路返回,去时每小时行驶60千米,返程时间比去时少1 小时.\n\n(1)甲,乙两地的距离是多少。\n(2)返回时的货车每小时比去时多行驶多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "240 千米;20 千米",
+        "solution":  "【分析】(1)根据题意,上午8:00从甲地出发,中午12:00到达乙地,经过时间是12时-8时,再乘去时每小时行驶的60千米,求出甲、乙两地的距离。(2)因为返程时间比去时少1小时,用经过时间减去1,求出返程时间,用甲、乙两地的距离除以返程时间,就是返程速度,再减去60,就是返回时每小时比去时多行驶多少千米。\n【详解】(1)12-8 = 4(小时),60×4 = 240(千米)。答:甲、乙两地的距离是240千米。(2)240÷(4-1)-60 = 240÷3-60 = 80-60 = 20(千米)。答:返回时的货车每小时比去时多行驶20千米。",
+        "knowledge_points":  [
+                                 "24 时计时法与普通计时法的互化",
+                                 "基础行程问题",
+                                 "24时计时法时间的计算"
+                             ],
+        "source":  "25-26四年级上·福建莆田·期中"
+    },
+    {
+        "id":  "199",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "游泳池长 50 米,小丽游了10个来回,她游了多少米。合多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "1000 米;1 千米",
+        "solution":  "【分析】1个来回是2个游泳池的长度,50乘2求出1个来回的长度,再求出10个来回的总长度,最后根据1千米 = 1000米将单位化成千米即可。\n【详解】50×2×10 = 100×10 = 1000(米),1000米 = 1千米。答:她游了1000米,合1千米。",
+        "knowledge_points":  [
+                                 "三位数与整十数的口算乘法",
+                                 "千米和米之间的进率与换算"
+                             ],
+        "source":  "25-26 三年级上·江西宜春·期中"
+    },
+    {
+        "id":  "200",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "学校操场跑道一圈长 400 米,小明晨练时跑了 5 圈,他一共跑了多少米。合多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "2000 米;2 千米",
+        "solution":  "【分析】已知跑道一圈长400米,小明跑了5圈,用一圈的长度乘圈数即可求出总长度。最后根据1千米 = 1000米,将单位换算为千米。\n【详解】400×5 = 2000(米),2000米 = 2千米。答:他一共跑了2000米,合2千米。",
+        "knowledge_points":  [
+                                 "三位数与一位数的口算乘法",
+                                 "千米和米之间的进率与换算"
+                             ],
+        "source":  "25-26 三年级上·浙江杭州·期中"
+    },
+    {
+        "id":  "201",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "2024 年我国\"嫦娥六号\"实现首次月球背面采样返回的任务,2025 年\"天问二号\"探测器开启小行星探测征程,两者均展现了中国航天的速度与精度.\"嫦娥六号\"从月球启程时,以 2700 米/秒的速度进入月地转移轨道,在该轨道段飞行时间约 4 天;\"天问二号\"发射入轨时,速度达到 11300 米/秒,其进入地火转移轨道前的加速飞行时间为 20 分钟。要解决\"\u0027嫦娥六号\u0027在月地转移轨道飞行的路程\"这个问题,需要用到的信息是( )。",
+        "options":  [
+                        "A. 2700 米/秒,11300 米/秒",
+                        "B. 2700 米/秒,20 分钟",
+                        "C. 2700 米/秒,4 天",
+                        "D. 11300 米/秒,20 分钟"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】路程 = 速度×时间,要知道\u0027嫦娥六号\u0027在月地转移轨道飞行的路程,就要用到\u0027嫦娥六号\u0027的飞行速度2700米/秒和飞行时间4天。据此选择即可。\n【详解】要解决\u0027嫦娥六号\u0027在月地转移轨道飞行的路程这个问题,需要用到的信息是2700米/秒和4天。故答案为:C。",
+        "knowledge_points":  [
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26 四年级上·福建泉州·期中"
+    },
+    {
+        "id":  "202",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列描述中的面积,与 1 公顷最接近的是( )。",
+        "options":  [
+                        "A. 边长 10 米的正方形花坛面积",
+                        "B. 长202米,宽50米的长方形操场面积",
+                        "C. 50 个教室的总面积(长 8 米,宽 6 米)",
+                        "D. 安溪县的总占地面积"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】1公顷 = 10000平方米,分别根据长方形的面积 = 长×宽、正方形的面积 = 边长×边长,计算A、B、C选项的面积,并与10000平方米比较;D选项根据生活经验判断即可。据此解答。\n【详解】A. 10×10 = 100(平方米),与10000平方米相差很大,不符合题意;B. 202×50 = 10100(平方米),与10000平方米只差100平方米,很接近,符合题意;C. 8×6 = 48(平方米),50个教室总面积是50×48 = 2400(平方米),与10000平方米相差较大,不符合题意;D. 安溪县的总占地面积远大于1公顷,不符合题意。故答案为:B。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法",
+                                 "长方形的面积",
+                                 "正方形的面积",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·福建泉州·期中"
+    },
+    {
+        "id":  "203",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一块长方形绿地面积是400平方米,宽8米,如果长不变,宽增加到24米,扩大后的绿地面积是( )平方米。",
+        "options":  [
+                        "A. 800",
+                        "B. 1200",
+                        "C. 9600",
+                        "D. 960"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】一块长方形绿地面积是400平方米,宽8米,扩大后长不变,宽增加到24米,24÷8 = 3,则宽扩大到原来的3倍。长方形面积 = 长×宽,根据积的变化规律,长不变、宽乘3,则面积也乘3,用400×3即可求出扩大后的绿地面积。\n【详解】400×3 = 1200(平方米)。面积是1200平方米。故答案为:B。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)",
+                                 "长方形的面积"
+                             ],
+        "source":  "25-26 四年级上·广东潮州·期中"
+    },
+    {
+        "id":  "204",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "在一个乘法算式中。两个因数(0 除外)都扩大到原数的 5 倍,此时计算的积与原来的积相比较,( )。",
+        "options":  [
+                        "A. 扩大到原来的5倍",
+                        "B. 扩大到原来的25倍",
+                        "C. 没有变化",
+                        "D. 扩大到原来的10倍"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】在乘法计算中,一个因数乘一个数,另一个因数也乘一个数,那么积就乘这两个数的积。\n【详解】例如:25×4 = 100,(25×5)×(4×5) = 125×20 = 2500,2500÷100 = 25。两个因数(0除外)都扩大到原数的5倍,此时计算的积扩大到原来的5×5 = 25倍。故答案为:B。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26四年级上·河北唐山·期中"
+    },
+    {
+        "id":  "205",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一套数学课外书有 12本,每本售价 15 元,某书店已经售出 216 套。用算式\" 12×216 \"解决的问题是( )。",
+        "options":  [
+                        "A. 216套课外书多少钱。",
+                        "B. 一共售出多少本课外书。",
+                        "C. 还剩多少本课外书还没售出。",
+                        "D. 每套课外书多少钱。"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据题意,已知一套数学课外书有12本,已经售出216套,因此算式12×216解决的问题是一共售出多少本课外书。以此选择即可。\n【详解】一套数学课外书有12本,某书店已经售出216套。用算式12×216解决的问题是一共售出多少本课外书。故答案为:B。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "25-26 四年级上·广东汕尾·期中"
+    },
+    {
+        "id":  "206",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面( )能说明\"三位数乘两位数的积一定是五位数\"是错误的。",
+        "options":  [
+                        "A. 24×280",
+                        "B. 99×999",
+                        "C. 540×87",
+                        "D. 70×700"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】计算出各选项的积,找出积不是五位数的即可。\n【详解】A. 24×280 = 6720,积是四位数;B. 99×999 = 98901,积是五位数;C. 540×87 = 46980,积是五位数;D. 70×700 = 49000,积是五位数。因此算式24×280能说明\u0027三位数乘两位数的积一定是五位数\u0027是错误的。故答案为:A。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法",
+                                 "三位数乘两位数,乘数末尾有0"
+                             ],
+        "source":  "25-26 四年级上·广东汕尾·期中"
+    },
+    {
+        "id":  "207",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列问题中,( )能用算式\"80-20×3\"解决。",
+        "options":  [
+                        "A. 篮球每个80元,实心球每个20元,各买3个,篮球 总价比实心球贵多少元。",
+                        "B. 衬衫原价 80 元,现降价 20 元,买了 3 件,一共花了多少元。",
+                        "C. 甲车速度为 80 千米/时,乙车每小时比甲车慢 20 千米,乙车 3 小时行多少千米。",
+                        "D. 扫地机器人要打扫 80 平方米的房间,每分钟打扫 20平方米,3 分钟还剩多少没有打扫。"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】逐项判断各情境对应的算式即可。\n【详解】A. 篮球每个80元,实心球每个20元,各买3个,篮球总价比实心球贵多少元,列式应为(80-20)×3。B. 衬衫原价80元,现降价20元,买了3件,一共花了多少元,列式应为(80-20)×3。C. 甲车速度为80千米/时,乙车每小时比甲车慢20千米,乙车3小时行多少千米,列式应为(80-20)×3。D. 扫地机器人要打扫80平方米的房间,每分钟打扫20平方米,3分钟还剩多少没有打扫,列式为80-20×3。故答案为:D。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "乘加,乘减混合运算",
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25 四年级上·浙江宁波·期中"
+    },
+    {
+        "id":  "208",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "2025年是意义非凡的一年,我国先后完成两项重大事件,彰显国家实力与民族自信。材料一:4 月 24 日 17 时 17 分,神舟二十号载人飞船在酒泉卫星发射中心成功发射,约10分钟后飞船与火箭分离,进入距地球约400千米的预定轨道。航天员团队在太空开展了为期18天的科学实验,完成了空间站设备维护,空间材料合成等多项任务。飞船返回舱着陆时,着陆场工作人员通过角度测量仪监测返回舱的着陆姿态,其中一个关键支撑腿与地面形成的角约 120 circ ,确保返回舱平稳落地。材料二:9月3日,纪念中国人民抗日战争暨世界反法西斯战争胜利80周年阅兵式在北京举行。此次阅兵设置徒步方队,装备方队和空中梯队。其中,徒步方队共12个,每个方队平均有352名官兵,他们迈着整齐的正步走过天安门广场,每步距离严格控制在75厘米;装备方队展示的新型坦克战车,车身长约8米,宽约3米,车顶的雷达天线可灵活旋转,能精准监测方圆20千米内的目标;阅兵训练场是一块长方形场地,占地面积约24公顷,为官兵训练提供了充足空间。根据材料信息,下列说法错误的是( )。",
+        "options":  [
+                        "A. 神舟二十号航天员在太空的时间约为 400 小时。",
+                        "B. 徒步方队总人数约为 3500 人。",
+                        "C. 装备方队坦克战车的占地面积约 24 平方米。",
+                        "D. 阅兵训练场的面积合 240000 平方米。"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】认真阅读题中资料,航天员团队在太空开展了为期18天的科学实验,一天有24小时,用18乘24计算总时间;徒步方队共12个,每个方队平均有352名官兵,用12乘352求总人数;坦克战车车身长约8米,宽约3米,占地面积就是长方形面积8×3;训练场占地面积约24公顷,1公顷 = 10000平方米,据此换算单位即可。\n【详解】18×24 = 432(小时),12×352 = 4224(人),8×3 = 24(平方米),24公顷 = 240000平方米。A. 432小时约是400小时,说法正确;B. 4224人说成约3500人,错误;C、D正确。故错误的是B。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法",
+                                 "长方形的面积",
+                                 "公顷,平方千米的进率与换算"
+                             ],
+        "source":  "25-26四年级上·内蒙古鄂尔多斯·期中"
+    },
+    {
+        "id":  "209",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一款运动服原价 368元,在国庆期间降价促销,每件249元,一支足球队要在国庆期间购买24套这种运动服,可以省( )元。",
+        "options":  [
+                        "A. 2086",
+                        "B. 2180",
+                        "C. 2056",
+                        "D. 2856"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】先用368-249求出一套运动服节省的钱数,再乘24即可求出买24套这种运动服可以节省多少钱。\n【详解】(368-249)×24 = 119×24 = 2856(元)。可以省2856元。故答案为:D。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "25-26 四年级上·陕西榆林·期中"
+    },
+    {
+        "id":  "210",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "明明用计算器计算\u0027678×56\u0027时,发现按键\u00275\u0027坏了,但他还是用这个计算器算出了正确的答案,他用的方法可能是( )。",
+        "options":  [
+                        "A. 678×60-4",
+                        "B. 678×7×8",
+                        "C. 678×40+16"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】按键\u00275\u0027损坏,需将56拆分为不需要使用\u00275\u0027的等值算式。原式678×56可以利用56 = 7×8来改写。\n【详解】678×56 = 37968。A. 678×60-4 = 40680-4 = 40676,不相等;B. 678×7×8 = 4746×8 = 37968,相等;C. 678×40+16 = 27120+16 = 27136,不相等。故答案为:B。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法",
+                                 "计算器的初步认识与使用"
+                             ],
+        "source":  "25-26四年级上·广东佛山·月考"
+    },
+    {
+        "id":  "211",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "林灵在用计算器计算 214×56 时,错把十位上的5按成了3,所得结果比正确结果小( )。",
+        "options":  [
+                        "A. 4280",
+                        "B. 7704",
+                        "C. 11984"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】计算214×56时,错把十位上的\u00275\u0027按成了\u00273\u0027,现在变成了214×36。结果比正确结果少了(56-36)个214,据此解答。\n【详解】(56-36)×214 = 20×214 = 4280。所得结果比正确结果小4280。故答案为:A。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "计算器的初步认识与使用"
+                             ],
+        "source":  "25-26 四年级上·湖北襄阳·期中"
+    },
+    {
+        "id":  "212",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "三位数乘两位数,积是( )位数。",
+        "options":  [
+                        "A. 四",
+                        "B. 五",
+                        "C. 四或五"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】判断三位数乘两位数积的位数,找出两组典型:最大的三位数×最大的两位数,以及最小的三位数×最小的两位数,然后进行选择即可。\n【详解】999×99 = 98901,积是五位数;100×10 = 1000,积是四位数。所以三位数乘两位数,积可能是四位数,也可能是五位数。故答案为:C。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "25-26四年级上·广东东莞·期中"
+    },
+    {
+        "id":  "213",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一个正方形果园的边长是 600 米,如果平均每公顷种树 250 棵,那么这个果园一共能种树( )棵。",
+        "options":  [
+                        "A. 90000",
+                        "B. 9000",
+                        "C. 1500",
+                        "D. 900"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】先用边长乘边长求出正方形果园的面积,根据10000平方米 = 1公顷把面积单位换算成公顷,再乘250即可解答。\n【详解】600×600 = 360000(平方米),360000平方米 = 36公顷,250×36 = 9000(棵)。这个果园一共能种树9000棵。故答案为:B。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有 0",
+                                 "正方形的面积",
+                                 "公顷,平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·广东东莞·期中"
+    },
+    {
+        "id":  "214",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "估算308与29的乘积大约是( )。",
+        "options":  [
+                        "A. 9000",
+                        "B. 8000",
+                        "C. 7000"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】三位数乘两位数的估算,可以分别把它们看作和它接近的整百数、整十数,然后再相乘。\n【详解】308×29≈300×30 = 9000。故答案为:A。",
+        "knowledge_points":  [
+                                 "三位数乘两位数的估算"
+                             ],
+        "source":  "25-26四年级上·广东潮州·期中"
+    },
+    {
+        "id":  "215",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "一条公路占地 11 万平方米,原来宽7米,现在要拓宽至21米,长不变。拓宽后这条公路的占地面积是( )万平方米。",
+        "options":  [
+                        "A. 70",
+                        "B. 34",
+                        "C. 33",
+                        "D. 28"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】根据长方形的面积 = 长×宽,再根据因数与积的变化规律,一个因数不变,另一个因数扩大到原来的几倍,积就扩大到原来的几倍。长不变,先求出拓宽后的宽是原来宽的几倍,进而求出拓宽后的面积。\n【详解】11×(21÷7) = 11×3 = 33(万平方米)。拓宽后这条公路的占地面积是33万平方米。故答案为:C。",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·广东东莞·期中"
+    },
+    {
+        "id":  "216",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "两个因数的积是 360,如果一个因数乘 5,另一个因数除以 5,积是( );如果一个因数不变,另一个因数乘 3,积是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "360;1080",
+        "solution":  "【分析】根据题意,第一个问题中,一个因数乘5,另一个因数除以5,相当于积乘5再除以5,积不变;第二个问题中,一个因数不变,另一个因数乘3,积也乘3。以此答题即可。\n【详解】360×3 = 1080。两个因数的积是360,如果一个因数乘5,另一个因数除以5,积是360;如果一个因数不变,另一个因数乘3,积是1080。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·广东潮州·期中"
+    },
+    {
+        "id":  "217",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个因数除以 3,另一个因数除以 2,积是 120,原来的积是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "720",
+        "solution":  "【分析】根据积的变化规律:一个因数除以3,另一个因数除以2,那么积就要先除以3,再除以2。现在积是120,要求原来的积,先用120乘2,再乘3即可。\n【详解】120×2 = 240,240×3 = 720。即一个因数除以3,另一个因数除以2,积是120,原来的积是720。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·湖北武汉·期中"
+    },
+    {
+        "id":  "218",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据37×58 = 2146,不计算就可以知道370×58 = ( )。",
+        "options":  [
+
+                    ],
+        "answer":  "21460",
+        "solution":  "【分析】在乘法算式里,两个因数都不为0时,一个因数不变,另一个因数乘几,积就乘几;依此计算并填空。\n【详解】37×10 = 370,2146×10 = 21460,即根据37×58 = 2146,不计算就可以知道370×58 = 21460。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·海南省直辖县级单位·期中"
+    },
+    {
+        "id":  "219",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "解答。(1)最初蜗牛在井的底部,从第一天白天开始,蜗牛每天白天向上爬 7 米,晚上下滑 2 米,蜗牛在第 10 天爬出井口,井最深有( )米。 (2)最初蜗牛在井的底部,井深30米,从第一天白天开始,蜗牛每天白天向上爬6米,每天晚上下滑固定的一段距离(整数米),最后蜗牛在第 6 天爬出井口。每天晚上蜗牛下滑( )米。 (3)最初蜗牛在井的底部,井深50米,从第一天白天开始,蜗牛每天白天向上爬固定的一段距离(整数米),晚上下滑2米,最后蜗牛在第10天爬出井口。每天白天蜗牛向上爬( )米。",
+        "options":  [
+
+                    ],
+        "answer":  "52;1;7",
+        "solution":  "【分析】最后一天爬出井口后,不再下滑,因此需先算最后一天的爬行距离,再算前几天每天实际爬行的距离(白天爬的距离-晚上下滑的距离),总深度 = 前几天实际爬行的总距离+最后一天爬行的距离。\n【详解】(1)7-2 = 5(米),前9天共爬行9×5 = 45(米),第10天白天再爬7米后出井,因此井的深度 = 45+7 = 52(米)。(2)设蜗牛每天晚上下滑x米,则前5天每天实际爬行(6-x)米,第6天白天爬6米后出井,所以5×(6-x)+6 = 30,解得x = 1。(3)设蜗牛每天白天向上爬y米,则前9天每天实际爬行(y-2)米,第10天白天爬y米后出井,所以9×(y-2)+y = 50。因为y取整数,代入验证可知y = 7时,9×(7-2)+7 = 52,首次达到并超过50,所以每天白天向上爬7米。",
+        "knowledge_points":  [
+                                 "基础行程问题"
+                             ],
+        "source":  "2024三年级·全国·竞赛"
+    },
+    {
+        "id":  "220",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "笑笑通过实验得出:一个水龙头如果没有拧紧,每分大约要浪费30克水。如果一个住宅小区有800个这样的水龙头,都没有拧紧,一分大约要浪费( )千克的水。",
+        "options":  [
+
+                    ],
+        "answer":  "24",
+        "solution":  "【分析】用一个水龙头没有拧紧时每分钟大约要浪费的水的质量乘住宅小区内的水龙头数,即可求得所有水龙头1分大约要浪费的水的质量,再根据1千克 = 1000克,将单位转换为千克即可。\n【详解】30×800 = 24000(克),24000克 = 24千克。所以一分大约要浪费24千克的水。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "克,千克之间的换算与比较"
+                             ],
+        "source":  "25-26四年级上·广东·期末"
+    },
+    {
+        "id":  "221",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个因数扩大到它的 3 倍,另一个因数不变,积扩大到原来的( )倍;一个因数扩大到它的3倍,另一个因数也扩大3倍,积扩大到原来的( )倍。",
+        "options":  [
+
+                    ],
+        "answer":  "3;9",
+        "solution":  "【分析】积的变化规律:两个数相乘,一个因数不变,另一个因数乘几或除以几(0除外),积也乘几或除以相同的数;两个数相乘,一个因数扩大到它的3倍,另一个因数也扩大到原来的 3 倍,积扩大到原来的( ( 3×3 ) )倍.\n【详解】根据积的变化规律可知:一个因数扩大到它的3倍,另一个因数不变,积扩大到原来的3倍;一个因数扩大到它的 3 倍,另一个因数也扩大 3 倍,积扩大到原来的 9 倍。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26四年级上·湖南张家界·期中"
+    },
+    {
+        "id":  "222",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "两个因数的积是250,其中一个因数乘 4,另一个因数也乘 4,积是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "4000",
+        "solution":  "【分析】根据题意,当两个因数同时乘或除以一个数(0除外)时,积就要乘或除以两次这个数。以此答题即可。\n【详解】250×4×4 = 1000×4 = 4000。两个因数的积是250,其中一个因数乘4,另一个因数也乘4,积是4000。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·广西河池·期中"
+    },
+    {
+        "id":  "223",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "聪聪在做一道三位数乘两位数的题目时,误把其中一个因数162写成了167,得到的积比正确的积多了 220,正确的积是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "7128",
+        "solution":  "【分析】根据题意,把其中一个因数162看成了167,也就是多了167-162 = 5个另一个因数。用220除以5,得出另一个因数,然后再乘上162即可。\n【详解】220÷(167-162) = 220÷5 = 44,162×44 = 7128,即正确的积是7128。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "25-26四年级上·广东汕尾·期中"
+    },
+    {
+        "id":  "224",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "要使□8×241的积是四位数,□里最大填( );23□4670≈235 万,□里应该填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "3;5",
+        "solution":  "【分析】最小的五位数是10000,可以从1开始尝试代入,计算积的位数,即可求出□里最大的数;省略\u0027万\u0027后面的尾数求近似数,要根据千位上的数字进行四舍五入,据此解答。\n【详解】当□ = 1时,18×241 = 4338(四位数);当□ = 2时,28×241 = 6748(四位数);当□ = 3时,38×241 = 9158(四位数);当□ = 4时,48×241 = 11568(五位数)。因此,要使积是四位数,□里最大填3。23□4670≈235万时,要看千位上的4,小于5,不需要向万位进位,所以□应填5。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法",
+                                 "亿以内数的改写",
+                                 "亿以内数的近似数"
+                             ],
+        "source":  "25-26 四年级上·广东汕尾·期中"
+    },
+    {
+        "id":  "225",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\"\u003e\"\"\u003c\"或\"=\"。450 万( )450000;30 公顷( )3 平方千米;3411205( )541205;78×130( )13×780。",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;\u003c;\u003e;=",
+        "solution":  "【分析】整数大小比较时,位数多的数大于位数少的数,位数相同的数,要从高位开始依次比较;不同计量单位的数先统一单位再比较;两个数相乘,一个因数乘10,另一个因数除以10,积不变,据此解答。\n【详解】450万 = 4500000,4500000＞450000,故450万＞450000;1平方千米 = 100公顷,3平方千米 = 300公顷,30公顷＜300公顷,故30公顷＜3平方千米;3411205是七位数,541205是六位数,故3411205＞541205;78×130 = 13×780。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "公顷,平方千米的进率与换算",
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·广东汕尾·期中"
+    },
+    {
+        "id":  "226",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "67□809≈68 万,□里最小填( );200×□2的积是五位数,□里最小填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "5;5",
+        "solution":  "【分析】省略\u0027万\u0027位后面的尾数就是四舍五入到万位,把万位后的千位上的数进行四舍五入,再在数的后面写上\u0027万\u0027字。67□809≈68万时,千位上的数可以是5,6,7,8,9;因为200×42 = 8400,200×52 = 10400,据此判断第二空。\n【详解】要使67□809≈68万,□里最小填5;200×42 = 8400,200×52 = 10400,所以要使200×□2的积是五位数,□里最小填5。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "三位数乘两位数的估算",
+                                 "亿以内数的近似数"
+                             ],
+        "source":  "24-25 四年级上·湖南益阳·期中"
+    },
+    {
+        "id":  "227",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "最小的自然数是(),自然数的个数是( )的;最小的三位数与最大的两位数相乘所得的积是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "0;无限;9900",
+        "solution":  "【分析】表示物体个数的0,1,2,3,4,5,6,7,8……都是自然数。最小的自然数是0,没有最大的自然数,自然数的个数是无限的。最小的三位数是100,最大的两位数是99,计算它们的乘积即可。\n【详解】100×99 = 9900。最小的自然数是0,自然数的个数是无限的;最小的三位数与最大的两位数相乘所得的积是9900。",
+        "knowledge_points":  [
+                                 "自然数的认识",
+                                 "三位数乘两位数,乘数末尾有0"
+                             ],
+        "source":  "24-25 四年级上·湖南益阳·期中"
+    },
+    {
+        "id":  "228",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "小丽沿着一个正方形菜园的四周走了一圈,刚好是 600 步,她的步长是 15 分米,这个菜园的边长是( )米,占地面积是( )平方米。",
+        "options":  [
+
+                    ],
+        "answer":  "225;50625",
+        "solution":  "【分析】本题考查正方形的周长和面积的计算。先根据小丽走的步数和步长,求出正方形菜园的周长,再根据周长求出边长,最后根据边长求出面积。\n【详解】600×15 = 9000(分米),9000分米 = 900米。正方形菜园的边长为900÷4 = 225(米),面积为225×225 = 50625(平方米)。所以这个菜园的边长是225米,占地面积是50625平方米。",
+        "knowledge_points":  [
+                                 "正方形的周长",
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "正方形的面积",
+                                 "毫米,厘米,分米,米之间的进率与换算"
+                             ],
+        "source":  "25-26四年级上·湖南永州·月考"
+    },
+    {
+        "id":  "229",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "学校开展国学经典读书活动,需要购买 218 本《诗经》,每本《诗经》的售价是 42 元,一共要花( )元。",
+        "options":  [
+
+                    ],
+        "answer":  "9156",
+        "solution":  "【分析】已知每本《诗经》的售价是42元,需要购买218本,根据单价×数量 = 总价,即可计算出一共要花多少钱。据此解答。\n【详解】218×42 = 9156(元)。所以一共要花9156元。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法",
+                                 "经济问题"
+                             ],
+        "source":  "25-26 四年级上·陕西榆林·期中"
+    },
+    {
+        "id":  "230",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号填上\"\u003e\"\"\u003c\"或\"=\"。3000 平方米( )3 公顷;6200620( )6200260;9 万( )90000;4000 公顷( )4 平方千米;200×13( )130×20;207×54( )270×45。",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003e;=;\u003e;=;\u003c",
+        "solution":  "【分析】根据单位换算和整数大小比较方法,先统一单位或先计算出结果,再比较大小。\n【详解】1公顷 = 10000平方米,所以3公顷 = 30000平方米,3000平方米＜30000平方米,故3000平方米＜3公顷。6200620和6200260都是七位数,前四位相同,百位上6＞2,故6200620＞6200260。9万 = 90000。1平方千米 = 100公顷,4平方千米 = 400公顷,4000公顷＞400公顷,故4000公顷＞4平方千米。200×13 = 2600,130×20 = 2600,故200×13 = 130×20。207×54 = 11178,270×45 = 12150,故207×54＜270×45。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "三位数乘两位数,乘数末尾有 0",
+                                 "三位数乘两位数,三位数中间有 0",
+                                 "公顷,平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·陕西榆林·期中"
+    },
+    {
+        "id":  "231",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上\"\u003e\"\"\u003c\"或\"=\"。603000( )63 万;120×30( )12×300;4500680( )450680。",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;=;\u003e",
+        "solution":  "【分析】①根据整数比较大小的方法:位数不同时,位数多的数大,位数相同时,从高位比起。用\u0027万\u0027作单位的数先化成用\u0027个\u0027作单位的数,再比较;②如果一个因数扩大到原来的若干倍,另一个因数缩小到原来的若干分之一(0除外),那么积不变;③直接比较大小即可。\n【详解】63万 = 630000,603000＜630000,所以603000＜63万;12×300 = (120÷10)×(30×10) = 120×30;4500680＞450680。",
+        "knowledge_points":  [
+                                 "亿以内数的改写",
+                                 "积的变化规律(整数乘法)",
+                                 "大数的比较"
+                             ],
+        "source":  "25-26 四年级上·湖北襄阳·期中"
+    },
+    {
+        "id":  "232",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\"\u003e\"\"\u003c\"或\"=\"。5629060( )5631060;8752080000( )875208000;6000 公顷( )6 平方千米;14 亿( )85000 万;24×500( )50×240;8 公顷 800 平方米( )8800 平方米。",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003e;\u003e;\u003e;=;\u003e",
+        "solution":  "【分析】(1)(2)整数比较大小时,先看位数,位数多的那个数就大;如果位数相同,就从高位依次比较。(3)(6)1平方千米 = 100公顷,1公顷 = 10000平方米,先化成相同的单位再比较。(4)分别将85000万、14亿改写成整数再比较。(5)计算出两边结果即可。\n【详解】5629060＜5631060;8752080000＞875208000;1平方千米 = 100公顷,6平方千米 = 600公顷,6000公顷＞600公顷,所以6000公顷＞6平方千米;14亿 = 1400000000,85000万 = 850000000,所以14亿＞85000万;24×500 = 12000,50×240 = 12000,所以24×500 = 50×240;8公顷 = 80000平方米,8公顷800平方米 = 80800平方米,80800平方米＞8800平方米,所以8公顷800平方米＞8800平方米。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "公顷,平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·福建福州·期中"
+    },
+    {
+        "id":  "233",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "林灵以每分钟220米的速度在洱海生态廊道上骑行,她骑车的速度可以记作( ),20分钟她骑行了( )米。",
+        "options":  [
+
+                    ],
+        "answer":  "220 米/分;4400",
+        "solution":  "【分析】常用的速度单位有千米/时、米/分、米/秒。已知速度是每分钟行220米,所以速度是220米/分;求路程,用速度乘时间即可,据此解答。\n【详解】220×20 = 4400(米)。林灵以每分钟220米的速度在洱海生态廊道上骑行,她骑车的速度可以记作220米/分,20分钟她骑行了4400米。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有 0",
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26 四年级上·湖北襄阳·期中"
+    },
+    {
+        "id":  "234",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\"\u003e\"\"\u003c\"或\"=\"。340608000( )34700800;5999 公顷( )6 平方千米;51×301( )15000。",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;\u003e;\u003e",
+        "solution":  "【分析】整数比较大小时,先看位数,位数多的那个数就大;如果位数相同,就从高位依次比较。1平方千米 = 100公顷,据此将6平方千米换算成公顷,再和5999公顷比较大小。计算出51×301的结果后再比较。\n【详解】340608000是九位数,34700800是八位数,所以340608000＞34700800;6平方千米 = 600公顷,5999公顷＞600公顷,所以5999公顷＞6平方千米;51×301 = 15351,15351＞15000,所以51×301＞15000。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "三位数乘两位数,三位数中间有0",
+                                 "公顷,平方千米的进率与换算"
+                             ],
+        "source":  "25-26四年级上·湖北襄阳·期中"
+    },
+    {
+        "id":  "235",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "飞机每小时行驶 380 千米,飞机的速度可以写作( ),照这样的速度,它 6 个小时可以飞行( )千米。",
+        "options":  [
+
+                    ],
+        "answer":  "380 千米/时;2280",
+        "solution":  "【分析】每小时行驶380千米可以写作380千米/时;路程 = 速度×时间,用380乘6小时,可算出它6个小时可以飞行多少千米。\n【详解】380×6 = 2280(千米)。所以飞机每小时行驶380千米,飞机的速度可以写作380千米/时;照这样的速度,它6个小时可以飞行2280千米。",
+        "knowledge_points":  [
+                                 "因数中间或末尾是 0 的乘法",
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26四年级上·海南省直辖县级单位·期中"
+    },
+    {
+        "id":  "236",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "学校要购进图书399本,每本图书5元,购置这些图书大约需要( )元。",
+        "options":  [
+
+                    ],
+        "answer":  "2000",
+        "solution":  "【分析】可以先将图书数量399本估算成400本,然后根据总价 = 单价×数量,用每本图书的单价5元乘400本,即可求得购置这些图书大约需要多少钱。\n【详解】399本可以估算成400本,400×5 = 2000(元)。因此,学校要购进图书399本,每本图书5元,购置这些图书大约需要2000元。",
+        "knowledge_points":  [
+                                 "经济问题"
+                             ],
+        "source":  "25-26 四年级上·海南省直辖县级单位·期中"
+    },
+    {
+        "id":  "237",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\"\u003e\"\"\u003c\"或\"=\"。204 万( )20450000;105×50( )210×30;150×50( )500×15;47 公顷( )3 平方千米。",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003c;=;\u003c",
+        "solution":  "【分析】本题考查大数比较、乘法计算以及面积单位换算。先将\u0027万\u0027转换成具体数字,再比较;计算乘法结果后再比较;第四组先统一单位再比较。\n【详解】204万 = 2040000,2040000＜20450000;105×50 = 5250,210×30 = 6300,所以105×50＜210×30;150×50 = 7500,500×15 = 7500,所以150×50 = 500×15;1平方千米 = 100公顷,3平方千米 = 300公顷,47公顷＜300公顷,所以47公顷＜3平方千米。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "三位数与两位数的乘法",
+                                 "公顷,平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·广东东莞·期中"
+    },
+    {
+        "id":  "238",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "计算 340×20 时,可以先算( )×( )=( ),再在积的末尾添上( )个 0 。",
+        "options":  [
+
+                    ],
+        "answer":  "34;2;68;2",
+        "solution":  "【分析】计算乘数末尾有0的乘法时,可以先忽略乘数末尾的0,计算0前面的数相乘,再看乘数末尾一共有几个0,就在积的末尾添上几个0。340末尾有1个0,20末尾有1个0,共有2个0。\n【详解】计算340×20时,先算34×2 = 68,再在积的末尾添上2个0。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0"
+                             ],
+        "source":  "25-26 四年级上·广东东莞·期中"
+    },
+    {
+        "id":  "239",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号内填入\"\u003e\"\"\u003c\"或\"=\"。64500008( )6 亿;200 公顷( )2 平方千米;702356( )89025;130×50( )13×500;45 公顷( )45000 平方米。",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;=;\u003e;=;\u003e",
+        "solution":  "【分析】比较大数时先看位数,位数相同再从高位依次比较;面积单位先统一后再比较;利用积不变规律比较两个乘法算式。\n【详解】6亿 = 600000000,64500008＜600000000,所以64500008＜6亿;2平方千米 = 200公顷,所以200公顷 = 2平方千米;702356＞89025;130×50 = 13×500;45公顷 = 450000平方米,450000＞45000,所以45公顷＞45000平方米。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "公顷,平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·河北保定·期中"
+    },
+    {
+        "id":  "240",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\"\u003e\"\"\u003c\"或\"=\"。直角( )钝角;40×600( )600×400;周角( )平角;198×70( )42×600。",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003c;\u003e;\u003c",
+        "solution":  "【分析】直角是90°的角,钝角是大于90°小于180°的角,平角是180°的角,周角是360°的角,据此比较大小。计算两个乘法算式的结果后再比较。\n【详解】直角＜钝角;40×600 = 24000,600×400 = 240000,所以40×600＜600×400;周角＞平角;198×70 = 13860,42×600 = 25200,所以198×70＜42×600。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "直角,钝角,锐角的认识及特征",
+                                 "平角,周角的认识及特征",
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·广东东莞·期中"
+    },
+    {
+        "id":  "241",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\"\u003e\"\"\u003c\"或\"=\"。3646789( )3654897;25×400( )40×250;29700 公顷( )30 平方千米;2790010( )27 万。",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;=;\u003e;\u003e",
+        "solution":  "【分析】比较大数时从高位开始逐位比较;计算两个乘法算式的积后再比较;将面积单位统一为公顷后再比较;将\u0027万\u0027改写成具体数字后再比较。\n【详解】3646789和3654897都是七位数,前两位相同,万位上4＜5,因此3646789＜3654897;25×400 = 10000,40×250 = 10000,因此25×400 = 40×250;1平方千米 = 100公顷,30平方千米 = 3000公顷,29700公顷＞3000公顷,因此29700公顷＞30平方千米;27万 = 270000,2790010＞270000,因此2790010＞27万。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "公顷,平方千米的进率与换算",
+                                 "亿以内数的改写"
+                             ],
+        "source":  "25-26四年级上·山西晋中·期中"
+    },
+    {
+        "id":  "242",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "□58×26,若积是五位数,则□里最小可以填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "4",
+        "solution":  "【分析】根据题意,分别计算出158×26、258×26、358×26、458×26,再根据积的位数情况进行解答。\n【详解】158×26 = 4108,积是四位数;258×26 = 6708,积是四位数;358×26 = 9308,积是四位数;458×26 = 11908,积是五位数。所以,□58×26若积是五位数,则□里最小可以填4。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "25-26 四年级上·福建莆田·期中"
+    },
+    {
+        "id":  "243",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "猎豹是陆地上跑得最快的动物,每分钟可以跑990米,它的速度可以记作( ),照这样的速度,猎豹奔跑24分钟,跑过的路程是( )米。",
+        "options":  [
+
+                    ],
+        "answer":  "990 米/分;23760",
+        "solution":  "【分析】速度的简记法:先写上米(或千米等)数,再写\u0027/分\u0027或\u0027/时\u0027等即可;根据路程 = 速度×时间,用猎豹每分钟跑的距离乘跑的时间,即可求出跑过的路程是多少米。\n【详解】990×24 = 23760(米)。猎豹每分钟可以跑990米,它的速度记作990米/分,照这样的速度,猎豹奔跑24分钟,跑过的路程是23760米。",
+        "knowledge_points":  [
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26四年级上·福建莆田·期中"
+    },
+    {
+        "id":  "244",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上 \"\u003e\"\"\u003c\"或\"=\"。(原卷此处比较式略,解析见【详解】)",
+        "options":  [
+
+                    ],
+        "answer":  "=;\u003c;=",
+        "solution":  "【分析】根据100公顷 = 1平方千米,把6平方千米换算成公顷再比较。2亿改写成数后再与200000100比较。根据积不变的规律:两个数相乘,一个因数乘或除以一个数(0除外),另一个因数同时除以或乘相同的数,积不变。\n【详解】6×100 = 600(公顷),所以600公顷 = 6平方千米。2亿 = 200000000,200000000＜200000100,所以2亿＜200000100。48×500 = (48×10)×(500÷10) = 480×50,所以48×500 = 480×50。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "公顷,平方千米的进率与换算",
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26四年级上·广东潮州·期中"
+    },
+    {
+        "id":  "245",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上合适的单位。 (1)一支铅笔长 20( )。 (2)一张邮票的封面面积大约是 6( )。 (3)一个课室面积约是 50( ),200个这样的课室面积约是1( )。",
+        "options":  [
+
+                    ],
+        "answer":  "厘米;平方厘米;平方米;公顷",
+        "solution":  "【分析】常用的长度单位有毫米、厘米、分米、米和千米。常用的面积单位有平方厘米、平方分米、平方米、公顷等。(1)一支铅笔的长度一般用厘米作单位;(2)一张邮票的面积较小,一般用平方厘米作单位;(3)一个课室的面积一般用平方米作单位。200个这样的课室面积可先求出总平方米数,再根据10000平方米 = 1公顷换算。\n【详解】(1)一支铅笔长20厘米。(2)一张邮票的封面面积大约是6平方厘米。(3)一个课室面积约是50平方米。50×200 = 10000(平方米),10000平方米 = 1公顷,所以200个这样的课室面积约是1公顷。",
+        "knowledge_points":  [
+                                 "三位数与整十数的口算乘法",
+                                 "长度单位的选择",
+                                 "面积单位的选择",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·广东潮州·期中"
+    },
+    {
+        "id":  "246",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "奇奇在做一道三位数乘两位数的题目时,误把其中一个因数 143 写成了 148,得到的结果比正确答案多了 320,正确答案是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "9152",
+        "solution":  "【分析】根据题意,乘法算式中的一个因数143错看为148,多了148 - 143 = 5,错看后的算式结果多了320,用320÷5就是另一个因数。再用求出的另一个因数乘143,即是正确答案。\n【详解】320÷(148 - 143)×143\n= 320÷5×143\n= 64×143\n= 9152\n所以正确答案是9152。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "25-26四年级上·福建厦门·期中"
+    },
+    {
+        "id":  "247",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在下面的括号里填上\"\u003e\"\"\u003c\"或\"=\"。80 公顷( )8 平方千米;1 周角( )4 直角;640×15( )320×30;48×120( )460×12。",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;=;=;\u003e",
+        "solution":  "【分析】根据题意,先明确单位间的进率,1平方千米 = 100公顷;周角是360°,直角是90°。先换算成统一单位,再比较大小;算式先计算,再比较结果大小。\n【详解】8平方千米 = 800公顷,80公顷＜800公顷,所以80公顷＜8平方千米;90×4 = 360,所以1周角 = 4直角;640×15 = 9600,320×30 = 9600,所以640×15 = 320×30;48×120 = 5760,460×12 = 5520,5760＞5520,所以48×120＞460×12。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "直角,钝角,锐角的认识及特征",
+                                 "平角,周角的认识及特征",
+                                 "公顷,平方千米的进率与换算"
+                             ],
+        "source":  "25-26四年级上·广东东莞·期中"
+    },
+    {
+        "id":  "248",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "398×29 的积接近12000。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】估算三位数乘两位数时,通常将两个因数看作最接近的整十或整百数进行计算。由题意得,估算398×29时,可以把398估成400,把29估成30,然后直接口算出结果即可。\n【详解】398×29≈400×30 = 12000,即398×29的积接近12000。原题说法正确。故答案为:√。",
+        "knowledge_points":  [
+                                 "三位数乘两位数的估算"
+                             ],
+        "source":  "25-26四年级上·湖北襄阳·期中"
+    },
+    {
+        "id":  "249",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "洛阳唐三彩文创店以单价 58 元从厂家购进了 250 个纪念瓷盘,牡丹文化节期间以 78 元单价售出 165 个,剩下的以 48 元单价在年底前全部售罄。该文创店销售这批纪念瓷盘是赚钱了还是亏钱了。",
+        "options":  [
+
+                    ],
+        "answer":  "赚钱了",
+        "solution":  "【分析】根据单价×数量 = 总价,先求出文创店购买250个纪念瓷盘的总价钱,再算78元单价售出165个的价钱和48元单价卖完剩下的价钱,然后用售出的总价钱和购买的价钱比较。如果售出的价钱多,则赚钱;如果售出的价钱少,则亏钱。\n【详解】250×58 = 14500(元)\n78×165 = 12870(元)\n48×(250 - 165) = 48×85 = 4080(元)\n12870 + 4080 = 16950(元)\n16950＞14500\n答:该文创店销售这批纪念瓷盘赚钱了。",
+        "knowledge_points":  [
+                                 "三位数乘两位数",
+                                 "乘数末尾有0",
+                                 "三位数乘两位数的实际问题",
+                                 "乘加,乘减混合运算"
+                             ],
+        "source":  "25-26四年级上·河南洛阳·期中"
+    },
+    {
+        "id":  "250",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "虢国博物馆地处三门峡市,馆内文创商店以单价 65 元购买了 300 个纪念徽章,国庆节前以 80 元单价售出 175 个,剩下的以 60 元单价年底前全部售罄。该文创商店销售这批纪念徽章是赚钱了还是亏钱了。",
+        "options":  [
+
+                    ],
+        "answer":  "赚钱了",
+        "solution":  "【分析】根据单价×数量 = 总价,先求出商店购买300个纪念徽章的价钱,再算80元单价售出175个的价钱和60元单价卖完剩下的价钱,然后用售出的总价钱和购买的价钱比较。如果售出的价钱多,则赚钱;如果售出的价钱少,则亏钱。\n【详解】300×65 = 19500(元)\n80×175 = 14000(元)\n60×(300 - 175) = 60×125 = 7500(元)\n14000 + 7500 = 21500(元)\n21500＞19500\n答:该文创商店销售这批纪念徽章赚钱了。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "三位数乘两位数的实际问题",
+                                 "乘加,乘减混合运算"
+                             ],
+        "source":  "25-26四年级上·河南三门峡·期中"
+    },
+    {
+        "id":  "251",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "陈老师从家步行去学校,6 分钟后手机软件显示已经走了 612 步,步行距离为432 米。照这样的速度,陈老师再步行 8 分钟就能到学校了。(1)要求\"陈老师每分钟步行多少米。\",我会选择的信息是。(填序号)①6分钟 ②已经走了612步 ③步行距离为432米 ④再步行8分钟 (2)陈老师的家到学校的距离多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "①、③;1008 米",
+        "solution":  "【分析】(1)根据题意,要求\"陈老师每分钟步行多少米\",需要用到对应的时间和路程,即步行6分钟(信息①)和步行距离为432米(信息③)。(2)先用432÷6求出陈老师步行的速度,再乘8求出后来8分钟走的路程,最后加上已经走的432米,就是陈老师的家到学校的距离。\n【详解】(1)要求\"陈老师每分钟步行多少米\",我会选择的信息是①、③。(2)432÷6 = 72(米)\n72×8 = 576(米)\n576 + 432 = 1008(米)\n答:陈老师的家到学校的距离是1008米。",
+        "knowledge_points":  [
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26四年级上·福建福州·期末"
+    },
+    {
+        "id":  "252",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "一列火车从 A 城市出发去 B 城市,平均每小时行 95 千米,经过 12 小时后,距离 B城市还有132千米。A,B两城市间铁路长多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "1272 千米",
+        "solution":  "【分析】用火车的速度乘时间,求出已经行驶的路程,再加上132千米,即可求出两地的铁路长度。\n【详解】95×12 + 132 = 1140 + 132 = 1272(千米)\n答:A、B两城市间铁路长1272千米。",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "乘加,乘减混合运算",
+                                 "两位数乘两位数的进位乘法"
+                             ],
+        "source":  "25-26 四年级上·广东·期末"
+    },
+    {
+        "id":  "253",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "小明外出旅游12天,每天预算食宿费270元,准备3000元够吗。如果不够,还要多带多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "不够;240元",
+        "solution":  "【分析】12天,每天预算食宿费270元,用270元乘12可算出需要花多少钱,再与3000元比较可知够不够;如果不够,用乘积减3000即可求出还要多带多少钱。\n【详解】270×12 = 3240(元)\n3240＞3000\n3240 - 3000 = 240(元)\n答:准备3000元不够,还要多带240元。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0"
+                             ],
+        "source":  "25-26四年级上·广东·期末"
+    },
+    {
+        "id":  "254",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "科技的发展给人们带来了无人售货机,只要一部手机进行相关操作,就可轻松支付。一个商场有25台无人售货机,一台无人售货机日均营业额为824元,25台无人售货机日均营业额一共是多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "20600 元",
+        "solution":  "【分析】由题意得,一台无人售货机日均营业额为824元,一个商场有25台无人售货机,直接用824乘25即可算出25台无人售货机日均营业额一共是多少元。\n【详解】824×25 = 20600(元)\n答:25台无人售货机日均营业额一共是20600元。",
+        "knowledge_points":  [
+                                 "三位数乘两位数的实际问题"
+                             ],
+        "source":  "24-25 四年级上·重庆秀山·期中"
+    },
+    {
+        "id":  "255",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "国庆期间,李叔叔一家自驾去旅游,走国道时车速为 75 千米/时,6 小时后到达目的地。从原路返回时,李叔叔选择走高速,用的时间比去时少了1小时。两地相距多少千米。返回时车速是多少。",
+        "options":  [
+
+                    ],
+        "answer":  "450 千米;90 千米/小时",
+        "solution":  "【分析】走国道时车速为75千米/时,6小时后到达目的地,根据路程 = 速度×时间可算出路程。从原路返回,说明路程不变,再根据路程÷时间 = 速度可计算出返回时车速,此时的时间是6 - 1 = 5(小时)。\n【详解】75×6 = 450(千米)\n450÷(6 - 1) = 450÷5 = 90(千米/小时)\n答:两地相距450千米,返回时车速是90千米/小时。",
+        "knowledge_points":  [
+                                 "两,三位数与一位数连续进位的乘法",
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25 四年级上·浙江宁波·期中"
+    },
+    {
+        "id":  "256",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "某体育中心的看台,一排能坐 31 人,一共有 195 排。请你算一算:这个体育中心能不能容纳 6000 人同时观看体育比赛。",
+        "options":  [
+
+                    ],
+        "answer":  "能",
+        "solution":  "【分析】先用每排能坐的人数乘排数,求出能坐的人数,然后与6000人比较即可。\n【详解】31×195 = 6045(人)\n6045＞6000\n答:这个体育中心能容纳6000人同时观看体育比赛。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "25-26 四年级上·广东汕尾·期中"
+    },
+    {
+        "id":  "257",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "垃圾分类是对传统垃圾收集处置方式的改革,是对垃圾进行有效处置的一种科学管理方法。一台智能垃圾回收柜每天的垃圾回收量在37 ~ 50 千克之间,某市投放了 151 台智能垃圾回收柜,它们一天最少回收多少千克垃圾。最多呢。",
+        "options":  [
+
+                    ],
+        "answer":  "最少 5587 千克;最多 7550 千克",
+        "solution":  "【分析】一台智能垃圾回收柜每天回收垃圾最少的质量乘台数,即可算出它们一天最少回收多少千克垃圾;一台智能垃圾回收柜每天回收垃圾最多的质量乘台数,即可算出它们一天最多回收多少千克垃圾。\n【详解】37×151 = 5587(千克)\n50×151 = 7550(千克)\n答:它们一天最少回收5587千克垃圾,最多回收7550千克垃圾。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法",
+                                 "三位数乘两位数,乘数末尾有 0"
+                             ],
+        "source":  "25-26四年级上·甘肃庆阳·期中"
+    },
+    {
+        "id":  "258",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "我国是最早发现并利用茶的国家,六安瓜片更是有悠久的历史,据《六安州志》记载:\"茶之极品,明朝始入贡\"。这说明六安瓜片在明朝就已经很出名了。现有一块长为2000米,宽为600米的长方形六安瓜片茶园。(1)这块茶园的面积是多少公顷。(2)若每公顷产茶叶 1600 千克,这块茶园可产多少吨六安瓜片茶叶。",
+        "options":  [
+
+                    ],
+        "answer":  "120 公顷;192 吨",
+        "solution":  "【分析】(1)根据长方形面积 = 长×宽计算,结果再根据1公顷 = 10000平方米换算成公顷作单位。\n(2)每公顷产茶叶的数量×茶园面积 = 这块茶园可产茶叶的总质量,再根据1吨 = 1000千克,把结果换算成吨作单位。\n【详解】(1)2000×600 = 1200000(平方米)\n1200000平方米 = 120公顷\n答:这块茶园的面积是120公顷。\n(2)120×1600 = 192000(千克)\n192000千克 = 192吨\n答:这块茶园可产192吨六安瓜片茶叶。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "长方形的面积",
+                                 "吨,千克之间的换算与比较",
+                                 "公顷,平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·广西柳州·期中"
+    },
+    {
+        "id":  "259",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "临近春节,某市为筹备年货节,计划在一块面积为1公顷的闲置空地上设置摊位。已知需要规划设置250个摊位,且每个摊位的标准占地面积为30平方米。那么这块空地的总面积是否足以容纳规划的所有摊位。",
+        "options":  [
+
+                    ],
+        "answer":  "足以",
+        "solution":  "【分析】根据1公顷 = 10000平方米,将公顷转换为平方米,然后用每个摊位的标准占地面积乘规划设置摊位的个数,计算出250个摊位所需要的总面积,最后再与空地的总面积作比较即可解答。\n【详解】1公顷 = 10000平方米\n250×30 = 7500(平方米)\n7500平方米＜10000平方米\n答:这块空地的总面积足以容纳规划的所有摊位。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26四年级上·河北石家庄·期中"
+    },
+    {
+        "id":  "260",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "聪聪和明明从网上查阅了有关植树造林美化环境与净化空气的相关资料。聪聪说:1公顷森林1年大约滞尘32吨。明明说:1公顷森林在生长季节每周大约可吸收 6300 千克二氧化碳。(1)3 平方千米的森林 1 年大约滞尘多少吨。(2)1 公顷的森林在 6 月份(生长季节)大约可以吸收多少千克二氧化碳。合多少吨。",
+        "options":  [
+
+                    ],
+        "answer":  "9600 吨;27000 千克;27 吨",
+        "solution":  "【分析】(1)3平方千米 = 300公顷,1公顷森林1年大约滞尘32吨,300公顷森林1年大约滞尘32×300吨。(2)1公顷森林在生长季节每周大约可吸收6300千克二氧化碳,用6300÷7求出平均每天大约可吸收二氧化碳的质量,然后再乘6月份的天数30即可。\n【详解】(1)3平方千米 = 300公顷\n32×300 = 9600(吨)\n答:3平方千米的森林1年大约滞尘9600吨。\n(2)6300÷7×30 = 900×30 = 27000(千克)\n27000千克 = 27吨\n答:1公顷的森林在6月份大约可以吸收27000千克二氧化碳,合27吨。",
+        "knowledge_points":  [
+                                 "三位数乘两位数的实际问题",
+                                 "吨,千克之间的换算与比较",
+                                 "公顷,平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·福建福州·期中"
+    },
+    {
+        "id":  "261",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "某商场举办月饼中秋节促销活动,李阿姨想买 12 盒月饼送给亲朋好友,她参加哪个活动更省钱。(每位顾客只能参加一种促销活动)原价:110 元/盒 中秋节促销活动活动一:买5盒送1盒活动二:每盒便宜 15 元",
+        "options":  [
+
+                    ],
+        "answer":  "活动一",
+        "solution":  "【分析】买5盒送1盒,即花5盒的钱,可以得到6盒月饼。据此先求12盒里有几组6盒,再求需要付钱的盒数,然后根据单价×数量 = 总价,求出活动一应付的钱数。活动二用每盒原价减去优惠后的单价,再乘12盒,求出总价。比较两种活动花的钱数即可。\n【详解】12÷(5 + 1) = 2(组)\n2×5 = 10(盒)\n10×110 = 1100(元)\n12×(110 - 15) = 12×95 = 1140(元)\n1100＜1140\n答:她参加活动一更省钱。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "25-26四年级上·福建莆田·期中"
+    },
+    {
+        "id":  "262",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "\"蛟龙号\"载人深潜器是我国首台自主设计研制的作业型深海载人潜水器,设计最大下潜深度为 7000 米级,也是目前世界上(同类型中)下潜能力最深的作业型载人潜水器.\"蛟龙号\"载人深潜器以 41 米/分的速度向海底下潜,2 小时下潜多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "4920 米",
+        "solution":  "【分析】根据1小时 = 60分,先把单位进行换算,即2小时 = 120分,所以2小时下潜的长度 = 深潜器下潜的速度×120,据此代入数值作答即可。\n【详解】2小时 = 120分\n41×120 = 4920(米)\n答:2小时下潜4920米。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "基础行程问题",
+                                 "时,分的认识"
+                             ],
+        "source":  "25-26四年级上·甘肃平凉·期中"
+    },
+    {
+        "id":  "263",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "学校准备给26个班发笔记本,每班发112本,还需要留98本备用。该学校应买多少本笔记本。",
+        "options":  [
+
+                    ],
+        "answer":  "3010 本",
+        "solution":  "【分析】根据题意,已知学校准备给26个班发笔记本,每班发112本,还需要留98本备用。先用112乘26,计算出发给26个班的总本数;再加上98,就是学校应买的总本数,列式计算即可.\n【详解】根据分析可知:\n112×26+98 \n= 2912 + 98 \n= 3010 (本)\n答:该学校应买3010本笔记本。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "24-25四年级上·湖南益阳·期中"
+    },
+    {
+        "id":  "264",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一辆洒水车洒水宽度是 4 米,它以每分钟 250 米的速度前进,1 个小时能洒水多少公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "6 公顷",
+        "solution":  "【分析】洒水车洒水的区域是长方形,每分钟洒水的面积 = 洒水宽度×速度。先计算每分钟洒水面积,再乘时间60分钟,最后将平方米转换为公顷。\n【详解】4×250 = 1000(平方米)\n1小时 = 60分钟\n1000×60 = 60000(平方米)\n60000平方米 = 6公顷\n答:1个小时能洒水6公顷。",
+        "knowledge_points":  [
+                                 "因数中间或末尾是0的乘法",
+                                 "公顷,平方千米的进率与换算"
+                             ],
+        "source":  "24-25四年级上·湖南益阳·期中"
+    },
+    {
+        "id":  "265",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一个没拧紧的水龙头每天会浪费 13 千克的水,这个水龙头 2024 年一共会浪费多少千克水。",
+        "options":  [
+
+                    ],
+        "answer":  "4758 千克",
+        "solution":  "【分析】先判断2024年是平年还是闰年,确定这一年的天数,再用每天浪费水的质量乘一年的天数,得到2024年浪费水的总质量。2024能被4整除且不能被100整除,所以是闰年。\n【详解】2024÷4 = 506,所以2024年是闰年,有366天。\n366×13 = 4758(千克)\n答:这个水龙头2024年一共会浪费4758千克水。",
+        "knowledge_points":  [
+                                 "平年,闰年的认识及判定方法",
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "24-25四年级上·湖南益阳·期中"
+    },
+    {
+        "id":  "266",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "学校买了 15 个篮球,每个篮球 120 元;又买了 8 个排球,每个排球 95 元。买篮球比买排球多花了多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "1040 元",
+        "solution":  "【分析】根据总价 = 单价×数量,分别算出两种球用了多少钱,再相减即可。\n【详解】120×15 - 95×8 = 1800 - 760 = 1040(元)\n答:买篮球比买排球多花了1040元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "三位数乘两位数,乘数末尾有 0"
+                             ],
+        "source":  "25-26 四年级上·河北邢台·期中"
+    },
+    {
+        "id":  "267",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一块正方形草坪的边长是 800 米,这块草坪的面积是多少公顷。如果每公顷施肥 200 千克,一共需要施肥多少千克。",
+        "options":  [
+
+                    ],
+        "answer":  "64 公顷;12800 千克",
+        "solution":  "【分析】根据正方形的面积 = 边长×边长,用800乘800求出草坪的面积,再根据1公顷 = 10000平方米将平方米换算成公顷,最后用草坪的面积乘200就可求出需要施肥多少千克。\n【详解】800×800 = 640000(平方米)\n640000平方米 = 64公顷\n64×200 = 12800(千克)\n答:这块草坪的面积是64公顷,一共需要施肥12800千克。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "正方形的面积",
+                                 "公顷,平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·河北邯郸·期中"
+    },
+    {
+        "id":  "268",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "一列高铁从武汉开往广州,速度是 300 千米/时,行驶了 3 小时后,距离广州还有 240 千米,武汉到广州的铁路长多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "1140 千米",
+        "solution":  "【分析】根据路程 = 速度×时间,用300乘3求出已经行驶的路程,再加上还剩下的距离240千米,就能得到武汉到广州的铁路总长。\n【详解】300×3 + 240 = 900 + 240 = 1140(千米)\n答:武汉到广州的铁路长1140千米。",
+        "knowledge_points":  [
+                                 "乘加,乘减混合运算",
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26 四年级上·河北邯郸·期中"
+    },
+    {
+        "id":  "269",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "学校要为 24 个班各配一套投影仪,每套投影仪售价 856 元,准备 2 万元够吗。",
+        "options":  [
+
+                    ],
+        "answer":  "不够",
+        "solution":  "【分析】学校要为24个班各配一套投影仪,每套投影仪售价856元,用24乘856求出总价格,再与2万元进行比较,据此解答。\n【详解】856×24 = 20544(元)\n2万元 = 20000元\n20544＞20000\n答:准备2万元不够。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "25-26 四年级上·河北邯郸·期中"
+    },
+    {
+        "id":  "270",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "小军和爸爸,妈妈一起到一座可容纳 380 人的音乐厅看文艺节目,每张入场券的价格是 18 元。他们赶到时已售出 119 张,余下的票按每张 13 元售出,这样余下的票又可以收入多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "3393 元",
+        "solution":  "【分析】根据题意,已知一座可容纳380人的音乐厅,每张入场券的价格是18元。他们赶到时已售出119张,余下的票按每张13元售出,先用380减去119,就是剩余票数,再乘13,就是余下票的总收入。\n【详解】(380 - 119)×13 = 261×13 = 3393(元)\n答:余下的票可以收入3393元。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "24-25四年级上·四川成都·期中"
+    },
+    {
+        "id":  "271",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "郑州博物馆新馆举办\u0027华夏文明展\u0027,展厅可容纳 460 人参观,每张门票的价格是 25 元。开展当天已售出 158 张,余下的票按每张 18 元售出,这样余下的票又可以收入多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "5436 元",
+        "solution":  "【分析】根据题意,已知展厅可容纳460人,每张门票原价25元。开展当天已售出158张,余下的票按每张18元售出。先用460减去158,求出剩余票数,再乘18,即可得到余下票的总收入。\n【详解】(460 - 158)×18 = 302×18 = 5436(元)\n答:余下的票可以收入5436元。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法",
+                                 "乘减混合运算"
+                             ],
+        "source":  "25-26四年级上·河南郑州·期中"
+    },
+    {
+        "id":  "272",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "两个工程队同时修一条公路,甲每天修 450 米,乙每天修 380 米,40 天完成,这条公路有多长。方法 1:先求 (),列式为:450+ 380 = 830 米再求 ()列式为() 方法二:先分别求出甲,乙工程队各修了多少米。再求这条公路的总长度。",
+        "options":  [
+
+                    ],
+        "answer":  "甲乙两队一天一共修多少米;40天一共修多少米;40×830 = 33200(米);33200 米",
+        "solution":  "【分析】根据题意,已知两个工程队同时修一条公路,甲每天修450米,乙每天修380米,40天完成。方法一:先用450加上380,求出甲乙两队一天一共修多少米,再乘40,就是这条公路有多长。方法二:先用450乘40,求出甲40天一共修多少米;再用380乘40,求出乙40天一共修多少米;最后把两数相加即可。\n【详解】方法1:先求甲乙两队一天一共修多少米,列式为:450 + 380 = 830(米)\n再求40天一共修多少米,列式为:40×830 = 33200(米)。\n方法2:450×40 + 380×40\n= 18000 + 15200\n= 33200(米)\n答:这条公路的总长度是33200米。",
+        "knowledge_points":  [
+                                 "两人合作的工程问题",
+                                 "三位数乘两位数,乘数末尾有 0"
+                             ],
+        "source":  "25-26 四年级上·湖北襄阳·期中"
+    },
+    {
+        "id":  "273",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一个长方形果园,长 200 米,宽 150 米。如果每公顷能收获苹果 25 吨,这个果园一共能收获苹果多少吨。",
+        "options":  [
+
+                    ],
+        "answer":  "75 吨",
+        "solution":  "【分析】长方形的面积 = 长×宽。由题意得,一个长方形果园的长为200米,宽为150米,直接用乘法可以算出长方形果园的面积,然后根据10000平方米 = 1公顷将得数转化为公顷。如果每公顷能收获苹果25吨,直接用25乘前面的公顷数即可算出这个果园一共能收获苹果多少吨。\n【详解】200×150 = 30000(平方米)\n10000平方米 = 1公顷,所以30000平方米 = 3公顷。\n25×3 = 75(吨)\n答:这个果园一共能收获苹果75吨。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "长方形的面积",
+                                 "三位数与整十数的口算乘法"
+                             ],
+        "source":  "25-26 四年级上·湖北襄阳·期中"
+    },
+    {
+        "id":  "274",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "商店从工厂批发了80 台复读机,每台 105 元,商店要付给工厂多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "8400 元",
+        "solution":  "【分析】根据题意可知,用每台复读机的价格乘复读机的台数,就是一共要支付的钱数。\n【详解】105×80 = 8400(元)\n答:商店要付给工厂8400元。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,三位数中间有 0"
+                             ],
+        "source":  "25-26 四年级上·湖北襄阳·期中"
+    },
+    {
+        "id":  "275",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "云锦是中国传统的丝制工艺品,中国国家地理标志产品,有\u0027寸锦寸金\u0027之称。赵阿姨去旅游时买了 23 盒云锦礼盒,每盒 362 元,赵阿姨带了 9000 元,够吗。",
+        "options":  [
+
+                    ],
+        "answer":  "够",
+        "solution":  "【分析】根据总价 = 单价×数量,用赵阿姨买云锦礼盒的数量乘每盒价格,即可求出买这些云锦礼盒一共需要花费多少钱,再与9000元比较。\n【详解】362×23 = 8326(元)\n8326＜9000\n答:赵阿姨带了9000元,够。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法",
+                                 "经济问题"
+                             ],
+        "source":  "25-26 四年级上·陕西榆林·期中"
+    },
+    {
+        "id":  "276",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一块长方形地长 450米,宽 200 米。这块地的面积是多少公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "9 公顷",
+        "solution":  "【分析】已知长方形地长450米,宽200米。根据长方形的面积 = 长×宽,用450×200即可求出面积;再根据1公顷 = 10000平方米,进行面积单位间的转换即可解答。\n【详解】450×200 = 90000(平方米)\n90000平方米 = 9公顷\n答:这块地的面积是9公顷。",
+        "knowledge_points":  [
+                                 "公顷,平方千米的进率与换算",
+                                 "长方形的面积",
+                                 "三位数乘两位数,乘数末尾有 0"
+                             ],
+        "source":  "25-26 四年级上·陕西榆林·期中"
+    },
+    {
+        "id":  "277",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "\u0027朝辞白帝彩云间,千里江陵一日还。两岸猿声啼不住,轻舟已过万重山。\u0027这是李白的《早发白帝城》。白帝城位于重庆市奉节县,江陵即湖北省荆州市,实际上两地的水路大约350千米。(1)如果李白当天06:00出发,乘船的速度是21千米/时,\u0027千里江陵一日还\u0027(还:一天就可以从白帝城到达江陵),他当天能到达江陵吗。(2)明明一家在假期自驾从奉节县到荆州市旅游,前3个小时的速度是75千米/时,中途休息了1小时,后2个小时的速度是72千米/时,从奉节县到荆州市驾车全程多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "能;369 千米",
+        "solution":  "【分析】(1)根据题意,如果李白当天06:00出发,乘船的速度是21千米/时,先用24减去6求出行驶的时间,再用21乘时间就是一天行驶的距离,再与350千米比较即可。(2)分别用3乘75和2乘72求出两个时间段行驶的路程,再求出它们的和即可。\n【详解】(1)24 - 6 = 18(小时)\n21×18 = 378(千米)\n378＞350\n答:他当天能到达江陵。\n(2)75×3 + 72×2 = 225 + 144 = 369(千米)\n答:从奉节县到荆州市驾车全程369千米。",
+        "knowledge_points":  [
+                                 "乘加,乘减混合运算",
+                                 "基础行程问题",
+                                 "时,分,秒时间的推算"
+                             ],
+        "source":  "25-26 四年级上·福建福州·期中"
+    },
+    {
+        "id":  "278",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "为激发学生的科技创新热情,学校组织学生去科技馆参加科技节活动。科技馆门票每张 48 元,同时租用的大巴车往返费用共计 4000 元。四年级 350 名学生全部参加,这次活动的门票和交通总费用是多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "20800 元",
+        "solution":  "【分析】根据题意,科技馆门票每张48元,四年级350名学生全部参加,先用48乘350求出门票的总价钱,然后再加上大巴车往返费用,即可求出这次活动的门票和交通总费用。\n【详解】48×350 + 4000 = 16800 + 4000 = 20800(元)\n答:这次活动的门票和交通总费用是20800元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "乘加,乘减混合运算"
+                             ],
+        "source":  "25-26四年级上·福建福州·期中"
+    },
+    {
+        "id":  "279",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "人口密度指特定面积土地上的人口数量,通常使用\u0027人/平方千米\u0027作单位。福建省土地面积约为 120000 平方千米,根据省统计局发布的《2024 年福建省国民经济和社会发展统计公报》,2024年福建省的人口密度约为 345 人/平方千米。2024 年末福建省常住人口约多少人。",
+        "options":  [
+
+                    ],
+        "answer":  "4140 万人",
+        "solution":  "【分析】求2024年末福建省常住人口约多少人,用120000乘345即可解答。\n【详解】120000×345 = 41400000(人)\n41400000人 = 4140万人\n答:2024年末福建省常住人口约4140万人。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有 0"
+                             ],
+        "source":  "25-26 四年级上·福建福州·期中"
+    },
+    {
+        "id":  "280",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "李老师想去买一些钢笔作为奖品发给同学们。①一盒钢笔有12支;②买一盒钢笔需要240元;③一共带了5000元;④李老师准备买15盒这样的钢笔;⑤四年级共有180人。(1)如果要解决\u0027李老师买完钢笔后还剩下多少钱。\u0027这个问题,需要用到的信息是 。(从上面的信息里选择,填序号)(2)根据你选择的信息,解答问题。",
+        "options":  [
+
+                    ],
+        "answer":  "②③④;1400 元",
+        "solution":  "【分析】(1)要解决\"李老师买完钢笔后还剩下多少钱\"这个问题,需要知道李老师一共带的钱数和买钢笔花的钱数。信息③是李老师带的钱数,根据信息②和信息④可以用乘法算出李老师买钢笔花的钱数,所以选择的信息为②③④。(2)先用240乘15算出李老师买钢笔花的钱数,然后再用5000减去前面的得数即可。\n【详解】(1)如果要解决\"李老师买完钢笔后还剩下多少钱\"这个问题,需要用到的信息是②③④。\n(2)5000 - 240×15 = 5000 - 3600 = 1400(元)\n答:李老师买完钢笔后还剩下1400元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "\"提问题\"\"填条件\"问题",
+                                 "三位数乘两位数的实际问题",
+                                 "乘加,乘减混合运算"
+                             ],
+        "source":  "25-26 四年级上·湖北襄阳·期中"
+    },
+    {
+        "id":  "281",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "一艘轮船从甲港到乙港,速度是32千米/小时,15小时到达。返回时共用了8小时,返回时的速度是多少。",
+        "options":  [
+
+                    ],
+        "answer":  "60 千米/时",
+        "solution":  "【分析】根据路程 = 速度×时间,用32乘15求出甲乙两港之间的距离,再根据速度 = 路程÷时间,即可解答此题。\n【详解】32×15 = 480(千米)\n480÷8 = 60(千米/时)\n答:返回时的速度是60千米/时。",
+        "knowledge_points":  [
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26四年级上·湖北襄阳·期中"
+    },
+    {
+        "id":  "282",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一个广场的长是 500米,宽是400米,它的面积是多少平方米。合多少公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "200000 平方米;20 公顷",
+        "solution":  "【分析】根据长方形的面积 = 长×宽,用广场的长500米乘宽400米,即可得到它的面积的平方米数;再根据1公顷 = 10000平方米,把平方米数化成公顷即可。\n【详解】500×400 = 200000(平方米)\n200000平方米 = 20公顷\n答:它的面积是200000平方米,合20公顷。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "长方形的面积",
+                                 "公顷,平方千米的进率与换算"
+                             ],
+        "source":  "25-26 四年级上·湖南长沙·期中"
+    },
+    {
+        "id":  "283",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一辆汽车从海口开往长沙,每小时行驶 98 千米,行驶了 15 小时后,距离长沙还有360千米,海口到长沙的路程是多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "1830 千米",
+        "solution":  "【分析】根据路程 = 速度×时间,用每小时行驶98千米乘时间15小时,即可求得行驶了15小时的路程,然后加上360千米,即可求得海口到长沙的路程。\n【详解】98×15 + 360 = 1470 + 360 = 1830(千米)\n答:海口到长沙的路程是1830千米。",
+        "knowledge_points":  [
+                                 "两位数乘两位数的进位乘法",
+                                 "乘加,乘减混合运算",
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26 四年级上·海南省直辖县级单位·期中"
+    },
+    {
+        "id":  "284",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "超市购进 120 箱牛奶,每箱 24 盒,每盒售价 3 元。这些牛奶全部卖出一共能收入多少元钱。",
+        "options":  [
+
+                    ],
+        "answer":  "8640 元",
+        "solution":  "【分析】先计算出牛奶的总盒数,即24×120 = 2880盒,再根据总价 = 数量×单价即可解答。\n【详解】24×120 = 2880(盒)\n2880×3 = 8640(元)\n答:这些牛奶全部卖出一共能收入8640元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "用两步连乘解决实际问题",
+                                 "三位数乘两位数,乘数末尾有0"
+                             ],
+        "source":  "25-26 四年级上·海南省直辖县级单位·期中"
+    },
+    {
+        "id":  "285",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "京沪高铁开通后,一辆300千米/时的动车组列车从北京南站到上海虹桥站全程票价:一等座 935 元/张,二等座 555 元/张。(1)小聪买了 11 张二等座的票,他需要用多少钱。(2)小聪妈妈准备 1 万元,够买 10 张一等座的票吗。",
+        "options":  [
+
+                    ],
+        "answer":  "6105 元;够",
+        "solution":  "【分析】(1)根据总价 = 单价×数量,用乘法计算出小聪买11张二等座的票需要用多少钱。(2)根据总价 = 单价×数量,用乘法计算出10张一等座的票的总价,再跟1万元进行比较。\n【详解】(1)555×11 = 6105(元)\n答:小聪买了11张二等座的票,需要用6105元。\n(2)935×10 = 9350(元)\n9350＜10000\n答:小聪妈妈准备1万元,够买10张一等座的票。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "三位数乘两位数的实际问题"
+                             ],
+        "source":  "25-26 四年级上·广东东莞·期中"
+    },
+    {
+        "id":  "286",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "一个电影院的某个放映厅共有 20 排座位,每排刚好 22 个座位,每张电影票30 元,电影开场前电影票全部卖完了,这场电影的卖票收入是多少。",
+        "options":  [
+
+                    ],
+        "answer":  "13200 元",
+        "solution":  "【分析】本题需要计算电影票的总收入,根据单价×数量 = 总价的关系,先用22乘20求出总座位数,再乘票价即可。\n【详解】22×20 = 440(个)\n440×30 = 13200(元)\n答:这场电影的卖票收入是13200元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "三位数乘两位数,乘数末尾有0"
+                             ],
+        "source":  "25-26四年级上·湖北襄阳·期中"
+    },
+    {
+        "id":  "287",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "一篇文章有 1710 个字,如果王阿姨平均每分钟能录入102个字,那么王阿姨18分钟能录入完成吗。",
+        "options":  [
+
+                    ],
+        "answer":  "能录入完成",
+        "solution":  "【分析】根据题意,已知一篇文章有1710个字,如果王阿姨平均每分钟能录入102个字,先用102乘18,计算王阿姨18分钟能录入的字数,再与文章总字数1710进行比较即可。\n【详解】102×18 = 1836(个)\n1836＞1710\n答:王阿姨18分钟能录入完成。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "25-26 四年级上·湖北襄阳·期中"
+    },
+    {
+        "id":  "288",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "我们这个新能源汽车充电停车场原来的宽是 7 米,面积是 504 平方米。现在要扩建,长不变,宽增加到 28 米。面积增加了多少平方米呢。",
+        "options":  [
+
+                    ],
+        "answer":  "1512 平方米",
+        "solution":  "【分析】由题意得,若长不变,宽增加到28米,28÷7 = 4,即长方形的宽扩大到原来的4倍。根据积的变化规律,一个因数不变,另一个因数乘4,积也乘4。所以这个停车场的面积扩大到原来的4倍,直接用504乘4即可算出扩建后的面积,再减去原来的面积就是增加的面积。\n【详解】504×(28÷7) = 504×4 = 2016(平方米)\n2016 - 504 = 1512(平方米)\n答:面积增加了1512平方米。",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·山西晋中·期中"
+    },
+    {
+        "id":  "289",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "双十一,永辉超市某品牌酸奶推出促销活动,原价每瓶 8 元,现在买三送一,妈妈买了 12 瓶,要花多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "72 元",
+        "solution":  "【分析】将\"买三送一\"中的3 + 1 = 4(瓶)看作一组,购买的12瓶里正好有3组,每组中有3瓶需要花钱购买,3组共需要购买9瓶,最后根据单价×数量 = 总价求出总价即可。\n【详解】12÷(3 + 1) = 12÷4 = 3(组)\n3×3×8 = 9×8 = 72(元)\n答:要花72元。",
+        "knowledge_points":  [
+                                 "经济问题"
+                             ],
+        "source":  "25-26四年级上·福建莆田·期中"
+    },
+    {
+        "id":  "290",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "2025 年是中国人民志愿军抗美援朝 75 周年,集邮公司发行相关珍藏邮票,邮票珍藏册每本208元,买25本要多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "5200 元",
+        "solution":  "【分析】根据总价 = 单价×数量,每本208元为单价,25本为数量,代入数据计算。\n【详解】208×25 = 5200(元)\n答:买25本要5200元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "25-26 四年级上·福建莆田·期中"
+    },
+    {
+        "id":  "291",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "王叔叔开车去省城学习,去的时候速度是 75 千米/时,开了 12 小时到达,原路返回时少用了 3 小时,返回时平均每小时行驶多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "100 千米/时",
+        "solution":  "【分析】路程 = 速度×时间,据此代入去时的速度和时间可算出路程,再用路程除以返回时的时间可算出返回时的速度。\n【详解】75×12 = 900(千米)\n900÷(12 - 3) = 900÷9 = 100(千米/时)\n答:返回时平均每小时行驶100千米/时。",
+        "knowledge_points":  [
+                                 "整十,整百,整千数除以一位数的除法",
+                                 "两位数乘两位数的进位乘法",
+                                 "带有小括号的混合运算",
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26四年级上·湖北襄阳·期中"
+    },
+    {
+        "id":  "292",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "两辆汽车同时从甲,乙两地相向而行,快车每小时行 65 千米,慢车每小时行 56千米,两车相遇时,慢车比快车少行了90千米。甲,乙两地相距多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "1210 千米",
+        "solution":  "【分析】首先用减法算出两车的速度差,根据路程差÷速度差 = 相遇时间,代入数值计算出相遇时间。再用加法算出两车的速度和,根据速度和×相遇时间 = 路程和,代入数值计算出甲、乙两地相距的距离。\n【详解】90÷(65 - 56) = 90÷9 = 10(小时)\n(65 + 56)×10 = 121×10 = 1210(千米)\n答:甲、乙两地相距1210千米。",
+        "knowledge_points":  [
+                                 "相遇问题"
+                             ],
+        "source":  "25-26 四年级上·福建厦门·期中"
+    },
+    {
+        "id":  "293",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一天早上 7:00,王叔叔开车从公司出发去省城,去时每小时行 80 千米,下午2:00 到达省城。原路返回公司王叔叔多用了 1 小时,返回时每小时行多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "70 千米/时",
+        "solution":  "【分析】根据题意,明确路程 = 速度×时间,速度 = 路程÷时间。已知早上7:00出发,下午2:00到达省城,先求出去时所用时间,再乘80求出总路程;返回时多用了1小时,用总路程除以返回时间,就是返回时每小时行多少千米。\n【详解】下午2:00 = 14:00\n14 - 7 = 7(小时)\n7×80÷(7 + 1) = 560÷8 = 70(千米/时)\n答:返回时每小时行70千米/时。",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "24时计时法时间的计算"
+                             ],
+        "source":  "25-26 四年级上·福建厦门·期中"
+    },
+    {
+        "id":  "294",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.4,
+        "question":  "甲,乙,丙三人,甲每分钟走 100 米,乙每分钟走 80 米,丙每分钟走 75 米,甲从东村,乙,丙从西村同时出发相向而行,途中甲,乙相遇三分钟后又与丙相遇,求东西两村的距离。",
+        "options":  [
+
+                    ],
+        "answer":  "18900 米",
+        "solution":  "【分析】根据题意,甲、乙相遇后三分钟甲又与丙相遇,这3分钟内甲与丙一共走了它们之间原来的距离,所以先用3×(100 + 75)求出甲、乙相遇时甲与丙之间的距离,这也是乙和丙3分钟走路程的差。再根据路程差÷速度差 = 时间,求出甲、乙相遇时间;最后用甲、乙的速度和乘相遇时间,求出东西两村的距离。\n【详解】甲、乙相遇时甲与丙相距:\n3×(75 + 100) = 3×175 = 525(米)\n甲、乙相遇时间:\n525÷(80 - 75) = 525÷5 = 105(分钟)\n东西两村的距离:\n105×(100 + 80) = 105×180 = 18900(米)\n答:东西两村的距离是18900米。",
+        "knowledge_points":  [
+                                 "相遇问题"
+                             ],
+        "source":  "2025·湖南长沙·小升初真题"
+    },
+    {
+        "id":  "295",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "某园林中有一个长方形的荷花池塘,宽 8 米,面积是 288 平方米。现保持池塘的长度不变,将宽度增加到 32 米。那么扩建后的花池塘面积是多少平方米。",
+        "options":  [
+
+                    ],
+        "answer":  "1152 平方米",
+        "solution":  "【分析】根据题意,长方形的面积 = 长×宽。长方形的长不变,宽扩大到原来的32÷8 = 4倍,根据积的变化规律可知,面积也扩大到原来的4倍,据此即可求出扩建后的荷花池塘面积。\n【详解】288×(32÷8) = 288×4 = 1152(平方米)\n答:扩建后的荷花池塘面积是1152平方米。",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·福建福州·期中"
+    },
+    {
+        "id":  "296",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一辆洒水车,每小时行驶9千米,洒水宽度是5米,这辆洒水车如果不间断工作 2 小时,洒水的面积是多少公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "9 公顷",
+        "solution":  "【分析】根据路程 = 速度×时间,求出洒水车行驶路程,即9×2千米,也就是洒水地面的长度。千米和米之间的进率是1000,据此将长度换算成米。再根据长方形的面积 = 长×宽,求出洒水地面面积。平方米和公顷之间的进率是10000,据此将面积换算成公顷。\n【详解】9×2 = 18(千米) = 18000(米)\n18000×5 = 90000(平方米)\n90000平方米 = 9公顷\n答:它洒水的地面面积是9公顷。",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "长方形的面积",
+                                 "千米和米之间的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26 四年级上·福建福州·期中"
+    },
+    {
+        "id":  "297",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.94,
+        "question":  "王老师从福州大戏院出发去三坊七巷游玩。他骑电动车每分钟行驶240米,15分钟后到达冰心故居北门,那么王老师从福州大戏院到冰心故居北门骑行的路程是多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "3600 米",
+        "solution":  "【分析】根据路程 = 速度×时间,已知速度是240米/分钟,时间是15分钟,直接相乘即可求出总路程。\n【详解】240×15 = 3600(米)\n答:王老师从福州大戏院到冰心故居北门骑行的路程是3600米。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法",
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26 四年级上·福建福州·期中"
+    },
+    {
+        "id":  "298",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "公园有一块长方形的运动场,长 30 米,宽 20 米。现在要扩建运动场,使扩建后的面积是原来的 2 倍。扩建后的面积是多少平方米。![](images/18aa709c6707284e599fbd983d8d1b9a.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "1200 平方米",
+        "solution":  "【分析】积的变化规律:一个乘数不变,另一个乘数乘或除以几(0除外),得到的积就等于原来的积乘或除以几。长方形的面积 = 长×宽。由题意得,长方形运动场的长是30米,宽是20米,先求出原来的面积。现在要扩建运动场,使扩建后的面积是原来的2倍,直接用原来的面积乘2即可求出扩建后的面积。要使面积变为原来的2倍,可以让长扩大到原来的2倍,宽不变,也可以让宽扩大到原来的2倍,长不变。\n【详解】30×20 = 600(平方米)\n600×2 = 1200(平方米)\n要使扩建后的面积是原来的2倍,可以让长方形运动场的长扩大到原来的2倍,宽不变(左图);也可以让长方形运动场的宽扩大到原来的2倍,长不变(右图)。\n方法一:30×2 = 60(米),即长方形的长增加到60米,宽不变。\n方法二:20×2 = 40(米),即长方形的宽增加到40米,长不变。\n答:扩建后的面积是1200平方米。要想扩建后的面积是原来的2倍,可以让长增加到60米,宽不变,或者宽增加到40米,长不变。",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "25-26 四年级上·江西赣州·期中"
+    },
+    {
+        "id":  "299",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "王叔叔开汽车回老家用了4小时,每小时大约行100千米。王叔叔返程时从原路返回,用了5小时,返程时汽车平均每小时行多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "80 千米/时",
+        "solution":  "【分析】已知王叔叔开汽车回老家用了4小时,每小时大约行100千米,根据路程 = 时间×速度,先求出路程。返程用了5小时,再用路程÷时间计算返程速度。\n【详解】4×100 = 400(千米)\n400÷5 = 80(千米/时)\n答:返程时汽车平均每小时行80千米/时。",
+        "knowledge_points":  [
+                                 "基础行程问题"
+                             ],
+        "source":  "25-26四年级上·江西赣州·期中"
+    },
+    {
+        "id":  "300",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "超市开展促销活动,牛奶买5箱送1箱。每箱牛奶52元,买30箱牛奶至少要多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "1300 元",
+        "solution":  "【分析】由题意得,超市牛奶买5箱送1箱,即买5箱牛奶实际得到6箱。求买30箱牛奶至少要多少元,先用30÷6算出一共需要买多少组,然后再乘5算出一共需要付钱买多少箱牛奶。每箱牛奶52元,最后再乘52即可。\n【详解】5 + 1 = 6(箱)\n30÷6 = 5(组)\n5×5 = 25(箱)\n52×25 = 1300(元)\n答:买30箱牛奶至少要1300元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "两位数乘两位数的进位乘法"
+                             ],
+        "source":  "25-26四年级上·江西赣州·期中"
+    },
+    {
+        "id":  "301",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "陈老师买了5筒羽毛球,每筒 12 个,一共花了 120 元。解决\u0027每筒多少元\u0027的问题,需要用到的信息是( )。",
+        "options":  [
+                        "A. 5 筒,12 个,120 元",
+                        "B. 5 筒,120 元",
+                        "C. 12 个,120 元"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】总价÷数量 = 单价,求每筒多少元,用一共花的钱数除以买的筒数即可。\n【详解】一共花的钱数为120元,买的筒数为5筒,120÷5 = 24(元/筒),所以解决\u0027每筒多少元\u0027的问题,需要用到的信息是\u00275筒,120元\u0027。故答案为:B。",
+        "knowledge_points":  [
+                                 "几百几十或几千几百除以一位数的除法",
+                                 "含有多余条件的实际问题",
+                                 "\u0027提问题\u0027\u0027填条件\u0027问题",
+                                 "经济问题"
+                             ],
+        "source":  "24-25四年级上·浙江温州·期末"
+    },
+    {
+        "id":  "302",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面三个速度中,最快的是( )。",
+        "options":  [
+                        "A. 每秒走 10 米",
+                        "B. 1 分钟走 85 米",
+                        "C. 8 分钟走了 1 千米"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】1分 = 60秒,用10乘60可以计算出选项A每分钟走的米数;1千米 = 1000米,根据路程÷时间 = 速度,用1000米除以8分钟,可以计算出选项C每分钟走的米数;最后将3个选项每分钟走的米数进行比较即可。\n【详解】A. 1分 = 60秒,10×60 = 600(米/分钟),所以速度为600米/分钟;B. 1分钟走85米,说明速度为85米/分钟;C. 1千米 = 1000米,1000÷8 = 125(米/分钟),所以速度为125米/分钟。600＞125＞85,所以三个速度中,最快的是每秒走10米。故答案为:A。",
+        "knowledge_points":  [
+                                 "两位数乘整十数的口算乘法",
+                                 "基础行程问题",
+                                 "时,分,秒之间的换算与比较",
+                                 "千米和米之间的进率与换算"
+                             ],
+        "source":  "24-25四年级上·浙江杭州·期末"
+    },
+    {
+        "id":  "303",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "在计算 234×56 的时候, 2×6 表示( )。",
+        "options":  [
+                        "A. 2×6",
+                        "B. 20×6",
+                        "C. 200×6",
+                        "D. 200×60"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】观察算式,看看数字2在234中的哪个数位,就代表多少,同时看看数字6在56中的哪个数位就代表几,依此可以解答。\n【详解】数字2在234中的百位上,因此2代表200;数字6在56中的个位上,代表6。因此在计算234×56时,2×6表示200×6。故答案为:C。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "24-25四年级上·浙江温州·期末"
+    },
+    {
+        "id":  "304",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "某小学四年级师生共298人,准备去秋游,旅行车包车的价格是39元/人,老师带了 12000 元钱包车,( )。",
+        "options":  [
+                        "A. 钱正好",
+                        "B. 钱够且有剩余",
+                        "C. 钱不够"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】先用四年级人数乘包车每人要付的钱数,求出包车的总价格,再和12000元进行比较即可。\n【详解】298×39 = 11622(元)\n11622＜12000\n所以老师带了12000元钱包车,钱够且有剩余。故答案为:B。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "24-25 四年级上·湖北黄冈·期末"
+    },
+    {
+        "id":  "305",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列问题中,能用90-21×3解决的是( )。\n①篮球每个90元,垒球每个21元,一个篮球比3个垒球贵多少元。②工程队翻修一条90千米的路,每天修21千米,3天后还剩多少千米。③篮球原价90元,现降价21元,买3个能省多少钱。④轿车每小时行90千米,卡车每小时比它少行21千米,卡车从甲地到乙地用了3小时,甲乙两地相距多少千米。",
+        "options":  [
+                        "A. ①②",
+                        "B. ③④",
+                        "C. ①②③",
+                        "D. ①②③④"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】该算式表示从90中减去21乘3的结果。选项①和②都符合这个算式:选项①是篮球价格减去3个垒球的价格,即90-21×3;选项②是总路程减去3天修的路程,即90-21×3。选项③应列式为21×3,选项④应列式为(90-21)×3,都不符合该算式。\n【详解】由分析可知,选项①和②都符合这个算式,能用90-21×3解决。故答案为:A。",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "乘加,乘减混合运算",
+                                 "经济问题"
+                             ],
+        "source":  "24-25四年级上·湖北武汉·期末"
+    },
+    {
+        "id":  "306",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "在用计算器计算380×76 时,错误地输入了 38×76 ,除了按清除键清除后再重新输入外,下列操作( )也可以弥补错误。",
+        "options":  [
+                        "A. *10",
+                        "B. /10",
+                        "C. *100"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】两个因数相乘,一个因数乘或除以一个数(0除外),另一个因数不变,积同时乘或除以相同的数;据此即可解答。\n【详解】计算380×76时,错误地输入了38×76,相当于把380除以10,另一个因数不变,错误的积等于正确的积除以10,要得到正确的积就要再乘10。故答案为:A。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "24-25 四年级上·重庆万州·期末"
+    },
+    {
+        "id":  "307",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "周末,小丽爸爸骑自行车去梁山龟山胜境公园游玩。他骑车的速度是 259 米/分,从家到公园骑行需要18分钟,从家到公园有多少米。小丽用竖式计算出了结果,下图竖式中圈出的\u0027259\u0027表示( ).\n 259×18\n ───────\n 2072 (259×8 = 2072)\n 259 (259×1 = 259,末位与十位对齐)\n ───────\n 4662",
+        "options":  [
+                        "A. 8 分钟骑行的路程",
+                        "B. 1 分钟骑行的路程",
+                        "C. 10 分钟骑行的路程",
+                        "D. 18 分钟骑行的路程"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】根据三位数乘两位数的计算,竖式中圈出的\u0027259\u0027实际是2590,是三位数259乘两位数十位上的1的结果,即 259×10 的结果,其中259是骑车的速度,10是骑车的时间,2590表示10分钟骑车的路程,据此选择即可.\n【详解】 259×10 = 2590 (米)竖式中圈出的\u0027259\u0027表示 10 分钟骑行的路程。故答案为:C",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "24-25 四年级上·山东济宁·期末"
+    },
+    {
+        "id":  "308",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "已知6□9×□4是一个三位数乘两位数的算式,那么下面三个数中,( )可能是它的得数。",
+        "options":  [
+                        "A. 6876",
+                        "B. 29876",
+                        "C. 65986"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】三位数和两位数相乘,这个三位数个位上的数是9,两位数个位上的数是4,9×4 = 36,所以积的个位数是6。6□9最大是699,最小是609,□4最大是94,最小是14,求出积可能的最大和最小结果,再判断即可。\n【详解】699×94 = 65706,609×14 = 8526。\n所以,6□9×□4的积在8526与65706之间,个位上是6。\nA. 6876＜8526,不符合;\nB. 8526＜29876＜65706,符合;\nC. 65986＞65706,不符合。\n由此可知,29876可能是它的得数。故答案为:B。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法",
+                                 "三位数乘两位数,三位数中间有0"
+                             ],
+        "source":  "24-25四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "309",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面说法正确的是( )。",
+        "options":  [
+                        "A. 读数时 50673002 要读出 2 个零。",
+                        "B. 三位数乘两位数,积有可能是六位数。",
+                        "C. 一个因数增加,另一个因数减少,积不变。"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】(1)大数的读法:先把数分级,从最高级开始一级一级往下读。每级末尾的零不读,其他数位上有一个零或连续几个零,都只读一个零。(2)三位数乘两位数,积可能是四位数,也可能是五位数。(3)一个因数增加,另一个因数减少,积不一定不变。\n【详解】A. 50673002读作:五千零六十七万三千零二,读出2个零,正确。B. 三位数乘两位数,积可能是五位数,也可能是四位数,原题说法错误。C. 一个因数增加,另一个因数减少,两数的乘积不一定不变,原题说法错误。故答案为:A。",
+        "knowledge_points":  [
+                                 "亿以内数的读,写法",
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "24-25 四年级上·湖南张家界·期末"
+    },
+    {
+        "id":  "310",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "两个因数相乘,积是 800,如果一个因数乘 5,要使积不变,另一个因数应该( )。",
+        "options":  [
+                        "A. 乘 5",
+                        "B. 除以 5",
+                        "C. 加 5",
+                        "D. 减 5"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】积的变化规律:如果一个乘数乘几(0除外),另一个乘数除以相同的数,那么积不变;据此解答。\n【详解】如果一个因数乘5,要使积不变,另一个因数应该除以5。故答案为:B。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "24-25 四年级上·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "311",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "两个因数的积是 360,其中一个因数扩大到原来的 4 倍,另一个因数不变,积是( )。",
+        "options":  [
+                        "A. 360",
+                        "B. 1240",
+                        "C. 1440",
+                        "D. 2880"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】根据积的变化规律:当一个因数乘或除以一个数(不为0),另一个因数不变时,积也随着乘或除以这个数。当一个因数扩大到原来的4倍,积也扩大到原来的4倍。\n【详解】360×4 = 1440。两个因数的积是360,其中一个因数扩大到原来的4倍,另一个因数不变,积是1440。故答案为:C。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "24-25 六年级下·福建福州·开学考试"
+    },
+    {
+        "id":  "312",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "张叔叔步行去公司上班,计划 40 分钟到达,实际每分钟走 60 米,只用了 30分钟。下班时,张叔叔改骑共享单车,从公司到家一共用了 10 分钟。张叔叔骑行速度是多少。解决这个问题需要用到的信息是( )。",
+        "options":  [
+                        "A. 60米,30分钟",
+                        "B. 60米,10分钟",
+                        "C. 40分钟,60米,10分钟",
+                        "D. 60 米,30 分钟,10 分钟"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】由题意可知,往返的路程不变。根据路程 = 速度×时间,先求出从家到公司的路程,再根据速度 = 路程÷时间,求出骑共享单车的速度。\n【详解】60×30÷10 = 1800÷10 = 180(米/分钟),即张叔叔骑行速度是180米/分钟。解决这个问题需要用到的信息是60米、30分钟、10分钟。故答案为:D。",
+        "knowledge_points":  [
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25四年级上·浙江温州·期末"
+    },
+    {
+        "id":  "313",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "超市推出优惠活动:某鲜牛奶 22 元/4 瓶。要求买 16 瓶这款鲜牛奶的价格,下面利用\u0027积的变化规律\u0027来列式解决的是( )。",
+        "options":  [
+                        "A. 22/4×16",
+                        "B. 16/4×22",
+                        "C. 22/4",
+                        "D. 22×16"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】由题可知,4瓶鲜牛奶22元。根据积的变化规律,看16是4的多少倍,那么花的钱数也是22元的多少倍,据此选择即可。\n【详解】16÷4×22 = 4×22 = 88(元)。利用\u0027积的变化规律\u0027来列式解决的是16÷4×22。故答案为:B。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "24-25四年级上·浙江温州·期末"
+    },
+    {
+        "id":  "314",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "一辆汽车从A地到B地,去的时候用了4小时,速度60千米/小时;返回时用了 3 小时,返回时速度是( )千米/小时。",
+        "options":  [
+                        "A. 240",
+                        "B. 720",
+                        "C. 420",
+                        "D. 80"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】根据题意,先用速度×时间 = 路程求出去时的路程。因为往返的路程是一样的,所以再用路程÷时间 = 速度,就可以求出返回时的速度。\n【详解】60×4 = 240(千米)\n240÷3 = 80(千米/小时)\n所以,返回时速度是80千米/小时。故答案为:D。",
+        "knowledge_points":  [
+                                 "因数中间或末尾是0的乘法",
+                                 "商末尾有0的除法",
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25四年级上·浙江杭州·期末"
+    },
+    {
+        "id":  "315",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "列竖式计算 234×56 的时候 3×5 表示( )。",
+        "options":  [
+                        "A. 30×50",
+                        "B. 30×5",
+                        "C. 3×50"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】观察算式,看看数字3在234中的哪个数位,就代表多少,同时看看数字5在56中的哪个数位就代表几,依此可以解答。\n【详解】数字3在234中的十位上,因此3代表30。数字5在56中的十位上,代表50。因此列竖式计算234×56的时候,3×5表示30×50。故答案为:A。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "24-25四年级上·湖北黄冈·期末"
+    },
+    {
+        "id":  "316",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "在计算 906×32 时,其中\u0027 9×3 \u0027表示( )。",
+        "options":  [
+                        "A. 3 个 9",
+                        "B. 3 个 90",
+                        "C. 30 个 90",
+                        "D. 30 个 900"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】本题考查整数乘法的笔算。906的9在百位上,表示9个百;32的3在十位上,表示3个十,因此9与3相乘实际是900×30。\n【详解】用竖式计算906×32时,9与3相乘表示900×30,即30个900。故答案为:D。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "24-25 四年级上·湖北黄冈·期末"
+    },
+    {
+        "id":  "317",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "王老师购买了 6 台打印机,每台打印机价格500多元。下面选项中,打印机总价钱可能是( )元。",
+        "options":  [
+                        "A. 3000",
+                        "B. 3300",
+                        "C. 3600"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据总价 = 单价×数量,用每个选项的总价钱除以购买的台数,求出对应的打印机单价,每台打印机价格500多元,则大于500元且小于600元,据此选择即可.\n【详解】A。 3000/6 = 500 (元),打印机的价格不是 500多元,不符合题意;B. 3300/6 = 550 (元),550 元\u003e500 元,打印机的价格是500 多元,符合题意;C. 3600/6 = 600 (元),打印机的价格不是 500 多元,不符合题意。打印机总价钱可能是3300元。故答案为:B",
+        "knowledge_points":  [
+                                 "整十,整百,整千数除以一位数的除法",
+                                 "经济问题"
+                             ],
+        "source":  "24-25 四年级上·湖北襄阳·期末"
+    },
+    {
+        "id":  "318",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一列高铁的平均速度是每小时 230 千米,这个速度可以写作( )。从北京到深圳大约需要 10 小时,距离约是( )千米。所用的数量关系是:( )",
+        "options":  [
+
+                    ],
+        "answer":  "230 千米/时;2300;速度×时间 = 路程",
+        "solution":  "【分析】速度指单位时间内所行的路程,其单位的表示方法是\u0027所走的路程单位/时间单位\u0027,一般用千米/时,米/分来表示。由题意得,一列高铁的平均速度是每小时 230 千米,这个速度写作:230千米/时。从北京到深圳大约需要10小时,那么直接用速度乘时间即可算出路程.\n【详解】 230×10 = 2300 (千米)一列高铁的平均速度是每小时230千米,这个速度可以写作230千米/时。从北京到深圳大约需要10小时,距离约是2300千米。所用的数量关系是:速度×时间 = 路程。",
+        "knowledge_points":  [
+                                 "三位数与整十数的口算乘法",
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25四年级上·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "319",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\u0027>\u0027\u0027<\u0027或\u0027=\u0027。\n160×60( )600×16;57万( )5700000;2直角( )1周角。",
+        "options":  [
+
+                    ],
+        "answer":  "=;\u003c;\u003c",
+        "solution":  "【分析】(1)先求出算式的结果,再比较大小;(2)先将57万写成570000,再进行比较;(3)1个直角 = 90°,2个直角 = 180°,1个周角 = 360°,据此比较。\n【详解】160×60 = 600×16 = 9600,所以160×60 = 600×16;57万 = 570000,570000＜5700000,所以57万＜5700000;2直角 = 180°,1周角 = 360°,180°＜360°,所以2直角＜1周角。",
+        "knowledge_points":  [
+                                 "亿以内数的改写",
+                                 "直角,钝角,锐角的认识及特征",
+                                 "三位数乘两位数,乘数末尾有 0",
+                                 "大数的比较"
+                             ],
+        "source":  "24-25 四年级上·山东临沂·期末"
+    },
+    {
+        "id":  "320",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小华步行的速度是70 米/分,他 1 小时走( )米。",
+        "options":  [
+
+                    ],
+        "answer":  "4200",
+        "solution":  "【分析】1小时 = 60分,根据速度×时间 = 路程,用70×60即可求出他1小时走多少米。\n【详解】70×60 = 4200(米)\n小华步行的速度是70米/分,他1小时走4200米。",
+        "knowledge_points":  [
+                                 "时,分的认识",
+                                 "基础行程问题",
+                                 "两位数乘整十数的口算乘法"
+                             ],
+        "source":  "24-25 四年级上·天津北辰·期末"
+    },
+    {
+        "id":  "321",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明每天早上跑步 12 分钟,他的速度是 180 米/分,他每天跑步( )米。",
+        "options":  [
+
+                    ],
+        "answer":  "2160",
+        "solution":  "【分析】小明每天早上跑步12分钟,他的速度是180米/分,他每天跑步的距离 = 跑步时间×跑步速度。\n【详解】12×180 = 2160(米)\n因此小明每天跑步2160米。",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "三位数乘两位数,乘数末尾有 0"
+                             ],
+        "source":  "23-24 四年级下·甘肃平凉·开学考试"
+    },
+    {
+        "id":  "322",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一辆汽车的速度是80 千米/时,11 小时可行驶( )千米;一只猎豹 1 秒钟可跑 32 米,它奔跑的速度可以写成( )。",
+        "options":  [
+
+                    ],
+        "answer":  "880;32 米/秒",
+        "solution":  "【分析】路程 = 速度×时间,依此计算出这辆汽车11小时行驶的路程;每小时或每秒行驶的路程,叫做速度;速度是单位时间内通过的路程,由长度单位和时间单位组成,长度单位与时间单位之间用\u0027/\u0027分开,依此计算并填空.\n【详解】 80×11 = 880 (千米)辆汽车的速度是 80 千米/时,11 小时可行驶 880 千米;一只猎豹1秒钟可跑32米,它奔跑的速度可以写成32米/秒。",
+        "knowledge_points":  [
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25 四年级上·广东东莞·期末"
+    },
+    {
+        "id":  "323",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "安徽黄山,古称黟山,山境南北长约40千米,东西宽约30千米。黄山总占地面积约( )平方千米,合( )公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "1200;120000",
+        "solution":  "【分析】根据题意可知,已知的是一个长方形的长和宽,根据长方形的面积 = 长×宽,代入数据可以求出面积;1平方千米 = 100 公顷,根据进率转换单位即可;把高级单位的名数换算成低级单位的名数,就乘单位间的进率,反之,则除以进率;据此解答.\n【详解】根据分析:30×40 = 1200 (平方千米)1200×100 = 120000 (公顷)所以黄山总占地面积约1200平方千米,合120000公顷。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "长方形的面积",
+                                 "公顷,平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·广东东莞·期末"
+    },
+    {
+        "id":  "324",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "四年级有一块长方形的学农基地,面积200平方米,宽8米,现在把基地扩大,长不变,宽增加到24米,那么扩大后的基地面积比原来多( )平方米。",
+        "options":  [
+
+                    ],
+        "answer":  "400",
+        "solution":  "【分析】长方形的面积 = 长×宽,宽由8米增加到24米,长不变,根据积的变化规律可知,现在的宽是原来宽的3倍,现在的面积就是原来面积的3倍,用现在的面积减原来的面积,就是扩大后的基地面积比原来多的面积。\n【详解】200×(24÷8) - 200 = 200×3 - 200 = 600 - 200 = 400(平方米)\n扩大后的基地面积比原来多400平方米。",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "24-25 四年级上·湖南张家界·期末"
+    },
+    {
+        "id":  "325",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\u0027>\u0027\u0027<\u0027或\u0027=\u0027。\n160×60( )600×16;57万( )5700000;2直角( )1周角。",
+        "options":  [
+
+                    ],
+        "answer":  "=;\u003c;\u003c",
+        "solution":  "【分析】(1)根据积的变化规律,如果一个因数扩大到原来的几倍,另一个因数缩小到原来的几分之一(0除外),积不变;(2)把57万改写成整数,再进行比较即可;(3)根据周角 = 360°,直角 = 90°,计算出2直角的度数,再进行比较即可。\n【详解】160×60 = 600×16;57万 = 570000,570000＜5700000,所以57万＜5700000;2直角 = 2×90° = 180°,180°＜360°,所以2直角＜1周角。",
+        "knowledge_points":  [
+                                 "直角,钝角,锐角的认识及特征",
+                                 "平角,周角的认识及特征",
+                                 "积的变化规律(整数乘法)",
+                                 "亿以内数的改写"
+                             ],
+        "source":  "24-25 四年级上·山东临沂·期末"
+    },
+    {
+        "id":  "326",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个乘法算式的积是 40,一个因数乘 6,另一个因数不变,积是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "240",
+        "solution":  "【分析】根据积的变化规律,一个因数不变,另一个因数乘几或除以几(0除外),积也乘或除以几。据此解答。\n【详解】40×6 = 240\n一个乘法算式的积是40,一个因数乘6,另一个因数不变,积是240。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "24-25 四年级上·天津河北·期末"
+    },
+    {
+        "id":  "327",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "已知 A×B = 120 ,如果 A 不变,B 乘 3,则积就变成( );如果 B 除以 2,A乘 2,此时的积是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "360;120",
+        "solution":  "【分析】两数相乘,一个因数不变,另一个因数乘或除以几(0除外),积也同时乘或除以这个数;一个因数乘几,另一个因数同时除以相同的数,积不变。\n【详解】A×B = 120。如果A不变,B乘3,积也乘3,120×3 = 360,所以积变成360。如果B除以2,A乘2,则积不变,积还是120。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "24-25 四年级上·天津滨海新·期末"
+    },
+    {
+        "id":  "328",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个人骑自行车的速度是 250 米/分,2 分钟可以行( )米。",
+        "options":  [
+
+                    ],
+        "answer":  "500",
+        "solution":  "【分析】根据路程 = 速度×时间,用骑自行车的速度乘2,即可求出2分钟可以行几米.\n【详解】 250×2 = 500 (米)即一个人骑自行车的速度是 250 米/分,2 分钟可以行 500 米。",
+        "knowledge_points":  [
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25 四年级上·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "329",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "甲数×乙数 = 48,甲数不变,乙数乘3,积是( );甲数乘4,乙数除以4,积是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "144;48",
+        "solution":  "【分析】积的变化规律:一个乘数不变,另一个乘数乘或除以几(0除外),得到的积就等于原来的积乘或除以几。甲数不变,乙数乘3,那么积也乘3;甲数乘4,乙数除以4,积不变。\n【详解】48×3 = 144\n甲数×乙数 = 48,甲数不变,乙数乘3,积是144;甲数乘4,乙数除以4,积是48。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "24-25 四年级上·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "330",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据 36×8 = 288 写出下面两题的积.\n360×8 = ( ) 36×800 = ( )",
+        "options":  [
+
+                    ],
+        "answer":  "2880;28800",
+        "solution":  "【分析】一个乘数不变,另一个乘数乘几或除以几(0除外),积也乘几或除以几(0除外)。\n【详解】360×8 = (36×10)×8 = 288×10 = 2880\n36×800 = 36×(8×100) = 288×100 = 28800\n根据36×8 = 288,可得上面两题的积分别是2880和28800。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "24-25四年级上·广东东莞·期末"
+    },
+    {
+        "id":  "331",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "电动车速度不超过 25 千米/小时,爷爷 10 分钟骑了 4 千米。他的速度是( )千米/时。",
+        "options":  [
+
+                    ],
+        "answer":  "24",
+        "solution":  "【分析】根据速度 = 路程/时间,1小时 = 60 分钟,已知爷爷骑了10分钟行驶4千米,且本题速度的单位是千米/时,故要先把分钟换算为小时;若爷爷骑了60分钟(1小时)也就是6个4千米,行驶的路程为: 4×6 = 24 (千米),再根据路程公式,代入数据计算即可解答.\n【详解】根据分析可得:速度 = 24/1 = 24 (千米/时)因此,他的速度是24千米/时。",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "时,分的认识"
+                             ],
+        "source":  "24-25 四年级上·浙江温州·期末"
+    },
+    {
+        "id":  "332",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上\u0027>\u0027\u0027<\u0027或\u0027=\u0027。\n7万( )69000;106×70( )160×70;520×30( )52×300;8平方千米( )80公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;\u003c;=;\u003e",
+        "solution":  "【分析】(1)把7万改写成整数再比较;(2)根据整数乘法计算方法,分别求出各算式的乘积,再比较大小;(3)在乘法中,一个因数扩大到原来的10倍,另一个因数缩小到原来的1/10,积不变;(4)根据1平方千米 = 100公顷,将左边的单位换算成公顷再比较。\n【详解】7万 = 70000,70000＞69000,因此7万＞69000;106×70 = 7420,160×70 = 11200,7420＜11200,因此106×70＜160×70;520×30 = 52×300;8平方千米 = 800公顷,800公顷＞80公顷,因此8平方千米＞80公顷。",
+        "knowledge_points":  [
+                                 "亿以内数的改写",
+                                 "积的变化规律(整数乘法)",
+                                 "公顷,平方千米的进率与换算",
+                                 "三位数乘两位数,三位数中间有 0"
+                             ],
+        "source":  "24-25 四年级上·浙江温州·期末"
+    },
+    {
+        "id":  "333",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "比较大小。\n981600( )2000050;108×35( )54×70;680×5( )580×6;478600( )45900;80058米( )95千米;45个千( )500个百。",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;=;\u003c;\u003e;\u003c;\u003c",
+        "solution":  "【分析】多位数的大小比较,数位不同时,位数多的就大;数位相同时,从最高位比起,依次比较。两个数相乘,一个乘数扩大到原来的几倍(0除外),另一个乘数缩小到原来的几分之一,积不变;也可以直接算出结果再比较。根据1千米 = 1000米,把95千米转化成95000米再比较。1个千是1000,1个百是100,据此比较。\n【详解】981600是六位数,2000050是七位数,所以981600＜2000050;108变成54,缩小到原来的一半,35变成70,扩大到原来的2倍,积不变,所以108×35 = 54×70;680×5 = 3400,580×6 = 3480,所以680×5＜580×6;478600＞45900;95千米 = 95000米,80058米＜95000米,所以80058米＜95千米;45个千是45000,500个百是50000,45000＜50000,所以45个千＜500个百。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "三位数乘两位数,三位数中间有0",
+                                 "千米和米之间的进率与换算",
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "24-25 四年级上·浙江杭州·期末"
+    },
+    {
+        "id":  "334",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一块长方形草地宽5米,占地面积78平方米。若长不变,宽增加到20米,那么扩大后这块长方形草地占地面积是( )平方米。",
+        "options":  [
+
+                    ],
+        "answer":  "312",
+        "solution":  "【分析】根据积的变化规律:一个因数不变,另一个因数乘几(0除外),积也乘几。这里长不变就相当于一个因数不变。宽增加到20米,与原来宽相比扩大到4倍,那么面积也应该扩大到原来的4倍。\n【详解】20÷5 = 4\n4×78 = 312(平方米)\n所以,扩大后这块长方形草地占地面积是312平方米。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)",
+                                 "长方形的面积",
+                                 "两,三位数与一位数连续进位的乘法"
+                             ],
+        "source":  "24-25 四年级上·浙江温州·期末"
+    },
+    {
+        "id":  "335",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "\u0027复兴号\u0027是中国自主研发,具有完全知识产权的新一代高速列车,它每小时能行驶350千米,这个速度可以简写成( );按这个速度,若乘坐\u0027复兴号\u0027从甲地到乙地需要2个小时,则甲,乙两地相距( )千米。",
+        "options":  [
+
+                    ],
+        "answer":  "350 千米/时;700",
+        "solution":  "【分析】速度的写法:先写单位时间内所行的路程,然后画一斜线,在斜线后写上时间单位;速度×时间 = 路程,用\u0027复兴号\u0027的速度乘2,可以算出2小时行驶的路程,也就是甲、乙两地的距离。\n【详解】350×2 = 700(千米)\n所以它每小时能行驶350千米,这个速度可以简写成350千米/时;按这个速度,若乘坐\u0027复兴号\u0027从甲地到乙地需要2个小时,则甲、乙两地相距700千米。",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "因数中间或末尾是 0 的乘法"
+                             ],
+        "source":  "24-25 四年级上·浙江温州·期末"
+    },
+    {
+        "id":  "336",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "文文做一道乘法题,他把其中一个因数 21 错看成了 12,结果得到的积比正确的积少了 180,正确的积是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "420",
+        "solution":  "【分析】一个因数21错看成12,少了9,结果得到的积比正确的积少了180,所以另一个因数是180÷9。再用这个因数乘21,即可求出正确的积。\n【详解】21 - 12 = 9\n180÷9 = 20\n20×21 = 420\n所以正确的积是420。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)",
+                                 "两位数乘两位数的不进位乘法",
+                                 "商末尾有0的除法"
+                             ],
+        "source":  "24-25 四年级上·浙江杭州·期末"
+    },
+    {
+        "id":  "337",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "杭州地铁 3 小时行驶 240 千米,地铁的速度可以写作( ),小明的步行速度是每分钟约 109 米他从家到学校走了 21 分钟,他家到学校大约( )千米。",
+        "options":  [
+
+                    ],
+        "answer":  "80 千米/时;2200",
+        "solution":  "【分析】根据路程÷时间 = 速度,用240÷3求出每小时行驶的速度。根据速度×时间 = 路程,用109×21即可求出他家到学校大约多少米,计算时把109看成110,21看成20进行估算即可。\n【详解】240÷3 = 80(千米/时)\n109×21≈110×20 = 2200(米)\n杭州地铁3小时行驶240千米,地铁的速度可以写作80千米/时;小明从家到学校走了21分钟,他家到学校大约2200米。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "三位数乘两位数的估算",
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25 四年级上·浙江杭州·期末"
+    },
+    {
+        "id":  "338",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "猎豹是陆地上跑得最快的动物,每分钟可以跑 990 米,它的速度记作( ),照这个速度,猎豹奔跑 15 分钟,跑过的路程是( )米。",
+        "options":  [
+
+                    ],
+        "answer":  "990 米/分;14850",
+        "solution":  "【分析】速度的简记法:先写上米(或千米等)数,再写上\u0027/分\u0027或\u0027/时\u0027等即可;根据路程 = 速度×时间,用猎豹每分钟跑的距离乘跑的时间,即可求出跑过的路程。\n【详解】990×15 = 14850(米)\n猎豹每分钟可以跑990米,它的速度记作990米/分,照这个速度,猎豹奔跑15分钟,跑过的路程是14850米。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25四年级上·湖北黄冈·期末"
+    },
+    {
+        "id":  "339",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\u0027>\u0027\u0027<\u0027或\u0027=\u0027。\n7000公顷( )7平方千米;98562145( )98562154;530×35( )350×53。",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;\u003c;=",
+        "solution":  "【分析】1平方千米 = 100公顷,先把7000公顷换算成平方千米再比较;根据大数的比较方法比较两个整数;根据三位数乘两位数的计算,分别计算出左右算式的结果再比较。\n【详解】7000公顷 = 70平方千米,70＞7,所以7000公顷＞7平方千米;98562145和98562154都是八位数,前面各位都相同,十位上4＜5,所以98562145＜98562154;530×35 = 18550,350×53 = 18550,所以530×35 = 350×53。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "公顷,平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·湖北黄冈·期末"
+    },
+    {
+        "id":  "340",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "已知 A×B = 180 ,如果 B 不变,A 乘 2,则积是( );如果 A 不变,B 除以5,则积是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "360;36",
+        "solution":  "【分析】根据积的变化规律,一个因数不变,另一个因数乘或除以一个不为0的数,积也要乘或除以这个数,据此填空即可。\n【详解】A×B = 180\n(A×2)×B = 180×2 = 360\nA×(B÷5) = 180÷5 = 36\n已知A×B = 180,如果B不变,A乘2,则积是360;如果A不变,B除以5,则积是36。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "24-25 四年级上·湖北孝感·期末"
+    },
+    {
+        "id":  "341",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "我国关于电瓶车的道路交通安全要求中规定:电动自行车的最高时速不得超越每小时 25 千米,这个速度可以记作( ),读作( )。",
+        "options":  [
+
+                    ],
+        "answer":  "25 千米/时;25 千米每时",
+        "solution":  "【分析】速度的写法:先写单位时间内所行的路程,然后画一斜线,在斜线后写上时间单位;斜线表示\u0027每\u0027,从左往右读出即可。\n【详解】电动自行车的最高时速不得超过每小时25千米,这个速度可以记作25千米/时,读作25千米每时。",
+        "knowledge_points":  [
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25四年级上·湖北武汉·期末"
+    },
+    {
+        "id":  "342",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用0,1,2,4,5组成一个三位数乘两位数的算式(每个数字只能用一次),使它们的乘积等于20600,这两个因数分别是( )和()。",
+        "options":  [
+
+                    ],
+        "answer":  "412;50",
+        "solution":  "【分析】根据题意可知,用0、1、2、4、5组成一个三位数乘两位数的算式且积是20600。积的最高位是2,只有4×5的积最高位是2,所以4和5适合作为这两个因数的最高位。积的末尾有2个0,其中2和5相乘积的末尾是0,所以可以组成的算式是412×50。\n【详解】412×50 = 20600。\n用0、1、2、4、5组成一个三位数乘两位数的算式(每个数字只能用一次),使它们的乘积等于20600,这两个因数分别是412和50。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有 0"
+                             ],
+        "source":  "24-25四年级上·湖北武汉·期末"
+    },
+    {
+        "id":  "343",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用0,3,6,7,8组成三位数乘两位数的乘法算式(每个数字不能重复使用),积最小的算式是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "678×30",
+        "solution":  "【分析】要使积最小,先不考虑最大的数字8,剩下的四个数字 0,3,6,7 组成的两位数乘两位数十位尽量放小数字,0不能放十位,十位可以分别放3和6,然后个位分别放0和7,这时可能组成 30×67 或 37×60 ,计算比较 30×67 最小,再看 8 跟在哪个两位数后面,通过计算可知.\n【详解】由分析可知剩下的四个数字0,3,6,7组成的两位数乘两位数中 30×67 = 2010 , 37×60 = 2220 ,比较 2010 \u003c 2220,可知 30×67 是0,3,6,7组成的两位数乘两位数中最小的积;再看8跟在哪个两位数后面,通过计算 308×67 = 20636 , 30×678 = 20340 ,比较 20340 \u003c20636 ,可知积最小的算式是 678×30 。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0"
+                             ],
+        "source":  "24-25四年级上·湖北黄冈·期末"
+    },
+    {
+        "id":  "344",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "两个因数的积是 72,如果一个因数不变,另一个因数乘 5,那么积是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "360",
+        "solution":  "【分析】根据积的变化规律,一个因数不变,另一个因数乘或除以一个不为0的数,积也要乘或除以这个数,据此填空即可。\n【详解】72×5 = 360\n两个因数的积是72,如果一个因数不变,另一个因数乘5,那么积是360。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "24-25 四年级上·湖北黄冈·期末"
+    },
+    {
+        "id":  "345",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一块长方形绿地宽 8米,面积是 320 平方米。若长不变,宽增加 16 米,则面积增加( )平方米。扩大后的绿地面积是( )平方米。",
+        "options":  [
+
+                    ],
+        "answer":  "640;960",
+        "solution":  "【分析】长方形的面积 = 长×宽,长不变,增加的宽16米是原来宽8米的2倍,则增加的面积就是原来面积的2倍,原来的面积加上增加的面积等于扩大后的绿地面积。\n【详解】320×(16÷8) = 320×2 = 640(平方米)\n640 + 320 = 960(平方米)\n一块长方形绿地宽8米,面积是320平方米。若长不变,宽增加16米,则面积增加640平方米,扩大后的绿地面积是960平方米。",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "24-25 四年级上·湖北襄阳·期末"
+    },
+    {
+        "id":  "346",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据 425×2 = 850 ,得到 4250×2 = ( ) 。",
+        "options":  [
+
+                    ],
+        "answer":  "8500",
+        "solution":  "【分析】根据积的变化规律,一个因数不变,另一个因数乘几(0除外),积也乘几。425×10 = 4250,另一个因数2不变,所以4250×2的乘积等于425×2的乘积乘10。\n【详解】425×10 = 4250\n850×10 = 8500\n所以4250×2 = 8500。",
+        "knowledge_points":  [
+                                 "三位数与整十数的口算乘法",
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "24-25 三年级上·湖北武汉·期末"
+    },
+    {
+        "id":  "347",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算306×□0,要使积是四位数,□里最大填( );要使积是五位数,□里最小填( )。",
+        "options":  [
+
+                    ],
+        "answer":  "3;4",
+        "solution":  "【分析】306×30 = 9180,306×40 = 12240,所以计算306×□0,要使积是四位数,□里最大填3;要使积是五位数,□里最小填4。\n【详解】根据分析可知,计算306×□0,要使积是四位数,□里最大填3;要使积是五位数,□里最小填4。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "三位数乘两位数,三位数中间有0"
+                             ],
+        "source":  "24-25 四年级上·湖北武汉·期末"
+    },
+    {
+        "id":  "348",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "用竖式计算 160×23 时,6 与 2 相乘,实际上是( )乘( )。",
+        "options":  [
+
+                    ],
+        "answer":  "60;20",
+        "solution":  "【分析】三位数乘两位数时,相同数位对齐,从个位乘起。160中的6在十位上,表示60;23中的2在十位上,表示20,据此填空即可。\n【详解】6是160十位上的6,代表60;2是23十位上的2,代表20,所以6与2相乘,实际上是60乘20。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有 0"
+                             ],
+        "source":  "24-25 四年级上·辽宁抚顺·期末"
+    },
+    {
+        "id":  "349",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一列高速列车的行驶速度是 350 千米/时,3 小时可行驶( )千米;李老师买 9个篮球用了 540 元,篮球的价格是( )元/个。",
+        "options":  [
+
+                    ],
+        "answer":  "1050;60",
+        "solution":  "【分析】根据路程 = 速度×时间,用列车行驶的速度乘列车行驶的时间,计算出列车行驶的路程;因为总价 = 单价×数量,所以单价 = 总价÷数量,用篮球的总价除以篮球的数量,计算出篮球的单价。\n【详解】350×3 = 1050(千米)\n540÷9 = 60(元)\n一列高速列车的行驶速度是350千米/时,3小时可行驶1050千米;李老师买9个篮球用了540元,篮球的价格是60元/个。",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "几百几十或几千几百除以一位数的除法",
+                                 "两,三位数与一位数连续进位的乘法",
+                                 "经济问题"
+                             ],
+        "source":  "24-25四年级上·江西吉安·期末"
+    },
+    {
+        "id":  "350",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\u0027>\u0027\u0027<\u0027或\u0027=\u0027。\n185746000( )2亿;120×20( )12×200;5平方千米( )50公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;=;\u003e",
+        "solution":  "【分析】先把2亿改写成整数后再比较;分别计算左右两个乘法算式的结果再比较;1平方千米 = 100公顷,把5平方千米换算成公顷后再比较。\n【详解】2亿 = 200000000,185746000和200000000都是九位数,亿位上1＜2,所以185746000＜2亿;120×20 = 2400,12×200 = 2400,所以120×20 = 12×200;5平方千米 = 500公顷,500＞50,所以5平方千米＞50公顷。",
+        "knowledge_points":  [
+                                 "亿以上数的改写",
+                                 "公顷,平方千米的进率与换算",
+                                 "三位数乘两位数,乘数末尾有 0",
+                                 "大数的比较"
+                             ],
+        "source":  "24-25 四年级上·江西吉安·期末"
+    },
+    {
+        "id":  "351",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "李叔叔骑自行车出行,8 分钟行驶了 2000 米,他的速度可以写成( ),用到的数量关系式是:速度=( )。",
+        "options":  [
+
+                    ],
+        "answer":  "250 米/分;速度 = 路程/时间",
+        "solution":  "【分析】每小时(或每分钟或每秒钟)行的路程,叫做速度;速度是单位时间内通过的路程,由长度单位和时间单位组成,长度单位与时间单位之间用\u0027/\u0027分开。一共行了多长的路,叫做路程;每小时(或每分钟等)行的路程,叫做速度;行了几小时(或几分钟等),叫做时间;每小时行驶的路程 = 行驶的总路程/所用的时间,依此解答.\n【详解】 2000/8 = 250 (米/分)李叔叔骑自行车出行,8 分钟行驶了 2000 米,他的速度可以写成250米/分,用到的数量关系式是:速度 = 路程/时间。",
+        "knowledge_points":  [
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25 四年级上·浙江宁波·期末"
+    },
+    {
+        "id":  "352",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "发簪是中国古代汉族人民用来固定和装饰头发的一种首饰。王老师步行 5 分钟,来到离家 300 米的吴都乔街内一家国风饰品店里购进 20 件发簪,准备发放给舞蹈团成员,每件 98 元,她的步行平均速度记作( ),一共大约花了( )元。",
+        "options":  [
+
+                    ],
+        "answer":  "60 米/分;2000",
+        "solution":  "【分析】距离是 300 米,时间是 5 分钟,用 300 除以 5,即可求出步行平均速度;买了 20 件发簪,每件 98 元最接近 100 元,估算就用 20 乘100 即可.\n【详解】 300/5 = 60 (米/分) 20×98 approx 20×100 = 2000 (元) 她的步行平均速度记作 60 米/分;一共大约花了 2000 元。",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "经济问题"
+                             ],
+        "source":  "24-25 四年级上·湖北鄂州·期末"
+    },
+    {
+        "id":  "353",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "20 □* 73 的积是()位数,如果积的个位上是 8,那么积是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "五;15038",
+        "solution":  "【分析】把20□看成与它接近的整百数200,用整百数乘73,算出的结果是几位数,积就是几位数。三位数乘两位数的竖式计算时,个位对齐。先用两位数个位上的数和三位数的每一位上的数相乘,乘得结果和个位对齐,再用两位数十位上的数和三位数的每一位上的数相乘,乘得结果和十位对齐。最后把两次乘得结果相加。末尾有0时,先将0不看,算出结果,再在积的后面添上没参加运算的几个0。中间有0时,这个0要参加运算。积的个位上是三位数的个位和两位数的的个位相乘得来的。积的个位上是8,就想□ * 3 的末尾是8,则□应该填6,据此解答即可.\n【详解】20□最接近200, 200×73 = 14600 ,积是五位数,所以 20 □* 73 的积是五位数。如果积的个位上是8,说明□ * 3 的末尾是 8,□应该填 6,所以是 206×73 = 15038 .\nnn 206 \nn×73 \nnhlinen 618 \nn1442phantom0 \nnhlinen15038nn",
+        "knowledge_points":  [
+                                 "三位数乘两位数,三位数中间有0",
+                                 "三位数乘两位数,乘数末尾有0"
+                             ],
+        "source":  "24-25 四年级上·山东菏泽·期末"
+    },
+    {
+        "id":  "354",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "一列高铁长200米,以 74 米/秒的速度通过一条隧道,从车头进隧道到车尾离隧道,一共用了 12 秒,这条隧道长 888 米。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】路程 = 速度×时间。由题意可知,从车头进隧道到车尾离开隧道,高铁行驶的总路程等于隧道长加高铁长。先用速度乘时间求出总路程,再减去高铁长即可求出隧道长。\n【详解】74×12 = 888(米)\n888 - 200 = 688(米)\n所以这条隧道长688米,而不是888米,原题说法错误。故答案为:×。",
+        "knowledge_points":  [
+                                 "两位数乘两位数的进位乘法",
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25四年级上·湖南张家界·期末"
+    },
+    {
+        "id":  "355",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "计算 465×32 与 465×2 + 465×30 的结果相等。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】根据题意,分别将两个算式的结果计算出来,再进行比较,即可得出答案。\n【详解】465×32 = 14880\n465×2 + 465×30 = 930 + 13950 = 14880\n所以465×32与465×2 + 465×30的结果相等。故答案为:√。",
+        "knowledge_points":  [
+                                 "乘加,乘减混合运算",
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "24-25 四年级上·天津滨海新·期末"
+    },
+    {
+        "id":  "356",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "小轿车每小时行 80 千米,即速度为 80 千米。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】速度的表示方法是\u0027距离/时间\u0027,小轿车每小时行80千米,可以改写成80千米/时。\n【详解】小轿车每小时行80千米,即速度为80千米/时。原题表述错误。故答案为:×。",
+        "knowledge_points":  [
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25四年级上·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "357",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "李老师家离学校3千米,他骑车的速度是 190 米/分钟,他从家骑车 15 分钟可以到校。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】根据路程 = 速度×时间,用李老师骑车的速度乘骑车的时间,求出一共可以骑多少米。再把3千米换算成米,与骑行距离比较即可判断。\n【详解】190×15 = 2850(米)\n3千米 = 3000米\n2850米＜3000米\n所以他从家骑车15分钟不可以到校。原题说法错误。故答案为:×。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "基础行程问题",
+                                 "千米和米之间的进率与换算"
+                             ],
+        "source":  "24-25四年级上·湖北孝感·期末"
+    },
+    {
+        "id":  "358",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "250×80 的积的末尾只有两个 0。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】根据三位数乘两位数的计算,计算出250×80的结果,据此判断积的末尾有几个0即可。\n【详解】250×80 = 20000\n250×80的积的末尾有4个0,原题说法错误。故答案为:×。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0"
+                             ],
+        "source":  "24-25 四年级上·湖北孝感·期末"
+    },
+    {
+        "id":  "359",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "因为 A×33 =B×44 (A,B 均大于 0),所以 A \u003e B 。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】A×33 = B×44,且A、B都大于0。因为33＜44,要使两个积相等,与33相乘的A就应大于与44相乘的B。\n【详解】33＜44,所以A＞B,原题说法正确。故答案为:√。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法",
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "24-25 四年级上·湖北武汉·期末"
+    },
+    {
+        "id":  "360",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "250×□0 是一个三位数乘两位数的算式,积的末尾最多有4个0。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】250×□0是一个三位数乘两位数的算式。因为5乘2、4、6、8这些数的积个位是0,所以分别计算250×20、250×40、250×60、250×80,看看积的末尾最多有几个0。\n【详解】250×20 = 5000,250×40 = 10000,250×60 = 15000,250×80 = 20000。这几个积中,250×40和250×80末尾最多有4个0。故答案为:√。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0"
+                             ],
+        "source":  "24-25 四年级上·湖北鄂州·期末"
+    },
+    {
+        "id":  "361",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "列竖式计算。\n246×57 = ;320×61 = ;460×50 = ;36×805 =",
+        "options":  [
+
+                    ],
+        "answer":  "14022;19520;23000;28980",
+        "solution":  "【分析】三位数乘两位数的笔算法则:先用两位数的个位分别与三位数的每一位数相乘,再用两位数的十位分别与三位数的每一位数相乘,最后把两次乘得的积相加。\n【详解】246×57 = 14022\n320×61 = 19520\n460×50 = 23000\n36×805 = 28980",
+        "knowledge_points":  [
+                                 "三位数乘两位数,三位数中间有0",
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "25-26 四年级上·江西赣州·期中"
+    },
+    {
+        "id":  "362",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "计算题",
+        "difficulty":  0.85,
+        "question":  "直接写出得数。\n38×2 = ;32×30 = ;7.9 + 4.5 = ;62×19≈\n60×40 = ;450÷3 = ;106×50 = ;398×51≈",
+        "options":  [
+
+                    ],
+        "answer":  "76;960;12.4;1200;2400;150;5300;20000",
+        "solution":  "【详解】38×2 = 76\n32×30 = 960\n7.9 + 4.5 = 12.4\n62×19≈60×20 = 1200\n60×40 = 2400\n450÷3 = 150\n106×50 = 5300\n398×51≈400×50 = 20000",
+        "knowledge_points":  [
+                                 "一位小数的进位加法,退位减法",
+                                 "三位数乘两位数的估算",
+                                 "两位数乘两位数的估算",
+                                 "几百几十或几千几百除以一位数的除法"
+                             ],
+        "source":  "25-26 四年级上·江西赣州·期中"
+    },
+    {
+        "id":  "363",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "180 的 63 倍是多少。",
+        "options":  [
+
+                    ],
+        "answer":  "11340",
+        "solution":  "【分析】要求一个数的几倍是多少,用这个数乘倍数即可,因此要求180的63倍是多少,用180乘63即可。\n【详解】180×63 = 11340\n180的63倍是11340。",
+        "knowledge_points":  [
+                                 "求一个数的几倍是多少",
+                                 "三位数乘两位数,乘数末尾有0"
+                             ],
+        "source":  "23-24 四年级下·甘肃平凉·开学考试"
+    },
+    {
+        "id":  "364",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "列式计算。405 乘 13 加 15 的和,积是多少。",
+        "options":  [
+
+                    ],
+        "answer":  "11340",
+        "solution":  "【分析】求积是多少,说明最后计算的是乘法。先计算出13加15的和,再用405乘这个和,注意将加法用小括号括起来。\n【详解】405×(13 + 15) = 405×28 = 11340\n所以积是11340。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,三位数中间有0",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25 四年级上·浙江杭州·期末"
+    },
+    {
+        "id":  "365",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "火锅已经成为成都的美食名片。春熙路某火锅店搞促销活动,限时套餐价 128 元/桌,中午卖出 16 桌,晚上又卖出 24 桌,这家店中午和晚上火锅套餐一共卖了多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "5120 元",
+        "solution":  "【分析】根据题意,用中午卖的桌数加上晚上卖的桌数,求出总桌数,再用每桌火锅套餐的限时价乘总桌数,即可求出一共卖了多少钱。\n【详解】128×(16 + 24)\n= 128×40\n= 5120(元)\n答:这家店中午和晚上火锅套餐一共卖了5120元。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "乘加,乘减混合运算",
+                                 "三位数乘两位数",
+                                 "经济问题"
+                             ],
+        "source":  "24-25 四年级上·四川成都·期末"
+    },
+    {
+        "id":  "366",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "烤鱼已经成为万州的美食名片。天生城文旅街区某烤鱼店搞促销活动,限时价 108 元/条,下午卖出 14 条烤鱼,晚上又卖出 21 条,这家店下午和晚上烤鱼一共卖了多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "3780 元",
+        "solution":  "【分析】根据题意,用下午卖的烤鱼条数加上晚上卖的烤鱼条数,求出总条数,再用每条烤鱼的限时价乘总条数,即可求出一共卖了多少钱。\n【详解】108×(14 + 21) = 108×35 = 3780(元)\n答:这家店下午和晚上烤鱼一共卖了3780元。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "乘加,乘减混合运算",
+                                 "三位数乘两位数,三位数中间有0",
+                                 "经济问题"
+                             ],
+        "source":  "24-25 四年级上·重庆万州·期末"
+    },
+    {
+        "id":  "367",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "妈妈在网上购买玩具车,套装包括一辆车和一组电池。单独买车每辆 198 元,单独买电池每组 10 元。她购买了 3 套套装,店家总共优惠了 15 元。妈妈实际一共花了多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "609 元",
+        "solution":  "【分析】先将一辆车和一组电池的价钱相加,求出一套玩具车的单价。因为妈妈买了3套,用单价乘数量即可求出总钱数,然后再减去总共优惠的15元。\n【详解】198 + 10 = 208(元)\n208×3 - 15 = 624 - 15 = 609(元)\n答:妈妈实际一共花了609元。",
+        "knowledge_points":  [
+                                 "乘加,乘减混合运算",
+                                 "经济问题"
+                             ],
+        "source":  "24-25 四年级上·天津河北·期末"
+    },
+    {
+        "id":  "368",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "果农李叔叔开车从果园往县城批发店送水果,去的时候速度是75千米/时,用了4小时,原路返回时用了3小时,返回时平均每小时行驶多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "100 千米/时",
+        "solution":  "【分析】根据路程 = 速度×时间,先用去时的速度乘所用时间,求出路程。再根据速度 = 路程÷时间,用路程除以返回时所用时间即可。\n【详解】75×4÷3 = 300÷3 = 100(千米/时)\n答:返回时平均每小时行驶100千米/时。",
+        "knowledge_points":  [
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25四年级上·天津滨海新·期末"
+    },
+    {
+        "id":  "369",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一辆汽车从甲城开往乙城,速度是每小时 60 千米,用了 12 小时到达。返回时只用了 9 小时。这辆汽车返回时的速度是多少。",
+        "options":  [
+
+                    ],
+        "answer":  "80 千米/时",
+        "solution":  "【分析】先用速度乘时间求出甲城到乙城的距离,再用路程除以返回时间即可求出返回时的速度。\n【详解】60×12÷9 = 720÷9 = 80(千米/时)\n答:这辆汽车返回时的速度是80千米/时。",
+        "knowledge_points":  [
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25 四年级上·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "370",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "星期天,爸爸开车带家人去七峰山游玩,使用导航软件查询路线,发现有两条路可以选择。路线一需要35分钟,路线二需要40分钟,导航软件默认汽车的速度是840米/分。若选择路线二,则开车要多行驶多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "4200 米",
+        "solution":  "【分析】用路线二的时间减去路线一的时间,求出路线二比路线一多用的时间,再乘汽车的速度,即可求出开车要多行驶多少米。\n【详解】(40 - 35)×840 = 5×840 = 4200(米)\n答:开车要多行驶4200米。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25 四年级上·河南南阳·期末"
+    },
+    {
+        "id":  "371",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "学校装饰校园,购置了一批盆花。每盆花30元,买5盆送一盆,一次购买24盆需要多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "600 元",
+        "solution":  "【分析】每盆花30元,买5盆送1盆,也就是买5盆实际得到6盆。先用24除以6算出需要买几个6盆,再算出每组需要的钱数,最后相乘即可。\n【详解】5 + 1 = 6(盆)\n24÷6 = 4(组)\n30×5 = 150(元)\n150×4 = 600(元)\n答:一次购买24盆花需要600元。",
+        "knowledge_points":  [
+                                 "因数中间或末尾是0的乘法",
+                                 "整十,整百,整千数与一位数的乘法",
+                                 "经济问题"
+                             ],
+        "source":  "24-25 四年级上·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "372",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "文具店搞促销活动,买3支钢笔送1支,每支钢笔25元。李老师带300元,最多能买多少支钢笔。",
+        "options":  [
+
+                    ],
+        "answer":  "16 支",
+        "solution":  "【分析】买3支送1支,即付3支的钱可以得到4支。先用300除以25算出能买多少支付费钢笔,再看能分成几个3支,就送几支,最后把买的和送的相加即可。\n【详解】300÷25 = 12(支)\n12÷3 = 4(组)\n12 + 4 = 16(支)\n答:最多能买16支钢笔。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "整十数除法",
+                                 "经济问题"
+                             ],
+        "source":  "24-25 四年级上·湖北武汉·期末"
+    },
+    {
+        "id":  "373",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "水果店运来 108 箱苹果和 96 箱梨,苹果每箱重 25 千克,梨每箱重 30 千克,请问这两种水果共重多少千克。",
+        "options":  [
+
+                    ],
+        "answer":  "5580 千克",
+        "solution":  "【分析】先用25乘108求出苹果的质量,再用30乘96求出梨的质量,最后相加即可。\n【详解】25×108 + 30×96 = 2700 + 2880 = 5580(千克)\n答:这两种水果共重5580千克。",
+        "knowledge_points":  [
+                                 "乘加,乘减混合运算",
+                                 "三位数与两位数的乘法",
+                                 "两位数乘两位数的进位乘法"
+                             ],
+        "source":  "24-25 四年级上·天津北辰·期末"
+    },
+    {
+        "id":  "374",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "陈叔叔开车从甲地送货去乙地,每小时行驶 65 千米,行驶了 11 个小时后(不考虑司机中途休息的时间),离乙地还有 30 千米。甲,乙两地相距多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "745 千米",
+        "solution":  "【分析】路程 = 速度×时间,先算出这辆车已经行驶了多少千米,再加上距乙地的距离,即可算出甲、乙两地相距多少千米。\n【详解】65×11 + 30 = 715 + 30 = 745(千米)\n答:甲、乙两地相距745千米。",
+        "knowledge_points":  [
+                                 "乘加,乘减混合运算",
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25 四年级上·广东东莞·期末"
+    },
+    {
+        "id":  "375",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "新华书店与出版社共同举办的\u0027阅读文化节\u0027活动在遗爱清风广场顺利开展。李老师骑共享单车去清风广场,去时他每分钟行 250 米,12分钟到达。李老师家距离清风广场多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "3000 米",
+        "solution":  "【分析】根据路程 = 速度×时间,用250×12即可求出李老师家距离清风广场多少米。\n【详解】250×12 = 3000(米)\n答:李老师家距离清风广场3000米。",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "三位数乘两位数,乘数末尾有0"
+                             ],
+        "source":  "24-25 四年级上·湖北黄冈·期末"
+    },
+    {
+        "id":  "376",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "学校准备给班级图书角配置一些新的图书,一共有18个班,每个班配120本图书,学校还要预留 50 本图书作为备用,学校应购买多少本图书。",
+        "options":  [
+
+                    ],
+        "answer":  "2210 本",
+        "solution":  "【分析】每班120本,18个班需要120×18本,然后再加上预留的50本即可。\n【详解】18×120 + 50 = 2160 + 50 = 2210(本)\n答:学校应购买2210本图书。",
+        "knowledge_points":  [
+                                 "乘加,乘减混合运算",
+                                 "三位数乘两位数,乘数末尾有 0"
+                             ],
+        "source":  "24-25 四年级上·广东广州·期末"
+    },
+    {
+        "id":  "377",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "九山书会南戏艺术空间小剧场有 40 个座位,南戏《凤冠》每张票价 299 元。王老师银行卡里有 5000 元,他抢购了 15 张票后,银行卡里余额还有多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "515 元",
+        "solution":  "【分析】先计算出王老师购买15张票所花费的总金额,再用5000元减去购票金额,就能得到银行卡里的余额。\n【详解】299×15 = 4485(元)\n5000 - 4485 = 515(元)\n答:银行卡里余额还有515元。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法",
+                                 "三位数乘两位数的实际问题"
+                             ],
+        "source":  "24-25 四年级上·浙江温州·期末"
+    },
+    {
+        "id":  "378",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "王老师散步,大约每分钟走125米。他绕着月湖的湖边走一圈共用了38分钟,月湖一周有多少米呢。",
+        "options":  [
+
+                    ],
+        "answer":  "4750 米",
+        "solution":  "【分析】根据速度×时间 = 路程,用王老师每分钟走的125米乘38分钟,即可求出月湖一周的长度。\n【详解】125×38 = 4750(米)\n答:月湖一周有4750米。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法",
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25四年级上·浙江·期末"
+    },
+    {
+        "id":  "379",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "一块长方形土地,长 600 米,宽 400 米,如果每公顷地可以收玉米 8 吨,那么这块地一共可以收玉米多少吨。",
+        "options":  [
+
+                    ],
+        "answer":  "192 吨",
+        "solution":  "【分析】先根据长方形的面积 = 长×宽,求出这块地的面积,再换算成公顷,最后乘每公顷的产量即可。\n【详解】600×400 = 240000(平方米)\n240000平方米 = 24公顷\n8×24 = 192(吨)\n答:这块地一共可以收玉米192吨。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "长方形的面积",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "24-25 四年级上·浙江杭州·期末"
+    },
+    {
+        "id":  "380",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "赏民宿:近几年,永嘉大力发展旅游业,民宿数量逐年增加。小永爸爸在山上开起了农家乐民宿,设有单人间和双人间各 10 间,单人间 158 元/间,双人间 242 元/间。在农家乐民宿满员的情况下,一晚上能收入多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "4000 元",
+        "solution":  "【分析】根据单价×数量 = 总价,分别用单人间和双人间的单价乘房间数量,再加起来即可。\n【详解】158×10 + 242×10 = 1580 + 2420 = 4000(元)\n答:一晚上能收入4000元。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "两,三位数的连续进位加法"
+                             ],
+        "source":  "24-25 四年级上·浙江温州·期末"
+    },
+    {
+        "id":  "381",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "塔克拉玛干沙漠是我国面积最大的沙漠,也是世界第二大流动沙漠。为了防风固沙,某志愿队要开垦 3 公顷的沙地植树造林。该志愿队现已经开垦出一块长 250 米,宽 80 米的长方形沙地,还要开垦多少公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "1 公顷",
+        "solution":  "【分析】先根据长方形的面积 = 长×宽,求出已经开垦出的沙地面积,再换算成公顷,最后用3公顷减去已开垦的面积即可。\n【详解】250×80 = 20000(平方米)\n20000平方米 = 2公顷\n3 - 2 = 1(公顷)\n答:还要开垦1公顷。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "长方形的面积",
+                                 "公顷,平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·浙江温州·期末"
+    },
+    {
+        "id":  "382",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "一条裤子原价180元,元旦期间商场举行优惠活动,买3条送1条,妈妈用原来3条裤子的钱买了4条裤子,每条裤子比原来便宜了多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "45 元",
+        "solution":  "【分析】先求出原来买3条裤子一共花多少钱,再用总钱数除以4,求出现在每条裤子多少钱,最后用原价减现价即可。\n【详解】180×3 = 540(元)\n540÷4 = 135(元)\n180 - 135 = 45(元)\n答:每条裤子比原来便宜45元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "两,三位数的连续退位减法",
+                                 "三位数除以一位数,首位除不尽",
+                                 "促销问题"
+                             ],
+        "source":  "24-25四年级上·浙江杭州·期末"
+    },
+    {
+        "id":  "383",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一艘游轮从A市到B 市,去时的速度是 30 千米/时,用了 12 小时,原路返回时顺流,所以时间少用了 3 小时,返回时的平均每小时行多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "40 千米/时",
+        "solution":  "【分析】先用速度×时间求出去时路程。因为往返路程相同,再用去时时间减去少用时间,算出返回时间,最后用路程÷时间求出返回速度。\n【详解】30×12 = 360(千米)\n12 - 3 = 9(小时)\n360÷9 = 40(千米/时)\n答:返回时的平均每小时行40千米/时。",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "两位数乘整十数的口算乘法",
+                                 "几百几十或几千几百除以一位数的除法"
+                             ],
+        "source":  "24-25四年级上·浙江杭州·期末"
+    },
+    {
+        "id":  "384",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "秋天到了,2 位教师带着 101 名学生外出秋游,每人需要车票和门票共 53 元,老师带了 5000 元钱够吗。",
+        "options":  [
+
+                    ],
+        "answer":  "不够",
+        "solution":  "【分析】101名学生和2位老师一共是103人,再用乘法求出需要多少钱,最后与5000元比较即可。\n【详解】(101 + 2)×53 = 103×53 = 5459(元)\n5459＞5000\n答:老师带了5000元不够。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,三位数中间有 0",
+                                 "经济问题"
+                             ],
+        "source":  "24-25 四年级上·浙江杭州·期末"
+    },
+    {
+        "id":  "385",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "绿鹰草莓君是一个综合性主题农场,农场内准备打造一座长280米,宽250米草莓生态园。若每公顷土地能产出大约 1500 千克的草莓,每千克草莓按 78 元计算,大约能卖多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "819000 元",
+        "solution":  "【分析】先根据长方形的面积 = 长×宽,计算出这块农场的面积,再换算成公顷。然后用公顷数乘每公顷产量,求出草莓总质量,最后根据单价×数量 = 总价,求出能卖多少钱。\n【详解】280×250 = 70000(平方米)\n70000平方米 = 7公顷\n1500×7 = 10500(千克)\n10500×78 = 819000(元)\n答:大约能卖819000元。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "长方形的面积",
+                                 "三位数乘两位数,乘数末尾有 0",
+                                 "经济问题"
+                             ],
+        "source":  "24-25四年级上·浙江·期末"
+    },
+    {
+        "id":  "386",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "某品牌钢笔的单价是14元/支,张阿姨要为单位采购106支,她准备了1200元够吗。",
+        "options":  [
+
+                    ],
+        "answer":  "不够",
+        "solution":  "【分析】根据总价 = 单价×数量,先求出一共需要付多少钱,再和准备的1200元比较即可。\n【详解】106×14 = 1484(元)\n1484元＞1200元\n答:她准备1200元不够。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,三位数中间有 0",
+                                 "经济问题"
+                             ],
+        "source":  "24-25 四年级上·湖北孝感·期末"
+    },
+    {
+        "id":  "387",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "王叔叔去水果批发市场买砂糖桔,他买了 28 箱,每箱砂糖桔 105 元。王叔叔应付多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "2940 元",
+        "solution":  "【分析】根据总价 = 单价×数量,用每箱砂糖桔的价格乘买的箱数,即可求出王叔叔应付多少钱。\n【详解】105×28 = 2940(元)\n答:王叔叔应付2940元。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,三位数中间有0",
+                                 "经济问题"
+                             ],
+        "source":  "24-25 四年级上·湖北襄阳·期末"
+    },
+    {
+        "id":  "388",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "老张家有一个长方形鱼塘。长 156 米,宽 64 米。小林家的鱼塘面积是 1 公顷,老张和小林谁家的鱼塘面积大。",
+        "options":  [
+
+                    ],
+        "answer":  "小林家",
+        "solution":  "【分析】先根据长方形面积 = 长×宽,计算出老张家鱼塘的面积,再把1公顷换算成10000平方米,据此比较即可。\n【详解】156×64 = 9984(平方米)\n1公顷 = 10000平方米\n9984平方米＜10000平方米\n答:小林家的鱼塘面积大。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "长方形的面积",
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "24-25四年级上·湖北襄阳·期末"
+    },
+    {
+        "id":  "389",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "《少儿百科》原价每套 150 元,\u0027双十一\u0027活动期间买 4 套送 1 套,5 个同学每人买一套,平均每套比原价便宜多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "30 元",
+        "solution":  "【分析】买4套送1套,实际得到5套。先用每套书的原价乘4,求出5套共付的钱数,再除以5,求出活动后每套的实际价格,最后用原价减去现价即可。\n【详解】4 + 1 = 5(套)\n150×4 = 600(元)\n600÷5 = 120(元)\n150 - 120 = 30(元)\n答:平均每套比原价便宜30元。",
+        "knowledge_points":  [
+                                 "经济问题"
+                             ],
+        "source":  "24-25 四年级上·湖北黄冈·期末"
+    },
+    {
+        "id":  "390",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "《寄小读者》每本 38 元,买 5 本送 1 本,学校要为四年级的 204 名同学每人买一本,最少要花多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "6460 元",
+        "solution":  "【分析】买5本送1本,则买6本只需要花5本的钱。用总人数除以6求出组数,再求出需要付多少本书的钱,最后乘每本书的价格即可。\n【详解】204÷(5 + 1) = 204÷6 = 34(组)\n34×5×38 = 170×38 = 6460(元)\n答:最少要花6460元。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "经济问题"
+                             ],
+        "source":  "24-25四年级上·湖北黄冈·期末"
+    },
+    {
+        "id":  "391",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "四年级学生折1200只千纸鹤布置\u0027庆元旦\u0027会场。已知有105名学生,每人折12只,这些学生折的千纸鹤数量够吗。",
+        "options":  [
+
+                    ],
+        "answer":  "够",
+        "solution":  "【分析】用每人折的只数乘学生人数,计算出这些学生共能够折多少只千纸鹤,再与1200进行比较即可。\n【详解】12×105 = 1260(只)\n1260＞1200\n答:这些学生折的千纸鹤数量够。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,三位数中间有 0"
+                             ],
+        "source":  "24-25四年级上·浙江宁波·期末"
+    },
+    {
+        "id":  "392",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "你道地球赤道有多长吗。假如某人绕地球赤道一圈,每天走 55 千米,大约两年(一年按365天计算)才能走完,算一算,地球赤道大约长多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "40150 千米",
+        "solution":  "【分析】先用55乘365计算出大约一年走多少千米,再乘2计算出两年大约走多少千米,也就是地球赤道大约的长度。\n【详解】55×365×2 = 20075×2 = 40150(千米)\n答:地球赤道大约长40150千米。",
+        "knowledge_points":  [
+                                 "用两步连乘解决实际问题",
+                                 "三位数乘两位数的实际问题",
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25 四年级上·湖北武汉·期末"
+    },
+    {
+        "id":  "393",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.94,
+        "question":  "武汉动物园的狮子每天大约要吃304千克的食物,11月份一个月大约要吃多少千克的食物。",
+        "options":  [
+
+                    ],
+        "answer":  "9120 千克",
+        "solution":  "【分析】11月份有30天,用每天吃的食物质量乘11月份的天数,即可求出11月份一个月要吃多少千克食物。\n【详解】304×30 = 9120(千克)\n答:11月份一个月大约要吃9120千克的食物。",
+        "knowledge_points":  [
+                                 "年,月,日的认识及换算",
+                                 "三位数乘两位数,乘数末尾有 0"
+                             ],
+        "source":  "24-25四年级上·湖北武汉·期末"
+    },
+    {
+        "id":  "394",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "上海大剧院是中国第一座国际性,现代化的综合剧院,主舞台面积 720 平方米,宽 16 米。如果想继续增强视听效果,将主舞台的宽度增加到32米,长不变,扩建后的主舞台面积达到多少平方米。",
+        "options":  [
+
+                    ],
+        "answer":  "1440 平方米",
+        "solution":  "【分析】主舞台的宽由16米增加到32米,长不变,宽扩大到原来的2倍。根据积的变化规律,面积也扩大到原来的2倍,即用720平方米乘2。\n【详解】(32÷16)×720 = 2×720 = 1440(平方米)\n答:扩建后的主舞台面积达到1440平方米。",
+        "knowledge_points":  [
+                                 "因数中间或末尾是0的乘法",
+                                 "长方形的面积",
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "24-25 四年级上·湖北黄冈·期末"
+    },
+    {
+        "id":  "395",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "人民公园为规范游客使用厕纸,使用\u0027人脸识别厕纸机\u0027,平均每人用纸量为 6 分米,比原来平均每人用纸量节约了 9 分米,原来一卷纸可供 60 人使用,现在同样一卷纸可供多少人使用。",
+        "options":  [
+
+                    ],
+        "answer":  "150 人",
+        "solution":  "【分析】现在平均每人用纸量6分米,比原来节约了9分米,所以原来平均每人用纸量为15分米。先算出一卷纸一共有多少分米,再除以现在每人的用纸量即可。\n【详解】(6 + 9)×60 = 15×60 = 900(分米)\n900÷6 = 150(人)\n答:现在同样一卷纸可供150人使用。",
+        "knowledge_points":  [
+                                 "归总问题",
+                                 "整十,整百,整千数除以一位数的除法",
+                                 "两位数乘两位数,乘数末尾有0",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25 四年级上·湖北武汉·期末"
+    },
+    {
+        "id":  "396",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "陈老师组织舞蹈社团的同学从学校乘坐客车前往剧院表演。客车每分钟行驶650米,行驶11分钟后,离中点还有500米。学校距剧院有多远。",
+        "options":  [
+
+                    ],
+        "answer":  "15300 米",
+        "solution":  "【分析】先算出客车已经行驶了多长的路程,再加上离中点的距离,就是学校到剧院的一半距离。然后再乘2即可求出学校到剧院的全程。\n【详解】650×11 + 500\n= 7150 + 500\n= 7650(米)\n7650×2 = 15300(米)\n答:学校距剧院有15300米。",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "乘加,乘减混合运算",
+                                 "三位数乘两位数,乘数末尾有 0"
+                             ],
+        "source":  "24-25四年级上·湖北武汉·期末"
+    },
+    {
+        "id":  "397",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "张叔叔从菜场出发去蔬菜基地批发一批货,去时的速度是 78 千米/时,用了 5小时,原路返回时多用了 1 小时。从菜场到蔬菜基地有多远。原路返回时的速度是多少。",
+        "options":  [
+
+                    ],
+        "answer":  "390 千米;65 千米/时",
+        "solution":  "【分析】先根据路程 = 速度×时间,求出从菜场到蔬菜基地的路程;再根据返回时比去时多用了1小时,求出返回时间;最后根据速度 = 路程÷时间,求出返回时的速度。\n【详解】78×5 = 390(千米)\n390÷(5 + 1) = 390÷6 = 65(千米/时)\n答:从菜场到蔬菜基地有390千米;原路返回时的速度是65千米/时。",
+        "knowledge_points":  [
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25 四年级上·江西吉安·期末"
+    },
+    {
+        "id":  "398",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "为了培养学生的实践能力,拓展学生的视野,弘扬乡土文化,某小学组织298名学生包车去桃源梯田参加研学旅行实践活动。包车价格是 29 元/人,准备 9000 元车费够吗。",
+        "options":  [
+
+                    ],
+        "answer":  "够",
+        "solution":  "【分析】根据总价 = 单价×数量,用每人包车的价格乘学生人数,即可求出一共需要多少元车费。这里可以通过估算比较。\n【详解】298×29≈300×30 = 9000(元)\n因为298＜300,29＜30,所以298×29＜9000。\n答:准备9000元车费够。",
+        "knowledge_points":  [
+                                 "三位数乘两位数的估算",
+                                 "经济问题"
+                             ],
+        "source":  "24-25 四年级上·江西吉安·期末"
+    },
+    {
+        "id":  "399",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "甲,乙两地相距 510千米,一辆货车以 85 千米/时的速度从甲地开往乙地,还有2 小时就要到达乙地,已经行驶了多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "340 千米",
+        "solution":  "【分析】用2乘85可以计算出还需要行驶的路程,再用甲、乙两地的距离减去还需要行驶的路程,就能算出已经行驶的路程。\n【详解】510 - 2×85 = 510 - 170 = 340(千米)\n答:已经行驶了340千米。",
+        "knowledge_points":  [
+                                 "乘加,乘减混合运算",
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25 四年级上·浙江宁波·期末"
+    },
+    {
+        "id":  "400",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "\u0027复兴号\u0027是中国自主研发,具有完全知识产权的新一代高速列车,是中国科技创新的又一重大成果,它每小时能行驶 350 千米。已知普通火车平均速度为 90 千米/时,则同样行驶 10 小时,\u0027复兴号\u0027比普通火车多行驶多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "2600 千米",
+        "solution":  "【分析】分别求出\u0027复兴号\u0027和普通火车10小时行驶的路程,再相减即可。\n【详解】350×10 = 3500(千米)\n90×10 = 900(千米)\n3500 - 900 = 2600(千米)\n答:\u0027复兴号\u0027比普通火车多行驶2600千米。",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "三位数与整十数的口算乘法",
+                                 "两位数乘整十数的口算乘法"
+                             ],
+        "source":  "24-25四年级上·广东广州·期末"
+    },
+    {
+        "id":  "401",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面四位同学描述的数学问题,能用算式60×12来解决的有( )个。\n①欢欢:妈妈买了12本书,每本60元,一共花了多少钱。②天天:小林每分钟走60米,他12分钟走了多少米。③笑笑:学校一个长方形花坛的面积是60平方米,长12米,宽是多少。④融融:两个因数的积是60,一个因数乘12,另一个因数不变,积是多少。",
+        "options":  [
+                        "A. 4",
+                        "B. 3",
+                        "C. 2",
+                        "D. 1"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】单价×数量 = 总价,速度×时间 = 路程,长方形的面积÷长 = 宽,积的变化规律是一个因数不变,另一个因数乘几,积也乘几。逐个分析即可。\n【详解】①60×12 = 720(元),能解决;②60×12 = 720(米),能解决;③应列式为60÷12,不能解决;④60×12 = 720,能解决。所以能用算式60×12来解决的有3个。故答案为:B。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)",
+                                 "长方形的面积",
+                                 "基础行程问题",
+                                 "经济问题"
+                             ],
+        "source":  "24-25四年级上·浙江宁波·期末"
+    },
+    {
+        "id":  "402",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "\"1□7×□8\"的积可能是( )。",
+        "options":  [
+                        "A. 1006",
+                        "B. 13026",
+                        "C. 20056",
+                        "D. 7645"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】先看第一个因数1□7,它最小是107,最大是197;再看第二个因数□8,它最小是18,最大是98。通过估算两个因数取值范围的乘积来判断选项。\n【详解】当1□7取最小107,□8取最小18时,107×18 = 1926;当1□7取最大197,□8取最大98时,197×98 = 19306。A. 1006＜1926,不符合;B. 1926＜13026＜19306,符合;C. 20056＞19306,不符合;D. 因为个位上7×8 = 56,所以积末尾是6,不可能是7645。故答案为:B。",
+        "knowledge_points":  [
+                                 "三位数乘两位数的估算",
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "24-25 四年级上·广东广州·期末"
+    },
+    {
+        "id":  "403",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面四个选项中,面积最接近 1 公顷的是( )。",
+        "options":  [
+                        "A. 200张单人课桌桌面",
+                        "B. 200间教室地面",
+                        "C. 2 个标准篮球场",
+                        "D. 10000 本数学课本封面"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】分别估计各选项的面积,再与1公顷 = 10000平方米进行比较。\n【详解】A. 1张单人课桌桌面的面积约50平方分米,200张是10000平方分米,与1公顷差距很大;B. 1间教室地面的面积通常约60平方米,200间约是12000平方米,最接近1公顷;C. 2个标准篮球场约840平方米,与1公顷差距很大;D. 10000本数学课本封面的总面积也远小于1公顷。故答案为:B。",
+        "knowledge_points":  [
+                                 "三位数与整十数的口算乘法",
+                                 "平方厘米,平方分米,平方米之间的进率与换算",
+                                 "公顷,平方千米的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·广东广州·期末"
+    },
+    {
+        "id":  "404",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "李阿姨从超市买了 4箱梨,每箱5千克。李阿姨买梨一共花了多少钱。根据已知的信息,还需要知道下面信息( )就可以解决问题了。",
+        "options":  [
+                        "A. 每千克梨有多少个",
+                        "B. 每个梨多少元",
+                        "C. 每千克梨是多少元",
+                        "D. 每箱梨有多少个"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】已知梨的数量,要求一共花了多少钱,还需要知道每千克梨的价格。根据总价 = 数量×单价即可求解。\n【详解】知道了买的箱数和每箱的千克数,可以求出买梨的总千克数,要求一共花了多少钱,还需要知道每千克梨是多少元。故答案为:C。",
+        "knowledge_points":  [
+                                 "经济问题"
+                             ],
+        "source":  "24-25 四年级上·广东广州·期末"
+    },
+    {
+        "id":  "405",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "\"小汽车每小时停车费 9 元。一辆小汽车停了 3 小时,求需要付费多少元.\"这里用到的数量关系是( )。",
+        "options":  [
+                        "A. 单价×数量 = 总价",
+                        "B. 速度×时间 = 路程",
+                        "C. 路程+时间 = 速度"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】每小时停车费9元是单价,3小时是数量,用9×3即可求出需要付费多少元,利用的数量关系是总价 = 单价×数量。\n【详解】9×3 = 27(元),这里用到的数量关系是单价×数量 = 总价。故答案为:A。",
+        "knowledge_points":  [
+                                 "经济问题"
+                             ],
+        "source":  "24-25四年级上·广东广州·期末"
+    },
+    {
+        "id":  "406",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "某公司买了 23 台打印机,每台打印机 370 元,共花了多少钱。下面说法正确的是( )。",
+        "options":  [
+                        "A. \u0027每台打印机 370 元\u0027是总价。",
+                        "B. \u002723 台打印机\u0027是单价。",
+                        "C. 这题已知单价和数量,求总价。",
+                        "D. 这题已知总价和单价,求数量。"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】每台打印机370元是单价,23台打印机是数量,共花的钱数是总价。根据单价×数量 = 总价可判断各项。\n【详解】A. \u0027每台打印机370元\u0027是单价,错误;B. \u002723台打印机\u0027是数量,错误;C. 这题已知单价和数量,求总价,正确;D. 错误。故答案为:C。",
+        "knowledge_points":  [
+                                 "经济问题"
+                             ],
+        "source":  "24-25 四年级上·广东广州·期末"
+    },
+    {
+        "id":  "407",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "在计算 268×37 的时候, 6×3 表示( )。",
+        "options":  [
+                        "A. 6×3",
+                        "B. 60×3",
+                        "C. 60×30"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】在268中,6在十位上表示60;在37中,3在十位上表示30,所以6×3表示60×30。\n【详解】在计算268×37的时候,6×3表示60×30。故答案为:C。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "24-25 四年级上·广东广州·期末"
+    },
+    {
+        "id":  "408",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "某区所有小学四年级共有290个班,每班人数在 38 ~ 45 人之间,这个区小学四年级学生的总人数可能是( )人。",
+        "options":  [
+                        "A. 6000",
+                        "B. 8700",
+                        "C. 12000",
+                        "D. 15000"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】先分别求出每班最少38人和最多45人时,290个班对应的总人数范围,再与各选项比较。\n【详解】290×38 = 11020(人)\n290×45 = 13050(人)\n所以总人数应在11020人与13050人之间。只有12000在这个范围内。故答案为:C。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有 0"
+                             ],
+        "source":  "24-25四年级上·广东广州·期末"
+    },
+    {
+        "id":  "409",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面与 240×18 的积不相等的算式是( )。",
+        "options":  [
+                        "A. 24×180",
+                        "B. 120×54",
+                        "C. 60×72",
+                        "D. 480×9"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】分别计算240×18和每个选项算式的积,再进行比较即可。\n【详解】240×18 = 4320。\nA. 24×180 = 4320,相等;\nB. 120×54 = 6480,不相等;\nC. 60×72 = 4320,相等;\nD. 480×9 = 4320,相等。\n故答案为:B。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "两位数乘两位数,乘数末尾有 0",
+                                 "两,三位数与一位数连续进位的乘法"
+                             ],
+        "source":  "24-25 四年级上·广东广州·期末"
+    },
+    {
+        "id":  "410",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "李叔叔开车每小时行驶 85 千米,3 小时行驶多少千米,求的是( )。",
+        "options":  [
+                        "A. 时间",
+                        "B. 速度",
+                        "C. 路程"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】每小时行驶85千米是速度,3小时是时间,85乘3求的是总距离,也就是路程。\n【详解】85×3 = 255(千米),求的是路程。故答案为:C。",
+        "knowledge_points":  [
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25四年级上·山东菏泽·期末"
+    },
+    {
+        "id":  "411",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面四道题中,()的积与其它三道题不相等。",
+        "options":  [
+                        "A. 18×70",
+                        "B. 180×70",
+                        "C. 7×180",
+                        "D. 18×7×10"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】分别判断四个算式的积是否相同即可。\n【详解】18×70 = 1260;180×70 = 12600;7×180 = 1260;18×7×10 = 1260。所以A、C、D结果相同,B与其他三道题不相等。故答案为:B。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "24-25 四年级上·辽宁抚顺·期末"
+    },
+    {
+        "id":  "412",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "小林每分钟走 60 米,他 15 分钟走多少米。求的是( )。",
+        "options":  [
+                        "A. 速度",
+                        "B. 路程",
+                        "C. 时间"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】每分钟走60米是速度,15分钟是时间,根据速度×时间 = 路程即可判断。\n【详解】求的是路程。故答案为:B。",
+        "knowledge_points":  [
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25 四年级上·辽宁抚顺·期末"
+    },
+    {
+        "id":  "413",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "与 302×19 的积最接近的数是( )。",
+        "options":  [
+                        "A. 5000",
+                        "B. 6000",
+                        "C. 7000"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】估算时,把302看成300,19看成20,再相乘即可。\n【详解】302×19≈300×20 = 6000,所以与302×19的积最接近的数是6000。故答案为:B。",
+        "knowledge_points":  [
+                                 "三位数乘两位数的估算"
+                             ],
+        "source":  "24-25 四年级上·辽宁铁岭·期末"
+    },
+    {
+        "id":  "414",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "\"小刚家和学校相距434米,他从家到学校走了7分钟,他每分钟走多少米。\"这道题是求( )。",
+        "options":  [
+                        "A. 路程",
+                        "B. 时间",
+                        "C. 速度"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】已知路程和时间,用434÷7求出每分钟走多少米,求得的是速度。\n【详解】434÷7 = 62(米),这道题是求速度。故答案为:C。",
+        "knowledge_points":  [
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25 四年级上·辽宁铁岭·期末"
+    },
+    {
+        "id":  "415",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "两个因数相乘,一个因数扩大为原来的 10 倍,另一个因数不变,积()。",
+        "options":  [
+                        "A. 不变",
+                        "B. 扩大为原来的 10 倍",
+                        "C. 缩小为原来的 1/10"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据积的变化规律,一个因数不变,另一个因数扩大到原来的10倍,积也扩大到原来的10倍。\n【详解】两个因数相乘,一个因数扩大为原来的10倍,另一个因数不变,积扩大为原来的10倍。故答案为:B。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "24-25 四年级上·河北邯郸·期末"
+    },
+    {
+        "id":  "416",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "狮子8小时跑640千米,鸵鸟4小时跑248千米,马5小时跑345千米,()跑得最快。",
+        "options":  [
+                        "A. 马",
+                        "B. 鸵鸟",
+                        "C. 狮子"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】根据速度 = 路程÷时间,分别算出狮子、鸵鸟和马的速度,再比较。\n【详解】640÷8 = 80(千米/时)\n248÷4 = 62(千米/时)\n345÷5 = 69(千米/时)\n80＞69＞62,即狮子跑得最快。故答案为:C。",
+        "knowledge_points":  [
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25四年级上·河北邯郸·期末"
+    },
+    {
+        "id":  "417",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "吸烟不仅有害健康而且费钱,如果一位吸烟者每天吸一包28元的香烟,那么他每年花在吸烟上的钱大约是( )元。",
+        "options":  [
+                        "A. 10000",
+                        "B. 7000",
+                        "C. 5000"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】一年可看成365天,用每天吸烟花的钱乘一年的天数即可。这里采用估算法:365天可看成400天,28元可看成25元。\n【详解】28×365≈25×400 = 10000(元),即他每年花在吸烟上的钱大约是10000元。故答案为:A。",
+        "knowledge_points":  [
+                                 "三位数乘两位数的估算"
+                             ],
+        "source":  "24-25四年级上·河北邯郸·期末"
+    },
+    {
+        "id":  "418",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列说法正确的一共有( )句。\n(1)已知5支同款钢笔的总价,可以求单价。\n(2)面积是1公顷的土地,一定是边长为100米的正方形。\n(3)将圆平均分成360份,其中1份所对的角的度数是角的单位。",
+        "options":  [
+                        "A. 1",
+                        "B. 2",
+                        "C. 3"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】(1)已知总价和数量,可以求单价;(2)面积是1公顷的土地不一定是边长为100米的正方形;(3)将圆平均分成360份,其中1份所对的角是1度,是角的度量单位。\n【详解】(1)说法正确。(2)例如长500米、宽20米的长方形面积也是1公顷,所以说法错误。(3)说法正确。综上,正确的说法有2句。故答案为:B。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "角的度量",
+                                 "公顷,平方千米的认识"
+                             ],
+        "source":  "24-25 四年级上·湖北武汉·期末"
+    },
+    {
+        "id":  "419",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "在计算 354×46 时,3×6 表示( )。",
+        "options":  [
+                        "A. 800×60",
+                        "B. 30×60",
+                        "C. 300×6"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】在354中,3在百位上表示300;在46中,6在个位上表示6,所以3×6表示300×6。\n【详解】在计算354×46时,3×6表示300×6。故答案为:C。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "24-25 四年级上·湖北武汉·期末"
+    },
+    {
+        "id":  "420",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "苹果每千克 5 元,妈妈共用了 20 元,买了多少千克。求的是( )。",
+        "options":  [
+                        "A. 单价",
+                        "B. 数量",
+                        "C. 总价"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】数量 = 总价÷单价,20元是总价,5元是单价。\n【详解】求的是数量。故答案为:B。",
+        "knowledge_points":  [
+                                 "经济问题"
+                             ],
+        "source":  "24-25 四年级上·河北邯郸·期末"
+    },
+    {
+        "id":  "421",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "两个数相乘得 870,一个因数不变,另一个因数除以 2,积是( )。",
+        "options":  [
+                        "A. 870",
+                        "B. 1740",
+                        "C. 435"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】一个因数不变,另一个因数除以2,积也除以2。\n【详解】870÷2 = 435,所以积是435。故答案为:C。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "24-25 四年级上·河北邯郸·期末"
+    },
+    {
+        "id":  "422",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "730×200 的乘积是 73×20 的乘积的( )倍。",
+        "options":  [
+                        "A. 10",
+                        "B. 20",
+                        "C. 100"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】730 = 73×10,200 = 20×10,两个因数都扩大到原来的10倍,所以积扩大到原来的100倍。\n【详解】730×200 = (73×10)×(20×10) = 73×20×100,所以730×200的乘积是73×20的乘积的100倍。故答案为:C。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)",
+                                 "三位数乘两位数,乘数末尾有 0"
+                             ],
+        "source":  "23-24 四年级上·内蒙古赤峰·期末"
+    },
+    {
+        "id":  "423",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面的动物中,奔跑速度最快的是( )。",
+        "options":  [
+                        "A. 叉角羚羊每分钟跑 1630 米",
+                        "B. 东北虎的奔跑速度是1300 米/分",
+                        "C. 灰狐 3 分钟跑 3 千米"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】把三种动物的奔跑速度都统一写成米/分,再比较大小。\n【详解】A. 叉角羚羊每分钟跑1630米,速度是1630米/分;B. 东北虎速度是1300米/分;C. 3千米 = 3000米,3000÷3 = 1000(米/分),灰狐速度是1000米/分。1000＜1300＜1630,所以奔跑速度最快的是叉角羚羊。故答案为:A。",
+        "knowledge_points":  [
+                                 "基础行程问题"
+                             ],
+        "source":  "23-24 四年级上·内蒙古鄂尔多斯·期末"
+    },
+    {
+        "id":  "424",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "在计算 469×28 时, \" 6×2 \" 表示( )。",
+        "options":  [
+                        "A. 60×2",
+                        "B. 60×20",
+                        "C. 600×20"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】6在469的十位上,表示60;2在28的十位上,表示20,所以6×2表示60×20。\n【详解】在计算469×28时,\u00276×2\u0027表示60×20。故答案为:B。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "23-24 四年级上·内蒙古鄂尔多斯·期末"
+    },
+    {
+        "id":  "425",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一辆货车 2小时行了 120 千米,货车的速度是( )。",
+        "options":  [
+                        "A. 60千米/时",
+                        "B. 60 千米",
+                        "C. 240 千米/时"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】根据路程÷时间 = 速度,用120除以2即可求出货车的速度。\n【详解】120÷2 = 60(千米/时),所以货车的速度是60千米/时。故答案为:A。",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "几百几十或几千几百除以一位数的除法"
+                             ],
+        "source":  "24-25 四年级上·内蒙古呼伦贝尔·期末"
+    },
+    {
+        "id":  "426",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "如果1平方米内最多能站15个人,那么2公顷大的操场最多能站( )人。",
+        "options":  [
+                        "A. 3 万",
+                        "B. 30 万",
+                        "C. 300 万"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】1公顷 = 10000平方米,先把2公顷换算成平方米,再乘1平方米内可站的人数即可。\n【详解】2公顷 = 20000平方米\n20000×15 = 300000(人)\n300000人 = 30万人\n所以2公顷大的操场最多能站30万人。故答案为:B。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "公顷,平方千米的进率与换算",
+                                 "亿以内数的改写"
+                             ],
+        "source":  "23-24 四年级上·内蒙古鄂尔多斯·期末"
+    },
+    {
+        "id":  "427",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "有一个长方形草坪,它的长是400米。如果小华以75米/分的平均速度绕该草坪步行一圈,大约需要12分钟,那么这个长方形草坪的面积大约是 公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "2",
+        "solution":  "【分析】小华的平均速度×时间 = 长方形草坪一圈的长度,也就是长方形的周长。先求出周长,再用周长÷2 - 长求出宽,最后求面积并换算成公顷。\n【详解】75×12 = 900(米)\n900÷2 - 400 = 450 - 400 = 50(米)\n400×50 = 20000(平方米)\n20000平方米 = 2公顷\n所以这个长方形草坪的面积大约是2公顷。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "长方形的面积",
+                                 "三位数乘两位数,乘数末尾有 0",
+                                 "长方形的周长"
+                             ],
+        "source":  "24-25四年级上·广东广州·期末"
+    },
+    {
+        "id":  "428",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一辆汽车3小时行驶252千米,它的速度是( );照这样的速度,这辆车15小时可以行驶的路程是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "84 千米/时;1260 千米",
+        "solution":  "【分析】根据路程÷时间 = 速度,先用252除以3求出这辆汽车的速度;再根据速度×时间 = 路程,用速度乘15即可求出15小时行驶的路程。\n【详解】252÷3 = 84(千米/时)\n84×15 = 1260(千米)\n所以它的速度是84千米/时,15小时可以行驶1260千米。",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "两位数乘两位数的进位乘法",
+                                 "三位数除以一位数,首位不够除"
+                             ],
+        "source":  "24-25四年级上·浙江宁波·期末"
+    },
+    {
+        "id":  "429",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "特快列车每小时可以行驶 160 千米,可以写作( )。",
+        "options":  [
+
+                    ],
+        "answer":  "160 千米/时",
+        "solution":  "【分析】速度的简记法:先写路程单位,再写\u0027/\u0027,最后写时间单位。\n【详解】特快列车每小时可以行驶160千米,可以写作160千米/时。",
+        "knowledge_points":  [
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25 四年级上·辽宁抚顺·期末"
+    },
+    {
+        "id":  "430",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上\"\u003e\"\"\u003c\"或\"=\"。\n20060( )20201;1亿( )91756420;210×4( )16×400;400( )78×5。",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003e;\u003c;\u003e",
+        "solution":  "【分析】先比较整数大小,再分别计算两个乘法算式的结果进行比较即可。\n【详解】20060＜20201;1亿 = 100000000,100000000＞91756420,所以1亿＞91756420;210×4 = 840,16×400 = 6400,所以210×4＜16×400;400 = 80×5,而80＞78,所以400＞78×5。",
+        "knowledge_points":  [
+                                 "亿以上数的改写",
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "两,三位数与一位数连续进位的乘法",
+                                 "大数的比较"
+                             ],
+        "source":  "24-25 四年级上·广东广州·期末"
+    },
+    {
+        "id":  "431",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "两个数相乘,积是 300,其中一个数乘 5,另一个数不变,积是( );如果其中的一个数除以 5,另一个数不变,积是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "1500;60",
+        "solution":  "【分析】根据积的变化规律,一个因数不变,另一个因数乘几或除以几(0除外),积也乘或除以同一个数。\n【详解】300×5 = 1500,所以其中一个数乘5,另一个数不变,积是1500;300÷5 = 60,所以其中一个数除以5,另一个数不变,积是60。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "24-25 四年级上·山东菏泽·期末"
+    },
+    {
+        "id":  "432",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小华从家去姥姥家是上坡路,骑自行车一共用了 20 分钟,速度为 126 米/分,回家时只用了 9 分钟,小华从姥姥家回家时的速度是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "280 米/分",
+        "solution":  "【分析】根据路程 = 速度×时间,先求出小华家到姥姥家的路程;再根据速度 = 路程÷时间,求出回家时的速度。\n【详解】126×20 = 2520(米)\n2520÷9 = 280(米/分)\n小华从姥姥家回家时的速度是280米/分。",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "三位数乘两位数,乘数末尾有0"
+                             ],
+        "source":  "24-25 四年级上·河北邯郸·期末"
+    },
+    {
+        "id":  "433",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个长方形的面积是 84 平方厘米,如果长扩大到原来的 2 倍,宽扩大到原来的 3 倍,那么面积扩大到原来的( )倍,面积变为( )平方厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "6;504",
+        "solution":  "【分析】根据积的变化规律,一个因数扩大到原来的2倍,另一个因数扩大到原来的3倍,则面积扩大到原来的2×3倍,再用原来的面积乘扩大后的倍数即可。\n【详解】2×3 = 6\n84×6 = 504(平方厘米)\n所以面积扩大到原来的6倍,面积变为504平方厘米。",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "24-25 四年级上·辽宁抚顺·期末"
+    },
+    {
+        "id":  "434",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "两个因数的积是120,一个因数乘 4,另一个因数不变,积是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "480",
+        "solution":  "【分析】根据积的变化规律,一个因数不变,另一个因数扩大到原来的几倍,积也扩大到原来的几倍。\n【详解】120×4 = 480,所以积是480。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "24-25 四年级上·辽宁铁岭·期末"
+    },
+    {
+        "id":  "435",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个篮球 80 元,买3 个需要多少钱。解决这个问题要用到的数量关系式是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "单价×数量 = 总价",
+        "solution":  "【分析】每件商品的价钱叫做单价,买了多少叫做数量,一共用的钱数叫做总价,它们之间的关系是单价×数量 = 总价。\n【详解】解决\u0027一个篮球80元,买3个需要多少钱\u0027这个问题用到的数量关系式是单价×数量 = 总价。",
+        "knowledge_points":  [
+                                 "经济问题"
+                             ],
+        "source":  "24-25 四年级上·湖北襄阳·期末"
+    },
+    {
+        "id":  "436",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "新一代标准动车组\u0027复兴号\u0027是中国自主研发,具有完全知识产权的新一代高速列车,最高时速可达 400 千米 / 小时,画线部分表示的含义是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "每小时行400千米",
+        "solution":  "【分析】速度是指单位时间内所行的路程。\n【详解】画线部分表示的含义是每小时行400千米。",
+        "knowledge_points":  [
+                                 "基础行程问题"
+                             ],
+        "source":  "23-24四年级上·内蒙古赤峰·期末"
+    },
+    {
+        "id":  "437",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "李叔叔从敖勒召其镇开车去大沟湾游玩,去的时候速度是60千米/小时,用了2小时,原路返回用了3小时。从敖勒召其镇到大沟湾有( )千米。",
+        "options":  [
+
+                    ],
+        "answer":  "120",
+        "solution":  "【分析】根据速度×时间 = 路程,由去的时候速度是60千米/小时,用了2小时,可求出从敖勒召其镇到大沟湾的距离。\n【详解】60×2 = 120(千米)\n从敖勒召其镇到大沟湾有120千米。",
+        "knowledge_points":  [
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25 四年级上·内蒙古鄂尔多斯·期末"
+    },
+    {
+        "id":  "438",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "张老师要为学校足球队购买运动服,每套运动服 148 元,张老师准备了3000 元。这些钱( )买19套运动服。(填\u0027够\u0027或\u0027不够\u0027)",
+        "options":  [
+
+                    ],
+        "answer":  "够",
+        "solution":  "【分析】用每套运动服的价格乘需要购买的套数,即可求出买19套运动服所需的总价钱,再与3000元进行比较即可。\n【详解】148×19 = 2812(元)\n3000＞2812,所以这些钱够买19套运动服。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "24-25 四年级上·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "439",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个长方形果园的长是 250 米,宽是 200 米,这个果园的面积是( )公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "5",
+        "solution":  "【分析】根据长方形的面积 = 长×宽,先求出果园的面积,再把平方米换算成公顷即可。\n【详解】250×200 = 50000(平方米)\n50000平方米 = 5公顷\n所以这个果园的面积是5公顷。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "长方形的面积",
+                                 "公顷,平方千米的进率与换算"
+                             ],
+        "source":  "23-24 四年级上·吉林松原·期末"
+    },
+    {
+        "id":  "440",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "蓝鲸在水中7分钟游4620千米,游的时间是( ),游的路程是( ),游的速度是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "7 分钟;4620 千米;660 千米/分钟",
+        "solution":  "【分析】已知蓝鲸所游的路程和时间,根据速度 = 路程÷时间计算即可。\n【详解】4620÷7 = 660(千米/分钟)\n所以游的时间是7分钟,游的路程是4620千米,游的速度是660千米/分钟。",
+        "knowledge_points":  [
+                                 "基础行程问题"
+                             ],
+        "source":  "23-24四年级上·内蒙古鄂尔多斯·期末"
+    },
+    {
+        "id":  "441",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "48个125的和是( ), 15×80 的积的末尾有( )个 0。",
+        "options":  [
+
+                    ],
+        "answer":  "6000;2",
+        "solution":  "【分析】求几个相同数的和是多少,用乘法计算;因数末尾有0的乘法可以先把0前面的数相乘,再看两个因数末尾一共有几个0,就在积的末尾添上几个0。\n【详解】48×125 = 6000\n15×80 = 1200\n所以48个125的和是6000,15×80的积的末尾有2个0。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法",
+                                 "两位数乘两位数,乘数末尾有 0"
+                             ],
+        "source":  "23-24四年级上·内蒙古鄂尔多斯·期末"
+    },
+    {
+        "id":  "442",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一身校服120元,小红、小李和小强分别买一身校服,他们三人总共花费( )元。",
+        "options":  [
+
+                    ],
+        "answer":  "360",
+        "solution":  "【分析】用一身校服的钱数乘3人,即可求出他们三人总共花费多少元。\n【详解】120×3 = 360(元)\n所以他们三人总共花费360元。",
+        "knowledge_points":  [
+                                 "因数中间或末尾是0的乘法",
+                                 "两,三位数与一位数的不进位乘法",
+                                 "经济问题"
+                             ],
+        "source":  "24-25 四年级上·新疆喀什·期末"
+    },
+    {
+        "id":  "443",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个面积为120平方米的长方形草坪,将它的长扩大到原来的5倍,宽不变,扩建后草坪的面积是( )平方米。",
+        "options":  [
+
+                    ],
+        "answer":  "600",
+        "solution":  "【分析】根据积的变化规律,长方形的长扩大到原来的5倍,宽不变,那么面积也扩大到原来的5倍。\n【详解】120×5 = 600(平方米)\n所以扩建后草坪的面积是600平方米。",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "24-25 四年级上·内蒙古呼伦贝尔·期末"
+    },
+    {
+        "id":  "444",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "401×59 大约等于(),实际结果为( )。",
+        "options":  [
+
+                    ],
+        "answer":  "24000;23659",
+        "solution":  "【分析】估算时,把401看作400,59看作60;再依据三位数乘两位数的计算方法求出实际结果。\n【详解】401×59≈400×60 = 24000\n401×59 = 23659\n所以401×59大约等于24000,实际结果为23659。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,三位数中间有0",
+                                 "三位数乘两位数的估算"
+                             ],
+        "source":  "23-24 四年级上·吉林松原·期末"
+    },
+    {
+        "id":  "445",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "某工地给一位工人结算工资时,不小心把他工作的天数 36 天看成了 30天(工资按天算),事后核对时发现了不对,又给这位工人叔叔 1440 元。这位工人叔叔结算的总工资是( )元。",
+        "options":  [
+
+                    ],
+        "answer":  "8640",
+        "solution":  "【分析】少给的1440元对应的是36天与30天相差的6天工资。先求出一天工资,再乘36即可求出总工资。\n【详解】1440÷(36 - 30) = 1440÷6 = 240(元)\n240×36 = 8640(元)\n故这位工人叔叔结算的总工资是8640元。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "三位数乘两位数,乘数末尾有 0",
+                                 "商末尾有 0 的除法"
+                             ],
+        "source":  "24-25四年级上·内蒙古鄂尔多斯·期末"
+    },
+    {
+        "id":  "446",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\u0027>\u0027\u0027<\u0027或\u0027=\u0027。\n40300( )43000;40万( )390000;530×30( )53×300。",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003e;=",
+        "solution":  "【分析】先比较两个整数大小;再把40万改写成整数比较;最后分别计算两个算式的结果再比较。\n【详解】40300＜43000;40万 = 400000,400000＞390000,所以40万＞390000;530×30 = 15900,53×300 = 15900,所以530×30 = 53×300。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "大数的比较"
+                             ],
+        "source":  "23-24四年级上·内蒙古赤峰·期末"
+    },
+    {
+        "id":  "447",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\u0027>\u0027\u0027<\u0027或\u0027=\u0027。\n70万( )701234;430×30( )43×430;100平方千米( )100公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003c;\u003e",
+        "solution":  "【分析】先比较大数大小;再利用积的变化规律或直接计算比较两个乘法算式;最后根据1平方千米 = 100公顷进行单位换算后比较。\n【详解】70万 = 700000,700000＜701234,所以70万＜701234;430×30 = 43×300,而43×300＜43×430,所以430×30＜43×430;100公顷 = 1平方千米,100平方千米＞1平方千米,所以100平方千米＞100公顷。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "公顷,平方千米的进率与换算",
+                                 "积的变化规律(整数乘法)",
+                                 "亿以内数的改写"
+                             ],
+        "source":  "24-25四年级上·吉林四平·期末"
+    },
+    {
+        "id":  "448",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用2,1,9,7,5 这 5 个数字组成一个最大的三位数是( ),组成一个最小的两位数是( ),它们的积是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "975;12;11700",
+        "solution":  "【分析】用2、1、9、7、5这5个数字组成一个最大的三位数,把最大的三个数字按从大到小排列;组成一个最小的两位数,把最小的两个数字按从小到大排列;再把这两个数相乘即可。\n【详解】最大的三位数是975,最小的两位数是12,975×12 = 11700。所以它们的积是11700。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法",
+                                 "1000 以内数的组成"
+                             ],
+        "source":  "24-25 四年级上·内蒙古呼伦贝尔·期末"
+    },
+    {
+        "id":  "449",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "每件上衣 160 元,买 12 件用多少元。是求总价。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】总价 = 单价×数量,160元是单价,12件是数量,用160×12即可求出总价。\n【详解】\u0027每件上衣160元,买12件用多少元\u0027是求总价,这句话正确。故答案为:√。",
+        "knowledge_points":  [
+                                 "经济问题"
+                             ],
+        "source":  "24-25 四年级上·河北沧州·期末"
+    },
+    {
+        "id":  "450",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "\u0027每箱葡萄 56 元,买 8 箱要用多少钱。\u0027这道题求的是总价。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】每件商品的价钱叫做单价,买了多少叫做数量,一共用的钱数叫做总价,总价 = 单价×数量。56元是每箱葡萄的单价,8箱是购买数量。\n【详解】56×8 = 448(元)\n\u0027每箱葡萄56元,买8箱要用多少钱\u0027这道题求的是总价。说法正确。故答案为:√。",
+        "knowledge_points":  [
+                                 "经济问题"
+                             ],
+        "source":  "24-25 四年级上·辽宁铁岭·期末"
+    },
+    {
+        "id":  "451",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "\u0027小明家和图书馆相聚 800 米,他从家到图书馆走了 15 分钟,他每分钟走多少米。\u0027这道题是求路程。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】题中小明家和图书馆相距800米,表示路程;走了15分钟,表示时间;要求每分钟走多少米,求的是速度。\n【详解】这道题是求速度,原题说法错误。故答案为:×。",
+        "knowledge_points":  [
+                                 "基础行程问题"
+                             ],
+        "source":  "23-24 四年级上·内蒙古赤峰·期末"
+    },
+    {
+        "id":  "452",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "三位数乘两位数的计算方法和两位数乘两位数完全相同。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】三位数乘两位数和两位数乘两位数的笔算方法本质相同,都是相同数位对齐,先用个位去乘,再用十位去乘,最后把两次乘得的积相加。\n【详解】三位数乘两位数的计算方法与两位数乘两位数的计算方法是相同的。故答案为:√。",
+        "knowledge_points":  [
+                                 "两位数乘两位数的进位乘法",
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "24-25 四年级上·内蒙古呼伦贝尔·期末"
+    },
+    {
+        "id":  "453",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "买一种商品,买的数量越多,总价就越多,单价不变。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】单价×数量 = 总价。当单价不变时,数量越多,总价就越多。\n【详解】原题说法正确。故答案为:√。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)",
+                                 "经济问题"
+                             ],
+        "source":  "24-25 四年级上·内蒙古呼伦贝尔·期末"
+    },
+    {
+        "id":  "454",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "闪电和打雷是同时发生的,但是人们却不是同时看到闪电和听到雷声,这是由于光传播的时间可以忽略不计,而声音在空气中传播的速度大约是每秒 340 米。如果见到远处的闪电 12 秒后听到雷声,所见到的闪电大约有多远。",
+        "options":  [
+
+                    ],
+        "answer":  "4080 米",
+        "solution":  "【分析】根据速度×时间 = 路程,用声音在空气中的传播速度乘12秒,即可求出闪电有多远。\n【详解】340×12 = 4080(米)\n答:所见到的闪电大约有4080米远。",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "三位数乘两位数,乘数末尾有0"
+                             ],
+        "source":  "24-25 四年级上·山东菏泽·期末"
+    },
+    {
+        "id":  "455",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "不解答,只写出下面各题已知的是什么。要求的是什么。需要用到的公式是什么。每套校服 120 元,买 5 套要用多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "单价;数量;总价;单价×数量 = 总价",
+        "solution":  "【分析】每件商品的价钱叫做单价;买了多少叫做数量;一共用的钱叫做总价。\n【详解】已知的是单价和数量,要求的是总价,需要用到的公式是单价×数量 = 总价。",
+        "knowledge_points":  [
+                                 "经济问题"
+                             ],
+        "source":  "24-25 四年级上·山东菏泽·期末"
+    },
+    {
+        "id":  "456",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "洒水车的洒水宽度是 4 米,每分钟行驶 125 米,这台洒水车每小时能洒水多少平方米。合多少公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "30000 平方米;3 公顷",
+        "solution":  "【分析】用洒水宽度乘每分钟行驶的距离,就是洒水车每分钟洒水的面积。1小时是60分钟,再乘60即可求出每小时洒水多少平方米,最后换算成公顷。\n【详解】125×4 = 500(平方米)\n500×60 = 30000(平方米)\n30000平方米 = 3公顷\n答:这台洒水车每小时能洒水30000平方米,合3公顷。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "长方形的面积",
+                                 "基础行程问题",
+                                 "三位数乘两位数,乘数末尾有0"
+                             ],
+        "source":  "24-25 四年级上·山东菏泽·期末"
+    },
+    {
+        "id":  "457",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一辆高铁的最高时速可达每小时 350 千米,照这样的速度计算,12 小时可以行驶多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "4200 千米",
+        "solution":  "【分析】根据路程 = 速度×时间,用高铁的最高时速乘行驶的时间,即可求出12小时可以行驶多少千米。\n【详解】350×12 = 4200(千米)\n答:12小时可以行驶4200千米。",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "三位数乘两位数,乘数末尾有0"
+                             ],
+        "source":  "24-25 四年级上·山东菏泽·期末"
+    },
+    {
+        "id":  "458",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "小林每天早上以 70米/分的速度步行上学,18分钟可以到达学校。放学时妈妈骑自行车接他回家,只用了 6 分钟就到家了。妈妈骑自行车的速度是多少。",
+        "options":  [
+
+                    ],
+        "answer":  "210 米/分",
+        "solution":  "【分析】先根据速度×时间 = 路程,求出小林家与学校之间的距离,再根据路程÷时间 = 速度,求出妈妈骑自行车的速度。\n【详解】70×18 = 1260(米)\n1260÷6 = 210(米/分)\n答:妈妈骑自行车的速度是210米/分。",
+        "knowledge_points":  [
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25 四年级上·浙江宁波·期末"
+    },
+    {
+        "id":  "459",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一艘轮船从甲城开往乙城逆风行驶,每小时行驶 36 千米,11 小时到达,返回时顺风行驶 9 小时到达,那么顺风每小时行驶多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "44 千米/时",
+        "solution":  "【分析】先根据路程 = 速度×时间,求出甲城到乙城的距离,再用距离除以顺风行驶时间,即可求出顺风每小时行驶多少千米。\n【详解】36×11 = 396(千米)\n396÷9 = 44(千米/时)\n答:顺风每小时行驶44千米/时。",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "两位数乘两位数的不进位乘法",
+                                 "三位数除以一位数,首位不够除"
+                             ],
+        "source":  "24-25 四年级上·江西吉安·期末"
+    },
+    {
+        "id":  "460",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "只列式不计算。\u0027凡事预则立,不预则废\u0027小明计划在 2025 年每天背 25 个单词,按计划这一整年他可以背多少个单词。",
+        "options":  [
+
+                    ],
+        "answer":  "25×365",
+        "solution":  "【分析】2025年不是闰年,全年有365天。因为题目要求只列式不计算,所以只列出每天背单词数量乘全年天数的算式即可。\n【详解】25×365",
+        "knowledge_points":  [
+                                 "有具体量的工程问题",
+                                 "三位数与两位数的乘法",
+                                 "平年,闰年的认识及判定方法"
+                             ],
+        "source":  "24-25 四年级上·江西吉安·期末"
+    },
+    {
+        "id":  "461",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "十一期间,王刚一家乘火车到北京旅游,晚上 11 点从沈阳北站出发,第二天早上 8 点到达。火车每小时行驶 120 千米,火车行驶了多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "1080 千米",
+        "solution":  "【分析】从晚上11点到第二天早上8点共经过9小时,再根据路程 = 速度×时间求出火车行驶的千米数。\n【详解】12 - 11 = 1(小时)\n8 - 0 = 8(小时)\n1 + 8 = 9(小时)\n9×120 = 1080(千米)\n答:火车行驶了1080千米。",
+        "knowledge_points":  [
+                                 "时,分,秒有关的计算",
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25 四年级上·辽宁抚顺·期末"
+    },
+    {
+        "id":  "462",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "某园林中有一个长方形的荷花池塘,宽16米,面积是288平方米。现保持池塘的长度不变,将宽度增加到 32 米。那么扩建后的荷花池塘面积是多少平方米。 (应用积的变化规律解决问题)",
+        "options":  [
+
+                    ],
+        "answer":  "576 平方米",
+        "solution":  "【分析】长方形的长不变,宽由16米增加到32米,宽扩大到原来的2倍,所以面积也扩大到原来的2倍。\n【详解】288×(32÷16) = 288×2 = 576(平方米)\n答:扩建后的荷花池塘面积是576平方米。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)",
+                                 "长方形的面积"
+                             ],
+        "source":  "24-25 四年级上·广东广州·期末"
+    },
+    {
+        "id":  "463",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "生活中经常有瓶装矿泉水没喝完就被扔掉,造成了很大的浪费。某体育馆一天有 35 瓶 500 毫升的矿泉水只喝了一半,就被扔掉了。这个体育馆一天浪费了多少毫升水。",
+        "options":  [
+
+                    ],
+        "answer":  "8750 毫升",
+        "solution":  "【分析】500毫升的矿泉水只喝了一半,先求出每瓶浪费多少毫升,再乘扔掉的数量即可。\n【详解】500÷2 = 250(毫升)\n250×35 = 8750(毫升)\n答:这个体育馆一天浪费了8750毫升水。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "三位数与两位数的乘法",
+                                 "整十,整百,整千数除以一位数的除法"
+                             ],
+        "source":  "24-25四年级上·广东广州·期末"
+    },
+    {
+        "id":  "464",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "有一块长方形苗圃,长 300 米,宽 200 米,平均每公顷地栽树苗 1500 棵,那么该苗圃一共能栽多少棵树苗。",
+        "options":  [
+
+                    ],
+        "answer":  "9000 棵",
+        "solution":  "【分析】先根据长方形的面积 = 长×宽,计算出苗圃的面积,再换算成公顷,最后乘每公顷能栽的棵数即可。\n【详解】300×200 = 60000(平方米)\n60000平方米 = 6公顷\n6×1500 = 9000(棵)\n答:该苗圃一共能栽9000棵树苗。",
+        "knowledge_points":  [
+                                 "面积单位间的进率及换算",
+                                 "长方形的面积",
+                                 "因数中间或末尾是 0 的乘法"
+                             ],
+        "source":  "24-25 四年级上·广东广州·期末"
+    },
+    {
+        "id":  "465",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一辆汽车从甲地开往乙地,速度是 85 千米/时,用了 6 小时,返回时只用了 5小时,返回时的速度是多少。",
+        "options":  [
+
+                    ],
+        "answer":  "102 千米/时",
+        "solution":  "【分析】先根据速度×时间 = 路程,求出甲乙两地的路程;再根据路程÷时间 = 速度,求出返回时的速度。\n【详解】85×6÷5 = 510÷5 = 102(千米/时)\n答:返回时的速度是102千米/时。",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "多位数与一位数的乘除混合运算"
+                             ],
+        "source":  "24-25 四年级上·河北沧州·期末"
+    },
+    {
+        "id":  "466",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "王爷爷想扩大长方形菜地的面积,菜地的面积是 360 平方米,宽是 9 米,如果长不变,宽增加 18 米,扩大后菜地的面积是多少平方米。",
+        "options":  [
+
+                    ],
+        "answer":  "1080 平方米",
+        "solution":  "【分析】先用面积÷宽求出菜地的长。再用原来的宽加上增加的18米得到现在的宽,最后根据长方形的面积 = 长×宽求出扩大后的面积。也可以根据积的变化规律解答。\n【详解】360÷9 = 40(米)\n18 + 9 = 27(米)\n27×40 = 1080(平方米)\n或:27÷9 = 3,360×3 = 1080(平方米)\n答:扩大后菜地的面积是1080平方米。",
+        "knowledge_points":  [
+                                 "因数中间或末尾是0的乘法",
+                                 "两位数乘两位数,乘数末尾有0",
+                                 "长方形的面积",
+                                 "积的变化规律(整数乘法)"
+                             ],
+        "source":  "24-25 四年级上·河北沧州·期末"
+    },
+    {
+        "id":  "467",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "王老师家到公园大约有 7000 米。如果他骑车的速度是 198 米/分,他从家到公园骑车半小时能到达吗。",
+        "options":  [
+
+                    ],
+        "answer":  "不能到达",
+        "solution":  "【分析】先把半小时换算成30分钟,再根据路程 = 速度×时间求出半小时骑行的路程,最后与7000米比较。\n【详解】1小时 = 60分\n60÷2 = 30(分)\n198×30 = 5940(米)\n5940＜7000\n答:他从家到公园骑车半小时不能到达。",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "三位数乘两位数,乘数末尾有0"
+                             ],
+        "source":  "24-25 四年级上·河北邯郸·期末"
+    },
+    {
+        "id":  "468",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "在旧城改造工作落实中,高新区坚持\u0027以环境为主导,让绿色连成片\u0027的绿色理念。高新区人民广场有一块长方形绿地,宽是 8 米,面积是1200平方米。现在要扩建,如果长不变,宽增加48米,扩建后的绿地面积是多少平方米。",
+        "options":  [
+
+                    ],
+        "answer":  "8400 平方米",
+        "solution":  "【分析】先根据长方形的长 = 面积÷宽求出长,再用原来的宽加上增加的宽度得到现在的宽,最后根据长方形的面积 = 长×宽求出扩建后的面积。\n【详解】1200÷8 = 150(米)\n8 + 48 = 56(米)\n150×56 = 8400(平方米)\n答:扩建后的绿地面积是8400平方米。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有 0",
+                                 "长方形的面积"
+                             ],
+        "source":  "24-25 四年级上·湖北咸宁·期末"
+    },
+    {
+        "id":  "469",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "在科学课上,同学们学习了声音在空气中的传播速度大约是每秒 340 米。小明在看到远处闪电后,经过 6 秒听到了雷声。已知光的传播速度很快,大约每秒 300000 千米,在计算闪电与小明的距离时可忽略光传播的时间。那么闪电发生的地方距离小明有多远。",
+        "options":  [
+
+                    ],
+        "answer":  "2040 米",
+        "solution":  "【分析】根据速度×时间 = 路程,用声音在空气中的传播速度乘小明听到雷声经过的时间即可。\n【详解】340×6 = 2040(米)\n答:闪电发生的地方距离小明有2040米。",
+        "knowledge_points":  [
+                                 "因数中间或末尾是 0 的乘法",
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25 四年级上·河北邯郸·期末"
+    },
+    {
+        "id":  "470",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一块长方形果园,长300米,宽200米,每公顷收水果60吨,这块地收水果多少吨。",
+        "options":  [
+
+                    ],
+        "answer":  "360 吨",
+        "solution":  "【分析】先根据长方形的面积 = 长×宽求出果园面积,再换算成公顷,最后乘每公顷收水果的吨数即可。\n【详解】300×200 = 60000(平方米)\n60000平方米 = 6公顷\n6×60 = 360(吨)\n答:这块地收水果360吨。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "长方形的面积",
+                                 "面积单位间的进率与换算"
+                             ],
+        "source":  "24-25 四年级上·河北邯郸·期末"
+    },
+    {
+        "id":  "471",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "张叔叔自驾去西安,汽车平均每小时行85千米,4小时后,剩下的路程比全程的一半少15千米,张叔叔全程要行多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "650 千米",
+        "solution":  "【分析】4小时后,剩下的路程比全程的一半少15千米,说明4小时行驶的路程等于全程的一半多15千米。先求出4小时行驶的路程,再减15千米求出全程的一半,最后乘2即可。\n【详解】(85×4 - 15)×2 = (340 - 15)×2 = 325×2 = 650(千米)\n答:张叔叔全程要行650千米。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25 四年级上·湖北武汉·期末"
+    },
+    {
+        "id":  "472",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "亮亮家今年第一季度3个月一共缴纳电费321元,如果照这样计算,亮亮家今年一年需要缴纳电费多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "1284 元",
+        "solution":  "【分析】可以先求1个月缴费多少元,再乘一年12个月;也可以先求12个月是3个月的几倍,再用3个月的电费乘这个倍数。\n【详解】321÷3×12 = 107×12 = 1284(元)\n或321×(12÷3) = 321×4 = 1284(元)\n答:亮亮家今年一年需要缴纳电费1284元。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法",
+                                 "三位数除以一位数,首位能除尽",
+                                 "两,三位数与一位数的一次进位乘法"
+                             ],
+        "source":  "24-25四年级上·辽宁鞍山·期末"
+    },
+    {
+        "id":  "473",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一辆汽车3小时行了 210 千米,乘汽车从 A 城到 B 城行 6 小时,A 城到 B 城有多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "420 千米",
+        "solution":  "【分析】先根据路程÷时间 = 速度求出汽车速度,再根据速度×时间 = 路程求出A城到B城的路程。\n【详解】210÷3×6 = 70×6 = 420(千米)\n答:A城到B城有420千米。",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "多位数与一位数的乘除混合运算"
+                             ],
+        "source":  "24-25四年级上·湖北襄阳·期末"
+    },
+    {
+        "id":  "474",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "学校组织四年级学生去科技馆参观,科技馆的门票价格为每人 25 元。四年级共有 4 个班,每个班有 45 名学生。学校租了 6 辆大巴车送学生去科技馆,每辆大巴车的租金是 400 元。在参观过程中,学生们参与了一个科技制作活动,每套制作材料的费用是15元,共有一半的学生参加了该活动。另外,老师还为每位学生买了一瓶3元的矿泉水。请问,这次学校组织学生去科技馆参观一共花费了多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "8790元",
+        "solution":  "【分析】先求出四年级学生总人数,再分别求门票费、大巴车租金、制作材料费和矿泉水费用,最后把各项费用相加即可。\n【详解】门票:45×4×25 = 180×25 = 4500(元)\n大巴车租金:400×6 = 2400(元)\n制作材料费:45×4÷2×15 = 180÷2×15 = 90×15 = 1350(元)\n矿泉水费用:45×4×3 = 180×3 = 540(元)\n4500 + 2400 + 1350 + 540 = 8790(元)\n答:这次学校组织学生去科技馆参观一共花费了8790元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "多位数与一位数的连乘",
+                                 "多位数与位数的乘除混合运算",
+                                 "三位数乘两位数,乘数末尾有0"
+                             ],
+        "source":  "24-25 四年级上·河北邯郸·期末"
+    },
+    {
+        "id":  "475",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "一辆汽车以 60 千米/小时的速度从一座城市驶向另一座城市,用了 7 小时,返回时少用了 1 小时。(1)两座城市之间相距多少。 (2)返回时这辆汽车的速度是多少千米/小时。",
+        "options":  [
+
+                    ],
+        "answer":  "420 千米;70 千米/时",
+        "solution":  "【分析】(1)根据路程 = 速度×时间,用去时的速度乘去时的时间,计算出两座城市之间的距离。(2)先用7 - 1算出返回时的时间,再根据速度 = 路程÷时间求出返回速度。\n【详解】(1)60×7 = 420(千米)\n答:两座城市之间相距420千米。\n(2)7 - 1 = 6(小时)\n420÷6 = 70(千米/时)\n答:返回时这辆汽车的速度是70千米/时。",
+        "knowledge_points":  [
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25 四年级上·河北邯郸·期末"
+    },
+    {
+        "id":  "476",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "一本《小学生字典》26元,四年级要买405本,一共花多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "10530 元",
+        "solution":  "【分析】根据总价 = 单价×数量,用26元乘405本计算即可。\n【详解】405×26 = 10530(元)\n答:一共花10530元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "三位数乘两位数,三位数中间有0"
+                             ],
+        "source":  "24-25四年级上·河北邯郸·期末"
+    },
+    {
+        "id":  "477",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一辆汽车从甲城开往乙城,去时每小时 66 千米的速度行驶了 4 小时,返回时行驶了 3 小时,这辆汽车返回时的速度是多少。",
+        "options":  [
+
+                    ],
+        "answer":  "88 千米/时",
+        "solution":  "【分析】先根据路程 = 时间×速度,用66乘4求出甲乙两地的距离;再根据速度 = 路程÷时间,用求出的路程除以3即可求出返回速度。\n【详解】66×4÷3 = 264÷3 = 88(千米/时)\n答:这辆汽车返回时的速度是88千米/时。",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "多位数与一位数的乘除混合运算"
+                             ],
+        "source":  "24-25 四年级上·黑龙江齐齐哈尔·期末"
+    },
+    {
+        "id":  "478",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一个篮球 80 元,一个排球 75 元,买 15 个篮球和 12 个排球一共要多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "2100 元",
+        "solution":  "【分析】根据总价 = 单价×数量,篮球的单价×数量加上排球的单价×数量,就是一共需要的钱数。\n【详解】80×15 + 75×12 = 1200 + 900 = 2100(元)\n答:买15个篮球和12个排球一共要2100元。",
+        "knowledge_points":  [
+                                 "乘加,乘减混合运算",
+                                 "经济问题"
+                             ],
+        "source":  "24-25 四年级上·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "479",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "植树造林是构建绿色生态屏障的重要行动。森林公园有一片长方形的柏树林,长 500 米,宽 300 米,这片柏树林的面积是多少公顷。1 公顷柏树林 1 天大约能分泌 40 千克植物杀菌素,这片柏树林 30 天大约分泌植物杀菌素多少千克。",
+        "options":  [
+
+                    ],
+        "answer":  "15 公顷;18000 千克",
+        "solution":  "【分析】先根据长方形的面积 = 长×宽求出柏树林面积,再换算成公顷,最后用每公顷每天分泌的植物杀菌素质量乘公顷数和天数即可。\n【详解】500×300 = 150000(平方米)\n150000平方米 = 15公顷\n40×15×30 = 600×30 = 18000(千克)\n答:这片柏树林30天大约分泌植物杀菌素18000千克。",
+        "knowledge_points":  [
+                                 "公顷,平方千米的进率与换算",
+                                 "长方形的面积",
+                                 "用两步连乘解决实际问题"
+                             ],
+        "source":  "24-25四年级上·黑龙江鸡西·期末"
+    },
+    {
+        "id":  "480",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "小红家前 3 个月的电费是 312 元,平均每个月的电费是多少元。照这样计算,一年的电费是多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "104 元;1248 元",
+        "solution":  "【分析】先用前三个月的电费除以3,求出平均每个月的电费,再乘12即可求出一年的电费。\n【详解】312÷3 = 104(元)\n1年 = 12个月\n104×12 = 1248(元)\n答:平均每个月的电费是104元;照这样计算,一年的电费是1248元。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,三位数中间有 0",
+                                 "经济问题"
+                             ],
+        "source":  "24-25 四年级上·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "481",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "2024年5月3日,我国发射了\u0027嫦娥六号\u0027探月卫星,对月球开展地貌探测和地质勘察等工作。搭载\u0027嫦娥六号\u0027的火箭飞行速度约是11千米/秒,发射5分钟后它飞行了多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "3300 千米",
+        "solution":  "【分析】先把5分钟换算成300秒,再用速度乘时间即可求出飞行路程。\n【详解】5分钟 = 5×60 = 300秒\n300×11 = 3300(千米)\n答:发射5分钟后它飞行了3300千米。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0",
+                                 "时,分,秒之间的换算与比较"
+                             ],
+        "source":  "24-25四年级上·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "482",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "学校今年为同学们定制校服。每套校服定价 120 元,5 套校服要用多少元钱。",
+        "options":  [
+
+                    ],
+        "answer":  "600 元",
+        "solution":  "【分析】根据总价 = 单价×数量,用每套校服的价格乘5即可求出买5套要用多少钱。\n【详解】120×5 = 600(元)\n答:5套校服要用600元钱。",
+        "knowledge_points":  [
+                                 "经济问题"
+                             ],
+        "source":  "23-24 四年级上·内蒙古赤峰·期末"
+    },
+    {
+        "id":  "483",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "学校为四年级学生统一购买冬季棉服。每件棉服定价 150 元,买 4 件需要多少元钱。",
+        "options":  [
+
+                    ],
+        "answer":  "600 元",
+        "solution":  "【分析】根据总价 = 单价×数量,用每件棉服的价格乘4,求出买4件棉服需要多少钱即可。\n【详解】150×4 = 600(元)\n答:买4件棉服需要600元钱。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "因数末尾有0的乘法"
+                             ],
+        "source":  "23-24 四年级上·河北石家庄·期末"
+    },
+    {
+        "id":  "484",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "四年级师生共204人,准备包车去研学。旅行社包车的价格是39元/人,老师带8000元钱够吗。",
+        "options":  [
+
+                    ],
+        "answer":  "够",
+        "solution":  "【分析】根据单价×数量 = 总价,先算出204人需要的总钱数,再与8000元比较即可。\n【详解】39×204 = 7956(元)\n7956元＜8000元,够。\n答:老师带8000元钱够。",
+        "knowledge_points":  [
+                                 "三位数乘两位数的实际问题",
+                                 "三位数乘两位数,三位数中间有 0",
+                                 "经济问题"
+                             ],
+        "source":  "24-25 四年级上·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "485",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "某小学举行\u0027相约十四冬,魅力内蒙古\u0027系列活动,其中绘画比赛评选出96名同学,学校要为他们每人购买一个奖品,购买奖品需要多少元。(1)购买奖品需要多少元,就是求总价,总价等于( )乘( )。题目中缺少的信息是( )。a。每个奖品25元 b。购买25个奖品 c。购买96个奖品(2)请你用选择的信息求出购买奖品需要多少元。 (3)\u0027双十二\u0027期间商场举行\u0027买三送一\u0027活动,学校如果在\u0027双十二\u0027期间购买这些奖品,只需买多少个奖品就可以。",
+        "options":  [
+
+                    ],
+        "answer":  "单价;数量;a;2400 元;72 个",
+        "solution":  "【分析】(1)求购买奖品需要多少元,就是求总价,总价等于单价乘数量。题目中已知数量是96名同学,缺少的信息是单价。(2)已知数量是96个,单价是每个奖品25元,根据总价 = 单价×数量求解。(3)\u0027买三送一\u0027表示每付3个奖品的钱可得到4个奖品,把96个奖品按4个一组进行计算即可。\n【详解】(1)购买奖品需要多少元,就是求总价,总价等于单价乘数量。题目中缺少的信息是a。\n(2)25×96 = 2400(元)\n答:购买奖品需要2400元。\n(3)3 + 1 = 4(个)\n96÷4×3 = 24×3 = 72(个)\n答:只需买72个奖品就可以。",
+        "knowledge_points":  [
+                                 "除加,除减混合运算",
+                                 "两位数乘两位数的进位乘法",
+                                 "经济问题"
+                             ],
+        "source":  "24-25 四年级上·内蒙古鄂尔多斯·期末"
+    },
+    {
+        "id":  "486",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一列火车以每时118千米的速度从甲地开往乙地,6时后,距乙地还有335千米,甲地距乙地多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "1043 千米",
+        "solution":  "【分析】先根据路程 = 速度×时间求出6小时行驶的距离,再加上还距乙地的距离,就是甲乙两地相距的距离。\n【详解】118×6 + 335 = 708 + 335 = 1043(千米)\n答:甲地距乙地1043千米。",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "乘加,乘减混合运算"
+                             ],
+        "source":  "23-24四年级上·吉林松原·期末"
+    },
+    {
+        "id":  "487",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一个坏掉的水龙头每分钟要浪费 208 克的水,这个水龙头 1 小时要浪费多少克的水。",
+        "options":  [
+
+                    ],
+        "answer":  "12480 克",
+        "solution":  "【分析】1小时 = 60分钟,用每分钟浪费的水的质量乘时间,即可求出1小时浪费的水有多少克。\n【详解】1小时 = 60分钟\n208×60 = 12480(克)\n答:这个水龙头1小时要浪费12480克的水。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "23-24 四年级上·吉林松原·期末"
+    },
+    {
+        "id":  "488",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "李叔叔从甲村往乙村送货,去的时候平均速度是 62 千米/时,用了 3 小时。原路返回用了 2 小时。甲村到乙村的路程是多少。",
+        "options":  [
+
+                    ],
+        "answer":  "186 千米",
+        "solution":  "【分析】根据路程 = 速度×时间,用去时的速度乘去时用的时间即可求出总路程。\n【详解】62×3 = 186(千米)\n答:甲村到乙村的路程是186千米。",
+        "knowledge_points":  [
+                                 "基础行程问题"
+                             ],
+        "source":  "23-24 四年级上·内蒙古巴彦淖尔·期末"
+    },
+    {
+        "id":  "489",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "学校买 4 个同样的篮球,花了1160元,每个篮球多少钱。请你先写出数量关系,再解答。数量关系:",
+        "options":  [
+
+                    ],
+        "answer":  "单价 = 总价÷数量;290 元",
+        "solution":  "【分析】根据单价 = 总价÷数量,代入数值即可求出每个篮球的价格。\n【详解】数量关系:单价 = 总价÷数量\n1160÷4 = 290(元)\n答:每个篮球290元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "整十,整百,整千数除以一位数的除法"
+                             ],
+        "source":  "23-24 四年级上·内蒙古巴彦淖尔·期末"
+    },
+    {
+        "id":  "490",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "超市共购进 120 箱纯牛奶,每箱 24 盒,超市一共购进纯牛奶多少盒。",
+        "options":  [
+
+                    ],
+        "answer":  "2880 盒",
+        "solution":  "【分析】用每箱纯牛奶的盒数乘箱数,即可求出一共购进纯牛奶的盒数。\n【详解】120×24 = 2880(盒)\n答:超市一共购进纯牛奶2880盒。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有0"
+                             ],
+        "source":  "23-24 四年级上·吉林白城·期末"
+    },
+    {
+        "id":  "491",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一列火车的速度是160 千米/时,这列火车从甲地去乙地,经过 12 时到达,甲地到乙地的距离是多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "1920 千米",
+        "solution":  "【分析】根据路程 = 速度×时间,用速度160千米/时乘12小时即可。\n【详解】160×12 = 1920(千米)\n答:甲地到乙地的距离是1920千米。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法",
+                                 "基础行程问题"
+                             ],
+        "source":  "23-24四年级上·吉林松原·期末"
+    },
+    {
+        "id":  "492",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "小明 6 分钟可以打 750 个字。照这样计算,15 分钟他可以打多少个字。",
+        "options":  [
+
+                    ],
+        "answer":  "1875 个",
+        "solution":  "【分析】先用750除以6,求出小明1分钟可以打多少个字,再乘15即可求出15分钟可以打多少个字。\n【详解】750÷6 = 125(个)\n125×15 = 1875(个)\n答:15分钟他可以打1875个字。",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法",
+                                 "单归一问题"
+                             ],
+        "source":  "23-24 四年级上·内蒙古赤峰·期末"
+    },
+    {
+        "id":  "493",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "\u0027我国是全球人均水资源最贫乏的国家之一,全国 600 多个城市中,缺水城市已达 300 多个,珍惜水资源,是我们共同的责任!\u0027据统计如果一个坏掉的水龙头每分钟要白白流掉168克水,那么一刻钟(15分钟)会浪费掉多少克水。1小时又会浪费多少克水呢。对此你有什么想法。",
+        "options":  [
+
+                    ],
+        "answer":  "2520 克;10080 克;珍惜水资源",
+        "solution":  "【分析】用每分钟浪费水的质量乘15分钟,即可计算出一刻钟会浪费多少克水;再根据1小时 = 60分钟,用每分钟浪费水的质量乘60,即可计算出1小时会浪费多少克水。最后写出合理想法即可。\n【详解】168×15 = 2520(克)\n168×60 = 10080(克)\n答:一刻钟会浪费掉2520克水,1小时又会浪费10080克水。我们要珍惜水资源。",
+        "knowledge_points":  [
+                                 "时,分的认识",
+                                 "三位数乘两位数的实际问题"
+                             ],
+        "source":  "23-24 四年级上·内蒙古赤峰·期末"
+    },
+    {
+        "id":  "494",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "学校要为图书馆增添两种新书,一种是《儿童百科》,每套 136 元,另一种是《数学猜想》,每套 27 元,每种各买 3 套,一共要花多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "489 元",
+        "solution":  "【分析】根据单价×数量 = 总价,可以先用两种书的单价分别乘数量3套,求得两种书的总价,再把它们的总价相加得到一共要花的钱;也可以先把两种书的单价相加,求得单价和,再用单价和乘数量3套,求得一共要花的钱。据此解答。n【详解】方法一: 136×3 + 27×3 = 408 + 81 = 489 (元) (136+27)×3 = 163×3 = 489 (元) 答:一共要花 489 元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "乘加,乘减混合运算",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25 四年级上·内蒙古呼伦贝尔·期末"
+    },
+    {
+        "id":  "495",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一座大桥全长约9270 米,一列动车通过大桥时每分钟约行 4840 米,从车头开上桥到车尾离开桥共需 2 分钟。这列动车长约多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "410 米",
+        "solution":  "【分析】动车行驶的总路程等于大桥全长加上动车长度。先根据路程 = 速度×时间求出这列动车行驶的路程,再减去大桥全长即可。\n【详解】4840×2 - 9270 = 9680 - 9270 = 410(米)\n答:这列动车长约410米。",
+        "knowledge_points":  [
+                                 "乘加,乘减混合运算",
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25 四年级上·吉林四平·期末"
+    },
+    {
+        "id":  "496",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一辆汽车从 A 地开往 B 地,每小时行驶 80 千米,7 小时到达。原路返回时少用了 2 小时,这辆汽车返回时的速度是多少。",
+        "options":  [
+
+                    ],
+        "answer":  "112 千米/时",
+        "solution":  "【分析】先根据路程 = 速度×时间求出AB两地的距离。返回时用的时间是7 - 2小时,再根据速度 = 路程÷时间求出返回速度。\n【详解】80×7÷(7 - 2) = 560÷5 = 112(千米/时)\n答:这辆汽车返回时的速度是112千米/时。",
+        "knowledge_points":  [
+                                 "多位数与一位数的乘除混合运算",
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25 四年级上·吉林四平·期末"
+    },
+    {
+        "id":  "497",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "小楠放学步行回家每分钟走64米,她走了5分钟后,发现钥匙忘在教室了,立即返回取钥匙,她比平时回家多走了多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "640 米",
+        "solution":  "【分析】忘带钥匙后返回教室,相当于多走了2个5分钟的路程。先求出5分钟走的路程,再乘2即可。\n【详解】64×5×2 = 640(米)\n答:她比平时回家多走了640米。",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "多位数与一位数的连乘"
+                             ],
+        "source":  "24-25四年级上·内蒙古呼伦贝尔·期末"
+    },
+    {
+        "id":  "498",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "342 元可以买 6 个汽车模型,2个汽车模型的价钱和1个火车模型的价钱相等。1 个火车模型的价钱是多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "114 元",
+        "solution":  "【分析】先根据单价 = 总价÷数量求出一个汽车模型的价钱,再乘2即可求出一个火车模型的价钱。\n【详解】342÷6×2 = 57×2 = 114(元)\n答:1个火车模型的价钱是114元。",
+        "knowledge_points":  [
+                                 "多位数与一位数的乘除混合运算",
+                                 "经济问题"
+                             ],
+        "source":  "23-24 四年级上·北京石景山·期末"
+    },
+    {
+        "id":  "499",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "李叔叔送一批货物去县城,去时用了 5 小时,原路返回时少用了 1 小时,去时平均每小时行驶 48 千米,原路返回时平均每小时行驶多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "60 千米/时",
+        "solution":  "【分析】先根据路程 = 时间×速度求出去时路程,再用路程除以返回时间,即可求出返回时平均每小时行驶多少千米。\n【详解】48×5 = 240(千米)\n240÷(5 - 1) = 240÷4 = 60(千米/时)\n答:返回时平均每小时行驶60千米/时。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "基础行程问题"
+                             ],
+        "source":  "23-24 四年级上·内蒙古赤峰·期末"
+    },
+    {
+        "id":  "500",
+        "unit":  "第四单元 三位数乘两位数",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "某玩具车间原来每天生产 62 个玩具,改进技术之后,现在每天生产玩具的个数是原来的 2 倍。现在每个月生产多少个玩具。(按30 天计算)",
+        "options":  [
+
+                    ],
+        "answer":  "3720 个",
+        "solution":  "【分析】现在每天生产玩具的个数是原来的2倍,先用原来每天生产的个数乘2,求出现在每天能生产多少个玩具,再乘30即可求出每个月生产多少个玩具。\n【详解】62×2×30 = 124×30 = 3720(个)\n答:现在每个月生产玩具的个数是3720个。",
+        "knowledge_points":  [
+                                 "三位数乘两位数,乘数末尾有 0",
+                                 "用两步连乘解决实际问题"
+                             ],
+        "source":  "23-24 四年级上·内蒙古鄂尔多斯·期末"
+    }
+]
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+[
+    {
+        "id":  "1",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.4,
+        "question":  "先写出一个两位数35，接着在35右端写这两个数字的和8，得到358，再写末两位数字5和8的和13，得到35813，用上述方法得到一个有2025位的整数。则这个整数的数字之和是()。",
+        "options":  [
+                        "A. 7070",
+                        "B. 7090",
+                        "C. 7089",
+                        "D. 7094"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】先根据题意多写几位这个整数，直至找出规律：这个整数是以\"3581347112\"每10个数字为一个循环;先用除法求出2025里有几个10，然后根据余数的情况，判断2025位的最后一个数字;求这个整数的2025位的数字之和是多少，先求出一个循环周期各个数位上的数字之和，再乘循环的次数，最后加上余数中出现的几个数字即可求解。本题考查周期性问题，关键是根据写数的方法找出这个整数的循环规律，再用除法确定2025位最后一个数字是几，进而求出2025位数字之和.\n【详解】接着往下写:35813471123581347112。发现是以\"3581347112\"每10个数字为一个循环;2025÷10=202。5，余数是5表示是一个循环里的第5个数，即3;(3+5+8+1+3+4+7+1+1+2)×202+3+5+8+1+3=35×202+3+5+8+1+3=7070+3+5+8+1+3=7090，则这个整数的数字之和是7090。故答案为:B",
+        "knowledge_points":  [
+                                 "整数排列的规律",
+                                 "周期问题"
+                             ],
+        "source":  "2025·广东广州·小升初模拟"
+    },
+    {
+        "id":  "2",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列说法正确的是()。",
+        "options":  [
+                        "A. 200-98÷7×4的运算顺序是先乘后除最后减。",
+                        "B. 在没有括号的加减混合运算中，要从左往右依次计算。",
+                        "C. 在一道有括号的算式里，应先算中括号里的，再算小括号里的。"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】在没有括号的算式中，先乘除后加减，同级运算从左到右;有括号时，先算小括号里的，再算中括号里的。200-98÷7×4的运算顺序：根据运算规则，应先算除法98÷7=14，再算乘法14×4=56，最后算减法200-56=144。因此，顺序是先除后乘最后减，不是先乘后除最后减。在没有括号的加减混合运算中，加法和减法是同级运算，应从左往右依次计算。例如，计算5+3-2，先算5+3=8，再算8-2=6。在一道有括号的算式里，应先算小括号里的，再算中括号里的。以此选择即可.\n【详解】根据分析可知:A。200-98÷7×4的运算顺序是先乘后除最后减。错误。B。在没有括号的加减混合运算中，要从左往右依次计算。正确。C。在一道有括号的算式里，应先算中括号里的，再算小括号里的。错误。故答案为:B",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "带有中括号的混合运算",
+                                 "无括号的运算顺序"
+                             ],
+        "source":  "25-26四年级上·河南濮阳·期末"
+    },
+    {
+        "id":  "3",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一个减法算式中，被减数，减数，差的和是80，且差是减数的3倍。差是()。",
+        "options":  [
+                        "A. 10",
+                        "B. 20",
+                        "C. 30"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】被减数+减数+差=80，被减数=减数+差，差=减数*3，据此求出差.\n【详解】减数+差:80÷2=40;减数:40÷(1+3)=10;差:10×3=30，故答案为:C",
+        "knowledge_points":  [
+                                 "和差倍问题",
+                                 "加，减法的意义和各部分间的关系"
+                             ],
+        "source":  "25-26六年级·全国·随堂练习"
+    },
+    {
+        "id":  "4",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "0不能作()。",
+        "options":  [
+                        "A. 被除数",
+                        "B. 除数",
+                        "C. 乘数",
+                        "D. 积"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】由题意得，可以举例与0相关的算式，然后来判断每个选项说法的正确性.\n【详解】A。0÷1=0,0÷7=0，所以0可以做被除数;B。0不能作除数,0作除数没有意义，该选项说法正确;C。0×1=0,0×7=0，所以0可以做乘数;D。0×1=0,0×7=0，所以0可以做积。故答案选:B",
+        "knowledge_points":  [
+                                 "0不能作除数",
+                                 "乘，除法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "5",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "在计算18+(18×21-28)÷25时，最后一步是求()。",
+        "options":  [
+                        "A. 和",
+                        "B. 差",
+                        "C. 积",
+                        "D. 商"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】按照运算顺序逐步计算，先算括号内的乘法，再算括号内的减法，接着算括号外的除法，最后算括号外的加法，确定最后一步的运算类型.\n【详解】根据分析可得：先算乘法再算减法接着算除法最后算加法，最后算加法所以应该是求和，所以A正确;故正确答案:A",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "6",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "已知□÷24=2，求□里的数。下面方法正确的是()。",
+        "options":  [
+                        "A. 24÷2",
+                        "B. 24-2",
+                        "C. 24×2",
+                        "D. 24+2"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】根据题意可以算出里的数，进而选择正确的结果 □.\n【详解】根据分析可知：被除数=除数×商，故 □=24×2=48。故正确答案为:C",
+        "knowledge_points":  [
+                                 "乘，除法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "7",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "已知327-□=16，这个算式中 □ 的求法是()。",
+        "options":  [
+                        "A. 被减数-减数=差",
+                        "B. 加数=和-另一个加数",
+                        "C. 减数=被减数-差",
+                        "D. 和=加数+另一个加数"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】在减法算式327-□=16中,327是被减数,□是减数,16是差，要求□里的数，就是求减数，根据减数=被减数-差解答即可.\n【详解】□是算式中的减数，减数的求法：减数=被减数-差，故答案为:C",
+        "knowledge_points":  [
+                                 "加，减法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "8",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "丽丽用计算器计算一道乘法算式时，将一个因数24输成了4，要想得到正确结果，丽丽应再()。",
+        "options":  [
+                        "A. 加20",
+                        "B. 乘6",
+                        "C. 除以6"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据积的变化规律：一个因数不变，另一个因数乘几，积也会随之乘相同的数;据此解答.\n【详解】将一个因数24输成了4，即:4×6=24，即要想得到正确结果，丽丽应再乘6。故答案为:B",
+        "knowledge_points":  [
+                                 "乘，除法的意义和各部分间的关系"
+                             ],
+        "source":  "25-26六年级上·山西晋中·期中"
+    },
+    {
+        "id":  "9",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "利民工厂某生产小组6小时生产108个零件。照这样计算，生产1800个零件需要多少小时。正确的算式是()。",
+        "options":  [
+                        "A. 1800×(108÷6)",
+                        "B. 1800÷(108/6)",
+                        "C. 1800÷(108×6)",
+                        "D. 1800÷6×108"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】已知6小时生产108个零件，先求出每小时生产的零件数，即108÷6=18(个)。生产1800个所需时间为1800÷18=100小时。正确算式应为1800÷(108/6)。选项B的结构符合此思路.\n【详解】每小时生产量:108÷6=18(个)。生产1800个所需时间:1800÷18=100(小时)。综合算式:1800÷(108/6)。故答案为:B",
+        "knowledge_points":  [
+                                 "单归一问题",
+                                 "有具体量的工程问题",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25五年级上·江西赣州·期中"
+    },
+    {
+        "id":  "10",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面加减法各部分之间的关系中，错误的是()。",
+        "options":  [
+                        "A. 减数=被减数+差",
+                        "B. 和=加数+加数",
+                        "C. 被减数=减数+差"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】在加法算式中，加数+加数=和，一个加数=和-另一个加数;在减法算式中，被减数-减数=差，被减数-差=减数，差+减数=被减数。据此解答.\n【详解】A。减数=被减数-差，该选项说法错误。B。和=加数+加数，该选项说法正确。C。被减数=减数+差，该选项说法正确。故答案为:A",
+        "knowledge_points":  [
+                                 "加，减法的意义和各部分间的关系"
+                             ],
+        "source":  "25-26三年级上·云南玉溪·期中"
+    },
+    {
+        "id":  "11",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "对比这两道算式:60-20÷5与(60-20)÷5，下面表达正确的是()。",
+        "options":  [
+                        "A. 运算顺序不同，结果相同",
+                        "B. 运算顺序相同，结果不同",
+                        "C. 运算顺序相同，结果相同",
+                        "D. 运算顺序不同，结果不同"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】混合运算的运算顺序：如果只有加减法或只有乘除法，按照从左到右的顺序计算;如果既有加减法又有乘除法，要先算乘除法，再算加减法;如果有括号，要先算括号里面的，再算括号外面的。据此判断运算顺序并计算出结果，选择即可.\n【详解】60-20÷5=60-4=56，先算除法，再算减法，结果是56;(60-20)÷5=40÷5=8，先算括号里面的减法，再算括号外面的除法，结果是8。因此这两道算式运算顺序不同，结果也不同。故答案为:D",
+        "knowledge_points":  [
+                                 "无括号的运算顺序",
+                                 "有括号的运算顺序"
+                             ],
+        "source":  "25-26三年级上·陕西安康·阶段练习"
+    },
+    {
+        "id":  "12",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "某品牌奶粉营养成分表标注\"钙含量\u003e=1200mg/100g\"，冲调一杯25g奶粉的牛奶，钙至少()mg。",
+        "options":  [
+                        "A. 120",
+                        "B. 240",
+                        "C. 300",
+                        "D. 360"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】已知\"钙含量\u003e=1200mg/100g\"，先用1200÷100求出每克奶粉的钙含量，再乘25，即是冲调一杯25g奶粉的牛奶中的钙含量.\n【详解】1200÷100×25=12×25=300(mg),钙至少300mg。故答案为:C",
+        "knowledge_points":  [
+                                 "多位数与两位数的乘除混合运算"
+                             ],
+        "source":  "24-25六年级下·广东东莞·期末"
+    },
+    {
+        "id":  "13",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "某小学师生一起租车去春游。学校共有教师14人，学生326人。大车可坐40人，租金900元;小车可坐20人，租金500元，则租()最省钱。",
+        "options":  [
+                        "A. 7辆大车和3辆小车",
+                        "B. 6辆大车和5辆小车",
+                        "C. 8辆大车和1辆小车",
+                        "D. 5辆大车和7辆小车"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】根据题意，学校共有教师14人，学生326人。大车可坐40人，租金900元;小车可坐20人，租金500元，先用326加上14，求出总人数;需计算各选项的总租金。先验证各选项座位数是否足够，再计算费用.\n【详解】A。7辆大车和3辆小车，座位数:7×40+3×20=280+60=340(人),费用:7×900+3×500=6300+1500=7800(元)。B。6辆大车和5辆小车，座位数:6×40+5×20=240+100=340(人),费用:6×900+5×500=5400+2500=7900(元)。C。8辆大车和1辆小车：座位数:8×40+1×20=320+20=340(人),费用:8×900+1×500=7200+500=7700(元)。D。5辆大车和7辆小车：座位数:5×40+7×20=200+140=340(人),费用:5×900+7×500=4500+3500=8000(元)。7700\u003c7800\u003c7900\u003c8000。故答案为:C",
+        "knowledge_points":  [
+                                 "租船问题"
+                             ],
+        "source":  "24-25四年级下·陕西安康·期末"
+    },
+    {
+        "id":  "14",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "某校四年级210名师生乘车去春游。大车每辆租金720元，限乘客40人;小车每辆租金600元，限乘客30人。租()最省钱。",
+        "options":  [
+                        "A. 6辆大车",
+                        "B. 4辆大车和2辆小车",
+                        "C. 5辆大车和1辆小车",
+                        "D. 7辆小车"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】用租大车的数量乘租金加上租小车的数量乘租金即为一共花的钱数。计算出各选项的钱数再比较即可解答.\n【详解】A。6×720=4320(元);B。4×720+600×2=2880+1200=4080(元);C。5×720+600=3600+600=4200(元);D。7×600=4200(元),4320\u003e4200=4200\u003e4080，租4辆大车2辆小车最省钱。故答案为:B",
+        "knowledge_points":  [
+                                 "租船问题"
+                             ],
+        "source":  "24-25四年级下·浙江宁波·期末"
+    },
+    {
+        "id":  "15",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "张老师买了18本《童话选》,付给收银员600元，找回60元。花花列式60/[(600-60)÷18],她想解决的问题是()。",
+        "options":  [
+                        "A. 《童话选》单价是多少",
+                        "B. 18本《童话选》共花了多少钱",
+                        "C. 剩下的钱还能买几本《童话选》",
+                        "D. 600元一共能买多少本《童话选》"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】算式\"60/[(600-60)÷18]\"，是先算小括号里面的减法，再算中括号里面的除法，最后算括号外的除法，第一步小括号里面600-60是用总钱数减去找回的钱数，表示18本《童话选》花了多少钱，第二步18本的总价除以18，表示每一本的价格是多少元，最后一步，用剩下的60元除以每一本的价格，就表示剩下的钱数还能买几本《童话选》,由此求解.\n【详解】根据分析可知：张老师买了18本《童话选》,付给收银员600元，找回60元。花花列式60/[(600-60)÷18],她想解决的问题是剩下的钱还能买几本《童话选》。故答案为:C",
+        "knowledge_points":  [
+                                 "带有中括号的混合运算",
+                                 "\"提问题\"\"填条件\"问题",
+                                 "经济问题"
+                             ],
+        "source":  "24-25四年级下·河北唐山·期末"
+    },
+    {
+        "id":  "16",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "小红带了80元零花钱买了一筒羽毛球，共12个，还剩32元。下面算式\"32/[(80-32)÷12]\"解决的问题是()。",
+        "options":  [
+                        "A. 每个羽毛球要多少钱。",
+                        "B. 剩下的钱还能买几个羽毛球。",
+                        "C. 小红花了多少钱。",
+                        "D. 一筒羽毛球要多少钱。"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】由题意得，在算式\"32/[(80-32)÷12]\"中，要先算\"80-32\".\"80\"表示小红带了80元零花钱，\"32\"表示还剩32元，所以\"80-32\"算的是12个羽毛球一共花了多少钱，算式(80-32)÷12算的是每个羽毛球多少钱，所以算式\"32/[(80-32)÷12]\"解决的问题是剩下的钱还能买几个羽毛球.\n【详解】由分析得，算式32/[(80-32)÷12]解决的问题是剩下的钱还能买几个羽毛球。故答案为:B",
+        "knowledge_points":  [
+                                 "带有中括号的混合运算",
+                                 "\"提问题\"\"填条件\"问题"
+                             ],
+        "source":  "24-25四年级下·浙江杭州·期末"
+    },
+    {
+        "id":  "17",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "某文具店\"六一\"促销，原价每本25元的《数学练习册》推出\"买4本送1本\"活动，王老师要为班级购买20本作为期末奖品，实际每本比原价便宜()元。",
+        "options":  [
+                        "A. 4",
+                        "B. 5",
+                        "C. 6",
+                        "D. 7"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】\"买4本送1本\"意味着花买4本的钱可以得到5本练习册。因为每5本为一组，那么20本里面有20÷5=4组。每组需要花钱买4本，每本原价25元，所以每组花费25×4=100元。一共4组，所以实际总花费为100×4=400元。实际总花费400元，一共买了20本，所以实际每本价格为400÷20=20元。原价每本25元，实际每本20元，所以每本比原价便宜25-20=5元.\n【详解】20÷(4+1)=20÷5=4(组);25×4=100(元);100×4=400(元);400÷20=20(元);25-20=5(元),实际每本比原价便宜5元。故答案为:B",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "促销问题"
+                             ],
+        "source":  "24-25六年级下·广东东莞·期末"
+    },
+    {
+        "id":  "18",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把125+85=210,110-80=30,210÷30=7，写成一个综合算式是()。",
+        "options":  [
+
+                    ],
+        "answer":  "(125+85)÷(110-80)=7",
+        "solution":  "【分析】在分步算式中，先算加法和减法，最后算除法，为了先计算加法和减法，需要给加法和减法算式分别加上小括号，再将两个小括号的结果用除号连接，得到综合算式.\n【详解】由分析可知综合算式为:(125+85)÷(110-80)=7",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "19",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "学校组织四年级14名教师和326名学生到七星湖研学，租()辆大车和()辆小车最省钱，一共需要()元。\n![](images/2026-03-04%20205017.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "6;2;7600",
+        "solution":  "【详解】根据题意可知，去七星湖研学的师生总人数为:326+14=340(名),每辆大客车可坐50人，租金1100元，每人平均费用为1100÷50=22(元),每辆小客车可坐20人，租金500元，每人平均费用为500÷20=25(元),22\u003c25，所以尽量租大客车，并尽量满载。用师生总人数除以每辆大客车可坐人数，求出需要大客车的数量，再结合余数判断是否需要小客车;据此计算解答即可.\n[解答]326+14=340(名),1100÷50=22(元),500÷20=25(元),22\u003c25。所以尽量租大客车。340÷50=6(辆)。40(人),40÷20=2(辆),租6辆大客车和2辆小客车，需要费用:1100×6+500×2=6600+1000=7600(元),答：租6辆大客车和2辆小客车最省钱;一共需要7600元。",
+        "knowledge_points":  [
+                                 "乘加，乘减混合运算",
+                                 "租船问题"
+                             ],
+        "source":  "25-26四年级下·全国·课前预习"
+    },
+    {
+        "id":  "20",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小强，小华，小丽，小平和小松五个人是好朋友。如果他们每两人之间通一次电话，一共要通()次电话;如果他们互相寄一张新年贺卡，一共要寄()张贺卡。",
+        "options":  [
+
+                    ],
+        "answer":  "10;20",
+        "solution":  "【分析】两人通电话时，小强给小华打和小华给小强打是同一次通话，只算1种结果，说明搭配无先后顺序，计算时需去掉重复的情况，可以用公式：人数×(人数-1)÷2来求解。两人互寄贺卡时，小强给小华寄和小华给小强寄是两张不同的贺卡，算2种结果，说明搭配有先后顺序，计算时无需剔除重复，可以用公式：人数×(人数-1)来求解.\n【详解】5×(5-1)÷2=5×4÷2=20÷2=10(次);5×(5-1)=5×4=20(张),如果他们每两人之间通一次电话，一共要通10次电话;如果他们互相寄一张新年贺卡，一共要寄20张贺卡。",
+        "knowledge_points":  [
+                                 "搭配问题",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "25-26五年级上·湖南邵阳·期末"
+    },
+    {
+        "id":  "21",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "图书馆一共有128本故事书，借出了56本，还剩多少本故事书。\n(1)题中已知总量()和部分量(),求另一部分量;另一部分量为(),所以用()法计算.\n(2)列式计算:().\n(3)我发现：已知两个数的()与其中一个()求另一个()的运算，叫做减法。在减法中，已知的和叫做(),已知的一个加数叫做(),求得的另一个加数叫做()。",
+        "options":  [
+
+                    ],
+        "answer":  "图书馆共有故事书的本数;借出的本数;剩下的本数;减;128-56=72(本);和;加数;加数;被减数;减数;差",
+        "solution":  "【分析】(1)已知\"图书馆一共有128本故事书，借出了56本\"，要求\"还剩多少本故事书\"，用128减去56即可;(2)根据题意，用故事书的总本数减去借出的本数，即可求出还剩下多少本，列式为:128-56=72(本)。(3)根据减法算式各部分的名称可得：被减数-减数=差.\n【详解】(1)根据分析可得：已知总量图书馆共有故事书128本和部分量借出的本数，求另一部分量;另一部分量为剩下的本数，所以用减法计算。(2)根据分析可知，列式为:128-56=72(本)。(3)根据分析我发现：已知两个数的和与其中一个加数求另一个加数的运算，叫做减法。在减法中，已知的和叫做被减数，已知的一个加数叫做减数，求得的另一个加数叫做差。",
+        "knowledge_points":  [
+                                 "加，减法的意义和各部分间的关系",
+                                 "总量与分量之间的关系"
+                             ],
+        "source":  "25-26四年级下·全国·课前预习"
+    },
+    {
+        "id":  "22",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "新华书店举办\"书香国庆·阅读同行\"主题图书展销活动，全场图书实行阶梯满减：\"满100元立减20元，满200元立减50元，满300元立减80元\"，可将购买总额按上述不同档位自由拆分。李老师购买了48本《三字经》,一本《三字经》10元，李老师购买这些书，实际最少应付()元。",
+        "options":  [
+
+                    ],
+        "answer":  "380",
+        "solution":  "【分析】根据单价×数量=总价，计算出48本《三字经》的总钱数是480元。方式一：满300元立减80元，还有180元继续进行满100元立减20元，总钱数-80元-20元=实际应付钱数;方式二：直接参加满200元立减50元，总钱数包含几个200元，就从总钱数里减去几个50元是实际应付钱数.\n【详解】10×48=480(元)。方式一:480÷300=1(个)。180(元),180÷100=1(个)。80(元)。方式二:480-80-20=380(元),480-50×2=480-100=380(元),实际最少应付380元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "乘加，乘减混合运算",
+                                 "促销问题"
+                             ],
+        "source":  "25-26五年级上·福建莆田·期末"
+    },
+    {
+        "id":  "23",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "超市水果大促销，如图，妈妈买9千克苹果需要()元;买10千克梨需要()元.\n![](images/2026-03-04%20205516.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "60;65",
+        "solution":  "【分析】①用苹果的千克数9千克除以3再乘20元，即可求出妈妈买9千克苹果需要的钱数;②用梨的千克数10千克除以2再乘13元，即可求出买10千克梨需要的钱数.\n【详解】①(9÷3)×20=3×20=60(元),即妈妈买9千克苹果需要60元。②(10÷2)×13=5×13=65(元),即买10千克梨需要65元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "25-26四年级上·广东珠海·期中"
+    },
+    {
+        "id":  "24",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "两个数的差是15，如果被减数增加2，减数减少2，差应该是()。",
+        "options":  [
+
+                    ],
+        "answer":  "19",
+        "solution":  "【分析】根据数量关系：被减数-减数=差，当减数不变的情况下，被减数增加或减少多少，差也会同时增加或者减少相同数量;当被减数不变的情况下，减数增加差反而会减少，减数减少差反而会增加，变化的具体数量相同;两个数的差是15，如果被减数增加2，差也会增加2，即差是15+2=17;减数减少2，差反而增加2，所以此时差应该是17+2=19.\n【详解】15+2+2=19，所以两个数的差是15，如果被减数增加2，减数减少2，差应该是19。",
+        "knowledge_points":  [
+                                 "加，减法的意义和各部分间的关系"
+                             ],
+        "source":  "25-26四年级上·广东汕尾·期中"
+    },
+    {
+        "id":  "25",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "被减数，减数与差的和是240，被减数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "120",
+        "solution":  "【分析】根据被减数=减数+差，被减数，减数与差的和是240，则被减数+被减数=240，被减数为240÷2，据此解答即可.\n【详解】240÷2=120，所以被减数，减数与差的和是240，被减数是120。",
+        "knowledge_points":  [
+                                 "三位数除以一位数，首位能除尽",
+                                 "加，减法的意义和各部分间的关系"
+                             ],
+        "source":  "25-26四年级上·广东汕尾·期中"
+    },
+    {
+        "id":  "26",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "算式180/[(6+3)×5]要先算()法，再算()法，最后算()法，结果是()。",
+        "options":  [
+
+                    ],
+        "answer":  "加;乘;除;4",
+        "solution":  "【分析】根据四则混合运算的运算顺序，有小括号先算小括号里面的，再算中括号里面的，最后算括号外面的.\n【详解】180/[(6+3)×5]=180÷(9×5)=180÷45=4，所以，算式180/[(6+3)×5]要先算加法，再算乘法，最后算除法，结果是4。",
+        "knowledge_points":  [
+                                 "带有中括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "27",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一道综合算式的计算顺序是①1080÷72=15;②120+6=126;③126×15=1890。这道综合算式是()。",
+        "options":  [
+
+                    ],
+        "answer":  "(1080÷72)×(120+6)=1890",
+        "solution":  "【分析】根据题意可知，此题是先算除法，再算加法，最后算乘法，根据混合运算的计算顺序列出综合算式即可.\n【详解】混合运算的计算顺序是：先算乘，除法，再算加，减法，有括号时应先算括号里面的，再算括号外面的，因此综合算式是:(1080÷72)×(120+6)=1890。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "28",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "根据714÷17=42，写出一道乘法算式和一道除法算式分别是()和()。",
+        "options":  [
+
+                    ],
+        "answer":  "17×42=714;714÷42=17",
+        "solution":  "【分析】先明确除法算式中各部分的名称，再根据乘除法的互逆关系，分别写出对应的乘法算式和除法算式.\n【详解】除数×商=被除数:17×42=714;被除数/商=除数:714÷42=17。所以：根据714÷17=42，写出一道乘法算式和一道除法算式分别是17×42=714和714÷42=17。",
+        "knowledge_points":  [
+                                 "乘，除法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "29",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在一道减法算式中，被减数，减数与差的和是540，已知差是59，被减数是(),减数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "270;211",
+        "solution":  "【分析】被减数=减数+差，被减数，减数，差三个数的和是540，也就是被减数+减数+差=540，被减数+被减数=540，所以用540除以2即可计算出被减数，再根据关系：减数=被减数-差，计算出减数，据此解题.\n【详解】被减数:540÷2=270;减数:270-59=211。综上所述，被减数是270，减数是211。",
+        "knowledge_points":  [
+                                 "加，减法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "30",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.4,
+        "question":  "小梦在某购物平台上购买甲，乙，丙三种商品，当购物车内选择3件甲,2件乙,1件丙时显示的价格为420元;当购物车内选择2件甲,3件乙,4件丙时显示的价格为580元，则当她购买甲，乙，丙各2件时，应该付款()元。",
+        "options":  [
+
+                    ],
+        "answer":  "400",
+        "solution":  "【分析】已知购买3件甲,2件乙,1件丙时价格为420元;购买2件甲,3件乙,4件丙时价格为580元。将这两种购买情况相加，此时购买甲的数量为3+2=5件，购买乙的数量为2+3=5件，购买丙的数量为1+4=5件，总共花费420+580=1000元。即购买5件甲,5件乙,5件丙一共需要1000元，所以购买甲，乙，丙各1件的价格为1000÷5=200元。那么购买甲，乙，丙各2件的价格就是购买甲，乙，丙各1件价格的2倍，即200×2=400元。据此作答。本题可通过将两种购买情况相加，求出购买5件甲,5件乙,5件丙的总价格，进而求出购买甲，乙，丙各2件时的付款金额.\n【详解】(420+580)÷(3+2)×2=1000÷5×2=200×2=400(元),因此购买甲，乙，丙各2件时应该付款400元。",
+        "knowledge_points":  [
+                                 "解决多步计算的实际问题",
+                                 "和差倍问题",
+                                 "经济问题"
+                             ],
+        "source":  "2025·重庆北碚·小升初真题"
+    },
+    {
+        "id":  "31",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明在做一道乘法题时，错把一个因数19看成了16，结果得到的积比正确的积少63，正确的积是()。",
+        "options":  [
+
+                    ],
+        "answer":  "399",
+        "solution":  "【分析】把一个因数19看成了16，这个因数减少19-16=3，积减少3个另一个因数;则3个另一个因数是63，另一个因数是63÷3=21，再计算21×19即可;据此解答.\n【详解】根据分析:63÷(19-16)×19=63÷3×19=21×19=399，小明在做一道乘法题时，错把一个因数19看成了16，结果得到的积比正确的积少63，正确的积是(399)。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "25-26四年级上·湖北襄阳·期中"
+    },
+    {
+        "id":  "32",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "a÷b=6……12(a,b为非0自然数),b最小是(),此时a=()。",
+        "options":  [
+
+                    ],
+        "answer":  "13;90",
+        "solution":  "【分析】根据除法规则，余数必须小于除数。因为b是非零自然数,b\u003e12，由此即可求出b的最小值。根据\"被除数=除数×商+余数\"，利用a=b×6+12即可求出a的值.\n【详解】b\u003e12，即b最小是13;a=13×6+12=78+12=90，此时a=90。",
+        "knowledge_points":  [
+                                 "余数与除数的关系",
+                                 "有余数的除法验算"
+                             ],
+        "source":  "2025·河北唐山·小升初真题"
+    },
+    {
+        "id":  "33",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "学校购买同样价格的课桌，第一次买16张，第二次买25张，两次花的钱相差990元。课桌的单价是()元，两次买课桌一共花了()元。",
+        "options":  [
+
+                    ],
+        "answer":  "110;4510",
+        "solution":  "【分析】单价×数量=总价，两次花的钱数差/买的数量差=课桌的单价;课桌的单价×两次买的总数量=两次花的总钱数，据此列式计算.\n【详解】990÷(25-16)=990÷9=110(元),110×(16+25)=110×41=4510(元),课桌的单价是110元，两次买课桌一共花了4510元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "25-26六年级上·河北唐山·期中"
+    },
+    {
+        "id":  "34",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "在一个减法算式里，差加减数，结果等于()。",
+        "options":  [
+
+                    ],
+        "answer":  "被减数",
+        "solution":  "【分析】根据公式被减数-减数=差，可得到另一个公式：差+减数=被减数，据此解答.\n【详解】由分析可知，在一个减法算式里，差加减数，结果等于被减数。",
+        "knowledge_points":  [
+                                 "加，减法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25三年级上·湖南永州·期中"
+    },
+    {
+        "id":  "35",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "要使算式\"360÷12×3-24\"的运算顺序为先算乘法，再算减法，最后算除法，请给这个算式添上括号:()。",
+        "options":  [
+
+                    ],
+        "answer":  "360÷(12×3-24)",
+        "solution":  "【分析】根据四则运算规则，先乘除，后加减，有括号要先算，原式按从左到右顺序计算除法和乘法，最后算减法。题目要求先乘，再减，最后除，需通过括号调整运算顺序。将\"12×3-24\"用括号括起，使乘法和减法优先于除法.\n【详解】要使算式\"360÷12×3-24\"的运算顺序为先算乘法，再算减法，最后算除法，请给这个算式添上括号:360÷(12×3-24)。",
+        "knowledge_points":  [
+                                 "无括号的运算顺序",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·河北唐山·期末"
+    },
+    {
+        "id":  "36",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把算式300-108=192,23+9=32,192÷32=6改写成综合算式是()。",
+        "options":  [
+
+                    ],
+        "answer":  "(300-108)÷(23+9)=6",
+        "solution":  "【分析】由题意得，观察算式300-108=192,23+9=32,192÷32=6可知，第一个算式的结果在第三个算式中作被除数，第二个算式的结果在第三个算式中作除数，据此列出综合算式为:300-108÷23+9。要想先算减法和加法，需要在算式300-108和算式23+9的两边分别加上小括号，即综合算式为:(300-108)÷(23+9)。\n【详解】(300-108)÷(23+9)=192÷32=6，把算式300-108=192,23+9=32,192÷32=6改写成综合算式是(300-108)÷(23+9)=6。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25三年级上·河南郑州·期中"
+    },
+    {
+        "id":  "37",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据3345+678=4023，可以直接得出4023-3345=()。根据314×26=8164，可以直接得出8164÷26=()。",
+        "options":  [
+
+                    ],
+        "answer":  "678;314",
+        "solution":  "【分析】根据加减法和乘除法的互逆关系，和减去一个加数等于另一个加数;积除以一个因数等于另一个因数;可以直接写出对应的减法算式和除法算式的答案.\n【详解】根据分析可知：根据3345+678=4023，可以直接得出4023-3345=678。根据314×26=8164，可以直接得出8164÷26=314。",
+        "knowledge_points":  [
+                                 "乘，除法的意义和各部分间的关系",
+                                 "加，减法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·河北廊坊·期末"
+    },
+    {
+        "id":  "38",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "计算960/[(24+6)×16]时，应先算()法，再算()法，最后算()法，得数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "加;乘;除;2",
+        "solution":  "【分析】根据整数四则混合运算的运算顺序：在没有括号的算式里，如果只含有乘除法，或只含有加减法，要从左往右依次计算;如果既含有乘除法又含有加减法，要先算乘除法再算加减法;如果有小括号，要先算小括号里面的，再算小括号外面的，据此解答.\n【详解】960/[(24+6)×16]=960/[30×16]=960÷480=2，计算960/[(24+6)×16]时，应先算加法，再算乘法，最后算除法，得数是2。",
+        "knowledge_points":  [
+                                 "有括号的运算顺序",
+                                 "带有中括号的混合运算",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·安徽淮北·期末"
+    },
+    {
+        "id":  "39",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "27加上18的和是(),再除以9得(),列成综合算式是()。",
+        "options":  [
+
+                    ],
+        "answer":  "45;5;(27+18)÷9",
+        "solution":  "【分析】(1)计算27加上18的和，用加法计算出结果。(2)用(1)所得的和除以9，计算出结果。(3)列综合算式时，因为要先算加法再算除法，所以需要给加法加上括号.\n【详解】(1)27+18=45;(2)45÷9=5;(3)(27+18)÷9",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25三年级上·黑龙江大庆·期中"
+    },
+    {
+        "id":  "40",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "两个数相加，如果其中的一个加数增加160，另一个加数减少110，那么和()。",
+        "options":  [
+
+                    ],
+        "answer":  "增加50",
+        "solution":  "【分析】和的变化规律：一个加数不变，另一个加数加上几或减去几，和就等于原来的和加上几或减去几。由题意得，两个数相加，其中的一个加数增加160，另一个加数减少110.160\u003e110，即增加的比减少的值多。160-110=50，所以和会增加50.\n【详解】两个数相加，如果其中的一个加数增加160，另一个加数减少110，那么和增加50。",
+        "knowledge_points":  [
+                                 "和与差的变化规律"
+                             ],
+        "source":  "24-25四年级下·河南驻马店·期末"
+    },
+    {
+        "id":  "41",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算660/[(160-127)×2]时，要先算()法，再算()法，最后算()法。",
+        "options":  [
+
+                    ],
+        "answer":  "减;乘;除",
+        "solution":  "【分析】根据四则混合运算的规则，有括号时先算括号内的，多层括号则先算小括号里的，再算中括号里的，最后进行括号外的运算。计算660/[(160-127)×2]时，运算顺序如下：先算小括号内的减法:160-127=33;再算中括号内的乘法:33×2=66;最后算括号外的除法:660÷66=10。以此答题即可.\n【详解】根据分析可知:660/[(160-127)×2]=660/[33×2]=660÷66=10，计算660/[(160-127)×2]时，要先算减法，再算乘法，最后算除法。",
+        "knowledge_points":  [
+                                 "带有中括号的混合运算",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "42",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "算式32×75-25÷25，如果要改变运算顺序，先算减法，再算除法，最后一步算乘法，那么应把算式改成(),得到的积是()。",
+        "options":  [
+
+                    ],
+        "answer":  "32*[(75-25)÷25];64",
+        "solution":  "【分析】32×75-25÷25，是先算乘法和除法，再算减法，改写成先算减法，再算除法，最后一步算乘法，是把减法提前到第一步，所以需要在减法上加小括号来改变运算顺序，算式变为32×(75-25)÷25;再把除法提前到乘法之前计算，则需要在除法上加中括号，算式变成32*[(75-25)÷25],由此求解.\n【详解】32*[(75-25)÷25]=32*[50÷25]=32×2=64，算式32×75-25÷25，如果要改变运算顺序，先算减法，再算除法，最后一步算乘法，那么应把算式改成32*[(75-25)÷25],得到的积是64。",
+        "knowledge_points":  [
+                                 "有括号的运算顺序",
+                                 "带有中括号的混合运算",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·浙江宁波·期末"
+    },
+    {
+        "id":  "43",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "王师傅驾车从甲地开往乙地交货。如果他往返都以每小时60千米的速度行驶，正好可以按时返回甲地。可是，当到达乙地时，他发现从甲地到乙地的速度只有每小时50千米。如果他想按时返回甲地，他应以每小时()千米的速度往回开。",
+        "options":  [
+
+                    ],
+        "answer":  "75",
+        "solution":  "【分析】根据题意，可以设甲，乙两地的距离是300千米。如果往返都以每小时60千米的速度行驶，那么行驶的路程是((300×2))千米，根据\"路程/速度=时间\"求出往返的时间;实际从甲地到乙地的速度是每小时50千米，根据\"路程/速度=时间\"求出实际从甲地到乙地的时间;用往返的时间减去实际从甲地到乙地的时间，即是实际返回甲地所需的时间;再根据\"速度=路程/时间\"求出按时返回甲地往回开所需的速度.\n【详解】设甲，乙两地的距离是300千米。300×2÷60=600÷60=10(小时),300÷50=6(小时)300÷(10-6)=300÷4=75(千米),他应以每小时75千米的速度往回开。",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "2025·重庆渝北·小升初真题"
+    },
+    {
+        "id":  "44",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "计算12*[(540-270)÷45]时，应先算()法，再算()法，最后算()法。",
+        "options":  [
+
+                    ],
+        "answer":  "减;除;乘",
+        "solution":  "【分析】根据四则混合运算顺序，从左往右依次计算，先算乘，除法，再算加，减法，有括号的先算括号里面的。计算12*[(540-270)÷45]时，先算小括号里的减法，再算中括号里的除法，最后算乘法，以此答题即可.\n【详解】根据分析可知:\n12*[(540-270)÷45]\n=12×(270÷45)\n=12×6\n=72\n计算12*[(540-270)÷45]时，应先算减法，再算除法，最后算乘法。",
+        "knowledge_points":  [
+                                 "有括号的运算顺序",
+                                 "带有中括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·广东汕头·期末"
+    },
+    {
+        "id":  "45",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "○ *11=132,□-17=128，那么 ○+□=() 。",
+        "options":  [
+
+                    ],
+        "answer":  "157",
+        "solution":  "【分析】根据四则运算各部分名称和关系\"因数×因数=积，因数=积/另一个因数，被减数-减数=差，被减数=差+减数\"，据此解答.\n【详解】○=132÷11=12,□=128+17=145,12+145=157，所以 ○+□=157 。",
+        "knowledge_points":  [
+                                 "乘，除法的意义和各部分间的关系",
+                                 "加，减法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·河北邢台·期末"
+    },
+    {
+        "id":  "46",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把180×72+32÷8的运算顺序变成：先算加法，再算除法，最后算乘法。此时算式应为:()。",
+        "options":  [
+
+                    ],
+        "answer":  "180*[(72+32)÷8]",
+        "solution":  "【分析】180×72+32÷8的运算顺序是可以同时先算乘法和除法，最后算加法。要想把算式变成先算加法，再算除法，最后算乘法，要给加法加上小括号，给除法加上中括号.\n【详解】把180×72+32÷8的运算顺序变成：先算加法，再算除法，最后算乘法。180*[(72+32)÷8]=180*[104÷8]=180×13=2340，此时算式应为:180*[(72+32)÷8]。",
+        "knowledge_points":  [
+                                 "有括号的运算顺序",
+                                 "带有中括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·河北唐山·期末"
+    },
+    {
+        "id":  "47",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "(159-105÷5)×30应先算()法，再算()法，最后算()法。",
+        "options":  [
+
+                    ],
+        "answer":  "除;减;乘",
+        "solution":  "【分析】四则混合运算的运算顺序，同级运算，从左往右依次计算，既有乘除，又有加减，先算乘除，后算加减，有括号的先算括号里面的.\n【详解】根据分析可知,(159-105÷5)×30应先算除法，再算减法，最后算乘法。",
+        "knowledge_points":  [
+                                 "有括号的运算顺序"
+                             ],
+        "source":  "24-25四年级下·辽宁盘锦·期末"
+    },
+    {
+        "id":  "48",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "1040减去一个数得746，这个数是();被除数是283，商是16，余数是11，则除数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "294;17",
+        "solution":  "【分析】求减数，根据被减数-差=减数，解答即可;求除数，根据被除数=商×除数+余数，除数=(被除数-余数)/商解答即可.\n【详解】1040-746=294,(283-11)÷16=272÷16=17，所以1040减去一个数得746，这个数是294;被除数是283，商是16，余数是11，则除数是17。",
+        "knowledge_points":  [
+                                 "乘，除法的意义和各部分间的关系",
+                                 "加，减法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·四川南充·期末"
+    },
+    {
+        "id":  "49",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "两个数的差是125，如果被减数减少15，减数减少15，这时差是();如果被减数增加15，减数减少20，这时差是()。",
+        "options":  [
+
+                    ],
+        "answer":  "125;160",
+        "solution":  "【分析】两个数的差是125，当被减数和减数同时减少相同的数，差不变，还是125;两个数的差是125，当被减数增加15，则差也增加15，这时减数再减少20，则差再增加20，所以这时的差是125+15+20.\n【详解】当被减数和减数同时减少相同的数，差不变，所以如果被减数减少15，减数减少15，这时差是125;125+15+20=140+20=160，所以如果被减数增加15，减数减少20，这时差是160。",
+        "knowledge_points":  [
+                                 "和与差的变化规律"
+                             ],
+        "source":  "24-25三年级下·广东汕头·期末"
+    },
+    {
+        "id":  "50",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把下面几个算式改写成一个综合算式，并计算出得数。48÷8=6,35×6=210,210-86=124。综合算式:()。",
+        "options":  [
+
+                    ],
+        "answer":  "35×(48÷8)-86=124",
+        "solution":  "【分析】此题的运算顺序是：先算除法，再算乘法，最后算减法，所以48除以8应带上小括号，小括号外面是乘法，然后用这个积减86即可.\n【详解】综合式:35×(48÷8)-86=124。\n35×6-86=210-86=124。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·河南郑州·期末"
+    },
+    {
+        "id":  "51",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在一个加法算式中，两个加数相等，和是214，其中一个加数是();在一个除法算式中，被除数和除数相等(都不为0),商是()。",
+        "options":  [
+
+                    ],
+        "answer":  "107;1",
+        "solution":  "【分析】两个加数相等，和是214，所以用214÷2即可计算出其中的一个加数是几;两个相同的数相除，商是1，被除数和除数相等(都不为0),商是1，据此解题,\n【详解】214÷2=107，在一个加法算式中，两个加数相等，和是214，其中一个加数是107;在一个除法算式中，被除数和除数相等(都不为0),商是1。",
+        "knowledge_points":  [
+                                 "乘，除法的意义和各部分间的关系",
+                                 "加，减法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·河北唐山·期末"
+    },
+    {
+        "id":  "52",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在计算65*[(70-42)÷7]时，应先算()法，再算()法，最后算()法，计算结果是()。",
+        "options":  [
+
+                    ],
+        "answer":  "减;除;乘;260",
+        "solution":  "【分析】同级运算，从左到右依次计算;既有乘除又有加减的，先算乘除，后算加减;有括号时，先算括号里面的;有多层括号时，先算小括号里的，再算中括号里面的;据此即可解答.\n【详解】65*[(70-42)÷7]=65*[28÷7]=65×4=260，所以在计算65*[(70-42)÷7]时，应先算减法，再算除法，最后算乘法，计算结果是260。",
+        "knowledge_points":  [
+                                 "有括号的运算顺序",
+                                 "带有中括号的混合运算",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·湖南衡阳·期末"
+    },
+    {
+        "id":  "53",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "将算式840-15×24÷6的运算顺序改成先算乘法，再算减法，最后算除法，那么这个算式应改写为(),按照这样的运算顺序算出的得数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "(840-15×24)÷6;80",
+        "solution":  "【分析】算式840-15×24÷6的运算顺序是先算乘法，再算除法，最后算减法;要使其运算顺序改成先算乘法，再算减法，最后算除法，要用小括号把840-15×24括起来，据此即可解答.\n【详解】(840-15×24)÷6=(840-360)÷6=480÷6=80，将算式840-15×24÷6的运算顺序改成先算乘法，再算减法，最后算除法，那么这个算式应改写为(840-15×24)÷6，按照这样的运算顺序算出的得数是80。",
+        "knowledge_points":  [
+                                 "有括号的运算顺序",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·山东日照·期末"
+    },
+    {
+        "id":  "54",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "已知两个因数的积是120，其中一个因数是15，另一个因数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "8",
+        "solution":  "【分析】已知两个因数的积和其中一个因数，求另一个因数，根据\"积/一个因数=另一个因数\"可直接列除法算式计算.\n【详解】根据因数与积的关系，另一个因数为:120÷15=8所以另一个因数是8。",
+        "knowledge_points":  [
+                                 "乘，除法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·陕西安康·期末"
+    },
+    {
+        "id":  "55",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果要把算式864-12×6÷9的运算顺序改写成先算乘法，再算减法，最后算除法，那么该算式应改为(),结果是()。",
+        "options":  [
+
+                    ],
+        "answer":  "(864-12×6)÷9;88",
+        "solution":  "【分析】原算式要求改写为先算乘法，再算减法，最后算除法。根据运算顺序规则，有括号先算括号内，所以需将减法和乘法部分用括号括起来，以保证先算乘法，再算减法，最后算除法;计算改写后的算式即可得到结果。本题要先根据要求确定运算顺序调整后的算式，再按照新运算顺序分步计算结果.\n【详解】根据要求改写算式为(864-12×6)÷9,(864-12×6)÷9=(864-72)÷9=792÷9=88，所以算式应改为(864-12×6)÷9，结果是88。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·四川自贡·期末"
+    },
+    {
+        "id":  "56",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "给算式480÷16+24添上括号，使其运算顺序变为先加后除，算式应变为()。",
+        "options":  [
+
+                    ],
+        "answer":  "480÷(16+24)",
+        "solution":  "【分析】原算式的运算顺序是先算除法，再算加法，要使先算加法，后算除法，应将16加24写在小括号里，小括号外面是除法，据此解答.\n【详解】给算式480÷16+24添上括号，使其运算顺序变为先加后除，算式应变为480÷(16+24)。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·湖南长沙·期末"
+    },
+    {
+        "id":  "57",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "130-30÷5×4如果想改变运算顺序，先算减法，则必须使用小括号，算式是(),计算的结果是()。",
+        "options":  [
+
+                    ],
+        "answer":  "(130-30)÷5×4;80",
+        "solution":  "【分析】原式130-30÷5×4按运算顺序先算除法，乘法，再算减法。若想先算减法，需用小括号将130-30括起来，再依次计算除法和乘法.\n【详解】添加小括号改变运算顺序:(130-30)÷5×4;(130-30)÷5×4=100÷5×4=20×4=80;130-30÷5×4如果想改变运算顺序，先算减法，则必须使用小括号，算式是(130-30)÷5×4，计算的结果是80。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·安徽淮北·期末"
+    },
+    {
+        "id":  "58",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在算式9+42÷7-4中添上括号，使运算顺序为\"减→除→加\"，添上括号后的算式是();在算式36÷4-2×3中添上括号，使计算结果为6，添上括号后的算式是()。",
+        "options":  [
+
+                    ],
+        "answer":  "9+42÷(7-4);36÷[(4-2)×3]",
+        "solution":  "【分析】使运算顺序为\"减除加\"，减法要加上小括号，添上括号后是9+42÷(7-4);观察36÷4-2×3中数字特点，想:36÷6=6，则可以变为36÷[(4-2)×3]。\n【详解】使运算顺序为\"减→除→加\"，添上括号后是9+42÷(7-4);观察36÷4-2×3中数字特点，想:36÷6=6，则可以变为36÷[(4-2)×3]。36÷[(4-2)×3]=36÷[2×3]=36÷6=6，所以添上括号后的算式是:36÷[(4-2)×3]。",
+        "knowledge_points":  [
+                                 "带有中括号的混合运算",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·浙江杭州·期末"
+    },
+    {
+        "id":  "59",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "158*[(27+54)÷9]的运算顺序是先算加法，再算除法，最后算乘法。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】在四则混合运算中，有括号要先算括号里面的，再算括号外面的。根据四则运算的规则,158*[(27+54)÷9],要先计算小括号内的加法，再计算中括号内的除法，最后进行括号外的乘法。据此解答.\n【详解】由分析可知,158*[(27+54)÷9],要先算加法，再算除法，最后算乘法。题目说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "带有中括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·河北邢台·期末"
+    },
+    {
+        "id":  "60",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "158*[(27+54)÷9]的运算顺序是先算加法，再算除法，最后算乘法。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】根据四则运算的规则，先算括号内的运算，括号内先算小括号里的加法，再算中括号里的除法，最后进行括号外的乘法，据此解答即可.\n【详解】根据分析可知,158*[(27+54)÷9]的运算顺序是先算加法，再算除法，最后算乘法。故答案为:√",
+        "knowledge_points":  [
+                                 "带有中括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·河北邢台·期末"
+    },
+    {
+        "id":  "61",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "12×3÷12×3=1。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】一个算式中既有乘法，又有除法，要按照从左往右的顺序依次计算。由题意得，在算式12×3÷12×3中，要先算最左边的乘法，再算中间的除法，最后算最右边的乘法。据此解答.\n【详解】12×3÷12×3=36÷12×3=3×3=9，即算式12×3÷12×3的结果是9。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "无括号的运算顺序"
+                             ],
+        "source":  "24-25四年级下·河北唐山·期末"
+    },
+    {
+        "id":  "62",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "两数相减，被减数增加8，减数减少8，差不变。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】被减数增加8，如果减数不变，差就增加8;如果减数减少8，被减数不变，差也增加8。被减数增加8，减数减少8，差增加16.\n【详解】假设原被减数为30，减数为10，则差为30-10=20。变化后：被减数:30+8=38;减数:10-8=2;新差:38-2=36;原差20变为36，差增大16，所以差不变是错误的。故答案为: ×",
+        "knowledge_points":  [
+                                 "和与差的变化规律"
+                             ],
+        "source":  "24-25四年级下·四川南充·期末"
+    },
+    {
+        "id":  "63",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "脱式计算。①864÷12/8;②(608-173)÷15;③256÷16×12;④(347+459)÷62",
+        "options":  [
+
+                    ],
+        "answer":  "9;29;192;13",
+        "solution":  "【分析】①计算864÷12/8，根据四则运算顺序，从左到右计算即可。②计算(608-173)÷15，根据四则运算顺序，先算括号内减法，再算括号外除法。③计算256÷16×12，根据四则运算顺序，按照从左到右的顺序依次计算。④计算(347+459)÷62，根据四则运算顺序，先算括号内的加法，再算除法.\n【详解】①864÷12/8=72÷8=9;②(608-173)÷15=435÷15=29;③256÷16×12=16×12=192;④(347+459)÷62=806÷62=13",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "25-26四年级上·广西贵港·期中"
+    },
+    {
+        "id":  "64",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "脱式计算。14×70+30÷5;243/[(21-18)×3];14×(46+54)-568",
+        "options":  [
+
+                    ],
+        "answer":  "986;27;832",
+        "solution":  "【分析】(1)先算乘法和除法，最后算加法;(2)先算小括号内的减法，再算中括号内的乘法，最后算中括号外的除法;(3)先算小括号内的加法，再算乘法，最后算减法.\n【详解】14×70+30÷5=980+6=986;243/[(21-18)×3]=243/[3×3]=243÷9=27;14×(46+54)-568=14×100-568=1400-568=832",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "带有中括号的混合运算"
+                             ],
+        "source":  "25-26四年级上·山西晋中·期中"
+    },
+    {
+        "id":  "65",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "计算题",
+        "difficulty":  0.85,
+        "question":  "计算下面各题，并利用加，减法各部分间的关系进行验算。290+576=();894-487=()",
+        "options":  [
+
+                    ],
+        "answer":  "866;407",
+        "solution":  "【分析】整数加法计算法则：相同数位对齐，从个位算起;哪一位上的数相加满十，就向前一位进一;得数的数位也要对齐;整数减法计算法则：相同数位对齐，从低位减起，哪一位上的数不够减，就从它的前一位退一当十，和本位上的数合并在一起，再减;加减法各部分间的关系：加数+加数=和，一个加数=和-另一个加数;被减数-减数=差，被减数=差+减数，减数=被减数-差;据此进行验算即可.\n【详解】290+576=866\n![](images/2026-03-04%20213830.jpg)\n894-487=407\n![](images/2026-03-04%20213847.jpg)",
+        "knowledge_points":  [
+                                 "两，三位数的连续进位加法",
+                                 "两，三位数的连续退位减法",
+                                 "加，减法验算",
+                                 "加，减法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "66",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "计算题",
+        "difficulty":  0.85,
+        "question":  "列式计算。16乘316与59的和除以3的商，积是多少。",
+        "options":  [
+
+                    ],
+        "answer":  "2000",
+        "solution":  "【分析】根据题意可知，先算出316与59的和，再算求出的和除以3的商，最后用16乘求出的商即可，计算时按照\"先算和，再算商，最后算积\"的顺序列式计算(注意添加括号保证运算顺序)\n【详解】16×[(316+59)÷3]=16×[375÷3]=16×125=2000，所以16乘316与59的和除以3的商，积是2000。",
+        "knowledge_points":  [
+                                 "带有中括号的混合运算",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·辽宁盘锦·期末"
+    },
+    {
+        "id":  "67",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "一个数除以30，商15余23，这个数是多少。",
+        "options":  [
+
+                    ],
+        "answer":  "473",
+        "solution":  "【分析】在有余数的除法中，被除数=商×除数+余数。由题意得，一个数除以30，商15余23，那么直接用15乘上30再加上23即可算出这个数是多少.\n【详解】15×30+23=450+23=473，故这个数是473。",
+        "knowledge_points":  [
+                                 "乘，除法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级上·重庆九龙坡·期中"
+    },
+    {
+        "id":  "68",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "列式计算。625除以25的商加上16乘29的积，和是多少。",
+        "options":  [
+
+                    ],
+        "answer":  "489",
+        "solution":  "【分析】根据题意，先用625除以25，求出商，再计算16乘29，求出乘积，最后把两数相加，求出和，列式计算即可.\n【详解】根据分析可知:625÷25+16×29=25+464=489，答：和是489。",
+        "knowledge_points":  [
+                                 "乘加",
+                                 "乘减混合运算",
+                                 "除加",
+                                 "除减混合运算"
+                             ],
+        "source":  "24-25四年级下·广东汕头·期末"
+    },
+    {
+        "id":  "69",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "西湖，位于浙江省杭州市，是中国首批国家重点风景名胜区。2011年被列入世界遗产名录。暑假里，张老师带着69名同学来西湖泛舟.\n| 船型 | 限乘坐人数 | 租金 |\n| :--- | :--------: | :--: |\n| 大船 | 6 | 30 |\n| 小船 | 4 | 24 |\n(1)比较哪种租金便宜：租大船平均每人()元，租小船平均每人()元，租()船便宜.\n(2)制定最省钱的租船方案：一共()人，需要租大船()条，还剩()人，剩余的人租()条小船，这样保证不空座位，会更省钱.\n(3)计算此方案需要的费用：租大船需要()元，租小船需要()元，一共需要()元.\n我发现：解决如何租船\"最省钱\"的问题：先计算哪种船的每人租金便宜，则考虑尽量()租这种船,()租另一种船，再调整到没有空座或空座最少。",
+        "options":  [
+
+                    ],
+        "answer":  "5元;6元;大;70;11;4;1;330;24;354;多;少",
+        "solution":  "【分析】(1)租大船平均每人多少钱可以通过除法来列式解答，大船的租金÷每船限坐人数;同理可求得租小船平均每人的价格，然后两者进行比较;(2)总人数是张老师加上学生数，用总人数÷大船每船限坐人数，商就是大船的数量，余数的人数再安排小船;(3)大船租用费用=大船的租金×大船的数量，同理可求得小船的租用费用，两者相加就是总共的费用。解决如何租船\"最省钱\"的问题：先计算哪种船的每人租金便宜，则考虑尽量多租这种船，少租贵的船。\n【详解】(1)根据分析，租大船平均每人多少钱可列式为:30÷6=5(元);租小船平均每人多少钱可列式为:24÷4=6(元);5元\u003c6元，即租大船便宜。(2)总人数:1+69=70(人);70÷6=11(条)……4(人)，即需要租大船11条，还剩4人去租小船;4÷4=1(条)，即需要租小船1条，这样保证不空座位，会更省钱。(3)根据分析，大船租用费用:11×30=330(元);小船租用费用:1×24=24(元);总费用:330+24=354(元)。即解决如何租船\"最省钱\"的问题：先计算哪种船的每人租金便宜，则考虑尽量多租这种船，少租另一种船。",
+        "knowledge_points":  [
+                                 "有余数除法的实际应用",
+                                 "租船问题"
+                             ],
+        "source":  "25-26四年级下·全国·课前预习"
+    },
+    {
+        "id":  "70",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "小明寒假看书，原计划每天看12页,25天看完;实际每天比原计划多看3页，实际多少天看完了全书。",
+        "options":  [
+
+                    ],
+        "answer":  "20天",
+        "solution":  "【分析】用原计划每天看的页数×计划看的天数，即12×25，求出这本书的总页数;再用原计划每天看的页数+3，求出实际每天看的页数，再用这本书的总页数/实际每天看的页数，即可求出实际看的天数.\n【详解】12×25÷(12+3)=12×25÷15=300÷15=20(天),答：实际20天看完了全书。",
+        "knowledge_points":  [
+                                 "归总问题",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "25-26五年级上·内蒙古巴彦淖尔·期末"
+    },
+    {
+        "id":  "71",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "小明家和小红家相距4千米，两人同时从自家骑自行车相向而行。已知小明每分钟骑180米，小红每分钟骑220米，两人多久后相遇。为了直观地寻找数量关系，请你根据题意，画出线段图。",
+        "options":  [
+
+                    ],
+        "answer":  "10分钟;图见详解",
+        "solution":  "【分析】把4千米换算成4000米，小明和小红从两地同时出发，相向而行，两人行驶的路程之和就是总路程4000米。根据\"相遇时间=总路程/速度和\"这个核心公式，我们先算出两人的速度和是180+220=400(米/分钟),再用总路程除以速度和，就能求出相遇时间了。掌握相遇问题的核心公式\"相遇时间=总路程/速度和\"，并能正确进行单位换算和线段图的绘制.\n【详解】线段图如下:\n![](images/d1e825f8f6d44ec93658978086fbb4c29ac5e77f7da9de2a7fbe58de56bff195.jpg)\n4千米=4×1000=4000米,4000÷(180+220)=4000÷400=10(分钟),答：两人10分钟后相遇。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "相遇问题",
+                                 "千米和米之间的进率与换算"
+                             ],
+        "source":  "25-26六年级上·福建厦门·期末"
+    },
+    {
+        "id":  "72",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "某超市在十一黄金周期间搞促销活动，店内牛奶实行买四送一活动。王阿姨买了16盒牛奶，每盒55元，一共花了多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "715元",
+        "solution":  "【分析】首先根据买四送一，可得花4盒牛奶的钱，可以买到5盒牛奶，据此求出买16盒牛奶一共需要多少盒牛奶的钱;然后根据总价=单价×数量，求出一共花了多少钱即可.\n【详解】16÷(1+4)=16÷5=3(组)。1(盒),4×3+1=12+1=13(盒),13×55=715(元),答：一共花了715元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "乘加",
+                                 "乘减混合运算",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "25-26五年级上·湖南永州·期末"
+    },
+    {
+        "id":  "73",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.4,
+        "question":  "兄妹家与学校相距1400米。哥哥骑自行车每分钟行200米，妹妹每分钟走80米。兄妹俩同时出发，哥哥到学校后立即返回接妹妹。哥哥接到妹妹时，妹妹离家有多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "800米",
+        "solution":  "【分析】由题知，哥哥与妹妹相遇时，他们所走的路程相当于从家到学校距离的2倍，也就是1400×2=2800(米)两人同时出发，根据相遇时间=路程/速度和，可以求出两人相遇的时间，再根据路程=速度×时间，用妹妹每分钟行的速度乘相遇时间，就是妹妹离家的距离。往返相遇问题的路程和=两地距离*2.\n【详解】1400×2÷(200+80)=1400×2÷280=2800÷280=10(分钟),80×10=800(米),答：哥哥接到妹妹时，妹妹离家有800米。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "相遇问题"
+                             ],
+        "source":  "24-25五年级上·山东济南·期末"
+    },
+    {
+        "id":  "74",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "春节快到了，贴春联是中国的传统习俗。老师准备写一些春联，原计划每天写60副,14天写完。实际每天比计划多写10副，照这样计算，实际提前几天完成。",
+        "options":  [
+
+                    ],
+        "answer":  "2天",
+        "solution":  "【分析】用60×14，求出准备写春联的数量，实际每天比计划多写10副，用原来每天写春联的数量+10，求出实际每天写春联的数量，再用准备写春联的数量/实际每天写春联的数量，求出实际写春联用的天数，再用原计划写春联用的天数-实际写春联用的天数，即可解答.\n【详解】60×14÷(60+10)=60×14÷70=840÷70=12(天),14-12=2(天),答：实际提前了2天完成。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25五年级上·河北唐山·期末"
+    },
+    {
+        "id":  "75",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "为响应全民健身计划，乐乐和爸爸沿400米环形跑道跑步锻炼。两人从同一起点同向出发，爸爸每分钟跑245米，乐乐每分钟跑205米。经过多少分钟爸爸能比乐乐多跑1圈。",
+        "options":  [
+
+                    ],
+        "answer":  "10分钟",
+        "solution":  "【分析】一圈的长度为400米。爸爸的速度为245米/分钟，乐乐的速度为205米/分钟，爸爸每分钟比乐乐多跑40米。多跑一圈即多跑400米，因此所需时间等于多跑的距离除以速度差.\n【详解】400÷(245-205)=400÷40=10(分钟) 答：经过10分钟爸爸能比乐乐多跑1圈。",
+        "knowledge_points":  [
+                                 "追及问题",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "2025五年级上·全国·专题练习"
+    },
+    {
+        "id":  "76",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.4,
+        "question":  "国庆节期间，四年级50名同学去野营。大帐篷每顶住5人，租金100元;小帐篷每顶住4人，租金96元。怎样租帐篷最省钱。",
+        "options":  [
+
+                    ],
+        "answer":  "租10顶大帐篷最省钱",
+        "solution":  "【分析】先比较两种帐篷的人均租金，确定优先租用大帐篷，再计算全租大帐篷的数量，确定最省钱方案。通过比较两种帐篷的人均租金，确定租哪种帐篷更划算.\n【详解】100÷5=20(元),96÷4=24(元),20\u003c24，应尽量多租大帐篷。50÷5=10(顶),答：租10顶大帐篷最省钱。",
+        "knowledge_points":  [
+                                 "租船问题",
+                                 "经济问题"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "77",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "李经理和公司45名员工去吃自助餐，中餐每人98元，西餐每人126元，他们吃西餐比吃中餐多花多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "1288元",
+        "solution":  "【分析】根据题意，已知李经理和公司45名员工去吃自助餐，中餐每人98元，西餐每人126元，先用126减去98，求出吃西餐比吃中餐每人差的钱数，再乘46，就是他们吃西餐比吃中餐多花多少钱，列式计算即可.\n【详解】根据分析可知:(126-98)×46=28×46=1288(元),答：他们吃西餐比吃中餐多花1288元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "25-26四年级上·江西宜春·期中"
+    },
+    {
+        "id":  "78",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "《琵琶行》是唐朝诗人白居易的一首诗，包括小序和正文两部分，其中正文有616个字，小序比正文少478个字。小序和正文一共有多少个字。",
+        "options":  [
+
+                    ],
+        "answer":  "754个",
+        "solution":  "【分析】小序比正文少478个字，可知正文字数-478=小序字数;再用小序字数+正文字数，求出一共有多少个字，据此解答.\n【详解】小序:616-478=138(个),一共:138+616=754(个),答：小序和正文一共有754个字。",
+        "knowledge_points":  [
+                                 "和差倍问题"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "79",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "市园林局计划新栽树苗76行，每行24棵，已经栽了956棵，以后每天栽31棵。还要栽多少天才能完成任务。",
+        "options":  [
+
+                    ],
+        "answer":  "28天",
+        "solution":  "【分析】根据行数和每行的棵数，利用乘法计算出一共要栽多少棵，用总棵数减去已经栽的棵数，可以求出剩下的棵数，用剩下的棵数除以每天栽的树苗的棵数，即可求出还要栽几天才能完成任务.\n【详解】(76×24-956)÷31=(1824-956)÷31=868÷31=28(天),答：还要栽28天才能完成任务。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "乘加",
+                                 "乘减混合运算",
+                                 "除数是两位数的笔算除法",
+                                 "归总问题"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "80",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "画线段图是解决数学问题，分析数量关系的好方法。你能借助画线段图把这道题的解法给大家讲清楚吗。玩具厂接到了生产180个汽车玩具的订单。前3天每天生产32个，剩下的要求2天完成，剩下的平均每天要生产多少个。",
+        "options":  [
+
+                    ],
+        "answer":  "图见详解;42个",
+        "solution":  "【分析】画线段图，用一条线段表示180个汽车玩具，前3天用3条小线段表示，每条小线段表示32个，剩下的用2条稍长一点的线段表示，求这两条线段每段表示多少个。用总的个数减去前3天已经生产的个数，等于剩下的个数。用剩下的个数除以2，就是剩下的平均每天要生产的个数.\n【详解】画线段图如下:\n![](images/4fdbdd7cc7a9a4d115f9b674e11159476d66af3504fe0e3de273f7f79753d776.jpg)\n(180-32×3)÷2=(180-96)÷2=84÷2=42(个),答：剩下的平均每天要生产42个。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "解决多步计算的实际问题"
+                             ],
+        "source":  "25-26五年级上·河北唐山·期中"
+    },
+    {
+        "id":  "81",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.4,
+        "question":  "妈妈从家到公司，要行1800米，若以45米/分的速度刚好赶上上班。某天，她在去公司途中遇到一位熟人耽误了3分钟，为了赶上上班，她必须在后面每分钟多行15米。她是在离家多远处遇到的熟人。",
+        "options":  [
+
+                    ],
+        "answer":  "1260米",
+        "solution":  "【分析】计算耽误3分钟少走的路程：原本速度是45米/分，耽误了3分钟，少走的路程就是速度乘耽误的时间，即45×3=135(米)。计算提速后弥补少走路程所需时间：提速后速度比原来每分钟多走15米。少走的135米，用少走的路程除以提速后比原来每分钟多走的路程，就是提速后弥补少走路程需要的时间。即135÷15=9(分钟)。计算提速后行走的路程：提速后速度为45+15=60(米/分),行走时间是9分钟，根据路程=速度×时间可算出这部分路程。提速后行走的路程为60×9=540(米)。计算遇到熟人时离家的距离：用总路程1800米减去提速后行走的540米，就是遇到熟人时离家的距离。据此解答。解题关键在于通过分析耽误时间导致少走的路程，以及提速后弥补少走路程的情况，利用速度，时间和路程的关系求出遇到熟人时离家的距离.\n【详解】45×3=135(米),135÷15=9(分钟),1800-(45+15)×9=1800-60×9=1800-540=1260(米),答：她是在离家1260米处遇到的熟人。",
+        "knowledge_points":  [
+                                 "解决多步计算的实际问题",
+                                 "基础行程问题",
+                                 "带有小括号的混合运算",
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "25-26五年级上·山东临沂·期中"
+    },
+    {
+        "id":  "82",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "星星小学三年级有三个综合实践小队，第一小队有46人，第二小队有24人，第三小队有20人。现在需要把第三小队的人全部分到第一，第二小队当中去，并且要求补充后，第一小队的人数是第二小队的2倍。第三小队有多少人要调到第一小队。",
+        "options":  [
+
+                    ],
+        "answer":  "14人",
+        "solution":  "【分析】总人数不变，所以将三小队的人数相加，计算出总人数，也就是(46+24+20)人，将第三小队的人全部分到第一，第二小队当中后，第一小队的人数是第二小队的2倍，用总人数/(1+2)即为第二小队的人数，然后再用第二小队的人数*2即可计算出第一小队的人数，然后用现在第一小队的人数减去原来第一小队的人数即可.\n【详解】46+24+20=70+20=90(人),90÷(1+2)=90÷3=30(人),30×2=60(人),60-46=14(人),答：第三小队有14人要调到第一小队。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "和差倍问题"
+                             ],
+        "source":  "25-26三年级上·全国·周测"
+    },
+    {
+        "id":  "83",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "武汉植物园举办花展，成人票每张40元，儿童票每张20元，团体票(10人及以上)每张30元。现有6位成人和8位儿童，怎样购票最省钱。最少需要多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "买10人团体票(6成人+4儿童)和4张儿童票最省钱;最少380元",
+        "solution":  "【分析】分别计算\"成人买成人票，儿童买儿童票\"\"全部买团体票\"\"部分买团体票，部分买儿童票\"三种方案的费用，再比较得出最省钱的方式.\n【详解】方案一：成人买成人票，儿童买儿童票6×40+8×20=240+160=400(元)。方案二：全部买团体票(6+8)×30=14×30=420(元)。方案三:6位成人和4位儿童买团体票，剩下4位儿童买儿童票10×30+(8-4)×20=10×30+4×20=300+80=380(元),380\u003c400\u003c420。答：买10人团体票(6成人+4儿童)和4张儿童票最省钱;最少需要380元。",
+        "knowledge_points":  [
+                                 "购票问题"
+                             ],
+        "source":  "24-25五年级上·湖北襄阳·期中"
+    },
+    {
+        "id":  "84",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.4,
+        "question":  "小丽和妈妈今年的年龄和是40岁,5年后，妈妈年龄是小丽的4倍。小丽和妈妈今年各有多少岁。",
+        "options":  [
+
+                    ],
+        "answer":  "5岁;35岁",
+        "solution":  "【分析】5年后，每人增长5岁，所以小丽和妈妈增长年龄5×2=10(岁),5年后，妈妈年龄是小丽的4倍，把5年后小丽的年龄看作1份，则妈妈的年龄就是4份，两人的年龄和就是(1+4)份，小丽和妈妈今年的年龄和是40岁,5年后两人的年龄和就是(40+10)岁，除以(1+4)即可得一份的年龄，即5年后小丽的年龄，再减去5年即可得小丽今年的年龄;用今年两人的年龄和减去小丽今年的年龄，即可得妈妈今年的年龄。据此列式解答。解题关键是通过计算5年后的年龄和，利用和倍问题的解法求出5年后小丽的年龄，再倒推今年两人的年龄.\n【详解】5×2=10(岁),(40+10)÷(1+4)=50÷5=10(岁),10-5=5(岁),40-5=35(岁),答：小丽今年有5岁，妈妈今年有35岁。",
+        "knowledge_points":  [
+                                 "年龄问题",
+                                 "带有小括号的混合运算",
+                                 "和差倍问题"
+                             ],
+        "source":  "25-26五年级上·湖南永州·阶段练习"
+    },
+    {
+        "id":  "85",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "甲，乙两只船同时从相距126千米的两个码头相向开出，甲船每小时行20千米，乙船每小时行15千米。几小时后两船相遇。",
+        "options":  [
+
+                    ],
+        "answer":  "3.6小时",
+        "solution":  "【分析】两个码头相距的总路程为126千米;甲船速度为20千米/小时;乙船速度为15千米/小时。相向而行时，两船的相对速度为速度之和为(20+15)(千米/小时)。根据相遇问题公式，相遇时间=总路程/速度和，把用126除以(20+15)计算即可.\n【详解】126÷(20+15)=126÷35=3.6(小时),答:3.6小时后两船相遇。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "相遇问题"
+                             ],
+        "source":  "24-25五年级上·浙江宁波·期中"
+    },
+    {
+        "id":  "86",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "小明周末走路去外婆家玩，每分钟行走50米,10分钟到达，下午他走路原路返回,15分钟到家，小明去外婆家往返的平均速度是多少米/分。",
+        "options":  [
+
+                    ],
+        "answer":  "40米/分",
+        "solution":  "【分析】已知小明每分钟行走50米,10分钟到达，根据\"路程=速度×时间\"求出小明家到外婆家的距离;下午他走路原路返回,15分钟到家，根据\"平均速度=路程÷时间\"，用往返路程除以往返时间，求出小明往返的平均速度.\n【详解】50×10=500(米),(500+500)÷(10+15)=1000÷25=40(米/分),答：小明去外婆家往返的平均速度是40米/分。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25五年级上·广东江门·期中"
+    },
+    {
+        "id":  "87",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "商店把每千克4元的奶糖5千克，每千克8元的水果糖5千克，每千克6元的软糖2千克，混合成什锦糖，什锦糖每千克多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "6元",
+        "solution":  "【分析】要求什锦糖每千克多少元，就要知道什锦糖的总千克数和总价钱，总价钱等于奶糖价钱加上水果糖价钱再加上软糖价钱，即4×5+8×5+6×2=72(元),总千克数等于5+5+2=12(千克),用总价钱除以总千克数就是什锦糖的单价.\n【详解】(5×4+8×5+6×2)÷(5+5+2)=(20+40+12)÷12=(60+12)÷12=72÷12=6(元),答：什锦糖每千克6元。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "经济问题"
+                             ],
+        "source":  "24-25四年级下·河南驻马店·期末"
+    },
+    {
+        "id":  "88",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "某游泳馆推出两种付费方式：方式一，单次卡，每次收费30元：方式二，办理会员年卡，一次缴纳240元会员费，每次游泳另外收费14元(一年内有效)。李叔叔游泳锻炼的计划是一年，每月两次。他选择哪种方式更划算。",
+        "options":  [
+
+                    ],
+        "answer":  "方式二",
+        "solution":  "【分析】方式一的总费用为单次费用乘次数;方式二的总费用为单次费用乘次数，再加上会员费;比较两种付费方式的总费用，最后选择总费用最少的付费方式即可.\n【详解】方式一：一年=12个月,12×2×30=24×30=720(元)。方式二：一年=12个月,12×2×14+240=24×14+240=336+240=576(元)。因为576元\u003c720元，所以方式二更划算。答：他选择方式二更划算。",
+        "knowledge_points":  [
+                                 "购票问题",
+                                 "乘加",
+                                 "乘减混合运算",
+                                 "年，月，日的认识及换算",
+                                 "经济问题"
+                             ],
+        "source":  "2025·四川内江·小升初真题"
+    },
+    {
+        "id":  "89",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "小明与妈妈今年的年龄和是42岁，妈妈比小明大28岁，妈妈今年的年龄是小明年龄的多少倍。",
+        "options":  [
+
+                    ],
+        "answer":  "5倍",
+        "solution":  "【分析】根据题意，小明与妈妈今年的年龄和是42岁，妈妈比小明大28岁，由和差公式分别求出今年妈妈和小明的年龄，然后再用妈妈的年龄除以小明的年龄即可。和差问题的基本数量关系:(和+差)÷2=较大数,(和-差)÷2=较小数.\n【详解】(42+28)÷2=70÷2=35(岁);(42-28)÷2=14÷2=7(岁);35÷7=5，答：妈妈今年的年龄是小明的5倍。",
+        "knowledge_points":  [
+                                 "和差倍问题",
+                                 "带有小括号的混合运算",
+                                 "求一个数是另一个数的几倍",
+                                 "年龄问题"
+                             ],
+        "source":  "2025·四川南充·小升初真题"
+    },
+    {
+        "id":  "90",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "王老师在周末进行徒步锻炼。他步行的速度是90米/分，如果每走40分钟休息5分钟，从8:00到9:30，他一共步行多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "7200米",
+        "solution":  "【分析】首先计算出从8:00到9:30，一共有多少个40+5=45分钟;然后根据速度×时间=路程，用王老师步行的速度乘走的时间，求出他一共步行多少米即可.\n【详解】9时30分-8时=1小时30分,1小时30分=90分钟,90÷(40+5)=90÷45=2(个),90×40×2=7200(米),答：他一共步行7200米。",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "时，分，秒有关的计算"
+                             ],
+        "source":  "24-25六年级下·重庆巴南·期末"
+    },
+    {
+        "id":  "91",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "某动物园门票有以下两种售卖方案。方案一：成人180元/人，儿童60元/人。方案二：团体6人及6人以上120元/人。(1)若有5个成人和2个儿童，怎样购票更划算。(2)若有2个成人和5个儿童，怎样购票更划算。",
+        "options":  [
+
+                    ],
+        "answer":  "5名成人和1名儿童买团体票，剩下1名儿童买儿童票;780元;2名成人买成人票，5名儿童买儿童票;660元",
+        "solution":  "【分析】由题意得，成人每人180元，儿童60元，而团体票每人120元，对于成人来讲，买团体票更划算，而对于儿童来讲，买儿童票更划算。所以，最优的方案应该是让成人尽可能享受到团体票的优惠，而儿童就买儿童票即可.\n【详解】(1)方案一:5名成人买成人票,2名儿童买儿童票;方案二:5名成人和2名儿童一起买团体票;方案三:5名成人和1名儿童买团体票，剩下1名儿童买儿童票;分别算出各种方案需要的钱数，然后比较大小即可。(2)方案一:2名成人买成人票,5名儿童买儿童票;方案二:2名成人和5名儿童一起买团体票;方案三:2名成人和4名儿童买团体票，剩下1名儿童买儿童票;分别算出各种方案需要的钱数，然后比较大小即可.\n答:5名成人和1名儿童买团体票，剩下1名儿童买儿童票最划算。(2)2名成人买成人票,5名儿童买儿童票最划算。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "乘加",
+                                 "乘减混合运算",
+                                 "带有小括号的混合运算",
+                                 "购票问题"
+                             ],
+        "source":  "24-25四年级下·河北廊坊·期末"
+    },
+    {
+        "id":  "92",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "同学们做早操，排成一个正方形的方阵，从前，后，左，右数，小明都是第5个，这个方阵共有多少人。",
+        "options":  [
+
+                    ],
+        "answer":  "81人",
+        "solution":  "【分析】根据题意可知，从前，后数，小明都是第5个，这样就把小明多数了一次，用5加5再减去1就是行数，因为从左，右数，小明也都是第5个，所以列数与行数相等，然后列数与行数相乘即可求出这个方阵共有多少人.\n【详解】(5+5-1)×(5+5-1)=(10-1)×(10-1)=9×9=81(人),答：这个方阵共有81人。",
+        "knowledge_points":  [
+                                 "方阵问题",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·河南驻马店·期末"
+    },
+    {
+        "id":  "93",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "王师傅和李师傅两人合作加工1600个零件,4小时加工完成。王师傅每小时加工218个，李师傅每小时加工多少个。",
+        "options":  [
+
+                    ],
+        "answer":  "182个",
+        "solution":  "【分析】已知两人合作加工1600个零件,4小时完成。根据工作效率=工作总量/工作时间，可得两人合作每小时加工零件的个数;已知王师傅每小时加工218个，用两人合作每小时加工的个数减去王师傅每小时加工的零件个数得出李师傅每小时加工的个数.\n【详解】1600÷4-218=400-218=182(个),答：李师傅每小时加工182个。",
+        "knowledge_points":  [
+                                 "除加",
+                                 "除减混合运算",
+                                 "两人合作的工程问题"
+                             ],
+        "source":  "24-25四年级下·安徽芜湖·期末"
+    },
+    {
+        "id":  "94",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "小明和小红一共有140枚邮票，如果小明给小红20枚，两人的邮票就同样多，小明和小红原来各有多少枚邮票。",
+        "options":  [
+
+                    ],
+        "answer":  "小明:90枚;小红:50枚",
+        "solution":  "【分析】小明给小红20枚后两人邮票同样多，这说明小明原来比小红多的邮票数为:20×2=40(枚);两人邮票总数是140枚，小明比小红多40枚。如果把总数加上多出来的40枚，那么此时的数量就相当于小明邮票数的2倍。所以小明原有的邮票数为:(140+40)÷2=90(枚);已知两人邮票总数是140枚，小明有90枚，那么小红原有的邮票数为:140-90=50(枚)。据此解答.\n【详解】(140+20×2)÷2=(140+40)÷2=180÷2=90(枚) 140-90=50(枚) 小明原来有90枚邮票，小红原来有50枚邮票。",
+        "knowledge_points":  [
+                                 "和差倍问题",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·安徽芜湖·期末"
+    },
+    {
+        "id":  "95",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "\"AA制\"是当下流行的消费分摊方式，适用于集体消费场景。15位同学采用\"AA制\"组织户外烧烤活动，总共花费750元。小轩先行垫付费用后，其余同学通过转账将各自应付的钱转给小轩，小轩一共会收到多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "700元",
+        "solution":  "【分析】依据总费用750元由15人平均分摊，可计算出每个应分摊的费用。因小轩自己不需要转账，所以只需计算出其余14人一共的转账费用即可.\n【详解】750÷15=50(元);(15-1)×50=14×50=700(元),答：小轩一共会收到700元。",
+        "knowledge_points":  [
+                                 "解决多步计算的实际问题"
+                             ],
+        "source":  "24-25四年级下·河北廊坊·期末"
+    },
+    {
+        "id":  "96",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "学校合唱队要购买48套演出服，其中上衣每件105元，裤子每条45元。一共要付多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "7200元",
+        "solution":  "【分析】一套演出服包含一件上衣和一条裤子，上衣每件105元，裤子每条45元，所以一套的价格为:105+45=150(元),需要购买48套，根据\"总价=一套价格×套数\"，可得总费用为:150×48=7200(元).\n【详解】(105+45)×48=150×48=7200(元),答：一共要付7200元。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "乘加",
+                                 "乘减混合运算",
+                                 "经济问题"
+                             ],
+        "source":  "24-25六年级下·浙江杭州·期末"
+    },
+    {
+        "id":  "97",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "六一儿童节，四(1)班男同学做了16个花环，女同学做了24个花环，每个花环用了25朵花，四(1)班同学做花环一共用了多少朵花。",
+        "options":  [
+
+                    ],
+        "answer":  "1000朵",
+        "solution":  "【分析】先算出全班同学一共做了多少个花环：男生16个加上女生24个，总共40个花环。每个花环用了25朵花，总花朵数量是40个花环乘25。这样分两步想：先算总花环数量，再算总花朵数量.\n【详解】(16+24)×25=40×25=1000(朵),答：四(1)班同学做花环一共用了1000朵花。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·河北邢台·期末"
+    },
+    {
+        "id":  "98",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "有两袋玻璃球，第一袋有164个，第二袋有36个。如果每次从第一袋中拿出4个放入第二袋，需要拿几次，才能使两个袋里的玻璃球同样多。",
+        "options":  [
+
+                    ],
+        "answer":  "16次",
+        "solution":  "【分析】先求第一袋比第二袋多多少个，再平均分成2份，用结果除以4即可知道需要拿几次，能使两个袋里的玻璃球同样多，据此解答.\n【详解】(164-36)÷2÷4=128÷2÷4=64÷4=16(次),答：需要拿16次，才能使两个袋里的玻璃球同样多。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·广东汕头·期末"
+    },
+    {
+        "id":  "99",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "工厂生产一批零件，如果每天生产120个,25天能完成任务。如果每天多生产30个，只需几天完成任务。",
+        "options":  [
+
+                    ],
+        "answer":  "20天",
+        "solution":  "【分析】每天生产的个数×完成任务的天数=这批零件的总个数，依此计算出这批零件的总个数，再用120个加30个计算出调整后每天生产的个数，然后再用这批零件的总个数除以调整后每天生产的个数即可，依此解答.\n【详解】(120×25)÷(120+30)=3000÷150=20(天),答：只需20天完成任务。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·安徽宣城·期末"
+    },
+    {
+        "id":  "100",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "王师傅要加工580个零件，已经加工了220个。如果每小时加工40个，还要几小时才能完成。",
+        "options":  [
+
+                    ],
+        "answer":  "9小时",
+        "solution":  "【分析】根据题意，先用要加工的零件总个数减去已经加工的个数，求出剩下多少零件需要加工，再除以每小时加工的个数，即可求出还要几小时能加工完.\n【详解】(580-220)÷40=360÷40=9(小时),答：还要9小时才能完成。",
+        "knowledge_points":  [
+                                 "有具体量的工程问题",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·河南驻马店·期末"
+    },
+    {
+        "id":  "101",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "小新在计算\u002796/[(6+□)×3]\u0027时忽略了题中所有的括号，得数是22，正确的得数是()。",
+        "options":  [
+                        "A. 2",
+                        "B. 4",
+                        "C. 22"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据题意，小新在计算\u002796/[(6+□)×3]\u0027时，忽略了所有括号的存在，即将算式的运算顺序看成了先算乘除法，再算加法;所以先算96除以6的商，用得数22减去上一步算得的商，那么计算所得的差就是 □*3得积，所以就用所得的差除以3，即可求出 □表示的数;然后再将 □的数代回原式按照正确的运算顺序进行计算.\n【详解】根据分析可得:96÷6=16,22-16=6,6÷3=2。所以 □表示的数是2;96/[(6+2)×3]=96/[8×3]=96÷24=4，正确的得数是4;故答案为:B",
+        "knowledge_points":  [
+                                 "带有中括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·河南郑州·期末"
+    },
+    {
+        "id":  "102",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "小区居委会用4000元购买了180本太极拳教学用书送给当地社区的居民，还剩400元。式子400/[(4000-400)÷180]要解决的问题是()。",
+        "options":  [
+                        "A. 每本太极拳教学用书多少元。",
+                        "B. 买180本要用多少钱。",
+                        "C. 一共买了多少本。",
+                        "D. 剩下的钱还能买多少本。"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】根据题意，已知小区居委会用4000元购买了180本太极拳教学用书送给当地社区的居民，还剩400元。4000减去400，求出180本太极拳教学用书的价格，除以180，就是1本书的价格，最后用400除以1本书的价格，就是剩下的400元钱还能买多少本;以此选择正确的答案即可.\n【详解】根据分析可知：小区居委会用4000元购买了180本太极拳教学用书送给当地社区的居民，还剩400元。式子400/[(4000-400)÷180]要解决的问题是剩下的钱还能买多少本。故答案为:D",
+        "knowledge_points":  [
+                                 "带有中括号的混合运算，\"提问题\"\"填条件\"问题"
+                             ],
+        "source":  "24-25四年级下·河南焦作·期末"
+    },
+    {
+        "id":  "103",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一只蜗牛用4分钟爬行了24米，照这样的速度，要爬行72米须用几分钟。列式是()。",
+        "options":  [
+                        "A. 24×(72÷4)",
+                        "B. 24÷(72/4)",
+                        "C. 72×(24÷4)",
+                        "D. 72÷(24/4)"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】4分钟爬行了24米，根据速度=路程/时间，用24米除以4分钟，求得每分钟爬行的距离，即求得速度，照这样的速度，要爬行72米须用几分钟，根据时间=路程/速度，用72米除以每分钟爬行的距离，即可求得爬行72米须用几分钟.\n【详解】爬行的速度:24÷4;爬行72米的时间:72÷(24/4)。因此，求爬行72米须用几分钟，列式是72÷(24/4)。故答案为:D",
+        "knowledge_points":  [
+                                 "基础行程问题，带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "104",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一辆小汽车4小时行驶280千米，照这样计算，要行驶420千米需要多少小时。下面列式正确的是()。",
+        "options":  [
+                        "A. 420÷(280/4)",
+                        "B. 280÷(420/4)",
+                        "C. 420÷4×280"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】一辆小汽车4小时行驶280千米，用280除以4求出小汽车的速度，再用420除以小汽车的速度求出行驶420千米需要多少小时;据此解答.\n【详解】速度:(280÷4)千米/小时;时间:420÷(280/4)小时。所以要行驶420千米需要420÷(280/4)小时。故答案为:A",
+        "knowledge_points":  [
+                                 "基础行程问题，带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·河南洛阳·期末"
+    },
+    {
+        "id":  "105",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "2名老师和24名学生去公园划船，大船租金30元，限乘6人;小船租金24元，限乘4人，租()只大船和()只小船最省钱。()",
+        "options":  [
+                        "A. 41",
+                        "B. 32",
+                        "C. 15"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】大船租金30元，限乘6人,30÷6=5(元/人),小船租金24元，限乘4人,24÷4=6(元/人),大船人均租金更便宜，所以尽量租大船。24+2=26(人),用总人数26÷6=4(只)。2(人),余下2人去坐小船，因为小船没有坐满人，所以减少一只大船,6+2=8(人),这8个人租小船,8÷4=2(只),正好租2只小船。将这两种租船方案的租金算出来再比较哪种最省钱.\n【详解】30÷6=5(元/人),24÷4=6(元/人),大船人均租金更便宜。24+2=26(人),26÷6=4(只)。2(人)。方案一：租4只大船,1只小船，租金为:30×4+24=120+24=144(元),减少一只大船4-1=3(只)6+2=8(人)8÷4=2(只)。方案二：租3只大船,2只小船，租金为:30×3+24×2=90+48=138(元),144\u003e138，所以租3只大船2只小船最便宜。故答案为:B",
+        "knowledge_points":  [
+                                 "租船问题"
+                             ],
+        "source":  "24-25三年级下·广东汕头·期末"
+    },
+    {
+        "id":  "106",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "牡丹文化节期间，四年级师生260人租车去看舞台剧《花木兰》。大车每辆限乘40人，租金800元。小车每辆限乘20人，租金500元。租()最省钱。",
+        "options":  [
+                        "A. 7辆大车",
+                        "B. 6辆大车1辆小车",
+                        "C. 5辆大车3辆小车"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据题意，用800÷40=20(元),500÷20=25(元),20\u003c25，所以尽量先安排乘坐大车。用总人数除以大车每辆限乘的人数等于需要几辆大车，剩下的人乘坐一辆小车即可.\n【详解】800÷40=20(元),500÷20=25(元),20\u003c25,260÷40=6(辆)。20(人),20÷20=1(辆)。所以，需要6辆大车和1辆小车。A。7辆大车,7×800=5600(元)。B。6辆大车1辆小车,6×800+500=4800+500=5300(元)。C。5辆大车3辆小车,5×800+3×500=4000+1500=5500(元)。5300\u003c5500\u003c5600，租6辆大车1辆小车最省钱。故答案为:B",
+        "knowledge_points":  [
+                                 "租船问题，乘加，乘减混合运算"
+                             ],
+        "source":  "24-25四年级下·河南洛阳·期末"
+    },
+    {
+        "id":  "107",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "甲仓库有大米40袋，乙仓库有大米170袋。从乙仓库运()袋给甲仓库，可使乙仓库的大米袋数正好是甲仓库的2倍。",
+        "options":  [
+                        "A. 100",
+                        "B. 65",
+                        "C. 70",
+                        "D. 30"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】根据题意，甲仓库有大米40袋，乙仓库有大米170袋。两个仓库共有40+170=210(袋),乙仓库的大米袋数正好是甲仓库的2倍，根据和倍问题，这时总袋数是甲仓库袋数的(2+1)倍，用大米的总袋数除以这时乙仓库大米的袋数是甲仓库有大米袋数的(2+1)倍，求出这时甲仓库大米的袋数，再用这时甲仓库大米的袋数减去原来有的40袋，就是从乙仓库运给甲仓库的袋数，据此解答即可.\n【详解】(40+170)÷(2+1)=210÷(2+1)=210÷3=70(袋);70-40=30(袋)。甲仓库有大米40袋，乙仓库有大米170袋。从乙仓库运30袋给甲仓库，可使乙仓库的大米袋数正好是甲仓库的2倍。故答案为:D",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算，和差倍问题"
+                             ],
+        "source":  "24-25四年级下·四川凉山·期末"
+    },
+    {
+        "id":  "108",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "在一个减法算式中，被减数，减数，差相加的和是8，则被减数是()。",
+        "options":  [
+                        "A. 2",
+                        "B. 4",
+                        "C. 6"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】在减法中，被减数=减数+差，减数=被减数差，差=被减数-减数;读题可知，被减数+减数+差=8，即减数+差+减数+差=8，减数+差=4。据此解答.\n【详解】由分析可知，减数+差=4，即被减数为4。在一个减法算式中，被减数，减数，差相加的和是8，则被减数是4。选项B正确。故答案为:B",
+        "knowledge_points":  [
+                                 "加，减法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "109",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "有52人要租景区观光车游玩1小时，大车限乘10人，每辆1小时40元;小车限乘6人，每辆1小时30元。下面租车方案最省钱的是()。",
+        "options":  [
+                        "A. 租5辆大车和1辆小车",
+                        "B. 租4辆大车和2辆小车",
+                        "C. 租6辆大车"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】首先要想最省钱，得先算每种车坐1个人花多少钱。大车能坐10人,40元，那1个人就是4元;小车坐6人,30元,1个人就是5元。所以尽量多租大车更划算，但也不能浪费座位，不然白花钱。然后看三个方案，分别算能不能坐下52人，再算花多少钱，最后比谁最便宜.\n【详解】A。能坐人数:5×10+1×6=50+6=56(人)(56\u003e52，能坐下)。总费用:5×40+1×30=200+30=230(元)。B。能坐人数:4×10+2×6=40+12=52(人)。总费用:4×40+2×30=160+60=220(元)。C。能坐人数:6×10=60(人)(60\u003e52，浪费多)。总费用:6×40=240(元)。220元\u003c230元\u003c240元，最省钱的是租4辆大车和2辆小车。故答案为:B",
+        "knowledge_points":  [
+                                 "乘加，乘减混合运算，租船问题"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "110",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "计算420+[(315-205)×4],运算顺序是()。",
+        "options":  [
+                        "A. 乘→减→加",
+                        "B. 减→加→乘",
+                        "C. 减→乘→加",
+                        "D. 乘→加→减"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】根据带有中括号的四则运算顺序：一个算式里，既有小括号，又有中括号，要先算小括号里面的，再算中括号里面的，最后算中括号外面的.\n【详解】在式子420+[(315-205)×4]中，根据四则混合运算的运算顺序，先算小括号里是减法，再算中括号里的乘法，最后再算加法。所以运算顺序是：减→乘→加。故答案为:C",
+        "knowledge_points":  [
+                                 "带有中括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·福建福州·期末"
+    },
+    {
+        "id":  "111",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "某超市运进50箱苹果，每箱15千克，第一天卖出270千克，剩下的4天才卖完。下面哪个问题可以用算式(50×15-270)÷4来解答。()",
+        "options":  [
+                        "A. 剩下多少千克苹果。",
+                        "B. 剩下的平均每天卖多少千克。",
+                        "C. 平均每天卖多少千克。"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据题意，分析三个选项所涉及的计算过程，看哪个选项的计算逻辑与给定算式一致即可.\n【详解】A。要求剩下多少千克苹果，首先要算出运进苹果的总重量，已知运进50箱，每箱15千克，那么总重量为50×15千克。然后用总重量减去第一天卖出的270千克，即50×15-270，这就是剩下苹果的重量。不符合要求。B。先算出运进苹果的总重量为50×15千克，再减去第一天卖出的270千克，得到剩下的重量(50×15-270)千克。因为剩下的4天卖完，要求剩下的平均每天卖多少千克，就是把剩下的重量平均分成4份，用除法计算，算式就是(50×15-270)÷4。符合要求。C。平均每天卖多少千克，首先要算出运进苹果的总重量，已知运进50箱，每箱15千克，那么总重量为50×15千克。再算出卖的天数，已知先卖了1天，剩下的4天卖完，一共卖了1+4=5天。最后用总重量除以卖的天数得出平均每天卖的千克数，列式为50×15÷(1+4)。不符合要求。故答案为:B",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·河南洛阳·期末"
+    },
+    {
+        "id":  "112",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列实际问题，可以用50+35×2这个算式解决的是()。",
+        "options":  [
+                        "A. 一件上衣50元，一条裤子35元。买了两件上衣和两条裤子一共花了多少钱。",
+                        "B. 文文看一本故事书，第一天看了50页，第二天看了35页，还剩35页没看。这本故事书一共多少页。",
+                        "C. 饼干每箱50元，一箱火腿比一箱饼干贵35元。买两箱火腿需要多少钱。"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】算式50+35×2的运算顺序是先乘后加，即先计算35×2，再加50。A。求买了两件上衣和两条裤子的花费，将两件上衣和两条裤子的价钱加在一起;B。总页数=第一天看的页数+第二天看的页数+剩余页数，据此解答;C。先求出一箱火腿多少钱，再乘2即可求出买两箱火腿需要多少钱。需找出符合此算式解决的实际问题.\n【详解】A。两件上衣费用为50×2元，两条裤子费用为35×2元，总费用为((50×2+35×2))元，不符合题意;B。这本故事书一共有:50+35+35=50+35×2=50+70=120(页),符合题意;C。一箱火腿的价钱为(50+35)元，两箱火腿的费用为(50+35)×2元，不符合题意。故答案为:B",
+        "knowledge_points":  [
+                                 "乘加，乘减混合运算，经济问题"
+                             ],
+        "source":  "24-25四年级下·河北邯郸·期末"
+    },
+    {
+        "id":  "113",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "王师傅制作零件，上午制作了32个，下午制作了24个，每4个零件包装成一盒，这些零件一共可以包装成多少盒。解决这个问题时，小强列出了这样的算式:32÷4+24÷4，他的解题思路是()。",
+        "options":  [
+                        "A. 先算出每天一共能制作的零件数，再计算能包装的盒数。",
+                        "B. 先计算上，下午分别制作的零件可以包装的盒数，再计算全天的总盒数。",
+                        "C. 先计算上午制作多少个包装盒，再计算下午制作多少个包装盒，最后算总盒数。",
+                        "D. 先计算上午制作多少个零件，再计算下午制作多少个零件，最后算总零件数。"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据题意，小强的算式是分别将上午和下午制作的零件数除以每盒的数量4，再将结果相加。这表示他先计算了上，下午各自能包装的盒数，再求和。以此选择即可.\n【详解】根据分析可知：小强的算式为32÷4+24÷4，即先计算上午的32个零件能包装成32÷4=8(盒),下午的24个零件能包装成24÷4=6(盒),再将两者相加得到总盒数8+6=14(盒)。A。先算总数再求盒数，对应算式应为(32+24)÷4，与小强的算式不符，此选项错误。B。先分别计算上，下午的盒数再相加，与小强的算式完全一致，此选项正确。C。提到\"制作包装盒\"，但题目中包装盒是由零件数量决定的，表述不准确，此选项错误。D。计算总零件数，但小强的算式结果已是盒数之和，此选项错误。故答案为:B",
+        "knowledge_points":  [
+                                 "除加，除减混合运算，带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·四川乐山·期末"
+    },
+    {
+        "id":  "114",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "对于没有余数的除法，下列说法正确的有()个。①被除数+除数+商=1;②被除数-(除数+商)=0;③除数一定比商大",
+        "options":  [
+                        "A. 0",
+                        "B. 1",
+                        "C. 2",
+                        "D. 3"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】在没有余数的除法中，被除数÷除数=商，商×除数=被除数。据此举例说明即可判断给定的关系是否正确，然后再选择.\n【详解】例如:100÷5=20,5×20=100，被除数是100，除数是5，商是20。①100+5+20=125，被除数+除数+商=1不成立，说法错误;②100-(5+20)=100-25=75，被除数-(除数+商)=0不成立，说法错误;③5\u003c20，除数比商小，则除数一定比商大不成立，说法错误;说法正确的有0个。故答案为:A",
+        "knowledge_points":  [
+                                 "乘，除法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·四川乐山·期末"
+    },
+    {
+        "id":  "115",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据先算17-8=9，再算45÷9=5，最后算74×5=370，列出综合算式为()。",
+        "options":  [
+
+                    ],
+        "answer":  "74*[45÷(17-8)]=370",
+        "solution":  "【分析】观察可知，有减法，除法和乘法，且要先算减法，再算除法，最后算乘法，要用小括号改变运算顺序，将减法括起来;由于乘法中的\"5\"是除法的商，乘法在除法的左侧，要用中括号将除法括起来。据此解答.\n【详解】74*[45÷(17-8)]=74*[45÷9]=74×5=370，列出的综合算式为:74*[45÷(17-8)]=370。",
+        "knowledge_points":  [
+                                 "带有中括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·河北邢台·期末"
+    },
+    {
+        "id":  "116",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "把45-26=19,19×20=380列成一个综合算式是()。",
+        "options":  [
+
+                    ],
+        "answer":  "(45-26)×20=380",
+        "solution":  "【分析】分析算式的关系，算式19×20=380中的19是由算式45-26=19得来的，由于要先算减法，所以应该列综合算式应加小括号，据此列综合算式.\n【详解】由分析知：把45-26=19,19×20=380列成一个综合算式是(45-26)×20=380。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25三年级下·安徽淮北·期末"
+    },
+    {
+        "id":  "117",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把81-66=15,15×6=90,90÷30=3合并成综合算式是()。",
+        "options":  [
+
+                    ],
+        "answer":  "(81-66)×6÷30=3",
+        "solution":  "【分析】根据分析可知：将分步算式合并成综合算式时，需按照运算顺序逐步替换。首先用81-66替换第一个等式中的15，再将结果乘以6得到第二个等式中的90，最后用这个乘积除以30得到3。为确保运算顺序正确，需使用括号将减法的部分括起来。以此答题即可.\n【详解】根据分析可知:(81-66)×6÷30=15×6÷30=90÷30=3，把81-66=15,15×6=90,90÷30=3合并成综合算式是(81-66)×6÷30=3。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·辽宁盘锦·期末"
+    },
+    {
+        "id":  "118",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据576+290=886，可以直接得到886-290=(),886-576=()。像这样已知两个数的和与其中一个加数，求另一个加数的运算，叫做()。",
+        "options":  [
+
+                    ],
+        "answer":  "576;290;减法",
+        "solution":  "【分析】在加法算式中，加数+加数=和，一个加数=和-另一个加数。由题意得,576+290=886，根据加法各部分之间的关系可知,886-290=576,886-576=290;已知两个数的和与其中一个加数，求另一个加数的运算，叫做减法.\n【详解】根据576+290=886，可以直接得到886-290=576,886-576=290。像这样已知两个数的和与其中一个加数，求另一个加数的运算，叫做减法。",
+        "knowledge_points":  [
+                                 "加，减法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·辽宁盘锦·期末"
+    },
+    {
+        "id":  "119",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在一道减法算式中，被减数，减数与差的和是2，减数是0.4，差是()。",
+        "options":  [
+
+                    ],
+        "answer":  "0.6",
+        "solution":  "【分析】根据被减数-减数=差可知，差+减数=被减数。已知被减数+减数+差=2，将(差+减数)替换为被减数，那么可得出被减数的2倍是2，故用2÷2计算出被减数是多少，又知减数为0.4，故可继续求出差。据此作答.\n【详解】2÷2=1,1-0.4=0.6，因此，在一道减法算式中，被减数，减数与差的和是2，减数是0.4，差是0.6。",
+        "knowledge_points":  [
+                                 "一位小数的进位加法，退位减法，加，减法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25三年级下·河北邢台·期末"
+    },
+    {
+        "id":  "120",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "括号在数学中扮演着至关重要的角色，它能够改变运算的顺序，影响最终的结果。请在算式320÷16+24×2中添上小括号和中括号，使运算顺序变为先算加法，再算乘法，最后算除法。这个新的算式是()。",
+        "options":  [
+
+                    ],
+        "answer":  "320÷[(16+24)×2]",
+        "solution":  "【分析】原算式中除法在前面，乘法在后面，加法在中间，要想先算加法需要在加法上加括号，又需要让后面的乘法优于前面的除法先算，因此需要在加法和乘法外面再加一级括号，需要两级括号，就需要用到小括号和中括号.\n【详解】由分析可得：为了先算加法应在加法上加小括号，又要让乘法优于除法计算，还需要在加法和乘法外面加中括号。结果为:320÷[(16+24)×2]。括号在数学中扮演着至关重要的角色，它能够改变运算的顺序，影响最终的结果。请在算式320÷16+24×2中添上小括号和中括号，使运算顺序变为先算加法，再算乘法，最后算除法。这个新的算式是320÷[(16+24)×2]。",
+        "knowledge_points":  [
+                                 "带有中括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·河南省直辖县级单位·期末"
+    },
+    {
+        "id":  "121",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "填一填。(1)检验839-474=365的计算结果是否正确，可以用474+365=839进行检验，验算的依据是.\n(2)检验365+474=839的计算结果是否正确，可以用474+365=839进行检验，这种验算方法的依据是。",
+        "options":  [
+
+                    ],
+        "answer":  "被减数=差+减数;加法交换律",
+        "solution":  "【分析】差+减数=被减数，对于839-474=365的计算结果是否正确，用474+365=839进行检验。加法的交换律：两个数相加，交换加数的位置，和不变。对于365+474=839的计算结果是否正确，用474+365=839进行检验.\n【详解】(1)差+减数=被减数。在这个式子中,365是差,474是减数,839是被减数。当差与减数相加等于被减数时，说明原减法计算正确。(2)加法的交换律：两个数相加，交换加数的位置，和不变。即a+b=b+a，在这个式子中a=365,b=474，交换位置后和不变，说明原加法计算正确。",
+        "knowledge_points":  [
+                                 "整数加法交换律，加，减法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·河南许昌·期末"
+    },
+    {
+        "id":  "122",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "①359-334=25②25×28=700③1400÷700=2\n(1)根据第①个式子，写出下面算式的得数:334+25=().\n(2)根据第①个，第②个和第③个算式列出一个综合算式:()。",
+        "options":  [
+
+                    ],
+        "answer":  "359;1400÷[(359-334)×28]=2",
+        "solution":  "【分析】(1)根据被减数-减数=差，减数+差=被减数，即可解答;(2)根据题意可知，先计算359-334=25，求出差，再用求出的差乘28，即25×28=700，求出积，最后计算1400除以25×28的积，即1400÷700=2，求出商;根据混合运算的运算顺序，先算减法，给359-334加上小括号，再算乘法，给(359-334)×28加上中括号，最后计算除法，据此列式为:1400÷[(359-334)×28]=2.\n【详解】(1)根据第①个式子，写出下面算式的得数:334+25=359。(2)1400÷[(359-334)×28]=1400÷[25×28]=1400÷700=2。根据第①个，第②个和第③个算式列出一个综合算式:1400÷[(359-334)×28]=2。",
+        "knowledge_points":  [
+                                 "加，减法的意义和各部分间的关系，带有中括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·河南濮阳·期末"
+    },
+    {
+        "id":  "123",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "给算式840÷45-39×20添上括号，使运算顺序变为先算减法，再算乘法，最后算除法，算式为:()。",
+        "options":  [
+
+                    ],
+        "answer":  "840/[(45-39)×20]",
+        "solution":  "【分析】由题意得，在算式840÷45-39×20中，要先算除法，再算乘法，最后算减法。要想先算减法，再算乘法，最后算除法，需要在算式45-39的两边加上小括号，在算式45-39×20的两边加上中括号，即综合算式为:840/[(45-39)×20].\n【详解】由分析可知，给算式840÷45-39×20添上括号，使运算顺序变为先算减法，再算乘法，最后算除法，算式为:840/[(45-39)×20]。",
+        "knowledge_points":  [
+                                 "有括号的运算顺序"
+                             ],
+        "source":  "24-25四年级下·河南焦作·期末"
+    },
+    {
+        "id":  "124",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算180+25×(40-36)时，应先算()法，再算()法，最后算()法，计算结果是()。",
+        "options":  [
+
+                    ],
+        "answer":  "减;乘;加;280",
+        "solution":  "【分析】根据四则混合运算顺序，从左往右依次计算，先算乘，除法，再算加，减法，有括号的先算括号里面的。计算180+25×(40-36)时，根据运算规则，先算括号内的减法，再算括号外的乘法，最后算加法。以此答题即可.\n【详解】根据分析可知:180+25×(40-36)=180+25×4=180+100=280，计算180+25×(40-36)时，应先算减法，再算乘法，最后算加法，计算结果是280。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·河南许昌·期末"
+    },
+    {
+        "id":  "125",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "甲乙两地相距365千米，一辆汽车从甲地开往乙地，途中在某服务区休息，此时剩下的路程比已经行驶的路程多45千米。这辆汽车已经行了()千米。",
+        "options":  [
+
+                    ],
+        "answer":  "160",
+        "solution":  "【分析】根据题意，行驶途中剩下的路程比已经行驶的路程多45千米，先用两地的距离减去多的45千米，即可求出已经行驶的路程的2倍是多少千米，除以2即可求出这辆汽车已经行了多少千米.\n【详解】(365-45)÷2=320÷2=160(千米),甲乙两地相距365千米，一辆汽车从甲地开往乙地，途中在某服务区休息，此时剩下的路程比已经行驶的路程多45千米。这辆汽车已经行了160千米。",
+        "knowledge_points":  [
+                                 "和差倍问题，带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·河南洛阳·期末"
+    },
+    {
+        "id":  "126",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把下面三个算式写成一个综合算式是:()。(只列式，不计算)18-16=2;2×23=46;920÷46=20",
+        "options":  [
+
+                    ],
+        "answer":  "920÷[(18-16)×23]",
+        "solution":  "【分析】根据四则混合运算顺序，从左往右依次计算，先算乘、除法，再算加、减法，有括号的先算括号里面的。先算减法，需要给18-16加上小括号，再算乘法，最后算除法，把乘法加上中括号，三个算式写成一个综合算式是:920÷[(18-16)×23]。以此答题即可.\n【详解】根据分析可知:920÷[(18-16)×23]=920÷[2×23]=920÷46=20，综上可知，把题中三个算式写成一个综合算式是:920÷[(18-16)×23]。",
+        "knowledge_points":  [
+                                 "带有中括号的混合运算，带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·河南郑州·期末"
+    },
+    {
+        "id":  "127",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "先写出解决问题的思路，再列出综合算式。只列式不解答。在创意编程挑战赛中评委老师需要在一天内完成编程任务的设计。评委老师共分了3个小组，每组12位老师，平均每位老师设计5个编程任务。这次创意编程挑战赛一共需要设计多少个编程任务。\n①解题思路：先求:()再求:().\n②根据以上思路列出综合算式:()。",
+        "options":  [
+
+                    ],
+        "answer":  "每组12位老师一共需要设计多少个编程任务;这次创意编程挑战赛一共需要设计多少个编程任务;12×5×3",
+        "solution":  "【分析】每组12位老师，平均每位老师设计5个编程任务，用12×5求出每组12位老师一共需要设计多少个编程任务;评委老师共分了3个小组，再乘3，就是这次创意编程挑战赛一共需要设计多少个编程任务.\n【详解】①解题思路：先求：每组12位老师一共需要设计多少个编程任务，再求：这次创意编程挑战赛一共需要设计多少个编程任务。②根据以上思路列出综合算式:12×5×3(答案不唯一)",
+        "knowledge_points":  [
+                                 "多位数与一位数的连乘"
+                             ],
+        "source":  "24-25四年级下·山东日照·期末"
+    },
+    {
+        "id":  "128",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "被减数是190，那么被减数，减数，差的和是()。",
+        "options":  [
+
+                    ],
+        "answer":  "380",
+        "solution":  "【分析】根据题意，在减法算式中，被减数，减数和差的关系为：被减数-减数=差，即被减数=减数+差，据此解答.\n【详解】被减数+减数+差=被减数+被减数=190+190=380，被减数是190，那么被减数，减数，差的和是380。",
+        "knowledge_points":  [
+                                 "加，减法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·河南驻马店·期末"
+    },
+    {
+        "id":  "129",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "四一班5位老师带45名同学去划船。大船坐6人，租金30元;小船坐4人，租金24元。租()条大船,()条小船最省钱，最少要花()元。",
+        "options":  [
+
+                    ],
+        "answer":  "7;2;258",
+        "solution":  "【分析】已知老师有5位，同学有45名，所以总人数为5+45=50(人)。大船坐6人，租金30元，那么大船每人租金为30÷6=5(元)。小船坐4人，租金24元，那么小船每人租金为24÷4=6(元)。因为5\u003c6，所以尽量多租大船更省钱。计算50人租大船需要租多少条:50÷6=8(条)。2(人),全部坐下需要租9条大船。方案一：租9条大船,0条小船。方案二：租8条大船,1条小船。方案三：租7条大船,2条小船。比较各方案租金，即可判断怎样租船最省钱，据此解答即可.\n【详解】5+45=50(人),30÷6=5(元),24÷4=6(元),因为5\u003c6，所以尽量多租大船更省钱。50÷6=8(条)。2(人),全部坐下需要租9条大船。方案一：租9条大船,0条小船。可坐人数9×6=54(人),租金9×30=270(元)。方案二：租8条大船,1条小船。可坐人数:8×6+4=48+4=52(人)。租金:8×30+24=240+24=264(元)。方案三：租7条大船,2条小船。可坐人数:7×6+2×4=42+8=50(人)此时刚好每条船都可以坐满。租金:7×30+2×24=210+48=258(元)。比较各方案租金,258\u003c264\u003c270。所以租7条大船,2条小船最省钱，最少要花258元。",
+        "knowledge_points":  [
+                                 "租船问题"
+                             ],
+        "source":  "24-25四年级下·河北廊坊·期末"
+    },
+    {
+        "id":  "130",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "根据下面的算式和乘，除法的关系，写出一个除法算式和一个乘法算式.\n\n△÷25=□",
+        "options":  [
+
+                    ],
+        "answer":  "△÷□=25;□×25=△",
+        "solution":  "【分析】根据除法各部分间的关系：被除数÷除数=商，可得被除数=商×除数，除数=被除数÷商。原式△÷25=□中,△是被除数,25是除数,□是商，因此可推导出对应的乘法算式和另一个除法算式.\n【详解】△÷25=□;△÷□=25;□×25=△。",
+        "knowledge_points":  [
+                                 "乘，除法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·河北廊坊·期末"
+    },
+    {
+        "id":  "131",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "10+5×()=90;60-()÷8=50",
+        "options":  [
+
+                    ],
+        "answer":  "16;80",
+        "solution":  "【分析】把5×□看作一个整体，是加数，加数=和-另一个加数，那么5×□=90-10=80，然后求出□=80÷5=16。把□÷8看作一个整体，是减数，减数=被减数-差，那么□÷8=60-50=10，然后求出□=10×8=80。\n【详解】(90-10)÷5=80÷5=16;(60-50)×8=10×8=80，所以10+5×16=90,60-80÷8=50。",
+        "knowledge_points":  [
+                                 "加，减法的意义和各部分间的关系，带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·浙江杭州·期末"
+    },
+    {
+        "id":  "132",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用2,3,4,5四个数，经过怎样的运算能得到\"24\"。在括号里列出算式:()。",
+        "options":  [
+
+                    ],
+        "answer":  "(3+4+5)×2=24",
+        "solution":  "【分析】运用四则运算的法则，添加运算符号，设置一定的运算顺序，使计算结果等于24即可。3+4+5的和是12,12乘2等于24。据此填写.\n【详解】由分析可知，用2,3,4,5四个数，经过怎样的运算能得到\"24\"。在括号里列出算式:(3+4+5)×2=24。(答案不唯一)",
+        "knowledge_points":  [
+                                 "24点，带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·湖北十堰·期末"
+    },
+    {
+        "id":  "133",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把算式42-34=8,8×5=40,120÷40=3，写成一个综合算式是:()。",
+        "options":  [
+
+                    ],
+        "answer":  "120÷[(42-34)×5]=3",
+        "solution":  "【分析】观察可知，\"8×5=40\"中的\"8\"是\"42-34=8\"的差，\"120÷40=3\"中的\"40\"是\"8×5=40\"的积，也就是42与34的差乘5，所得的积再去除120，要先算减法，再算乘法，最后算除法，需要运用到括号改变运算顺序。据此解答.\n【详解】把算式42-34=8,8×5=40,120÷40=3，写成一个综合算式是:120÷[(42-34)×5]=3。",
+        "knowledge_points":  [
+                                 "带有中括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·福建龙岩·期末"
+    },
+    {
+        "id":  "134",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "()-139=448;120÷()=10",
+        "options":  [
+
+                    ],
+        "answer":  "587;12",
+        "solution":  "【分析】先根据减法中被减数与减数，差的关系求出第一个空的数值，再根据除法中除数与被除数，商的关系求出第二个空的数值.\n【详解】在减法算式中，各部分之间的关系为：被减数=减数+差。本题中，已知减数是139，差是448，所以用减数加上差即可求出被减数139+448=587;在除法算式中，各部分之间的关系为：除数=被除数/商。本题中，已知被除数是120，商是10，所以用被除数除以商即可求出除数120÷10=12。",
+        "knowledge_points":  [
+                                 "乘，除法的意义和各部分间的关系",
+                                 "加，减法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·陕西西安·期末"
+    },
+    {
+        "id":  "135",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "把33-25=8,8×3=24,120÷24=5合并成一个综合算式，这个综合算式是:",
+        "options":  [
+
+                    ],
+        "answer":  "120÷[(33-25)×3]=5",
+        "solution":  "【分析】我们看到第一个算式33-25=8，算出了一个结果8;第二个算式8×3=24，这里的8就是第一个算式的结果，它把第一个算式的结果乘3得到24;第三个算式120÷24=5，这里的24就是第二个算式的结果。要先算33-25，再用得到的差去乘3，最后用120除以乘得的积。根据四则运算规则，为了保证先算减法，需要给33-25加上小括号(33-25);然后为了保证先算小括号里减法的结果再乘3，并且最后算除法，需要给(33-25)×3加上中括号.\n【详解】根据分析可知，把33-25=8,8×3=24,120÷24=5合并成一个综合算式，这个综合算式是:120÷[(33-25)×3]=5。",
+        "knowledge_points":  [
+                                 "带有中括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·福建龙岩·期末"
+    },
+    {
+        "id":  "136",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "计算40×[(27+54)÷9]时，要先算()法，再算()法，最后算()法，结果是()。",
+        "options":  [
+
+                    ],
+        "answer":  "加;除;乘;360",
+        "solution":  "【分析】根据四则混合运算的顺序，没有小括号时，先算乘除，再算加减，有括号先算括号内的数，所以40×[(27+54)÷9]先算小括号里的加法，再算中括号里的除法，最后算乘法.\n【详解】40×[(27+54)÷9]=40×[81÷9]=40×9=360。计算40×[(27+54)÷9]时，要先算加法，再算除法，最后算乘法，结果是360。",
+        "knowledge_points":  [
+                                 "带有中括号的混合运算",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·福建龙岩·期末"
+    },
+    {
+        "id":  "137",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "小东在计算一道乘法算式时，把一个因数63错看成了36，计算结果是1008，正确的积应该是()。",
+        "options":  [
+
+                    ],
+        "answer":  "1764",
+        "solution":  "【分析】由题意得，小东在计算一道乘法算式时，把其中个因数63错写成了36，得到的积是1008，根据乘法各部分间的关系，错误的积等于错误因数与另一个正确因数的乘积，可以先用1008除以36算出另一个因数。然后再用这个因数乘上63即可算出正确的积.\n【详解】1008÷36×63=28×63=1764，所以正确的积应该是1764。",
+        "knowledge_points":  [
+                                 "乘，除法的意义和各部分间的关系",
+                                 "除数是两位数的笔算除法",
+                                 "两位数乘两位数的进位乘法"
+                             ],
+        "source":  "24-25四年级下·福建漳州·期末"
+    },
+    {
+        "id":  "138",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "计算25×[(72-48)÷6]时，应先算()法，再算()法，最后算()法，结果是()。",
+        "options":  [
+
+                    ],
+        "answer":  "减;除;乘;100",
+        "solution":  "【分析】根据四则混合运算的顺序，有括号的先算括号里面的，先小括号，再中括号，最后算括号外面的乘法进行。计算25×[(72-48)÷6]时，先算小括号里的减法，再计算中括号里的除法，最后再算乘法.\n【详解】25×[(72-48)÷6]=25×[24÷6]=25×4=100，所以计算25×[(72-48)÷6]时，应先算减法，再算除法，最后算乘法，结果是100。",
+        "knowledge_points":  [
+                                 "带有中括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·河北石家庄·期末"
+    },
+    {
+        "id":  "139",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把218-177=41,82÷41=2,38×2=76这三个算式改写成一个综合算式是()。",
+        "options":  [
+
+                    ],
+        "answer":  "38×[82÷(218-177)]=76",
+        "solution":  "【分析】根据算式可知,38×2中2是算式82÷41的结果，而算式中的41是218-177的结果，因此先将41换成218-177，因为要先算减法，所以把218-177加上小括号，再把38×2中的2换成82÷(218-177),因为最后算乘法，所以在82÷(218-177)外添上中括号，变成38×[82÷(218-177)]=76;据此解答.\n【详解】根据分析可知：把218-177=41,82÷41=2,38×2=76这三个算式改写成一个综合算式是38×[82÷(218-177)]=76。",
+        "knowledge_points":  [
+                                 "带有中括号的混合运算",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·河北石家庄·期末"
+    },
+    {
+        "id":  "140",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小高和小新是学校机器人社团的成员，他们需要为4间教室的\u0027智能温控系统\u0027升级控制芯片。王老师给他们一共提供了16块\u0027节能控制芯片\u0027,每块芯片的功率是48瓦。他们要把这些芯片平均分配给4间教室的温控系统，每间教室的温控系统能分到多少瓦的功率。下面是三位学生的解答过程.\n小高:48×16÷4=192(瓦) 小新:48÷4×16=192(瓦) 小美:48×(16÷4)=192(瓦)\n(1)小美的解决方法中，第一步16÷4解决的问题是();第二步48×4解决的问题是()\n(2)根据上面三位学生解决问题的几种方法，我们可以发现:a×b/c=()。(用对应的字母表示)",
+        "options":  [
+
+                    ],
+        "answer":  "每间教室分到几块节能控制芯片每间教室的温控系统能分到功率的瓦数;a/c×b或者a×(b/c)",
+        "solution":  "【分析】(1)16块节能控制芯片分给4个教室，所以16÷4，结果是4，求的是一个教室分4块节能控制芯片，每块芯片的功率是48瓦，用48×4，就是每间教室的温控系统能分到功率的瓦数。(2)根据观察48×16÷4=48÷4×16=48×(16÷4),48,16,4分别用字母a,b,c表示,a×b/c=a/c×b=a×(b/c),据此解题.\n【详解】(1)小美的解决方法中，第一步16÷4解决的问题是每间教室分到几块节能控制芯片;第二步48×4解决的问题是每间教室的温控系统能分到功率的瓦数。(2)根据上面三位学生解决问题的几种方法，我们可以发现:a×b/c=a/c×b=a×(b/c)。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·河南郑州·期末"
+    },
+    {
+        "id":  "141",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "两个加数的和是23.5，如果其中一个加数减少4.7，另一个加数不变，和变为()。",
+        "options":  [
+
+                    ],
+        "answer":  "18.8",
+        "solution":  "【分析】根据加数+加数=和，一个加数减少4.7，另一个加数不变，则和减少4.7，用两个加数的和减去4.7，即可解答.\n【详解】23.5-4.7=18.8，两个加数的和是23.5，如果其中一个加数减少4.7，另一个加数不变，和变为18.8。",
+        "knowledge_points":  [
+                                 "加，减法的意义和各部分间的关系",
+                                 "一位小数的进位加法",
+                                 "退位减法",
+                                 "和与差的变化规律"
+                             ],
+        "source":  "24-25四年级下·河北邯郸·期末"
+    },
+    {
+        "id":  "142",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "如果把137+72÷3×8的运算顺序改成先算乘法，再算除法，最后算加法，那么算式是(),最后结果是()。",
+        "options":  [
+
+                    ],
+        "answer":  "137+72÷(3×8);140",
+        "solution":  "【分析】题目要求先算乘法，再算除法，最后算加法。在原算式137+72÷3×8中，乘法是3×8，根据四则混合运算先乘除后加减，有括号先算括号内的运算顺序，需要给3×8加上括号，得到137+72÷(3×8).\n【详解】137+72÷(3×8)=137+72÷24=137+3=140，所以，运算顺序改变后得到的算式为:137+72÷(3×8),最后结果是140。",
+        "knowledge_points":  [
+                                 "有括号的运算顺序",
+                                 "无括号的运算顺序"
+                             ],
+        "source":  "24-25四年级下·河北邯郸·期末"
+    },
+    {
+        "id":  "143",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "已知△÷15=☆,请根据除法各部分间的关系写出一个算式()。",
+        "options":  [
+
+                    ],
+        "answer":  "☆×15=△或△÷☆=15",
+        "solution":  "【分析】根据被除数÷除数=商，被除数=除数×商，被除数÷商=除数，解答此题即可.\n【详解】由分析可知：已知△÷15=☆,根据除法各部分间的关系,△是被除数,15是除数,☆是商，可以写出算式是:☆×15=△，△÷☆=15，15×☆=△。(答案不唯一)",
+        "knowledge_points":  [
+                                 "乘，除法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·河南省直辖县级单位·期末"
+    },
+    {
+        "id":  "144",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "把212+460=672,672÷24=28,20×28=560合并成综合算式是()。",
+        "options":  [
+
+                    ],
+        "answer":  "20×[(212+460)÷24]=560",
+        "solution":  "【分析】观察这些算式的共同数值，发现首先计算加法，得到的和再算除法，得到的商再算乘法，合理添加括号，据此作答.\n【详解】把212+460=672,672÷24=28,20×28=560合并成综合算式是20×[(212+460)÷24]=560。",
+        "knowledge_points":  [
+                                 "带有中括号的混合运算",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·江西上饶·期末"
+    },
+    {
+        "id":  "145",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据115+29=144,144÷3=48,960÷48=20列出一个综合算式()。",
+        "options":  [
+
+                    ],
+        "answer":  "960÷[(115+29)÷3]=20",
+        "solution":  "【分析】分析算式的关系，算式960÷48=20中的48是由算式144÷3=48得来的，而144是由115+29得来的，因为要先计算115+29=144，所以给115+29加上小括号，接着要计算144÷3，所以给(115+29)÷3加上中括号，据此列综合算式.\n【详解】根据115+29=144,144÷3=48,960÷48=20列出一个综合算式为:960÷[(115+29)÷3]=960÷[144÷3]=960÷48=20。",
+        "knowledge_points":  [
+                                 "带有中括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·江西九江·期末"
+    },
+    {
+        "id":  "146",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "验算592÷37=16，可以用乘法验算，这是根据(),也可以用除法()验算，这是根据()。",
+        "options":  [
+
+                    ],
+        "answer":  "16×37=592;商×除数=被除数;592÷16=37;被除数÷商=除数",
+        "solution":  "【分析】根据乘法与除法的各部分之间的关系，可知除法可以用乘法验算，即商乘除数等于被除数，也可以用除法验算，即被除数除以商等于除数.\n【详解】由分析可得：验算592÷37=16，可以用乘法16×37=592验算，这是根据商×除数=被除数，也可以用除法592÷16=37验算，这是根据被除数÷商=除数。",
+        "knowledge_points":  [
+                                 "乘，除法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·河南郑州·期末"
+    },
+    {
+        "id":  "147",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "甲，乙两个油桶中共有100千克油，将乙桶中的15千克油注入甲桶，此时甲桶中的油是乙桶中的油的4倍，那么原来甲桶中的油比乙桶中的油多()千克。",
+        "options":  [
+
+                    ],
+        "answer":  "30",
+        "solution":  "【分析】根据题意，把甲乙两个油桶共有的油看作5份，可以计算出每份是多少千克油，将乙桶中的15千克油注入甲桶后，甲桶占了其中的4份，乙桶占了其中的1份,1份即100÷5=20千克，可以计算出注入后各桶中油的质量，再用乙桶的油加上15千克，甲桶的油减去15千克，即是甲乙两桶原有油的质量，再用甲桶原有油的质量减去乙桶原有油的质量，列式解答即可.\n【详解】100÷(1+4)=100÷5=20(千克),20×4=80(千克),80-15=65(千克),20+15=35(千克),65-35=30(千克),所以原来甲桶中的油比乙桶中的油多30千克。",
+        "knowledge_points":  [
+                                 "和差倍问题",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·江西上饶·期末"
+    },
+    {
+        "id":  "148",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一道减法算式中，被减数，减数，差的和是480，减数是差的2倍，被减数是(),减数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "240;160",
+        "solution":  "【分析】在减法中，被减数-减数=差，差+减数=被减数，由题可知，被减数+减数+差=480，根据减法各部分间关系可知，减数+差+减数+差=480，用480除以2，算出减数与差的和是240，即被减数也是240;再根据题目可知，减数是差的2倍，即差+差+差=240，由此可以算出差，再用差乘2，即可算出减数。据此解答.\n【详解】480÷2=240,240÷3=80,80×2=160，这道减法算式为:240-160=80。被减数是240，减数是160。",
+        "knowledge_points":  [
+                                 "加，减法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·河南安阳·期末"
+    },
+    {
+        "id":  "149",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "明明在计算除法时，错把除数36写成了63，得到的商是12，那么被除数是(),正确的商是()。",
+        "options":  [
+
+                    ],
+        "answer":  "756;21",
+        "solution":  "【分析】先用错误的除数63乘商12，求出原来的被除数，再用被除数除以除数36，求出正确的商，据此解答即可.\n【详解】63×12=756,756÷36=21，明明在计算除法时，错把除数36写成了63，得到的商是12，那么被除数是756，正确的商是21。",
+        "knowledge_points":  [
+                                 "乘，除法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·河南许昌·期末"
+    },
+    {
+        "id":  "150",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "\u0027洛阳唐三彩\u0027是中国陶瓷烧制工艺的珍品，哪吒买了4个，敖丙买了6个，这样哪吒比敖丙少花358元。那么一个洛阳唐三彩工艺品的价钱是()元。",
+        "options":  [
+
+                    ],
+        "answer":  "179",
+        "solution":  "【分析】先计算敖丙与哪吒买的数量差，再用总价之差除以数量差，可以计算出一个洛阳唐三彩工艺品的价钱是多少元.\n【详解】358÷(6-4)=358÷2=179(元),\"洛阳唐三彩\"是中国陶瓷烧制工艺的珍品，哪吒买了4个，敖丙买了6个，这样哪吒比敖丙少花358元。那么一个洛阳唐三彩工艺品的价钱是179元。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·河南洛阳·期末"
+    },
+    {
+        "id":  "151",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "这几个数1,2,4,8经过怎样的运算才能得到24。请写出两种不同的算式;算式①(),算式②()。",
+        "options":  [
+
+                    ],
+        "answer":  "(4+1-2)×8=24;2×1×(4+8)=24",
+        "solution":  "【分析】可以用先4+1=5，再用5-2=3，最后用3×8=24;还可以用4+8=12，再用2×1=2，最后用2×12=24.\n【详解】根据分析列出综合算式如下：算式①:(4+1-2)×8=24;算式②:2×1×(4+8)=24。(答案不唯一)",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "24点"
+                             ],
+        "source":  "24-25四年级下·四川乐山·期末"
+    },
+    {
+        "id":  "152",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用3,5,6,8经过怎样的运算能得到24，列式是()。",
+        "options":  [
+
+                    ],
+        "answer":  "3×8×(6-5)=24",
+        "solution":  "【分析】要使用3,5,6,8通过运算得到24，可考虑将24分解为3×8，再通过剩余的数5和6构造出1，使得3×8×1=24。通过观察,6-5=1，因此将3,8与(6-5)相乘即可得到结果.\n【详解】用3,5,6,8经过怎样的运算能得到24，列式是3×8×(6-5)=24。具体运算过程：先计算括号内的减法:6-5=1;再依次相乘:3×8=24;最后乘1:24×1=24。因此,3×8×(6-5)=24满足条件。(答案不唯一)",
+        "knowledge_points":  [
+                                 "24点",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·河南郑州·期末"
+    },
+    {
+        "id":  "153",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "根据3527-685=2842，直接写出下面两道题的得数.\n3527-2842=();2842+685=()",
+        "options":  [
+
+                    ],
+        "answer":  "685;3527",
+        "solution":  "【分析】被减数-减数=差，被减数-差=减数，差+减数=被减数，据此解答.\n【详解】根据3527-685=2842，直接写出下面两道题的得数:3527-2842=685,2842+685=3527。",
+        "knowledge_points":  [
+                                 "加，减法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·河南郑州·期末"
+    },
+    {
+        "id":  "154",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把21+13=34,34÷17=2,50-2=48组成一个综合算式是()。",
+        "options":  [
+
+                    ],
+        "answer":  "50-(21+13)÷17=48",
+        "solution":  "【分析】根据题意可知，这个算式先算加法求出和，再用和作为被除数计算除法求出商，最后用商作为减数计算减法，如果列成一个综合算式那么需要把加法放进小括号内再计算.\n【详解】50-(21+13)÷17=50-34÷17=50-2=48，把21+13=34,34÷17=2,50-2=48组成一个综合算式是50-(21+13)÷17=48。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·四川凉山·期末"
+    },
+    {
+        "id":  "155",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "某书店搞促销活动，同款图书\u0027买四送一\u0027,小明选中了一套20本的图书，每本25元，小明购买这套图书实际应付()元。",
+        "options":  [
+
+                    ],
+        "answer":  "400",
+        "solution":  "【分析】同款图书\u0027买四送一\u0027,即花费4本图书的价钱可以得到5本图书，先用一套图书的本数除以5，求出几组，就是花费几个4本图书的价钱，根据总价=单价×数量，解答即可.\n【详解】20÷(4+1)×4=20÷5×4=4×4=16(本),16×25=400(元)。小明购买这套图书实际应付400元。",
+        "knowledge_points":  [
+                                 "两位数乘两位数的进位乘法",
+                                 "经济问题"
+                             ],
+        "source":  "24-25四年级下·江西赣州·期末"
+    },
+    {
+        "id":  "156",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "两个不为零的数的差是0，则商一定是1。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】两个不为零的数的差为0，说明这两个数相等。根据除法的意义，一个非零数除以它本身，商为1。依此举例并判断即可.\n【详解】假如两个不为零的数为10,10-10=0,10÷10=1，两个不为零的数的差是0，则商一定是1。原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "乘，除法的意义和各部分间的关系",
+                                 "加，减法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·四川凉山·期末"
+    },
+    {
+        "id":  "157",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "6.3-5.24+2.48与102÷17×42的运算顺序相同，都是从左往右依次计算。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】四则混合运算的运算顺序是同级运算时，从左到右依次计算;两级运算时，先算乘除，后算加减。有括号时，先算括号里面的。据此解答即可.\n【详解】6.3-5.24+2.48由减法和加法组成，属于同级运算，应从左往右依次计算;102÷17×42由除法和乘法组成，属于同级运算，应从左往右依次计算;则6.3-5.24+2.48与102÷17×42的运算顺序相同，都是从左往右依次计算。原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "无括号的运算顺序"
+                             ],
+        "source":  "24-25四年级下·四川凉山·期末"
+    },
+    {
+        "id":  "158",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "被减数=减数+差，除数=商/被除数。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】在减法算式中，被减数-减数=差，差+减数=被减数;在除法算式中，被除数/除数=商，被除数/商=除数。据此解答.\n【详解】由分析得，被减数=减数+差，除数=被除数/商。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "乘，除法的意义和各部分间的关系",
+                                 "加，减法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·陕西商洛·期末"
+    },
+    {
+        "id":  "159",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "两个数相乘的积一定大于两个数相加的和。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】两个数相乘的积可能大于，等于或小于两个数相加的和。可以通过举反例的方法，例如1×1=1,1+1=2，而1\u003c2。由此反驳两个数相乘的积一定大于两个数相加的和.\n【详解】两个数相乘的积不一定大于这两个数相加的和，例如1×1=1,1+1=2，而1\u003c2。原说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "乘，除法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25二年级上·河南周口·期末"
+    },
+    {
+        "id":  "160",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "牛牛在计算872+ □ ÷8时先算了加法，计算出结果是125，这个算式正确的结果是128。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】根据题意，在计算872+□÷8时先算了加法，结果是125，先用125×8求出872+□的结果，再减去872即可求出□是多少，代入算式后先计算除法再计算加法，据此计算出正确的结果后填空即可.\n【详解】(872+□)÷8=125,872+□=125×8=1000,□=1000-872=128,872+128÷8=872+16=888，牛牛在计算872+□÷8时先算了加法，计算出结果是125，这个算式正确的结果是888，所以原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "除加，除减混合运算",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·四川乐山·期末"
+    },
+    {
+        "id":  "161",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "列综合算式计算。368除以一个数，商是15，余数是8，这个数是多少。",
+        "options":  [
+
+                    ],
+        "answer":  "24",
+        "solution":  "【分析】有余数的除法算式为：被除数/除数=商。余数，根据除数=(被除数-余数)/商，计算即可.\n【详解】(368-8)÷15=360÷15=24，这个数是24。",
+        "knowledge_points":  [
+                                 "乘，除法的意义和各部分间的关系",
+                                 "三位数除以两位数的有余数除法"
+                             ],
+        "source":  "24-25四年级下·四川凉山·期末"
+    },
+    {
+        "id":  "162",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "列综合算式计算。75与25的差除它们的和，商是多少。",
+        "options":  [
+
+                    ],
+        "answer":  "(75+25)÷(75-25)=2;2",
+        "solution":  "【分析】先求出75加25的和，再求出75减去25的差，求和与差都需要加上括号，然后用和除以差即可.\n【详解】(75+25)÷(75-25)=100÷50=2。商是2。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·四川凉山·期末"
+    },
+    {
+        "id":  "163",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "李大伯是某农产品合作社的负责人，近期收获了一批24吨的蔬菜需要一次性从乡下运到县城的批发市场。运输公司提供两种车型：每辆大卡车一次能运5吨，运一次收运费110元，每辆小卡车一次能运2吨，运一次收运费50元。要使运费最省，请你设计一个运货方案。",
+        "options":  [
+
+                    ],
+        "answer":  "用大卡车运4次和小卡车运2次最省钱;需要运费540元。",
+        "solution":  "【分析】大卡车运1吨货物需要110÷5=22(元),小卡车运1吨货物需要50÷2=25(元),因此要使运费最省，尽量租用大卡车,24÷5=4(辆)。4(吨),剩下的尽量租用小卡车能装满,4÷2=2(辆);据此解答.\n【详解】110÷5=22(元/吨),50÷2=25(元/吨),22元\u003c25元，所以尽量多用大卡车而且尽量装满比较省钱。24÷5=4(辆)。4(吨),这4吨如果用1辆大卡车，需要110元，如果用2辆小卡车，需要:50×2=100(元),110元\u003e100元，用2辆小卡车比较省钱。则用大卡车运4次和小卡车运2次需要运费:4×110+2×50=440+100=540(元),答：用大卡车运4次和小卡车运2次最省钱;需要运费540元。",
+        "knowledge_points":  [
+                                 "租船问题",
+                                 "经济问题"
+                             ],
+        "source":  "24-25四年级下·陕西商洛·期末"
+    },
+    {
+        "id":  "164",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "四年级共有360名师生去红色革命基地研学。每辆大车可坐50人，租金500元;每辆小车可坐30人，租金390元。怎样租车最省钱。最少要花多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "租6辆大车和2辆小车;3780元",
+        "solution":  "【分析】根据两种车所坐人数和租金，用\u0027租金/可坐人数\u0027分别计算出每人所需钱数，比较即可找到最省钱的是哪种车，然后多租这种车，空位越少越省钱，综合考虑计算即可.\n【详解】390÷30=13(元),500÷50=10(元),13\u003e10，所以大车更便宜;360÷50=7(辆)。10(人),租7辆大车和1辆小车，此时有空余,500×7+390=3500+390=3890(元),租6辆大车，此时还剩:360-50×6=360-300=60(人),60÷30=2(辆),剩余人数再租2辆小车,500×6+390×2=3000+780=3780(元),3780元\u003c3890元。答：租6辆大车和2辆小车最省钱;最少要花3780元。",
+        "knowledge_points":  [
+                                 "租船问题",
+                                 "乘加，乘减混合运算",
+                                 "经济问题"
+                             ],
+        "source":  "24-25四年级下·河南省直辖县级单位·期末"
+    },
+    {
+        "id":  "165",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "李叔叔家上个月用水10吨，交水费50元;王叔叔家比李叔叔家多交了25元水费，王叔叔家上个月用了多少吨水。",
+        "options":  [
+
+                    ],
+        "answer":  "15吨",
+        "solution":  "【分析】根据\u0027总费用/数量=单价\u0027,用李叔叔家上个月交的50元水费除以吨数，求出单价，再根据\u0027总费用/单价=数量\u0027,先用李叔叔家上个月交的水费加上王叔叔家比李叔叔家多交的25元水费，求出上个月王叔叔家交的水费，再除以单价即可解答.\n【详解】(50+25)÷(50/10)=75÷5=15(吨),答：王叔叔家上个月用了15吨水。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "经济问题"
+                             ],
+        "source":  "2025·山东·小升初真题"
+    },
+    {
+        "id":  "166",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "某景区门票价格方案.\n方案一：成人每人100元，儿童每人50元;方案二：团体10人以上(含10人)每人80元.\n(1)成人10人，儿童4人，怎样买票比较划算。\n(2)成人4人，儿童10人，怎样买票比较省钱。",
+        "options":  [
+
+                    ],
+        "answer":  "10个成人购团体票,4个儿童购儿童票;4个成人购买成人票,10个儿童购买儿童票",
+        "solution":  "【分析】(1)10个成人购买成人票,10×100=1000(元)求出10个成人购买成人票的钱数,4个儿童购买儿童票,4×50=200(元),求出4个儿童购买儿童票的钱数，再把两者相加。方案二可按总人数买团体票，也可10个成人买团体票,4个儿童买儿童票，比较三种方式费用。(2)4个成人和10个儿童同理分别计算：全按原价，全按团体，部分团体部分儿童票，再比较即可.\n【详解】(1)方案一:10×100+4×50=1200(元)。方案二：购买团体票(10+4)×80=1120(元);10个成人购买团体票和4个儿童购买儿童票,10×80+4×50=1000(元)。1200\u003e1120\u003e1000，答:10个成人购团体票,4个儿童购儿童票比较划算。(2)方案一:4×100+10×50=900(元)。方案二：购买团体票,(10+4)×80=1120(元);6个儿童和4个成人购团体票,4个儿童购儿童票,80×(6+4)+(10-6)×50=800+200=1000(元)。900\u003c1000\u003c1120，答:4个成人购买成人票和10个儿童购买儿童票比较划算。",
+        "knowledge_points":  [
+                                 "购票问题",
+                                 "带有小括号的混合运算",
+                                 "乘加，乘减混合运算"
+                             ],
+        "source":  "24-25四年级下·河南濮阳·期末"
+    },
+    {
+        "id":  "167",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "A超市和B超市卖同款篮球.\nA超市：买5个送1个，每个48元.\nB超市：每个45元，每满200元减20元.\n体育老师要买12个篮球，在哪家超市购买更省钱。请计算说明。",
+        "options":  [
+
+                    ],
+        "answer":  "A超市;计算见详解",
+        "solution":  "【分析】A超市采用\u0027买5送1\u0027优惠，以6个为一组，用12除以6求出组数(也就是赠送个数),用12减去赠送个数求出实际需购买数量，再乘单价求总价;B超市先按单价求原价，再看满200元可减几次20元，求出优惠后价格，最后比较两家总费用.\n【详解】A超市:12÷6=2(组),(12-2)×48=10×48=480(元)。B超市:12×45=540(元),540÷200=2(个)。140(元),即优惠2次20元,540-20×2=540-40=500(元)。480元\u003c500元，答：在A超市购买更省钱。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·河南许昌·期末"
+    },
+    {
+        "id":  "168",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "某科技馆举办\u0027中国人工智能发展成果展\u0027,门票有两种价格方案，阳光小学四年级有2名老师和8名学生准备去参观，怎样购票最划算。最少需要多少钱。\n方案一：成人每人40元，儿童每人20元;方案二：团体票(10人及以上)每人25元。",
+        "options":  [
+
+                    ],
+        "answer":  "方案一;240元",
+        "solution":  "【分析】由题意得，四年级有2名老师和8名学生准备去参观。若选方案一，用成人票价乘2再加儿童票价乘8;若选方案二，总人数2+8=10人，直接按25元/人计算，再比较两种方案总价.\n【详解】方案一:40×2+20×8=80+160=240(元)。方案二:(2+8)×25=10×25=250(元),240\u003c250。答：选择方案一购票更划算，最少需要240元。",
+        "knowledge_points":  [
+                                 "购票问题"
+                             ],
+        "source":  "24-25四年级下·河南郑州·期末"
+    },
+    {
+        "id":  "169",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "龙门石窟是我国丰富的石窟艺术宝库，被联合国教科文组织评为\u0027中国石刻艺术的最高峰\u0027,位居中国各大石窟之首。周末学校组织同学们去龙门石窟研学。16名老师带领168名同学乘坐客车去研学。怎样租车最省钱。需要花费多少元。\n大客车限坐40人，每辆车租金520元;小客车限坐16人，每辆车租金240元。",
+        "options":  [
+
+                    ],
+        "answer":  "租3辆大客车和4辆小客车，花费为2520元",
+        "solution":  "【分析】根据题意，分别用除法计算出大，小客车的人均租金，租金低的多租;用总人数除以较便宜车型的载人量，求出车辆数后结合余数调整其它车型，列出多种方案并比较总费用，选择最低方案即可.\n【详解】大客车:520÷40=13(元),小客车:240÷16=15(元),因此尽量租大客车，且尽量满载。16+168=184(人)。①184÷40=4(辆)。24(人),全租大客车至少5辆，费用5×520=2600(元)。②租4辆大客车，余24人坐2辆小客车，费用4×520+2×240=2560(元)。③租3辆大客车，可坐120人，余64人正好坐4辆小客车，费用3×520+4×240=2520(元)。2600\u003e2560\u003e2520，即租3辆大客车和4辆小客车花费最小。答：租3辆大客车和4辆小客车，花费为2520元。",
+        "knowledge_points":  [
+                                 "租船问题"
+                             ],
+        "source":  "24-25四年级下·河南郑州·期末"
+    },
+    {
+        "id":  "170",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "甲，乙两桶汽油共重200千克(不算桶重)。如果从甲桶倒出30千克给乙桶，那么甲桶剩余汽油的质量是现在乙桶中汽油质量的3倍。原来甲桶里有汽油多少千克。",
+        "options":  [
+
+                    ],
+        "answer":  "180千克",
+        "solution":  "【分析】根据题意可知，两桶汽油的总质量不变。甲桶剩余汽油的质量是现在乙桶中汽油质量的3倍，把乙桶中汽油质量看作1份，则甲桶剩余汽油的质量是3份，两桶共重200千克，用200除以4可算出1份的质量，也就是现在乙桶中汽油质量，再乘3，求出甲桶剩余汽油，再加上30千克，就是原来甲桶中汽油的质量.\n【详解】3+1=4(份) 200÷4=50(千克) 50×3+30=150+30=180(千克) 答：原来甲桶里有汽油180千克。",
+        "knowledge_points":  [
+                                 "乘加，乘减混合运算",
+                                 "和差倍问题"
+                             ],
+        "source":  "24-25四年级下·四川乐山·期末"
+    },
+    {
+        "id":  "171",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "某修路队要修一条长6000米的公路，已经修了6天，共修了900米。照这样的速度，该修路队还需要修多少天才能完成任务。",
+        "options":  [
+
+                    ],
+        "answer":  "34天",
+        "solution":  "【分析】根据题意，用900除以6，计算出每天修的长度，用总长度减去已经修了的长度，求出剩下没有修的长度，再用剩下没有修的长度除以每天修的长度，即可求出该修路队还需要修多少天才能完成任务.\n【详解】(6000-900)÷(900/6)=5100÷(900/6)=5100÷150=34(天),答：该修路队还需要修34天才能完成任务。",
+        "knowledge_points":  [
+                                 "有具体量的工程问题",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·河南安阳·期末"
+    },
+    {
+        "id":  "172",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "在旅游途中，一个旅行团有24名游客想穿古装拍照。有三类古装可选，租金分别为63元,60元和58元;有三类造型可选，费用分别为42元,50元和57元。选24套古装加造型，最少需要花多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "2400元",
+        "solution":  "【分析】要花费最少的钱，需要分别选择租金最便宜的古装和费用最便宜的造型。古装租金分别为63元,60元,58元，其中58元是最便宜的。造型费用分别为42元,50元,57元，其中42元是最便宜的。然后计算总花费即可解答.\n【详解】(58+42)×24=100×24=2400(元),答：最少需要花2400元。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "经济问题"
+                             ],
+        "source":  "24-25四年级下·河南洛阳·期末"
+    },
+    {
+        "id":  "173",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "材料：为迎接\"六一\"儿童节，星光小学计划为三年级200名学生每人准备一份节日礼物。学校安排王老师到惠民超市采购，超市现有A,B两种礼盒可供选择。A礼盒内含笔记本1本，铅笔2支，橡皮1块，单价15元;B礼盒内含笔记本2本，铅笔1支，尺子1把，单价18元。超市正在进行促销活动：购买A礼盒满100盒立减100元(满200盒立减200元，以此类推);购买B礼盒每满50盒赠送5盒(不满50盒不赠送，赠送的礼盒不参与满赠)。王老师带了3500元钱，需要根据学生需求和预算选择合适的礼盒.\n(1)如果全部购买A礼盒，需要支付多少元。\n(2)如果全部购买B礼盒，实际需要购买多少盒。\n(3)比较两种方案的总费用，王老师带的钱是否足够购买更贵的那种礼盒。",
+        "options":  [
+
+                    ],
+        "answer":  "2800元;185盒;足够",
+        "solution":  "【分析】(1)根据总价=单价×数量，即可算出总价为3000元，购买A礼盒满100盒立减100元，所以购买了200盒，需减去200元，即可计算出需要支付多少元。(2)因为购买B礼盒每满50盒赠送5盒，所以若赠送了3个5盒，需购买200-5×3=185(盒)。将50盒看作一组，若购买185盒，则185÷50=3(组)。35(盒),符合赠送3个5盒，即求得实际需要购买盒数。(3)由(1)已算出买A礼盒的价格，由(2)及总价=单价×数量，可求得此时总价。二者的总价与3500比较大小，即可求得结果.\n【详解】(1)200×15=3000(元),3000-200=2800(元),答：如果全部购买A礼盒，需要支付2800元。(2)若赠送了3个5盒，需购买200-5×3=200-15=185(盒),验证:185÷50=3(组)。35(盒),即赠3×5=15(盒),符合假设。答：如果全部购买B礼盒，实际需要购买185盒。(3)由(1)知,A方案费用2800元，由(2)知B方案费用为18×185=3330(元),2800元\u003c3330元\u003c3500元。答：王老师带的钱足够购买更贵的那种礼盒。",
+        "knowledge_points":  [
+                                 "促销问题"
+                             ],
+        "source":  "24-25六年级下·广东东莞·期末"
+    },
+    {
+        "id":  "174",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "绿色发展，低碳出行。某公司计划在新城区投放1800辆共享单车,()。照这样的速度，剩下的共享单车还要几天能投放完。(1)\"()\"处需要补充的信息是()(填序号)。(2)根据补充的信息，算一算剩下的共享单车还要几天能投放完。",
+        "options":  [
+                        "A. 已经投放了540辆。",
+                        "B. 每天投放180辆。",
+                        "C. 3天投放了540辆。",
+                        "D. 一共需要10天能投放完共享单车。"
+                    ],
+        "answer":  "C;7天",
+        "solution":  "【分析】(1)剩下的共享单车还要几天能投放完，通过\"剩下的共享单车\"需要知道前几天一共已经投放了多少辆，总数量减去已投放数量求出剩下的数量。通过\"还要几天能投放完\"需要知道前几天平均每天投放多少辆，用剩下的数量除以每天的投放数量。(2)总数量减去已投放数量求出剩下的数量，再用剩下的数量除以每天的投数量(前几天一共投放的量数除以天数，就是平均每天的投放数量。),求出剩下的共享单车还要几天能投放完.\n【详解】(1)A。已经投放了540辆，只知道已经投放的量数，不能求出平均每天投放的量数，不能求出剩下的共享单车还要几天能投放完。B。每天投放180辆，只知道平均每天投放的数量，不知道已经投放的数量，不能求出剩下的共享单车还要几天能投放完。C。3天投放了540辆，知道前3天一共投放了540辆，再用540÷3，求出平均每天投放的数量，能求出剩下的共享单车还要几天能投放完。D。一共需要10天能投放完共享单车，如果知道一共需要10天投完，但是题中没有说已经投放了几天了，不能求出剩下的共享单车还要几天能投放完。故答案为:C。(2)(1800-540)÷(540/3)=1260÷180=7(天),答：剩下的共享单车还要7天能投放完。",
+        "knowledge_points":  [
+                                 "\"提问题\"\"填条件\"问题",
+                                 "带有小括号的混合运算",
+                                 "单归一问题"
+                             ],
+        "source":  "24-25四年级下·山东日照·期末"
+    },
+    {
+        "id":  "175",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "乐乐要看一本450页的《海底两万里》,前6天看了90页，照这样的速度，还需要多少天才能把这本书看完。(1)小明列算式为450÷(90/6)-6，你知道他每一步计算解决的是什么问题吗。请将他的思路补充完整.\n①先算90÷6=15(页),求的是();②再算(),求的是();③最后算(),求的是().\n(2)你还有不同的思路吗。请列出综合算式并解答。",
+        "options":  [
+
+                    ],
+        "answer":  "①每天看多少页;②450÷15=30(天);看完这本书需要多少天;③30-6=24(天);还需要多少天才能看完。;(450-90)÷(90/6);24天",
+        "solution":  "【分析】(1)这个算式先计算小括号里的除法，接着算小括号外的除法，最后算小括号外的减法。①前6天看了90页，所以90除以6解答的是乐乐每天看多少页，这个商是15。②450是这本书的总页数，接下来计算的是450除以15，求出商，所以这个商表示的是看完这本书需要多少天。③最后算30减6，计算出差是24，其中的30是看完这本书需要的时间，而6是已经看了的天数，所以这个差就是看完剩下的还需要的天数。(2)可以先求出乐乐每天看多少页，用90除以6得15，即他每天看15页,450减90求出看了6天后还剩下360页没有看，最后用360除以每天看的页数，即可求出看完剩下的页数还需要多少天.\n【详解】(1)450÷(90/6)-6=450÷15-6=30-6=24(天)。①先算90÷6=15(页),求的是每天看多少页;②再算450÷15=30(天),求的是看完这本书需要多少天;③最后算30-6=24(天),求的是还需要多少天才能看完。(2)(450-90)÷(90/6)=360÷15=24(天),答：还需要24天。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·四川广元·期末"
+    },
+    {
+        "id":  "176",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "只列综合算式，不计算。李阿姨借了一本书，如果每天看40页，要13天才能看完。为了按时归还，必须提前3天看完，那么她平均每天应该看多少页。",
+        "options":  [
+
+                    ],
+        "answer":  "40×13÷(13-3)",
+        "solution":  "【分析】根据题意可知，每天看40页,13天看完，所以40乘13等于总页数，再除以提前3天看的天数，即等于提前3天平均每天看的页数，据此即可解答.\n【详解】40×13÷(13-3)=520÷10=52(页),答：她平均每天应该看52页。",
+        "knowledge_points":  [
+                                 "归总问题",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·广东广州·期末"
+    },
+    {
+        "id":  "177",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "只列综合算式，不计算.\n某品牌的新能源汽车在2022年销量达到176万辆,2024年全年销量比2022年的2倍还要多75万辆,2024年的销量达到多少万辆。",
+        "options":  [
+
+                    ],
+        "answer":  "176×2+75",
+        "solution":  "【分析】根据题意可知,2022年的销量乘2，再加75万辆等于2024年的销量，据此列式计算即可解答.\n【详解】176×2+75=352+75=427(万辆),答:2024年的销量达到427万辆。",
+        "knowledge_points":  [
+                                 "乘加，乘减混合运算",
+                                 "求一个数的几倍是多少"
+                             ],
+        "source":  "24-25四年级下·广东广州·期末"
+    },
+    {
+        "id":  "178",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "游泳馆有A,B两类卡,A类卡:150元，限使用6次;B类卡:180元，限使用8次。即将要上五年级的华华在暑假期间准备去游泳馆游泳30次.\n(1)怎样购卡最合算。需花费多少元。\n(2)爸爸，妈妈想各陪华华游7次，这时他们全家怎样购卡最合算。",
+        "options":  [
+
+                    ],
+        "answer":  "购买3张B类卡和1张A类卡最合算，需花费690元;这时他们全家购买2张A类卡和4张B类卡最合算",
+        "solution":  "【分析】(1)根据题意，已知A类卡:150元，限使用6次;B类卡:180元，限使用8次。首先，用除法计算，比较两类卡的单次使用成本;优先使用单次成本更低的B类卡;华华需要30次，用30除以8，求出商就是B类卡的张数，如果有余数，合理安排A类卡，最后根据价格，计算出总花费即可。(2)爸爸，妈妈想各陪华华游7次，先计算出全家进入总次数;用总人数除以8，求出B类卡的张数，如果有余数，合理安排A类卡，分别制定方案，最后根据卡的价格，计算出总费用，比较各个方案的总费用，选择费用最低的即可.\n【详解】根据分析可知:(1)A类卡:150÷6=25(元),B类卡:180÷8=22.5(元),B类卡更便宜,30÷8=3(张)……6(次),6÷6=1(张),3×180+1×150=540+150=690(元)。答：购买3张B类卡和1张A类卡最合算，需花费690元。(2)30+7×2=30+14=44(次),44÷8=5(张)……4(次),5×180+1×150=900+150=1050(元),(4+8)÷6=12÷6=2(张),(5-1)×180+2×150=4×180+300=720+300=1020(元),1020\u003c1050。答：这时他们全家购买2张A类卡和4张B类卡最合算。",
+        "knowledge_points":  [
+                                 "租船问题"
+                             ],
+        "source":  "24-25四年级下·湖北十堰·期末"
+    },
+    {
+        "id":  "179",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "5只鸡和4只鸭子的重量相等,1只鸡和1只鸭子共重9千克,3只鸭子和4只鸡共重多少千克。",
+        "options":  [
+
+                    ],
+        "answer":  "31千克",
+        "solution":  "【分析】根据题意,1只鸡和1只鸭子共重9千克，那么5只鸡和5只鸭子共重((9×5))千克。而5只鸡和4只鸭子的重量相等，说明(5+4)只鸭子的重量是((9×5))千克，据此算出1只鸭子的重量。再用9千克减去1只鸭子的重量，就是1只鸡的重量。最后用鸡和鸭各自的重量乘只数，算出4只鸡,3只鸭重多少千克，再相加即可.\n【详解】1只鸭子重:9×5÷(4+5)=45÷9=5(千克),1只鸡重:9-5=4(千克),3×5+4×4=15+16=31(千克)。答:3只鸭子和4只鸡共重31千克。",
+        "knowledge_points":  [
+                                 "乘加，乘减混合运算",
+                                 "带有小括号的混合运算",
+                                 "等量代换"
+                             ],
+        "source":  "24-25四年级下·湖南永州·期末"
+    },
+    {
+        "id":  "180",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "李阿姨要加工一批机器零件，计划每小时加工25个,160小时刚好可以完成任务，实际每小时比计划多加工15个。实际比计划提前几小时完成任务。",
+        "options":  [
+
+                    ],
+        "answer":  "60小时",
+        "solution":  "【分析】用每小时计划加工个数乘需要时间，即可算出共需要加工多少个，再算出实际每小时加工多少个，然后用一共需要加工的个数除以实际每小时加工个数，即可算出实际需要加工多少个小时，最后再用计划加工需要的时间减去实际加工需要的时间，即可算出实际比计划提前了几个小时完成任务。据此解答.\n【详解】25×160÷(25+15)=4000÷40=100(小时),160-100=60(小时)。答：实际比计划提前60小时完成任务。",
+        "knowledge_points":  [
+                                 "归总问题",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·湖南永州·期末"
+    },
+    {
+        "id":  "181",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "厦门市鼓浪屿有\"万国建筑博览\"之称,2017年7月8日正式被列入《世界遗产名录》。幸福小学6名老师带领80名三年级学生前往鼓浪屿参观，请你帮忙设计一个最省钱的购票方案.\n![](images/2026-03-05%20113021.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "76张学生票,10张团体票;3580元",
+        "solution":  "【分析】观察可知，学生票40元一人，成人票65元一人,10人以上可以购买团体票。购票时，可以学生买学生票，老师买成人票;也可以学生和老师一起买团体票;还可以6名老师和4名学生组成一个团体，购买团体票，其余学生购买学生票。分别计算出这三种购票方式需要的钱，再进行比较，即可解答.\n【详解】方案一：买80张学生票,6张成人票。40×80=3200(元),65×6=390(元),3200+390=3590(元)。方案二：买86张团体票。54×86=4644(元)。方案三：买76张学生票,10张团体票。40×76=3040(元),54×10=540(元),3040+540=3580(元),3580元\u003c3590元\u003c4644元。答：购买76张学生票,10张团体票最省钱，需要3580元。",
+        "knowledge_points":  [
+                                 "购票问题"
+                             ],
+        "source":  "24-25三年级下·福建厦门·期末"
+    },
+    {
+        "id":  "182",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "泰山经典登山路线：从红门出发至玉皇顶，全程大约7500米。爸爸和小明步行一段时间后，发现已经走的米数比剩下的米数多2100米。他们已经走了多少米。(先补充线段图，再解答)剩下的米数：已走的米数:",
+        "options":  [
+
+                    ],
+        "answer":  "作图见详解;4800米",
+        "solution":  "【分析】由题意得，从红门出发至玉皇顶，全程大约7500米。爸爸和小明步行一段时间后，发现已经走的米数比剩下的米数多2100米。可以用7500米减去2100米算出剩下的米数的2倍是多少，然后再除以2算出还剩下多少米。最后再用得数加上2100米即可算出他们已经走了多少米.\n【详解】\n![](images/3eebe46c5d2ee526b616203617aaf8dd33712dc33f6b3976259ab0dece82a86c.jpg)\n(7500-2100)÷2=5400÷2=2700(米)\n2700+2100=4800(米)\n答：他们已经走了4800米。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "和差倍问题"
+                             ],
+        "source":  "24-25四年级下·河南新乡·期末"
+    },
+    {
+        "id":  "183",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "小明和小红同时从家出发相向而行，小明每分钟走60米，小红每分钟走50米，经过8分钟两人相遇，他们两家相距多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "880米",
+        "solution":  "【分析】根据路程和=速度和×时间，用小明每分钟走的速度加上小红每分钟走的速度，求出他俩的速度和，再乘相遇的时间，即可求出他们两家的距离.\n【详解】(60+50)×8=110×8=880(米),答：他们两家相距880米。",
+        "knowledge_points":  [
+                                 "乘加，乘减混合运算",
+                                 "相遇问题"
+                             ],
+        "source":  "24-25五年级下·贵州铜仁·期末"
+    },
+    {
+        "id":  "184",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "林阿姨编60个中国结送给养老院,(她3天编了15个),根据括号内的信息可以知道:(),算式\"60÷(15/3)\"是求出了(),得数是()天。",
+        "options":  [
+
+                    ],
+        "answer":  "她平均每天编5个中国结;编60个中国结需要的总天数;12",
+        "solution":  "【分析】3天编了15个，用15÷3可以计算出她平均每天编5个中国结，然后用60除以平均每天编的个数，即可计算出编60个中国结需要的总天数，据此解题.\n【详解】15÷3=5(个),60÷(15/3)=60÷5=12(天),林阿姨编60个中国结送给养老院,(她3天编了15个),根据括号内的信息可以知道：她平均每天编5个中国结，算式\"60÷(15/3)\"是求出了编60个中国结需要的总天数，得数是12天。",
+        "knowledge_points":  [
+                                 "单归一问题",
+                                 "乘，除法的意义和各部分间的关系",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·福建福州·期末"
+    },
+    {
+        "id":  "185",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "四年级同学去植树，一班植了42棵，二班植的棵数是一班的2倍，三班植的棵数比二班少15棵。三班植了多少棵树。",
+        "options":  [
+
+                    ],
+        "answer":  "69棵",
+        "solution":  "【分析】已知二班植的棵数是一班的2倍，用一班植的棵数乘2，求出二班植的棵数;已知三班植的棵数比二班少15棵，用二班植的棵数减去15，即是三班植的棵数.\n【详解】42×2-15=84-15=69(棵),答：三班植了69棵树。",
+        "knowledge_points":  [
+                                 "求一个数的几倍是多少",
+                                 "乘加，乘减混合运算"
+                             ],
+        "source":  "25-26五年级上·黑龙江鹤岗·开学考试"
+    },
+    {
+        "id":  "186",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "实验学校为了丰富在校学生的课外知识，学校图书室新增文艺类书籍750本，比新增的科技类书籍本数的3倍少了24本，新增科技类书籍有多少本。",
+        "options":  [
+
+                    ],
+        "answer":  "258本",
+        "solution":  "【分析】学校图书室新增文艺类书籍750本，比新增的科技类书籍本数的3倍少了24本，先用750加上24，求出新增科技书的3倍是多少，再除以3，即可求出新增科技类书籍有多少本，据此作答.\n【详解】(750+24)÷3=774÷3=258(本),答：新增科技类书籍有258本。",
+        "knowledge_points":  [
+                                 "三位数除以一位数，首位除不尽",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·河南许昌·期末"
+    },
+    {
+        "id":  "187",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "AI实验室需将112个智能垃圾分类机器人从研发区运输至测试区，有两种专用运输车可选:\nA型运输车：每次运8个机器人，运输费96元/次.\nB型运输车：每次运12个机器人，运输费132元/次.\n如何安排运输最省钱。运输费最少多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "安排8次B型运输车和2次A型运输车;1248元",
+        "solution":  "【分析】要确定最省钱的运输方案，需先对比两种车型的单位运输成本(即运输1个机器人的费用),即用运输费除以运的机器人个数;优先选择成本更低的车型;再通过\"满载优先，少留余数\"的原则调整运输次数.\n【详解】96÷8=12(元/个),132÷12=11(元/个),所以B型车单位成本更低，应优先安排B型车运输，剩余机器人用A型车补充。112÷12=9(次)。4(个),9×132+1×96=1188+96=1284(元),若安排9次B型车，剩余4个机器人，需1次A型车(但A型车每次能运8个，仅运4个会浪费运力，运费仍需96元),此时总运费为1284元;减少1次B型车(即8次B型车):9-1=8(次),8×12=96(个),112-96=16(个),16÷8=2(次),16个机器人恰好能装满2次A型车，无浪费运力;8×132+2×96=1056+192=1248(元)。若全用A型车:112÷8=14(次),14×96=1344(元)。若全用B型车:112÷12≈10(次),10×132=1320(元),1344\u003e1320\u003e1284\u003e1248。答：最省钱的方案是安排8次B型运输车和2次A型运输车，运输费最少为1248元。",
+        "knowledge_points":  [
+                                 "乘加，乘减混合运算",
+                                 "租船问题"
+                             ],
+        "source":  "24-25四年级下·河南郑州·期末"
+    },
+    {
+        "id":  "188",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "玩具加工厂要制作一批吉祥物，原计划每天制作50个,12天完成。改进工艺后提前2天完成任务，平均每天制作多少个。",
+        "options":  [
+
+                    ],
+        "answer":  "60个",
+        "solution":  "【分析】用计划每天制作的个数×计划的天数计算出总个数，改进工艺后提前2天完成任务，用12-2计算出改进工艺后需要的天数，用总个数/天数即可解题.\n【详解】(50×12)÷(12-2)=600÷10=60(个),答：平均每天制作60个。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·河南省直辖县级单位·期末"
+    },
+    {
+        "id":  "189",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "甲，乙两车从相距660千米的A城和B城同时出发，相向而行，经过4小时相遇，已知乙车的速度是甲车的2倍，求甲车，乙车每小时分别行多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "甲车55千米，乙车110千米",
+        "solution":  "【分析】根据题意，用A城和B城之间相距的距离除以两车相遇用的时间，求出两车的速度和，已知乙车的速度是甲车的2倍，则两车的速度和是甲车的(2+1)倍，用两车的速度和除以两车的速度和是甲车的(2+1)倍，求出甲车的速度，再用甲车的速度乘2，即可求出乙车每小时行驶多少千米.\n【详解】660÷4/(2+1)=165÷(2+1)=165÷3=55(千米);55×2=110(千米)。答：甲车每小时行55千米，乙车每小时行110千米。",
+        "knowledge_points":  [
+                                 "相遇问题",
+                                 "和差倍问题"
+                             ],
+        "source":  "24-25四年级下·江西宜春·期末"
+    },
+    {
+        "id":  "190",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "学校礼堂有16排座位，每排24个座。学校安排三年级听心理讲座,8个班的学生刚好坐满。平均每个班有多少人。",
+        "options":  [
+
+                    ],
+        "answer":  "48人",
+        "solution":  "【分析】每排座位数乘座位的排数等于总座位数，总座位数/8即可算出平均每个班的学生人数.\n【详解】由分析可得:24×16÷8=384÷8=48(人),答：平均每个班有48人",
+        "knowledge_points":  [
+                                 "三位数除以一位数，首位不够除",
+                                 "两位数乘两位数的进位乘法"
+                             ],
+        "source":  "24-25三年级下·河南郑州·期末"
+    },
+    {
+        "id":  "191",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "小浪底风景区一周接待了许多旅行团。其中有20个成人团和12个学生研学团，平均每个团有30人。成人团比学生研学团多多少人。",
+        "options":  [
+
+                    ],
+        "answer":  "240人",
+        "solution":  "【分析】先用20-12可算出成人团比学生研学团多多少个团，再乘30可算出成人团比学生研学团多多少人.\n【详解】(20-12)×30=8×30=240(人)。答：成人团比学生研学团多240人。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·河南洛阳·期末"
+    },
+    {
+        "id":  "192",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "张大伯今年共榨了672千克菜油，第一天卖出176千克菜油，第二天卖出208千克菜油，张大伯还剩多少千克菜油。",
+        "options":  [
+
+                    ],
+        "answer":  "288千克",
+        "solution":  "【分析】根据题意，先将第一天和第二天卖出的菜油重量相加，求出一共卖出了多少千克菜油，再用今年榨的菜油总重量减去卖出的菜油重量，即可求出张大伯还剩多少千克菜油.\n【详解】672-(176+208)=672-384=288(千克)。答：张大伯还剩288千克菜油。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·四川乐山·期末"
+    },
+    {
+        "id":  "193",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "学校组织四年级去犁桥景区春游，师生共有280人。每辆燃油大巴可坐40人，租金560元;每辆新能源大巴能坐50人，租金600元。怎样租车最省钱。",
+        "options":  [
+
+                    ],
+        "answer":  "租4辆新能源大巴和2辆燃油大巴",
+        "solution":  "【分析】已知每辆燃油大巴可坐40人，租金560元;每人需要560÷40=14(元);每辆新能源大巴能坐50人，租金600元，每人需要600÷50=12(元)。对比可知，新能源大巴更便宜。在没有空位的情况下，租新能源大巴更便宜。师生共280人参加春游。直接用280除以50算出需要新能源大巴的数量以及多余的人数，然后再合理安排多余的人使得燃油大巴和新能源大巴尽量没有空位即可.\n【详解】560÷40=14(元),600÷50=12(元),14\u003e12，所以租新能源大巴便宜。方案一:280÷50=5(辆)。30(人),需要租5辆新能源大巴和1辆燃油大巴;5×600+1×560=3000+560=3560(元)。方案二:4×50+2×40=200+80=280(人),需要租4辆新能源大巴和2辆燃油大巴:4×600+2×560=2400+1120=3520(元),3560\u003e3520。答：租4辆新能源大巴和2辆燃油大巴更省钱。",
+        "knowledge_points":  [
+                                 "租船问题"
+                             ],
+        "source":  "24-25四年级下·安徽铜陵·期末"
+    },
+    {
+        "id":  "194",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "红旗小学共100名师生去郊游，怎样租车最省钱。写出最省钱的方法.\n大客车限载45人，租金300元\n小客车限载30人，租金220元",
+        "options":  [
+
+                    ],
+        "answer":  "租1辆大客车,2辆小客车。",
+        "solution":  "【分析】尽量租单价便宜的车，而且调整到少留空位，甚至无空位才最省钱.\n【详解】300÷45=6(元)⋯⋯30(元),220÷30=7(元)⋯⋯10(元),6\u003c7，尽量租大车。100÷45=2(辆)⋯⋯10(人),租2辆大客车,1辆小客车:2×300+220=600+220=820(元),租1辆大客车,2辆小客车:300+2×220=300+440=740(元),820\u003e740。答：租一辆大客车和2辆小客车最省钱。",
+        "knowledge_points":  [
+                                 "租船问题"
+                             ],
+        "source":  "24-25四年级下·河南郑州·期末"
+    },
+    {
+        "id":  "195",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "奇思和妙想玩猜数游戏：奇思先在心里想一个数，把心里想的数乘4，再加上90，最后的结果是210，妙想很快就猜到奇思心里想的是什么数了。你知道奇思心里想的数是多少吗。",
+        "options":  [
+
+                    ],
+        "answer":  "30",
+        "solution":  "【分析】根据题意，奇思想的数先乘4，再加90得到210。逆推时，先减去90，再除以4，即可求出原数。以此答题即可.\n【详解】根据分析可知:(210-90)÷4=120÷4=30。答：奇思心里想的数是30。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·内蒙古包头·期末"
+    },
+    {
+        "id":  "196",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "小明计划8天看完一本240页的书，实际每天比原计划少看6页，实际多少天看完这本书。",
+        "options":  [
+
+                    ],
+        "answer":  "10天",
+        "solution":  "【分析】8天看完一本240页的书,240除以8即可求出1天看30页，而实际每天比原计划少看6页，用30减6即可求出实际每天看24页，再用总页数240除以24，即可求出实际多少天看完.\n【详解】240÷8=30(页),240÷(30-6)=240÷24=10(天),答：实际10天看完这本书。",
+        "knowledge_points":  [
+                                 "除数是整十数的口算除法",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·四川凉山·期末"
+    },
+    {
+        "id":  "197",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "四(1)班44名同学去公园划船。一条大船限坐游客6人，租金120元;一条小船限坐游客4人，租金100元。怎样租船最省钱。需要多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "租6条大船和2条小船最省钱;920元",
+        "solution":  "【分析】根据题意，首先用120除以6,100除以4，求出大船和小船的人均费用;要使租船最省钱，应尽量选人均费用低的船，剩下的人坐其他船型，用总人数除以省钱的船型求出船的数量，如果有余数，就合理安排其他的船型，并且使每条船都坐满才最省钱。然后用租大船的条数×租每条大船的钱+租小船的条数×租每条小船的钱=需要的钱.\n【详解】120÷6=20(元),100÷4=25(元),25\u003e20,44÷6=7(条)。2(人),7-1=6(条),(6+2)÷4=8÷4=2(条),6×120+2×100=720+200=920(元)。答：租6条大船和2条小船最省钱，需要920元。",
+        "knowledge_points":  [
+                                 "租船问题",
+                                 "乘加，乘减混合运算"
+                             ],
+        "source":  "24-25四年级下·江西上饶·期末"
+    },
+    {
+        "id":  "198",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "妈妈给爷爷奶奶，外公外婆每人买了2套同一品牌同一价格的保暖内衣，每套保暖内衣的价格是125元，妈妈买保暖内衣一共花了多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "1000元",
+        "solution":  "【分析】根据题意可知，妈妈给4个人，每人买2套保暖内衣，一共买了((2×4))套保暖内衣。用每套保暖内衣的价钱乘购买保暖内衣的数量，求出花费总钱数.\n【详解】2×4×125=8×125=1000(元),答：妈妈买保暖内衣一共花了1000元。",
+        "knowledge_points":  [
+                                 "多位数与一位数的连乘",
+                                 "经济问题"
+                             ],
+        "source":  "24-25四年级下·江西赣州·期末"
+    },
+    {
+        "id":  "199",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一辆汽车从甲地开往乙地，前3小时行驶了210千米，后2小时行驶了150千米。这辆汽车平均每小时行驶多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "72千米",
+        "solution":  "【分析】先用前3小时行驶的距离加上后2小时行驶的距离，求出一共行驶了多少千米，用3+2求出一共行驶了多少小时，根据速度=路程/时间，用一共行驶的距离除以行驶的时间，即可求出这辆汽车平均每小时行驶多少千米.\n【详解】(210+150)÷(3+2)=360÷5=72(千米),答：这辆汽车平均每小时行驶72千米。",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·广西贵港·期末"
+    },
+    {
+        "id":  "200",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "某校四年级师生去研学，有老师15人，学生245人。每辆大车可坐乘客40人，租金800元;每辆小车可坐乘客20人，租金500元。怎样租车最省钱。",
+        "options":  [
+
+                    ],
+        "answer":  "租6辆大车和1辆小车最省钱",
+        "solution":  "【分析】根据题意，首先用800除以40,500除以20，计算出大车和小车的人均费用，比较一下，选择便宜一点的多租;再用245加上15，求出总人数;用总人数除以人均费用较低的车型的载人量，求出租车的数量，如果有余数，合理安排其它车型使之尽量没空座才最省钱;最后根据车辆的租金，计算出租车的总费用;比较选出最省钱的即可.\n【详解】大车每人租金:800÷40=20(元),小车每人租金:500÷20=25(元),20\u003c25，应尽量多租大车，且应尽量没有空座才能最省钱;15+245=260(人),260÷40=6(辆)。20(人),剩余20人租小车:20÷20=1(辆),6×800+500=4800+500=5300(元),剩余20人租大车:20\u003c40，租1辆大车\n(6+1)×800=7×800=5600(元),5300\u003c5600。答：租6辆大车和1辆小车最省钱。",
+        "knowledge_points":  [
+                                 "租船问题",
+                                 "经济问题"
+                             ],
+        "source":  "24-25四年级下·贵州遵义·期末"
+    },
+    {
+        "id":  "201",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.94,
+        "question":  "计算360÷(12+6)×5时，先算()。",
+        "options":  [
+                        "A. 加法",
+                        "B. 除法",
+                        "C. 乘法"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】根据四则混合运算的运算法则：在有括号的算式里，要先算括号里面的，再算括号外面的。据此解答.\n【详解】360÷(12+6)×5=360÷18×5=20×5=100，计算360÷(12+6)×5时，先算加法。故答案为:A.",
+        "knowledge_points":  [
+                                 "有括号的运算顺序"
+                             ],
+        "source":  "24-25五年级下·新疆克孜勒苏·期中"
+    },
+    {
+        "id":  "202",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "计算6000/[65+(65-10)]*5，第二步应算()。",
+        "options":  [
+                        "A. 加法",
+                        "B. 减法",
+                        "C. 乘法"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】当算式里有小括号，中括号，加减法和乘除法时，要先计算小括号里的，再计算中括号里的，最后计算括号外面的。如果括号外面只有乘除法或者只有加减法，则按照从左到右的顺序进行计算，据此解答.\n【详解】题目中有小括号和中括号，并且括号外面只有乘除法，所以第一步先计算小括号里的减法;第二步计算中括号里的加法;第三步计算括号外的除法，最后计算乘法。故答案为:A.",
+        "knowledge_points":  [
+                                 "带有中括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·河南新乡·期末"
+    },
+    {
+        "id":  "203",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "实验小学的4名老师带52名学生去公园划船，大船每条限乘6人，租金30元，小船每条限乘4人，租金24元，下面()种租船方式最省钱。",
+        "options":  [
+                        "A. 9条大船",
+                        "B. 14条小船",
+                        "C. 7条大船和3条小船",
+                        "D. 8条大船和2条小船"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】先求总人数56人，再比较人均费用：大船30÷6=5元，小船24÷4=6元，优先租大船并调整到总费用最小.\n【详解】56÷6=9(条)。2(人)。方案一:9条大船和1条小船，费用9×30+24=294(元);方案二:8条大船和2条小船，费用8×30+2×24=288(元)。288\u003c294，最省钱的是8条大船和2条小船。故答案为:D.",
+        "knowledge_points":  [
+                                 "租船问题"
+                             ],
+        "source":  "24-25四年级下·陕西西安·期末"
+    },
+    {
+        "id":  "204",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "有40袋黄豆，每袋30千克，每千克黄豆做4千克豆腐，每千克豆腐卖7元钱，一袋黄豆做多少千克豆腐。解决这个问题要用的信息是()。",
+        "options":  [
+                        "A. 40袋,30千克,4千克,7元",
+                        "B. 40袋,30千克,4千克",
+                        "C. 40袋,30千克,7元",
+                        "D. 30千克,4千克"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】求\"一袋黄豆做多少千克豆腐\"，只需\"每袋黄豆质量\"和\"每千克黄豆可做豆腐质量\".\n【详解】30×4=120(千克)。只需30千克和4千克这两个信息。故答案为:D.",
+        "knowledge_points":  [
+                                 "含有多余条件的实际问题"
+                             ],
+        "source":  "24-25四年级下·河北邯郸·期末"
+    },
+    {
+        "id":  "205",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "计算42+(42×42-42)÷42，最后一步算的是()法。",
+        "options":  [
+                        "A. 加",
+                        "B. 减",
+                        "C. 乘",
+                        "D. 除"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】有小括号先算括号内，再算除法，最后算加法.\n【详解】42+(42×42-42)÷42：先算42×42，再减42，再除以42，最后与42相加，因此最后一步是加法。故答案为:A.",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·福建福州·期末"
+    },
+    {
+        "id":  "206",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "小学举办阅读活动，需要采购一批笔记本。李老师购买了22箱笔记本，每箱笔记本中有7摞，每摞中有12本笔记本。算式22×(7×12)中\"7×12\"算的是()。",
+        "options":  [
+                        "A. 一共买了多少本笔记本",
+                        "B. 每箱中有多少本笔记本",
+                        "C. 每箱笔记本多少元",
+                        "D. 22箱笔记本一共有多少摞"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】7表示每箱7摞,12表示每摞12本，因此7×12表示每箱的本数.\n【详解】\"7×12\"算的是每箱中有多少本笔记本。故答案为:B.",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·广西百色·期末"
+    },
+    {
+        "id":  "207",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "文具店进了30捆笔记本，每捆25本。第一天卖出180本，剩下的又卖了4天才卖完。下面问题中可以用算式(30×25-180)÷4解答的是()。",
+        "options":  [
+                        "A. 剩下多少本笔记本。",
+                        "B. 剩下的平均每天卖多少本。",
+                        "C. 一共有多少本笔记本。",
+                        "D. 平均每天卖多少本。"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】30×25求总本数，减180求剩余本数，再除以4求剩余4天的平均每天卖出本数.\n【详解】该算式对应\"剩下的平均每天卖多少本\"。故答案为:B.",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·广西南宁·期末"
+    },
+    {
+        "id":  "208",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "小贵和奶奶需剪198张窗花，奶奶每小时剪18张，小贵每小时剪12张。奶奶先剪1小时，然后和小贵一起剪，全部剪完还要几小时。下面算式正确的是()。",
+        "options":  [
+                        "A. 198÷(18+12)",
+                        "B. (198-12)÷(18+12)",
+                        "C. (198-18)÷(18+12)"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】奶奶先剪1小时，先完成18张，剩下198-18张再由两人合作完成，效率为18+12.\n【详解】所需时间为(198-18)÷(18+12)。故答案为:C.",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "有具体量的工程问题"
+                             ],
+        "source":  "24-25四年级下·贵州黔南·期末"
+    },
+    {
+        "id":  "209",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "明明在计算□-30÷3时，先算减法，再算除法，得到的结果是5，那么正确的结果是()。",
+        "options":  [
+                        "A. 60",
+                        "B. 55",
+                        "C. 35",
+                        "D. 40"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】错误计算对应(□-30)÷3=5，先求□,再代回正确算式□-30÷3.\n【详解】(□-30)÷3=5→□-30=15→□=45。正确计算:45-30÷3=45-10=35。故答案为:C.",
+        "knowledge_points":  [
+                                 "除加，除减混合运算",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·河南开封·期末"
+    },
+    {
+        "id":  "210",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "已知a×b=c,那么下面算式中不正确的是()。",
+        "options":  [
+                        "A. a/c=b",
+                        "B. c/b=a",
+                        "C. b×a=c",
+                        "D. c/a=b"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】乘法关系是\"乘数×乘数=积\"，并有\"积/一个乘数=另一个乘数\".\n【详解】B,C,D都符合上述关系;A\"乘数/积=另一个乘数\"不成立。故答案为:A.",
+        "knowledge_points":  [
+                                 "乘，除法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·浙江台州·期末"
+    },
+    {
+        "id":  "211",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面算式中，去掉括号后结果不变的是()。",
+        "options":  [
+                        "A. 3×(6+360÷2)",
+                        "B. 960÷5+(12×8)",
+                        "C. (42+56)÷7×4"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】分别比较去括号前后运算顺序与结果.\n【详解】A去括号后变为3×6+360÷2，结果变化;C去括号后变为42+56÷7×4，结果变化;B去括号前后均为960÷5+12×8，结果不变。故答案为:B.",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "无括号的运算顺序",
+                                 "有括号的运算顺序"
+                             ],
+        "source":  "24-25四年级下·黑龙江绥化·期末"
+    },
+    {
+        "id":  "212",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "一个除法算式中被除数是120，则\"被除数+除数×商\"的结果是()。",
+        "options":  [
+                        "A. 143",
+                        "B. 162",
+                        "C. 240",
+                        "D. 241"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】除法中有\"被除数=除数×商\".\n【详解】被除数+除数×商=120+120=240。故答案为:C.",
+        "knowledge_points":  [
+                                 "乘，除法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·浙江杭州·期末"
+    },
+    {
+        "id":  "213",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "小明不小心把243-(128+65),错算成了243-128+65，结果和原来相比相差()。",
+        "options":  [
+                        "A. 130",
+                        "B. 128",
+                        "C. 65",
+                        "D. 大小不变"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】分别算出正确结果与错误结果，再求差.\n【详解】正确:243-(128+65)=243-193=50;错误:243-128+65=115+65=180;相差180-50=130。故答案为:A.",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·浙江·期末"
+    },
+    {
+        "id":  "214",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "甲，乙两班共有96名学生，如果调甲班的4名学生去乙班，两班的学生就一样多了。甲班原来有()人，乙班原来有()人。",
+        "options":  [
+
+                    ],
+        "answer":  "52;44",
+        "solution":  "【分析】调4人后相等，说明原来甲比乙多8人;总和96，用和差问题公式求两班人数.\n【详解】差=4×2=8(人);甲班=(96+8)÷2=52(人),乙班=(96-8)÷2=44(人)。",
+        "knowledge_points":  [
+                                 "和差倍问题"
+                             ],
+        "source":  "24-25六年级下·河南南阳·期末"
+    },
+    {
+        "id":  "215",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "飞飞在玩24点游戏，抽到四张扑克牌上面的分别是:A,2,3,10,(A代表1)。这四个数通过四则运算(每个数仅用一次，可以使用小括号)得到24。请你写一道相应的算式:()。",
+        "options":  [
+
+                    ],
+        "answer":  "(10-2)×3×1=24",
+        "solution":  "【分析】先凑出8，再与3相乘得到24，同时用到A=1.\n【详解】(10-2)×3×1=8×3×1=24。",
+        "knowledge_points":  [
+                                 "24点",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·福建福州·期末"
+    },
+    {
+        "id":  "216",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据128+147=275,275÷25=11,32×11=352，可以列出综合算式()。",
+        "options":  [
+
+                    ],
+        "answer":  "32*[(128+147)÷25]=352",
+        "solution":  "【分析】按\"先加，再除，后乘\"把中间结果依次代入，并用括号保证顺序.\n【详解】综合算式可写为32*[(128+147)÷25]=352。",
+        "knowledge_points":  [
+                                 "带有中括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·河南新乡·期末"
+    },
+    {
+        "id":  "217",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "根据32×14=448，可以知道448÷14=( ),还可以根据商的变化规律再写出一个除法算式:()。",
+        "options":  [
+
+                    ],
+        "answer":  "32;224÷7=32(答案不唯一)",
+        "solution":  "【分析】利用乘除互逆：积/一个因数=另一个因数;再用商不变规律构造等值除法式.\n【详解】448÷14=32;被除数和除数同时除以2，可得224÷7=32。",
+        "knowledge_points":  [
+                                 "商的变化规律及应用",
+                                 "乘，除法的意义和各部分间的关系",
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "24-25四年级下·河南新乡·期末"
+    },
+    {
+        "id":  "218",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把102-83=19,19×3=57,684÷57=12这三道算式合并成一道算式是:()。",
+        "options":  [
+
+                    ],
+        "answer":  "684/[(102-83)×3]=12",
+        "solution":  "【分析】先减后乘再除，把前一步结果作为后一步输入并用括号锁定顺序.\n【详解】合并后为684/[(102-83)×3]=12。",
+        "knowledge_points":  [
+                                 "带有中括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·河南新乡·期末"
+    },
+    {
+        "id":  "219",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "有三个算式14+26=40,800÷40=20,20×302=6040，将这三个算式合并成一个综合算式是()。",
+        "options":  [
+
+                    ],
+        "answer":  "800÷(14+26)×302=6040",
+        "solution":  "【分析】先算14+26，再用其结果做除法，最后乘302.\n【详解】综合算式:800÷(14+26)×302=6040。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·陕西西安·期末"
+    },
+    {
+        "id":  "220",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "15×28-12÷4，如果要使用加括号的方法改变运算顺序，先算减法，再算除法，最后算乘法，算式应该为(),此时它的计算结果是()。",
+        "options":  [
+
+                    ],
+        "answer":  "15*[(28-12)÷4];60",
+        "solution":  "【分析】把\"28-12\"放入小括号，再把\"(28-12)÷4\"放入中括号，可实现指定顺序.\n【详解】15*[(28-12)÷4]=15×(16÷4)=15×4=60。",
+        "knowledge_points":  [
+                                 "带有中括号的混合运算",
+                                 "有括号的运算顺序"
+                             ],
+        "source":  "24-25四年级下·河南开封·期末"
+    },
+    {
+        "id":  "221",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据851-240=611，可得加法算式();根据5400÷45=120，可得乘法算式()。",
+        "options":  [
+
+                    ],
+        "answer":  "611+240=851(或240+611=851);45×120=5400(或120×45=5400)",
+        "solution":  "【分析】减法与加法互逆，除法与乘法互逆.\n【详解】由851-240=611得611+240=851;由5400÷45=120得45×120=5400。",
+        "knowledge_points":  [
+                                 "加，减法的意义和各部分间的关系",
+                                 "乘，除法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·河北沧州·期末"
+    },
+    {
+        "id":  "222",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "\"未来星\"书店进行促销活动，所有图书\"买三送一\"，小华当日选好了13本书，每本15元，他需要支付()元。",
+        "options":  [
+
+                    ],
+        "answer":  "150",
+        "solution":  "【分析】\"买三送一\"表示每4本付3本钱。13本可看作3组4本再加1本.\n【详解】3组共付9本，再付1本，共付10本。10×15=150(元)。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "乘加，乘减混合运算",
+                                 "带有小括号的混合运算",
+                                 "促销问题"
+                             ],
+        "source":  "24-25四年级下·贵州遵义·期末"
+    },
+    {
+        "id":  "223",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在一道减法算式中，被减数，减数与差的和是24，若减数是8，则差是()。",
+        "options":  [
+
+                    ],
+        "answer":  "4",
+        "solution":  "【分析】被减数=减数+差，所以\"被减数+减数+差\"=2*被减数.\n【详解】被减数=24÷2=12，差=12-8=4。",
+        "knowledge_points":  [
+                                 "加，减法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·河北邯郸·期末"
+    },
+    {
+        "id":  "224",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明有26枚邮票，小红有38枚邮票，小红给小明()枚后，两人的邮票数就一样多了。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】先求两人相差38-26=12枚，要相等需转移一半.\n【详解】12÷2=6(枚)。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·河北邯郸·期末"
+    },
+    {
+        "id":  "225",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "被减数，减数和差的和是150，被减数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "75",
+        "solution":  "【分析】由被减数=减数+差，可得三者和=2*被减数.\n【详解】被减数=150÷2=75。",
+        "knowledge_points":  [
+                                 "加，减法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·江西赣州·期末"
+    },
+    {
+        "id":  "226",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把20×7=140,140-95=45,180÷45=4这三个算式组成一个综合算式是()。",
+        "options":  [
+
+                    ],
+        "answer":  "180÷(20×7-95)=4",
+        "solution":  "【分析】按\"先乘再减后除\"把前两步嵌入最后一步.\n【详解】综合算式:180÷(20×7-95)=4。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·江西宜春·期末"
+    },
+    {
+        "id":  "227",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果把240÷8×34-264的运算顺序改为先算乘法，再算减法，最后算除法，那么算式应改为(),结果是()。",
+        "options":  [
+
+                    ],
+        "answer":  "240÷(8×34-264);30",
+        "solution":  "【分析】用括号把\"8×34-264\"整体作为除数.\n【详解】240÷(8×34-264)=240÷(272-264)=240÷8=30。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·江西上饶·期末"
+    },
+    {
+        "id":  "228",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "粗心的小明在计算一个数加190时错看成一个数减去109，得到的结果是246，正确的结果应是()。",
+        "options":  [
+
+                    ],
+        "answer":  "545",
+        "solution":  "【分析】先由错误式求\"这个数\"，再计算正确式.\n【详解】设这个数为x,x-109=246，所以x=355;正确结果为355+190=545。",
+        "knowledge_points":  [
+                                 "加，减法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·江西赣州·期末"
+    },
+    {
+        "id":  "229",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "根据13×27=351，直接写出两个除法算式:()。",
+        "options":  [
+
+                    ],
+        "answer":  "351÷13=27;351÷27=13",
+        "solution":  "【分析】利用积与因数关系：积/一个因数=另一个因数.\n【详解】可得351÷13=27,351÷27=13。",
+        "knowledge_points":  [
+                                 "乘，除法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·贵州黔西·期末"
+    },
+    {
+        "id":  "230",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "某超市购进120箱牛奶和若干面包，牛奶比面包的2倍少40箱，则该超市购进牛奶和面包一共()箱。",
+        "options":  [
+
+                    ],
+        "answer":  "200",
+        "solution":  "【分析】\"牛奶比面包2倍少40\"即120=2*面包-40，先求面包箱数，再求总数.\n【详解】面包=(120+40)÷2=80(箱),总数=120+80=200(箱)。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·贵州铜仁·期末"
+    },
+    {
+        "id":  "231",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "做课间操时,288名学生排成16队，求平均每队有多少人，可以列出算式(),可以这样验算()。",
+        "options":  [
+
+                    ],
+        "answer":  "288÷16=18(人);16×18=288(人)",
+        "solution":  "【分析】平均每队人数=总人数/队数;无余数除法可用\"商×除数=被除数\"验算.\n【详解】288÷16=18(人),验算16×18=288(人)。",
+        "knowledge_points":  [
+                                 "乘，除法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·河南新乡·期末"
+    },
+    {
+        "id":  "232",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在一道减法算式中，被减数，减数，差的和是634。已知差是159，那么被减数是(),减数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "317;158",
+        "solution":  "【分析】减法中被减数=减数+差，因此三者和=2*被减数.\n【详解】被减数=634÷2=317;减数=317-159=158。",
+        "knowledge_points":  [
+                                 "加，减法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·河南周口·期末"
+    },
+    {
+        "id":  "233",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明同学在计算时，最后一步应减去22，他却除以22，得到的错误结果是15，正确结果应是()。",
+        "options":  [
+
+                    ],
+        "answer":  "308",
+        "solution":  "【分析】错误结果15对应\"最后一步之前的数/22=15\"，先反推该数，再按正确操作减22.\n【详解】最后一步之前的数=15×22=330，正确结果=330-22=308。",
+        "knowledge_points":  [
+                                 "乘加，乘减混合运算",
+                                 "乘，除法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·江西南昌·期末"
+    },
+    {
+        "id":  "234",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "乐乐在玩\"24点\"游戏时，抽到\"4\"\"5\"\"7\"\"8\"的4张扑克牌，用这4张扑克牌上的数算得24，算式为:()。",
+        "options":  [
+
+                    ],
+        "answer":  "(8-7+5)×4=24",
+        "solution":  "【分析】计算\"24\"点时，我们可以想\"三八二十四\"，\"四六二十四\"等。观察可知，\"4\"\"5\"\"7\"\"8\"这几个数，有数字\"4\"，且用\"5\"\"7\"\"8\"可以通过\"8-7+5\"运算得出\"6\"。据此解答.\n【详解】8-7+5=6;6×4=24。综合算式:(8-7+5)×4=6×4=24，乐乐在玩\"24点\"游戏时，抽到\"4\"\"5\"\"7\"\"8\"的4张扑克牌，用这4张扑克牌上的数算得24，算式为:(8-7+5)×4=24。",
+        "knowledge_points":  [
+                                 "24点",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25三年级下·河南新乡·期末"
+    },
+    {
+        "id":  "235",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "算式12×210-36÷3，如果想改变运算顺序，先算减法，算式是(),计算的结果是()。",
+        "options":  [
+
+                    ],
+        "answer":  "12×(210-36)÷3;696",
+        "solution":  "【分析】根据题意，算式12×210-36÷3，要先算减法的话，则需要在减法加上小括号，即12×(210-36)÷3，根据整数的四则运算法则，则该算式先算减法，再算乘法，最后算除法即可算出结果.\n【详解】12×(210-36)÷3=12×174÷3=2088÷3=696，即算式12×210-36÷3，如果想改变运算顺序，先算减法，算式是12×(210-36)÷3，计算的结果是696。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "有括号的运算顺序"
+                             ],
+        "source":  "24-25四年级下·河南漯河·期末"
+    },
+    {
+        "id":  "236",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "将以下三个分步算式合并成一个综合算式:460-70=390,390÷13=30,48-30=18;综合算式:()。",
+        "options":  [
+
+                    ],
+        "answer":  "48-(460-70)÷13=18",
+        "solution":  "【分析】460-70=390的差为390÷13=30中的被除数，那么将两个算式变为(460-70)÷13=30，又因为30为48-30=18中的减数，那么将算式变为48-(460-70)÷13=18，据此解答即可.\n【详解】由分析可知，综合算式为48-(460-70)÷13=18。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·浙江宁波·期末"
+    },
+    {
+        "id":  "237",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "给算式600÷50+14×2添上括号，使得运算顺序变成：先算加法，再算乘法，最后算除法，那么变化后的算式为()(不用计算)。",
+        "options":  [
+
+                    ],
+        "answer":  "600/[(50+14)×2]",
+        "solution":  "【分析】由题意得，在算式600÷50+14×2中，要想先算加法，再算乘法，最后算除法，需要在算式50+14的两边加上小括号，在算式50+14×2的两边加上中括号，即综合算式为:600/[(50+14)×2].\n【详解】根据分析，给算式600÷50+14×2添上括号，使得运算顺序变成：先算加法，再算乘法，最后算除法，那么变化后的算式为600/[(50+14)×2]。",
+        "knowledge_points":  [
+                                 "带有中括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·陕西安康·期末"
+    },
+    {
+        "id":  "238",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把672÷29×48-2的运算顺序变为先求差，再求积，最后求商，那么算式为()。",
+        "options":  [
+
+                    ],
+        "answer":  "672/[29×(48-2)]",
+        "solution":  "【分析】一个算式中既有乘除法，又有减法，要先算乘除法，再算减法。在算式672÷29×48-2中，要先算除法，再算乘法，最后算减法。要想先求差，再求积，最后求商，也就是先算减法，再算乘法，最后算除法，需要在算式48-2的两边加上小括号，在算式29×48-2的两边加上中括号，即综合算式为672/[29×(48-2)].\n【详解】由分析可得：把672÷29×48-2的运算顺序变为先求差，再求积，最后求商，那么算式为:672/[29×(48-2)]。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "带有中括号的混合运算",
+                                 "无括号的运算顺序",
+                                 "有括号的运算顺序"
+                             ],
+        "source":  "24-25四年级下·贵州黔南·期末"
+    },
+    {
+        "id":  "239",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一道除法算式中，除数是9，商是7，余数是2，被除数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "65",
+        "solution":  "【分析】在有余数的除法中，被除数=商×除数+余数。由题意得，一道除法算式中，除数是9，商是7，余数是2，直接将数据代入即可算出被除数.\n【详解】7×9+2=63+2=65，一道除法算式中，除数是9，商是7，余数是2，被除数是65。",
+        "knowledge_points":  [
+                                 "乘，除法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·陕西渭南·期末"
+    },
+    {
+        "id":  "240",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算180÷(30-2×5),应先算()法，再算()法，最后算()法。",
+        "options":  [
+
+                    ],
+        "answer":  "乘;减;除",
+        "solution":  "【分析】同级运算，从左往右依次进行计算;既有加减，又有乘除，先算乘除，再算加减;有括号，先算括号里面的，如果既有小括号又中括号的，先算小括号里面的，再算中括号里面的.\n【详解】180÷(30-2×5)=180÷(30-10)=180÷20=9，计算180÷((30-2×5)),应先算乘法，再算减法，最后算除法。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "有括号的运算顺序"
+                             ],
+        "source":  "24-25四年级下·新疆克孜勒苏·期末"
+    },
+    {
+        "id":  "241",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "天天和亮亮玩\"算式大变身\"游戏，他们给算式730-35+16×7添上括号，这个算式的运算顺序就变成先算加法，再算减法，最后算乘法.\"变身\"后的算式是()。",
+        "options":  [
+
+                    ],
+        "answer":  "[730-(35+16)]*7",
+        "solution":  "【分析】一个算式中既有加，减法，又有乘法，要先算乘法，再算加，减法。由题意得，在算式730-35+16×7中，要先算乘法，再算减法，最后算加法。要想先算加法，再算减法，最后算乘法，需要在算式35+16的两边加上小括号，在算式730-35+16的两边加上中括号，即\"变身\"后的算式是[730-(35+16)]*7.\n【详解】天天和亮亮玩\"算式大变身\"游戏，他们给算式730-35+16×7添上括号，这个算式的运算顺序就变成先算加法，再算减法，最后算乘法.\"变身\"后的算式是[730-(35+16)]*7。",
+        "knowledge_points":  [
+                                 "带有中括号的混合运算",
+                                 "有括号的运算顺序"
+                             ],
+        "source":  "24-25四年级下·四川德阳·期末"
+    },
+    {
+        "id":  "242",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "把算式(201÷3)+((12×11))的括号去掉后，运算顺序改变。",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】原算式(201÷3)+(12×11)的运算顺序是先算括号内的除法和乘法，再算加法。去掉括号后变为201÷3+12×11。根据四则运算规则，乘除法优先级高于加法，因此去掉括号后仍需先计算除法和乘法，再相加，运算顺序未改变.\n【详解】由分析可知，原算式(201÷3)+(12×11)去掉括号后，运算顺序不变。题目中\"运算顺序改变\"的说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "有括号的运算顺序"
+                             ],
+        "source":  "24-25四年级下·河北沧州·期末"
+    },
+    {
+        "id":  "243",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "在一道没有余数除法算式中，被除数，除数和商的和是49，商是4，被除数是32。",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】这道题中，除法没有余数，商是4，三个数的和是49。我们先减去商，得到被除数和除数的和是45。因为被除数是除数的4倍，所以除数和它的4倍加起来是5倍除数，等于45，那么除数就是9。被除数就是9×4=36。所以被除数不是32.\n【详解】被除数，除数，商的和是49，商是4，所以被除数+除数=49-4=45;在没有余数的除法中，被除数=除数×商，因为商是4，所以被除数=除数*4;被除数+除数=除数*4+除数=除数*5=45;因此，除数:45÷5=9;被除数:9×4=36;所以，被除数应该是36，而不是32degree故答案为: ×",
+        "knowledge_points":  [
+                                 "乘，除法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·河南新乡·期末"
+    },
+    {
+        "id":  "244",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "某小学计划组织四年级90名学生租车到实践基地参加研学活动，租车公司给出小客车和大客车两种车型的价格，小客车限乘25人，每辆550元，大客车限乘40人，每辆800元，则租1辆大客车和2辆小客车最省钱。",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】要判断租1辆大客车和2辆小客车是否最省钱，需根据总价格=大客车辆数×大客车单价+小客车辆数×小客车单价计算所有可能的租车方案的总费用，并比较最小值。首先确保总座位数\u003e=90人，再分别计算单独租大客车，单独租小客车以及混合租车的费用，最终确定最优方案，据此解答.\n【详解】①单独租大客车：需租3辆，可乘人数:40×3=120(人),总费用:3×800=2400元。②单独租小客车：需租4辆，可乘人数:25×4=100(人),总费用:4×550=2200元。③混合租车：方案1:1辆大客车和2辆小客车，可乘人数:40+25×2=40+50=90(人),总费用:800+550×2=800+1100=1900(元)。方案2:2辆大客车和1辆小客车，可乘人数:40×2+25=80+25=105(人),总费用:800×2+550=1600+550=2150(元),其他方案(如3辆大客车或更多小客车)费用均高于1900元。综上，租1辆大客车和2辆小客车最省钱。故答案为:√",
+        "knowledge_points":  [
+                                 "租船问题"
+                             ],
+        "source":  "24-25四年级下·陕西渭南·期末"
+    },
+    {
+        "id":  "245",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "在减法算式中，被减数和减数同时增加5，差不变。",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】在减法运算中，被减数和减数同时增加相同的数，差保持不变。据此判断.\n【详解】例如：算式10-3=7，被减数和减数同时加5后变为15-8=7，差仍为7。题干说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "和与差的变化规律",
+                                 "加，减法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·云南红河·期末"
+    },
+    {
+        "id":  "246",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "算式138×8-117÷3中可以同时计算乘除法。",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】一个算式中既有减法，又有乘，除法，要先算乘，除法，再算减法。在算式138×8-117÷3中，乘法和除法属于同级运算且两者被减法分隔，所以乘，除法可以同时计算.\n【详解】由分析得，算式138×8-117÷3中可以同时计算乘除法。原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "无括号的运算顺序"
+                             ],
+        "source":  "24-25四年级下·云南昭通·期末"
+    },
+    {
+        "id":  "247",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "列式计算。从780里面减去12个45，差是多少。",
+        "options":  [
+
+                    ],
+        "answer":  "240",
+        "solution":  "【分析】先算12乘45的积，再用780减去12乘45的积求出差，列式为780-12×45，先算乘法，再算减法，据此解答.\n【详解】780-12×45=780-540=240，所以差是240。",
+        "knowledge_points":  [
+                                 "乘加",
+                                 "乘减混合运算"
+                             ],
+        "source":  "24-25四年级下·江西吉安·期末"
+    },
+    {
+        "id":  "248",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "列式计算。68与42的和乘它们的差，积是多少。",
+        "options":  [
+
+                    ],
+        "answer":  "2860",
+        "solution":  "【分析】根据题意,68与42的和即68+42,68与42的差即68-42，最后用\"和\"*·\"差\"计算，据此列式计算.\n【详解】(68+42)×(68-42)=110×26=2860,68与42的和乘它们的差，积是2860。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·江西吉安·期末"
+    },
+    {
+        "id":  "249",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "列式计算。从780里面减去12个45，差是多少。",
+        "options":  [
+
+                    ],
+        "answer":  "240",
+        "solution":  "【分析】先求出12个45是多少，用乘法，再用780减去这个积得到差。本题考查整数的四则混合运算，要明确先算什么，再算什么.\n【详解】780-12×45=780-540=240，所以从780里面减去12个45，差是240。",
+        "knowledge_points":  [
+                                 "乘加",
+                                 "乘减混合运算"
+                             ],
+        "source":  "24-25四年级下·江西吉安·期末"
+    },
+    {
+        "id":  "250",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "列式计算。68与42的和乘它们的差，积是多少。",
+        "options":  [
+
+                    ],
+        "answer":  "2860",
+        "solution":  "【分析】根据题意，先用68加上42，求出两数的和;再计算出两数的差，最后把两数的和与差相乘，求出乘积，列式计算即可.\n【详解】(68+42)×(68-42)=110×26=2860，积是2860。",
+        "knowledge_points":  [
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·江西吉安·期末"
+    },
+    {
+        "id":  "251",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "请给下面的算式添上括号，使等式成立。9+3×6-1=24;3×6+4-2=24",
+        "options":  [
+
+                    ],
+        "answer":  "9+3×(6-1)=24;3×(6+4-2)=24或3*[6+(4-2)]=24",
+        "solution":  "【分析】根据整数四则混合运算规律，有小括号先计算小括号里的,6-1=5,3×5=15,9+15=24，所以在6-1两侧加上小括号;6+4-2=8,3×8=24，所以在6+4-2两侧加上小括号;4-2=2,6+2=8,3×8=24，也可以先在4-2两侧加上小括号，再在6+(4-2))两侧加上中括号.\n【详解】9+3×(6-1)=9+3×5=9+15=24;3×(6+4-2)=3×8=24;3*[6+(4-2)]=3*[6+2]=3×8=24;9+3×(6-1)=243×(6+4-2)=24或3*[6+(4-2)]=24",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·浙江嘉兴·期末"
+    },
+    {
+        "id":  "252",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "开车去外地运货，去时速度是68千米/时，用了3小时到达。装满货物后按原路返回，比去时多用了1小时。回来时的速度是多少。",
+        "options":  [
+
+                    ],
+        "answer":  "51千米/时",
+        "solution":  "【分析】根据路程，速度，时间的关系，先用去时速度×时间=路程求出总路程，回来时所用时间多了1小时即为用了4小时，再用总路程/回来所用时间=速度求出回来的速度，据此列式并作答.\n【详解】68×3÷(3+1)=68×3÷4=204÷4=51(千米/时),答：回来时的速度是51千米/时。",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·浙江杭州·期末"
+    },
+    {
+        "id":  "253",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "阳光洒满社区活动中心，赵奶奶和刘奶奶正为福利院孩子们筹备惊喜--编织260个小竹筐当收纳篮。赵奶奶每小时编3个，刘奶奶每小时编2个。赵奶奶先编35个后，赵奶奶和刘奶奶开始合作编制，她们还要多少小时才能完成任务。",
+        "options":  [
+
+                    ],
+        "answer":  "45小时",
+        "solution":  "【分析】根据题意，用总任务量减去赵奶奶已编的个数，先计算出剩余需要编织的小竹筐数量，再计算两人合作每小时编织的数量(即3+2=5个),最后用剩余数量除以合作效率得到所需时间.\n【详解】(260-35)÷(3+2)=225÷5=45(小时),答：她们还要45小时才能完成任务。",
+        "knowledge_points":  [
+                                 "两人合作的工程问题",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·贵州铜仁·期末"
+    },
+    {
+        "id":  "254",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "莆田市历史底蕴深厚，素有\"海滨邹鲁\"，\"文献名邦\"之美称，因其丰富的自然资源和独特的地理环境，孕育了丰富多样的莆仙文化。今年端午节期间，百艘龙舟巡游木兰溪10公里活动，吸引来自福州的小朋友明明一家人过来观光。明明爸爸根据百度地图查询知道，福州到莆田全程大约是120千米。而明明爸爸开车行驶30千米，要用去2升汽油。照这样计算，他们全家人自驾从福州到莆田一趟需要多少升汽油。",
+        "options":  [
+
+                    ],
+        "answer":  "8升",
+        "solution":  "【分析】根据题意，先用福州到莆田全程大约的路程120千米除以30千米，得到总路程是每30千米的几倍，则全程用汽油的升数也是2升的几倍，即再用2升与得到的倍数相乘。据此解答.\n【详解】120÷30×2=4×2=8(升),答：他们全家人自驾从福州到莆田一趟需要8升汽油。",
+        "knowledge_points":  [
+                                 "求一个数的几倍是多少",
+                                 "求一个数是另一个数的几倍",
+                                 "多位数与两位数的乘除混合运算"
+                             ],
+        "source":  "24-25四年级下·福建莆田·期末"
+    },
+    {
+        "id":  "255",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "两辆汽车从相距392千米的两地同时开出，相向而行。甲车每小时行56千米，乙车每小时行42千米，经过几小时两车相遇。",
+        "options":  [
+
+                    ],
+        "answer":  "4小时",
+        "solution":  "【分析】根据相遇问题中，路程/速度和=相遇时间，据此代入数值进行计算即可.\n【详解】392÷(56+42)=392÷98=4(小时),答：经过4小时两车相遇。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "相遇问题"
+                             ],
+        "source":  "24-25四年级下·江西吉安·期末"
+    },
+    {
+        "id":  "256",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "为支援洪涝受灾学校小学生的学习，某小学学生开展捐书活动，四年级捐116本，五年级比四年级多捐42本，六年级捐的本数是五年级的3倍。某小学六年级学生捐书多少本。",
+        "options":  [
+
+                    ],
+        "answer":  "474本",
+        "solution":  "【分析】由题意得，某小学学生开展捐书活动，四年级捐116本，五年级比四年级多捐42本，可以先用116加上42算出五年级捐书多少本。六年级捐的本数是五年级的3倍，再用前面的得数乘上3即可算出某小学六年级学生捐书多少本.\n【详解】(116+42)×3=158×3=474(本),答：某小学六年级学生捐书474本。",
+        "knowledge_points":  [
+                                 "求一个数的几倍是多少",
+                                 "带有小括号的混合运算",
+                                 "乘加",
+                                 "乘减混合运算"
+                             ],
+        "source":  "24-25四年级下·福建福州·期末"
+    },
+    {
+        "id":  "257",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "东风小学发起\"为山区孩子点亮书架\"公益活动，总共要捐出1000本书，第一批捐出242本，第二批捐出350本，剩下的是第三批捐书，第三批要捐出多少本书。",
+        "options":  [
+
+                    ],
+        "answer":  "408本",
+        "solution":  "【分析】用总本数减去第一批和第二批的本数和即可解答.\n【详解】1000-(242+350)=1000-592=408(本),答：第三批要捐出408本书。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·福建福州·期末"
+    },
+    {
+        "id":  "258",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "鲜果园有梨624千克，比黄桃的3倍多27千克，鲜果园有梨和黄桃共多少千克。",
+        "options":  [
+
+                    ],
+        "answer":  "823千克",
+        "solution":  "【分析】根据题意，用梨的千克数减去27千克，得到的是黄桃千克数的3倍。所以算出的结果再除以3就是黄桃的千克数。最后用黄桃的千克数加上梨的千克数就是共多少千克.\n【详解】(624-27)÷3+624=597÷3+624=199+624=823(千克),答：梨和黄桃共823千克。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "已知一个数的几倍是多少，求这个数"
+                             ],
+        "source":  "24-25四年级下·江西吉安·期末"
+    },
+    {
+        "id":  "259",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "四年级一班42名学生去永叔公园划船，大船限乘10人，每条船的租金是60元;小船限乘6人，每条船的租金是42元。怎样租船最省钱。需要多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "租3条大船和2条小船;264元",
+        "solution":  "【分析】由题意得，大船限乘10人，每条船的租金是60元，每人需要:60÷10=6(元);小船限乘6人，每条船的租金是42元，每人需要:42÷6=7(元)。对比可知，大船更便宜。在没有空位的情况下，应该尽量多租大船;四年级一班42名学生去永叔公园划船，直接用42除以10算出需要租大船的数量以及多余的人数，然后合理调整大船的数量和小船的数量使得总钱数最少即可.\n【详解】60÷10=6(元),42÷6=7(元),6\u003c7，所以租大船更便宜。42÷10=4(条)。2(人),即租4条大船后还剩下2名同学，这2名同学可以租1条小船。也可以和1条大船上的同学一起租船。60×4+42=240+42=282(元)。方案二:4-1=3(条),10+2=12(人),12÷6=2(条),即租3条大船和2条小船,60×3+42×2=180+84=264(元),282\u003e264。答：租3条大船和2条小船最省钱，一共需要264元。",
+        "knowledge_points":  [
+                                 "租船问题"
+                             ],
+        "source":  "24-25四年级下·江西吉安·期末"
+    },
+    {
+        "id":  "260",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "两辆汽车从相距392千米的两地同时开出，相向而行。甲车每小时行56千米，乙车每小时行42千米，经过几小时两车相遇。",
+        "options":  [
+
+                    ],
+        "answer":  "4小时",
+        "solution":  "【分析】由题意得，两辆汽车从相距392千米的两地同时开出，相向而行。甲车每小时行56千米，乙车每小时行42千米，可以先用加法算出两车的速度之和。相遇时间=总路程/速度之和，直接用392千米除以两车的速度之和即可算出经过几小时两车相遇.\n【详解】392÷(56+42)=392÷98=4(小时),答：经过4小时两车相遇。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "相遇问题"
+                             ],
+        "source":  "24-25四年级下·江西吉安·期末"
+    },
+    {
+        "id":  "261",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "欢欢买了12瓶可乐和18瓶矿泉水，共花了84元。1瓶可乐的价钱相当于2瓶矿泉水的价钱，买1瓶可乐需要多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "4元",
+        "solution":  "【分析】由题意得，欢欢买了12瓶可乐和18瓶矿泉水，共花了84元。1瓶可乐的价钱相当于2瓶矿泉水的价钱，那么2瓶矿泉水的价钱就等于1瓶可乐的价钱。18÷2=9，所以18瓶可乐的价钱等于9瓶可乐的价钱。欢欢买12瓶可乐和18瓶矿泉水共花了84元，相当于买了((12+9))瓶可乐共花了84元，直接用84除以((12+9))即可算出买1瓶可乐需要多少钱.\n【详解】84÷(12+18÷2)=84÷(12+9)=84÷21=4(元),答：买1瓶可乐需要4元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·江西南昌·期末"
+    },
+    {
+        "id":  "262",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "\"行走河南，读懂中国\"，河南文旅不仅为中外游客提供了丰富多彩的旅游体验，更带动了许多河南文化创意的出圈。五一期间，神舟旅行社接待了一个52人的旅行团要租车到河南旅游，他们怎么租车最省钱。至少要付多少钱。商务车：限乘8人，租金每辆440元 中巴车：限乘12人，租金每辆600元",
+        "options":  [
+
+                    ],
+        "answer":  "中巴车3辆，商务车2辆;2680元",
+        "solution":  "【分析】根据题意，先求出商务车，中巴车人均租金，比较后可知租中巴车便宜;设计方案时尽可能租中巴车，且没有空位;根据单价×数量=总价，计算各方案的费用，最后比较得出最省钱的租车方案。列式计算即可.\n【详解】根据分析可知:440÷8=55(元);600÷12=50(元);50\u003c55;52÷12=4(辆)。4(人)\n租4辆中巴车,1辆商务车。600×4+440=2400+440=2840(元);52-12×3=52-36=16(人);16÷8=2(辆),租3辆中巴车,2辆商务车。600×3+440×2=1800+880=2680(元),2680\u003c2840。答：租3辆中巴车,2辆商务车最省钱，至少要付2680元钱。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "租船问题"
+                             ],
+        "source":  "24-25四年级下·河南周口·期末"
+    },
+    {
+        "id":  "263",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "学校需要购置一批课桌椅。原来买30套课桌椅需要4740元。现在每套课桌椅优惠28元。优惠后，购买30套课桌椅需要多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "3900元",
+        "solution":  "【分析】由题意得，原来买30套课桌椅需要4740元。单价=总价/数量，直接用4740除以30先算出每套桌椅需要多少元。现在每套课桌椅优惠28元，再用前面的得数减去28算出现在每套课桌椅需要多少钱。最后再乘30即可算出优惠后购买30套课桌椅需要多少元.\n【详解】(4740÷30-28)×30=(158-28)×30=130×30=3900(元),答：优惠后购买30套课桌椅需要3900元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·河南新乡·期末"
+    },
+    {
+        "id":  "264",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "星光小学5位老师带领190名学生也在端午节那天去世博园游玩，他们有两种车型可供选择租用，怎样租车最省钱。最少花多少钱。\n![](images/2026-03-05%20124604.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "租3辆大客车和2辆小客车最省钱;1230元",
+        "solution":  "【分析】星光小学5位老师带领190名学生也在端午节那天去世博园游玩，据此用加法先算出总人数;先分别计算出大客车，小客车平均每人的钱数，然后进行比较，看哪种车便宜，在设计方案时尽量租便宜的车，而且不留空位时费用最低;因此用总人数除以最便宜的一种车型可坐的人数，再根据计算出的结果进行解答.\n【详解】190+5=195(人),270÷45=6(元/人),210÷30=7(元/人),6\u003c7，所以租大客车省钱，尽量租用大客车，并且最好不要有空座位;195÷45=4(辆)。15(人),4-1=3(辆),45+15=60(人),60÷30=2(辆),3×45+2×30=135+60=195(人),即租3辆大客车和2辆小客车最省钱。3×270+2×210=810+420=1230(元)。答：租3辆大客车和2辆小客车最省钱;最少花1230元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "乘加",
+                                 "乘减混合运算",
+                                 "租船问题"
+                             ],
+        "source":  "24-25四年级下·陕西安康·期末"
+    },
+    {
+        "id":  "265",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "小王在图书馆借了一本224页的书，借阅时间是一个星期。如果前3天看90页，照这样的速度，他能按时看完吗。",
+        "options":  [
+
+                    ],
+        "answer":  "不能",
+        "solution":  "【分析】前3天看的除以3算出平均每天看的页数，每天看的页数乘7算出一个星期能看的页数，与书的总页数相比，大于书的总页数则能看完，反之看不完.\n【详解】90÷3×7=30×7=210(页),210页\u003c224页。答：他不能按时看完。",
+        "knowledge_points":  [
+                                 "多位数与一位数的乘除混合运算",
+                                 "单归一问题"
+                             ],
+        "source":  "24-25四年级下·浙江嘉兴·期末"
+    },
+    {
+        "id":  "266",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "李老师带领学生去体验宇航员训练项目，租6套服装，其中航天服31元/套，鞋子19元/双。租6套需要多少元钱。(一套服装包括一套航天服和一双鞋子)小明用算式(31+19)×6解决这个问题.\"31+19\"求的是()。想一想，除了上面的方法，你还有其他方法解决这个问题吗。请写出相应的算式并解答。",
+        "options":  [
+
+                    ],
+        "answer":  "每套服装的价格;31×6+19×6=300(元),租6套需要300元。",
+        "solution":  "【分析】租6套服装，一套航天服31元，一双鞋子19元，求租6套需要多少元钱。(31+19)×6中的(31+19)是将一套航天服31元加上了一双鞋子19元，即\"31+19\"计算的是每套服装的价格。还可以分别计算6套航天服和6双鞋子的价钱，然后将两者的价钱加起来就是租6套服装的总价，据此解答.\n【详解】(31+19)×6=50×6=300(元),因此，\"31+19\"求的是每套服装的价格。31×6+19×6=186+114=300(元)。答:31×6+19×6=300(元)也可以解决租6套需要多少元钱的问题，求得租6套需要300元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "267",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "商店里有两种规格的卷布：大卷布，每卷布能做8件衣服，售价400元;小卷布，每卷布能做5件衣服，售价300元。王叔叔要定制50件衣服，怎样买最省钱。最少要花多少钱。(只能买整卷)",
+        "options":  [
+
+                    ],
+        "answer":  "买5卷大卷布和2卷小卷布;2600元",
+        "solution":  "【分析】根据题意，已知大卷布，每卷布能做8件衣服，售价400元;小卷布，每卷布能做5件衣服，售价300元。用400除以8,300除以5，分别计算出每件布的价格，选择价格低的布多买;分别制定方案，再根据总价=单价×数量，求出每种方案的总价，再选择费用最低的方案即可.\n【详解】根据分析可知:400÷8=50(元);300÷5=60(元)尽量多的买大卷布。50÷8=6(卷)。2(件)\n方案一：买6卷大卷布,1卷小卷布,400×6+300=2400+300=2700(元),50-5×8=50-40=10(件),10÷5=2(卷)。方案二：买5卷大卷布,2卷小卷布,5×400+2×300=2000+600=2600(元),2600\u003c2700。答：买5卷大卷布,2卷小卷布最省钱，最少要花2600元钱。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "租船问题"
+                             ],
+        "source":  "24-25四年级下·浙江台州·期末"
+    },
+    {
+        "id":  "268",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "工人师傅要组装960辆自行车，前3天组装了360辆。照这样计算，剩下的还要几天才能组装完。",
+        "options":  [
+
+                    ],
+        "answer":  "5天",
+        "solution":  "【分析】用3天组装自行车数量除以3，求出平均每天组装自行车数量。先用自行车总数量减去已经组装自行车数量，求出余下的自行车数量，再除以平均每天组装自行车数量，求出还需要的天数.\n【详解】(960-360)÷(360/3)=600÷120=5(天),答：照这样计算，剩下的还要5天才能组装完。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "单归一问题"
+                             ],
+        "source":  "24-25四年级下·河北邯郸·期末"
+    },
+    {
+        "id":  "269",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "有2位教师带领57名学生去公园划船。大船限乘7人，租金42元;小船限乘3人，租金21元，怎样租船最省钱。",
+        "options":  [
+
+                    ],
+        "answer":  "租8条大船和1条小船最省钱",
+        "solution":  "【分析】先分别求出租大船和租小船每人需要钱，租大船每人需要:42÷7=6(元),租小船每人需要:21÷3=7(元),因为6\u003c7，所以租大船省钱，要尽量多租大船，最好不要有空位，最省钱。用学生的人数加上老师的人数为总人数:57+2=59(人),再用总人数除以每条大船限乘的人数,59÷7=8(条)。3(人),因为余下3人，刚好可以租一条小船，所以租8条大船和1条小船，刚好没有空位，即最省钱;用每条大船的租金乘8和每条小船的租金乘1，分别求出租8条大船和1条小船的钱数，然后将两个钱数相加即可，据此解答.\n【详解】大船每人租金:42÷7=6(元),小船每人租金:21÷3=7(元),所以尽可能地租大船。57+2=59(人),59÷7=8(条)。3(人),此时租8条大船和1条小船，刚好没有空位。8×42+21=336+21=357(元)。答：租8条大船和1条小船最省钱。",
+        "knowledge_points":  [
+                                 "租船问题"
+                             ],
+        "source":  "24-25四年级下·河北邯郸·期末"
+    },
+    {
+        "id":  "270",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "认真阅读下面的资料并解决问题.\n中国是世界上最早养蚕织绸的国家，\"丝绸之路\"就是因丝绸外销而得名的重要贸易通道。蚕桑养殖每年五月初开始，十月末结束，一年可养10批蚕，一只雌蛾一生可产400-500粒蚕卵,10000粒蚕卵重约5克。贵州省某县去年第一批共收获296吨蚕茧，其中一等茧每千克售价53元，二等茧每千克售价17元，蚕茧的丰收给蚕农生活带来了巨大改变.\n(1)现在有5辆小货车和7辆大货车，正好把某县去年收获的第一批蚕茧全部装满拉走，每辆大货车载重量28吨，剩下的正好装满小货车，每辆小货车载重量是多少吨。\n(2)小货车和大货车都从某蚕桑养殖地出发开往销售地，已知数学信息如下.\n①小货车已经行驶了2小时。②小货车距离终点还剩120千米。③小货车的速度是大货车的2倍。④小货车每小时行驶50千米.\n要求出某蚕桑养殖地到销售地相距多少千米，需要知道()数学信息(填序号)。选择后请列式解答。",
+        "options":  [
+
+                    ],
+        "answer":  "20吨;①②④;220千米",
+        "solution":  "【分析】(1)先用7×28求出7辆大货车运蚕茧的吨数，用第一批蚕茧的数量减去7辆大货车运蚕茧的吨数，即为小货车运的吨数，再除以5即可求出每辆小货车可以装多少吨蚕茧。(2)先用小货车每小时行驶的距离乘时间，求出小货车已经行驶的路程;再加上小货车距离终点还剩的距离，即可求出某蚕桑养殖地到销售地相距多少千米，据此解答.\n【详解】(1)7×28=196(吨),(296-196)÷5=100÷5=20(吨)。答：每辆小货车载重量是20吨。(2)根据分析可知，要求出某蚕桑养殖地到销售地相距多少千米，需要知道①②④数学信息。2×50=100(千米)100+120=220(千米)。答：某蚕桑养殖地到销售地相距220千米。",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "带有小括号的混合运算",
+                                 "提问题",
+                                 "填条件问题"
+                             ],
+        "source":  "24-25四年级下·江西赣州·期末"
+    },
+    {
+        "id":  "271",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "上饶某小学组织了四年级60名优秀学生前往\"上饶集中营\"进行参观研学，怎样租车最省钱。费用是多少。小型车：限乘客6人，租金:90元 中型车：限乘客8人，租金:100元",
+        "options":  [
+
+                    ],
+        "answer":  "租6辆中型车,2辆小型车比较划算;需要780元",
+        "solution":  "【分析】根据题意，先用除法计算出两种车型的人均费用，人均费用低的车型多租;用60除以8，求出需要的中型车的车辆数，如果有余数，合理安排其它车型，再根据租金计算出总费用即可.\n【详解】根据分析可知:90÷6=15(元),100÷8=12(元)。4(元),15\u003e12，尽量多租中型车。60÷8=7(辆)。4(人),租7辆中型车,1辆小型车,7×100+90=700+90=790(元)。中:7-1=6(辆),小:8+4=12(人),12÷6=2(辆),租6辆中型车,2辆小型车。6×100+2×90=600+180=780(元),780\u003c790。答：租6辆中型车,2辆小型车比较划算，需要780元。",
+        "knowledge_points":  [
+                                 "租船问题"
+                             ],
+        "source":  "24-25四年级下·江西上饶·期末"
+    },
+    {
+        "id":  "272",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "小芳和小美相约一起从家里出发去\"博物馆\"参观，已知小芳家，小美家和\"博物馆\"活动现场在同一条直线上。小芳家距离\"博物馆\"538米，小美家距离\"博物馆\"962米，小芳家和小美家可能相距多少米。(画一画，列出所有情况)",
+        "options":  [
+
+                    ],
+        "answer":  "图见详解;小芳家和小美家可能相距424米或1500米",
+        "solution":  "【分析】根据题意，小芳家，小美家和博物馆在同一条直线上，存在两种位置关系：两家在博物馆的两侧：此时两家距离为各自到博物馆的距离之和。两家在博物馆的同侧：此时两家距离为两段距离的差.\n【详解】如图:\n情况一：同侧\n![](images/5f5c92f5aee084349f71683592104b0bb8222b9464ddb71a471a041191695eb3.jpg)\n情况二：两侧\n![](images/297249ca8c79bef5e20e9f00808003bc48e515fbdf6232a3ed26cc56144a5dfd.jpg)\n当两家在博物馆的同一侧时，两家的距离为:962-538=424(米);当两家在博物馆的两侧时，两家的距离为:538+962=1500(米)。答：小芳家和小美家可能相距424米或1500米。",
+        "knowledge_points":  [
+                                 "加，减法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·江西上饶·期末"
+    },
+    {
+        "id":  "273",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "小明在图书馆借了一本书，如果每天看40页,12天才能看完。但是为了按时归还，他必须提前2天看完，那么平均每天应该看多少页。",
+        "options":  [
+
+                    ],
+        "answer":  "48页",
+        "solution":  "【分析】根据题意可知，计划每天看的页数×看完的天数=这本书的总页数，计划看完的天数-2天=实际应看完的天数，这本书的总页数/实际应看完的天数=平均每天看的页数，依此列式并计算即可.\n【详解】根据分析可知:40×12÷(12-2)=480÷10=48(页),答：那么平均每天应该看48页。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·江西宜春·期末"
+    },
+    {
+        "id":  "274",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "原来爸爸在超市有6030个积分，爸爸3月份用1022个积分兑换了一个鼠标,8月份爸爸又攒了3178个积分。现在的积分刚好可以兑换一台电子秤，兑换一台电子秤需要多少个积分。",
+        "options":  [
+
+                    ],
+        "answer":  "8186个",
+        "solution":  "【分析】根据题意，爸爸原有6030个积分,3月份用掉1022个，剩余积分需先减去用掉的积分，再加上8月份攒的3178个，总和即为兑换电子秤所需的积分.\n【详解】6030-1022+3178=5008+3178=8186(个),答：兑换一台电子秤需要8186个积分。",
+        "knowledge_points":  [
+                                 "加，减法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·江西上饶·期末"
+    },
+    {
+        "id":  "275",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "学校图书室里有连环画140本，故事书100本。两种书每天各借出4本，多少天后剩下的连环画是故事书的3倍。",
+        "options":  [
+
+                    ],
+        "answer":  "20天",
+        "solution":  "【分析】先求出连环画比故事书多的本数，再根据剩下的数量关系求出剩下的故事书的本数，进而求出天数。已知连环画有140本，故事书有100本，那么连环画比故事书多的本数为140-100=40(本)。因为每天两种书借出的本数相同，所以它们的数量差始终不变，还是40本。又因为剩下的连环画是故事书的3倍，那么多出来的40本就是剩下的故事书的3-1=2倍。所以剩下的故事书的本数为40÷2=20(本)。故事书原本有100本，剩下20本，那么借出的故事书的本数为100-20=80(本)。因为每天借出4本故事书，所以天数为80÷4=20(天),据此解答即可.\n【详解】140-100=40(本);40÷(3-1)=40÷2=20(本);100-20=80(本);80÷4=20(天)。答:20天后剩下的连环画是故事书的3倍。",
+        "knowledge_points":  [
+                                 "两位数除以一位数，首位能除尽",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·江西上饶·期末"
+    },
+    {
+        "id":  "276",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "学校需要购置一批课桌椅。原来买30套课桌椅需要4740元。现在每套课桌椅优惠28元。优惠后，购买30套课桌椅需要多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "3900元",
+        "solution":  "【分析】先计算原来每套课桌椅的价钱，用30套课桌椅的钱数除以套数，再求出优惠后的单价，用原来每套课桌椅的价格减去28元，最后计算购买30套的总费用，用优惠后课桌椅的价钱乘套数即可.\n【详解】(4740÷30-28)×30=(158-28)×30=130×30=3900(元),答：优惠后购买30套课桌椅需要3900元。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·河南新乡·期末"
+    },
+    {
+        "id":  "277",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "学校举行运动会，四年级有125人参加，五年级参加的人数比四年级多18人，六年级参加的人数是四，五年级总人数的2倍。六年级有多少人参加运动会。",
+        "options":  [
+
+                    ],
+        "answer":  "536人",
+        "solution":  "【分析】根据题意，要计算六年级参加运动会的人数，需先求出五年级参加的人数，再求出四，五年级的总人数，最后依据六年级人数是四，五年级总人数的2倍这一关系来计算。先通过四年级的125人加18人得到五年级人数，再将四，五年级人数相加后乘2得到六年级人数，据此解答.\n【详解】(125+18+125)×2=(143+125)×2=268×2=536(人),答：六年级有536人参加运动会。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "求一个数的几倍是多少"
+                             ],
+        "source":  "24-25五年级下·新疆克孜勒苏·期中"
+    },
+    {
+        "id":  "278",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "张晶在看《世界上下五千年》,第一周读了126页。第二周比第一周多读了74页，两周一共读了多少页。",
+        "options":  [
+
+                    ],
+        "answer":  "326页",
+        "solution":  "【分析】根据题意，要算两周一共读的页数，先求第二周读的页数，再把两周读的页数相加。运用加法先求出第二周页数，再用加法求出两周总页数，据此解答.\n【详解】(126+74)+126=200+126=326(页),答：两周一共读了326页。",
+        "knowledge_points":  [
+                                 "两位数与两位数的进位加法",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25五年级下·新疆克孜勒苏·期中"
+    },
+    {
+        "id":  "279",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "学校组织62名师生赴城郊开展踏青活动，近中午，饭菜的香气将大家引至一家农家餐厅，原木桌椅摆放整齐，墙上色彩鲜艳的特色菜海报令人垂涎欲滴。饥肠辘辘的同学们簇拥在李老师身旁，看着菜单犯了愁：大桌最多能坐10人，每桌280元;小桌最多能坐6人，每桌180元。怎样订餐最省钱。最少要多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "订5个大桌,2个小桌最省钱，最少需要1760元。",
+        "solution":  "【分析】280÷10=28(元/人)180÷6=30(元/人),可知大桌人均更便宜，所以优先订大桌，用62÷10=6(桌)。2(人),所以订6个大桌，剩下的人订1个小桌，这是方案一;因为小桌没有坐满人，所以再减少1个大桌,6-1=5(桌),10+2=12(人)12÷6=2(桌),所以此方案是订5个大桌,2个小桌;分别将桌数乘每桌的钱数可算出两种方案各需要多少钱，对比可知哪种方案更便宜.\n【详解】280÷10=28(元/人),180÷6=30(元/人),28\u003c30，所以优先订大桌。62÷10=6(桌)。2(人)。方案一：订6个大桌,1个小桌。6×280+180=1680+180=1860(元)。方案二:6-1=5(桌)10+2=12(人),12÷6=2(桌),订5个大桌,2个小桌。5×280+180×2=1400+360=1760(元)1860\u003e1760。所以方案二更划算。答：订5个大桌,2个小桌最省钱，最少需要1760元。",
+        "knowledge_points":  [
+                                 "租船问题"
+                             ],
+        "source":  "24-25四年级下·贵州铜仁·期末"
+    },
+    {
+        "id":  "280",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "四年级一班42名学生去永叔公园划船，大船限乘10人，每条船的租金是60元;小船限乘6人，每条船的租金是42元。怎样租船最省钱。需要多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "租3条大船和2条小船，需要264元。",
+        "solution":  "【分析】根据题意，乘坐大船每人需要:60÷10=6(元),乘坐小船每人需要:42÷6=7(元),从中可以看出租大船便宜，所以尽量租大船，少租小船，而且保证空位最少，这样租金会最少.\n【详解】60÷10=6(元)42÷6=7(元) 6\u003c7 所以尽量租大船，且没空位时最省钱。42÷10=4(条)。2(人)租4条大船和1条小船，花费:4×60+42=240+42=282(元) 10×3+6×2=30+12=42(人) 租3条大船和2条小船，花费:3×60+2×42=180+84=264(元) 282\u003e264 答：租3条大船和2条小船最省钱，需要264元。",
+        "knowledge_points":  [
+                                 "租船问题"
+                             ],
+        "source":  "24-25四年级下·江西吉安·期末"
+    },
+    {
+        "id":  "281",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一本故事书，小明计划每天看48页,20天可以看完。实际每天比计划多看12页，实际几天可以看完。",
+        "options":  [
+
+                    ],
+        "answer":  "16天",
+        "solution":  "【分析】由题意得，一本故事书，小明计划每天看48页,20天可以看完，可以先用48乘20算出这本书一共有多少页。实际每天比计划多看12页，可以用48加上12算出实际每天看多少页书。最后再用总页数除以实际每天看的页数即可算出实际几天可以看完.\n【详解】48×20÷(48+12)=48×20÷60=960÷60=16(天)答：实际16天可以看完。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·江西吉安·期末"
+    },
+    {
+        "id":  "282",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一本故事书，小明计划每天看48页,20天可以看完。实际每天比计划多看12页，实际几天可以看完。",
+        "options":  [
+
+                    ],
+        "answer":  "16天",
+        "solution":  "【分析】根据题意，已知一本故事书，小明计划每天看48页,20天可以看完。用48乘20，就是故事书的总页数;实际每天比计划多看12页，用48加上12，就是实际每天看的页数，用总页数除以实际每天看的页数，就是实际看的天数;列式计算即可.\n【详解】根据分析可知:48×20÷(48+12)=960÷60=16(天)答：实际16天可以看完。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·江西吉安·期末"
+    },
+    {
+        "id":  "283",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "鲜果园有梨624千克，比黄桃的3倍多27千克，鲜果园有梨和黄桃共多少千克。",
+        "options":  [
+
+                    ],
+        "answer":  "823千克",
+        "solution":  "【分析】根据题意，已知鲜果园有梨624千克，比黄桃的3倍多27千克，用624减27，再除以3，就是鲜果园黄桃的重量，再加上624，就是鲜果园有梨和黄桃共多少千克。据此解答即可.\n【详解】根据分析可知:(624-27)÷3+624=597÷3+624=199+624=823(千克)。答：鲜果园有梨和黄桃共823千克。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "求一个量增加/减少多少能成为另一个量的几倍"
+                             ],
+        "source":  "24-25四年级下·江西吉安·期末"
+    },
+    {
+        "id":  "284",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "商场周年庆搞促销，原价126元的篮球，现在售价89元，体育学校买80个这样的篮球，少花了多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "2960元",
+        "solution":  "【分析】由题意得，篮球原价为126元，现在售价89元，可以先用126减去89算出一个篮球便宜了多少钱。然后再乘上80即可算出买80个这样的篮球少花了多少钱.\n【详解】(126-89)×80=37×80=2960(元)。答：体育学校买80个这样的篮球，少花了2960元。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·江西南昌·期末"
+    },
+    {
+        "id":  "285",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "为了保护水资源，北京市采用阶梯水价的方法收取水费，该市居民用水阶梯水价表如下:\n| 阶梯 | 每户年用水量 / 吨 | 水价 / (元 / 吨) |\n| :----------- | :-------------------- | :--------------: |\n| **第一阶梯** | 0 - 180 (含) | 5 |\n| **第二阶梯** | 180 (不含) - 260 (含) | 7 |\n| **第三阶梯** | 260 以上 | 9 |\n根据上面的信息，回答问题:\n(1)居住在本市的李阿姨去年家庭用水总量为120吨，她家去年共交水费多少元。\n(2)居住在本市的王先生去年家庭用水总量为221吨，他家去年共交水费多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "600元;1187元",
+        "solution":  "【分析】(1)李阿姨家去年用水总量为120吨,120吨在第一阶梯(0-180含)范围内，水价为5元/吨。用5乘120计算即可解答。(2)王先生家去年用水总量为221吨，其中180吨按第一阶梯水价5元/吨计算，超出180吨的部分为221-180=41吨，这41吨按第二阶梯水价7元/吨计算。即:(180×5+41×7).\n【详解】(1)5×120=600(元)答：她家去年共交水费600元。(2)221-180=41(吨),5×180+7×41=900+287=1187(元)。答：他家去年共交水费1187元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "乘加",
+                                 "乘减混合运算"
+                             ],
+        "source":  "24-25六年级下·北京昌平·期末"
+    },
+    {
+        "id":  "286",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "学校组织\"书香润童心\"读书活动，购入一批课外书。全校共有25个班，给每个班发放4套课外书。要求平均每套书多少钱，还缺少一条信息。(1)需要补充的信息是()。(2)根据所选信息列式解答。",
+        "options":  [
+                        "A. 平均每班有40人",
+                        "B. 买书一共花了800元",
+                        "C. 全校一共有1000名学生"
+                    ],
+        "answer":  "B;8元",
+        "solution":  "【分析】(1)根据单价=总价/数量，已知全校共有25个班，给每个班发放4套课外书，据此用乘法，可以求出购买的总套数，也就是数量，因此要求平均每套书多少钱，需要知道买这些书一共花了多少钱;据此结合选项选择即可。(2)根据题意，用全校总班数乘每班发的套数，先算出需要购买书的套数，用买这些书花的总钱数除以购买书的套数，即可求出平均每套书多少钱.\n【详解】(1)A。平均每班有40人，不是购买这些书花的钱数，不符合题意;B。买书一共花了800元，符合题意;C。全校一共有1000名学生，不是购买这些书花的钱数，不符合题意。故答案为:B。(2)800÷(25×4)=800÷100=8(元)。答：平均每套需要8元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "提问题",
+                                 "填条件问题",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "287",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "小华从家里去学校，每分钟走46米,20分钟可以到达。照这样的速度，小华已经走了16分钟，离学校还有多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "184米",
+        "solution":  "【分析】已知小华从家到学校速度是每分钟46米，时间是20分钟。速度不变，小华已经走了16分钟，说明小华距离学校还有20-16=4(分钟)的路程。根据路程=速度×时间，得到离学校的剩余距离:46×4=184(米),据此解答即可.\n【详解】46×(20-16)=46×4=184(米)。答：离学校还有184米。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·河南开封·期末"
+    },
+    {
+        "id":  "288",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "体育器材店新进了3个篮球和5副羽毛球拍，共用去275元，买一个篮球的钱可以买两副羽毛球拍，一副羽毛球拍多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "25元",
+        "solution":  "【分析】买一个篮球的钱可以买两副羽毛球拍，买3个篮球相当于买6副羽毛球拍，即相当于买11副羽毛球拍用了275元,275除以11即可求出一副羽毛球拍的价格.\n【详解】3个篮球的价格相当于羽毛球拍:3×2=6(副);总价相当于羽毛球拍的数量:6+5=11(副),一副羽毛球拍的价格为:275÷11=25(元)。答：一副羽毛球拍25元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "24-25四年级下·江西南昌·期中"
+    },
+    {
+        "id":  "289",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "A城到B城的公路长600千米。如果爸爸开车的平均速度是120千米/时，妈妈开车的平均速度是100千米/时。从A城到B城爸爸比妈妈快多长时间。(列综合算式解答)",
+        "options":  [
+
+                    ],
+        "answer":  "1小时",
+        "solution":  "【分析】根据速度=路程/时间，用600分别除以爸爸以及妈妈开车的平均速度，得到的得数即为爸爸以及妈妈开车从A城到B城的时间，相减即可得知爸爸比妈妈快多长时间.\n【详解】600÷100-600÷120=6-5=1(小时)。答：从A城到B城爸爸比妈妈快1小时。",
+        "knowledge_points":  [
+                                 "无括号的运算顺序",
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25四年级下·江西赣州·期末"
+    },
+    {
+        "id":  "290",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "新能源汽车是指采用非常规的车用燃料作为动力来源(或使用常规的车用燃料，采用新型车载动力装置),综合车辆的动力控制和驱动方面的先进技术，形成的技术原理先进，具有新技术，新结构的汽车。某品牌新能源电动汽车，原来充足电可以行驶480千米，技术改进后，现在充足电后行驶的路程比原来的2倍还多70千米，现在充足电后可以行驶多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "1030千米",
+        "solution":  "【分析】某品牌新能源电动汽车，原来充足电可以行驶480千米，技术改进后，现在充足电后行驶的路程比原来的2倍还多70千米，用480乘2，再加上70可求出现在充足电后可以行驶多少千米，据此解答.\n【详解】480×2+70=960+70=1030(千米)。答：现在充足电后可以行驶1030千米。",
+        "knowledge_points":  [
+                                 "乘加",
+                                 "乘减混合运算"
+                             ],
+        "source":  "24-25四年级下·贵州遵义·期末"
+    },
+    {
+        "id":  "291",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "商店举办促销活动：每双儿童袜子6元，买4双送1双。现在用120元拿到了几双袜子。",
+        "options":  [
+
+                    ],
+        "answer":  "25双",
+        "solution":  "【分析】根据\"数量=总价/单价\"，可得不考虑赠送时能买的袜子数量;再根据\"买4双送1双\",20双里面有几个4双，就会送几双。最后将不考虑赠送时买的袜子数量和赠送的袜子数量相加，可得总共拿到的袜子数.\n【详解】120÷6=20(双),20÷4=5(双),20+5=25(双)。答：现在用120元拿到了25双袜子。",
+        "knowledge_points":  [
+                                 "解决多步计算的实际问题"
+                             ],
+        "source":  "24-25四年级下·浙江嘉兴·期末"
+    },
+    {
+        "id":  "292",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "和合公司要给74位员工制作工作服，要按整卷买布。大卷布每卷480元，可以做8套工作服;小卷布每卷350元，可以做5套工作服。怎样买布最省钱。至少需要多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "购买8卷大卷布和2卷小卷布最省钱;至少需要4540元",
+        "solution":  "【分析】计算大卷布和小卷布的单套成本：大卷布每套60元，小卷布每套70元。因此尽可能多买大卷布，剩余套数用小卷布补充;尽可能刚好够最省钱。用总套数除以每卷大卷布可做的8套工作服，求出需要用的大卷布数，结合余数判断用小卷布数，并求出所需要的钱数.\n【详解】大卷布每套成本:480÷8=60(元),小卷布每套成本:350÷5=70(元),大卷布每套成本少，所以尽可能用大卷布，剩余套数再用小卷布。74÷8=9(卷)。2(套),可用9卷大卷布,1卷小卷布;费用:480×9+350×1=4670(元)。可考虑少用1卷大卷布，用8卷大卷布时需小卷布套数:74-(9-1)×8=10(套),10÷5=2(卷),费用:480×8+350×2=4540(元)。4670\u003e4540。答：购买8卷大卷布和2卷小卷布最省钱，至少需要4540元。",
+        "knowledge_points":  [
+                                 "租船问题"
+                             ],
+        "source":  "24-25四年级下·浙江台州·期末"
+    },
+    {
+        "id":  "293",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "万里长城的总长度不止万里，据统计，大约有21196.18千米，堪称世界奇迹，世界各地的人们纷纷去长城游玩。现有180人一起乘车去长城进行研学旅行，一辆大客车限乘40人，租金为1000元/辆;一辆小客车限乘30人，租金为900元/辆。怎样租车最省钱。最少花多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "租3辆大客车和2辆小客车;4800元",
+        "solution":  "【分析】首先用每辆车的租金/限乘人数，分别计算出两种车型人均哪种较便宜，尽可能多的选择较便宜的车，且满载时最便宜.\n【详解】1000÷40=25(元),900÷30=30(元),30\u003e25，所以大客车较便宜。180÷40=4(辆)。20(人),方案一:4辆大客车+1辆小客车，费用1000×4+900=4900(元)。方案二:3辆大客车后剩60人，再租2辆小客车，费用1000×3+900×2=4800(元)。4900\u003e4800。答：租3辆大客车和2辆小客车，最少花4800元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "乘加",
+                                 "乘减混合运算",
+                                 "租船问题"
+                             ],
+        "source":  "24-25四年级下·内蒙古鄂尔多斯·期末"
+    },
+    {
+        "id":  "294",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "五一商场促销，按套出售课桌椅。①课桌每张67元;②椅子每把33元;③上午卖了12套;④下午卖了3400元。(1)解决\"下午卖了多少套课桌椅。\"必须用到的信息是()。列式解答:(2)这一天一共卖了多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "34套;①②④;3400÷(67+33)=3400÷100=34(套);4600元",
+        "solution":  "【分析】(1)根据题意，四个信息中只知道下午卖了3400元，根据数量=总价/单价，还需要知道一套桌椅的价格，用下午卖出的总价除以一套桌椅的价格，即可求出下午卖了多少套课桌椅。(2)用上午卖出的套数乘一套桌椅的价格，求出上午卖了多少元，再加上下午卖的钱数，即可求出这一天一共卖了多少元.\n【详解】(1)解决\"下午卖了多少套课桌椅。\"必须用到的信息是①②④。3400÷(67+33)=3400÷100=34(套)。答：下午卖了34套课桌椅。(2)12×(67+33)+3400=12×100+3400=1200+3400=4600(元)。答：这一天一共卖了4600元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "乘加",
+                                 "乘减混合运算",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·内蒙古鄂尔多斯·期末"
+    },
+    {
+        "id":  "295",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "树木的年龄不仅是数字，更是一部记录岁月风雨的史诗。①银杏树被称为长寿树，其寿命可以达到2000岁。②银杏树的寿命比樟树寿命的2倍还多400岁。③樟树的寿命比樱桃树长600岁。④樟树的寿命是樱桃树的4倍。要求：樱桃树的寿命是多少岁。需要()数学信息。(填写序号)",
+        "options":  [
+
+                    ],
+        "answer":  "③④;200岁",
+        "solution":  "【分析】根据题意,③樟树的寿命比樱桃树长600岁。④樟树的寿命是樱桃树的4倍。樟树和樱桃树寿命相差600岁，樱桃树的寿命看作1份，樟树的寿命就是4份，根据差倍公式可知,600除以4减1的差，即等于樱桃树的寿命，据此即可解答.\n【详解】根据分析可知，求樱桃树的寿命是多少岁。需要(③④)数学信息。(填写序号)600÷(4-1)=600÷3=200(岁)。答：樱桃树的寿命是200岁。",
+        "knowledge_points":  [
+                                 "和差倍问题",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "296",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "甲队每天修路85米，乙队每天修路115米。两队合作20天，共修路多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "4000米",
+        "solution":  "【分析】甲队每天修路85米，乙队每天修路115米，两队合作20天，求共修路多少米，可以先计算甲队和乙队每天修路的长度，则用甲队每天修的路长加上乙队每天修的路长，即85+115=200(米),然后用甲队和乙队每天修路的长度乘合作修路的天数，即可求得共修路多少米，据此解答.\n【详解】(85+115)×20=200×20=4000(米),答：共修路4000米。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "有具体量的工程问题"
+                             ],
+        "source":  "24-25四年级下·四川德阳·期末"
+    },
+    {
+        "id":  "297",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "学校新购买了42套单人课桌椅。每张桌子55元，每把椅子45元，王老师带5000元购买这些课桌椅，够吗。",
+        "options":  [
+
+                    ],
+        "answer":  "够",
+        "solution":  "【分析】根据题意可知，先用一张桌子的价钱加上一把椅子的价钱，算出一套桌椅的价钱，再用一套桌椅的价钱乘42套，求出一共需要的钱数，与5000元比较大小;小于或等于5000元，则够;大于5000元，则不够;据此解答.\n【详解】(55+45)×42=100×42=4200(元),4200\u003c5000。答：王老师带5000元够。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·甘肃兰州·期中"
+    },
+    {
+        "id":  "298",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "四年级三个班的同学为庆祝元旦做纸花，一共做了85朵，二班比一班多做了4朵，三班和一班做的同样多。三个班各做了多少朵。(先根据题意画出线段图，再解答)",
+        "options":  [
+
+                    ],
+        "answer":  "见详解;一班27朵;二班31朵;三班27朵",
+        "solution":  "【分析】根据题意，一共做了85朵，二班比一班多做了4朵，先用85-4即可求出一班做的朵数的2倍加上三班做的朵数是多少，三班和一班做的同样多，则用85-4的结果除以3，即可求出一班做了多少朵;用一班做的朵数加4即可求出二班做了多少朵;三班做的朵数和一班一样多.\n【详解】![](images/3245d7599aeda0db1a46fbaa83554427bb490324b4e81ec577e45432741d83b0.jpg)\n如图：一班:(85-4)÷3=81÷3=27(朵);二班:27+4=31(朵)。答：一班做了27朵，二班做了31朵，三班做了27朵。",
+        "knowledge_points":  [
+                                 "和差倍问题",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·河南新乡·期末"
+    },
+    {
+        "id":  "299",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "某花店以每枝9.9元的价格购入一批向日葵花，然后以每枝19.9元的价格出售，共赢利450元。花店共购入多少枝向日葵花。",
+        "options":  [
+
+                    ],
+        "answer":  "45枝",
+        "solution":  "【分析】根据题目可知，每枝向日葵的进价是9.9元，售价是19.9元，每枝利润为售价与进价的差，即19.9-9.9=10(元)。总利润为450元，因此用总利润除以每枝利润即可得到购入的数量.\n【详解】450÷(19.9-9.9)=450÷10=45(枝)。答：花店共购入45枝向日葵花。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·浙江杭州·期末"
+    },
+    {
+        "id":  "300",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "郑老师买了30个环保布袋,25套再生纸笔记本，一共花费1060元。其中每个环保布袋12元，每套再生纸笔记本多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "28元",
+        "solution":  "【分析】已知环保布袋的数量和单价，以及再生纸笔记本的数量和总花费，要求再生纸笔记本的单价。首先计算环保布袋的总费用，再用总花费减去环保布袋的费用，得到再生纸笔记本的总费用，最后用总费用除以数量即可求出再生纸笔记本的单价.\n【详解】(1060-30×12)÷25=(1060-360)÷25=700÷25=28(元)。答：每套再生纸笔记本28元。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "解决多步计算的实际问题"
+                             ],
+        "source":  "24-25四年级下·浙江·期末"
+    },
+    {
+        "id":  "301",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "计算470*[280-(132+36)]/4时，第二步运算是()。",
+        "options":  [
+                        "A. 加法",
+                        "B. 减法",
+                        "C. 乘法"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据题意，此算式中有小括号，也有中括号，先算小括号里的加法，再算中括号里的减法，接着算中括号外的乘法，最后算中括号外的除法。以此选择即可.\n【详解】根据分析可知:\n470*[280-(132+36)]/4\n:=470*[280-168]/4\n:=470×112÷4\n:=52640÷4\n:=13160\n计算470*[280-(132+36)]/4时，第二步运算是减法。故答案为:B",
+        "knowledge_points":  [
+                                 "有括号的运算顺序"
+                             ],
+        "source":  "24-25四年级下·内蒙古呼伦贝尔·期末"
+    },
+    {
+        "id":  "302",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "水果超市运进40箱枇杷，每箱10kg。第一天卖出160kg,剩下的又卖了3天才卖完。下面()问题可以用算式(40×10-160)÷3解答。",
+        "options":  [
+                        "A. 剩下的平均每天卖多少千克。",
+                        "B. 剩下多少千克枇杷。",
+                        "C. 一共有多少千克枇杷。",
+                        "D. 平均每天卖多少千克。"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】分析算式(40×10-160)÷3,40×10是40箱枇杷的总重量,-160是总重量减去第一天卖出的重量,/3是剩余的重量分了三天卖，每天卖的重量，据此作答.\n【详解】A。剩余平均每天卖的量，算式符合，此选项正确。B。剩余量是240kg,算式为40×10-160，此选项错误。C。总重量为400kg,算式为40×10，此选项错误。D。平均每天卖多少千克，需计算总量400kg除以总共卖的天数4天，算式为40×10÷4，算式不符，此选项错误。故答案为:A",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "经济问题"
+                             ],
+        "source":  "24-25四年级下·浙江宁波·期中"
+    },
+    {
+        "id":  "303",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "计算1250+(526-204÷4×6))时，第一步应算()。",
+        "options":  [
+                        "A. 1250+526",
+                        "B. 526-204",
+                        "C. 4×6",
+                        "D. 204÷4"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】根据四则运算顺序，先算括号内的运算。括号内有减，除，乘，需先进行乘除运算，且乘除同级按从左到右顺序计算，因此第一步应算204÷40\n【详解】计算1250+(526-204÷4×6)时，第一步应算204÷4。故答案为:D",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·贵州遵义·期末"
+    },
+    {
+        "id":  "304",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "计算18*[(26+14)÷2]的运算顺序正确的是()。",
+        "options":  [
+                        "A. 除法乘法加法",
+                        "B. 加法乘法除法",
+                        "C. 加法除法乘法",
+                        "D. 乘法加法除法"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】算式18*[(26+14)÷2]中既有小括号，又有中括号，要先算小括号里面的加法，再算中括号里面的除法，最后算中括号外面的乘法.\n【详解】18*[(26+14)÷2]\n:=18*[40÷2]\n:=18×20\n:=360\n运算顺序正确的是加法除法乘法。故答案为:C",
+        "knowledge_points":  [
+                                 "带有中括号的混合运算",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·福建漳州·期末"
+    },
+    {
+        "id":  "305",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "和合超市运进50箱雪糕，每箱20支，第1天卖出85支，剩下的9天才卖完。下面可以用算式(50×20-85)÷9来解决的问题是()。",
+        "options":  [
+                        "A. 还剩多少支雪糕。",
+                        "B. 平均每天卖出多少支雪糕。",
+                        "C. 剩下的雪糕平均每天卖出多少支。",
+                        "D. 一共有多少支雪糕。"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】(50×20-85)÷9中,50表示雪糕的箱数,20表示每箱雪糕的支数，则50×20表示雪糕的总支数,85支表示第一天卖出雪糕的支数，则50×20-85表示剩下雪糕的支数,9表示卖出剩下雪糕卖的天数，则(50×20-85)÷9表示剩下的雪糕平均每天卖出的支数;据此解答.\n【详解】A。还剩多少支雪糕。,应该用箱数×每箱的支数-卖出的支数，列算式是50×20-85。不符合题意。B。平均每天卖出多少支雪糕。用总支数除以天数就是平均每天卖的支数。列算式是50×20÷(9+1)。不符合要求。C。剩下的雪糕平均每天卖出多少支。用总支数减去卖出的支数，算出结果再除以剩下的天数。列算式是(50×20-85)÷9。符合要求。D。一共有多少支雪糕。应该用箱数×每箱的支数=总支数。列算式是50×20，不符合题意。故答案为:C",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·内蒙古鄂尔多斯·期末"
+    },
+    {
+        "id":  "306",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "妈妈买了(),一共花了48元。其中1千克樱桃20元。每千克苹果多少钱。算式是:(48-20)÷2=14(元),请你根据算式来推断，括号内的条件可能是()。",
+        "options":  [
+                        "A. 2千克樱桃和1千克草莓",
+                        "B. 2千克樱桃和1千克苹果",
+                        "C. 1千克樱桃和1千克苹果",
+                        "D. 1千克樱桃和2千克苹果"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】由题意得，在算式(48-20)÷2=14中，\"48\"表示一共花了48元，\"20\"表示1千克樱桃需要20元，那么算式\"48-20\"表示苹果一共花了多少钱。然后再用前面的得数除以2算出每千克苹果多少钱，所以\"2\"表示买了2千克苹果。所以妈妈应该买了1千克樱桃和2千克苹果.\n【详解】由分析得，横线上的条件可能是1千克樱桃和2千克苹果。故答案为:D",
+        "knowledge_points":  [
+                                 "\"提问题\"\"填条件\"问题",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·贵州黔南·期末"
+    },
+    {
+        "id":  "307",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "某商城搞促销活动：全场买四送一。抽纸每包6元，妈妈支付了120元，她可以买到()抽纸。",
+        "options":  [
+                        "A. 20包",
+                        "B. 24包",
+                        "C. 25包",
+                        "D. 30包"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】根据题意，总价/单价=数量，用支付的价格除以每包纸的价格，等于妈妈原来买了几包。再用原来买的包数除以4，算出有这样的几组。用组数乘1就是赠送几包。最后用原来买的加上赠送的就是可以买到的包数.\n【详解】120÷6+(120÷6)÷4×1\n:=120÷6+20÷4×1\n:=20+5\n:=25(包)\n所以，她可以买到25包抽纸。故答案为:C",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·广东珠海·期末"
+    },
+    {
+        "id":  "308",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(400+50×40)÷200，正确的运算顺序是()。",
+        "options":  [
+                        "A. 加→乘→除",
+                        "B. 乘→除→加",
+                        "C. 除→加→乘",
+                        "D. 乘→加→除"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】根据四则混合运算的运算法则，在没有括号的算式里，既有乘，除，又有加，减，要先算乘除，后算加减;在含有括号的算式里，要先算括号里面的;这里有小括号，且括号里有乘法和加法，那么我们应该先算括号里的乘法，再算括号里的加法，最后算括号外的除法，据此解答.\n【详解】根据分析可得:\n(400+50×40)÷200\n=(400+2000)÷200\n:=2400÷200\n:=12\n以上算式的运算顺序是：先乘法，再加法，最后除法。故答案为:D",
+        "knowledge_points":  [
+                                 "有括号的运算顺序"
+                             ],
+        "source":  "24-25四年级下·广东珠海·期末"
+    },
+    {
+        "id":  "309",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "两个数相加，一个数增加8，另一个数减少3，和()。",
+        "options":  [
+                        "A. 增加5",
+                        "B. 增加4",
+                        "C. 减少5"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】加数+加数=和，一个加数不变，另一个加数增加多少或者减少多少，和就增加多少或者减少多少。据此解答.\n【详解】两个数相加，另一个数不变，一个数增加8，那么和也增加8。此时，一个数不变，另一个数减少3，和在增加8的基础上再减少3,8-3=5，此时的和增加5。故答案为:A",
+        "knowledge_points":  [
+                                 "和与差的变化规律",
+                                 "加，减法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·广东阳江·期末"
+    },
+    {
+        "id":  "310",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "计算(210-45×2)÷5时，正确的运算顺序是()。",
+        "options":  [
+                        "A. 减法→乘法→除法",
+                        "B. 乘法→减法→除法",
+                        "C. 除法→乘法→减法",
+                        "D. 除法→减法→乘法"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】四则混合运算的顺序：一个算式里，如果只有加减法或者只有乘除法，按照从左到右的顺序依次计算;如果既有加减法，又有乘除法，先算乘除法，再算加减法;如果有括号，先算括号里面的.\n【详解】根据分析:(210-45×2)÷5=(210-90)÷5=120÷5=24，所以计算(210-45×2)÷5时，正确的运算顺序是乘法→减法→除法。故答案为:B",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·福建龙岩·期末"
+    },
+    {
+        "id":  "311",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "在一道减法算式中，被减数，减数以及差的和是100，则被减数是()。",
+        "options":  [
+                        "A. 100",
+                        "B. 80",
+                        "C. 50",
+                        "D. 无法确定"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】根据\"被减数-减数=差\"可得：减数+差=被减数，所以被减数，减数与差的和是被减数的2倍，据此用100除以2即可求出被减数是多少.\n【详解】由分析可知:100÷2=50，所以在一道减法算式中，被减数，减数以及差的和是100，则被减数是50。故答案为:C",
+        "knowledge_points":  [
+                                 "加，减法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·云南昆明·期末"
+    },
+    {
+        "id":  "312",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一个水果超市进了35箱荔枝，每箱12千克。第一天卖出120千克，剩下的三天卖完。算式(35×12-120)÷3解决的问题是()。",
+        "options":  [
+                        "A. 平均每天卖多少千克荔枝",
+                        "B. 一共进多少千克荔枝",
+                        "C. 剩下的平均每天卖多少千克",
+                        "D. 还剩下多少千克荔枝"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】由题意得，在算式(35×12-120)÷3中，要先算35×12*。在算式35×12中，\"35\"表示一个水果超市进了35箱荔枝，\"12\"表示每箱荔枝12千克，所以35×12算的是超市一共进了多少千克荔枝.\"120\"表示第一天卖出120千克，所以算式35×12-120算的是还剩下多少千克荔枝。剩下的荔枝要3天卖完，所以算式(35×12-120)÷3解决的问题是剩下的平均每天卖多少千克.\n【详解】由分析得，算式(35×12-120)÷3解决的问题是剩下的平均每天卖多少千克。故答案为:C",
+        "knowledge_points":  [
+                                 "\"提问题\"\"填条件\"问题",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "23-24四年级下·河南驻马店·期末"
+    },
+    {
+        "id":  "313",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "有关0的运算，下面算式错误的是()。",
+        "options":  [
+                        "A. b×0=0",
+                        "B. b+0=b",
+                        "C. b-b=0",
+                        "D. b/0=0"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】0乘任何数都得0。任何数加0都得原数。两个相同的数相减得0.0除以任何不为0的数都得0,0不能做除数。据此解答.\n【详解】A.b×0=0，该算式计算正确。B.b+0=b,该算式计算正确。C.b-b=0，该算式计算正确。D。0不能做除数，所以算式b/0=0计算错误。故答案为:D",
+        "knowledge_points":  [
+                                 "0不能作除数",
+                                 "有关0的乘法",
+                                 "有关0的加减法"
+                             ],
+        "source":  "24-25四年级下·广东潮州·期末"
+    },
+    {
+        "id":  "314",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "在算式a/b=c中，不能为0的数是()。",
+        "options":  [
+                        "A. a",
+                        "B. b",
+                        "C. c"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据除法算式里，除数不能为0这个知识进行解答.\n【详解】在算式a/b=c中,b是除数，除数不能为0，即b不能为0.\n故答案为:B",
+        "knowledge_points":  [
+                                 "乘，除法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·新疆阿勒泰·期末"
+    },
+    {
+        "id":  "315",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "□*(72÷9)=88,□里应该填的数是()。",
+        "options":  [
+                        "A. 8",
+                        "B. 9",
+                        "C. 11",
+                        "D. 15"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】在乘法算式中，乘数×乘数=积，一个乘数=积/另一个乘数。一个算式中，有小括号的，要先算小括号里面的，再算小括号外面的。由题意得，在算式 □*(72÷9)=88中，先算72÷9=8，即 □*8=88。在算式 □*8=88中，要求 □的值，直接用88除以8即可解答.\n【详解】□*(72÷9)=88，那么 □*8=88，所以 □=88÷8=11。即 □里应该填的数是11。故答案为:C",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "乘，除法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·广东江门·期末"
+    },
+    {
+        "id":  "316",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面的说法错误的是()。",
+        "options":  [
+                        "A. 0与任何数相加，结果还得原数",
+                        "B. 0与任何数相乘，结果还得原数",
+                        "C. 任何数减去0，结果还得原数",
+                        "D. 0除以任何不是0的数，结果还是0"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】0与任何数相加都得原数;任何数减去0都得原数;0除以任何不为0的数，都等于0;0乘任何数都得0;据此解答即可.\n【详解】A。0与任何数相加，结果还得原数。说法正确，不符合题意;B。0与任何数相乘都得0;原题说法错误，符合题意;C。任何数减去0，结果还得原数。说法正确，不符合题意;D。0除以任何不是0的数，结果还是0。说法正确，不符合题意。故答案为:B",
+        "knowledge_points":  [
+                                 "有关0的加减法",
+                                 "有关0的乘法",
+                                 "被除数是0的除法运算"
+                             ],
+        "source":  "24-25四年级下·吉林·期末"
+    },
+    {
+        "id":  "317",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "大客车限乘30人，租金是50元，小客车限乘20人，租金是35元，国粒小学两名老师带着258名学生去淇澳岛游玩，租()最划算。",
+        "options":  [
+                        "A. 租8辆大客车和1辆小客车",
+                        "B. 租7辆大客车和3辆小客车",
+                        "C. 租9辆大客车"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】每辆大客车限乘人数乘租的辆数，可以算出大客车乘坐多少人;每辆小客车限乘人数乘租的辆数，可以算出小客车乘坐多少人。大客车乘坐人数加上小客车乘坐人数，即可算出一共能坐多少人。每辆大客车的租金乘租的辆数，可以算出租大客车需要多少钱;每辆小客车的租金乘租的辆数，可以算出租小客车需要多少钱;大客车租金加上小客车租金，即可算出一共需要多少钱。据此算出各选项中租车方案可以乘坐的人数和需要的钱数，再选择最省钱的方案.\n【详解】258+2=260(人)\nA。30×8+20=240+20=260(人)\n50×8+35=400+35=435(元)\nB。30×7+20×3=210+60=270(人)\n50×7+35×3=350+105=455(元)\nC。30×9=270(人)\n50×9=450(元)\n435\u003c450\u003c455\n租8辆大客车和1辆小客车最划算.\n故答案为:A",
+        "knowledge_points":  [
+                                 "乘加，乘减混合运算",
+                                 "租船问题"
+                             ],
+        "source":  "24-25四年级下·广东阳江·期末"
+    },
+    {
+        "id":  "318",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "欢欢和乐乐玩猜数游戏，欢欢说：\"我先把这个数乘34，再把这个数乘66，然后再把两次乘得的积相加，得到的数是1800.\"原来的数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "18",
+        "solution":  "【分析】这个数的34倍和66倍相加是这个数的100倍，即这个数的100倍是1800，据此作答.\n【详解】1800÷(34+66)=1800÷100=18，因此，原来的数是18。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "被除数和除数末尾都有0的除法",
+                                 "倍的认识"
+                             ],
+        "source":  "24-25四年级下·云南昭通·期中"
+    },
+    {
+        "id":  "319",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在计算(78+24)×(364-278)时，可将()法和()法同时计算，最后的结果是();计算360/[3×(8+12)]时，第一步计算()法，最后一步计算()法。",
+        "options":  [
+
+                    ],
+        "answer":  "加;减;8772;加;除",
+        "solution":  "【分析】同级运算，从左往右依次计算，既有加减，又有乘除的，先算乘除，再算加减，有括号的，先算括号里面的，既有中括号，又有小括号的，先算小括号里面的，再算中括号里面的，据此即可解答.\n【详解】(78+24)×(364-278)\n:=102×86\n:=8772\n360/[3×(8+12)]\n:=360÷(3×20)\n:=360÷60\n:=6\n在计算((78+24)×(364-278))时，可将加法和减法同时计算，最后的结果是8772;计算360/[3×(8+12)]时，第一步计算加法，第二步算乘法，最后一步计算除法。",
+        "knowledge_points":  [
+                                 "带有中括号的混合运算",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·云南昭通·期中"
+    },
+    {
+        "id":  "320",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小马虎在计算 96/[(6+□)×3] 时，漏看了括号，结果得到22，这道题目正确结果是()。",
+        "options":  [
+
+                    ],
+        "answer":  "4",
+        "solution":  "【分析】小马虎漏看了括号，将原式错误地计算为 96÷6+□*3，结果是22。□=(22-96÷6)÷3，然后按照先算小括号内的加法，再算中括号内的乘法，最后算除法，据此解题.\n【详解】(22-96÷6)÷3\n:=(22-16)÷3\n:=6÷3\n:=2\n96/[(6+2)×3]\n:=96/[8×3]\n:=96÷24\n:=4\n小马虎在计算 96/[(6+□)×3] 时，漏看了括号，结果得到22，这道题目正确结果是4。",
+        "knowledge_points":  [
+                                 "带有中括号的混合运算",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·浙江宁波·期中"
+    },
+    {
+        "id":  "321",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "有5,2,3,6四个数，可选用+,-,*,/及括号组成算式，每个数只用一次，使结果等于24，请写出综合算式()。",
+        "options":  [
+
+                    ],
+        "answer":  "5×6-2×3=24",
+        "solution":  "【分析】本题需用5,2,3,6四个数，通过加减乘除和括号组成结果为24的算式。可通过尝试不同运算顺序和括号组合，寻找符合条件的表达式，重点考虑乘法与加减法的结合，可先用5乘6得30，减去2乘3的积算出24.\n【详解】根据分析，先计算5乘6，再计算2乘3，再将两积相减，列综合算式为:5×6-2×3=24。(答案不唯一)",
+        "knowledge_points":  [
+                                 "24点"
+                             ],
+        "source":  "24-25四年级下·浙江杭州·期末"
+    },
+    {
+        "id":  "322",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "已知150×36=5400，请写出下面两题的得数.\n5400÷36=();5402÷36=()",
+        "options":  [
+
+                    ],
+        "answer":  "150;150。2",
+        "solution":  "【分析】根据已知的乘法算式150×36=5400，积/因数=另一个因数，可以推导出除法算式5400÷36=150。对于5402÷36，可以将5402拆分为5400+2,5400=150×36，所以150×36+2=5402，根据有余数除法，商×除数+余数=被除数，所以5402÷36=150。2.\n【详解】根据分析，积/因数=另一个因数，可以推导出除法算式5400÷36=150;5402=5400+2;5400=150×36，所以150×36+2=5402，根据商×除数+余数=被除数，所以5402÷36=150。2。",
+        "knowledge_points":  [
+                                 "有余数除法的认识",
+                                 "乘，除法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·浙江杭州·期末"
+    },
+    {
+        "id":  "323",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把\"67-25=42,42÷6=7\"合并成一道综合算式是(),用3,5,4,7\n算\"24点\":()(列综合算式)。",
+        "options":  [
+
+                    ],
+        "answer":  "(67-25)÷6=7;(3+5)×(7-4)=24",
+        "solution":  "【分析】观察两个算式，发现42÷6=7中的42是由67-25=42得到的，所以写成综合算式时，要先算减法，再算除法，由于减法和除法在一起，先算的是减法，所以要添加小括号改变运算顺序。据此解答.\"算24点\"是一种扑克牌智力游戏，需要利用加，减，乘，除(可加括号)把牌面上的数算成24。以下为一些常用的算24点的方法：乘法:3×8=24,4×6=24,2×12=24，把牌面上的四个数想办法凑成3和8,4和6,2和12，再相乘求解。加减法:25-1=24,27-3=24,28-4=24,30-6=24，把牌面上的四个数想办法凑成上述几个减法的减数和被减数。利用0,11的运算特性求解：如3,4,4,8可组成3×8=24,24+4=28,28-4=24等。公式法：在有解的牌组中，用得最为广泛的是以下六种解法:\n(a-b)×(c+d):如(10-4)×(2+2)=6×4=24\n(a+b)/c×d:如(10+2)÷2×4=12÷2×4=6×4=24\n(a-b/c)×d:如(3-2÷2)×12=(3-1)×12=24\n(a+b-c)×d:如(9+5-2)×2=12×2=24等.\na×b+c-d:如11×3+1-10=33+1-10=24\n(a-b)×c+d:如(4-1)×6+6=3×6+6=18+6=24\n【详解】由分析可知：把＂67-25=42,42÷6=7\"合并成一道综合算式是(67-25)÷6=7;用3,5,4,7算＂24点＂:(3+5)×(7-4)=24",
+        "knowledge_points":  [
+                                 "24点",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25三年级下·河南新乡·期末"
+    },
+    {
+        "id":  "324",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据156-86=70,12+7=19,70×19=1330，列成综合算式是:()。",
+        "options":  [
+
+                    ],
+        "answer":  "(156-86)×(12+7)=1330",
+        "solution":  "【分析】156-86=70,12+7=19,70×19=1330，先用156减86求出差，再用12加7求出和，最后用156减86的差乘12加7的和，列式为(156-86)×(12+7)=1330，据此解答.\n【详解】根据分析可知，列成综合算式是:(156-86)×(12+7)=1330",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·贵州遵义·期末"
+    },
+    {
+        "id":  "325",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据82-45=37,37×20=740,740+109=849，组成一道综合算式是()。",
+        "options":  [
+
+                    ],
+        "answer":  "(82-45)×20+109=849",
+        "solution":  "【分析】根据题意可知，先算减法求出差，再用差作为乘数计算乘法求出积，最后用积作为加数计算加法求出和，即综合算式先算减法，再算乘法，最后算加法，列综合算式时要把减法放在小括号内.\n【详解】(82-45)×20+109\n:=37×20+109\n:=740+109\n:=849\n根据82-45=37,37×20=740,740+109=849，组成一道综合算式是(82-45)×20+109=849。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·陕西安康·期末"
+    },
+    {
+        "id":  "326",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "根据35×15=525，请你直接写出525÷35=(),525÷15=()。",
+        "options":  [
+
+                    ],
+        "answer":  "15;35",
+        "solution":  "【分析】根据乘法算式中各部分的关系可知：一个因数=积/另一个因数。据此解答.\n【详解】525÷35=15;525÷15=35。",
+        "knowledge_points":  [
+                                 "乘，除法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·吉林·期末"
+    },
+    {
+        "id":  "327",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个减法算式中，被减数，减数和差的和是180，被减数是(),如果差是30，减数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "90;60",
+        "solution":  "【分析】一个减法算式中，被减数，减数和差的和是180，根据被减数=减数+差，被减数=180÷2;根据减数=被减数-差，将被减数减30求出减数，据此解答.\n【详解】被减数:180÷2=90;差:90-30=60。一个减法算式中，被减数，减数和差的和是180，被减数是90，如果差是30，减数是60",
+        "knowledge_points":  [
+                                 "加，减法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·甘肃兰州·期中"
+    },
+    {
+        "id":  "328",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "根据4986+3294=8280可以得到两道减法算式()或()。",
+        "options":  [
+
+                    ],
+        "answer":  "8280-4986=3294;8280-3294=4986",
+        "solution":  "【分析】根据加法与减法的互逆关系，已知一个加法算式，可以写出对应的两个减法算式：和减去一个加数等于另一个加数;因此，由4986+3294=8280，可以得到8280-4986=3294和8280-3294=4986.\n【详解】根据分析可得：根据4986+3294=8280可以得到两道减法算式8280-4986=3294或8280-3294=4986。",
+        "knowledge_points":  [
+                                 "加，减法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·辽宁抚顺·期末"
+    },
+    {
+        "id":  "329",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把300÷15=20,20+13=33,6×33=198三个算式列成一个综合算式是:()。",
+        "options":  [
+
+                    ],
+        "answer":  "6×(300÷15+13)=198",
+        "solution":  "【分析】根据题意可知，先求300除以15的商，再求商加13的和，最后求6乘和的积，先算除法，再算加法，最后算乘法，由于先算加法后算乘法，所以加法要用小括号括起来，综合算式是:6×(300÷15+13)=198;据此即可解答.\n【详解】根据分析可知，把300÷15=20,20+13=33,6×33=198三个算式列成一个综合算式是:6×(300÷15+13)=198。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·辽宁葫芦岛·期末"
+    },
+    {
+        "id":  "330",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明计算两个数相乘将其中一个乘数123看成了132，计算结果比正确答案大540，则正确答案是()。",
+        "options":  [
+
+                    ],
+        "answer":  "7380",
+        "solution":  "【分析】误把一个乘数123写成了132，相当于把另一个乘数多算了(132-123)倍，即另一个乘数的(132-123)倍是540，然后用除法算出另一个因数，再将两个正确的乘数相乘，即可得出正确答案.\n【详解】540÷(132-123)\n:=540÷9\n:=60\n:=60×123=7380，正确答案是7380",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "三位数与整十数的口算乘法",
+                                 "和差倍问题"
+                             ],
+        "source":  "2025·重庆江北·小升初真题"
+    },
+    {
+        "id":  "331",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "观察下面算式,3+0=3,3-0=3,0×3=0,0÷3=0。获得结论：一个数加上或减去0，得();一个数乘0得();0除以一个数(非0)得()。",
+        "options":  [
+
+                    ],
+        "answer":  "原数;0;0",
+        "solution":  "【分析】由算式3+0=3,3-0=3可知,3加上0或减去0，还得3，所以一个数加上或减去0，得原数;由算式0×3=0可知,3乘0等于0，所以0乘任何数都等于0，因此一个数乘0也得0;由算式0÷3=0可知,0除以3等于0，所以0除以一个数(非0)得0.\n【详解】观察下面算式,3+0=3,3-0=3,0×3=0,0÷3=0。获得结论：一个数加上或减去0，得原数;一个数乘0得0;0除以一个数(非0)得0。",
+        "knowledge_points":  [
+                                 "有关0的加减法",
+                                 "有关0的乘法",
+                                 "被除数是0的除法运算"
+                             ],
+        "source":  "24-25四年级下·内蒙古呼伦贝尔·期末"
+    },
+    {
+        "id":  "332",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算25×(4+8)时，先算()法，再算()法，结果是()。",
+        "options":  [
+
+                    ],
+        "answer":  "加;乘;300",
+        "solution":  "【分析】整数四则混合运算的运算顺序是同级运算时，从左往右依次计算;既有乘除，又有加减的，先算乘除，后算加减。有括号时，先算括号里面的。据此解答.\n【详解】25×(4+8)\n:=25×12\n:=300\n所以，计算25×(4+8)时，先算加法，再算乘法，结果是300。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·广东肇庆·期中"
+    },
+    {
+        "id":  "333",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "马小虎在做减法题时，把被减数十位上的3错写成8，结果得到的差是284，正确的差是()。",
+        "options":  [
+
+                    ],
+        "answer":  "234",
+        "solution":  "【分析】把被减数十位上的3看成了8，也就是说被减数多了50，即差多了50，用所得的差减50即可解答.\n【详解】8-3=5,5×10=50,284-50=234，所以正确的差应该是234",
+        "knowledge_points":  [
+                                 "乘，除法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·云南昭通·期中"
+    },
+    {
+        "id":  "334",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "根据208×67=13936写出一个除法算式()。",
+        "options":  [
+
+                    ],
+        "answer":  "13936÷208=67",
+        "solution":  "【分析】根据乘数×乘数=积，积/乘数=另一个乘数，可以写出除法算式.\n【详解】根据208×67=13936写出一个除法算式可以是:13936÷208=67(或13936÷67=208))。",
+        "knowledge_points":  [
+                                 "乘，除法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·山西晋中·期末"
+    },
+    {
+        "id":  "335",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "根据20×35=700写出另外两个除法算式(),()。",
+        "options":  [
+
+                    ],
+        "answer":  "700÷20=35700÷35=20",
+        "solution":  "【分析】根据乘除法的意义和各部分的关系，已知两个因数的积与其中一个因数，求另一个因数的运算，用积除以一个因数得到另一个因数。据此写出算式.\n【详解】根据分析可知：根据20×35=700，写出另外两个算式:700÷20=35。,700÷35=20。",
+        "knowledge_points":  [
+                                 "乘，除法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·吉林白城·期末"
+    },
+    {
+        "id":  "336",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算380+(420-180÷3)时，应先算()法，再算()法，最后算()法，这道题的结果是()。",
+        "options":  [
+
+                    ],
+        "answer":  "除;减;加;740",
+        "solution":  "【分析】四则混合运算的顺序：一个算式里，如果只有加减法或者只有乘除法，按照从左往右的顺序依次计算;如果既有加减法，又有乘除法，先算乘除法，再算加减法;如果有括号，先算括号里面的.\n【详解】380+(420-180÷3)\n:=380+(420-60)\n:=380+360\n:=740\n计算380+(420-180÷3)时，应先算除法，再算减法，最后算加法，这道题的结果是740",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·云南玉溪·期末"
+    },
+    {
+        "id":  "337",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "请将405+595=1000,1000÷125=8,20-8=12合并成综合算式()。",
+        "options":  [
+
+                    ],
+        "answer":  "20-(405+595)÷125=12",
+        "solution":  "【分析】根据题意，分步算式的最后一步是减法，分别找出被减数和减数，被减数是20，减数是8，被减数不变，减数是由1000÷125得到,1000是由405+595得到，由于减数8是由两个算式先加法后除法算出,405+595必须加上小括号，所以综合算式是:20-(405+595)÷125=12.\n【详解】先算加法，要加上小括号，再算除法，最后算减法，综合算式是:20-(405+595)÷125=12。故答案为:20-(405+595)÷125=12。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·浙江宁波·期末"
+    },
+    {
+        "id":  "338",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "有四张扑克牌，上面的点数分别为2,3,4,6，经过怎样的运算才能得到24。()",
+        "options":  [
+
+                    ],
+        "answer":  "(3-2)×4×6=24",
+        "solution":  "【分析】仔细观察几个数据可知,4×6=24，而24×1=24，所以可以用剩下的3减去2得到1,24×1=24，即综合算式为:4×6×(3-2).\n【详解】4×6×(3-2)=4×6×1=24×1=24，有四张扑克牌，上面的点数分别为2,3,4,6，可以通过以下的算式得到24:4×6×(3-2)=24。(答案不唯一)",
+        "knowledge_points":  [
+                                 "24点，带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·浙江温州·期末"
+    },
+    {
+        "id":  "339",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "将\"720÷24=30,45-30=15,25×15=375prime,这三道算式合并成一道综合算式是()。",
+        "options":  [
+
+                    ],
+        "answer":  "25×(45-720÷24)=375",
+        "solution":  "【分析】第一道算式720÷24=30，其结果30在第二道算式45-30=15中作为减数，所以可以把第二道算式里30用720÷24替换，得到45-720÷24。;然后，第二道算式的结果15在第三道算式25×15=375中作为乘数，所以把第三道算式里的15用45-720÷24替换，为了保证先算减法，需要给45-720÷24加上小括号，最终得到综合算式25×(45-720÷24)=375。验证一下：先算括号里的除法720÷24=30，再算括号里的减法45-30=15，最后算乘法25×15=375，与原来三道算式的计算结果一致.\n【详解】由分析可知，将\u0027720÷24=30,45-30=15,25×15=375\u0027,这三道算式合并成一道综合算式是25×(45-720÷24)=375。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·浙江·期末"
+    },
+    {
+        "id":  "340",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在计算12×(240-240÷6)时，应先算(),再算(),最后算(),计算结果是()。",
+        "options":  [
+
+                    ],
+        "answer":  "小括号内的除法;小括号内的减法;小括号外的乘法;2400",
+        "solution":  "【分析】整数四则混合运算的运算顺序是同级运算时，从左往右依次计算;既有乘除，又有加减，先算乘除，后算加减。有括号时，先算括号里面的;有多层括号时，先算小括号里的，再算中括号里面的。据此算式中有小括号，先算小括号内的除法，再算小括号内的减法，最后算小括号外的乘法即可;据此解答.\n【详解】12×(240-240÷6)\n:=12×(240-40)\n:=12×200\n:=2400\n即在计算12×(240-240÷6)时，应先算小括号内的除法，再算小括号内的减法，最后算小括号外面的乘法，计算结果是2400。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "有括号的运算顺序"
+                             ],
+        "source":  "24-25四年级下·广东肇庆·期末"
+    },
+    {
+        "id":  "341",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把11+34=45,45÷3=15,8×15=120这三个算式合并成一个综合算式是()。",
+        "options":  [
+
+                    ],
+        "answer":  "8*[(11+34)÷3]=120",
+        "solution":  "【分析】根据题意，先算加法，再算除法，最后算乘法。将45÷3里面的45替换成11+34，要使加法先算，这里应该加上小括号。然后把8×15里面的15替换成(11+34)÷3，这里应该加上中括号，最后算乘法。据此解答.\n【详解】根据分析，把11+34=45,45÷3=15,8×15=120这三个算式合并成一个综合算式是8*[(11+34)÷3]=120。",
+        "knowledge_points":  [
+                                 "带有中括号的混合运算"
+                             ],
+        "source":  "23-24四年级下·吉林松原·期末"
+    },
+    {
+        "id":  "342",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "学校买回5个篮球和5个足球，一共花了1000元，一个篮球比一个足球贵20元，足球的单价是()元。",
+        "options":  [
+
+                    ],
+        "answer":  "90",
+        "solution":  "【分析】一个篮球比一个足球贵20元，买5个篮球就比足球贵了20×5=100元，用总钱数减去篮球比足球贵的总差价再除以篮球与足球的总个数即可求出足球的单价.\n【详解】(1000-20×5)÷(5+5)\n=(1000-100)÷10\n:=900÷10\n:=90(元)\n足球的单价是90元。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "343",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把\"20-8=12,56÷7=8,12+8=20\"个算式合并成一个综合算式是()。",
+        "options":  [
+
+                    ],
+        "answer":  "(20-8)+56÷7=20",
+        "solution":  "【分析】先用20减去8求出差，再用56除以7求出商，最后用求出的差加上求出的商即可.\n【详解】(20-8)+56÷7\n:=12+56÷7\n:=12+8\n:=20\n把\"20-8=12,56÷7=8,12+8=20\"三个算式合并成个综合算式是:(20-8)+56÷7=20。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "344",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据278+313=591,721-719=2,591×2=1182列出综合算式是()。",
+        "options":  [
+
+                    ],
+        "answer":  "(278+313)×(721-719)=1182",
+        "solution":  "【分析】仔细观察三个算式可知，第一个算式的结果在第三个算式做第一个乘数，第二个算式的结果在第三个算式做第二个乘数，即278与313的和乘上721减去719的差，据此列出综合算式为278+313×721-719。在算式278+313×721-719中，要先算乘法，再算加法，最后算减法。要想先算加法和减法，最后算乘法，需要在算式278+313和算式721-719的两边都加上小括号，即综合算式为(27+313)×(721-719)\n【详解】(278+313)×(721-719)=591×2=1182根据278+313=591,721-719=2,591×2=1182列出综合算式是:(278+313)×(721-719)=1182",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "23-24四年级下·河南安阳·期末"
+    },
+    {
+        "id":  "345",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据1000÷125=8,8+600=608组成一个综合算式是()。",
+        "options":  [
+
+                    ],
+        "answer":  "1000÷125+600=608",
+        "solution":  "【分析】根据题目给出的两个算式，需要将1000÷125的结果8代入到第二个算式8+600中，形成综合算式。由于除法优先级高于加法，无需添加括号.\n【详解】1000÷125+600\n:=8+600\n:=608\n根据1000÷125=8,8+600=608组成一个综合算式是1000÷125+600=608。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "乘加，乘减混合运算"
+                             ],
+        "source":  "23-24四年级下·江西吉安·期末"
+    },
+    {
+        "id":  "346",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把75-50=25,25×2=50,600÷50=12改写成一道综合算式是()。",
+        "options":  [
+
+                    ],
+        "answer":  "600/[(75-50)×2]=12",
+        "solution":  "【分析】仔细观察几个算式可知，第一个算式的结果在第二个算式中做第一个乘数，第二个算式的结果在第三个算式中做除数，据此列出综合算式为:600÷75-50×2。在算式600÷75-50×2中，要想先算减法，再算乘法，需要在算式75-50的两边加上小括号，在算式75-50×2的两边加上中括号，即综合算式为:600/[(75-50)×2].\n【详解】600/[(75-50)×2]\n:=600/[25×2]\n:=600÷50\n:=12\n故把75-50=25,25×2=50,600÷50=12改写成一道综合算式是600/[(75-50)×2]=12。",
+        "knowledge_points":  [
+                                 "带有中括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "347",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用2,3,5,6这四个数(每个数只能用一次),经过怎样的运算可以得到24，写出算式:()。",
+        "options":  [
+
+                    ],
+        "answer":  "(5+3)÷2×6=24",
+        "solution":  "【分析】可以借助一些典型乘法算式，如:3×8=24,4×6=24,12×2=24。这里有6，可以凑成6×4=24。那么就是5+3=8,8÷2=40。要使加法先算，加上小括号即可.\n【详解】根据分析，写出的算式是(5+3)÷2×6=24。(答案不唯一)",
+        "knowledge_points":  [
+                                 "24点",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "348",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "2,4,5和8四个数经过怎样的运算可以得到24。请你写出算式:(任写一种即可)",
+        "options":  [
+
+                    ],
+        "answer":  "(5-4+2)×8=24",
+        "solution":  "【分析】在这几个数中加入\"+\"，\"-\"，\"*\"，\"/\"符号，以及括号，数字位置可交换，只要使计算结果等于24即可，可以借助一些典型乘法算式，如:3×8=24,4×6=24,12×2=24，再根据题目所给数字来凑，最后结果等于24。根据混合运算的计算顺序填空即可.\n【详解】根据分析可知:(5-4+2)×8=(1+2)×8=3×8=24\n2,4,5和8四个数经过怎样的运算可以得到24。请你写出算式:(任写一种即可)\n(5-4+2)×8=24(答案不唯一)",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "24点"
+                             ],
+        "source":  "23-24四年级下·湖南长沙·期末"
+    },
+    {
+        "id":  "349",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "植树节当天，学校要种植280棵树苗，如果把这些树苗平均分给12个班，每班可分得23棵，还剩4棵。请根据题意写出一道综合算式:()。",
+        "options":  [
+
+                    ],
+        "answer":  "(280-4)÷12=23(棵)",
+        "solution":  "【分析】由题意得，学校要种植280棵树苗，如果把这些树苗平均分给12个班，每班可分得23棵，还剩4棵。可以用23乘12算出12个班一共分到多少棵树苗，然后再用280减去前面的得数即可算出还剩多少棵树苗，列式为:280-23×12=4(棵);也可以先用23乘12算出12个班一共分到多少棵树苗，然后加上剩下的4棵树苗即可算出一共有多少棵树苗，列式为:23×12+4=280(棵);280棵减去4棵等于12个班要种植的树苗棵数，再除以12，即等于每班可分得的树苗棵数，列式为:(280-4)÷12=23(棵);280棵减去4棵等于分出的树苗棵数，再除以每班可分得的树苗棵数，即等于班数，列式为:(280-4)÷23=12(班);据此即可解答.\n【详解】植树节当天，学校要种植280棵树苗，如果把这些树苗平均分给12个班，每班可分得23棵，还剩4棵。根据题意写出一道综合算式为:280-23×12=4(棵),或23×12+4=280(棵),或(280-4)÷12=23(棵),或(280-4)÷23=12(班)。(答案不唯一)",
+        "knowledge_points":  [
+                                 "乘加，乘减混合运算",
+                                 "除加，除减混合运算",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·广东潮州·期末"
+    },
+    {
+        "id":  "350",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "请根据加，减法各部分间的关系，写出两道与564-482=82相关联的算式:(),()。",
+        "options":  [
+
+                    ],
+        "answer":  "482+82=564;564-82=482",
+        "solution":  "【分析】根据减法各部分间的关系，已知被减数-减数=差，可以推导出被减数=减数+差，减数=被减数-差.\n【详解】564是被减数,482是减数,82是差，则相关联的算式分别为:482+82=564\ncdot,564-82=482。",
+        "knowledge_points":  [
+                                 "加，减法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·广东潮州·期末"
+    },
+    {
+        "id":  "351",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "已知10千克花生可以榨油4千克，照这样计算,1吨花生可以榨油()千克。",
+        "options":  [
+
+                    ],
+        "answer":  "400",
+        "solution":  "【分析】根据质量单位换算规则，将1吨转化为1000千克，统一质量单位。先算1000千克是10千克的1000÷10倍，再根据10千克花生榨油4千克，推出1吨花生榨油4×(1000÷10)千克.\n【详解】1吨=1000千克 4×(1000÷10)=4×100=400(千克) 所以1吨花生可以榨油400千克。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "吨，千克之间的换算与比较"
+                             ],
+        "source":  "24-25四年级下·广东潮州·期末"
+    },
+    {
+        "id":  "352",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "600除以75减去60的差，商是多少。列综合算式是()。",
+        "options":  [
+
+                    ],
+        "answer":  "600÷(75-60)",
+        "solution":  "【分析】根据题意，先计算75-60，再用600除以它们的差，则计算减法的时候需要加上小括号，最后计算除法，据此列式计算即可.\n【详解】600÷(75-60)=600÷15=40。因此，列综合算式是600÷(75-60)=40。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·广东潮州·期末"
+    },
+    {
+        "id":  "353",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上合适的数.\n()÷183=15;()×18=1368",
+        "options":  [
+
+                    ],
+        "answer":  "2745;76",
+        "solution":  "【分析】被除数=除数×商,183乘15，即可算出几除以183等于15。因数=积/另一个因数,1368除以18，即可算出几乘18等于1368.\n【详解】183×15=2745;1368÷18=76;2745÷183=15;76×18=1368",
+        "knowledge_points":  [
+                                 "乘，除法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·湖南岳阳·期末"
+    },
+    {
+        "id":  "354",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "被减数，减数，差的和是24。被减数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "12",
+        "solution":  "【分析】被减数-减数=差，减数+差=被减数，所以被减数+减数+差=被减数*2=24，据此作答.\n【详解】24÷2=12，所以被减数，减数，差的和是24。被减数是12。",
+        "knowledge_points":  [
+                                 "加，减法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·海南省直辖县级单位·期末"
+    },
+    {
+        "id":  "355",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据320÷16=20，写出一道乘法算式是(),一道除法算式是()。",
+        "options":  [
+
+                    ],
+        "answer":  "20×16=320或16×20=320;320÷20=16",
+        "solution":  "【分析】根据\"被除数/除数=商\"可得：商×除数=被除数，被除数/商=除数，据此解答.\n【详解】根据320÷16=20，写出一道乘法算式是16×20=320，一道除法算式是320÷20=16。",
+        "knowledge_points":  [
+                                 "乘，除法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·海南省直辖县级单位·期末"
+    },
+    {
+        "id":  "356",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "160+63÷3-36中没有括号，要从左往右依次计算。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】此题考查混合运算规则：在没有括号的算式里，应先算乘除，后算加减，同级运算从左往右依次进行。题目中的算式包含加，减和除法，需先计算除法，再按顺序处理加减.\n【详解】根据运算顺序,160+63÷3-36应先计算除法部分:63÷3=21，此时算式变为160+21-36。接着按从左往右的顺序计算加减:160+21=181,181-36=145。若直接从左往右依次计算，会先算160+63=223，导致后续步骤错误。因此原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "无括号的运算顺序",
+                                 "除加",
+                                 "除减混合运算"
+                             ],
+        "source":  "24-25四年级下·河南漯河·期末"
+    },
+    {
+        "id":  "357",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "用竖式计算，并用加，减，乘，除法各部分之间的关系进行验算.\n540-282=();768÷64=();475+689=();204×37=()",
+        "options":  [
+
+                    ],
+        "answer":  "258;12;1164;7548",
+        "solution":  "【分析】三位数减三位数的竖式计算，相同数位对齐，从个位减起。当哪一位上的数不够减，要从前一位退1当10，和这一位数合起来继续减。如果前一位上是0，则继续向前一位借。验算时，可以用差+减数=被减数，或者被减数-差=减数。除数是两位数的除法的笔算法则：从被除数的高位除起，先看被除数的前两位;如果前两位比除数小，就要看前三位;除到被除数的哪一位，商就写在那一位的上面;余下的数必须比除数小。验算方法是商×除数=被除数，或被除数/商=除数。三位数加三位数的竖式计算，相同数位对齐，从个位加起。哪一位相加满十就向前一位进1。验算时用和-一个加数=另一个加数。三位数乘两位数的竖式计算时，个位对齐。先用两位数个位上的数和三位数的每一位上的数相乘，乘得结果和个位对齐，再用两位数十位上的数和三位数的每一位上的数相乘，乘得结果和十位对齐。最后把两次乘得结果相加。据此解答即可。验算时，可以用积/一个因数=另一个因数.\n【详解】540-282=258\n![](images/2026-03-05%20144042.jpg)\n768÷64=12\n![](images/2026-03-05%20144054.jpg)\n475+689=1164\n![](images/2026-03-05%20144104.jpg)\n204×37=7548\n![](images/2026-03-05%20144127.jpg)",
+        "knowledge_points":  [
+                                 "乘，除法的意义和各部分间的关系",
+                                 "加，减法的意义和各部分间的关系",
+                                 "除数是两位数的笔算除法",
+                                 "三位数乘两位数，三位数中间有0"
+                             ],
+        "source":  "24-25四年级下·云南昭通·期中"
+    },
+    {
+        "id":  "358",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "列出算式并用递等式计算。24加上126的和，除以35与20的差，商是多少。",
+        "options":  [
+
+                    ],
+        "answer":  "10",
+        "solution":  "【分析】24加上126的和为(24+126),35与20的差为(35-20),最终的商为(24+126)÷(35-20),先计算两边括号里的加减法，再算除法.\n【详解】(24+126)÷(35-20)\n:=150÷15\n:=10\n所以算式为(126+24)÷(35-20),商为10。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·浙江宁波·期中"
+    },
+    {
+        "id":  "359",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "列出算式并用递等式计算。63减45的差与9所得的积除以3，商是多少。",
+        "options":  [
+
+                    ],
+        "answer":  "54",
+        "solution":  "【分析】先求出63减45的差，也就是63-45，然后再用求得的差乘9，最后用求得的积除以3，据此解题.\n【详解】(63-45)×9÷3=18×9÷3=162÷3=54。商是54。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·浙江宁波·期中"
+    },
+    {
+        "id":  "360",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一个32人的旅行团要租观光车游园。现在有大，小两种型号的观光车.\n| **观光车价目表** | |\n| :-------------------------- | :------------------------- |\n| **大观光车** | **小观光车** |\n| 限乘 6 人 租车费 80 元/辆 | 限乘 4 人 租车费 60 元/辆 |\n(1)小福列出算式\"32÷4×60=480(元)\"。解决的问题是()。\n(2)旅行团怎样租车最省钱，最少需要多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "如果全部租小观光车，一共需要多少钱。;租4辆大观光车和2辆小观光车;440元",
+        "solution":  "【分析】(1)在算式\"6632÷4×60=480(元)\"中，\"32\"表示旅行团一共有32人，\"4\"表示每辆小观光车可以坐4人，所以算式\"32÷4\"算的是32人需要租几辆小观光车.\"60\"表示每辆小观光车需要60元，所以算式\"32÷4×60=480(元)\"解决的问题是如果全部租小观光车，一共需要多少钱。(2)由题意得，大观光车每辆80元，限乘6人，那么每人需要:80÷6=13(元)。2(元)。小观光车每辆60元，限乘4人，那么每人需要:60÷4=15(元)。对比可知，大观光车更便宜。在没有空位的情况下，应该尽量多租大观光车，直接用32除以6算出需要租大观光车的数量以及多余的人数。然后再合理调整大观光车的数量和小观光车的数量使得总钱数最少即可.\n【详解】(1)小福列出算式\"32÷4×60=480(元)\"。解决的问题是如果全部租小观光车，一共需要多少钱.\n(2)80÷6=13(元)。2(元)\n60÷4=15(元)\n13\u003c15，即大观光车更便宜。在没有空位的情况下，应该尽量多租大观光车.\n32÷6=5(辆)。2(人),即租5辆大观光车后，还剩2人。这2人可以租1辆小观光车，也可以和1辆大观光车上的人一起租小观光车.\n方案一：租5辆大观光车和1辆小观光车\n80×5+60=400+60=460(元)\n方案二:5-1=4(辆)\n6+2=8(人)\n8÷4=2(辆)\n租4辆大观光车和2辆小观光车\n80×4+60×2=320+120=440(元)\n460\u003e440\n答：租4辆大观光车和2辆小观光车最省钱，最少需要440元。",
+        "knowledge_points":  [
+                                 "租船问题",
+                                 "\"提问题\"\"填条件\"问题"
+                             ],
+        "source":  "24-25四年级下·四川宜宾·期末"
+    },
+    {
+        "id":  "361",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "甲，乙两袋糖果共有120颗。若从一袋取出13颗糖果放入另一袋，则甲袋的糖果颗数是乙袋的2倍。甲乙两袋中原来各有多少颗糖果。",
+        "options":  [
+
+                    ],
+        "answer":  "甲袋原来有67颗，乙袋原来有53颗;或甲袋原来有93颗，乙袋原来有27颗。",
+        "solution":  "【分析】已知从一袋取出13颗糖果放入另一袋，需分两种情况讨论：从乙袋取13颗放入甲袋，或从甲袋取13颗放入乙袋。若从乙袋取13颗放入甲袋，把乙袋的糖果看作1份，则甲袋糖果有2份，两袋共(2+1)份，糖果共120颗不变，此时用120÷(2+1)求出1份糖果的数量，也就是现在乙袋糖果数量，再加上13颗，求出乙袋原来糖果数量，糖果共有的颗数减去乙袋原来糖果数量等于甲袋原来糖果数量。若从甲袋取13颗放入乙袋，同理先求出现在乙袋现在的糖果数量，再减去13颗，即可求出乙袋原来的糖果，糖果共有的颗数减去乙袋原来糖果数量等于甲袋原来糖果数量.\n【详解】若从乙袋取13颗放入甲袋:120÷(2+1)=120÷3=40(颗) 乙袋原来有:40+13=53(颗) 甲袋原来有:120-53=67(颗)若从甲袋取13颗放入乙袋:120÷(2+1)=120÷3=40(颗) 乙袋原来有:40-13=27(颗) 甲袋原来有:120-27=93(颗)答：甲袋原来有67颗糖果，乙袋原来有53颗糖果;或是甲袋原来有93颗糖果，乙袋原来有27颗糖果。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "和差倍问题"
+                             ],
+        "source":  "24-25四年级下·浙江宁波·期中"
+    },
+    {
+        "id":  "362",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "四年级有学生和老师共298人，现在准备租车去春游，有两种车可供选择:40座的大客车，每辆租金800元;20座的小客车，每辆租金500元。怎样租车租金最少。最少多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "租7辆大客车和1辆小客车租金最少;6100元",
+        "solution":  "【分析】先根据\"单价=总价/数量\"分别计算出租大客车和小客车时每人需要的钱，要使租车最省钱，则应尽量租最便宜的一种车型，并且使每辆车都坐满，尽量没有空位;因此用总人数除以最便宜的一种车型可坐的人数，再根据计算出的结果进行解答即可。共需要的租金=租小客车的辆数×每辆小客车的租金+租大客车的辆数×每辆大客车的租金，依此计算并解答.\n【详解】大客车:800÷40=201(元)小客车:500÷20=25(元) 25元\u003e20元，因此尽量多租大客车。298÷40=7(辆)。18(人)20-18=2(个)40÷20=2 可以租7辆大客车，剩余18人再租1辆小客车，此时有2个空位。大客车的座位数是小客车座位数的2倍，因此每少租1辆大巴车就要多租2辆小客车，租金就多了，所以只能留2个空位。7×800+500=5600+500=6100(元)答：租7辆大客车和1辆小客车租金最少，最少是6100元。",
+        "knowledge_points":  [
+                                 "租船问题",
+                                 "经济问题"
+                             ],
+        "source":  "24-25四年级下·浙江宁波·期中"
+    },
+    {
+        "id":  "363",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "请你根据\"78+22×4\"这道算式编写一道生活中的实际问题并解答.\n我编写的实际问题是:\n我的解答过程是:",
+        "options":  [
+
+                    ],
+        "answer":  "见详解;166元",
+        "solution":  "【分析】根据题意,78+22×4中的\"22×4\"表示有4个22\n相加，然后用78加上乘积，求和;据此解答即可.\n【详解】实际问题是：小芳到粮油店买了78元的大米一袋，又买了标价22元/瓶的麻油4瓶，一共要付多少钱。=78+88=166(元),答：一共要付166元。",
+        "knowledge_points":  [
+                                 "\"提问题\"\"填条件\"问题",
+                                 "乘加",
+                                 "乘减混合运算"
+                             ],
+        "source":  "24-25四年级下·江西宜春·期末"
+    },
+    {
+        "id":  "364",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "新冠肺炎疫情期间，某校组织五，六年级学生参加抗击疫情线上宣传活动创作宣传标语。六年级学生创作了228条宣传标语，比五年级学生创作的3倍少42条，五年级学生创作了多少条宣传标语。",
+        "options":  [
+
+                    ],
+        "answer":  "90条",
+        "solution":  "【分析】由题意得，六年级学生创作了228条宣传标语，比五年级学生创作的3倍少42条，可以先用228加上42算出五年级学生创作的标语数量的3倍是多少条，再除以3，即可算出五年级学生创作了多少条宣传标语.\n【详解】(228+42)÷3=270÷3=90(条)答：五年级学生创作了90条宣传标语。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·江西南昌·期中"
+    },
+    {
+        "id":  "365",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "书法是一门古老的书写艺术。东晋名家王羲之所作《兰亭集序》被誉为\"天下第一行书\"，全文共324字。李老师临摹《兰亭集序》,每天临摹18字，已经临摹了9天。剩下的计划6天临摹完，剩下的平均每天临摹多少字。",
+        "options":  [
+
+                    ],
+        "answer":  "27个",
+        "solution":  "【分析】由题意得,《兰亭集序》全文共324个字。李老师临摹《兰亭集序》,每天临摹18个字，已经临摹了9天，可以用18乘9算出李老师一共临摹了多少字，再用总字数减去前面的得数算出还剩多少个字。剩下的计划6天临摹完，直接用前面的得数除以6即可算出剩下的平均每天临摹多少个字.\n【详解】(324-18×9)÷6=(324-162)÷6=162÷6=27(个)答：剩下的平均每天临摹27个字。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·河南漯河·期末"
+    },
+    {
+        "id":  "366",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "李老师要批改98篇作文。第一天批改了2篇，第二天批改的比第一天多8篇，还有多少篇没有批改。",
+        "options":  [
+
+                    ],
+        "answer":  "86篇",
+        "solution":  "【分析】先用2加8求出第二天批改的作文数量，再把第天批改的作文数量加上第二天批改的作文数量求出两天共批改的数量，最后用总数减去已批改的数量，即可得到剩余未批改的数量.\n【详解】98-[2+(2+8)]\n:=98-[2+10]\n:=98-12\n:=86(篇)\n答：还有86篇没有批改。",
+        "knowledge_points":  [
+                                 "带有中括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·云南玉溪·期末"
+    },
+    {
+        "id":  "367",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "某车间计划30天生产900个零件，实际每天多生产6个，实际需要多少天完成。",
+        "options":  [
+
+                    ],
+        "answer":  "25天",
+        "solution":  "【分析】计划30天生产900个零件，根据工作总量/工作时间=工作效率，先求出计划每天生产的零件数;再加上实际每天多生产的6个得到实际每天生产的零件数，最后用总零件数除以实际每天生产的零件数得到实际需要的天数.\n【详解】900÷(900/30+6)\n:=900÷(30+6)\n:=900÷36\n:=25(天)\n答：实际需要25天完成。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·河南洛阳·期末"
+    },
+    {
+        "id":  "368",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "明明看一本330页的故事书，前5天平均每天看30页，剩下的准备每天看20页，剩下的几天能看完。",
+        "options":  [
+
+                    ],
+        "answer":  "9天",
+        "solution":  "【分析】根据题意，用5×30计算出前5天看的页数，再用总页数减去前5天已看的页数得到剩下的页数，然后用剩下的页数除以每天看的页数，得到需要的天数，据此解答.\n【详解】根据分析可得:\n330-5×30=330-150=180(页)\n180÷20=9(天)\n答：剩下的9天能看完。",
+        "knowledge_points":  [
+                                 "被除数和除数末尾都有0的除法",
+                                 "乘加，乘减混合运算"
+                             ],
+        "source":  "24-25四年级下·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "369",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "佛跳墙是闽菜中的第一代表菜。张师傅准备用2600元购买13份佛跳墙，结果还剩130元。每份佛跳墙多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "190元",
+        "solution":  "【分析】根据题意，已知总钱数2600元，购买13份后剩余130元。先用2600减去130，求出实际花费的钱数，再除以份数即可得到每份的价格。列式计算即可.\n【详解】根据分析可知:\n(2600-130)÷13=2470÷13=190(元)\n答：每份佛跳墙190元。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·福建漳州·期末"
+    },
+    {
+        "id":  "370",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "游泳馆的泳池长50米，王老师去锻炼游了8个来回，王老师一共游了多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "800米",
+        "solution":  "【分析】首先明确1个来回的距离，游1个来回是游了泳池长度的2倍，即50×2=100米;王老师游了8个来回，所以他游的总距离是1个来回距离的8倍，即100×8=800米.\n【详解】50×2×8\n:=100×8\n:=800(米)\n答：王老师一共游了800米。",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "整十，整百，整千数与一位数的乘法"
+                             ],
+        "source":  "24-25四年级下·新疆克孜勒苏·期末"
+    },
+    {
+        "id":  "371",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "一辆汽车从甲地开往乙地，每小时行驶65千米，行驶了3小时后，距离乙地还有100千米。甲，乙两地相距多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "295千米",
+        "solution":  "【分析】已知汽车每小时行驶65千米，行驶了3小时，根据路程=速度×时间，可得出汽车已经行驶的路程，因为距离乙地还有100千米，所以甲，乙两地的距离为已经行驶的路程加上剩下的路程.\n【详解】65×3+100\n:=195+100\n:=295(千米)\n答：甲，乙两地相距295千米。",
+        "knowledge_points":  [
+                                 "乘加，乘减混合运算",
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25四年级下·广东阳江·期末"
+    },
+    {
+        "id":  "372",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "社区推行垃圾分类积分制，小明家3月收集可回收物获得245个积分,4月收集可回收物获得355个积分。已知50个积分能兑换1个环保购物袋，小明家这两个月的积分能兑换几个购物袋。",
+        "options":  [
+
+                    ],
+        "answer":  "12个",
+        "solution":  "【分析】先用245加上355，求出两个月的总积分;求能兑换几个购物袋就是求总积分里面有几个50，用除法解决，即再用两个月的总积分除以50即可求解.\n【详解】(245+355)÷50\n:=600÷50\n:=12(个)\n答：小明家这两个月的积分能兑换12个购物袋。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·吉林·期末"
+    },
+    {
+        "id":  "373",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "育林小学体育社团计划购买一些体育用品，一筒羽毛球15元，一根跳绳21元，一副羽毛球拍的价格是一筒羽毛球和一根跳绳价格和的2倍,576元可以买几副羽毛球拍。",
+        "options":  [
+
+                    ],
+        "answer":  "8副",
+        "solution":  "【分析】先算一副羽毛球拍的价格=(一筒羽毛球的价格+一根跳绳价格)×2，再用576元除以一副羽毛球拍的价格.\n【详解】(15+21)×2\n:=36×2\n:=72(元)\n576÷72=8(副)\n答:576元可以买8副羽毛球拍。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·甘肃兰州·期中"
+    },
+    {
+        "id":  "374",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "学校运动会上，四年级同学表演团体操，共排成了5个方队，每个方队排成8行，每行8人。最外圈的同学穿红色运动服，其余同学穿白色运动服，一共要准备两种颜色的运动服各多少套。(先画图表示1个方队的示意图，再计算)",
+        "options":  [
+
+                    ],
+        "answer":  "见详解;红色运动服140套;白色运动服180套",
+        "solution":  "【分析】根据题意，每个方队排成8行，每行8人，最外圈的同学穿红色运动服，用8×4再减去四个角多算了一次的4人，即可求出最外圈有多少人;用8×8求出每个方阵的人数再减去外圈的人数，即可求出剩下多少人穿白色运动服，再乘方队的个数，即可求出一共要准备两种颜色的运动服各多少套.\n【详解】如图:![](images/2026-03-05%20220259.jpg)\n8×4-4=32-4=28(人)\n8×8-28=64-28=36(人)\n红色运动服:28×5=140(套)\n白色运动服:36×5=180(套)\n答：一共要准备红色运动服140套，白色运动服180套。",
+        "knowledge_points":  [
+                                 "方阵问题",
+                                 "乘加，乘减混合运算"
+                             ],
+        "source":  "24-25四年级下·河南新乡·期末"
+    },
+    {
+        "id":  "375",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "一幅潮绣作品的长是80厘米，比宽的2倍少20厘米，这幅潮绣作品的宽是多少厘米。(只列式不计算)",
+        "options":  [
+
+                    ],
+        "answer":  "(80+20)÷2",
+        "solution":  "【分析】潮绣作品的长是80厘米，比宽的2倍少20厘米，那么(80+20)厘米是这幅作品的宽的2倍，然后用(80+20)除以2即可得到宽的长度。所以列式为(80+20)÷2.\n【详解】(80+20)÷2\n:=100÷2\n:=50(厘米)\n这幅潮绣作品的宽是50厘米，列式为(80+20)÷2。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "2025·广东潮州·小升初真题"
+    },
+    {
+        "id":  "376",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "一辆汽车从甲地开往乙地，前3小时行驶了180千米，照这样的速度，再行驶2小时到达乙地。甲，乙两地相距多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "300千米",
+        "solution":  "【分析】已知汽车前3小时行驶了180千米，根据速度=路程/时间，计算出汽车的速度;汽车从甲地开往乙地，前3小时行驶后，又行驶了2小时到达乙地，那么总共行驶的时间就是之前的3小时加上后来的2小时;最后根据速度×时间=路程，计算出甲，乙两地的距离.\n【详解】(180÷3)×(2+3)\n:=60×5\n:=300(千米)\n答：甲，乙两地相距300千米。",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·黑龙江佳木斯·期末"
+    },
+    {
+        "id":  "377",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "\"618\"购物节期间，某电器商场联合国家\"以旧换新\"政策，推出笔记本电脑惠民活动：原价6868元/台的热门机型，可同时享受商场满减与国家补贴双重优惠。商场活动：每满1000元立减150元(不满1000元部分不参与满减);国补政策：完成以旧换新流程后，可享受最高500元的补贴(补贴金额基于商场优惠后的价格计算)。小临爸爸购买1台该笔记本电脑，先参与商场满减，再申请国补，至少需支付多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "5468元",
+        "solution":  "【分析】根据题意可知，每满1000元立减150元(不满1000元部分不参与满减),6868元可以享受6个150元的减免，用6868元减去6个150元等于享受商场优惠后笔记本电脑的价钱，再减去国补最高补贴500元，即等于小临爸爸至少需支付的钱，据此即可解答.\n【详解】6868-150×6-500\n:=6868-900-500\n:=5968-500\n:=5468(元)\n答：至少需支付5468元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "乘加，乘减混合运算"
+                             ],
+        "source":  "24-25四年级下·浙江·期末"
+    },
+    {
+        "id":  "378",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "学校要给运动会获奖的同学发奖品，商店正好有促销活动，一种笔记本买5本送1本，这种笔记本每本7元，老师共买回12本，花了多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "70元",
+        "solution":  "【分析】由题意得，商店的一种笔记本买5本送1本，也就是买5本实际得到6本。老师共买回12本，可以先用12除以6算出12本里面有几个6本，然后再乘上5算出需要付多少本的钱。最后再用前面的得数乘上每本书的价钱即可算出老师一共花了多少钱.\n【详解】12÷(5+1)×5×7\n:=12÷6×5×7\n:=2×5×7\n:=10×7\n:=70(元)\n答：老师一共花了70元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "23-24四年级下·河南安阳·期末"
+    },
+    {
+        "id":  "379",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "王老师要批改58篇作文。第一天批改了12篇，第二天比第一天多批改8篇，还有多少篇没批。",
+        "options":  [
+
+                    ],
+        "answer":  "26篇",
+        "solution":  "【分析】先算第二天批改的篇数=第一天批改的篇数+8，再用总篇数依次减去第一天和第二天批改的篇数。据此解答.\n【详解】58-12-(12+8)\n:=58-12-20\n:=26(篇)\n答：还有26篇没批。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·吉林白城·期末"
+    },
+    {
+        "id":  "380",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "粮店运进大豆油和花生油各20桶，每桶大豆油25千克，每桶花生油10千克，运进的大豆油和花生油一共有多少千克。",
+        "options":  [
+
+                    ],
+        "answer":  "700千克",
+        "solution":  "【分析】由题意得，每桶大豆油25千克，每桶花生油10千克，可以先用加法算出1桶大豆油和1桶花生油一共有多少千克。然后乘上20即可算出运进的大豆油和花生油一共有多少千克.\n【详解】(25+10)×20\n:=35×20\n:=700(千克)\n答：运进的大豆油和花生油一共有700千克。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "381",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "2名老师带领48名学生去划船，怎样租船最省钱。共花多少钱。\n小船：限乘4人，租金40元\n大船：限乘6人，租金50元",
+        "options":  [
+
+                    ],
+        "answer":  "租7只大船,2只小船最合算;430元",
+        "solution":  "【分析】用老师加学生计算出总人数，用一只船的租金/限乘人数计算出两种船每人的租金，然后再比较大小，尽可能的选择较便宜的船，船满载时最省钱.\n【详解】2+48=50(人)\n乘大船每人:50÷6=8(元)。2(元)\n乘小船每人:40÷4=10(元)\n8。2元\u003c10元，所以尽量租大船\n50÷6=8(只)。2(人)\n租8只大船，剩余2人租小船，此时有空位;8×50+40=400+40=440(元)\n8-1=7(只)\n50-6×7=50-42=8(人)\n8÷2=4(只)\n所以租7只大船,2只小船最合算，共需要:\n7×50+40×2=350+80=430(元)\n440\u003e430;答：租7只大船,2只小船最合算，需要430元。",
+        "knowledge_points":  [
+                                 "租船问题"
+                             ],
+        "source":  "23-24四年级下·河南安阳·期末"
+    },
+    {
+        "id":  "382",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "四年级有220名学生要去研学旅行，大客车限乘乘客50人，租金为300元;小客车限乘乘客35人，租金为200元。他们怎么租车最省钱。最少需要花多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "租5辆小客车和1辆大客车或租2辆小客车和3辆大客车;1300元",
+        "solution":  "【分析】由题意得，大客车限乘乘客50人，租金为300元，每人需要:300÷50=6(元)。小客车限乘乘客35人，租金为200元，每人需要:200÷35=5(元)。25(元)。对比可知，小客车更便宜。在没有空位的情况下，应该尽量多租小客车。一共有220名学生要去研学旅行，直接用220除以35算出需要租小车的数量以及多余的人数，然后合理调整小客车和大客车的数量使得总钱数最少即可.\n【详解】300÷50=6(元)\n200÷35=5(元)。25(元)\n6\u003e5，所以小客车更便宜。在没有空位的情况下，应该尽量多租小客车.\n220÷35=6(辆)。10(人),即租6辆小客车后，还剩余10人。这10人可以租1辆小客车。也可以让这10人和前面小客车上的人一起租大客车.\n第一种方案:6+1=7(辆),即租7辆小客车.\n200×7=1400(元)\n第二种方案:6-1=5(辆)\n35+10=45(人),45\u003c50，即租5辆小客车和1辆大客车.\n5×200+300=1000+300=1300(元)\n第三种方案:6-4=2(辆)\n35×4+10=140+10=150(人)\n150÷50=3(辆),即租2辆小客车和3辆大客车.\n2×200+300×3=400+900=1300(元)\n1400\u003e1300=1300\n答：租5辆小客车和1辆大客车或租2辆小客车和3辆大客车最省钱，最少需要1300元。",
+        "knowledge_points":  [
+                                 "租船问题"
+                             ],
+        "source":  "24-25四年级下·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "383",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "学校买了240个足球，平均分到12个班，每个班再分给5个小组，平均每个小组分到几个足球。",
+        "options":  [
+
+                    ],
+        "answer":  "4个",
+        "solution":  "【分析】已知有12个班，每个班有5个小组，用班级数乘每个班的小组数，得出一共有多少个小组;已知要把240个足球平均分给这些小组，求每组分到的足球个数，就用足球总个数除以要分的小组总数.\n【详解】240÷(12×5)=240÷60=4(个),答：平均每个小组分到4个足球。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "23-24四年级下·河南安阳·期末"
+    },
+    {
+        "id":  "384",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "有32人要租船游玩1小时。大船限乘6人，每小时租金30元，小船限乘4人，每小时租金24元。怎样租船最省钱。",
+        "options":  [
+
+                    ],
+        "answer":  "4条大船和2条小船",
+        "solution":  "【分析】比较大船和小船的人均租金，确定大船更划算。尽可能多租大船后，若剩余人数无法坐满大船，改用小船填补，计算小船也无法坐满，则减去一条大船，让这条大船的人和剩余的人都坐小船，计算不同组合的总费用，选择最小的方案.\n【详解】30÷6=5(元/人)\n24÷4=6(元/人)\n5\u003c6，所以大船更便宜.\n32÷6=5(条)。2(人)\n坐5条大船，余下2人坐一条小船，费用为:\n5×30+24=150+24=174(元)\n因为小船没有坐满，所以减去一条大船,6+2=8(人)\n8÷4=2(条)\n8人坐2条小船\n费用为:4×30+2×24=120+48=168(元)\n174\u003e168\n所以租4条大船和2条小船最省钱.\n答：租4条大船和2条小船最省钱。",
+        "knowledge_points":  [
+                                 "租船问题"
+                             ],
+        "source":  "23-24四年级下·吉林松原·期末"
+    },
+    {
+        "id":  "385",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "甲地和乙地相距720千米，一辆汽车从甲地出发3小时行驶了240千米，照这样的速度，这辆汽车行完全程一共要多少小时。",
+        "options":  [
+
+                    ],
+        "answer":  "9小时",
+        "solution":  "【分析】根据题意，已知甲地和乙地相距720千米，一辆汽车从甲地出发3小时行驶了240千米，这辆汽车行完全程共要用的时间=甲地和乙地相距的路程/(3小时行驶的路程/3),以此列式计算即可.\n【详解】根据分析可知:\n720÷(240/3)=720÷80=9(小时)\n答：这辆汽车行完全程一共要9小时。",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "23-24四年级下·浙江嘉兴·期末"
+    },
+    {
+        "id":  "386",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "明明家距离图书馆1800米，明明8:30从家出发去图书馆，每分钟走60米.\n(1)按照这个速度，明明几点可以到达图书馆。\n(2)明明出发10分钟后从家遇到骑车来接他去图书馆的爸爸，爸爸以200米/分的车速行驶。他们实际到达时间是多少。",
+        "options":  [
+
+                    ],
+        "answer":  "9:00;8:46",
+        "solution":  "【分析】(1)首先根据时间=路程/速度，计算出明明从家到图书馆需要花费的时间，再根据结束时间=开始时间+经过时间，即可求出明明到达图书馆的时间。(2)首先根据路程=速度×时间，计算出明明10分钟走的路程;再用明明家距离图书馆的距离减去明明10分钟走的路程，再除以爸爸行驶的速度，就是他们骑车需要花费的时间，最后用明明从家出发的时间加上10分钟，再加上他们骑车需要花费的时间，即可求出他们实际到达的时间.\n【详解】(1)1800÷60=30(分钟)8:30+30分=9:00 答：明明9:00可以到达图书馆.\n(2)60×10=600(米) 1800-600=1200(米) 1200÷200=6(分钟) 8:30+10分+6分=8:46\n答：他们实际到达时间是8:46。",
+        "knowledge_points":  [
+                                 "时，分，秒时间的推算",
+                                 "时，分，秒有关的计算",
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25四年级下·广东江门·期末"
+    },
+    {
+        "id":  "387",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "学校有16名老师和319名学生去研学，每辆大车可以坐45人，租金900元，每辆小车可以坐20人，租金500元。怎样租车最省钱。",
+        "options":  [
+
+                    ],
+        "answer":  "租7辆大车和1辆小车最省钱",
+        "solution":  "【分析】由题意得，每辆大车可以坐45人，租金900元，那么每人需要:900÷45=20(元)。每辆小车可以坐20人，租金500元，那么每人需要:500÷20=25(元)。对比可知，大车更便宜。在没有空位的情况下，应该尽量租大车。16名老师和319名学生去研学，一共有335个人，可以直接用335除以45算出需要租大车的数量以及多余的人数。然后合理调整大车和小车的数量使得总租金最少即可.\n【详解】900÷45=20(元)500÷20=25(元) 20\u003c25，所以大车更便宜。在没有空位的情况下，应该尽量租大车.\n16+319=335(名) 335÷45=7(辆)。20(人),即租7辆大车后还剩20人，这20人可以租1辆大车或1辆小车。第一种方案:7+1=8(辆),租8辆大车。900×8=7200(元)第二种方案：租7辆大车和1辆小车900×7+500=6300+500=6800(元) 7200\u003e6800 答：租7辆大车和1辆小车最省钱。",
+        "knowledge_points":  [
+                                 "租船问题"
+                             ],
+        "source":  "24-25四年级下·海南省直辖县级单位·期末"
+    },
+    {
+        "id":  "388",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "学校买了两种花要栽在校园的花坛里.\n①月季花买了128棵,②绣球花买了64棵,③由四年级的6个班负责栽花,④把这些花平均栽在4个花坛里。每个花坛里栽多少棵花呢。(1)要解决这个问题，需要选择信息()。(填上选择的序号)(2)请根据你选择的信息，求出每个花坛里栽多少棵花。",
+        "options":  [
+
+                    ],
+        "answer":  "①②④;48棵",
+        "solution":  "【分析】(1)要求每个花坛里栽多少棵花，需要知道要栽的花的棵数和花坛的个数，所以要选择的信息有①月季花买了128棵,②绣球花买了64棵,④把这些花平均栽在4个花坛里;(2)用月季花的棵数加绣球花的棵数，再除以花坛的个数，即等于每个花坛里栽花的棵数，据此即可解答.\n【详解】(1)根据分析可知，要解决这个问题，需要选择信息①②④。(填上选择的序号)\n(2)(128+64)÷4=192÷4=48(棵),答：每个花坛里栽48棵花。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "\"提问题\"\"填条件\"问题"
+                             ],
+        "source":  "24-25四年级下·广东江门·期末"
+    },
+    {
+        "id":  "389",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "为了美化校园环境，共建绿色校园，学校倡导学生将空水瓶收集起来。四(1)班收集了85个空水瓶，四(2)班收集了77个空水瓶，两个班收集的空水瓶的总数是四(3)班的2倍。四(3)班收集了多少个空水瓶。",
+        "options":  [
+
+                    ],
+        "answer":  "81个",
+        "solution":  "【分析】四(1)班，四(2)班收集的空水瓶的总数是四(3)班的2倍，所以四(1)班，四(2)班收集的空水瓶的个数之和除以2，即等于四(3)班收集的空水瓶个数，据此即可解答.\n【详解】(85+77)÷2=162÷2=81(个)答：四(3)班收集了81个空水瓶。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·广东江门·期末"
+    },
+    {
+        "id":  "390",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "某校计划举行研学活动，共有老师14人，学生326人。每辆大车租金900元，可以坐40人;每辆小车租金500元，可以坐20人。怎样租车最省钱。",
+        "options":  [
+
+                    ],
+        "answer":  "可以租8辆大车,1辆小车",
+        "solution":  "【分析】根据题意，先求出大车和小车的人均租车费用，尽量租费用低的车;用326加上14，求出总人数;再除以费用低的车型，如果有余数，合理安排其它车型;最后根据租金，求出租车的总费用;列式计算即可.\n【详解】900÷40=22(元)⋯⋯20(元)\n500÷20=25(元)\n22\u003c25，大车的人均费用更便宜，尽量租大车.\n326+14=340(人)\n340÷40=8(辆)。20(人)\n可以租8辆大车,1辆小车.\n所需费用为:\n8×900+500=7200+500=7700(元)\n答：租8辆大车,1辆小车最省钱。",
+        "knowledge_points":  [
+                                 "租船问题"
+                             ],
+        "source":  "23-24四年级下·湖南长沙·期末"
+    },
+    {
+        "id":  "391",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "四年级58名同学和4名老师准备去秋游，有两种车可供租用：大车限坐9人，每辆租金:120元;小车限坐4人，每辆租金60元。怎样租车最划算。最少需要多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "租6辆大车和2辆小车;840元。",
+        "solution":  "【分析】根据题意先分别计算出两种车平均每人所需的钱数，即大车平均每人的价钱是:120÷9=13(元)。3(元),小车平均每人的价钱是:60÷4=15(元),13\u003c15，所以尽量租大车，而且没有空位或空位最少则最省钱。用师生总人数(58+4)人除以一辆大车可乘坐的人数，求出需要大车的数量，再结合余数判断是否需要小车;据此计算解答即可.\n【详解】120÷9=13(元)。3(元)\n60÷4=15(元)\n(58+4)÷9=62÷9=6(人)。8(人)\n8÷4=2(辆)\n可以租6辆大车和2辆小车，正好没有空座，说明花的钱数最少\n需要的租金:6×120+2×60=720+120=840(元)\n答：租6辆大车和2辆小车最划算，最少需要840元。",
+        "knowledge_points":  [
+                                 "租船问题"
+                             ],
+        "source":  "24-25四年级下·广东阳江·期末"
+    },
+    {
+        "id":  "392",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一张桌子85元，一张椅子65元，学校准备买50套这样的桌椅，一共要花多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "7500元",
+        "solution":  "【分析】由题意可知：用一张桌子的价钱加上一张椅子的价钱求出一套桌椅的价钱，再乘50，即可求出一共要花多少钱.\n【详解】(85+65)×50\n:=150×50\n:=7500(元)\n答：一共要花7500元。",
+        "knowledge_points":  [
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·广东阳江·期末"
+    },
+    {
+        "id":  "393",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "四年级有12名老师带领258名学生参观红色教育基地。每辆大车可坐50人，租金800元;每辆小车可坐25人，租金500元。怎样租车最省钱。最少需要多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "租5辆大车和1辆小车;4500元",
+        "solution":  "【分析】由题意得，每辆大车可坐50人，租金800元，每人需要:800÷50=16(元)。每辆小车可坐25人，租金500元，每人需要:500÷25=20(元)。对比可知，大车更便宜。在没有空位的情况下，应该尽量多租大车。一共有12名老师和258名学生，那么一共是270人，直接用270除以50算出需要租大车的数量和多余的人数。然后再合理调整大车和小车的数量使得总费用最少即可.\n【详解】800÷50=16(元)\n500÷25=20(元)\n16\u003c20，所以大车更便宜。在没有空位的情况下，应该尽量多租大车.\n12+258=270(名)\n270÷50=5(辆)。20(名),即租5辆大车后，还剩20人，这20人可以租1辆小车。也可以让1辆大车上的人和剩下的20人一起租小车.\n5-1=4(辆)\n50+20=70(人)\n70÷25=2(辆)。20(人)\n2+1=3(辆)\n第一种方案：租5辆大车和1辆小车\n800×5+500×1=4000+500=4500(元)\n第二种方案：租4辆大车和3辆小车\n800×4+500×3=3200+1500=4700(元)\n4500\u003c4700\n答：租5辆大车和1辆小车最省钱，最少需要4500元。",
+        "knowledge_points":  [
+                                 "租船问题"
+                             ],
+        "source":  "24-25四年级下·广东潮州·期末"
+    },
+    {
+        "id":  "394",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "某工厂生产一批零件，加工12天后，还剩280个未完成。已知这批零件总共1600个，平均每天加工多少个零件。",
+        "options":  [
+
+                    ],
+        "answer":  "110个",
+        "solution":  "【分析】由题意可知：用1600-280求出已经加工零件的总个数，再除以12，即可求出平均每天加工多少个零件.\n【详解】(1600-280)÷12\n:=1320÷12\n:=110(个)\n答：平均每天加工110个零件。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "除加，除减混合运算"
+                             ],
+        "source":  "24-25四年级下·广东肇庆·期末"
+    },
+    {
+        "id":  "395",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "学校组织四年级8名老师带领80名学生到航模中心研学。已知航模中心体验馆的大飞船每艘可以坐12人，租金200元;小飞船每艘可以坐8人，租金180元。如果全体师生都要体验，怎样租飞船比较省钱。至少要花多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "6艘大飞船和2艘小飞船;1560元",
+        "solution":  "【分析】租大飞船每人需要:200÷12=16(元)。8(元),租小飞船每人需要:180÷8=22(元)。4(元),16\u003c22，所以租大飞船省钱，要尽量多租大飞船，最好不要有空位，最省钱。首先,80+8=88(人),求出老师和学生的总人数，再用总人数除以每艘大飞船限乘的人数,88÷12=7(艘)。4(人),租7艘大飞船，剩余的人数租1艘小飞船，用每艘大飞船的租金乘7，求出租7艘大飞船的钱数，再用租7艘大飞船的钱数加上租1艘小飞船的租金，求出租7艘大飞船和1艘小飞船需要的钱数。租6艘大飞船,2艘小飞船,6×12=72(人),88-72=16(人),租6艘大飞船，剩余人数刚好租2艘小飞船，且没有空位，用每艘大飞船的租金乘6，求出租6艘大飞船的钱数，用每艘小飞船的租金乘2，求出租2艘小飞船的钱数，再把租6艘大飞船的钱数和租2艘小飞船的钱数相加，求出租6艘大飞船和租2艘小飞船需要的钱数，最后再把两种方案所需要的钱数进行比较，即可解答.\n【详解】200÷12=16(元)。8(元)\n180÷8=22(元)。4(元)\n16\u003c22\n尽量租大飞船\n8+80=88(人)\n88÷12=7(艘)。4(人)\n租7艘大飞船和1艘小飞船,\n租7艘大飞船和1艘小飞船需要的钱数:\n200×7+180×1=1400+180×1=1400+180=1580(元)\n租6艘大飞船,2艘小飞船,\n6×12=72(人)\n88-72=16(人)\n租6艘大飞船，剩余人数刚好租2艘小飞船，且没有空位,\n租6艘大飞船和租2艘小飞船需要的钱数:\n200×6+180×2=1200+180×2=1200+360=1560(元)\n1580\u003e1560\n答：租6艘大飞船和2艘小飞船比较省钱，至少要花1560元钱。",
+        "knowledge_points":  [
+                                 "租船问题"
+                             ],
+        "source":  "24-25四年级下·广东潮州·期末"
+    },
+    {
+        "id":  "396",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "某校三，四年级师生参加研学活动，共租用12辆大巴车，每辆有49个座位(含司机座位1个),已知只有一辆大巴车剩下2个空位，其余全都坐满，参加活动的师生共有多少人。",
+        "options":  [
+
+                    ],
+        "answer":  "574人",
+        "solution":  "【分析】先用每辆车的座位数减司机座位，求出一辆大巴车坐满时的师生人数;再用每辆车师生座位数乘12，求出12辆车师生座位数;最后减去剩下的2个空位，即可求出参加活动的师生人数.\n【详解】(49-1)×12-2\n:=48×12-2\n:=576-2\n:=574(人)\n答：参加活动的师生共有574人。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·广东潮州·期末"
+    },
+    {
+        "id":  "397",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "机器人和女孩一起在舞台上扭秧歌，已知机器人排成4行，每行4位，女孩排成5行，每行4位。这个节目包含机器人一共有多少位演员。",
+        "options":  [
+
+                    ],
+        "answer":  "36位",
+        "solution":  "【分析】由题可知，机器人4行，每行4位，女孩5行，每行4位。先根据乘法的意义，用行数×每行人数，分别算出机器人和女孩的数量，再根据加法的意义，把机器人和女孩的数量相加得到演员总人数.\n【详解】4×4=16(位),5×4=20(位),16+20=36(位)\n答：这个节目包含机器人一共有36位演员。",
+        "knowledge_points":  [
+                                 "两位数与整十数的加法",
+                                 "5的乘法口诀及应用"
+                             ],
+        "source":  "24-25四年级下·广东潮州·期末"
+    },
+    {
+        "id":  "398",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "光明小学新购买68套课桌椅，每张课桌58元，每把椅子32元，购买这些课桌椅共花了多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "6120元",
+        "solution":  "【分析】用58元加上32元，可以算出1套课桌椅的价格，再用和乘68即可算出购买这些课桌椅共花了多少元.\n【详解】(58+32)×68\n:=90×68\n:=6120(元)\n答：购买这些课桌椅共花了6120元。",
+        "knowledge_points":  [
+                                 "两位数乘两位数，乘数末尾有0",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·湖南岳阳·期末"
+    },
+    {
+        "id":  "399",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "修一条路，已经修了350米，没修的比已修的2倍少50米，这条路有多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "1000米",
+        "solution":  "【分析】由题意可知：用已经修了的米数乘2，再减去50，即可求出没修的米数，最后用没修的米数加上已经修了的米数，即可求出这条路有多少米.\n【详解】350×2-50=700-50=650(米),650+350=1000(米)\n答：这条路有1000米。",
+        "knowledge_points":  [
+                                 "乘加，乘减混合运算"
+                             ],
+        "source":  "24-25四年级下·海南省直辖县级单位·期末"
+    },
+    {
+        "id":  "400",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "学校把600本练习本分给四年级5个班的学生，每个班分115本，还剩多少本练习本。",
+        "options":  [
+
+                    ],
+        "answer":  "25本",
+        "solution":  "【分析】由题意可知：用每个班分的本数乘四年级班级数，求出总共分出的本数，再用600减去总共分出的本数，即可求出还剩多少本练习本.\n【详解】600-115×5=600-575=25(本),答：还剩25本练习本。",
+        "knowledge_points":  [
+                                 "乘加，乘减混合运算"
+                             ],
+        "source":  "24-25四年级下·海南省直辖县级单位·期末"
+    },
+    {
+        "id":  "401",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "中山公园里的大船每条限乘6人，小船每条限乘4人，每条大船租金10元，每条小船租金8元，有46名师生租船，租()最省钱。",
+        "options":  [
+                        "A. 6条大船,3条小船",
+                        "B. 7条大船,1条小船",
+                        "C. 8条大船"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】先比较大船，小船平均每人的租金大小，租金小的更便宜，租船时尽量租便宜的船，并且不留空位时最省钱.\n【详解】10÷6=1(元)。4(元)\n8÷4=2(元)\n2\u003e1，大船便宜些.\n46÷6=7(条)。4(人)\n租7条大船还剩下4人，剩下的4人正好再租1条小船，这样师生全部坐下，并且没有空位，所以租7条大船,1条小船最省钱。故答案为:B",
+        "knowledge_points":  [
+                                 "租船问题",
+                                 "经济问题"
+                             ],
+        "source":  "24-25四年级下·广东汕头·期末"
+    },
+    {
+        "id":  "402",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "420减去19与12的积，所得的差乘5，积是多少。正确的列式是()。",
+        "options":  [
+                        "A. 420-(19×12)×5",
+                        "B. (420-19×12)×5",
+                        "C. 420-19×12×5",
+                        "D. [(420-19)×12)]*5"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据题意，正确的运算顺序应为：先计算19与12的积，再用420减去该积得到差，最后将差乘5。需通过括号确保运算顺序正确。根据整数四则运算法则:(1)在一个没有括号的算式里，如果只含同一级运算，按照从左往右的顺序依次计算;如果含有两级运算，要先算第二级运算乘除法，再算第一级运算加减法。(2)在一个有括号的算式里，要先算小括号里的，再算中括号里的，最后算括号外的.\n【详解】A。420-((19×12))×5先算19×12，再用该积乘5，最后算减法，不符合题意。此选项错误。B。(420-19×12)×5先算19×12，再求差后乘5，符合题意。此选项正确。C。420-19×12×5无括号，先算19×12×5，再算减法，此选项错误。D。[(420-19)×12]*5先算小括号的420-19，再算中括的乘法，最后算括号外的乘法，不符合题意，此选项错误。故答案为:B",
+        "knowledge_points":  [
+                                 "乘加，乘减混合运算",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·山东临沂·期末"
+    },
+    {
+        "id":  "403",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "420减去19与12的积，所得的差乘5，积是多少。正确的列式是()。",
+        "options":  [
+                        "A. 420-(19×12)×5",
+                        "B. (420-19×12)×5",
+                        "C. 420-19×12×5",
+                        "D. (420-19)×12×5"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】420减去19与12的积，所得的差乘5，运算顺序是先算乘法，再算减法，最后再算乘法。据此逐项分析解答即可.\n【详解】A。420-(19×12)×5，先算乘法，再算乘法，最后再算减法，不符合题意;B。(420-19×12)×5，先算乘法，再算减法，最后再算乘法，符合题意;C。420-19×12×5，先算乘法，再算乘法，最后再算减法，不符合题意;D。[(420-19)×12]*5，先算减法，再算乘法，最后再算乘法，不符合题意。故答案为:B",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·山东临沂·期末"
+    },
+    {
+        "id":  "404",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "张老师购买了5盒钢笔，每盒有10支,-共花600元钱。根据以上信息，不能解决的问题是()。",
+        "options":  [
+                        "A. 张老师一共购买多少支钢笔。",
+                        "B. 张老师还剩多少钱。",
+                        "C. 每支钢笔多少元。",
+                        "D. 每盒钢笔多少元。"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】A。张老师购买了5盒钢笔，每盒有10支，用5×10即可算出张老师-共购买多少支钢笔;B。张老师一共花了600元，没有剩下，故无法算出张老师还剩多少钱;C。一共有5×10=50支钢笔，用600元除以50支钢笔即可求出每支钢笔多少元;D。5盒钢笔一共600元，用600除以5即可算出每盒钢笔120元.\n【详解】A。5×10=50(支),张老师一共购买50支钢笔;B。张老师一共花了600元，没有剩下，故无法算出张老师还剩多少钱;C。600÷(5×10)=600÷50=12(元),每支钢笔12元;D。600÷5=120(元),每盒钢笔120元。所以不能解决的问题是张老师还剩多少钱。故答案为:B",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "\"提问题\"\"填条件\"问题",
+                                 "经济问题"
+                             ],
+        "source":  "24-25四年级下·贵州黔东南·期末"
+    },
+    {
+        "id":  "405",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "计算\"25*[(6+69)÷3]\"时，最后一步应该算()。",
+        "options":  [
+                        "A. 加法",
+                        "B. 减法",
+                        "C. 乘法",
+                        "D. 除法"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】根据整数四则混合运算规律，同级运算时，从左到右依次计算;二级运算时，先算乘除，后算加减。有括号时，先算括号里面的，据此可知，题中应先算小括号内的加法，再算中括号内的除法，最后算括号外的乘法;据此解答.\n【详解】25*[(6+69)÷3]=25*[75÷3]=25×25=625，最后一步应该算乘法。故答案为:C",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "带有中括号的混合运算",
+                                 "有括号的运算顺序"
+                             ],
+        "source":  "24-25四年级下·湖北武汉·期末"
+    },
+    {
+        "id":  "406",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "关于0，下面的说法中错误的是()。",
+        "options":  [
+                        "A. 除数不能为0",
+                        "B. 0不是整数",
+                        "C. 两个数相减差是0，则这两个数相等"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据0的意义，比如相同两个数相减的结果是0;一个数与0相加的和是它本身;一个数减0的差是它本身;0除以任何一个不为0的数商是0;0与任何数相乘的积是0。一个非0的数除以0无意义;0是整数;逐一判断，可得答案.\n【详解】A。一个非0的数除以0无意义，该选项正确;B。0是整数，该选项错误;C。相同两个数相减的结果是0，该选项正确;故答案为:B",
+        "knowledge_points":  [
+                                 "0的认识和读写",
+                                 "有关0的加减法",
+                                 "乘，除法的意义和各部分间的关系",
+                                 "0不能作除数"
+                             ],
+        "source":  "24-25四年级下·陕西西安·期末"
+    },
+    {
+        "id":  "407",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一辆卡车3次运苹果600箱，照这样计算，再运6次，一共可以运多少箱。下面算式错误的是()。",
+        "options":  [
+                        "A. 600÷3×6",
+                        "B. 600÷3×(6+3)",
+                        "C. 600×(6÷3+1)",
+                        "D. 600×(6÷3)+600"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】整数的四则混合运算顺序是：有小括号先算小括号里的算式，再乘除，最后加减;如果没括号，先乘除再加减;如果括号里面的算式也是混合运算，还是先乘除再加减;先用600除以3计算出每次运的箱数，再乘6计算出6次可以运多少箱，最后加上600可以计算出一共运的箱数;逐项分析后进行选择，据此解答.\n【详解】根据分析:\nA。600÷3×6=1200(箱),600÷3表示每次运的箱数，再乘6计算出了6次运的箱数，所以这里计算的是6次运1200箱，而不是一共可以运多少箱;B。600÷3×(6+3)=200×9=1800(箱),6+3计算的是运的总次数,600÷3计算的是每次运的箱数，用每次运的箱数×运的总次数，计算出的是一共可以运多少箱;C。600×(6÷3+1)=600×(2+1)=600×3=1800(箱)\n6÷3计算的是6次中有几个3次，再加上1就是再加上1个3次，计算出一共几个3次,1个3次运600箱，用600乘3次的总个数，可以计算出一共可以运多少箱;D。600×(6÷3)+600=600×2+600=1200+600=1800(箱)\n6÷3计算的是6次中有几个3次，用600乘计算出的3次的个数，可以计算出6次一共运多少箱，再加上3次运的600箱，计算出一共可以运多少箱;所以算式错误的是600÷3×6。故答案为:A",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "单归一问题"
+                             ],
+        "source":  "24-25四年级下·贵州遵义·期末"
+    },
+    {
+        "id":  "408",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面各题中，直接去掉括号后，运算顺序没发生改变的是()。",
+        "options":  [
+                        "A. (18×5)-20",
+                        "B. 375-(49+21)×5",
+                        "C. 69×6÷(30-12)",
+                        "D. 200-336÷(98-95)"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】四则混合运算运算顺序：同级运算，从左到右依次计算;既有乘除又有加减的，先算乘除，后算加减;有括号时，先算括号里面的;有多层括号时，先算小括号里的，再算中括号里面的，据此解答.\n【详解】A。在算式(18×5)-20中，要先算小括号里面的乘法，再算小括号外面的减法。去掉小括号后，算式变为18×5-20，要先算乘法，再算减法。即括号去掉后，这个算式的运算顺序没发生改变。满足题意.\nB。在算式375-(49+21)×5中，要先算小括号里面的加法，再算小括号外面的乘法，最后算小括号外面的减法。去掉小括号后，算式变为375-49+21×5。在算式375-49+21×5中，要先算乘法，再算减法，最后算加法。即括号去掉后，这个算式的运算顺序发生了改变。不满足题意.\nC。在算式69×6÷(30-12)中，要先算小括号里面的减法，再算小括号外面的乘法，最后算小括号外面的除法。去掉小括号后，算式变为69×6÷30-12。在算式69×6÷30-12中，要先算乘法，再算除法，最后算减法。即括号去掉后，这个算式的运算顺序发生了改变。不满足题意.\nD。在算式200-336÷(98-95)中，要先算小括号里面的减法，再算小括号外面的除法，最后算小括号外的减法。去掉小括号后，算式变为200-336÷98-95。在算式200-336÷98-95中，要先算除法，再从左往右算减法。即括号去掉后，这个算式的运算顺序发生了改变。不满足题意。故答案为:A",
+        "knowledge_points":  [
+                                 "有括号的运算顺序",
+                                 "无括号的运算顺序"
+                             ],
+        "source":  "24-25四年级下·广东广州·期末"
+    },
+    {
+        "id":  "409",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "算式5*[3876÷(36-17)]中，第一步应先算:()。",
+        "options":  [
+                        "A. 乘法",
+                        "B. 除法",
+                        "C. 减法"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】根据整数四则混合运算的运算顺序：在没有括号的算式里，如果只含有乘除法，或只含有加减法，要从左往右依次计算;如果既含有乘除法又含有加减法，要先算乘除法再算加减法;如果有小括号，要先算小括号里面的，再算小括号外面的，据此解答.\n【详解】根据分析可知，算式5*[3876÷(36-17)]中，第步应先算小括号里面的减法。故答案为:C",
+        "knowledge_points":  [
+                                 "有括号的运算顺序",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "410",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面说法错误的是()。",
+        "options":  [
+                        "A. 一个数加上0还得原数",
+                        "B. 当被减数等于减数时，差是0",
+                        "C. 0除以一个非0的数，还得0",
+                        "D. 一个数和0相乘，仍得原数"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】0加任何数都得原数;任何数减0还是原数，被减数等于减数时，差是0;0除以任何不是0的数都得0;0乘任何数都得0;可以分别举例说明.\n【详解】A。5+0=5,100+0=100,0+25=25。一个数加上0还得原数。说法正确;B。5-5=0,10-10=0,36-36=0。当被减数等于减数时，差是0。说法正确;C。0÷6=0,0÷10=0,0÷88=0.0除以一个非0的数，还得0。说法正确;D。0×0=0,0×5=0,20×0=0。一个数和0相乘，都得0。原说法错误。故答案为:D",
+        "knowledge_points":  [
+                                 "乘，除法的意义和各部分间的关系",
+                                 "加，减法的意义和各部分间的关系",
+                                 "被除数是0的除法运算",
+                                 "有关0的乘法"
+                             ],
+        "source":  "24-25四年级下·重庆·期末"
+    },
+    {
+        "id":  "411",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "两个数的和是216，其中一个加数是200，另一个加数是();减数是75，差是15，被减数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "16;90",
+        "solution":  "【分析】在加法算式中，加数+加数=和，加数=和-另一个加数;在减法算式中，被减数-减数=差，因此被减数=差+减数，依此填空.\n【详解】216-200=16,15+75=90，两个数的和是216，其中一个加数是200，另一个加数是16;减数是75，差是15，被减数是90。",
+        "knowledge_points":  [
+                                 "加，减法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·广东阳江·期末"
+    },
+    {
+        "id":  "412",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小强在做一道减法计算时，把减数45看成了54，算出来的结果是84。正确的结果是()。",
+        "options":  [
+
+                    ],
+        "answer":  "93",
+        "solution":  "【分析】被减数-减数=差，则被减数=差+减数，因此用54加84计算出被减数，然后用被减数减去45即可求出正确的结果，据此列式解答.\n【详解】被减数:54+84=138，正确的结果:138-45=93，小强在做一道减法计算时，把减数45看成了54，算出来的结果是84。正确的结果是93。",
+        "knowledge_points":  [
+                                 "错中求解问题(加减法)",
+                                 "加，减法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·内蒙古包头·期末"
+    },
+    {
+        "id":  "413",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据12×6=72,150-72=78,468÷78=6三个算式合并成一个综合算式()。",
+        "options":  [
+
+                    ],
+        "answer":  "468÷468/(150-12×6)=6",
+        "solution":  "【分析】根据题意，最后计算除法，被除数是468，除数是150减12乘6的积，要先算出150减12乘6的积，所以要给150减12乘6的积加上小括号.\n【详解】468÷(150-12×6)\n:=468÷(150-72)\n:=468÷78\n:=6\n三个算式合并成一个综合算式为:468÷(150-12×6)=6。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·广东佛山·期末"
+    },
+    {
+        "id":  "414",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据8.2-5.74=2.46，改写成另一道不同的减法算式是()和一道加法算式是()。",
+        "options":  [
+
+                    ],
+        "answer":  "8.2-2.46=5.74;5.74+2.46=8.2÷2.46+5.74=8.2",
+        "solution":  "【分析】根据\"被减数-减数=差\"可知：\"被减数-差=减数\"，\"差+减数=被减数\"，据此解答.\n【详解】由分析可知：根据8.2-5.74=2.46，改写成另一道不同的减法算式是8.2-2.46=5.74和一道加法算式是5.74+2.46=8.2。",
+        "knowledge_points":  [
+                                 "加，减法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·广东汕头·期末"
+    },
+    {
+        "id":  "415",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "①438÷73=6,②920+6=926,③926×2=1852，把上面的三道分步算式改写成综合算式是()。",
+        "options":  [
+
+                    ],
+        "answer":  "(920+438÷73)×2=1852",
+        "solution":  "【分析】先算438除以73，求出商，再用920加上求出的商，求出和，然后再用求出的和乘2，根据四则混合运算顺序，给920+438÷73，加上小括号，列式为:(920+438÷73)×2=1852.\n【详解】(920+438÷73)×2\n=(920+6)×2\n:=926×2\n:=1852\n①438÷73=6,②920+6=926,③926×2=1852，把三道分步算式改写成综合算式是(920+438÷73)×2=1852。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·广东佛山·期末"
+    },
+    {
+        "id":  "416",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据16+59=75,75÷5=15,15×28=420写出的综合算式是()。",
+        "options":  [
+
+                    ],
+        "answer":  "(16+59)÷5×28=420",
+        "solution":  "【分析】先用16加59求和，再用求出的和除以5求出商，最后用求出的商乘28即可，据此解答.\n【详解】根据分析可得:(16+59)÷5×28=420\n所以根据16+59=75,75÷5=15,15×28=420，写出的综合算式是(16+59)÷5×28=420。",
+        "knowledge_points":  [
+                                 "两位数乘两位数的进位乘法",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·广东佛山·期末"
+    },
+    {
+        "id":  "417",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在计算60+240÷15×6时，先算()法，再算()法，最后算()法。如果要改变运算顺序，先算除法，再算加法，最后算乘法，算式应该是()。",
+        "options":  [
+
+                    ],
+        "answer":  "(60+240÷15)×6",
+        "solution":  "【分析】根据运算规则，无括号时，先算乘除，后算加减，同级运算从左往右开始算，在计算60+240÷15×6的运算顺序是先除法，再乘法，最后加法。要改为先除法，再加法，最后乘法，需通过括号将除法和加法括起来，使加法在乘法前运算即可.\n【详解】(60+240÷15)×6\n:=(60+16)×6\n:=76×6\n:=456\n在计算60+240÷15×6时，先算除法，再算乘法，最后算加法。如果要改变运算顺序，先算除法，再算加法，最后算乘法，算式应该是(60+240÷15)×6。",
+        "knowledge_points":  [
+                                 "有括号的运算顺序",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "418",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "吴老师买了4个足球和8根跳绳，已知每根跳绳12元，共花了396元，则每个足球()元。",
+        "options":  [
+
+                    ],
+        "answer":  "75",
+        "solution":  "【分析】总价=单价×数量，把数据代入求出8根跳绳的钱，总共花的钱减去8根跳绳的钱等于4个足球的钱，再除以4，即等于每个足球的价钱，据此即可解答.\n【详解】(396-12×8)÷4\n:=(396-96)÷4\n:=300÷4\n:=75(元)\n每个足球75元。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "经济问题"
+                             ],
+        "source":  "24-25四年级下·贵州黔东南·期末"
+    },
+    {
+        "id":  "419",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "算式56÷8=7,7×11=77,126-77=49\"合并成一道综合算式为()。",
+        "options":  [
+
+                    ],
+        "answer":  "126-56÷8×11=49",
+        "solution":  "【分析】根据题意可知，先求56除以8的商，再求商乘11的积，最后求126减去积的差，先算除法，再算乘法，最后再算减法，据此列式即可解答.\n【详解】根据分析可知，算式\"56÷8=7,7×11=77,126-77=49\"合并成一道综合算式为:126-56÷8×11=49。",
+        "knowledge_points":  [
+                                 "无括号的运算顺序"
+                             ],
+        "source":  "24-25四年级下·贵州黔东南·期末"
+    },
+    {
+        "id":  "420",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "算式420+1125÷78-53若按减除加的运算顺序进行计算，则需要添上小括号改变运算顺序，变化后的算式是:()。",
+        "options":  [
+
+                    ],
+        "answer":  "420+1125÷(78-53)=465",
+        "solution":  "【分析】按照减，除，加的运算顺序，把减法提到第一步计算，需要在计算78-53时加上小括号，其余不变，然后进行计算即可.\n【详解】420+1125÷(78-53)\n:=420+1125÷25\n:=420+45\n:=465\n变化后的算式是:420+1125÷(78-53)=465",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·云南昆明·期末"
+    },
+    {
+        "id":  "421",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "160+24×(15-7),应先算()法，再算()法，最后算()法，结果是()。",
+        "options":  [
+
+                    ],
+        "answer":  "减;乘;加;352",
+        "solution":  "【分析】根据四则混合运算顺序，先算小括号里的减法，再算括号外的乘法，最后计算加法，据此计算出结果，即可解答.\n【详解】160+24×(15-7)\n:=160+24×8\n:=160+192\n:=352\n160+24×(15-7),应先算减法，再算乘法，最后算加法，结果是352。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "有括号的运算顺序"
+                             ],
+        "source":  "24-25四年级下·河北张家口·期末"
+    },
+    {
+        "id":  "422",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把54-48=6,6×7=42,126÷42=3合并成一个综合算式是()。",
+        "options":  [
+
+                    ],
+        "answer":  "126/[(54-48)×7]=3",
+        "solution":  "【分析】根据分步算式，先计算54-48得到6，再用6×7得到42，最后用126÷42得到3。合并时需要先算减法，再算乘法，最后算除法，因此需添加括号确保运算顺序.\n【详解】126/[(54-48)×7]\n:=126/[6×7]\n:=126÷42\n:=3\n把54-48=6,6×7=42,126÷42=3合并成一个综合算式是126÷(54-48)×7=3。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "带有中括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·广东广州·期末"
+    },
+    {
+        "id":  "423",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "请将330-80=250,750-250=500,500×3=1500，改写成一个综合算式为:()。",
+        "options":  [
+
+                    ],
+        "answer":  "[750-(330-80)]*3=1500",
+        "solution":  "【分析】根据题意，先计算330-80的结果，再用750减去前面求出的差，最后乘3，则计算第一个减法的时候需要加上小括号，计算第二个减法的时候需要加上中括号，最后计算乘法，据此列式即可.\n【详解】[750-(330-80)]*3\n:=[750-250]*3\n:=500×3\n:=1500\n请将330-80=250,750-250=500,500×3=1500，改写成一个综合算式为:[750-(330-80)]*3=1500。",
+        "knowledge_points":  [
+                                 "带有中括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·山东日照·期末"
+    },
+    {
+        "id":  "424",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "8.4+□=12.7，方框里应填();52÷□=13，方框里应填()。",
+        "options":  [
+
+                    ],
+        "answer":  "4.34",
+        "solution":  "【分析】根据加法算式各部分之间的关系，加数+加数=和，和-一个加数=另一个加数。根据除法算式各部分之间的关系，被除数/除数=商，被除数/商=除数，商×除数=被除数。据此解答.\n【详解】8.4+□=12.7,12.7-8.4=4.3，方框里应填4.3.52÷□=13,52÷13=4，方框里应填4",
+        "knowledge_points":  [
+                                 "加，减法的意义和各部分间的关系",
+                                 "乘，除法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·吉林通化·期末"
+    },
+    {
+        "id":  "425",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "已知被减数，减数与差的和是128，差是45，减数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "19",
+        "solution":  "【分析】根据题意，明确被减数-减数=差，被减数=减数+差，已知被减数，减数与差的和是128，差是45，用128除以2，求出减数与差的和，再减去45，就是减数的值，列式计算即可.\n【详解】根据分析可知:128÷2-45=64-45=19，已知被减数，减数与差的和是128，差是45，减数是19。",
+        "knowledge_points":  [
+                                 "加，减法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·黑龙江佳木斯·期末"
+    },
+    {
+        "id":  "426",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "将72+28=100,100×39=3900合成一道综合算式(),如果要变成先算乘法，再算加法，该综合算式可以变成()。",
+        "options":  [
+
+                    ],
+        "answer":  "(72+28)×39=3900;72+28×39=1164",
+        "solution":  "【分析】已知两个分步算式72+28=100和100×39=3900，因为在第二个算式100×39=3900中,100是由第一个算式72+28得到的。根据四则运算顺序，在有括号的算式里，要先算括号里面的。所以把72+28用括号括起来替换第二个算式中的100。得到综合算式(72+28)×39。要求先算乘法，再算加法，即先算28×39，再与72相加，去掉括号即可.\n【详解】将72+28=100,100×39=3900合成一道综合算式为(72+28)×39;如果变成先算乘法，再算加法，该综合算式可以变成72+28×39。",
+        "knowledge_points":  [
+                                 "乘加，乘减混合运算",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·广东东莞·期末"
+    },
+    {
+        "id":  "427",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "360÷(70-4×16)时，先算()法，再算()法，最后算()法。如果要先算减法，再算乘法，最后算除法，算式应改为()。",
+        "options":  [
+
+                    ],
+        "answer":  "乘;减;除;360÷((70-4)×16)",
+        "solution":  "【分析】四则运算的运算顺序为：有括号的先算括号里面的，括号里面先算乘除，后算加减;没有括号的先算乘除，后算加减。要改变运算顺序，需要合理添加括号.\n【详解】根据四则运算顺序，对于360÷(70-4×16),括号内有乘法和减法，先算乘法:4×16=64;再算括号内的减法:70-64=6;最后算括号外的除法:360÷6=60。所以先算乘法，再算减法，最后算除法。如果要先算减法，再算乘法，最后算除法，需要给减法加上小括号，这样就会先算小括号里的减法，再算括号外的乘法，最后除法。原算式中70-4需要先算，所以添加括号后算式应改为360/[(70-4)×16]。360÷(70-4×16))时，先算乘法，再算减法，最后算除法。如果要先算减法，再算乘法，最后算除法，算式应改为360/[(70-4)×16]。",
+        "knowledge_points":  [
+                                 "有括号的运算顺序",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·河南许昌·期末"
+    },
+    {
+        "id":  "428",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "\"六一\"儿童节，某书店推出\"买五送一\"促销活动。在这个书店购买12本单价是25元的故事书，一共需要()元。",
+        "options":  [
+
+                    ],
+        "answer":  "250",
+        "solution":  "【分析】\"买五送一\"也就是买(5+1)本只要付5本的钱，用12除以(5+1)求出需要买几个(5+1)本，每本的单价是25元,25乘5即等于买一个((5+1))本的钱，再乘买(5+1))本的个数，即等于买12本故事书实际需要付的钱，据此即可解答.\n【详解】12÷(5+1)\n:=12÷6\n:=2(个)\n25×5×2=250(元)\n在这个书店购买12本单价是25元故事书，一共需要250元。",
+        "knowledge_points":  [
+                                 "促销问题",
+                                 "带有小括号的混合运算",
+                                 "经济问题"
+                             ],
+        "source":  "24-25四年级下·云南昭通·期末"
+    },
+    {
+        "id":  "429",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "请你用3,6,8,9这四个数写出一道得数是24的综合算式(可以添括号)()。",
+        "options":  [
+
+                    ],
+        "answer":  "(9-8+3)×6=24÷(9-8+3)×6=24",
+        "solution":  "【分析】要使用3,6,8,9这四个数，通过加，减，乘，除运算(可添括号)得到24。需结合数字特点，尝试不同组合与运算顺序，利用24的常见构成方式(如3×8=24,4×6=24,12×2=24等)进行推导.\n【详解】观察数字3,6,8,9，可先尝试围绕\"3×8=24\"，构建算式：需用6-3得到3,(9-6)×8=24，还有数字3没有用，因为3÷3=1,(9-6)×8再去乘3÷3不影响最后结果，因此算式为:(9-6)×8×(3÷3)=24。围绕\"4×6=24\"尝试：需用3,8,9得到4，因为4×6=24。计算9-8+3=4，因此算式为:(9-8+3)×6=24.4×6=24。得数是24的综合算式可以是(9-6)×8×(3÷3)=24或(9-8+3)×6=240",
+        "knowledge_points":  [
+                                 "24点"
+                             ],
+        "source":  "24-25四年级下·贵州遵义·期末"
+    },
+    {
+        "id":  "430",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据14+26=40,80÷40=2,100-2=98，合并成综合算式是()。",
+        "options":  [
+
+                    ],
+        "answer":  "100-80÷(14+26)=98",
+        "solution":  "【分析】分步算式的最后一步是减法，分别找出被减数和减数，被减数是100，减数是2，由80÷(14+26)得到，综合算式是100-80÷(14+26)=98.\n【详解】根据分析：先算加法，再算除法，最后算减法，综合算式是100-80÷(14+26)=98。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·浙江宁波·期末"
+    },
+    {
+        "id":  "431",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据17×45=765写出两道除法算式是()和()。",
+        "options":  [
+
+                    ],
+        "answer":  "765÷17=45;765÷45=17",
+        "solution":  "【分析】根据乘法各部分之间的关系：积=因数×因数;因数=积/另一个因数;据此可解此题.\n【详解】由分析可知：根据17×45=765写出两道除法算式是765÷17=45和765÷45=17。",
+        "knowledge_points":  [
+                                 "乘，除法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·内蒙古赤峰·期末"
+    },
+    {
+        "id":  "432",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在一次排查检修中，电工师傅需要排查728户的线路，上午排查了139户，下午排查了161户，还剩()户没有排查。",
+        "options":  [
+
+                    ],
+        "answer":  "428",
+        "solution":  "【分析】总共有728户需要排查，上午和下午分别排查了139户和161户。要求剩下的户数，需要用总户数减去上午和下午排查的户数.\n【详解】728-139-161\n:=589-161\n:=428(户)\n在一次排查检修中，电工师傅需要排查728户的线路，上午排查了139户，下午排查了161户，还剩428户没有排查。",
+        "knowledge_points":  [
+                                 "两，三位数的加减混合运算",
+                                 "加，减法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·甘肃定西·期末"
+    },
+    {
+        "id":  "433",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "李阿姨买了三袋同样的大米共重60千克，共花费300元。这种大米的单价可以表示为()。",
+        "options":  [
+
+                    ],
+        "answer":  "5元/千克",
+        "solution":  "【分析】根据总价=单价×数量，三袋同样的大米共重60千克，共花费300元，用总价300元除以60千克，即可求得这种大米的单价.\n【详解】300÷60=5(元/千克),因此，这种大米的单价可以表示为5元/千克。",
+        "knowledge_points":  [
+                                 "经济问题"
+                             ],
+        "source":  "24-25四年级下·河南洛阳·期末"
+    },
+    {
+        "id":  "434",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把228÷19=12,15×12=180,180-42=138，列成综合算式是()。",
+        "options":  [
+
+                    ],
+        "answer":  "15×(228÷19)-42=138",
+        "solution":  "【分析】根据题意，先算除法，再算乘法，最后算减法。把15×12里面的12替换成228÷19，因为要使除法先算，所以这里加上小括号。然后再把180-42里的180替换成15×(228÷19)。据此解答.\n【详解】根据分析，把228÷19=12,15×12=180,180-42=138，列成综合算式是15×(228÷19)-42=138。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·辽宁盘锦·期末"
+    },
+    {
+        "id":  "435",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "红红在计算8+ □ *5时，弄错了运算顺序，按先算加法后算乘法的顺序计算，结果得数是100，正确的得数应该是()。",
+        "options":  [
+
+                    ],
+        "answer":  "68",
+        "solution":  "【分析】错误的运算是(8+□)×5=100，由此用100除以5，再减去8求出方框里面的数，然后按照正确的运算顺序求出正确的得数即可.\n【详解】错误的运算是(8+□)×5=100，所以□|=100÷5-8。□:100÷5-8=20-8=12;8+12×5=8+60=68所以正确的得数应该是68。",
+        "knowledge_points":  [
+                                 "有括号的运算顺序",
+                                 "除加，除减混合运算",
+                                 "乘加，乘减混合运算"
+                             ],
+        "source":  "24-25四年级下·浙江绍兴·期末"
+    },
+    {
+        "id":  "436",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据:42-35=7,7×5=35,420÷35=12，可以列出综合算式:()。",
+        "options":  [
+
+                    ],
+        "answer":  "420/[(42-35)×5]=12",
+        "solution":  "【分析】先算减法，再算乘法，最后计算除法,420是被除数。列综合算式时给减法加上小括号，给减法和乘法加上中括号即可.\n【详解】420/[(42-35)×5] = 420/[7×5] = 420÷35 = 12。因此根据:42-35=7,7×5=35,420÷35=12，可以列出综合算式:420/[(42-35)×5]=12。",
+        "knowledge_points":  [
+                                 "带有中括号的混合运算",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·浙江绍兴·期末"
+    },
+    {
+        "id":  "437",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "20个小朋友可以平分一些苹果，后来又来了5个小朋友，每人分到的苹果比原来少了2个。这些苹果一共有()个。",
+        "options":  [
+
+                    ],
+        "answer":  "200",
+        "solution":  "【分析】根据题意，已知20个小朋友可以平分一些苹果，后来又来了5个小朋友，每人分到的苹果比原来少了2个。原来20个小朋友每人少分2个，共少分20×2=40(个)苹果。这些苹果被新来的5个小朋友平分，每人分到40÷5=8(个),因此用现在的人数乘8，就是总苹果数，列式计算即可.\n【详解】根据分析可知:20×2÷5×(20+5) = 40÷5×25 = 8×25 = 200(个)。这些苹果一共有200个。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·云南楚雄·期末"
+    },
+    {
+        "id":  "438",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "将算式4×16=64,90-64=26,390÷26=15改写成一个综合算式是()。",
+        "options":  [
+
+                    ],
+        "answer":  "390÷(90-4×16)",
+        "solution":  "【分析】分析这三个算式，是要先计算4×16的积，再用90减4×16的积，先把这两个算式合并为90-4×16，最后用390除以这个算式的结果，需给这个算式加上括号.\n【详解】根据分析，写成综合算式为:390÷(90-4×16)。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "23-24四年级下·云南昭通·期末"
+    },
+    {
+        "id":  "439",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把①5×7=105,②420-105=315,③315÷5=63三个算式合并成一个综合算式是()。",
+        "options":  [
+
+                    ],
+        "answer":  "(420-15×7)÷5=63",
+        "solution":  "【分析】观察三个算式，我们发现第一个算式15×7=105的结果105在第二个算式420-105=315中作为减数。第二个算式的结果315又在第三个算式315÷5=63中作为被除数。为了保证运算顺序正确，先算乘法15×7，再用420减去其结果，最后将所得差除以5。所以合并后的综合算式为(420-15×7)÷5=63.\n【详解】根据题目要求，将三个算式合并时需要保持原有运算顺序，先算乘法，再算减法，最后算除法。因此综合算式为:(420-15×7)÷5=63。所以,①5×7=105,②420-105=315,③315÷5=63，三个算式合并成一个综合算式是((420-15×7)÷5=630(答案不唯一)",
+        "knowledge_points":  [
+                                 "乘加，乘减混合运算",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·云南大理·期末"
+    },
+    {
+        "id":  "440",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在快递驿站凭借取件码精准地领取快递。已知取件码\"1-3-6050\"表示\"星期六的第50个快递在第1个货柜的第3层\"。有一份星期一的第19个快递，在第2个货柜的第4层，这份快递的取件码是:()。",
+        "options":  [
+
+                    ],
+        "answer":  "2-4-1019",
+        "solution":  "【分析】已知取件码\"1-3-6050\"表示\"星期六的第50个快递在第1个货柜的第3层\"，即第一个数字表示货柜号，第二个数字表示层数，第三个数字表示星期几，第四个数字是0，第五，六个数字表示第几个快递，所以有一份星期的第19个快递，在第2个货柜的第4层中的2是第一个数字,4是第二个数字，星期一代表第三个数字是1，第19个快递代表第五，六个数字是19，因此取件码是2-4-1019.\n【详解】取件码的第一个数字是2，第二个数字是4，第三个数字是1，第五，六个数字是19。因此，有一份星期一的第19个快递，在第2个货柜的第4层，这份快递的取件码是:2-4-1019。",
+        "knowledge_points":  [
+                                 "数字编码问题"
+                             ],
+        "source":  "24-25四年级下·河南洛阳·期末"
+    },
+    {
+        "id":  "441",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "乐乐在计算三位数乘两位数时，把两位数\"62\"十位上的\"6\"错看成了\"9\"，结果比正确的积多了3780，这个三位数是(),正确的积是()。",
+        "options":  [
+
+                    ],
+        "answer":  "126;7812",
+        "solution":  "【分析】根据题意，乐乐把两位数\"62\"十位上的\"6\"看成了\"9\"，导致了这个两位数增加了92-62=30，而积增加了3780，那么这多出来的积就是增加的30与三位数相乘的结果，即这个三位数=多出来的积/因数增加的30，算出三位数后再用正确的三位数乘62，即可算出正确的积，据此解答.\n【详解】3780÷(92-62)=3780÷30=126126×62=7812 即乐乐在计算三位数乘两位数时，把两位数\"62\"十位上的\"6\"错看成了\"9\"，结果比正确的积多了3780，这个三位数是(126),正确的积是(7812)",
+        "knowledge_points":  [
+                                 "三位数与两位数的乘法",
+                                 "除数是整十数的笔算除法",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·河南洛阳·期末"
+    },
+    {
+        "id":  "442",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "某品牌家电周年庆活动：购买家电单价满4000元减325元。国家节能家电补贴政策：节能空调每台还可补贴(优惠)275元。该品牌某款节能空调原价4699元，王阿姨想购买一台这款空调，需付()元。",
+        "options":  [
+
+                    ],
+        "answer":  "40;99",
+        "solution":  "【分析】购买家电单价满4000元减325元，且节能空调每台还可补贴275元，只需要将4699元减325元与275元的和即可.\n【详解】4699-(325+275)=4699-600=4099(元)答：阿姨想购买一台这款空调，需付4099元。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·河南郑州·期末"
+    },
+    {
+        "id":  "443",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算360/[3×(101-89)]时，应先算()法，再算()法，最后算()法，结果是()。",
+        "options":  [
+
+                    ],
+        "answer":  "减;乘;除;10",
+        "solution":  "【分析】整数四则混合运算顺序：先算小括号内的减法，再算中括号内的乘法，最后算中括号外的除法.\n【详解】360/[3×(101-89)]=360/[3×12]=360÷36=10。计算360/[3×(101-89)]时，应先算减法，再算乘法，最后算除法，结果是10。",
+        "knowledge_points":  [
+                                 "有括号的运算顺序",
+                                 "带有中括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·河南信阳·期末"
+    },
+    {
+        "id":  "444",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据算式16-8=8,24÷8=3,3×5=15，列出综合算式为()。",
+        "options":  [
+
+                    ],
+        "answer":  "24÷(16-8)×5=15",
+        "solution":  "【分析】根据题目中的三个算式，需要将减法，除法和乘法合并成一个综合算式。首先，第二个算式中的8由第一个算式得到，因此用(16-8)代替;第三个算式中的3由第二个算式得到，因此用24÷(16-8)代替。最后将这三个步骤整合，注意运算顺序，确保减法优先计算，再依次进行除法和乘法.\n【详解】根据16-8=8，将第二个算式中的8替换为(16-8),得到24÷(16-8)=324/;再将第三个算式中的3替换为24÷(16-8),得到24÷(16-8)×5。运算顺序为：先算括号内的减法，再算除法，最后算乘法，结果为15。因此，综合算式为:24÷(16-8)×5=15。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·吉林四平·期末"
+    },
+    {
+        "id":  "445",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "商场促销，一种单价3元的笔记本，活动期间买3本送1本;朱老师想买20本这样的笔记本，最少需要花()元。",
+        "options":  [
+
+                    ],
+        "answer":  "45",
+        "solution":  "【分析】笔记本买3本送1本，也就是买(3+1)本只需要付3本的钱，那就把(3+1)本看作一组，先用20除以(3+1)求出需要买几组，再用组数乘3求出需要付钱的本数，然后再乘笔记本的价格即可求出要花的钱数.\n【详解】20本中包含的组数:20÷(3+1)=20÷4=5(组)。需付钱的笔记本数量:5×3=15(本)。总费用:15×3=45(元)。所以最少需要花45元。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "经济问题"
+                             ],
+        "source":  "24-25四年级下·重庆·期末"
+    },
+    {
+        "id":  "446",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把\"84-24=60,420÷60=7,8×7=56\"改写成综合算式是()。",
+        "options":  [
+
+                    ],
+        "answer":  "8*[420÷(84-24)]=56或8*[420÷(84-24)]",
+        "solution":  "【分析】根据题意可知，这道综合算式先算减法，再算除法，最后算乘法，那么把减法放在小括号内，再把小括号和除法放在中括号内即可.\n【详解】8*[420÷(84-24)]=8*[420÷60]=8×7=56。把\"84-24=60,420÷60=7,8×7=56\"改写成综合算式是8*[420÷(84-24)]=56。",
+        "knowledge_points":  [
+                                 "带有中括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·重庆·期末"
+    },
+    {
+        "id":  "447",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "已知两个因数的积是504，其中一个因数是36，另一个因数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "14",
+        "solution":  "【分析】已知两个因数的积和其中一个因数，根据\"因数=积/另一个因数\"的关系，用除法计算即可.\n【详解】504÷36=14，所以已知两个因数的积是504，其中一个因数是36，另一个因数是14。",
+        "knowledge_points":  [
+                                 "乘，除法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·重庆·期末"
+    },
+    {
+        "id":  "448",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "用2,4,6,7通过运算不可能得到24。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】把24先分解成两个数的和，差，商，积的形式，然后再通过已知的数字，尝试调整凑成得数是24，即可解答.\n【详解】7×4-(6-2)=28-4=24。即用2,4,6,7通过运算可以得到24;原题说法不正确。故答案为: ×",
+        "knowledge_points":  [
+                                 "24点",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·浙江宁波·期末"
+    },
+    {
+        "id":  "449",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "知夏在计算(10+4)×8时错写成10+4×8，得到的结果与正确结果相差70。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】根据题意，正确计算(10+4)×8时，先算括号内的加法，再算乘法;错误的算式10+4×8则先算乘法，再算加法。分别计算两种算式的结果，再求差值即可判断.\n【详解】(10+4)×8=14×8=112;10+4×8=10+32=42;112-42=70。知夏在计算(10+4)×8时错写成10+4×8，得到的结果与正确结果相差70。说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "乘加，乘减混合运算",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·山东日照·期末"
+    },
+    {
+        "id":  "450",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "在有意义且没有余数的除法算式里，如果被除数不为0，那么被除数/(除数×商)=1。()。",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】根据题意，在整除的除法中，被除数=除数×商。将这一关系代入被除数/(除数×商)中，即可得出结果。以此判断即可.\n【详解】根据分析可知：根据除法算式的关系，被除数=除数×商。因此，被除数/(除数×商)=(除数×商)÷(除数×商)=1。说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "乘，除法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·云南楚雄·期末"
+    },
+    {
+        "id":  "451",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "先列出综合算式，再计算。用182除以13的商，去乘28与13的差，积是多少。",
+        "options":  [
+
+                    ],
+        "answer":  "(182÷13)×(28-13)=210",
+        "solution":  "【分析】问积是多少，也就是最后算乘法。一个乘数是182除以13的商，另一个乘数是28与13的差。因为要先计算出商和差，所以给182÷13和28-13都加上括号.\n【详解】根据分析列式如下:(182÷13)×(28-13)=14×15=210。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "452",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "只列综合算式不计算.\n\"鼓声起，龙舟动，端午至\",2025年端午节，永州各地开展了赛龙舟比赛，道县非遗龙舟比赛中一共有146支参赛队伍，比零陵区参赛队伍的10倍少14支，零陵区有多少支参赛队伍。",
+        "options":  [
+
+                    ],
+        "answer":  "(146+14)÷10",
+        "solution":  "【分析】根据题意，道县非遗龙舟比赛中一共有146支参赛队伍，比零陵区参赛队伍的10倍少14支。给道县参赛队伍加上14，则正好是零陵区参赛队伍的10倍，用除法求出零陵区参赛队伍.\n【详解】(146+14)÷10=160÷10=16(支)。答：零陵区有16支参赛队伍。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25六年级下·湖南永州·期末"
+    },
+    {
+        "id":  "453",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一列火车通过440米的桥需要50秒，以同样的速度穿过310米的隧道需要40秒。这列火车的速度和车长各是多少。",
+        "options":  [
+
+                    ],
+        "answer":  "速度:13米/秒;车长:210米",
+        "solution":  "【分析】通过440米的桥和通过310米的隧道都要计算车长，抵消掉。过桥比过隧道多用了50-40=10(秒),多走了440-310=130(米)。用多走的路程除以多用的时间求速度;再用速度乘通过桥所用的时间得通过桥的路程，与桥长作差得到车长.\n【详解】(440-310)÷(50-40)=130÷10=13(米/秒)。13×50-440=650-440=210(米)。答：这列火车的速度是13米/秒;车长是210米。",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "带有小括号的混合运算",
+                                 "乘加，乘减混合运算"
+                             ],
+        "source":  "24-25四年级下·山东临沂·期末"
+    },
+    {
+        "id":  "454",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "龙门书店举办\"六一\"促销活动,《小学生优秀作文选》买5本送1本，这种《小学生优秀作文选》每本12元，万老师想买13本，需要花多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "132元",
+        "solution":  "【分析】买5本送1本相当于6本为一组，付钱的是5本的价格。先用13÷6确定有多少组，再根据余下的本数单独计算，最后乘每本12元得到总价.\n【详解】13÷(5+1)=13÷6=2(组)。1(本)。2×5×12+12=10×12+12=120+12=132(元)。答：需要花132元。",
+        "knowledge_points":  [
+                                 "两位数除以一位数的有余数除法",
+                                 "乘加，乘减混合运算",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·辽宁鞍山·期末"
+    },
+    {
+        "id":  "455",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一位消费者通过电商\"爱心助农\"活动下单了34千克沃柑。怎样包装最省运费。运费最少是多少钱。大号纸箱：每箱8千克，运费24元 小号纸箱：每箱5千克，运费20元",
+        "options":  [
+
+                    ],
+        "answer":  "3个大号纸箱和2个小号纸箱;112元",
+        "solution":  "【分析】分别求大号纸箱和小号纸箱的单位运费，运费少的更省钱。包装时尽量选择单位运费更低的纸箱，并使装箱组合刚好覆盖34千克且运费最小.\n【详解】24÷8=3(元),20÷5=4(元),4\u003e3，所以大号纸箱更省钱。34÷8=4(个)。2(千克)。若用4个大号纸箱和1个小号纸箱装不满，比较后选择3个大号纸箱和2个小号纸箱:24×3+20×2=72+40=112(元)。答：用3个大号纸箱和2个小号纸箱最省运费，运费最少112元。",
+        "knowledge_points":  [
+                                 "租船问题",
+                                 "经济问题"
+                             ],
+        "source":  "24-25四年级下·山东临沂·期末"
+    },
+    {
+        "id":  "456",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一列火车通过440米的桥需要50秒，以同样的速度穿过310米的隧道需要40秒。这列火车的速度和车长各是多少。",
+        "options":  [
+
+                    ],
+        "answer":  "13米/秒;210米",
+        "solution":  "【分析】通过桥比穿过隧道多走440-310=130(米),多用50-40=10(秒),所以速度=130÷10=13(米/秒)。车长=通过桥所行驶路程-桥长=13×50-440.\n【详解】440-310=130(米),130÷(50-40)=130÷10=13(米/秒)。13×50-440=650-440=210(米)。答：这列火车的速度是13米/秒，车长是210米。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "火车过桥问题"
+                             ],
+        "source":  "24-25四年级下·山东临沂·期末"
+    },
+    {
+        "id":  "457",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "为助力社区净化空气，绿植店特推出两款吸碳绿植：白掌和吊兰。已知店内现有白掌35盆，每盆售价18元;吊兰18盆，每盆售价15元。这两种绿植一共能卖多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "900元",
+        "solution":  "【分析】分别计算白掌总价和吊兰总价，再相加得到总售卖额.\n【详解】35×18+18×15=630+270=900(元)。答：这两种绿植一共能卖900元。",
+        "knowledge_points":  [
+                                 "乘加，乘减混合运算",
+                                 "经济问题"
+                             ],
+        "source":  "24-25四年级下·山东临沂·期末"
+    },
+    {
+        "id":  "458",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一列火车通过440米的桥需要50秒，以同样的速度穿过310米的隧道需要40秒。这列火车的速度和车长各是多少。",
+        "options":  [
+
+                    ],
+        "answer":  "火车的速度13米/秒，车长210米",
+        "solution":  "【分析】火车通过桥比穿过隧道多走440-310=130(米),多用50-40=10(秒),速度=130÷10=13(米/秒)。再用速度乘桥上用时50秒求路程，再减去桥长得到车长.\n【详解】(440-310)÷(50-40)=130÷10=13(米/秒)。13×50-440=650-440=210(米)。答：火车的速度是13米/秒，车长是210米。",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·山东临沂·期末"
+    },
+    {
+        "id":  "459",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "修路队要修一条长800米的路，原计划每天修45米，修了10天后，接到通知要提前完成，剩下的要5天修完，剩下的平均每天要比原计划多修多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "25米",
+        "solution":  "【分析】先求剩下的路程=全长-原计划每天修的米数×已修天数。再用剩下路程/剩下天数求平均每天修多少米，最后与原计划每天修的45米比较求多修多少.\n【详解】(800-45×10)÷5-45=(800-450)÷5-45=350÷5-45=70-45=25(米)。答：剩下的平均每天要比原计划多修25米。",
+        "knowledge_points":  [
+                                 "有具体量的工程问题",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·湖北孝感·期末"
+    },
+    {
+        "id":  "460",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "玩具厂接到一批玩具加工订单，原计划每天生产150个，可以按时完成任务。实际每天多生产30个，结果只用25天就完成订单任务。原计划完成订单任务需要多少天。",
+        "options":  [
+
+                    ],
+        "answer":  "30天",
+        "solution":  "【分析】先算实际每天生产数=150+30，再算总数量=实际每天生产数*25;最后用总数量/原计划每天生产数求原计划天数.\n【详解】(150+30)×25÷150=180×25÷150=4500÷150=30(天)。答：原计划完成订单任务需要30天。",
+        "knowledge_points":  [
+                                 "有具体量的工程问题",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25六年级下·河北唐山·期末"
+    },
+    {
+        "id":  "461",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一位消费者通过电商\"爱心助农\"活动下单了34千克沃柑。怎样包装最省运费。运费最少是多少钱。大号纸箱：每箱8千克，运费24元 小号纸箱：每箱5千克，运费20元\n![](images/2026-03-06%20102152.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "装3箱大号纸箱,2箱小号纸箱最省运费;112元",
+        "solution":  "【分析】先算大号纸箱和小号纸箱的单位运费，比较可知大号更合算。再考虑余数，分别尝试组合装箱，比较总运费，选择最小方案.\n【详解】24÷8=3(元),20÷5=4(元),4\u003e3，所以大号纸箱更合算。34÷8=4(箱)。2(千克)。装4箱大号纸箱,1箱小号纸箱:4×24+20=116(元);装3箱大号纸箱,2箱小号纸箱:3×24+2×20=72+40=112(元)。116\u003e112。答：装3箱大号纸箱,2箱小号纸箱最省运费，运费最少112元。",
+        "knowledge_points":  [
+                                 "租船问题"
+                             ],
+        "source":  "24-25四年级下·山东临沂·期末"
+    },
+    {
+        "id":  "462",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "六一文艺演出，学校采购了35套演出服。其中一件上衣需要56元，一条裤子需要44元，一共需要花费多少钱。\n(1)算式(56+44)×35，先算56+44，也就是先求出().\n(2)除了上述方法，你还有别的方法解决问题吗。请列式计算并解答。",
+        "options":  [
+
+                    ],
+        "answer":  "一套演出服的价格;见详解;3500元",
+        "solution":  "【分析】(1)用上衣价格加裤子价格，求出一套演出服的单价;再乘35套即可。(2)也可以分别求上衣总价和裤子总价，再相加.\n【详解】(1)56+44=100(元)。算式:(56+44)×35，先算56+44，也就是先求出一套演出服的价格.\n(2)56×35+44×35=1960+1540=3500(元)。答：可以用上衣和裤子的价分别乘35套，分别求出购买上衣和裤子的总价，再相加，即可求出一共需要多少元钱。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "乘加，乘减混合运算",
+                                 "经济问题"
+                             ],
+        "source":  "24-25四年级下·江西南昌·期末"
+    },
+    {
+        "id":  "463",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "学校要做42套演出服，要按整卷买布。大卷布，每卷300元，可以做8套演出服;小卷布，每卷200元，可以做5套演出服。怎么买布最省钱。",
+        "options":  [
+
+                    ],
+        "answer":  "4卷大卷布,2卷小卷布",
+        "solution":  "【分析】先比较大卷布和小卷布做一套演出服平均所需的钱，钱少的便宜。购买时尽量选择便宜的布，并保证数量刚好能做42套且不多余.\n【详解】300÷8=37(元)。4(元);200÷5=40(元)。40\u003e37，大卷更便宜。42÷8=5(卷)。2(套)。若买5卷大卷布再买1卷小卷布会有多余;买4卷大卷布再买2卷小卷布刚好做42套:8×4+5×2=32+10=42(套),300×4+200×2=1200+400=1600(元)。答：买4卷大卷布,2卷小卷布最省钱。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "租船问题"
+                             ],
+        "source":  "24-25四年级下·内蒙古赤峰·期末"
+    },
+    {
+        "id":  "464",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "张叔叔开车从甲城去乙城，平均每小时行驶110千米。他行驶了2小时后，距离中点还有70千米(还未到中点)。甲，乙两城相距多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "580千米",
+        "solution":  "【分析】行驶2小时共走110×2=220千米，距离中点还剩70千米，所以甲到乙一半距离是220+70，再乘2即为总路程.\n【详解】(110×2+70)×2=(220+70)×2=290×2=580(千米)。答：甲，乙两城相距580千米。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25四年级下·贵州黔东南·期末"
+    },
+    {
+        "id":  "465",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "某车间要生产一批零件，原计划每天生产200个,15天完成任务。实际每天生产250个，实际完成的天数比原计划少多少天。",
+        "options":  [
+
+                    ],
+        "answer":  "3天",
+        "solution":  "【分析】先算总量=原计划每天生产数×原计划天数;再算实际天数=总量/实际每天生产数;最后求天数差=原计划天数-实际天数.\n【详解】15-200×15÷250=15-3000÷250=15-12=3(天)。答：实际完成的天数比原计划少3天。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·江西南昌·期末"
+    },
+    {
+        "id":  "466",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "六一儿童节，四(1)班10个家庭共18个大人和12个小孩到某游乐场游玩。门票价格：成人35元/人，儿童15元/人，团体票(20人及以上)25元/人。他们怎样购票合算。最少要花多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "18个大人和2个小孩购买团体票，剩余的购买儿童票;650元",
+        "solution":  "【分析】比较三种方案：方案一分别买成人票和儿童票;方案二全部买团体票;方案三让18个大人和2个小孩凑够团体票20人，剩余儿童买儿童票，比较总价找最省钱.\n【详解】方案一:18×35=630(元),12×15=180(元),总费用=630+180=810(元).\n方案二:18+12=30(人),总费用=30×25=750(元).\n方案三:18+2=20(人),20×25=500(元)。剩余儿童12-2=10(人),10×15=150(元)。总费用=500+150=650(元)。810\u003e750\u003e650。答:18个大人和2个小孩购买团体票，剩余的购买儿童票，这样购票合算，共要花650元。",
+        "knowledge_points":  [
+                                 "购票问题"
+                             ],
+        "source":  "24-25四年级下·湖北孝感·期末"
+    },
+    {
+        "id":  "467",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "某旅行社推出\"一日游\"两种方案。方案一：成人每人150元，儿童每人60元;方案二：团体10人以上(含10人),每人100元。现有成人4人，儿童6人，怎样购票最省钱。",
+        "options":  [
+
+                    ],
+        "answer":  "方案一成人购买成人票和儿童购买儿童票最省钱。",
+        "solution":  "【分析】方案一分别购买成人票和儿童票;方案二全部购买团体票，需要的人数为成人4人+儿童6人=10人，比较两种方案花费哪种少.\n【详解】方案一:150×4+60×6=600+360=960(元)。方案二:100×10=1000(元)。960元\u003c1000元，所以选方案一。答：方案一成人购买成人票和儿童购买儿童票最省钱。",
+        "knowledge_points":  [
+                                 "经济问题"
+                             ],
+        "source":  "24-25四年级下·河北张家口·期末"
+    },
+    {
+        "id":  "468",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "学校图书馆有科技书280本，故事书的本数比科技书的3倍还多20本。学校图书馆有故事书多少本。",
+        "options":  [
+
+                    ],
+        "answer":  "860本",
+        "solution":  "【分析】故事书=科技书的3倍+20。先算280×3得到3倍，再加20求故事书数量.\n【详解】280×3+20=840+20=860(本)。答：学校图书馆有故事书860本。",
+        "knowledge_points":  [
+                                 "乘加，乘减混合运算",
+                                 "求一个数的几倍是多少"
+                             ],
+        "source":  "24-25四年级下·黑龙江佳木斯·期末"
+    },
+    {
+        "id":  "469",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "师徒二人加工同一种零件，师傅每小时加工56个，徒弟每小时加工44个，师傅工作了4小时，徒弟工作了6小时，师徒二人一共做了多少个零件。",
+        "options":  [
+
+                    ],
+        "answer":  "488个",
+        "solution":  "【分析】分别求师傅做的零件数和徒弟做的零件数，再相加.\n【详解】56×4+44×6=224+264=488(个)。答：师徒二人一共做了488个零件。",
+        "knowledge_points":  [
+                                 "有具体量的工程问题"
+                             ],
+        "source":  "24-25四年级下·贵州黔南·期末"
+    },
+    {
+        "id":  "470",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "甲，乙两个\"啦啦操\"队共有队员150人，如果从乙队调15人到甲队，则甲队人数正好是乙队的2倍，甲，乙两队原来各有队员多少人。",
+        "options":  [
+
+                    ],
+        "answer":  "甲队85人;乙队65人",
+        "solution":  "【分析】根据题意，已知甲，乙两个\"啦啦操\"队共有队员150人，如果从乙队调15人到甲队，则甲队人数正好是乙队的2倍，也就是150是乙队的2+1=3倍，用150除以3，求出此时乙队的人数，再加上15，就是原乙队的人数;用150减去原乙队的人数，就是原甲队的人数，列式计算即可.\n【详解】根据分析可知:150÷(2+1)=150÷3=50(人) 50+15=65(人) 150-65=85(人) 答：甲队原来有队员85人，乙队原来有队员65人。",
+        "knowledge_points":  [
+                                 "和差倍问题",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·湖北武汉·期末"
+    },
+    {
+        "id":  "471",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "工厂生产一批零件，原计划每天生产210个,25天可以完成。实际上采用了新技术，每天可以多生产40个，实际需要多少天完成。",
+        "options":  [
+
+                    ],
+        "answer":  "21天",
+        "solution":  "【分析】根据工作量/工作效率=工作时间，工作量=工作效率×工作时间，用计划每天生产零件的个数乘天数求出零件总数，实际每天生产零件的个数即210+40，再用总零件数除以实际每天生产零件的个数，可求出实际生产的天数.\n【详解】总零件:210×25=5250(个)实际天数:5250÷(210+40)=5250÷250=21(天)答：实际需要21天。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "有具体量的工程问题"
+                             ],
+        "source":  "24-25四年级下·湖北武汉·期末"
+    },
+    {
+        "id":  "472",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一辆汽车从甲地开往乙地，每小时行60千米,4小时到达。返回时每小时行80千米，需要几小时。",
+        "options":  [
+
+                    ],
+        "answer":  "3小时",
+        "solution":  "【分析】用路程=速度×时间可以算出甲地到乙地的路程，用路程除以返回时的速度即可得到返程的时间.\n【详解】60×4÷80=240÷80=3(小时)答：需要3小时。",
+        "knowledge_points":  [
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25四年级下·黑龙江佳木斯·期末"
+    },
+    {
+        "id":  "473",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "学校图书馆新购了科技书和故事书共480本，科技书的本数是故事书的3倍。科技书和故事书各有多少本。",
+        "options":  [
+
+                    ],
+        "answer":  "科技书360本;故事书120本",
+        "solution":  "【分析】因为科技书的本数是故事书的3倍，所以把故事书的数量看作1份，科技书的数量就是3份，那么两种书的总数就是故事书的:3+1=4倍;已知两种书共480本，且总数是故事书的4倍，所以用总本数除以4求出故事书的数量;因为科技书的本数是故事书的3倍，所以用故事书的本数乘3求出科技书的数量.\n【详解】480÷(3+1)=480÷4=120(本)120×3=360(本)答：科技书是360本，故事书是120本。",
+        "knowledge_points":  [
+                                 "和差倍问题",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·黑龙江佳木斯·期末"
+    },
+    {
+        "id":  "474",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "在六一游园活动中，乐乐和欢欢共获得积分150分，如果乐乐送给欢欢15个积分，两人的积分就同样多了。原来乐乐和欢欢各有多少分。",
+        "options":  [
+
+                    ],
+        "answer":  "乐乐90分;欢欢60分",
+        "solution":  "【分析】乐乐送给欢欢15分，两人的积分就同样多了，说明原来乐乐比欢欢多15×2=30分;原来欢欢的积分=(两人总积分-原来乐乐比欢欢多的积分)÷2，乐乐原来的积分=共获得的积分-原来欢欢的积分，据此作答.\n【详解】(150-15×2)÷2\n=(150-30)÷2\n=120÷2\n=60(分)\n150-60=90(分)\n答：原来乐乐有90分，欢欢有60分。",
+        "knowledge_points":  [
+                                 "和差倍问题",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·浙江杭州·期末"
+    },
+    {
+        "id":  "475",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "甲，乙两个工程队同时从两端开始修一条长5600米的公路，修了5天后还剩200米没有修，已知甲队每天修600米，乙队每天修多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "480米",
+        "solution":  "【分析】首先，公路总长5600米，还剩200米没修，那么用总长度减去没修的长度，就能得到两队5天一共修的长度，即5600-200=5400米。然后，已知两队修了5天，根据\"工作效率=工作总量/工作时间\"，用两队5天修的总长度除以5天，可算出两队每天一共修的长度为5400÷5=1080米。最后，因为已知甲队每天修600米，那么用两队每天一共修的长度减去甲队每天修的长度，就得到乙队每天修的长度.\n【详解】(5600-200)÷5\n:=5400÷5\n:=1080(米)\n1080-600=480(米)\n答：乙队每天修480米。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "两人合作的工程问题"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "476",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "优优旅行社推出\"风景区一日游\"的两种价格方案，方案一：成人每人120元，儿童每人60元;方案二：团体8人以上(包括8人),每人80元。现在有成人3人，儿童5人，选择哪种方案更合算。",
+        "options":  [
+
+                    ],
+        "answer":  "方案二",
+        "solution":  "【分析】方案一的总价=成人的人数×成人票单价+儿童人数×儿童票单价，方案二的总价=(成人的人数+儿童人数)×团体票单价;分别计算出两种方案的总价后再比较大小.\n【详解】方案一:120×3+60×5\n:=360+300\n:=660(元)\n方案二:(3+5)×80\n:=8×80\n:=640(元)\n660\u003e640\n答：选择方案二合算。",
+        "knowledge_points":  [
+                                 "购票问题"
+                             ],
+        "source":  "24-25四年级下·浙江嘉兴·期末"
+    },
+    {
+        "id":  "477",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "使用DeepSeekR1模型可以辅助编程，已知该模型在处理复杂编程任务时，平均每分析10行代码需要120秒。知夏爸爸用它分析150行代码，大约需要多少秒。",
+        "options":  [
+
+                    ],
+        "answer":  "1800秒",
+        "solution":  "【分析】已知平均每分析10行代码需要120秒，用除法先计算出分析1行代码需要的时间，再乘要分析的150行代码，即可得到总时间.\n【详解】120÷10×150\n:=12×150\n:=1800(秒)\n答：大约需要1800秒。",
+        "knowledge_points":  [
+                                 "乘，除法的意义和各部分间的关系",
+                                 "单归一问题"
+                             ],
+        "source":  "24-25四年级下·山东日照·期末"
+    },
+    {
+        "id":  "478",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "福泉市双谷果园场的王叔叔家去年收获了梨和葡萄共2680千克，梨每50千克装一筐，共装了28筐。葡萄每40千克装一筐，葡萄要装多少筐。",
+        "options":  [
+
+                    ],
+        "answer":  "32筐",
+        "solution":  "【分析】已知梨和葡萄总重量为2680千克，梨每筐装50千克，共装28筐。用乘法先计算出梨的总重量，再用总重量减去梨的重量得到葡萄的重量，最后用葡萄的重量除以每筐装的40千克，得到葡萄需要装的筐数.\n【详解】(2680-50×28)÷40\n=(2680-1400)÷40\n=1280÷40\n=32(筐)\n答：葡萄要装32筐。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·贵州黔南·期末"
+    },
+    {
+        "id":  "479",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "学校图书馆购进科技书180本，比购进的故事书的4倍少20本。学校图书馆购进故事书多少本。",
+        "options":  [
+
+                    ],
+        "answer":  "50本",
+        "solution":  "【分析】根据题意，已知科技书180本，比购进的故事书的4倍少20本。先用180加上20，求出故事书的4倍的数量，再除以4，就是故事书的数量;列式计算即可.\n【详解】根据分析可知:\n(180+20)÷4\n:=200÷4\n:=50(本)\n答：学校图书馆购进故事书50本。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·黑龙江佳木斯·期末"
+    },
+    {
+        "id":  "480",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "某超市第一季度平均每月盈利12万元，第二季度比第一季度多盈利8万元，全年预计盈利多少万元。",
+        "options":  [
+
+                    ],
+        "answer":  "80万元",
+        "solution":  "【分析】根据题意，明确1季度有3个月，已知某超市第季度平均每月盈利12万元，第二季度比第一季度多盈利8万元，用12乘3，求出第一季度的盈利;用12加上8，求出第二季度的盈利;最后把两个季度的盈利相加，列式计算即可.\n【详解】根据分析可知:\n12×3+(12×3+8)\n:=36+(36+8)\n:=36+44\n:=80(万元)\n答：全年预计盈利80万元。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·黑龙江佳木斯·期末"
+    },
+    {
+        "id":  "481",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "丽丽和文文到文具店买文具，丽丽买了7个笔记本花了63元，文文买了4个同样的笔记本和一个35元的文具盒，文文一共花了多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "71元",
+        "solution":  "【分析】根据题意，丽丽买了7个笔记本花了63元，先用63÷7求出一本笔记本的价格，文文买了4个同样的笔记本和一个35元的文具盒，用一本笔记本的价格乘4再加上文具盒的价格，即可求出文文一共花了多少钱.\n【详解】63÷7×4+35\n:=9×4+35\n:=36+35\n:=71(元)\n答：文文一共花了71元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "乘加",
+                                 "乘减混合运算"
+                             ],
+        "source":  "24-25四年级下·甘肃陇南·期末"
+    },
+    {
+        "id":  "482",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "四年级225名学生和10名老师研学活动，需要租车。每辆大客车限载客50人，租金450元;每辆小客车限载客40人，租金400元。怎样租车最省钱。最少需要多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "租4辆大客车和1辆小客车最省钱;2200元",
+        "solution":  "【分析】根据题意，先用450÷50求出大客车的人均租金，再用400÷40求出小客车的人均租金，尽量租便宜的，用225加上10，求出总人数;用总人数除以限乘人数，求出的商即为需要租的辆数，余数即为剩下的人数，看是否正好可以租另一种，用每种车的辆数乘每辆车的租金，最后相加即可求出需要多少钱。以此答题即可.\n【详解】根据分析可知:\n450÷50=9(元)\n400÷40=10(元)\n9\u003c10，尽量租大客车\n225+10=235(人)\n235÷50=4(辆)。35(人)\n租4辆大客车,1辆小客车\n450×4+400×1=1800+400=2200(元)\n答：租4辆大客车和1辆小客车最省钱，最少需要2200元。",
+        "knowledge_points":  [
+                                 "租船问题"
+                             ],
+        "source":  "24-25四年级下·湖北武汉·期末"
+    },
+    {
+        "id":  "483",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "每年的3月12日是植树节，某小学准备组织三，四年级的学生到婺源县江湾镇荷田村参加社会实践。已知三年级13个班，四年级17个班，每班有学生45人。如果每名学生种植一棵树苗，学校一共要准备多少棵树苗。",
+        "options":  [
+
+                    ],
+        "answer":  "1350棵",
+        "solution":  "【分析】根据题意，每名学生种植一棵树苗，先用三年级班级的个数加上四年级班级的个数，求出一共有多少个班，再乘每班有的学生人数，即可求出学校一共要准备多少棵树苗.\n【详解】(13+17)×45\n:=30×45\n:=1350(棵)\n答：学校一共要准备1350棵树苗。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·江西上饶·期末"
+    },
+    {
+        "id":  "484",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "博物馆买票，方案一：成人每人40元，儿童每人20元。方案二：团体票10人或10人以上可以买团体票，每人30元。(1)如果有7个大人和3个儿童，按哪种方案买票合算。(2)如果有3个大人和7个儿童，该怎样买票合算。",
+        "options":  [
+
+                    ],
+        "answer":  "方案二;方案一",
+        "solution":  "【分析】(1)如果有7个大人和3个儿童，方案一：需要的钱数=大人人数×成人票单价+儿童人数×儿童票单价;方案二：总人数为7+3=10(人),满足团体票要求，则需要的钱数=总人数×团体票单价;比较这两种方案需要钱数的大小即可解答。(2)如果有3个大人和7个儿童，方案一：需要的钱数=大人人数×成人票单价+儿童人数×儿童票单价;方案二：总人数为3+7=10(人),满足团体票要求，则需要的钱数=总人数×团体票单价;比较这两种方案需要钱数的大小即可解答.\n【详解】(1)方案一:40×7+20×3\n:=280+60\n:=340(元)\n方案二:(7+3)×30\n:=10×30\n:=300(元)\n340元\u003e300元\n答：如果有7个大人和3个儿童，按方案二买票合算.\n(2)方案一:40×3+20×7\n:=120+140\n:=260(元)\n方案二:(3+7)×30\n:=10×30\n:=300(元)\n260元\u003c300元\n答：如果有3个大人和7个儿童，按方案一买票合算。",
+        "knowledge_points":  [
+                                 "购票问题",
+                                 "带有小括号的混合运算",
+                                 "乘加",
+                                 "乘减混合运算"
+                             ],
+        "source":  "24-25四年级下·黑龙江佳木斯·期末"
+    },
+    {
+        "id":  "485",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "中国结艺是我国特有的民间手工编结艺术，具有独特的东方神韵和丰富多彩的变化，充分体现了人民的智慧和深厚的文化底蕴。手工小组要制作90个同样的中国结，有两种规格的绳子可以选择，大卷每卷20元，正好可以做8个中国结;小卷每卷15元，正好可以做5个中国结。怎样购买绳子最省钱。(按整卷购买)",
+        "options":  [
+
+                    ],
+        "answer":  "买10大卷绳子和2小卷绳子最省钱",
+        "solution":  "【分析】由题意得，大卷每卷20元，可以做8个中国结，每个中国结需要:20÷8=2(元)。4(元)。小卷每卷15元，可以做5个中国结，每个中国结需要:15÷5=3(元)。对比可知，在没有剩余的情况下，用大卷绳子做中国结更划算。一共要做90个中国结，直接用90除以8算出需要买大卷绳子的数量以及剩余的中国结数量。然后合理调整买大卷绳子的数量和小卷绳子的数量使得总价钱最便宜即可.\n【详解】20÷8=2(元)。4(元)15÷5=3(元)对比可知，在没有剩余的情况下，用大卷绳子做中国结更划算。90÷8=11(卷)。2(个),即买11大卷绳子后，还剩余2个中国结没做。这2个中国结可以买1小卷绳子来做。也可以少买1大卷绳子，让剩余的2个中国结和8个中国结一起用小卷绳子来做。方案一：买11大卷绳子和1小卷绳子，所需钱数为:20×11+15=220+15=235(元)方案二:11-1=10(卷)8+2=10(个)10÷5=2(卷),即买10大卷绳子和2小卷绳子。所需钱数为:20×10+15×2=200+30=230(元)235\u003e230 答：买10大卷绳子和2小卷绳子最省钱。",
+        "knowledge_points":  [
+                                 "租船问题"
+                             ],
+        "source":  "24-25四年级下·贵州遵义·期末"
+    },
+    {
+        "id":  "486",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "在学校体育节比赛中，四(1)班的学生表现出色，在篮球，跑步等多个项目中斩获佳绩，张老师准备买一些体育用品奖励给同学们。①买了6副羽毛球拍，花了480元。②每筒羽毛球36元。③又买了25筒羽毛球，每筒12个。④营业员找他20元。请你根据上面的信息提出一个两步或两步以上的数学问题并解答。(问题不唯一)我选择的信息是: ( )(填序号),提出的问题是: ( )。解答:",
+        "options":  [
+
+                    ],
+        "answer":  "①②③;买6副羽毛球拍和25筒羽毛球一共花了多少钱(问题不唯一);1380元",
+        "solution":  "【分析】由题意可知：要提出一个两步或两步以上的数学问题，可以选择①②③三个条件，根据这三个条件可算出的问题是：买6副羽毛球拍和25筒羽毛球一共花多少钱。用每筒羽毛球的钱×买羽毛球的筒数=买25筒羽毛球的钱;用买6副羽毛球拍花的钱+买25筒羽毛球花的钱=一共花的钱;据此解答.\n【详解】由分析可得:\n我选择的信息是:①②③\n提出的问题是：买6副羽毛球拍和25筒羽毛球一共花了多少钱。(问题不唯一)\n480+25×36=480+900=1380(元)\n答：买6副羽毛球拍和25筒羽毛球一共花了1380元钱。",
+        "knowledge_points":  [
+                                 "\"提问题\"\"填条件\"问题",
+                                 "两位数乘两位数的进位乘法",
+                                 "乘加",
+                                 "乘减混合运算"
+                             ],
+        "source":  "24-25四年级下·贵州遵义·期末"
+    },
+    {
+        "id":  "487",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "甲，乙两辆车分别从A,B两地同时相对开出，甲车每小时行驶54千米，乙车每小时行驶65千米，经过3小时后两车还相距170千米,A,B两地相距多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "527千米",
+        "solution":  "【分析】首先根据速度×时间=路程，用两车的速度之和乘3，求出两车3小时行驶的路程之和是多少;然后用它加上170千米，求出A,B两地相距多少千米即可.\n【详解】A,B相距距离:\n(54+65)×3+170\n:=119×3+170\n:=357+170\n:=527(千米)\n答:A,B两地相距527千米。",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·甘肃平凉·期末"
+    },
+    {
+        "id":  "488",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "为了更好地开展垃圾分类，某社区规定：每次正确投放垃圾可获得10分积分，每次错误投放垃圾倒扣5分积分。今年三月份，平平家每天投放一次垃圾，正确投放了28次，共获得多少分积分。",
+        "options":  [
+
+                    ],
+        "answer":  "265分",
+        "solution":  "【分析】三月份有31天，平平家每天投放一次垃圾，正确投放了28次，所以错误投放的次数=三月份的总天数-投放正确的次数。每次正确投放垃圾可获得10分积分，每次错误投放垃圾倒扣5分积分。所以共获得积分数=投放正确的次数*10-投放错误的次数*5;据此列式解答即可.\n【详解】31-28=3(次)\n28×10-3×5=280-15=265(分)\n答：共获得265分积分。",
+        "knowledge_points":  [
+                                 "年，月，日的认识及换算",
+                                 "乘加",
+                                 "乘减混合运算"
+                             ],
+        "source":  "24-25四年级下·内蒙古赤峰·期末"
+    },
+    {
+        "id":  "489",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "商店有105个羽毛球，把75个羽毛球平均装进5个盒子里，照这样计算，装完剩下的羽毛球还需要几个盒子。",
+        "options":  [
+
+                    ],
+        "answer":  "2个",
+        "solution":  "【分析】根据题意，已知商店有105个羽毛球，把75个羽毛球平均装进5个盒子里,75除以5等于一盒装羽毛球的个数,105除以一盒装羽毛球的个数等于需要的总盒数，再减已经装的5盒，即等于还需要的盒数，列式计算即可.\n【详解】根据分析可知:\n105÷(75/5)-5=105÷15-5=7-5=2(个)\n答：装完剩下的羽毛球还需要2个盒子。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·内蒙古赤峰·期末"
+    },
+    {
+        "id":  "490",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "星光超市是位于市中心的一家大型连锁超市，为了优化商品陈列，超市用分层货架设计，其中饮料区有一个三层立体货架，每层根据销售数据动态调整库存。已知上层摆放了157瓶饮料，中层摆放了126瓶饮料，下层摆放了174瓶饮料。这个货架上一共摆放了多少瓶饮料。",
+        "options":  [
+
+                    ],
+        "answer":  "457瓶",
+        "solution":  "【分析】题目要求计算货架三层饮料的总数量，只需将上层，中层，下层的饮料瓶数相加即可.\n【详解】157+126+174\n:=283+174\n:=457(瓶)\n答：这个货架上一共摆放了457瓶饮料。",
+        "knowledge_points":  [
+                                 "两，三位数的连续进位加法",
+                                 "两，三位数的连加运算"
+                             ],
+        "source":  "24-25四年级下·甘肃定西·期末"
+    },
+    {
+        "id":  "491",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "纪录片《脉动泰山》中记录了挑山工攀登十八盘的故事。泰山十八盘台阶约有1600级，远远望去，恰似天门云梯。在攀登时，挑山工15分钟爬了300级台阶，按照这样的速度，爬完1600级台阶还需要多少分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "65分钟",
+        "solution":  "【分析】首先计算挑山工的速度，挑山工15分钟爬了300级台阶，总台阶数为1600级，已爬300级，求出剩余台阶数，最后用剩余台阶数除以速度得到所需时间.\n【详解】(1600-300)÷(300/15)=1300÷20=65(分钟),答：爬完1600级台阶还需要65分钟。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·甘肃定西·期末"
+    },
+    {
+        "id":  "492",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "四年级师生共105人去参加社会实践活动，租大客车，每辆最多坐40人，租金1000元，小客车每辆最多坐25人，租金650元，怎样租车最省钱。最少需要多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "租两辆大客车和一辆小客车;2650元",
+        "solution":  "【分析】由题意得，大客车每辆最多坐40人，租金1000元，那么每人需要:1000÷40=25(元)。小客车每辆最多坐25人，租金650元，那么每人需要:650÷25=26(元)。对比可知，大客车更便宜。在没有空位的情况下，应该尽量多租大客车，直接用总人数除以40算出需要租大客车的数量以及多余的人数。然后合理安排大客车和小客车的数量使得总租金最少即可.\n【详解】1000÷40=25(元)\n650÷25=26(元)\n25\u003c26，所以大客车更便宜.\n105÷40=2(辆)。25(人),即租2辆大客车后还剩下25人。25÷25=1(辆),这25人可以坐1辆大客车，也可以坐1辆小客车。但因为小客车比大客车便宜，所以租1辆小客车更划算.\n1000×2+650=2000+650=2650(元)\n答：租两辆大客车和一辆小客车最省钱，最少需要2650元。",
+        "knowledge_points":  [
+                                 "租船问题"
+                             ],
+        "source":  "24-25四年级下·黑龙江佳木斯·期末"
+    },
+    {
+        "id":  "493",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "某超市举行促销活动，牛奶3元1瓶，买4瓶送1瓶，妈妈选购了26瓶，需要付多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "63元",
+        "solution":  "【分析】根据题意，买4瓶送1瓶，得到5瓶的里面是有1瓶送的。用26瓶除以5算出有这样的几组，就有几瓶是送的。再用26减去送的瓶数，就是实际付钱的瓶数。单价×数量=总价，最后用实际付钱的瓶数乘3就是实际需要付多少钱.\n【详解】4+1=5(瓶)\n26÷5=5(组)。1(瓶)\n5×1=5(瓶)\n(26-5)×3=21×3=63(元)\n答：需要付63元。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "经济问题"
+                             ],
+        "source":  "24-25四年级下·贵州黔西·期末"
+    },
+    {
+        "id":  "494",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "动物园推出\"一日游\"活动，有以下两种购票方案：方案一：成人每人140元，儿童每人60元。方案二：团体5人以上(包括5人),每人100元。现有成人4人，儿童6人，选择哪种方案划算。",
+        "options":  [
+
+                    ],
+        "answer":  "方案一",
+        "solution":  "【分析】分别计算两种方案的总费用并进行比较。方案按成人和儿童票价分开计算，分别用相应的人数乘相应的票价，再相加求出总费用;方案二按团体票价计算，先将成人与儿童人数相加求出总人数，看是否达到或超过5人(包括5人),够的话再用总人数乘团体票价求出总费用。最后比较两种方案的费用，哪个低就是哪个划算.\n【详解】方案一费用：成人费用:140×4=560(元)儿童费用:60×6=360(元)总费用:560+360=920(元)\n方案二费用：总人数:4+6=10(人)团体费用:100×10=1000(元)920元\u003c1000元答：选择方案一划算。",
+        "knowledge_points":  [
+                                 "购票问题",
+                                 "经济问题"
+                             ],
+        "source":  "24-25四年级下·内蒙古通辽·期末"
+    },
+    {
+        "id":  "495",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "辽西走廊是连接东北与华北的咽喉要道，也是中国历史上农耕文明，游牧文明与渔猎文明的交汇地带，长约200公里。王叔叔和李叔叔骑行穿越辽西走廊，王叔叔5小时骑行完全程，李叔叔每小时比王叔叔少骑行15公里，李叔叔骑完全程需要多少小时。",
+        "options":  [
+
+                    ],
+        "answer":  "8小时",
+        "solution":  "【分析】根据路程=时间×速度，速度=路程/时间，用200÷5可以求出王叔叔每小时骑行的公里，李叔叔每小时比王叔叔少骑行15公里，再用王叔叔的速度减15可以求出李叔叔的速度，用时间=路程/速度，求出李叔叔骑完全程的时间.\n【详解】200÷(200/5-15)=200÷(40-15)=200÷25=8(小时)答：李叔叔骑完全程需要8小时。",
+        "knowledge_points":  [
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25四年级下·云南楚雄·期末"
+    },
+    {
+        "id":  "496",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "阳光小学的大礼堂有40排座位，每排都有36个座位，学校对大礼堂进行了改造，改造后每排的座位增加到45个，一共增加了多少个座位。",
+        "options":  [
+
+                    ],
+        "answer":  "360个",
+        "solution":  "【分析】用改造后每排的座位数减去改造前每排的座位数，求出每排增加的座位数，再用每排增加的座位数乘总排数，即可求出一共增加了多少个座位.\n【详解】(45-36)×40=9×40=360(个) 答：一共增加了360个座位。",
+        "knowledge_points":  [
+                                 "乘加",
+                                 "乘减混合运算"
+                             ],
+        "source":  "24-25四年级下·云南昭通·期末"
+    },
+    {
+        "id":  "497",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一本科普书亮亮用了11天看完，前5天每天看16页，后6天每天看20页。这本科普书共有多少页。",
+        "options":  [
+
+                    ],
+        "answer":  "200页",
+        "solution":  "【分析】由题意得，一本科普书亮亮用了11天看完。前5天每天看16页，后6天每天看20页，可以用16乘5算出前5天看的页数，然后再用20乘6算出后6天看的页数，最后再把两部分的页数加起来即可算出这本科普书共有多少页.\n【详解】16×5+20×6=80+120=200(页) 答：这本科普书共有200页。",
+        "knowledge_points":  [
+                                 "乘加",
+                                 "乘减混合运算"
+                             ],
+        "source":  "24-25四年级下·云南昭通·期末"
+    },
+    {
+        "id":  "498",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "(1)\"粽子\"是节庆食物之一。端午前夕，学校\"味典小厨\"社团学生邀请老师，家长和他们一起包粽子。从下午4:20开始，到5:40结束。包的粽子放了102盆，平均每盆放15个。算一算，一共包了多少个粽子。(2)她们给敬老院送去八宝粽和蛋黄粽一共225个，其中蛋黄粽比八宝粽少45个。两种粽子各送了多少个。",
+        "options":  [
+
+                    ],
+        "answer":  "1530个;蛋黄粽:90个;八宝粽:135个",
+        "solution":  "【分析】(1)由题意得，包的粽子放了102盆，平均每盆放15个，那么直接用102乘15即可算出一共包了多少个粽子。(2)由题意得，她们给敬老院送去八宝粽和蛋黄粽一共225个，其中蛋黄粽比八宝粽少45个。可以先用225减去45算出送出蛋黄粽个数的2倍是多少，然后再除以2即可算出送出蛋黄粽有多少个。最后再用225减去蛋黄粽的个数即可算出送出八宝粽有多少个.\n【详解】(1)102×15=1530(个)答：一共包了1530个粽子.\n(2)(225-45)÷2\n:=180÷2\n:=90(个)\n225-90=135(个)\n答：蛋黄粽送了90个，八宝粽送了135个。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "三位数乘两位数，三位数中间有0",
+                                 "和差倍问题"
+                             ],
+        "source":  "24-25四年级下·河南洛阳·期末"
+    },
+    {
+        "id":  "499",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "商场推出了两种蛋糕购买方案。方案一：巧克力蛋糕250元/个，水果蛋糕200元/个。方案二：购买8个以上(含8个，不分种类)220元/个。如果要购买巧克力蛋糕5个，水果蛋糕3个，选哪种方案更省钱。省了多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "方案二;90元",
+        "solution":  "【分析】根据总价=单价×数量，假设选方案一，用巧克力蛋糕每个的价格乘买的个数，求出买巧克力蛋糕用多少元，用水果蛋糕每个的价格乘买的个数，求出买水果蛋糕用多少元，相加即可求出一共需要多少元;假设选方案二,5+3=8(个),一共购买8个蛋糕，购买8个以上(含8个，不分种类)220元/个，用220×8即可求出一共需要多少元，比较两个方案用的钱数即可找出哪种方案最省钱，用多的减去少的即可求出省了多少钱.\n【详解】方案一:250×5+200×3\n:=1250+600\n:=1850(元)\n方案二:(5+3)×220\n:=8×220\n:=1760(元)\n1850-1760=90(元)\n答：选方案二更省钱，省了90元。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "乘加",
+                                 "乘减混合运算",
+                                 "经济问题"
+                             ],
+        "source":  "24-25四年级下·甘肃天水·期末"
+    },
+    {
+        "id":  "500",
+        "unit":  "第一单元 四则运算",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "山陕会馆在节假日对于购票方案做了一些调整。导游李阿姨带了一个小型旅游团，其中有6个大人和4个孩子，以下的两种购票方案哪一种更划算。(导游持证不用购票)方案一：成人每人80元，儿童每人50元。方案二:10人以上(含10人)可购买团体票，每人60元。",
+        "options":  [
+
+                    ],
+        "answer":  "方案二",
+        "solution":  "【分析】根据题意，用成人票的价格×成人人数+儿童票的价格×儿童的人数，计算出方案一的钱数;用总人数×团体票价格，计算出方案二的钱数，再比较即可.\n【详解】方案一:6×80+4×50\n:=480+200\n:=680(元)\n方案二:(6+4)×60\n:=10×60\n:=600(元)\n680元\u003e600元\n答：方案二更划算。",
+        "knowledge_points":  [
+                                 "购票问题",
+                                 "乘加",
+                                 "乘减混合运算"
+                             ],
+        "source":  "24-25四年级下·河南南阳·期末"
+    }
+]
diff --git "a/\345\233\233\345\271\264\347\272\247\351\242\230\351\233\206/\345\233\233\345\271\264\347\272\247\344\270\213/\345\233\233\345\271\264\347\272\247\344\270\213\347\254\254\344\270\203\345\215\225\345\205\203\345\233\276\345\275\242\347\232\204\350\277\220\345\212\250.json" "b/\345\233\233\345\271\264\347\272\247\351\242\230\351\233\206/\345\233\233\345\271\264\347\272\247\344\270\213/\345\233\233\345\271\264\347\272\247\344\270\213\347\254\254\344\270\203\345\215\225\345\205\203\345\233\276\345\275\242\347\232\204\350\277\220\345\212\250.json"
new file mode 100644
index 0000000000000000000000000000000000000000..8a874912f8ed34366967671c19392b184de5be70
--- /dev/null
+++ "b/\345\233\233\345\271\264\347\272\247\351\242\230\351\233\206/\345\233\233\345\271\264\347\272\247\344\270\213/\345\233\233\345\271\264\347\272\247\344\270\213\347\254\254\344\270\203\345\215\225\345\205\203\345\233\276\345\275\242\347\232\204\350\277\220\345\212\250.json"
@@ -0,0 +1,8405 @@
+[
+    {
+        "id":  "1",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "如图，图形A可通过()得到图形B。![](images/eab15f8c1001bd6169c6109e2a8eb01767bb193d45e93171a6300bca94d77429.jpg)",
+        "options":  [
+                        "A. 向下平移3格，再向右平移5格",
+                        "B. 向右平移3格，再向下平移3格",
+                        "C. 向左平移3格，再向上平移3格"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】从图形A某一个顶点开始数起，到对应的图形B另一个顶点的位置，就是图形平移的格数.\n【详解】图形A通过向右平移3格，再向下平移3格得到图形B。故答案选:B",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "2",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "在方格纸上平移长方形ABCD,如果顶点A平移了4格，那么顶点B平移了()格。",
+        "options":  [
+                        "A. 4",
+                        "B. 6",
+                        "C. 8"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】在平面内，把一个图形整体沿某条直线方向平行移动的过程，称为平移。物体或图形平移后，它们的形状，大小，方向都不改变，只是位置发生了变化.\n【详解】根据平移的定义，原图形与平移后的图形各个对应点的移动距离是相同的，所以长方形顶点A平移了4格，顶点B也平移了4格。故答案为:A",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "3",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "轴对称在古代哲学中反映了\u0027天人合一\u0027思想，下列具有轴对称的汉字是()。",
+        "options":  [
+                        "A. 中",
+                        "B. 国",
+                        "C. 制",
+                        "D. 造"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】轴对称图形的定义：如果一个图形沿着一条直线对折后两部分完全重合，那么这个图形就是轴对称图形.\n【详解】A。沿竖直中线对折，左右两部分能完全重合，是轴对称图形。B。内部\u0027玉\u0027部分笔画不对称，对折后无法完全重合，不是轴对称图形。C。左边的和右边的\u0027刂\u0027对折后无法重合，不是轴对称图形。D。结构不对称，对折后无法完全重合，不是轴对称图形。故答案为:A",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "24-25五年级上·重庆九龙坡·期末"
+    },
+    {
+        "id":  "4",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列图形中对称轴数量最少的是()。",
+        "options":  [
+                        "A. 长方形",
+                        "B. 正方形",
+                        "C. 底角为30°的等腰三角形",
+                        "D. 等边三角形"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】将图形沿着一条直线对折，如果直线两侧的部分能够完全重合，折痕所在的直线叫做它的对称轴，根据对称轴定义找出各图形对称轴的数量.\n【详解】A。长方形有2条对称轴;B。正方形有4条对称轴;C。底角为30°的等腰三角形有1条对称轴;D。等边三角形有3条对称轴。综上，对称轴数量最少的是底角为30°的等腰三角形。故答案为:C",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "25-26五年级上·内蒙古包头·期末"
+    },
+    {
+        "id":  "5",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "在下面图形中，总共有4条对称轴的图形是()。",
+        "options":  [
+                        "A. 圆",
+                        "B. 正方形",
+                        "C. 等边三角形",
+                        "D. 平行四边形"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】轴对称图形：一个图形沿一条直线折叠，直线两旁的部分能够完全重合的图形。直线叫做对称轴，据此确定各项对称轴的数量.\n【详解】A。圆有无数条对称轴;B。正方形有4条对称轴;C。等边三角形有3条对称轴;D。平行四边形没有对称轴。综上，总共有4条对称轴的图形是正方形。故答案为:B",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "24-25五年级上·重庆合川·期末"
+    },
+    {
+        "id":  "6",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面的图形中对称轴最多的是()。",
+        "options":  [
+                        "A. 等边三角形",
+                        "B. 梯形",
+                        "C. 正方形",
+                        "D. 长方形"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】根据轴对称图形定义，比较各图形对称轴数量.\n【详解】A。等边三角形有3条;B。梯形最多1条(仅等腰梯形);C。正方形有4条;D。长方形有2条。对称轴最多的是正方形。故答案为:C",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "24-25五年级上·重庆梁平·期中"
+    },
+    {
+        "id":  "7",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "选择题",
+        "difficulty":  0.94,
+        "question":  "下面数字中，有两条对称轴的是()。",
+        "options":  [
+                        "A. 3",
+                        "B. 6",
+                        "C. 8"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】依据对称轴定义，逐项判断对称轴条数.\n【详解】A。3有1条;B。6没有;C。8有2条。故答案为:C",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "25-26五年级上·河南商丘·期中"
+    },
+    {
+        "id":  "8",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "把一张长方形的纸沿着同一方向对折，一共对折3次，展开后这张长方形纸上有()条折痕。",
+        "options":  [
+                        "A. 8",
+                        "B. 7",
+                        "C. 6"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】可通过实际操作画图解答.\n【详解】第一次对折:![](images/7800d29cbb01ba4a927db7fac89abddbec089546544d386918a85eb14265e333.jpg) 第二次对折:![](images/b6af01ff4ff4eb018c7b735c77d590795c2cf83b904178e03012c1a6908837b5.jpg) 第三次对折:![](images/8d48c130d5a82af5ffbad9a4b54fde5f46929904337b3ccd864ebcece6f0be81.jpg) 把一张长方形的纸沿着同一方向对折，一共对折3次，展开后有7条折痕。故答案为:B",
+        "knowledge_points":  [
+                                 "图形的折叠问题"
+                             ],
+        "source":  "25-26五年级上·河南商丘·期中"
+    },
+    {
+        "id":  "9",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "选择题",
+        "difficulty":  0.94,
+        "question":  "等边三角形有()条对称轴。",
+        "options":  [
+                        "A. 4",
+                        "B. 3",
+                        "C. 2",
+                        "D. 1"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】依据轴对称图形概念判断.\n【详解】画图可知:![](images/79c3110a1d0b959c90353de3878b2874db64ba5c1774640b2e62df3b499f6726.jpg) 等边三角形有3条对称轴。故答案为:B",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "25-26五年级上·重庆潼南·期中"
+    },
+    {
+        "id":  "10",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "选择题",
+        "difficulty":  0.94,
+        "question":  "下面图形中对称轴最多的是()。",
+        "options":  [
+                        "A. 圆",
+                        "B. 正方形",
+                        "C. 长方形"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】根据对称轴定义分别判断数量.\n【详解】A。圆有无数条;B。正方形有4条;C。长方形有2条。因此对称轴最多的是圆。故答案为:A",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "25-26五年级上·河南驻马店·月考"
+    },
+    {
+        "id":  "11",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下列日常生活情景中，不属于平移的是()。",
+        "options":  [
+                        "A. 飞机在跑道上加速滑行",
+                        "B. 运行中的电梯",
+                        "C. 工作中的电扇",
+                        "D. 运动员在雪地上滑翔"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】平移是沿直线移动且形状，大小，方向不变;旋转是绕点或轴做圆周运动.\n【详解】A,B,D都属于平移;C电扇扇叶绕中心轴转动，属于旋转，不属于平移。故答案为:C",
+        "knowledge_points":  [
+                                 "旋转与旋转现象",
+                                 "平移与平移现象"
+                             ],
+        "source":  "24-25五年级上·广东揭阳·期中"
+    },
+    {
+        "id":  "12",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "在\u0027A,C,J,R\u0027这四个字母中，属于轴对称图形的共有()个。",
+        "options":  [
+                        "A. 1",
+                        "B. 2",
+                        "C. 3",
+                        "D. 4"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】将图形沿一条直线对折，若两侧完全重合，则为轴对称图形.\n【详解】如图:![](images/f8e99c5d943483d830ae811f89e942d586175fffb3e54c2ddac7bfb696fe1747.jpg) 在\u0027A,C,J,R\u0027中,A,C是轴对称图形，共2个。故答案为:B",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "25-26五年级上·广东佛山·期中"
+    },
+    {
+        "id":  "13",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "五角星有()条对称轴，等腰三角形有()条对称轴，长方形有()条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "5;1;2",
+        "solution":  "【分析】依据对称轴定义判断各图形对称轴数量.\n【详解】五角星有5条;等腰三角形有1条;长方形有2条。",
+        "knowledge_points":  [
+                                 "长方形的概念及特点",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "2026六年级下·全国·专题练习"
+    },
+    {
+        "id":  "14",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "长方形有()条对称轴，正方形有()条对称轴，圆有()条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "2;4;无数",
+        "solution":  "【分析】利用\u0027折叠后完全重合\u0027定义判断.\n【详解】长方形2条，正方形4条，圆有无数条。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "25-26六年级·全国·随堂练习"
+    },
+    {
+        "id":  "15",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小丽将一张纸对折后，剪出了一个轴对称的三角形，将三角形完全展开后，经测量发现，其中两条边的长度分别是7厘米和15厘米，另外一条边的长度是()厘米。(边长为整厘米。)",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】另一边可能为7或15;结合三角形三边关系判断。若为7，则7+7=14\u003c15，不成立;若为15，则满足两边和大于第三边，两边差小于第三边.\n【详解】因此另外一条边是15厘米。",
+        "knowledge_points":  [
+                                 "轴对称的剪纸问题",
+                                 "三角形三边关系",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "2025·浙江宁波·小升初模拟"
+    },
+    {
+        "id":  "16",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "如下图，图形()和图形()的面积相等，都是()cm²。![](images/e1031c56cacbcf48fe6673c025803acc16ac1c07c6d6b0deabc84ad7c1272b84.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "A;B;8",
+        "solution":  "【分析】数方格并结合面积公式:A可拼成长4宽2的长方形，面积8;B由8个1平方厘米小方格组成，面积8;C是三角形面积6.\n【详解】图形A和图形B面积相等，都是8cm²。",
+        "knowledge_points":  [
+                                 "利用平移法巧算面积",
+                                 "三角形面积的计算",
+                                 "长方形的面积"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "17",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "下面图形有几条对称轴。请把答案写在括号里。长方形有()条，正方形有()条，圆有()条。",
+        "options":  [
+
+                    ],
+        "answer":  "2;4;无数",
+        "solution":  "【分析】依据轴对称图形定义判断.\n【详解】如图:![](images/6f1351c88e1c0d9bc05b788fc7ac98b4.png) 长方形2条，正方形4条，圆无数条。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "25-26五年级上·广东韶关·期中"
+    },
+    {
+        "id":  "18",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "等腰三角形有()条对称轴，圆有()条对称轴，拉抽屉是()现象。",
+        "options":  [
+
+                    ],
+        "answer":  "1;无数;平移",
+        "solution":  "【分析】依据对称轴定义确定数量;沿直线运动属于平移.\n【详解】等腰三角形1条，圆无数条，拉抽屉是平移现象。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "平移与平移现象"
+                             ],
+        "source":  "24-25五年级上·江西赣州·期末"
+    },
+    {
+        "id":  "19",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在\u0027创意七巧板\u0027拼图比赛中，要求拼出的是轴对称图形。希希拼出一个有4条对称轴的图形，她拼的可能是()形;乐乐拼的图形只有1条对称轴，他拼的可能是()形。",
+        "options":  [
+
+                    ],
+        "answer":  "正方;等腰三角",
+        "solution":  "【分析】根据对称轴定义和常见图形对称轴条数判断。\n【详解】4条对称轴可能是正方形；1条对称轴可能是等腰三角形。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "25-26五年级上·河南商丘·期中"
+    },
+    {
+        "id":  "20",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "轴对称图形沿()对折，两侧的图形能够完全重合，长方形有()条对称轴。等边三角形有()条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "对称轴;2;3",
+        "solution":  "【分析】轴对称图形沿对称轴对折后两侧重合.\n【详解】如图:![](images/1ab4ad44d8eed5642cb50992a8c78f09.png) 轴对称图形沿对称轴对折，长方形有2条对称轴，等边三角形有3条对称轴。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "轴对称的认识及辨认",
+                                 "长方形的概念及特点"
+                             ],
+        "source":  "25-26五年级上·河南驻马店·月考"
+    },
+    {
+        "id":  "21",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "等边三角形与正方形都是轴对称图形，其中等边三角形有()条对称轴，正方形有()条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "3;4",
+        "solution":  "【分析】等边三角形每个顶点到对边中点可形成一条对称轴;正方形有两条中线和两条对角线为对称轴.\n【详解】等边三角形有3条，正方形有4条。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "正方形的概念及特点"
+                             ],
+        "source":  "25-26五年级上·广东佛山·期中"
+    },
+    {
+        "id":  "22",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个平面图形对折后两边完全重合，这个平面图形就是()图形，正方形就是这样的图形，它共有()条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "轴对称;4或四",
+        "solution":  "【分析】依据轴对称图形定义；正方形有4条对称轴（两条对边中点连线和两条对角线）。\n【详解】应填：轴对称;4或四。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "轴对称的认识及辨认",
+                                 "正方形的概念及特点"
+                             ],
+        "source":  "25-26五年级上·广东佛山·阶段练习"
+    },
+    {
+        "id":  "23",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在长方形，正方形，等边三角形，等腰梯形中，对称轴最多的是(),对称轴最少的是()。",
+        "options":  [
+
+                    ],
+        "answer":  "正方形;等腰梯形",
+        "solution":  "【分析】比较各图形对称轴数量：长方形2条，正方形4条，等边三角形3条，等腰梯形1条.\n【详解】对称轴最多的是正方形，最少的是等腰梯形。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "24-25五年级上·广东佛山·期中"
+    },
+    {
+        "id":  "24",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小智用AI绘画工具生成图案，他输入关键词“轴对称图形并且只有一条对称轴”后可能会出现什么图案。()",
+        "options":  [
+
+                    ],
+        "answer":  "等腰三角形、等腰梯形、半圆",
+        "solution":  "【分析】题目要求只有1条对称轴。等腰三角形、等腰梯形、半圆都符合。\n【详解】可能出现等腰三角形、等腰梯形、半圆。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "24-25四年级下·山西晋中·期末"
+    },
+    {
+        "id":  "25",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计数器上拨珠子，珠子的运动是();剪窗花的时候经常用到我们数学中()的方法。(填\u0027平移\u0027\u0027旋转\u0027或\u0027轴对称\u0027)",
+        "options":  [
+
+                    ],
+        "answer":  "平移;轴对称",
+        "solution":  "【分析】拨珠子沿直线移动属于平移;剪窗花常利用对折重合特征，属于轴对称.\n【详解】应填：平移;轴对称。",
+        "knowledge_points":  [
+                                 "平移与平移现象",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "26",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在字母\u0027L,M,N,D\u0027中，有()个是轴对称图形。",
+        "options":  [
+
+                    ],
+        "answer":  "2",
+        "solution":  "【分析】沿某直线对折两侧完全重合的是轴对称图形.\n【详解】如图:![](images/f0a196c510ed52786d31f35393e9a99c.png) 在\u0027L,M,N,D\u0027中，有2个是轴对称图形。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "24-25五年级下·河北保定·期中"
+    },
+    {
+        "id":  "27",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "等边三角形有()条对称轴，圆有()条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "3或三;无数",
+        "solution":  "【分析】等边三角形每个顶点到对边中点可形成一条对称轴;圆任意过圆心的直线都是对称轴.\n【详解】等边三角形有3条，圆有无数条。",
+        "knowledge_points":  [
+                                 "圆的概念及特点",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "24-25四年级下·广东云浮·期末"
+    },
+    {
+        "id":  "28",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "把五边形向左平移4格后，再向下平移4格，仍是一个()形。",
+        "options":  [
+
+                    ],
+        "answer":  "五边",
+        "solution":  "【分析】平移只改变位置，不改变图形大小，形状和方向.\n【详解】平移后仍是一个五边形。",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "24-25四年级下·云南红河·期末"
+    },
+    {
+        "id":  "29",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "三角形ABC和三角形A′B′C′是关于对称轴l对称的一组图形，三角形ABC中的点A到三角形A′B′C′中的对应点A′的距离是6厘米，那么三角形A′B′C′中的点A′到对称轴l的距离是()厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "3",
+        "solution":  "【分析】对称点连线被对称轴垂直平分。已知AA′=6厘米，则A′到对称轴距离是其一半。示意图:![](images/2026-03-13%20201550.jpg)\n【详解】6÷2=3(厘米),所以A′到对称轴l的距离是3厘米。",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "24-25四年级下·云南楚雄·期末"
+    },
+    {
+        "id":  "30",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "四边形的内角和是();正方形有()条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "360°;4",
+        "solution":  "【分析】四边形内角和为360°;正方形是轴对称图形，有4条对称轴.\n【详解】应填:360°;4。",
+        "knowledge_points":  [
+                                 "多边形的内角和",
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "24-25四年级下·河南信阳·期末"
+    },
+    {
+        "id":  "31",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "\u0027锄禾日当午，汗滴禾下土。\u0027古诗中\u0027土\u0027字如果看作一个图形，它有()条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "1",
+        "solution":  "【分析】沿某直线对折后两侧完全重合则该直线为对称轴.\n【详解】\u0027土\u0027字只有沿竖线对折可重合(图示:![](images/2026-03-13%20201651.jpg)),所以有1条对称轴。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "24-25四年级下·贵州铜仁·期末"
+    },
+    {
+        "id":  "32",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果一个轴对称图形沿对称轴对折后是一个直角三角形，这个轴对称图形可能是()形或()形。",
+        "options":  [
+
+                    ],
+        "answer":  "等腰三角;正方(答案不唯一)",
+        "solution":  "【分析】对折后为直角三角形，说明原图形由两个全等直角三角形拼成;可能情形对应等腰三角形或正方形，也可能出现其他四边形(示意:![](images/915aedf11204ae97b29e18ef53f07237fe85daf4c806f838f6a8f272bda61d22.jpg)).\n【详解】可能是等腰三角形或正方形。(答案不唯一)",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "24-25四年级下·湖北武汉·期末"
+    },
+    {
+        "id":  "33",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "“王”“中”“日”“目”都是轴对称图形，请你再写一个这样的汉字()。",
+        "options":  [
+
+                    ],
+        "answer":  "田",
+        "solution":  "【分析】轴对称图形是沿某条直线对折后两侧能完全重合的图形。\n【详解】可写“田”。",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "24-25四年级下·湖北咸宁·期末"
+    },
+    {
+        "id":  "34",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "三角形ABC在平移运动时，三个顶点都向相同的()移动了相同的()。",
+        "options":  [
+
+                    ],
+        "answer":  "方向;距离",
+        "solution":  "【分析】在平面内，把一个图形整体沿某条直线方向平行移动一定距离的过程，称为平移。根据题意作图如下:![](images/2026-03-13%20201855.jpg) 由图可知，三角形的三个顶点A、B、C都向右平移了8格，所以三个顶点移动的方向和距离都相同。\n【详解】三角形ABC在平移运动时，三个顶点都向相同的方向移动了相同的距离。",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "24-25四年级下·湖北武汉·期末"
+    },
+    {
+        "id":  "35",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "下面图形中，是轴对称图形的有()。(填写序号)①长方形②平行四边形③等腰三角形④直角梯形⑤圆",
+        "options":  [
+
+                    ],
+        "answer":  "①③⑤",
+        "solution":  "【分析】对折后两边完全重合的图形是轴对称图形，据此解答即可.\n【详解】长方形，等腰三角形，圆对折后能完全重合，是轴对称图形，平行四边形，直角梯形对折后不能完全重合，不是轴对称图形。所以是轴对称图形的有①③⑤。",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "24-25四年级上·湖南邵阳·期末"
+    },
+    {
+        "id":  "36",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在数字0,3,5,7,8中,()有对称轴,()没有对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "0,3,8;5,7",
+        "solution":  "【分析】对称轴是指将一个图形分成两部分，使得这两个部分关于这条轴线对称。这样的图叫轴对称图形.\n【详解】0,3,8是轴对称图形,5,7不是轴对称图形。所以0,3,8有对称轴,5,7没有对称轴。",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "24-25四年级下·山东菏泽·期末"
+    },
+    {
+        "id":  "37",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在轴对称图形中，对称轴两边对应点的连线和对称轴互相()。",
+        "options":  [
+
+                    ],
+        "answer":  "垂直",
+        "solution":  "【分析】如果一个图形沿着一条直线对折后两部分完全重合，这样的图形叫做轴对称图形，这条直线叫做对称轴;对称轴两边的对称点的连线和对称轴互相垂直。例如:A与A1是对称点,A与A1的连线和对称轴互相垂直。![](images/df3337829d1eb40979af02aee9371ac85e29325c0f5be4119ffdbf25df5b2d13.jpg)\n【详解】根据分析可知：在轴对称图形中，对称轴两边对应点的连线和对称轴互相垂直。",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认",
+                                 "垂直的特征"
+                             ],
+        "source":  "24-25四年级下·山东菏泽·期末"
+    },
+    {
+        "id":  "38",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在等腰梯形，等边三角形，正方形中，对称轴数量最多的是(),对称轴数量最少的是()。",
+        "options":  [
+
+                    ],
+        "answer":  "正方形;等腰梯形",
+        "solution":  "【分析】根据轴对称的定义可知，等腰梯形有1条对称轴，等边三角形有3条对称轴，正方形有4条对称轴。再比较大小.\n【详解】4\u003e3\u003e1，在等腰梯形，等边三角形，正方形中，对称轴数量最多的是正方形，对称轴数量最少的是等腰梯形。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "24-25四年级下·云南文山·期末"
+    },
+    {
+        "id":  "39",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "我们学过的平面图形中，等边三角形有()条对称轴，圆形有()条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "3;无数",
+        "solution":  "【分析】根据轴对称图形的定义：如果一个图形沿一条直线对折后，直线两旁的部分能够完全重合，那么这个图形叫做轴对称图形，这条直线就是它的对称轴.\n【详解】如图所示:![](images/61a833aa8a211c4a623e3dce8014e919163f9ab1d831b66a6564b63bb80096a1.jpg) 我们学过的平面图形中，等边三角形有3条对称轴，圆形有无数条对称轴。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "24-25四年级下·山东菏泽·期末"
+    },
+    {
+        "id":  "40",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "将一张正方形白纸连续对折3次，然后在中央剪出一个\u0027♡\u0027空洞，展开后共可得到()个\u0027♡\u0027。",
+        "options":  [
+
+                    ],
+        "answer":  "8",
+        "solution":  "【分析】将正方形白纸每对折一次，得到的层数就是对折前的2倍，连续对折3次，共得到(2×2×2)层，即8层。在中央剪出一个\u0027♡\u0027空洞后，每层均会留下一个\u0027♡\u0027,因此展开后共有8个\u0027♡\u0027.\n【详解】2×2×2=4×2=8(个)。所以展开后共可得到8个\u0027♡\u0027。",
+        "knowledge_points":  [
+                                 "图形的折叠问题"
+                             ],
+        "source":  "24-25四年级下·内蒙古兴安盟·期末"
+    },
+    {
+        "id":  "41",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形，若一个底角是45°,它的顶角是()°,它有()条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "90;1或一",
+        "solution":  "【分析】三角形的内角和为180°。等腰三角形的两个底角的度数相等。由题意得，等腰三角形的一个底角是45°，那么另一个底角也是45°，直接用180°减去两个底角的度数即可算出顶角的度数；一个图形沿一条直线对折，直线两旁的图形完全重合，这样的图形叫做轴对称图形，折痕所在的直线就是对称轴。\n【详解】180°-45°-45°=90°。作图如下:![](images/f853818ce72a05cf3180cc588083b3e1769a2a54525c4d5161c501c7e6e341a0.jpg) 由图可知，这个三角形有1条对称轴。所以顶角是90°，它有1条对称轴。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "22-23四年级下·新疆乌鲁木齐·期末"
+    },
+    {
+        "id":  "42",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "等腰三角形有()条对称轴，等边三角形有()条对称轴，长方形有()条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "1;3;2",
+        "solution":  "【分析】根据轴对称图形的定义判断。图示:![](images/10dd0a7a46391256cb26fb6c67422069a69a4b98dd656ab172a0bf2fc0dd3070.jpg)\n【详解】等腰三角形有1条对称轴，等边三角形有3条对称轴，长方形有2条对称轴。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "长方形的概念及特点"
+                             ],
+        "source":  "23-24四年级下·广东湛江·期末"
+    },
+    {
+        "id":  "43",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "长方形的内角和是(),它有()条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "360°(或360度);2",
+        "solution":  "【分析】多边形内角和为(n-2)×180°，长方形n=4，故内角和360°；依据轴对称定义判断长方形对称轴条数。\n【详解】(4-2)×180°=360°。长方形的对称轴如图:![](images/959255c9fa373f689348bc3aea0706ba4fa3147237f2de01fe531244dd4a1923.jpg) 所以长方形内角和是360°，有2条对称轴。",
+        "knowledge_points":  [
+                                 "多边形的内角和",
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "23-24四年级下·河北廊坊·期末"
+    },
+    {
+        "id":  "44",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "一个图形沿着一条直线对折后，两边的部分能够完全(),这个图形就是轴对称图形。",
+        "options":  [
+
+                    ],
+        "answer":  "重合",
+        "solution":  "【分析】根据轴对称图形的意义：如果一个图形沿着一条直线对折后两部分完全重合，这样的图形叫做轴对称图形，这条直线叫做对称轴.\n【详解】由分析可知，应填\u0027重合\u0027。",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "24-25四年级下·海南海口·单元测试"
+    },
+    {
+        "id":  "45",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "正三角形有()条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "3",
+        "solution":  "【分析】正三角形一般指等边三角形。等边三角形是轴对称图形.\n【详解】如图:![](images/ef834ac38251990e9d342b43781067939a2161a768fa4e8373fed58e3e06b903.jpg) 正三角形有3条对称轴。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "22-23四年级下·四川绵阳·期末"
+    },
+    {
+        "id":  "46",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形的底角是70°,它的顶角是()°,按角分它是个()三角形;如果这个等腰三角形的两条边分别是5cm,11cm,它的周长是()cm,它的对称轴有()条。",
+        "options":  [
+
+                    ],
+        "answer":  "40;锐角;27;1",
+        "solution":  "【分析】等腰三角形两个底角相等，内角和180°；周长为三边和，并用三角形三边关系判断边长组合；等腰三角形只有1条对称轴。\n【详解】顶角=180°-(70°×2)=40°，故按角分是锐角三角形。边长应为5cm、11cm、11cm，周长=5+11+11=27cm。对称轴有1条。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "三角形的周长",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的分类"
+                             ],
+        "source":  "22-23四年级下·四川德阳·期末"
+    },
+    {
+        "id":  "47",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在常见的平面图形中，等边三角形有()条对称轴,()有无数条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "3;圆",
+        "solution":  "【分析】依据轴对称图形定义判断.\n【详解】如图:![](images/dd695bc8b5864cff2b52255d43d8cd391274449e18b20f12429f104187030986.jpg) 等边三角形有3条对称轴，圆有无数条对称轴。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "23-24四年级下·广东肇庆·期末"
+    },
+    {
+        "id":  "48",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "长方形是()图形，有()条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "轴对称;2",
+        "solution":  "【分析】一个图形沿一条直线对折后，直线两旁部分完全重合就是轴对称图形.\n【详解】如图:![](images/b27d1dc6d59f9191cd454c02d0a818e27c12b85c20c664006c75906509114ba1.jpg) 长方形是轴对称图形，有2条对称轴。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "23-24四年级下·广东阳江·期末"
+    },
+    {
+        "id":  "49",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在等腰梯形，长方形，正方形，等边三角形中，只有1条对称轴的是(),有3条对称轴的是(),有4条对称轴的是()。",
+        "options":  [
+
+                    ],
+        "answer":  "等腰梯形;等边三角形;正方形",
+        "solution":  "【分析】依据轴对称图形定义，分别判断各图形对称轴数量.\n【详解】如图所示：等腰梯形1条![](images/e7ebb056cf87587f0d40ad370f795e213c74d8f7a9fc7f88820133a60bc7354c.jpg);长方形2条![](images/9159bdd437fe6578df74d49e7b373163519d3c9393e176d3356c3a3639b21391.jpg);正方形4条![](images/d0d50c00e8c92afaee53ff69d8739746d3c5efe4842a2ecf50ec89f14ad4b0cd.jpg);等边三角形3条![](images/292c4833c3554529c919a3e1e281a8a652663bf3886f369de4b661f8e36b72c1.jpg)。所以对应填：等腰梯形，等边三角形，正方形。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "23-24四年级下·河北承德·期末"
+    },
+    {
+        "id":  "50",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "把纸对折后剪出的图形是()图形，对折时留下的这道线是()。",
+        "options":  [
+
+                    ],
+        "answer":  "轴对称;对称轴",
+        "solution":  "【分析】沿折痕对折后两边可完全重合的是轴对称图形;折痕所在直线是对称轴.\n【详解】把纸对折后剪出的图形是轴对称图形，对折时留下的这道线是对称轴。",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认",
+                                 "轴对称的剪纸问题"
+                             ],
+        "source":  "23-24四年级下·河北沧州·期末"
+    },
+    {
+        "id":  "51",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形，其中一个底角是38°,则顶角是()°;按角分，它是一个()三角形，它有()条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "104;钝角;1",
+        "solution":  "【分析】等腰三角形底角相等，利用内角和180°求顶角;按角分类判断;等腰三角形有1条对称轴.\n【详解】180°-38°*2=104°,故是钝角三角形，有1条对称轴。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的分类"
+                             ],
+        "source":  "23-24四年级下·河北保定·期末"
+    },
+    {
+        "id":  "52",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在等边三角形，长方形，正方形，圆中，对称轴最少的是(),最多的是()。",
+        "options":  [
+
+                    ],
+        "answer":  "长方形;圆",
+        "solution":  "【分析】根据轴对称定义并比较数量.\n【详解】图示:![](images/a9d5a06b821c40c1b6e499906bfdffc27c891689b43e7a3685b99983d68bcaa2.jpg) 等边三角形3条，长方形2条，正方形4条，圆无数条，因此最少是长方形，最多是圆。",
+        "knowledge_points":  [
+                                 "长方形，正方形，三角形和圆的初步认识",
+                                 "对称轴的画法及数量",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "23-24四年级下·广西河池·期末"
+    },
+    {
+        "id":  "53",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "等腰梯形有()条对称轴，正方形有()条对称轴，长方形有()条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "1;4;2",
+        "solution":  "【分析】等腰梯形上，下底中点连线是对称轴，故1条;长方形有2条;正方形有4条.\n【详解】如图:![](images/f2445e2eec459bd08f63448a240ab492.png) 等腰梯形有1条对称轴，正方形有4条对称轴，长方形有2条对称轴。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "23-24四年级下·云南楚雄·期末"
+    },
+    {
+        "id":  "54",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "正方形有()条对称轴，长方形有()条对称轴，圆有()条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "4;2;无数",
+        "solution":  "【分析】依据轴对称图形定义判断.\n【详解】如图:![](images/d0444bf07f90660a19fde6cec5708bfdd0f3ad119870b2d04e7b89cadea57174.jpg) 正方形4条，长方形2条，圆无数条。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "15-16二年级上·全国·课后作业"
+    },
+    {
+        "id":  "55",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在“中、国、梦”这几个汉字中，是轴对称图形的汉字是()。",
+        "options":  [
+
+                    ],
+        "answer":  "中",
+        "solution":  "【分析】判断轴对称图形，关键是看能否找到对称轴。\n【详解】在“中、国、梦”中，“中”(![](images/2026-03-13%20202206.jpg))可找到对称轴，“国、梦”不能，所以轴对称图形是“中”。",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "23-24四年级下·贵州黔南·期末"
+    },
+    {
+        "id":  "56",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "拉动三角形框架，我们发现它具有()性。有三条对称轴的三角形是()三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "稳定;等边(或正)",
+        "solution":  "【分析】三角形框架不易变形，具有稳定性;等边三角形有3条对称轴.\n【详解】应填：稳定;等边(或正)三角形。",
+        "knowledge_points":  [
+                                 "三角形三边关系",
+                                 "三角形的稳定性及应用",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "23-24四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "57",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形的顶角是80°,沿底边上的高把它对折后，得到两个直角三角形，每个直角三角形中的两个锐角从小到大依次是()°和()°。",
+        "options":  [
+
+                    ],
+        "answer":  "40;50",
+        "solution":  "【分析】等腰三角形两底角相等，先求底角:(180°-80°)÷2=50°;沿高对折后顶角平分，得40°.\n【详解】80°/2=40°,底角为50°,所以两个锐角从小到大是40°和50°。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "23-24四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "58",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在一个轴对称图形中，如果点A到对称轴的距离是5厘米，那么它的对称点A′到对称轴的距离是()厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "5",
+        "solution":  "【分析】轴对称图形中，对称点到对称轴的距离相等.\n【详解】A′到对称轴的距离也是5厘米。",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "23-24四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "59",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在汉字\u0027田，日，口，水\u0027中有()个是轴对称图形。",
+        "options":  [
+
+                    ],
+        "answer":  "3",
+        "solution":  "【分析】沿某直线对折两侧能完全重合的图形是轴对称图形.\n【详解】\u0027田，日，口\u0027是轴对称图形，共3个。",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "23-24五年级下·河北邢台·期末"
+    },
+    {
+        "id":  "60",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "等腰三角形中有一个角是60°,这个三角形有()条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "3或三",
+        "solution":  "【分析】等腰三角形若有一个角是60°,无论是顶角还是底角，最终三个角都为60°,即为等边三角形;等边三角形有3条对称轴.\n【详解】所以这个三角形有3条对称轴。图示:![](images/b24efbc8f72f72306ccffc35e219c41fedd371ad882b1f3ded3a6cc0dfddbf62.jpg)",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "23-24四年级下·新疆和田·期末"
+    },
+    {
+        "id":  "61",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "图形平移时只有()变化，图形的形状,()和()都不变。",
+        "options":  [
+
+                    ],
+        "answer":  "位置;大小;方向",
+        "solution":  "【分析】平移后图形形状，大小，方向不变，只改变位置.\n【详解】应填：位置;大小;方向。",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "23-24四年级下·江西抚州·期末"
+    },
+    {
+        "id":  "62",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等边三角形的边长是6cm,它的内角和是()°,它有()条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "180;3",
+        "solution":  "【分析】任意三角形内角和都是180°;等边三角形有3条对称轴.\n【详解】如图:![](images/505a77312212e2ad9229fc54eb81c9c9ce1207095114da8e2f9e75a11249ff08.jpg) 内角和是180°,有3条对称轴。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "23-24四年级下·湖南长沙·期末"
+    },
+    {
+        "id":  "63",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "长方形，正方形都是()图形;苹果落地这种运动是属于()。",
+        "options":  [
+
+                    ],
+        "answer":  "轴对称;平移",
+        "solution":  "【分析】长方形和正方形沿某直线对折可重合，属于轴对称图形;苹果落地沿直线下落，属于平移.\n【详解】应填：轴对称;平移。",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认",
+                                 "平移与平移现象"
+                             ],
+        "source":  "23-24四年级下·湖南怀化·期末"
+    },
+    {
+        "id":  "64",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "汽车行驶在一条平直的公路上，车轮的运动是(),车身的运动是()。",
+        "options":  [
+
+                    ],
+        "answer":  "旋转;平移",
+        "solution":  "【分析】车身整体沿直线移动是平移;车轮绕轴转动是旋转.\n【详解】汽车行驶时，车轮的运动是旋转，车身的运动是平移。",
+        "knowledge_points":  [
+                                 "旋转与旋转现象",
+                                 "平移与平移现象"
+                             ],
+        "source":  "23-24三年级下·湖北省直辖县级单位·期末"
+    },
+    {
+        "id":  "65",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "“王”“中”“日”都是轴对称图形，请你再写一个这样的汉字()。",
+        "options":  [
+
+                    ],
+        "answer":  "甲",
+        "solution":  "【分析】轴对称图形是沿一条直线对折后两侧完全重合的图形。\n【详解】可写“甲”。",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "23-24四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "66",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形的底角是55°,顶角是()°,它有()条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "70;1",
+        "solution":  "【分析】等腰三角形两个底角相等，内角和180°.\n【详解】180°-55°-55°=70°。图示:![](images/d9a451b69eb258a73ee1ada3478e9f5662980f498b2335d3de6d0ef22c515eab.jpg) 该等腰三角形有1条对称轴。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "23-24四年级下·黑龙江齐齐哈尔·期末"
+    },
+    {
+        "id":  "67",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "等腰直角三角形的一个底角是()°,可以画()条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "45;1",
+        "solution":  "【分析】等腰直角三角形顶角90°,两个底角相等.\n【详解】(180°-90°)÷2=45°。图示:![](images/154da0a4d2ecbd10ffddd4274f11478fea9cabbebe48a464ca165d16f80767bd.jpg) 可以画1条对称轴。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "23-24四年级下·湖北武汉·期末"
+    },
+    {
+        "id":  "68",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "直角三角形有()条高，等边三角形有()条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "3;3",
+        "solution":  "【分析】三角形有3个顶点，对应可作3条高;等边三角形三条高所在直线都可作为对称轴.\n【详解】直角三角形有3条高，等边三角形有3条对称轴。",
+        "knowledge_points":  [
+                                 "三角形的高及画法",
+                                 "对称轴的画法及数量",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "23-24四年级下·湖北武汉·期末"
+    },
+    {
+        "id":  "69",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "长方形、正方形、平行四边形和圆形都是轴对称图形。() 理由:()。",
+        "options":  [
+
+                    ],
+        "answer":  "×;平行四边形不是轴对称图形",
+        "solution":  "【分析】根据图形特征判断：长方形，正方形，圆是轴对称图形;一般平行四边形不是轴对称图形.\n【详解】原题说法错误，平行四边形不是轴对称图形。故答案为: ×。",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认",
+                                 "正方形的概念及特点",
+                                 "长方形的概念及特点",
+                                 "平行四边形的概念及特点"
+                             ],
+        "source":  "23-24六年级下·山西长治·期末"
+    },
+    {
+        "id":  "70",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "鸟儿直线飞行是()现象。",
+        "options":  [
+
+                    ],
+        "answer":  "平移",
+        "solution":  "【分析】平移是图形沿某个方向作相同距离的直线移动，平移后位置改变，形状，大小，方向不变;旋转是绕固定点转动.\n【详解】由分析可得，鸟儿直线飞行是平移现象。",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "23-24四年级下·广东广州·期末"
+    },
+    {
+        "id":  "71",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "等边三角形有()条对称轴，电梯上升或者下降属于()现象。",
+        "options":  [
+
+                    ],
+        "answer":  "3;平移",
+        "solution":  "【分析】轴对称图形沿某条直线对折后两侧完全重合;平移是沿某方向移动一定距离，大小，形状，方向不变，位置改变.\n【详解】根据分析，等边三角形有3条对称轴，电梯上升或下降属于平移现象。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "平移与平移现象"
+                             ],
+        "source":  "22-23四年级下·湖北十堰·期末"
+    },
+    {
+        "id":  "72",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等边三角形的周长是42厘米，那么它的每条边长是()厘米，每个角是()°,有()条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "14;60或60度;3",
+        "solution":  "【分析】等边三角形三边相等，三角均为60°;周长42厘米，边长=42÷3;等边三角形可作3条对称轴.\n【详解】42÷3=14(厘米),所以每条边长14厘米，每个角60°,有3条对称轴。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "三角形的周长",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "22-23四年级下·湖北黄冈·期末"
+    },
+    {
+        "id":  "73",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "两根小棒分别长3厘米和6厘米，第三根小棒最长()厘米，最短()厘米(两个空填整厘米数),就一定能摆成一个三角形。现有3厘米,6厘米的小棒若干，要摆成一个等腰三角形，这个等腰三角形的周长是()厘米，这个等腰三角形有()条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "8;4;15;1",
+        "solution":  "【分析】三角形三边关系：任意两边和大于第三边，差小于第三边。已知3和6，则第三边满足3\u003cx\u003c9，整厘米最短4，最长8。摆等腰三角形时，腰不能是3(3+3=6不满足大于),只能是6，三边为6,6,3.\n【详解】第三边最长8厘米，最短4厘米;等腰三角形周长=6+6+3=15(厘米);等腰三角形有1条对称轴。图示:![](images/905feebd77d5aa94cc30d0d268307bed49941e997ca34c88ed7e23dce334ad8f.jpg)",
+        "knowledge_points":  [
+                                 "三角形三边关系",
+                                 "对称轴的画法及数量",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "22-23四年级下·湖南怀化·期末"
+    },
+    {
+        "id":  "74",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "学校每周一举行升旗仪式，升国旗的这种运动是属于()。",
+        "options":  [
+
+                    ],
+        "answer":  "平移",
+        "solution":  "【分析】平移是图形沿某条直线方向平行移动一定距离。升国旗时，国旗沿旗杆直线上升.\n【详解】升国旗这种运动属于平移。",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "22-23四年级下·湖南怀化·期末"
+    },
+    {
+        "id":  "75",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在一个边长为24厘米的正方形四个角处各剪去一个边长为4厘米的小正方形后，那么所得图形的周长是()厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "96",
+        "solution":  "【分析】每个角剪去后，原边减少的长度与新增折线长度相等，周长保持不变.\n【详解】如图:![](images/00c93635b08de8fd6eabd28c205798472e8fa8b237bae6980f8528abec740395.jpg) 所得图形周长=24×4=96(厘米)。",
+        "knowledge_points":  [
+                                 "正方形的周长",
+                                 "利用平移巧算周长与面积",
+                                 "利用平移法巧算周长"
+                             ],
+        "source":  "23-24六年级下·四川绵阳·期末"
+    },
+    {
+        "id":  "76",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "确定一个图形平移后的位置，除了需要知道图形原来的位置外，还需要知道平移的()和()。",
+        "options":  [
+
+                    ],
+        "answer":  "方向;距离",
+        "solution":  "【分析】决定平移后位置的要素有两个：平移方向，平移距离.\n【详解】还需要知道平移的方向和距离。",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "23-24四年级下·全国·课后作业"
+    },
+    {
+        "id":  "77",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "平移后图形的(),(),()都没有发生改变，只是()发生了改变。",
+        "options":  [
+
+                    ],
+        "answer":  "形状;大小;方向;位置",
+        "solution":  "【分析】平移性质：位置改变，形状，大小，方向不变.\n【详解】平移后图形的形状，大小，方向都不变，只是位置改变。",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "23-24五年级下·河北邢台·期中"
+    },
+    {
+        "id":  "78",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "点a从(7,9)平移到(9,9)是向()移动了()格;b点用数对表示为(4,6)它在第()列，第()行。",
+        "options":  [
+
+                    ],
+        "answer":  "右;2;4;6",
+        "solution":  "【分析】数对中第一个数表示列，第二个数表示行。a点列数由7变9，行数不变，说明向右2格.\n【详解】点a向右移动2格;b点在第4列，第6行。",
+        "knowledge_points":  [
+                                 "根据数对找位置",
+                                 "用数对表示位置",
+                                 "平移与平移现象"
+                             ],
+        "source":  "22-23四年级下·河南周口·期中"
+    },
+    {
+        "id":  "79",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "下列运动中是平移的画\u0027√\u0027,是旋转的画\u0027○\u0027。电风扇叶片的运动。( ) 拉抽屉。( )",
+        "options":  [
+
+                    ],
+        "answer":  "○;√",
+        "solution":  "【分析】旋转是绕固定点转动;平移是沿直线方向移动且方向不变.\n【详解】电风扇叶片的运动是旋转(○);拉抽屉是平移(√)。",
+        "knowledge_points":  [
+                                 "旋转与旋转现象",
+                                 "平移与平移现象"
+                             ],
+        "source":  "22-23二年级下·湖南张家界·期中"
+    },
+    {
+        "id":  "80",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "电梯门的打开和关闭的运动是()。",
+        "options":  [
+
+                    ],
+        "answer":  "平移",
+        "solution":  "【分析】电梯门沿直线开合，不是绕点转动，因此属于平移.\n【详解】电梯门的打开和关闭是平移。",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "22-23六年级下·广东梅州·期中"
+    },
+    {
+        "id":  "81",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "利用图形的平移也可以解决问题，真有趣!我的收获是:()。",
+        "options":  [
+
+                    ],
+        "answer":  "平移前后两个图形完全相同",
+        "solution":  "【分析】平移得到的两个图形对应边平行且相等，对应角相等。\n【详解】可填写：平移前后两个图形完全相同。",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "23-24四年级下·全国·课后作业"
+    },
+    {
+        "id":  "82",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "等腰直角三角形有()条对称轴，正方形有()条对称轴，长方形有()条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "1;4;2",
+        "solution":  "【分析】根据轴对称图形定义判断：等腰直角三角形1条，正方形4条，长方形2条.\n【详解】应填:1;4;2。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "23-24五年级上·河南商丘·期中"
+    },
+    {
+        "id":  "83",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "抽动陀螺，陀螺的运动是()现象，推拉抽屉的运动是()现象。",
+        "options":  [
+
+                    ],
+        "answer":  "旋转;平移",
+        "solution":  "【分析】绕点或轴运动是旋转;沿直线运动且本身方向不变是平移.\n【详解】陀螺的运动是旋转现象，推拉抽屉的运动是平移现象。",
+        "knowledge_points":  [
+                                 "平移与平移现象",
+                                 "旋转与旋转现象"
+                             ],
+        "source":  "23-24五年级上·河南商丘·期中"
+    },
+    {
+        "id":  "84",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "我们学过的平面图形中，等边三角形有()条对称轴,()形有2条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "3;长方",
+        "solution":  "【分析】等边三角形有3条对称轴;长方形有2条对称轴.\n【详解】应填:3;长方。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "长方形的概念及特点"
+                             ],
+        "source":  "23-24六年级上·广东佛山·期中"
+    },
+    {
+        "id":  "85",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在字母L,M,N中,()是轴对称的;在轴对称图形中点A到对称轴的距离是5厘米，它的对应点到对称轴的距离是()厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "M;5",
+        "solution":  "【分析】轴对称图形沿对称轴对折后两侧重合;对称点到对称轴距离相等.\n【详解】L,M,N中M是轴对称的;对应点到对称轴距离也是5厘米。",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "22-23五年级上·河南周口·期中"
+    },
+    {
+        "id":  "86",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "等边三角形有()条对称轴，正方形有()条对称轴，等腰梯形有()条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "3;4;1",
+        "solution":  "【分析】依据各图形轴对称性质判断对称轴条数.\n【详解】等边三角形3条，正方形4条，等腰梯形1条。",
+        "knowledge_points":  [
+                                 "直角梯形和等腰梯形的概念及特点",
+                                 "对称轴的画法及数量",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "22-23五年级上·福建南平·期中"
+    },
+    {
+        "id":  "87",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "国旗升降属于()现象，在升降过程中只改变了国旗的(),而未改变国旗的()和()。",
+        "options":  [
+
+                    ],
+        "answer":  "平移;位置;形状;大小",
+        "solution":  "【分析】国旗沿直线方向升降，属于平移。平移只改变位置，不改变形状和大小.\n【详解】应填：平移;位置;形状;大小。",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "20-21五年级上·广东韶关·期中"
+    },
+    {
+        "id":  "88",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "等边三角形，长方形，正方形中，对称轴最多的图形是()。",
+        "options":  [
+
+                    ],
+        "answer":  "正方形",
+        "solution":  "【分析】比较对称轴数量：等边三角形3条，长方形2条，正方形4条.\n【详解】如图:![](images/90190a622bd2396e7d05d3d24293cbf7.png) 因为4\u003e3\u003e2，所以对称轴最多的是正方形。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "22-23四年级下·山东菏泽·期末"
+    },
+    {
+        "id":  "89",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在等腰梯形，长方形，正方形，平行四边形中，只有一条对称轴的是(),有两条对称轴的是(),不是轴对称图形的是()。",
+        "options":  [
+
+                    ],
+        "answer":  "等腰梯形;长方形;平行四边形",
+        "solution":  "【分析】依据轴对称图形定义分别判断：等腰梯形1条，长方形2条，正方形4条，平行四边形一般不是轴对称图形.\n【详解】图示：等腰梯形![](images/fadcb50591cd9cd638760d758b9d80dd.png),平行四边形![](images/40674b9f0d634a5fc4b476f8290987c3.png)。所以应填：等腰梯形;长方形;平行四边形。",
+        "knowledge_points":  [
+                                 "直角梯形和等腰梯形的概念及特点",
+                                 "对称轴的画法及数量",
+                                 "轴对称的认识及辨认",
+                                 "平行四边形的概念及特点"
+                             ],
+        "source":  "21-22五年级上·吉林辽源·期末"
+    },
+    {
+        "id":  "90",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计数器上拨珠子，珠子的运动是();抽奖时，幸运转盘的运动是(),剪窗花的时候经常用到我们数学中()的方法。(填平移，旋转或轴对称)",
+        "options":  [
+
+                    ],
+        "answer":  "平移;旋转;轴对称",
+        "solution":  "【分析】珠子上下直线移动是平移;转盘绕中心转动是旋转;剪窗花利用对折重合是轴对称.\n【详解】应填：平移;旋转;轴对称。",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认",
+                                 "平移与平移现象",
+                                 "旋转与旋转现象"
+                             ],
+        "source":  "20-21三年级下·浙江金华·期末"
+    },
+    {
+        "id":  "91",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "李老师给同学准备了3组小棒去围三角形：第一组:4cm,4cm,7cm;第二组:4cm,5cm,9cm;第三组:3cm,4cm,5cm。其中不能围成三角形的是第()组;其中能围成轴对称图形的是第()组;其中围成的三个内角都不相等的三角形是第()组。",
+        "options":  [
+
+                    ],
+        "answer":  "二;一;三",
+        "solution":  "【分析】三角形两边和大于第三边。第一组4+4\u003e7可成等腰三角形(轴对称);第二组4+5=9不能成三角形;第三组3,4,5可成不等边三角形，三个角不相等.\n【详解】不能围成三角形的是第二组;能围成轴对称图形的是第一组;三个内角都不相等的是第三组。图示:![](images/229c99a08fbe4764f06d2b2d9466e508375e3a312159057f37c4bc1f8e59810c.jpg)",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形三边关系"
+                             ],
+        "source":  "20-21四年级下·浙江温州·期末"
+    },
+    {
+        "id":  "92",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "车轮的运动方式是(),拉抽屉的运动方式是()。",
+        "options":  [
+
+                    ],
+        "answer":  "旋转;平移",
+        "solution":  "【分析】车轮绕轴转动是旋转;抽屉沿直线推进拉出是平移.\n【详解】车轮是旋转，拉抽屉是平移。",
+        "knowledge_points":  [
+                                 "平移与平移现象",
+                                 "旋转与旋转现象"
+                             ],
+        "source":  "21-22三年级上·贵州毕节·期末"
+    },
+    {
+        "id":  "93",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "公路上行驶的汽车，车轮的运动是()现象;升国旗时国旗的上升运动是()现象。",
+        "options":  [
+
+                    ],
+        "answer":  "旋转;平移",
+        "solution":  "【分析】车轮绕轴转动是旋转;国旗沿旗杆直线上升是平移.\n【详解】应填：旋转;平移。",
+        "knowledge_points":  [
+                                 "平移与平移现象",
+                                 "旋转与旋转现象"
+                             ],
+        "source":  "22-23二年级·全国·假期作业"
+    },
+    {
+        "id":  "94",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在ACHMNG中，轴对称字母有()。",
+        "options":  [
+
+                    ],
+        "answer":  "A,C,H,M",
+        "solution":  "【分析】轴对称图形沿某条直线对折后两侧完全重合.\n【详解】如图:![](images/2026-03-13%20202528.jpg) 在ACHMNG中，轴对称字母有A,C,H,M.",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "22-23五年级下·山东济宁·期中"
+    },
+    {
+        "id":  "95",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "甲数是32.78，比乙数少1.8，乙数是()。一个长方形有()条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "34.58;2",
+        "solution":  "【分析】由题意乙数=甲数+1.8;长方形有2条对称轴.\n【详解】32.78+1.8=34.58。图示:![](images/be8250d7392ccef3378f13e290ebffb8797bacc2a4fe1688b15dacbf201747e5.jpg) 所以乙数是34.58，长方形有2条对称轴。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "20-21四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "96",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.94,
+        "question":  "平行四边形有4条对称轴。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】一般平行四边形沿任何直线对折都不能完全重合，不是轴对称图形.\n【详解】平行四边形通常没有对称轴;特殊平行四边形如菱形，长方形也都不是4条。所以原说法错误，故答案为: ×。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "平行四边形的概念及特点"
+                             ],
+        "source":  "24-25五年级上·广东韶关·期中"
+    },
+    {
+        "id":  "97",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "所有的三角形都有3条对称轴。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】只有等边三角形有3条对称轴;等腰三角形1条;一般三角形没有对称轴.\n【详解】原说法错误，故答案为: ×。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "25-26五年级上·安徽宿州·期中"
+    },
+    {
+        "id":  "98",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "轴对称图形的两个对称点到对称轴的距离不一定相等。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】轴对称图形中，对称点关于对称轴成镜像关系，到对称轴距离一定相等.\n【详解】原说法错误，故答案为: ×。",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "24-25五年级上·河北邯郸·期中"
+    },
+    {
+        "id":  "99",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "长方形是特殊的平行四边形，所以长方形和平行四边形都有2条对称轴。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】长方形是轴对称图形，有2条对称轴;一般平行四边形不是轴对称图形.\n【详解】\u0027长方形和平行四边形都有2条对称轴\u0027说法错误，故答案为: ×。",
+        "knowledge_points":  [
+                                 "平行四边形的概念及特点",
+                                 "长方形的概念及特点",
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "24-25四年级下·安徽宣城·期末"
+    },
+    {
+        "id":  "100",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "甲骨文是汉字的源头和中华优秀传统文化的根脉。甲骨文![](images/2026-03-13%20202652.jpg)不能用其中一部分平移得到。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平移后图形应保持形状，大小，方向不变，只改变位置.\n【详解】根据图形特征，该甲骨文不能由其中一部分通过平移得到。故答案为: ×。",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "24-25四年级下·河北廊坊·期末"
+    },
+    {
+        "id":  "101",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "看图填一填。![](images/a1d3031fe5470c49ddf81234747bcebab362e1bcdbd1241c6005b2df31c36a11.jpg) (1)图①先向( )平移了( )格，再向( )平移了( )格。(2)图②先向( )平移了( )格，再向( )平移了( )格。",
+        "options":  [
+
+                    ],
+        "answer":  "下;4;右;6;上;4;左;5",
+        "solution":  "【分析】先找原图与对应图的关键点，再判断方向和格数.\n【详解】(1)图①先向下平移4格，再向右平移6格。(2)图②先向上平移4格，再向左平移5格。",
+        "knowledge_points":  [
+                                 "作平移后的图形",
+                                 "平移与平移现象"
+                             ],
+        "source":  "22-23四年级·全国·假期作业"
+    },
+    {
+        "id":  "102",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "点A和点B是轴对称图形上的对称点，点A到对称轴的距离是3厘米，点A和点B之间的距离是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】对称点到对称轴距离相等，且两点连线被对称轴垂直平分，所以两点距离是到对称轴距离的2倍.\n【详解】3×2=6(厘米)。",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "20-21四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "103",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "画一个图形的轴对称图形，可按以下四个步骤完成：第一步：确定所给图形的( )。第二步：确定关键点到对称轴的( )。第三步：确定关键点的( )。第四步：把描出的对应点按顺序( )。",
+        "options":  [
+
+                    ],
+        "answer":  "关键点;距离;对称点;连接各点",
+        "solution":  "【分析】补全轴对称图形的核心是先找关键点，再找对称点，最后连接.\n【详解】依次填：关键点，距离，对称点，连接各点。",
+        "knowledge_points":  [
+                                 "补全轴对称图形"
+                             ],
+        "source":  "22-23四年级下·全国·课前预习"
+    },
+    {
+        "id":  "104",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "我发现：画出平移后的图形的方法。(1)定点：确定所给图形的( )。(2)定方向：确定平移的( )。(3)数格子：按照要求数出对应点之间的( )。(4)连线：按照所给图形的形状( )各对应点，就画出了所给图形平移后的图形。",
+        "options":  [
+
+                    ],
+        "answer":  "关键点;方向;格子;顺次连接",
+        "solution":  "【分析】作平移图要抓住\u0027关键点，方向，距离\u0027三要素.\n【详解】依次填：关键点，方向，格子，顺次连接。",
+        "knowledge_points":  [
+                                 "作平移后的图形"
+                             ],
+        "source":  "22-23四年级下·全国·课前预习"
+    },
+    {
+        "id":  "105",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "平移时物体的形状和大小完全( )发生变化。",
+        "options":  [
+
+                    ],
+        "answer":  "没有",
+        "solution":  "【分析】平移只改变位置，不改变形状和大小.\n【详解】应填\u0027没有\u0027。图示:![](images/c8ee798e5f722c447e39c7c06eaf11171bac14f11bf65452682d1042cc15eebd.jpg)",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "22-23四年级下·全国·课前预习"
+    },
+    {
+        "id":  "106",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "轴对称图形都有( )。有的只有( )对称轴，有的图形有( )对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "对称轴;1条;无数条",
+        "solution":  "【分析】轴对称图形一定存在对称轴，不同图形对称轴条数不同.\n【详解】应填：对称轴;1条;无数条。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "22-23四年级下·全国·课前预习"
+    },
+    {
+        "id":  "107",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "要判断一个图形平移了几格，关键看( )平移了几格。",
+        "options":  [
+
+                    ],
+        "answer":  "关键点",
+        "solution":  "【分析】平移时所有点移动距离相同，抓任意关键点即可判断.\n【详解】关键看\u0027关键点\u0027平移了几格。",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "22-23四年级下·全国·课前预习"
+    },
+    {
+        "id":  "108",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "确定平移的方向：竖直有( )和( );水平有( )和( )。",
+        "options":  [
+
+                    ],
+        "answer":  "上;下;左;右",
+        "solution":  "【分析】平移方向按竖直和水平分类.\n【详解】竖直方向有上，下;水平方向有左，右。示意图:![](images/9a68d31de447882b29218a045a96a9b2d3cba98b475dd9c537e59945b66962e0.jpg)",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "22-23四年级下·全国·课前预习"
+    },
+    {
+        "id":  "109",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在平面内，将一个图形沿某一方向平行移动，不改变图形的( )和( )。",
+        "options":  [
+
+                    ],
+        "answer":  "形状;大小",
+        "solution":  "【分析】平移后图形形状，大小，方向不变，只是位置变化.\n【详解】应填\u0027形状\u0027和\u0027大小\u0027。",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "22-23四年级下·全国·课前预习"
+    },
+    {
+        "id":  "110",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "长方形的对称轴有( )条;正方形的对称轴有( )条，等边三角形的对称轴有( )条。",
+        "options":  [
+
+                    ],
+        "answer":  "2;4;3",
+        "solution":  "【分析】根据各常见图形的轴对称特征判断条数.\n【详解】长方形2条，正方形4条，等边三角形3条。图示:![](images/725b87fb108a7cca0104d6ba94a98d6f.png)",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "21-22四年级下·广东佛山·期末"
+    },
+    {
+        "id":  "111",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "长方形有( )条对称轴，正方形有( )条对称轴，圆有( )条对称轴，平行四边形有( )条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "2;4;无数;0",
+        "solution":  "【分析】逐个按轴对称定义判断.\n【详解】长方形2条，正方形4条，圆无数条，平行四边形0条。图示:![](images/a328eb0b4efe3cb310f6e867c4e15289.png)",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "22-23五年级上·江西南昌·期末"
+    },
+    {
+        "id":  "112",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "张老师早起出门锻炼时看到镜子中的时间是7:30，回到家看钟表时发现是6:30，那么张老师出门锻炼了( )小时。",
+        "options":  [
+
+                    ],
+        "answer":  "2",
+        "solution":  "【分析】镜中7:30对应实际4:30，再与6:30求时间差.\n【详解】6:30-4:30=2(小时)。",
+        "knowledge_points":  [
+                                 "时，分，秒有关的计算",
+                                 "镜面对称"
+                             ],
+        "source":  "22-23五年级上·江西南昌·期末"
+    },
+    {
+        "id":  "113",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "等边三角形有( )条对称轴，直角梯形有( )条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "3;0",
+        "solution":  "【分析】按轴对称定义判断：等边三角形是轴对称图形，直角梯形一般不是.\n【详解】等边三角形3条，直角梯形0条。图示:![](images/228979e83cf41ca187efeed44b8501eeea1207130f630daa57b4f262782fc861.jpg) ![](images/c9e1f9c4f396a629b304e07087bb1684bc026fc0222e1876fa27177bde443be4.jpg)",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "22-23四年级下·全国·单元测试"
+    },
+    {
+        "id":  "114",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "如果把一个图形沿着( )对折,( )的图形能够( ),这个图形就是( )图形,( )叫做对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "一条直线;直线两旁;完全重合;轴对称;这条直线",
+        "solution":  "【分析】此题直接考查轴对称图形定义.\n【详解】依次填：一条直线，直线两旁，完全重合，轴对称，这条直线。图示:![](images/39dda36dfafb60e950ba282d736c981312b61868af173a6b111bd3ef132450e8.jpg)",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "22-23四年级下·全国·单元测试"
+    },
+    {
+        "id":  "115",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明到中医院看望生病住院的奶奶，他乘电梯从一楼到七楼，这一运动过程属于( )现象。",
+        "options":  [
+
+                    ],
+        "answer":  "平移",
+        "solution":  "【分析】电梯沿直线方向移动，方向不变.\n【详解】属于平移现象。",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "21-22五年级上·重庆云阳·期末"
+    },
+    {
+        "id":  "116",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "长方形的对称轴有( )条，等腰梯形的对称轴有( )条。",
+        "options":  [
+
+                    ],
+        "answer":  "2;1",
+        "solution":  "【分析】长方形有两条中线对称轴，等腰梯形有1条中垂线对称轴.\n【详解】应填2;1。图示:![](images/f2278569f0cc228f86f1854378f80f4c7c770854377b37a4b57e05e2f372b0d8.jpg)",
+        "knowledge_points":  [
+                                 "直角梯形和等腰梯形的概念及特点",
+                                 "对称轴的画法及数量",
+                                 "长方形的概念及特点"
+                             ],
+        "source":  "21-22五年级上·重庆渝北·期末"
+    },
+    {
+        "id":  "117",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在方格图中，图形A可以先向右平移5格，再向下平移3格到图形B位置;也可以将图形A先向( )平移( )格，再向( )平移( )格到图形B位置。",
+        "options":  [
+
+                    ],
+        "answer":  "下;3;右;5",
+        "solution":  "【分析】同一终点下，先后顺序可交换，但方向和距离不变。图示:![](images/8afbb3090e1b7e0c374826f6c4f29205076a8b8a29fe706ee506192f46354cd3.jpg)\n【详解】可先向下3格，再向右5格。",
+        "knowledge_points":  [
+                                 "作平移后的图形"
+                             ],
+        "source":  "21-22五年级上·重庆彭水·期末"
+    },
+    {
+        "id":  "118",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "正方形有( )条对称轴，长方形有( )条对称轴，圆的对称轴有( )条，半圆的对称轴有( )条。",
+        "options":  [
+
+                    ],
+        "answer":  "4;2;无数;1",
+        "solution":  "【分析】按常见图形对称轴数量直接判断.\n【详解】正方形4条，长方形2条，圆无数条，半圆1条。图示:![](images/801882dd74e4bf636c60b87bddeaaf67.png)",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "21-22四年级下·吉林白城·期末"
+    },
+    {
+        "id":  "119",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "等腰三角形的一个底角是70°,它的顶角是( )°,这个三角形有( )条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "40°(或40度);1(或一)",
+        "solution":  "【分析】等腰三角形两底角相等，内角和180°.\n【详解】顶角=180°-70°-70°=40°;等腰三角形有1条对称轴。图示:![](images/5823a3d8380e0fbb06b8f9e844ccace40ca3bbf5f34c81b843b2ea85634a1a72.jpg)",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和",
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "21-22四年级下·云南曲靖·期末"
+    },
+    {
+        "id":  "120",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "轴对称图形上对称的两点到( )的距离( )。",
+        "options":  [
+
+                    ],
+        "answer":  "对称轴;相等",
+        "solution":  "【分析】轴对称性质：对应点到对称轴距离相等.\n【详解】应填\u0027对称轴\u0027\u0027相等\u0027。图示:![](images/609e951381907761422db03423eed1c7737fef875dd328eeda70089975c5d48a.jpg)",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "20-21五年级上·四川达州·期末"
+    },
+    {
+        "id":  "121",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "等腰三角形的对称轴有( )条。",
+        "options":  [
+
+                    ],
+        "answer":  "1或一",
+        "solution":  "【分析】等腰三角形沿底边上的高所在直线对折可重合.\n【详解】等腰三角形只有1条对称轴。图示:![](images/00b45adce906fa5ccc727dd1ad4a48beb158832f241f855d17caba8a529a39bd.jpg)",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "22-23六年级上·安徽宿州·期中"
+    },
+    {
+        "id":  "122",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个图形，如果沿着一条直线对折，两侧的部分能够( ),我们就说这个图是( )图形，这条直线就是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "完全重合;轴对称;对称轴",
+        "solution":  "【分析】本题直接考查轴对称图形定义.\n【详解】依次填：完全重合，轴对称，对称轴。",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "21-22五年级上·重庆云阳·期中"
+    },
+    {
+        "id":  "123",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "平移只改变图形的位置，不改变图形的( )和大小。",
+        "options":  [
+
+                    ],
+        "answer":  "形状",
+        "solution":  "【分析】平移特征：形状，大小不变，位置变.\n【详解】应填\u0027形状\u0027。",
+        "knowledge_points":  [
+                                 "作平移后的图形",
+                                 "平移与平移现象"
+                             ],
+        "source":  "21-22四年级下·河北承德·期末"
+    },
+    {
+        "id":  "124",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "我们学过的平面图形中，有两条对称轴的是( ),有三条对称轴的是( ),有四条对称轴的是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "长方形;等边三角形;正方形",
+        "solution":  "【分析】常见图形对称轴数量：长方形2条，等边三角形3条，正方形4条.\n【详解】依次填：长方形，等边三角形，正方形。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "21-22四年级下·河北保定·期末"
+    },
+    {
+        "id":  "125",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "( )三角形一定是轴对称图形，且只有一条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "等腰",
+        "solution":  "【分析】等腰三角形沿顶角平分线(底边中线)对折可重合，且只有1条对称轴.\n【详解】应填\u0027等腰\u0027。图示:![](images/48bbde2377a55a720e15caa157de7d38a4a7f65462597c5b602e005bbf7bcb26.jpg) ![](images/11c8bc8896a495b42b32d05dab3fdd513a6dd814eb3d5ffd0fbeac2399b93341.jpg)",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "21-22四年级下·河北保定·期末"
+    },
+    {
+        "id":  "126",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "小明说：“平行四边形一定是轴对称图形。”( )(正确打“√”，错误打“×” )，你的理由是:( )。",
+        "options":  [
+
+                    ],
+        "answer":  "×;平行四边形沿着一条直线对折后两部分不一定能完全重合",
+        "solution":  "【分析】一般平行四边形不满足沿某条直线对折后完全重合。\n【详解】判断为“×”；理由：平行四边形沿一条直线对折后两部分不一定能完全重合。图示:![](images/f2155e2945ee2996acf67e963cb4ccf580f48f212772f80a92b0009fb30fb1e2.jpg) ![](images/b62892de0a2000488104b666116a32772822cd88d56a49ad6003e187613bfe7a.jpg)",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认",
+                                 "平行四边形的概念及特点"
+                             ],
+        "source":  "21-22四年级下·河北唐山·期末"
+    },
+    {
+        "id":  "127",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "打开水龙头时，水龙头的运动是( )现象，关上大衣柜的推拉门，推拉门的运动是( )现象。(填平移或旋转)",
+        "options":  [
+
+                    ],
+        "answer":  "旋转;平移",
+        "solution":  "【分析】水龙头是绕轴转动，推拉门是沿直线移动.\n【详解】应填：旋转;平移。",
+        "knowledge_points":  [
+                                 "平移与平移现象",
+                                 "旋转与旋转现象"
+                             ],
+        "source":  "21-22三年级下·河北邯郸·期末"
+    },
+    {
+        "id":  "128",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "图形平移后位置变了,( )和( )都没有变。",
+        "options":  [
+
+                    ],
+        "answer":  "形状;大小",
+        "solution":  "【分析】平移只改变位置，不改变形状和大小.\n【详解】应填：形状;大小。",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "22-23五年级上·广东韶关·期中"
+    },
+    {
+        "id":  "129",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "现实中有很多有趣的图形运动，使我们的生活更丰富，请各举一例：轴对称:()平移:()旋转:()",
+        "options":  [
+
+                    ],
+        "answer":  "正方形;电梯的升降;荡秋千",
+        "solution":  "【分析】依据轴对称、平移、旋转的定义各举生活实例。\n【详解】可填：轴对称“正方形”；平移“电梯的升降”；旋转“荡秋千”。",
+        "knowledge_points":  [
+                                 "旋转与旋转现象",
+                                 "平移与平移现象",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "21-22三年级下·内蒙古赤峰·期末"
+    },
+    {
+        "id":  "130",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "轴对称图形上对称点到对称轴的( )相等。长方形有( )条对称轴，圆有( )条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "距离;2;无数",
+        "solution":  "【分析】轴对称图形的对应点到对称轴距离相等;长方形2条，圆无数条.\n【详解】应填：距离;2;无数。图示:![](images/1c500ead2c9cd48a3ee3eb9c9af960e0.png)",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认",
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "21-22四年级下·四川德阳·期末"
+    },
+    {
+        "id":  "131",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "正方形有( )条对称轴，等边三角形有( )对称轴，国旗上的大五角星代表中国共产党，这颗五角星有( )条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "4;3;5",
+        "solution":  "【分析】按常见图形对称轴条数判断.\n【详解】正方形4条，等边三角形3条，五角星5条。图示:![](images/0bcc95a947522c594f0a7bd75637de840c40b7ee4f84d54d587242649fbf6471.jpg)",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "21-22四年级下·湖南衡阳·期末"
+    },
+    {
+        "id":  "132",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "成轴对称的两个图形沿( )折叠后可以( )重合。",
+        "options":  [
+
+                    ],
+        "answer":  "对称轴;完全",
+        "solution":  "【分析】成轴对称的两个图形关于对称轴对折后应完全重合.\n【详解】应填：对称轴;完全。图示:![](images/ed343bc33960a99ff77032df71a67ab62861c163777179b3ecfb3b1abb3a30ff.jpg)",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "21-22四年级下·湖南怀化·期末"
+    },
+    {
+        "id":  "133",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在一个直角三角形里，一个锐角是38°,另一个锐角是( )。一个等边三角形，它的每个内角的度数都是( ),它有( )条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "52°;60°;3",
+        "solution":  "【分析】直角三角形两锐角和为90°,所以另一个锐角为90°-38°=52°;等边三角形三个内角都是60°,有3条对称轴.\n【详解】另一个锐角是52°;等边三角形每个内角60°,有3条对称轴。图示:![](images/c75f8eebfc24690db5719444b163a3ff36ef2a1294f556ee516d8858819b4718.jpg)",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "21-22四年级下·河南新乡·期末"
+    },
+    {
+        "id":  "134",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三角形最多有( )条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "3或三",
+        "solution":  "【分析】等腰三角形有1条对称轴，等边三角形有3条对称轴，一般三角形没有对称轴，所以三角形最多有3条对称轴.\n【详解】等腰三角形有1条对称轴:![](images/dc7418e4ed95ca170b5769df085a2970a628d18318442e4d9ebd3738caa77346.jpg);等边三角形有3条对称轴:![](images/f9845c9e515bd7bc11eecf56890c24d5d3181c4f64e3f802a99f0115519e6519.jpg);此三角形没有对称轴:![](images/2295780814e43b4592e93c32557dc8779516e028c7e2cba6b8fec4ea836b5e8d.jpg)。因此一个三角形最多有3条对称轴。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "21-22四年级下·河南开封·期末"
+    },
+    {
+        "id":  "135",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "电梯的运动属于( )现象，把篮球架上篮板的支架设计成三角形，是运用了三角形具有( )的特性。",
+        "options":  [
+
+                    ],
+        "answer":  "平移;稳定性",
+        "solution":  "【分析】电梯沿直线方向运动，属于平移;三角形不易变形，具有稳定性.\n【详解】电梯的运动属于平移现象，把篮球架上篮板的支架设计成三角形，是运用了三角形具有稳定性的特性。",
+        "knowledge_points":  [
+                                 "三角形的稳定性及应用",
+                                 "平移与平移现象"
+                             ],
+        "source":  "21-22四年级下·四川凉山·期末"
+    },
+    {
+        "id":  "136",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在长方形，正方形，等边三角形，等腰三角形这些图形中,( )的对称轴最多。",
+        "options":  [
+
+                    ],
+        "answer":  "正方形",
+        "solution":  "【分析】长方形有2条对称轴，正方形有4条，等边三角形有3条，等腰三角形有1条，比较可得最多的是正方形.\n【详解】在这些图形中，正方形的对称轴最多。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "21-22四年级下·河北秦皇岛·期末"
+    },
+    {
+        "id":  "137",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在常见的平面图形中，有无数条对称轴的图形是( ),等腰三角形有( )条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "圆;1",
+        "solution":  "【分析】圆的任意直径所在直线都可作对称轴，因此有无数条;等腰三角形只有底边高所在直线这一条对称轴.\n【详解】有无数条对称轴的图形是圆，等腰三角形有1条对称轴。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "21-22四年级下·山西太原·期末"
+    },
+    {
+        "id":  "138",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等边三角形有( )条对称轴,( )个0.01是0.59。",
+        "options":  [
+
+                    ],
+        "answer":  "3;59",
+        "solution":  "【分析】等边三角形三边相等，三角相等，所以有3条对称轴;0.59表示59个0.01.\n【详解】一个等边三角形有3条对称轴,59个0.01是0.59。",
+        "knowledge_points":  [
+                                 "小数的数位和计数单位的认识",
+                                 "小数的意义",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "21-22四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "139",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "圆有( )条对称轴，半圆有( )条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "无数;一",
+        "solution":  "【分析】圆的每条直径所在直线都是对称轴;半圆只有经过圆心并垂直于直径的那一条对称轴.\n【详解】圆有无数条对称轴，半圆有一条对称轴。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "18-19六年级上·全国·单元测试"
+    },
+    {
+        "id":  "140",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "正三角形有( )条对称轴，长方形最少有( )条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "3;2",
+        "solution":  "【分析】正三角形即等边三角形，有3条对称轴;长方形有2条对称轴.\n【详解】正三角形有3条对称轴，长方形最少有2条对称轴。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "21-22四年级下·陕西宝鸡·期末"
+    },
+    {
+        "id":  "141",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "图形的平移只改变图形的( ),不改变图形的( )。",
+        "options":  [
+
+                    ],
+        "answer":  "位置;形状和大小",
+        "solution":  "【分析】平移后图形的形状，大小，方向保持不变，仅位置改变.\n【详解】图形的平移只改变图形的位置，不改变图形的形状和大小。![](images/cc4f3d879b9c12b57175b81459b0d7cc5ec730bd62278e78b0bb2092cd41dabc.jpg)",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "21-22四年级下·陕西宝鸡·期末"
+    },
+    {
+        "id":  "142",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "三角形ABC经平移后，若点A向右平移了8格，则点B向右平移了( )格，点C向右平移了( )格。",
+        "options":  [
+
+                    ],
+        "answer":  "8;8",
+        "solution":  "【分析】平移时图形上所有点沿同一方向移动相同距离.\n【详解】点A向右平移8格，则点B,点C也都向右平移8格。",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "21-22五年级上·广东揭阳·期末"
+    },
+    {
+        "id":  "143",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "圆是( )图形，它有( )条对称轴。环形有( )条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "轴对称;无数;无数",
+        "solution":  "【分析】圆和环形都能沿经过圆心的直线对折重合，且这样的直线有无数条.\n【详解】圆是轴对称图形，它有无数条对称轴;环形也有无数条对称轴。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "22-23六年级上·全国·课后作业"
+    },
+    {
+        "id":  "144",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "点A到对称轴的距离是3小格，它的对称点A′到对称轴的距离是( )小格。",
+        "options":  [
+
+                    ],
+        "answer":  "3",
+        "solution":  "【分析】轴对称图形中，对称点到对称轴的距离相等.\n【详解】A′到对称轴的距离也是3小格。",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "21-22四年级下·广东阳江·期末"
+    },
+    {
+        "id":  "145",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在等边三角形，圆，长方形中，对称轴数量最多的是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "圆",
+        "solution":  "【分析】等边三角形有3条对称轴，长方形有2条，圆有无数条.\n【详解】对称轴数量最多的是圆。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "21-22四年级下·广东阳江·期末"
+    },
+    {
+        "id":  "146",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小学阶段学过的平面图形中，是轴对称图形的有( )(至少举3例),对称轴最多的图形是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "长方形，正方形，圆;圆",
+        "solution":  "【分析】满足沿某条直线对折后完全重合的图形是轴对称图形;常见图形中圆的对称轴最多.\n【详解】可举例：长方形，正方形，圆;对称轴最多的是圆。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "21-22六年级下·山东德州·期末"
+    },
+    {
+        "id":  "147",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在长方形，正方形，平行四边形和圆中,( )的对称轴最多。",
+        "options":  [
+
+                    ],
+        "answer":  "圆",
+        "solution":  "【分析】长方形2条，正方形4条，平行四边形0条，圆无数条.\n【详解】对称轴最多的是圆。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "20-21三年级上·贵州贵阳·期末"
+    },
+    {
+        "id":  "148",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "把一张长方形的纸对折3次，在中间剪下一个五角星，展开后这张纸上会出现( )个五角星。",
+        "options":  [
+
+                    ],
+        "answer":  "8",
+        "solution":  "【分析】对折3次后纸被平均分成2×2×2=8层，中间剪1个图案，展开得到8个.\n【详解】展开后会出现8个五角星。",
+        "knowledge_points":  [
+                                 "轴对称的剪纸问题"
+                             ],
+        "source":  "21-22四年级下·全国·假期作业"
+    },
+    {
+        "id":  "149",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "正五角星有( )条对称轴;等边三角形有( )条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "5;3",
+        "solution":  "【分析】正五角星和等边三角形均为轴对称图形，分别可数出对称轴条数.\n【详解】正五角星有5条对称轴，等边三角形有3条对称轴。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "21-22五年级下·全国·期末"
+    },
+    {
+        "id":  "150",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在轴对称图形中，对称轴两侧对应点到对称轴的距离( )。正方形有( )条对称轴。圆有( )条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "相等;4;无数",
+        "solution":  "【分析】轴对称图形对应点到对称轴的距离相等;正方形4条对称轴;圆有无数条.\n【详解】应填：相等,4，无数。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "21-22五年级下·全国·期末"
+    },
+    {
+        "id":  "151",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果在平面镜中看到钟表上时刻为![](images/3310777c09c1a8eb322e61d2f2af254604002c9d003488df768722bb57bab843.jpg),那么实际上现在是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "7:00",
+        "solution":  "【分析】镜面中时间与实际时间左右相反，图中镜像为5:00，则实际为7:00.\n【详解】实际上现在是7:00。",
+        "knowledge_points":  [
+                                 "镜面对称"
+                             ],
+        "source":  "19-20五年级·福建南平·期中"
+    },
+    {
+        "id":  "152",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "等腰梯形有( )条对称轴，长方形有( )条对称轴，正方形有( )条对称轴，圆有( )条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "1;2;4;无数",
+        "solution":  "【分析】按各图形轴对称性质判断.\n【详解】等腰梯形1条，长方形2条，正方形4条，圆无数条。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "2022六年级下·全国·专题练习"
+    },
+    {
+        "id":  "153",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "长方形有( )条对称轴，正方形有( )对称轴，等腰三角形有( )条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "2;4;1",
+        "solution":  "【分析】长方形2条，正方形4条，等腰三角形1条.\n【详解】如图:![](images/16d8ff7c2a8a7dcf7b45de542879e75413f340b4d7db817109f12a8a72d4897e.jpg);应填2,4,1。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "19-20四年级下·云南西双版纳·期末"
+    },
+    {
+        "id":  "154",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "平移不改变图形的( )和( ),只是图形的( )发生了变化。",
+        "options":  [
+
+                    ],
+        "answer":  "形状;大小;位置",
+        "solution":  "【分析】平移前后形状，大小不变，仅位置改变.\n【详解】应填：形状，大小，位置。",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "20-21四年级下·湖北荆门·期末"
+    },
+    {
+        "id":  "155",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在算盘上拨珠的运动是( )现象，时钟的时针，分针，秒针的运动是( )现象。(填\u0027平移\u0027或\u0027旋转\u0027)",
+        "options":  [
+
+                    ],
+        "answer":  "平移;旋转",
+        "solution":  "【分析】拨珠沿直线移动是平移;钟针绕中心转动是旋转.\n【详解】应填：平移;旋转。",
+        "knowledge_points":  [
+                                 "平移与平移现象",
+                                 "旋转与旋转现象"
+                             ],
+        "source":  "20-21三年级下·辽宁营口·期末"
+    },
+    {
+        "id":  "156",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "将\u0027平移\u0027或\u0027旋转\u0027填在下面括号里。①升降电梯的运动是( ) ②钟摆的运动是( ) ③拧开水龙头的运动是( ) ④抽屉拉出时的运动是( )",
+        "options":  [
+
+                    ],
+        "answer":  "平移;旋转;旋转;平移",
+        "solution":  "【分析】沿直线移动是平移，绕点转动是旋转.\n【详解】①平移;②旋转;③旋转;④平移。",
+        "knowledge_points":  [
+                                 "旋转与旋转现象",
+                                 "平移与平移现象"
+                             ],
+        "source":  "21-22三年级上·贵州毕节·期末"
+    },
+    {
+        "id":  "157",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "沿一条直线将某一图形对折后，两部分能完全重合，我们把这种图形叫( )图形，折痕所在的直线叫做( )。",
+        "options":  [
+
+                    ],
+        "answer":  "轴对称;对称轴",
+        "solution":  "【分析】本题是轴对称图形的定义.\n【详解】图形叫轴对称图形，折痕所在直线叫对称轴。图示:![](images/820748d6a63d9a6bb2e0c9ede35c5d43bb078db0fe62d7a737de3f1f0512e0f7.jpg)",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "21-22五年级上·四川宜宾·期末"
+    },
+    {
+        "id":  "158",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "拉抽屉，升国旗都属于( )运动。",
+        "options":  [
+
+                    ],
+        "answer":  "平移",
+        "solution":  "【分析】两者都是沿直线方向移动，方向与距离规律一致.\n【详解】都属于平移运动。",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "2020四年级下·湖南怀化·专题练习"
+    },
+    {
+        "id":  "159",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在超市乘电梯是( )运动。正方形有( )条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "平移;4",
+        "solution":  "【分析】乘电梯是直线移动，属平移;正方形有4条对称轴.\n【详解】在超市乘电梯是平移运动;正方形有4条对称轴。![](images/95e0f1b64ed33db61ba64e8d7e6119afaefb8303332f42cf51642c5076e05f9b.jpg)",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "平移与平移现象"
+                             ],
+        "source":  "17-18五年级上·河南商丘·期中"
+    },
+    {
+        "id":  "160",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "等边三角形的对称轴有( )条。",
+        "options":  [
+
+                    ],
+        "answer":  "3",
+        "solution":  "【分析】等边三角形三边三角都相等，可作3条对称轴.\n【详解】等边三角形:![](images/63a6355fba171a14ef17b539c6199f69be21feca6ae180fe3e70160db5a88463.jpg);对称轴有3条。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "18-19四年级下·湖南常德·期末"
+    },
+    {
+        "id":  "161",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "索道上运行的观光缆车是( )现象，推拉窗的移动也是( )现象。(填\u0027平移\u0027或\u0027旋转\u0027)",
+        "options":  [
+
+                    ],
+        "answer":  "平移;平移",
+        "solution":  "【分析】两者都沿直线方向移动，不绕点转动.\n【详解】都属于平移现象。",
+        "knowledge_points":  [
+                                 "平移与平移现象",
+                                 "旋转与旋转现象"
+                             ],
+        "source":  "2020四年级下·湖南怀化·专题练习"
+    },
+    {
+        "id":  "162",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "字母N,O,M,U,S,H中是轴对称图形的字母有( ),有无数条对称轴的字母是( ),有两条对称轴的字母是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "O,M,U,H;O;H",
+        "solution":  "【分析】按字母图形是否可沿直线对折重合判断，并统计对称轴条数.\n【详解】轴对称字母有O,M,U,H;其中O有无数条对称轴;H有两条对称轴。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "2022六年级上·全国·专题练习"
+    },
+    {
+        "id":  "163",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "圆的对称轴有( )条，正方形的对称轴有( )条。",
+        "options":  [
+
+                    ],
+        "answer":  "无数;4",
+        "solution":  "【分析】圆每条直径所在直线都是对称轴;正方形有2条中线和2条对角线共4条.\n【详解】圆有无数条对称轴，正方形有4条对称轴。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "2022六年级上·全国·专题练习"
+    },
+    {
+        "id":  "164",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "图形的平移是由( )和( )决定的。",
+        "options":  [
+
+                    ],
+        "answer":  "方向;距离",
+        "solution":  "【分析】平移位置由两个要素确定：平移方向和平移距离.\n【详解】图形的平移由方向和距离决定。",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "21-22五年级上·广东梅州·期末"
+    },
+    {
+        "id":  "165",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "圆是( )图形，有( )条对称轴，正方形有( )条对称轴，环形有( )条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "轴对称;无数;4;无数",
+        "solution":  "【分析】圆和环形都可沿过圆心直线对折重合，正方形有4条对称轴.\n【详解】圆是轴对称图形，有无数条对称轴;正方形有4条;环形有无数条。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "21-22六年级上·黑龙江七台河·期末"
+    },
+    {
+        "id":  "166",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "与时针旋转方向相同的是( )旋转，相反的是( )旋转。对折后两边能( )的图形是轴对称图形，折痕所在的直线叫作轴对称图形的( )。",
+        "options":  [
+
+                    ],
+        "answer":  "顺时针;逆时针;重合;对称轴",
+        "solution":  "【分析】与时针同向是顺时针，反向是逆时针;轴对称图形对折后两边重合，折痕是对称轴.\n【详解】依次填：顺时针，逆时针，重合，对称轴。",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认",
+                                 "旋转与旋转现象"
+                             ],
+        "source":  "21-22四年级下·全国·单元测试"
+    },
+    {
+        "id":  "167",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在平行四边形，等腰三角形和直角梯形中,( )是轴对称图形。",
+        "options":  [
+
+                    ],
+        "answer":  "等腰三角形",
+        "solution":  "【分析】等腰三角形沿底边高所在直线可对折重合;平行四边形和直角梯形一般不能.\n【详解】轴对称图形是等腰三角形。",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "21-22四年级下·全国·单元测试"
+    },
+    {
+        "id":  "168",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "圆是轴对称图形，有( )条对称轴，半圆有( )条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "无数;1",
+        "solution":  "【分析】圆的每条直径所在直线都可作对称轴;半圆只有1条对称轴.\n【详解】圆有无数条对称轴，半圆有1条对称轴。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "17-18六年级上·四川巴中·期末"
+    },
+    {
+        "id":  "169",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "请你根据图形对称轴的条数按照从多到少的顺序，在括号里填上适当的图形名称。圆,( ),( ),半圆。",
+        "options":  [
+
+                    ],
+        "answer":  "等边三角形;长方形",
+        "solution":  "【分析】按常见图形对称轴数量比较：圆无数条，等边三角形3条，长方形2条，半圆1条。\n【详解】从多到少可填：圆，等边三角形，长方形，半圆。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "20-21六年级上·江西宜春·期末"
+    },
+    {
+        "id":  "170",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "长方形，正方形和圆都是( )图形，半圆有( )条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "轴对称;1",
+        "solution":  "【分析】轴对称图形沿对称轴对折后两侧能完全重合。长方形，正方形，圆都满足这一特征;半圆只有1条对称轴.\n【详解】长方形，正方形和圆都是轴对称图形，半圆有1条对称轴。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "20-21六年级上·广东云浮·阶段练习"
+    },
+    {
+        "id":  "171",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "下图中(单位：米),阴影部分的面积是( )平方米。![](images/5b40d3d41fbc628e5bafe36143855e472bb9a3b29b1384eb09cb5f0b9cb28949.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "4600",
+        "solution":  "【分析】用平移思想把白色长方形移到右侧，阴影部分可看成长是117-17=100米，宽46米的长方形.\n【详解】117-17=100(米),100×46=4600(平方米),所以阴影部分面积是4600平方米。",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "阴影部分的周长和面积",
+                                 "利用平移法巧算面积"
+                             ],
+        "source":  "19-20四年级下·广东广州·期末"
+    },
+    {
+        "id":  "172",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "行驶中的车轮是( )运动，升国旗是( )运动。",
+        "options":  [
+
+                    ],
+        "answer":  "旋转;平移",
+        "solution":  "【分析】车轮绕轴转动属于旋转;升国旗沿直线方向上下移动属于平移.\n【详解】行驶中的车轮是旋转运动，升国旗是平移运动。",
+        "knowledge_points":  [
+                                 "旋转与旋转现象",
+                                 "平移与平移现象"
+                             ],
+        "source":  "20-21三年级下·河南商丘·期末"
+    },
+    {
+        "id":  "173",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个等腰三角形，它的一个底角是40°,顶角的度数是( )°;一个等边三角形，它每个内角的度数都是( )°,它有( )条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "100;60;3",
+        "solution":  "【分析】等腰三角形两个底角相等，顶角=180°-40°-40°。等边三角形三个内角都为60°,且有3条对称轴.\n【详解】180°-40°-40°=100°;等边三角形每个内角是60°,有3条对称轴。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "20-21四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "174",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "镜子中看到的钟面时间是3时，实际时间是( )时。",
+        "options":  [
+
+                    ],
+        "answer":  "9",
+        "solution":  "【分析】镜中时间与实际时间关于12点与6点连线对称，左右相反.\n【详解】镜中3时对应实际9时。示意图:![](images/db0738dd8e8c31cea404180bc882eb66d3aea64e4d06111c69379aac51913145.jpg)",
+        "knowledge_points":  [
+                                 "整点时间与钟面指针的位置",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "19-20三年级下·河南驻马店·期末"
+    },
+    {
+        "id":  "175",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "轴对称图形中两个对称点到( )的距离相等。图形平移前后,( )都不改变。",
+        "options":  [
+
+                    ],
+        "answer":  "对称轴;图形的形状、大小、方向",
+        "solution":  "【分析】轴对称图形对应点到对称轴距离相等；平移只改变位置，形状、大小、方向不变。\n【详解】应填：对称轴;图形的形状、大小、方向。",
+        "knowledge_points":  [
+                                 "平移与平移现象",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "20-21四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "176",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三角形最多有( )个锐角，一个正方形有( )条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "3;4",
+        "solution":  "【分析】等边三角形的三个角都是锐角，所以三角形最多有3个锐角;正方形有4条对称轴.\n【详解】一个三角形最多有3个锐角，一个正方形有4条对称轴。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "三角形的内角和"
+                             ],
+        "source":  "19-20四年级下·广西河池·期末"
+    },
+    {
+        "id":  "177",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把一个等边三角形沿着一条对称轴对折，然后沿折痕剪开，得到了两个相同的( )三角形，得到的三角形中最小的一个内角是( )°",
+        "options":  [
+
+                    ],
+        "answer":  "直角三角形;30°",
+        "solution":  "【分析】等边三角形每个角60°,沿对称轴对折后顶角被平分成30°,并形成直角三角形.\n【详解】得到两个相同的直角三角形，最小内角是30°。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "20-21四年级下·安徽芜湖·期末"
+    },
+    {
+        "id":  "178",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "学过的平面图形里，你最喜欢的轴对称图形是( ),它有( )条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "圆;无数条",
+        "solution":  "【分析】题目为开放填空，只要是轴对称图形即可。圆是常见示例，且有无数条对称轴。\n【详解】可填：圆;无数条。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "20-21四年级下·贵州贵阳·期末"
+    },
+    {
+        "id":  "179",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "长方形有( )条对称轴;国旗上的任意一个五角星都有( )条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "2;5",
+        "solution":  "【分析】长方形沿两条中线对折可重合;正五角星有5条对称轴.\n【详解】长方形有2条对称轴，五角星有5条对称轴。图示:![](images/11bff27a93ebfd35d74c3d278039fef329b636a536e0fe5ef7f64cfa4705803f.jpg) ![](images/6aa640673146885a6ad738a7d56c84889c9378bcc46a0fe35905131280964f69.jpg)",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "18-19四年级下·浙江杭州·期末"
+    },
+    {
+        "id":  "180",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "等边三角形有( )条对称轴，四边形的内角和是( )度。",
+        "options":  [
+
+                    ],
+        "answer":  "3;360",
+        "solution":  "【分析】等边三角形有3条对称轴。四边形可分成2个三角形，内角和为2×180°=360°.\n【详解】等边三角形有3条对称轴，四边形内角和是360度。",
+        "knowledge_points":  [
+                                 "多边形的内角和",
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "15-16四年级下·甘肃武威·期末"
+    },
+    {
+        "id":  "181",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "对称轴长方形有( )条，等腰梯形有( )条，正方形有( )条。",
+        "options":  [
+
+                    ],
+        "answer":  "2;1;4",
+        "solution":  "【分析】根据常见图形的对称轴数量：长方形2条，等腰梯形1条，正方形4条.\n【详解】应填2,1,4。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "15-16四年级下·甘肃平凉·期末"
+    },
+    {
+        "id":  "182",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个图形经过平移后，它的( )和( )没有变化，只有( )变化了。",
+        "options":  [
+
+                    ],
+        "answer":  "形状;大小;位置",
+        "solution":  "【分析】平移不改变形状，大小和方向，只改变位置.\n【详解】一个图形经过平移后，形状和大小没有变化，只有位置变化了。",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "18-19四年级下·甘肃天水·期末"
+    },
+    {
+        "id":  "183",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个正方形有( )条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "4",
+        "solution":  "【分析】正方形两条中线和两条对角线都可作对称轴.\n【详解】一个正方形有4条对称轴。![](images/a4a95c10116fcb73345780f1b27e8511d738f571ef48a5a87e0ca094be9910b0.jpg)",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "2014·全国·小升初模拟"
+    },
+    {
+        "id":  "184",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "用\u0027平移\u0027或\u0027旋转\u0027填空。拉抽屉是( )现象 电风扇的转动是( )现象 车轮转动是( )现象 直升机的螺旋桨转动是( )现象 升降国旗是( )现象 风车运动是( )现象",
+        "options":  [
+
+                    ],
+        "answer":  "平移;旋转;旋转;旋转;平移;旋转",
+        "solution":  "【分析】沿直线移动是平移，绕点或轴转动是旋转.\n【详解】依次填：平移，旋转，旋转，旋转，平移，旋转。",
+        "knowledge_points":  [
+                                 "平移与平移现象",
+                                 "旋转与旋转现象"
+                             ],
+        "source":  "20-21四年级下·全国·单元测试"
+    },
+    {
+        "id":  "185",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小强开始写作业了，看到镜子里的时间是5时整，则实际时间是( )时。",
+        "options":  [
+
+                    ],
+        "answer":  "7",
+        "solution":  "【分析】镜面对称下钟面左右相反，镜中5时对应实际7时.\n【详解】实际时间是7时。图示:![](images/5491ff9e70d8e539399c987c8d63cbe9b1d3ced710c9d2dfd6e3c4d80fc9f845.jpg)",
+        "knowledge_points":  [
+                                 "镜面对称"
+                             ],
+        "source":  "20-21四年级下·全国·单元测试"
+    },
+    {
+        "id":  "186",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "如果一个图形沿着一条直线对折，两侧的图形能够完全重合，这样的图形就叫( )图形，那条直线就是( )。长方形有( )条对称轴，等腰梯形有( )条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "轴对称;对称轴;2;1",
+        "solution":  "【分析】前两空是轴对称图形定义;后两空是常见图形对称轴数量.\n【详解】应填：轴对称，对称轴,2,1。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "20-21四年级下·全国·单元测试"
+    },
+    {
+        "id":  "187",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "等腰梯形有( )条对称轴，正方形有( )条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "1;4",
+        "solution":  "【分析】等腰梯形关于中垂线对称，有1条;正方形有4条对称轴.\n【详解】等腰梯形1条，正方形4条。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "19-20五年级下·河北·单元测试"
+    },
+    {
+        "id":  "188",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "我们从镜子里看到的时间是5时整，实际应是( )时。",
+        "options":  [
+
+                    ],
+        "answer":  "7",
+        "solution":  "【分析】镜中钟面与实际关于竖直轴对称，左右方向相反.\n【详解】镜子里5时整对应实际7时整。",
+        "knowledge_points":  [
+                                 "镜面对称"
+                             ],
+        "source":  "13-14三年级下·天津·期中"
+    },
+    {
+        "id":  "189",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "长方形有( )条对称轴，等腰三角形有( )条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "2;1",
+        "solution":  "【分析】长方形沿两条中线对折重合;等腰三角形沿底边高所在线对折重合.\n【详解】长方形有2条对称轴，等腰三角形有1条对称轴。",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "19-20四年级上·全国·期末"
+    },
+    {
+        "id":  "190",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在我们认识的平面图形中,( )只有两条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "长方形",
+        "solution":  "【分析】常见图形中：圆无数条，正方形4条，等边三角形3条，等腰三角形1条，长方形2条.\n【详解】只有两条对称轴的是长方形。",
+        "knowledge_points":  [
+
+                             ],
+        "source":  "19-20五年级上·河南郑州·期中"
+    },
+    {
+        "id":  "191",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "哪些是\u0027平移\u0027现象，哪些是\u0027旋转\u0027现象:(1)在算盘上拨珠的运动是( )现象;(2)自行车的踏脚运动是( )现象;(3)电梯里的上下运动是( )现象;(4)时钟上时针，分针，秒针的运动是( )现象。",
+        "options":  [
+
+                    ],
+        "answer":  "平移;旋转;平移;旋转",
+        "solution":  "【分析】沿直线移动是平移，绕固定点转动是旋转.\n【详解】(1)平移;(2)旋转;(3)平移;(4)旋转。",
+        "knowledge_points":  [
+
+                             ],
+        "source":  "2014五年级·全国·课后作业"
+    },
+    {
+        "id":  "192",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小红从镜子里看到的时间是七时整，实际应是( )时。",
+        "options":  [
+
+                    ],
+        "answer":  "五",
+        "solution":  "【分析】镜中与实际左右相反，镜中7时对应实际5时.\n【详解】实际应是五时整。图示:![](images/2026-03-13%20210935.jpg)",
+        "knowledge_points":  [
+                                 "镜面对称"
+                             ],
+        "source":  "13-14三年级下·甘肃酒泉·阶段练习"
+    },
+    {
+        "id":  "193",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "三角形在平移的过程中，三角形的( )不变,( )改变。",
+        "options":  [
+
+                    ],
+        "answer":  "形状和大小;位置",
+        "solution":  "【分析】平移前后形状，大小，方向不变，仅位置改变.\n【详解】三角形在平移中形状和大小不变，位置改变。",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "20-21五年级上·辽宁·单元测试"
+    },
+    {
+        "id":  "194",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "正方形有( )条对称轴，平行四边形有( )条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "4;0",
+        "solution":  "【分析】正方形是轴对称图形，有4条对称轴;一般平行四边形不是轴对称图形.\n【详解】正方形有4条对称轴，平行四边形有0条对称轴。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "19-20四年级下·江苏苏州·期末"
+    },
+    {
+        "id":  "195",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在平面内，如果一个图形沿着一条直线对折后两部分完全重合，这样的图形叫做( )。",
+        "options":  [
+
+                    ],
+        "answer":  "轴对称图形",
+        "solution":  "【分析】题目给出的正是轴对称图形定义.\n【详解】这样的图形叫做轴对称图形。",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "19-20三年级上·江苏·单元测试"
+    },
+    {
+        "id":  "196",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上\u0027平移\u0027或\u0027旋转\u0027。开关水龙头是( ) 电梯升降是( )",
+        "options":  [
+
+                    ],
+        "answer":  "旋转;平移",
+        "solution":  "【分析】开关水龙头绕轴转动是旋转;电梯上下直线移动是平移.\n【详解】开关水龙头是旋转，电梯升降是平移。",
+        "knowledge_points":  [
+
+                             ],
+        "source":  "19-20三年级下·福建南平·期中"
+    },
+    {
+        "id":  "197",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "黑体的汉字\u0027王\u0027\u0027田\u0027字母\u0027A\u0027都是轴对称图形，请你写出一个这样的汉字或字母( )。",
+        "options":  [
+
+                    ],
+        "answer":  "甲",
+        "solution":  "【分析】只要写出能沿某条直线对折后完全重合的字或字母即可。\n【详解】可写“甲”。",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "19-20四年级下·湖北武汉·期末"
+    },
+    {
+        "id":  "198",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "长方形有( )条对称轴，正方形有( )条对称轴，平行四边形有( )条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "2;4;0",
+        "solution":  "【分析】长方形2条，正方形4条，一般平行四边形0条.\n【详解】应填2,4,0。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "18-19四年级下·安徽合肥·期中"
+    },
+    {
+        "id":  "199",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "等边三角形有( )条对称轴，等腰梯形有( )条对称轴，圆有( )条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "3;1;无数",
+        "solution":  "【分析】等边三角形3条，等腰梯形1条，圆无数条.\n【详解】应填:3,1，无数。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "17-18六年级下·山东青岛·期末"
+    },
+    {
+        "id":  "200",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "梯形在平移前后，面积大小( )变化，圆形经过轴对称的转换得到的图形，与原图相比大小( )。",
+        "options":  [
+
+                    ],
+        "answer":  "没有;相等",
+        "solution":  "【分析】平移不改变图形大小;轴对称两侧图形完全重合，大小相等.\n【详解】梯形平移前后面积大小没有变化;圆形经轴对称转换后与原图大小相等。",
+        "knowledge_points":  [
+                                 "平移与平移现象",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "2020·四川·小升初模拟"
+    },
+    {
+        "id":  "201",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "线段AB经平移变换后，点A平移了5cm,则点B平移了( )cm。",
+        "options":  [
+
+                    ],
+        "answer":  "5",
+        "solution":  "【分析】平移时图形上的每一个点都沿相同方向移动相同距离.\n【详解】点A平移5cm,则点B也平移5cm。",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "19-20四年级下·江苏·单元测试"
+    },
+    {
+        "id":  "202",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明在镜子中看到钟面上是4:30，实际钟面上是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "7:30",
+        "solution":  "【分析】镜面对称下，钟面左右相反.\n【详解】镜中4:30对应实际7:30。",
+        "knowledge_points":  [
+                                 "镜面对称"
+                             ],
+        "source":  "19-20四年级下·全国·单元测试"
+    },
+    {
+        "id":  "203",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.4,
+        "question":  "阴影部分向左移动( )厘米，就可以使平行四边形变成长方形。![](images/f78b411f7ff14e646d0adbb2abc2323f3f1c64e007220df40d6ab866d86089c4.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】观察图形，需将阴影三角形向左平移，平移距离等于所需补齐的底边长度.\n【详解】需要向左平移15厘米。",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "19-20四年级下·全国·单元测试"
+    },
+    {
+        "id":  "204",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "图形平移后，图形的形状和大小( )。图形经过平移，对应线段( ),对应角( ),对应点所连的线段( )。平移的两个要素是平移的( )和平移的( )。",
+        "options":  [
+
+                    ],
+        "answer":  "不变;相等;相等;相等;方向;距离",
+        "solution":  "【分析】平移不改变形状和大小，只改变位置;对应关系保持一致.\n【详解】依次填：不变，相等，相等，相等，方向，距离。",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "19-20四年级下·全国·单元测试"
+    },
+    {
+        "id":  "205",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "轴对称图形至少有( )条对称轴。等腰三角形有( )条对称轴，等边三角形有( )条对称轴，长方形有( )条对称轴，正方形有( )条对称轴，圆有( )条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "1;1;3;2;4;无数",
+        "solution":  "【分析】按常见图形的对称轴数量填写.\n【详解】依次为1,1,3,2,4，无数。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "19-20四年级下·全国·单元测试"
+    },
+    {
+        "id":  "206",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "平移后的图形与原图形相比较，只改变( ),不改变( )和( )。",
+        "options":  [
+
+                    ],
+        "answer":  "位置;形状;大小",
+        "solution":  "【分析】平移是整体移动，形状大小保持不变.\n【详解】只改变位置，不改变形状和大小。",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "2020六年级下·全国·专题练习"
+    },
+    {
+        "id":  "207",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "等腰三角形有( )条对称轴;平行四边形有( )条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "1;0",
+        "solution":  "【分析】等腰三角形是轴对称图形，平行四边形一般不是.\n【详解】等腰三角形1条;平行四边形0条。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "18-19五年级上·广东深圳·期末"
+    },
+    {
+        "id":  "208",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "自行车车轮的转动是( )现象，学校伸缩大门的运动是( )现象",
+        "options":  [
+
+                    ],
+        "answer":  "旋转;平移",
+        "solution":  "【分析】车轮绕轴转动是旋转;伸缩门沿轨道直线移动是平移.\n【详解】依次为旋转，平移。",
+        "knowledge_points":  [
+                                 "旋转与旋转现象",
+                                 "平移与平移现象"
+                             ],
+        "source":  "19-20三年级上·全国·期末"
+    },
+    {
+        "id":  "209",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "钟表分针的运动是( )现象，活动推拉门是根据( )现象制成的。(填\u0027平移\u0027或\u0027旋转\u0027)",
+        "options":  [
+
+                    ],
+        "answer":  "旋转;平移",
+        "solution":  "【分析】分针绕中心转动是旋转;推拉门沿直线移动是平移.\n【详解】应填：旋转;平移。",
+        "knowledge_points":  [
+                                 "旋转与旋转现象",
+                                 "平移与平移现象"
+                             ],
+        "source":  "2020三年级上·江苏·专题练习"
+    },
+    {
+        "id":  "210",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "张叔叔开着小轿车行驶在一段笔直的公路上，这时车体的运动是( )现象。到达目的地后，张叔叔开门下车，车门的运动是( )现象。(填\u0027旋转\u0027或\u0027平移\u0027)",
+        "options":  [
+
+                    ],
+        "answer":  "平移;旋转",
+        "solution":  "【分析】车体直线前进是平移;车门绕铰链转动是旋转.\n【详解】依次为平移，旋转。",
+        "knowledge_points":  [
+                                 "旋转与旋转现象",
+                                 "平移与平移现象"
+                             ],
+        "source":  "18-19三年级上·山东德州·期末"
+    },
+    {
+        "id":  "211",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "经过平移，对应点所连的线段( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平行且相等",
+        "solution":  "【分析】平移性质：对应点连线方向一致且长度相同.\n【详解】对应点所连线段平行且相等。",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "19-20三年级上·山东·期中"
+    },
+    {
+        "id":  "212",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "☆它有( )条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "5",
+        "solution":  "【分析】五角星是轴对称图形，可找到5条对称轴.\n【详解】☆有5条对称轴。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "2019六年级下·全国·专题练习"
+    },
+    {
+        "id":  "213",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把一个长方形从中间对折，长方形的左右两边一定会完全( )。",
+        "options":  [
+
+                    ],
+        "answer":  "重合",
+        "solution":  "【分析】长方形是轴对称图形，沿对称轴对折两边重合.\n【详解】应填\u0027重合\u0027。",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认",
+                                 "长方形的概念及特点"
+                             ],
+        "source":  "19-20三年级上·全国·课后作业"
+    },
+    {
+        "id":  "214",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个图形对折后，两边能够完全( ),这个图形是( ),这条折痕所在的直线叫做( )。",
+        "options":  [
+
+                    ],
+        "answer":  "重合;轴对称图形;对称轴",
+        "solution":  "【分析】题目直接考查轴对称图形定义.\n【详解】依次填：重合，轴对称图形，对称轴。",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "18-19三年级下·甘肃张掖·期中"
+    },
+    {
+        "id":  "215",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "线段有( )条对称轴，圆有( )条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "1;无数",
+        "solution":  "【分析】线段有1条对称轴(垂直平分线);圆每条直径所在直线都是对称轴.\n【详解】线段1条，圆无数条。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "18-19四年级下·江苏连云港·期中"
+    },
+    {
+        "id":  "216",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "(填\u0027平移\u0027或\u0027旋转\u0027)升旗时国旗的运动是( )现象;钟面上时针的运动是( )现象。",
+        "options":  [
+
+                    ],
+        "answer":  "平移;旋转",
+        "solution":  "【分析】国旗沿旗杆直线升降是平移;时针绕中心转动是旋转.\n【详解】应填：平移;旋转。",
+        "knowledge_points":  [
+                                 "平移与平移现象",
+                                 "旋转与旋转现象"
+                             ],
+        "source":  "18-19三年级下·广东深圳·期中"
+    },
+    {
+        "id":  "217",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个图形的某条对称轴对称的图形和它的面积( )。",
+        "options":  [
+
+                    ],
+        "answer":  "相等",
+        "solution":  "【分析】轴对称后图形能完全重合，面积保持不变.\n【详解】面积相等。",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "2019六年级下·全国·专题练习"
+    },
+    {
+        "id":  "218",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在你认识常见的平面图形中，有无数条对称轴的图形是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "圆",
+        "solution":  "【分析】圆的任意直径所在直线都可作为对称轴.\n【详解】有无数条对称轴的是圆。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "2019·福建厦门·小升初真题"
+    },
+    {
+        "id":  "219",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "从0--9这些数字中，是轴对称图形的有( )。",
+        "options":  [
+
+                    ],
+        "answer":  "0,3,8",
+        "solution":  "【分析】按轴对称定义判断能否对折重合.\n【详解】是轴对称图形的有0,3,8。",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "18-19三年级下·辽宁·课后作业"
+    },
+    {
+        "id":  "220",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "坐升降电梯时，电梯的升降运动是( )现象。",
+        "options":  [
+
+                    ],
+        "answer":  "平移",
+        "solution":  "【分析】电梯整体沿竖直方向作直线移动.\n【详解】属于平移现象。",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "17-18四年级下·全国·期末"
+    },
+    {
+        "id":  "221",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在0,7,A,H,N,8这些数字或字母，是对称图形的有( )。",
+        "options":  [
+
+                    ],
+        "answer":  "0,A,H,8",
+        "solution":  "【分析】判断能否沿某直线对折后完全重合.\n【详解】对称图形有0,A,H,8。",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "17-18四年级下·广西贺州·期末"
+    },
+    {
+        "id":  "222",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "经过旋转的图形与原来图形的( )和( )完全相同。",
+        "options":  [
+
+                    ],
+        "answer":  "形状;大小",
+        "solution":  "【分析】旋转前后图形保持形状和大小不变.\n【详解】应填：形状，大小。",
+        "knowledge_points":  [
+                                 "旋转与旋转现象"
+                             ],
+        "source":  "14-15五年级·全国·期中"
+    },
+    {
+        "id":  "223",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "图形中的平行移动，简称平移，它是由移动的( )和( )决定的。",
+        "options":  [
+
+                    ],
+        "answer":  "方向;距离",
+        "solution":  "【分析】平移由两个核心要素决定：方向与距离.\n【详解】应填方向和距离。",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "18-19四年级下·全国·课后作业"
+    },
+    {
+        "id":  "224",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在一个三角形中,∠1=35°,∠2=45°,∠3=( ),这个三角形是( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "100°;钝角",
+        "solution":  "【分析】三角形内角和180°,∠3=180°-35°-45°。最大角大于90°即为钝角三角形.\n【详解】∠3=100°,这个三角形是钝角三角形。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "三角形的内角和"
+                             ],
+        "source":  "18-19四年级下·全国·单元测试"
+    },
+    {
+        "id":  "225",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "用3根同样长的小棒，摆成的三角形是( )三角形，每个角是( )度。",
+        "options":  [
+
+                    ],
+        "answer":  "等边;60",
+        "solution":  "【分析】三边相等是等边三角形，三个内角都为60°.\n【详解】摆成等边三角形，每个角是60度。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "18-19四年级下·全国·单元测试"
+    },
+    {
+        "id":  "226",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "7.81与3.39的和是( ),它们的差是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "11.2;4.42",
+        "solution":  "【分析】直接按小数加减法计算.\n【详解】7.81+3.39=11.2;7.81-3.39=4.42。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "18-19四年级下·全国·单元测试"
+    },
+    {
+        "id":  "227",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "8.11-0.73+1.89-2.77=8.11+( )-( )( )。",
+        "options":  [
+
+                    ],
+        "answer":  "1.89;0.73;2.77",
+        "solution":  "【分析】运用加法交换律和减法性质，把减去两个数写成减去它们的和.\n【详解】8.11-0.73+1.89-2.77=8.11+1.89-(0.73+2.77)。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算",
+                                 "整数减法的性质",
+                                 "小数加，减法简便运算"
+                             ],
+        "source":  "18-19四年级下·全国·单元测试"
+    },
+    {
+        "id":  "228",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "拨动算盘是( )现象。",
+        "options":  [
+
+                    ],
+        "answer":  "平移",
+        "solution":  "【分析】算珠沿算杆做直线往返运动，属于平移.\n【详解】拨动算盘是平移现象。",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "18-19三年级上·山东枣庄·期末"
+    },
+    {
+        "id":  "229",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "长方形的对称轴有( )条，等边三角形的对称轴有( )条。",
+        "options":  [
+                        "A. 2",
+                        "B. 3",
+                        "C. 4",
+                        "D. 5"
+                    ],
+        "answer":  "2;3",
+        "solution":  "【分析】长方形有2条对称轴，等边三角形有3条对称轴。\n【详解】所以长方形有2条对称轴，等边三角形有3条对称轴。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "18-19四年级下·全国·单元测试"
+    },
+    {
+        "id":  "230",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "圆有( )条对称轴。长方形有( )条对称轴，沿着对称轴至少旋转( )度与原来的图形重合。",
+        "options":  [
+
+                    ],
+        "answer":  "无数;2;180",
+        "solution":  "【分析】圆每条直径所在直线均为对称轴;长方形有2条对称轴，绕中心旋转180°可与原图重合.\n【详解】依次填：无数,2,180。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "18-19五年级下·全国·期中"
+    },
+    {
+        "id":  "231",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "等腰直角三角形有( )条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "1",
+        "solution":  "【分析】等腰直角三角形是等腰三角形的一种，仅有1条对称轴.\n【详解】等腰直角三角形有1条对称轴。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "18-19四年级下·全国·课后作业"
+    },
+    {
+        "id":  "232",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.4,
+        "question":  "在0-9这十个数中，是轴对称图形有( )。",
+        "options":  [
+
+                    ],
+        "answer":  "0,3,8",
+        "solution":  "【分析】按轴对称定义筛选数字.\n【详解】轴对称图形有0,3,8。",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "18-19四年级下·全国·课后作业"
+    },
+    {
+        "id":  "233",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个轴对称图形的点A离对称轴的距离为6厘米，它的对应点离对称轴是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】轴对称图形中，对应点到对称轴距离相等.\n【详解】对应点离对称轴也是6厘米。",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "18-19四年级下·全国·课后作业"
+    },
+    {
+        "id":  "234",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在方格纸上平移图形要先分别描出( )。",
+        "options":  [
+
+                    ],
+        "answer":  "对应点",
+        "solution":  "【分析】平移作图时先找关键点(对应点),再按方向和距离平移后连接.\n【详解】在方格纸上平移图形要先分别描出对应点。",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "18-19四年级下·全国·课后作业"
+    },
+    {
+        "id":  "235",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "等腰直角三角形有( )条对称轴，长方形有( )条对称轴，圆有( )条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "1;2;无数",
+        "solution":  "【分析】等腰直角三角形是特殊等腰三角形，有1条对称轴;长方形有2条;圆有无数条.\n【详解】应填:1,2，无数。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "2019·全国·小升初模拟"
+    },
+    {
+        "id":  "236",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "轴对称图形的特点是对称轴两侧的图形( ),对称点到对称轴的距离( )。",
+        "options":  [
+
+                    ],
+        "answer":  "完全重合;相等",
+        "solution":  "【分析】轴对称图形沿对称轴对折后两侧重合;对应点到对称轴距离相等.\n【详解】应填：完全重合，相等。",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "18-19四年级下·全国·课后作业"
+    },
+    {
+        "id":  "237",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "平移的特点:( ),( ),( )不变,( )变。",
+        "options":  [
+
+                    ],
+        "answer":  "大小;形状;方向;位置",
+        "solution":  "【分析】平移不改变图形大小，形状，方向，只改变位置.\n【详解】应填：大小，形状，方向，位置。",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "18-19四年级下·全国·单元测试"
+    },
+    {
+        "id":  "238",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "平移图形时要注意移动的方向和( )。",
+        "options":  [
+
+                    ],
+        "answer":  "距离",
+        "solution":  "【分析】平移由两个要素决定：方向和距离.\n【详解】应填：距离。",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "19-20四年级上·浙江·课后作业"
+    },
+    {
+        "id":  "239",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "线段AB是线段CD平移后得到的图形。点A为点C的对应点，说出点B的对应点D的位置:( )。",
+        "options":  [
+
+                    ],
+        "answer":  "在过B点与AC平行的直线上且点D在AC右侧,CD平行于AB,BD=AC",
+        "solution":  "【分析】平移性质：对应点连线平行且相等，对应线段也平行且相等.\n【详解】点D在过B点与AC平行的直线上，且在AC右侧;并满足CD平行于AB,BD=AC。",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "18-19四年级下·全国·课后作业"
+    },
+    {
+        "id":  "240",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在平移过程中，对应线段( )。",
+        "options":  [
+
+                    ],
+        "answer":  "平行且相等",
+        "solution":  "【分析】平移前后对应线段方向一致且长度不变.\n【详解】对应线段平行且相等。",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "18-19四年级下·全国·课后作业"
+    },
+    {
+        "id":  "241",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在平移过程中，平移后的图形与原来的图形( )和( )都相同，因此线段和对应角应该都( )。",
+        "options":  [
+
+                    ],
+        "answer":  "形状;大小;相同",
+        "solution":  "【分析】平移不改变形状和大小，对应线段和对应角保持不变.\n【详解】应填：形状，大小，相同。",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "18-19四年级下·全国·课后作业"
+    },
+    {
+        "id":  "242",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "填空(1)图形经过平移后，图形的( )位置，图形的( )形状，图形的( )大小。(填\u0027改变\u0027或\u0027不改变\u0027)(2)经过平移，每一组对应点所连成的线段( )。",
+        "options":  [
+
+                    ],
+        "answer":  "改变;不改变;不改变;平行而且相等",
+        "solution":  "【分析】平移改变位置，不改变形状和大小;对应点连线平行且相等.\n【详解】(1)位置改变，形状不改变，大小不改变。(2)对应点连线平行而且相等。",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "18-19四年级下·全国·课后作业"
+    },
+    {
+        "id":  "243",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在传送带上，如果传送带的某部分向前移动了20米，传送带上的其他部分向前移动了( )米。",
+        "options":  [
+
+                    ],
+        "answer":  "20",
+        "solution":  "【分析】传送带各部分作同方向同距离平移.\n【详解】其他部分也向前移动20米。",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "2019六年级下·全国·专题练习"
+    },
+    {
+        "id":  "244",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "轴对称图形沿它的( )对折后，两部分能完全重合。",
+        "options":  [
+
+                    ],
+        "answer":  "对称轴",
+        "solution":  "【分析】轴对称图形定义中，对折线即为对称轴.\n【详解】应填：对称轴。",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "15-16四年级上·浙江衢州·开学考试"
+    },
+    {
+        "id":  "245",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "物体平移后( )不变，改变的是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "大小，形状;位置",
+        "solution":  "【分析】平移只改变位置，不改变形状和大小.\n【详解】平移后大小，形状不变，改变的是位置。",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "19-20四年级上·浙江·课后作业"
+    },
+    {
+        "id":  "246",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "判断下列现象是否属于平移。(1)水管里水的流动。( ) (2)射出的子弹。( ) (3)火车在笔直的铁轨上行驶。( ) (4)滑雪运动员在平坦的雪地上滑行。( )",
+        "options":  [
+
+                    ],
+        "answer":  "不属于;不属于;属于;属于",
+        "solution":  "【分析】平移要求图形整体沿直线移动且形状方向不变.\n【详解】(1)不属于;(2)不属于;(3)属于;(4)属于。",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "18-19四年级下·全国·课后作业"
+    },
+    {
+        "id":  "247",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明的一本书共有104页，在这104页的页码中有两个数码的，并且这两个数码经过平移其中一个能得到另一个，这样的页码共有( )页。",
+        "options":  [
+
+                    ],
+        "answer":  "9",
+        "solution":  "【分析】两个数码能经平移重合，表示两个数字相同。两位页码中符合的是11,22,33,44,55,66,77,88,99.\n【详解】共有9页。",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "18-19四年级下·全国·课后作业"
+    },
+    {
+        "id":  "248",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "平移改变了图形的( )。",
+        "options":  [
+
+                    ],
+        "answer":  "位置",
+        "solution":  "【分析】平移不改变形状和大小，只改变位置.\n【详解】平移改变了图形的位置。",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "18-19四年级下·全国·课后作业"
+    },
+    {
+        "id":  "249",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个轴对称图形对称轴左右两边部分的面积( )。",
+        "options":  [
+
+                    ],
+        "answer":  "相等",
+        "solution":  "【分析】轴对称图形两侧能完全重合，面积相等.\n【详解】应填：相等。",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "18-19四年级下·全国·课后作业"
+    },
+    {
+        "id":  "250",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把一个图形从中间对折，左右两边完全重合，这样的图形就是( )图形。",
+        "options":  [
+
+                    ],
+        "answer":  "轴对称",
+        "solution":  "【分析】题干描述的是轴对称图形的定义。\n【详解】这样的图形就是轴对称图形。",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "18-19四年级下·全国·课后作业"
+    },
+    {
+        "id":  "251",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三角形先向上平移了8格，再向下平移了10格，这个三角形相对于原来的位置是向( )平移了( )格。",
+        "options":  [
+
+                    ],
+        "answer":  "下;2",
+        "solution":  "【分析】上下方向相反可相减:10-8=2.\n【详解】相对于原位置向下平移2格。",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "18-19三年级下·辽宁·课后作业"
+    },
+    {
+        "id":  "252",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "火车在一段笔直的铁路上开动，火车头以每小时60km的速度前行10分钟，则火车中间一节车厢走的路程是( )km。",
+        "options":  [
+
+                    ],
+        "answer":  "10",
+        "solution":  "【分析】火车整体平移，各节车厢路程相同。10分钟=1÷6小时.\n【详解】路程=60×1÷6=10(km)。",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "17-18四年级上·四川·课后作业"
+    },
+    {
+        "id":  "253",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "正方形有4条对称轴。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】正方形沿两条中线和两条对角线都可对折重合.\n【详解】说法正确，正方形有4条对称轴，故为√。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "正方形的概念及特点"
+                             ],
+        "source":  "16-17三年级·四川·课后作业"
+    },
+    {
+        "id":  "254",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "对折后两边能( )的图形是轴对称图形，折痕所在的直线叫作轴对称图形的( )。",
+        "options":  [
+
+                    ],
+        "answer":  "完全重合;对称轴",
+        "solution":  "【分析】题目直接考查轴对称定义.\n【详解】应填：完全重合，对称轴。",
+        "knowledge_points":  [
+                                 "图形与变换"
+                             ],
+        "source":  "16-17四年级下·河南新乡·期中"
+    },
+    {
+        "id":  "255",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "要画出某一图形平移后的图形，必须知道( )和( )。",
+        "options":  [
+
+                    ],
+        "answer":  "方向;距离",
+        "solution":  "【分析】平移作图的两个必要要素是方向和距离.\n【详解】必须知道平移方向和平移距离。",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "2014五年级·全国·课后作业"
+    },
+    {
+        "id":  "256",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "等边三角形有( )条对称轴，等腰三角形有( )条对称轴，等腰梯形有( )条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "3;1;1",
+        "solution":  "【分析】常见图形对称轴数量：等边三角形3条，等腰三角形1条，等腰梯形1条.\n【详解】应填:3,1,1。",
+        "knowledge_points":  [
+                                 "图形与变换"
+                             ],
+        "source":  "17-18四年级下·全国·课后作业"
+    },
+    {
+        "id":  "257",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "请你填一填。(1)正方形有( )条对称轴，圆有( )条对称轴。(2)在所在的轴对称图形中，最少有( )条对称轴，最多有( )条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "4;无数;1;无数",
+        "solution":  "【分析】正方形有4条，圆无数条;轴对称图形至少1条，对称轴最多可达无数条.\n【详解】应填:4，无数,1，无数。",
+        "knowledge_points":  [
+                                 "图形与变换"
+                             ],
+        "source":  "17-18四年级下·全国·课后作业"
+    },
+    {
+        "id":  "258",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在长方形，平行四边形，正方形中，一定是轴对称图形的是( )和( )。",
+        "options":  [
+
+                    ],
+        "answer":  "长方形;正方形",
+        "solution":  "【分析】长方形和正方形都能沿某条直线对折重合;一般平行四边形不能.\n【详解】一定是轴对称图形的是长方形和正方形。",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "15-16三年级·全国·期中"
+    },
+    {
+        "id":  "259",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "填一填。①将轴对称图形沿对称轴对折后，对称轴两边的图形( );②生活中常见的轴对称图形有( );③我们用手拧动水龙头属于( )现象;④物体从竿子的顶部滑下来属于( )现象。",
+        "options":  [
+
+                    ],
+        "answer":  "完全重合;水瓶;旋转;平移",
+        "solution":  "【分析】轴对称图形对折后两边重合；拧水龙头是旋转；沿竿滑下是平移。\n【详解】可填：完全重合;水瓶;旋转;平移。",
+        "knowledge_points":  [
+                                 "旋转与旋转现象",
+                                 "平移与平移现象",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "17-18三年级下·全国·课后作业"
+    },
+    {
+        "id":  "260",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "圆是( )图形，长方形有( )条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "轴对称;2",
+        "solution":  "【分析】圆可沿任意直径所在直线对折重合，是轴对称图形;长方形有2条对称轴.\n【详解】应填：轴对称;2。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "17-18二年级下·全国·期中"
+    },
+    {
+        "id":  "261",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "长方形有( )条对称轴，正方形有( )条对称轴，等边三角形有( )条对称轴，等腰三角形有( )条对称轴，正六边形有( )条对称轴，圆有( )条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "2;4;3;1;6;无数",
+        "solution":  "【分析】依据常见图形对称轴数量直接填写.\n【详解】依次为2,4,3,1,6，无数。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "17-18四年级下·全国·单元测试"
+    },
+    {
+        "id":  "262",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果一个图形沿着一条直线对折，两侧的图形能够完全重合，这样的图形就叫做( ),这条直线就是( )。对称点到对称轴的距离( ),对称点的连线与对称轴( )。",
+        "options":  [
+
+                    ],
+        "answer":  "轴对称图形;对称轴;相等;互相垂直",
+        "solution":  "【分析】题目综合考查轴对称图形定义及基本性质.\n【详解】应填：轴对称图形，对称轴，相等，互相垂直。",
+        "knowledge_points":  [
+                                 "图形与变换"
+                             ],
+        "source":  "17-18四年级下·全国·单元测试"
+    },
+    {
+        "id":  "263",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "等边三角形有( )条对称轴，长方形有( )条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "3;2",
+        "solution":  "【分析】等边三角形三条边对应三条对称轴;长方形沿两条中线对称.\n【详解】应填:3;2。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "17-18四年级下·全国·课后作业"
+    },
+    {
+        "id":  "264",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "等腰三角形有( )条对称轴，若它的一个底角是35°,则它的顶角是( )度。",
+        "options":  [
+
+                    ],
+        "answer":  "1;110",
+        "solution":  "【分析】等腰三角形有1条对称轴;顶角=180°-35°-35°.\n【详解】顶角是110°,对称轴1条。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "17-18四年级下·全国·课后作业"
+    },
+    {
+        "id":  "265",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明向前走了3米，是( )现象。(用\u0027平移\u0027或者\u0027旋转\u0027作答)",
+        "options":  [
+
+                    ],
+        "answer":  "平移",
+        "solution":  "【分析】人体整体沿某方向移动一定距离，属于平移.\n【详解】小明向前走是平移现象。",
+        "knowledge_points":  [
+                                 "图形与变换"
+                             ],
+        "source":  "17-18二年级下·全国·课后作业"
+    },
+    {
+        "id":  "266",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "如果一个图形沿着一条( )对折，两侧的图形能够完全重合，这个图形就是( )图形。折痕所在的这条( )叫做( )。",
+        "options":  [
+
+                    ],
+        "answer":  "直线;轴对称;直线;对称轴",
+        "solution":  "【分析】题目直接考查轴对称定义与名称.\n【详解】应填：直线，轴对称，直线，对称轴。",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "14-15三年级下·台湾新北·单元测试"
+    },
+    {
+        "id":  "267",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "数字和汉字是我们生活中离不开的\u0027朋友\u0027,仔细观察可以发现一些有对称现象的数字和汉字，请按要求写一写。①81033066388899 找出是轴对称图形的数字:( ) ②田具平字开图合 找出是轴对称图形的汉字:( )",
+        "options":  [
+
+                    ],
+        "answer":  "0、3、8;田、具、平、合",
+        "solution":  "【分析】按轴对称定义筛选能沿某直线对折重合的数字与汉字。\n【详解】轴对称的数字可选：0、3、8；轴对称的汉字可选：田、具、平、合。",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "17-18二年级下·全国·课后作业"
+    },
+    {
+        "id":  "268",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "等腰三角形有( )条对称轴，平行四边形有( )条高。",
+        "options":  [
+
+                    ],
+        "answer":  "一条或三条;无数",
+        "solution":  "【分析】一般等腰三角形有1条对称轴，特殊情况等边三角形(也是等腰三角形)有3条;平行四边形同一底可作无数条高.\n【详解】应填：一条或三条;无数。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "对称轴的画法及数量",
+                                 "平行四边形的高及画法"
+                             ],
+        "source":  "2014五年级·全国·课后作业"
+    },
+    {
+        "id":  "269",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "图形的变换方式有平移,( ),( )。",
+        "options":  [
+
+                    ],
+        "answer":  "旋转;轴对称",
+        "solution":  "【分析】小学阶段常见图形变换方式有三种：平移，旋转，轴对称.\n【详解】应填：旋转，轴对称。",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认",
+                                 "平移与平移现象",
+                                 "旋转与旋转现象"
+                             ],
+        "source":  "2014五年级·全国·课后作业"
+    },
+    {
+        "id":  "270",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "人们乘电梯的运动是( )现象，飞机飞行时，螺旋桨的运动是( )现象。",
+        "options":  [
+
+                    ],
+        "answer":  "平移;旋转",
+        "solution":  "【分析】乘电梯时人和电梯整体沿直线方向上下移动，属于平移;螺旋桨绕中心轴转动，属于旋转.\n【详解】人们乘电梯的运动是平移现象，飞机飞行时，螺旋桨的运动是旋转现象。",
+        "knowledge_points":  [
+                                 "平移与平移现象",
+                                 "旋转与旋转现象"
+                             ],
+        "source":  "2014五年级·全国·课后作业"
+    },
+    {
+        "id":  "271",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用\u0027平移\u0027或\u0027旋转\u0027填空。汽车在笔直的轨道上行驶车身是( )运动，它的方向盘运动是( )运动。",
+        "options":  [
+
+                    ],
+        "answer":  "平移;旋转",
+        "solution":  "【分析】车身沿直线前进属于平移;方向盘绕轴转动属于旋转.\n【详解】汽车车身是平移运动，方向盘是旋转运动。",
+        "knowledge_points":  [
+                                 "平移与平移现象",
+                                 "旋转与旋转现象"
+                             ],
+        "source":  "2014五年级·全国·课后作业"
+    },
+    {
+        "id":  "272",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "五星红旗缓缓升起，是一种( )现象。",
+        "options":  [
+
+                    ],
+        "answer":  "平移",
+        "solution":  "【分析】国旗升起时沿竖直方向直线移动，形状和大小不变.\n【详解】五星红旗缓缓升起是平移现象。",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "2014五年级·全国·课后作业"
+    },
+    {
+        "id":  "273",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "飞机降落到机场跑道到机身静止这一过程，对于整个机身而言，属于( )现象，而对于滚动的轮胎而言，它是( )现象。",
+        "options":  [
+
+                    ],
+        "answer":  "平移;旋转",
+        "solution":  "【分析】机身整体前进是平移;轮胎绕轴滚动是旋转.\n【详解】机身属于平移现象，轮胎属于旋转现象。",
+        "knowledge_points":  [
+                                 "平移与平移现象",
+                                 "旋转与旋转现象"
+                             ],
+        "source":  "2014五年级·全国·课后作业"
+    },
+    {
+        "id":  "274",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果三角形的两条边的长分别是4厘米和6厘米，那么第三条边的长必须大于( )厘米而小于( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "2;10",
+        "solution":  "【分析】由三角形三边关系可得：两边差\u003c第三边\u003c两边和.\n【详解】6-4\u003c第三边\u003c6+4，所以2\u003c第三边\u003c10。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "2014四年级·全国·课后作业"
+    },
+    {
+        "id":  "275",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三角形三条边的长度比是1∶1∶1，这个三角形有( )条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "3",
+        "solution":  "【分析】边长比1:1:1说明是等边三角形.\n【详解】等边三角形有3条对称轴。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "2014四年级·全国·课后作业"
+    },
+    {
+        "id":  "276",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三角形三条边的长度分别是5cm,7cm,5cm,这是一个( )三角形，有( )条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "等腰;1",
+        "solution":  "【分析】两边相等(5cm和5cm)是等腰三角形，等腰三角形有1条对称轴.\n【详解】这是一个等腰三角形，有1条对称轴。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "三角形的分类"
+                             ],
+        "source":  "2014四年级·全国·课后作业"
+    },
+    {
+        "id":  "277",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.64,
+        "question":  "图形的平移和旋转，相同点是图形的形状和大小不变，不同点是( )变了。",
+        "options":  [
+
+                    ],
+        "answer":  "方向",
+        "solution":  "【分析】平移过程中方向不变，旋转过程中方向会改变.\n【详解】不同点是方向变了。",
+        "knowledge_points":  [
+                                 "图形与变换"
+                             ],
+        "source":  "16-17四年级下·全国·期末"
+    },
+    {
+        "id":  "278",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.64,
+        "question":  "火箭升空是( )运动现象。",
+        "options":  [
+
+                    ],
+        "answer":  "平移",
+        "solution":  "【分析】火箭整体沿竖直方向移动，形状大小不变.\n【详解】火箭升空是平移运动现象。",
+        "knowledge_points":  [
+                                 "图形与变换"
+                             ],
+        "source":  "16-17五年级·全国·课后作业"
+    },
+    {
+        "id":  "279",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.64,
+        "question":  "( )梯形是轴对称图形。",
+        "options":  [
+
+                    ],
+        "answer":  "等腰",
+        "solution":  "【分析】梯形中只有等腰梯形可沿一条直线对折后重合.\n【详解】等腰梯形是轴对称图形。",
+        "knowledge_points":  [
+                                 "图形与变换"
+                             ],
+        "source":  "16-17五年级·全国·课后作业"
+    },
+    {
+        "id":  "280",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.64,
+        "question":  "轴对称图形是指( )。",
+        "options":  [
+
+                    ],
+        "answer":  "沿一条直线折叠，直线两旁的部分能够互相重合的图形",
+        "solution":  "【分析】这是轴对称图形的定义.\n【详解】轴对称图形是沿一条直线折叠后，直线两旁部分能够互相重合的图形。",
+        "knowledge_points":  [
+                                 "图形与变换"
+                             ],
+        "source":  "16-17五年级·全国·课后作业"
+    },
+    {
+        "id":  "281",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.64,
+        "question":  "写出一个轴对称图形:( )。",
+        "options":  [
+
+                    ],
+        "answer":  "等腰三角形",
+        "solution":  "【分析】能沿一条直线对折后两侧完全重合的图形都可以。\n【详解】可写：等腰三角形。",
+        "knowledge_points":  [
+                                 "图形与变换"
+                             ],
+        "source":  "16-17五年级·全国·课后作业"
+    },
+    {
+        "id":  "282",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "经过两次翻折(对称轴平行)后得到的图形，可以看成是原图形经过一次( )得到的。",
+        "options":  [
+
+                    ],
+        "answer":  "平移",
+        "solution":  "【分析】两次关于平行轴的对称变换等效于一次平移.\n【详解】可以看成经过一次平移得到。",
+        "knowledge_points":  [
+                                 "轴对称与平移的转换"
+                             ],
+        "source":  "17-18四年级上·四川·课后作业"
+    },
+    {
+        "id":  "283",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "每个轴对称图形至少有( )条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "1",
+        "solution":  "【分析】轴对称图形定义要求至少存在一条使两边重合的直线.\n【详解】每个轴对称图形至少有1条对称轴。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "16-17三年级下·全国·月考"
+    },
+    {
+        "id":  "284",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "等边三角形有( )条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "3",
+        "solution":  "【分析】等边三角形三边三角都相等，可作3条对称轴.\n【详解】等边三角形有3条对称轴。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "15-16三年级下·全国·月考"
+    },
+    {
+        "id":  "285",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "长方形有( )条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "2",
+        "solution":  "【分析】长方形沿两组对边中点连线对折可重合.\n【详解】长方形有2条对称轴。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "15-16六年级下·全国·课后作业"
+    },
+    {
+        "id":  "286",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "正方形有( )条对称轴;等腰三角形有( )条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "4;1",
+        "solution":  "【分析】正方形有4条对称轴;等腰三角形有1条对称轴.\n【详解】应填:4;1。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "14-15六年级·贵州铜仁·期末"
+    },
+    {
+        "id":  "287",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "等腰梯形有( )条对称轴，圆有( )条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "1;无数",
+        "solution":  "【分析】等腰梯形有且仅有1条对称轴;圆的任意直径所在直线都是对称轴.\n【详解】等腰梯形1条，圆无数条。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "11-12五年级下·湖南永州·期末"
+    },
+    {
+        "id":  "288",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "长方形有( )条对称轴，正方形有( )条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "2;4",
+        "solution":  "【分析】长方形有2条对称轴，正方形有4条对称轴.\n【详解】应填:2;4。",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "正方形的概念及特点",
+                                 "长方形的概念及特点"
+                             ],
+        "source":  "15-16四年级上·浙江衢州·开学考试"
+    },
+    {
+        "id":  "289",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "火车在铁轨上行驶是( )现象。",
+        "options":  [
+
+                    ],
+        "answer":  "平移",
+        "solution":  "【分析】火车沿轨道直线前进，形状大小方向不变.\n【详解】是平移现象。",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "14-15五年级上·安徽宿州·阶段练习"
+    },
+    {
+        "id":  "290",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "平行四边形( )轴对称图形。",
+        "options":  [
+
+                    ],
+        "answer":  "不是",
+        "solution":  "【分析】一般平行四边形无法沿某条直线对折后完全重合.\n【详解】平行四边形不是轴对称图形。",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "14-15四年级上·全国·课后作业"
+    },
+    {
+        "id":  "291",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "等腰三角形有( )条对称轴。",
+        "options":  [
+
+                    ],
+        "answer":  "1",
+        "solution":  "【分析】等腰三角形沿底边中线(高)对折可重合.\n【详解】等腰三角形有1条对称轴。![](images/ab007999a3c8b47ea2e5ae10821c6a072906c7ecf8411d7fc161ea156e5ef537.jpg)",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "15-16六年级下·全国·课后作业"
+    },
+    {
+        "id":  "292",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果把一个图形沿着( )对折，两侧的图形能够( ),这个图形就是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "一条直线;完全重合;轴对称图形",
+        "solution":  "【分析】这是轴对称图形定义.\n【详解】应填：一条直线，完全重合，轴对称图形。",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "15-16三年级·全国·期中"
+    },
+    {
+        "id":  "293",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "平移后得到的图形与原图形方向一致，大小一样。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】平移只改变位置，不改变方向与大小.\n【详解】题干说法正确，故为√。",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "23-24四年级下·河南焦作·期末"
+    },
+    {
+        "id":  "294",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "等边三角形有三条对称轴，等腰三角形有两条对称轴。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】等边三角形有3条对称轴，但等腰三角形通常只有1条.\n【详解】后半句错误，故整句为*。图示:![](images/a0737b4fc95819057c6c71170305ccfc.png)",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "23-24四年级下·河南焦作·期末"
+    },
+    {
+        "id":  "295",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "长方形，正方形，等边三角形和圆中，对称轴条数最多的是圆。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】长方形2条，正方形4条，等边三角形3条，圆无数条.\n【详解】圆对称轴最多，题干正确，故为√。",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "23-24四年级下·河南洛阳·期末"
+    },
+    {
+        "id":  "296",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "长方形，正方形，圆，平行四边形一定都是轴对称图形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】长方形，正方形，圆是轴对称图形，但一般平行四边形不是.\n【详解】题干错误，故为*。示意图:![](images/49d8af1fa9fea52b8949df27431cbfb8a86f991eaa7768d8989cdba637a577bf.jpg)",
+        "knowledge_points":  [
+                                 "长方形，正方形，三角形和圆的初步认识",
+                                 "轴对称的认识及辨认",
+                                 "平行四边形的概念及特点"
+                             ],
+        "source":  "23-24四年级下·河南信阳·期末"
+    },
+    {
+        "id":  "297",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "等边三角形有3条对称轴，平行四边形有2条对称轴。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】等边三角形有3条对称轴;一般平行四边形没有对称轴.\n【详解】后半句错误，故题干为*。图示:![](images/3faeafbecae0bfde8cae65fb87a8a80dadb552abb57be3436cd344a7394c1b2f.jpg)",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "轴对称的认识及辨认",
+                                 "平行四边形的概念及特点"
+                             ],
+        "source":  "23-24四年级下·河南新乡·期末"
+    },
+    {
+        "id":  "298",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "等腰三角形和等边三角形都只有1条对称轴。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】等腰三角形有1条对称轴，等边三角形有3条.\n【详解】说法错误，故为*。图示:![](images/a09621834e0f1f59c442b56297bab7d6e11d4bd540e9ec9c3fadb2f3b982696c.jpg) ![](images/80af77889547ee2e4be11815a2c26a88ad79fdb46913053c3fe6d88c75f8f872.jpg)",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "23-24四年级下·河南信阳·期末"
+    },
+    {
+        "id":  "299",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "在平移的过程中，图形的大小和形状不发生变化。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】平移保持图形大小和形状不变，仅位置改变.\n【详解】说法正确，故为√。",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "23-24四年级下·河南信阳·期末"
+    },
+    {
+        "id":  "300",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "正方形，平行四边形，等边三角形都是轴对称图形，其中正方形的对称轴条数最多。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平行四边形一般不是轴对称图形，所以前提错误.\n【详解】题干错误，故为*。图示:![](images/9679b05828b7db04d1586d44b6ecf809.png)",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "23-24四年级下·河南新乡·期末"
+    },
+    {
+        "id":  "301",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "平移后的图形和原图形相比大小，形状都不变，位置变了。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】平移后图形的位置改变，形状，大小，方向都不变.\n【详解】原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "23-24四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "302",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "轴对称，平移都是图形的运动。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】轴对称与平移都属于图形运动方式，图形大小形状不变，位置关系变化.\n【详解】原说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "平移与平移现象",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "23-24四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "303",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "![](images/2fd4449f51c237291dec10db495e675a8bc3d5ad8bac428b3e43cc0c850c34e1.jpg)是轴对称图形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】轴对称图形沿某条直线对折后两边完全重合.\n【详解】该图形左右可重合，是轴对称图形。故答案为:√",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "23-24四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "304",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "梯形至少有一条对称轴。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】只有等腰梯形有1条对称轴，一般梯形没有.\n【详解】原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "梯形的概念及特点",
+                                 "对称轴的画法及数量",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "24-25六年级上·全国·课后作业"
+    },
+    {
+        "id":  "305",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "平移改变了图形的形状和大小。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平移只改变位置，不改变图形形状和大小.\n【详解】原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "23-24四年级下·新疆吐鲁番·期末"
+    },
+    {
+        "id":  "306",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "梯形，正方形，长方形，等边三角形都是轴对称图形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】梯形中只有等腰梯形是轴对称图形.\n【详解】原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "直角梯形和等腰梯形的概念及特点",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "轴对称的认识及辨认",
+                                 "正方形的概念及特点"
+                             ],
+        "source":  "23-24四年级下·河北邯郸·期末"
+    },
+    {
+        "id":  "307",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "长方形，正方形，三角形，等腰梯形，圆都是轴对称图形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】三角形不一定是轴对称图形(一般三角形不是).\n【详解】原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认",
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "23-24四年级下·浙江宁波·期末"
+    },
+    {
+        "id":  "308",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "在所有的四边形中，对称轴条数最多的图形是正方形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】小学常见四边形中，正方形有4条对称轴，条数最多.\n【详解】原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "四边形的分类及关系"
+                             ],
+        "source":  "23-24五年级下·河北承德·期末"
+    },
+    {
+        "id":  "309",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "等腰三角形是轴对称图形，对称轴就是它底边上的高。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】等腰三角形沿底边上的高对折后两边重合.\n【详解】原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "三角形的高及画法",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "23-24四年级下·江西抚州·期末"
+    },
+    {
+        "id":  "310",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "图形平移前后的形状不变，位置变了。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】平移保持形状，大小不变，只改变位置.\n【详解】原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "23-24四年级下·江西赣州·期末"
+    },
+    {
+        "id":  "311",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "将5厘米长的线段向左平移3厘米，平移后的线段长8厘米。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平移不改变长度.\n【详解】平移后线段仍长5厘米，不是8厘米。故答案为: ×",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "23-24四年级下·湖南怀化·期末"
+    },
+    {
+        "id":  "312",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形中有一个角是60°,这个三角形有3条对称轴。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】等腰三角形有一个角为60°时，必为等边三角形;等边三角形有3条对称轴.\n【详解】原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "23-24四年级下·湖北鄂州·期末"
+    },
+    {
+        "id":  "313",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "一个正六边形有6条对称轴。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】正六边形关于过顶点与对边中点，过对顶点的直线均对称，共6条.\n【详解】原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "23-24四年级下·湖北随州·期末"
+    },
+    {
+        "id":  "314",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "三角形和梯形都是轴对称图形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】三角形，梯形都不一定是轴对称图形，只有部分特殊类型才是.\n【详解】原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "直角梯形和等腰梯形的概念及特点",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "23-24四年级下·山东日照·期末"
+    },
+    {
+        "id":  "315",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "长方形，等腰三角形，等腰梯形，平行四边形都是轴对称图形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平行四边形一般不是轴对称图形.\n【详解】原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "23-24四年级下·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "316",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "一个图形只能按上，下，左，右四个方向平移。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平移可以沿任意方向，不限于四个基本方向.\n【详解】原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "23-24四年级下·湖北武汉·期末"
+    },
+    {
+        "id":  "317",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "图形在平移运动时，图形上所有的点都朝着相同的方向平移了相同的距离。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】平移定义即图形上每个点同向，等距移动.\n【详解】原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "22-23四年级下·湖北武汉·期末"
+    },
+    {
+        "id":  "318",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "正方形，长方形，三角形，平行四边形都是轴对称图形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平行四边形不是轴对称图形，三角形也不一定是轴对称图形.\n【详解】原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "22-23四年级下·湖北十堰·期末"
+    },
+    {
+        "id":  "319",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "把水桶提起来是平移运动。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】提水桶时整体沿竖直方向移动，形状大小方向不变.\n【详解】属于平移运动。故答案为:√",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "22-23四年级下·湖北黄石·期末"
+    },
+    {
+        "id":  "320",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "等边三角形任意一条边上的高都是这个三角形的对称轴。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】等边三角形三条高，中线，角平分线重合，均为对称轴.\n【详解】原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "23-24六年级下·湖北·期末"
+    },
+    {
+        "id":  "321",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "平行四边形不是轴对称图形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】一般平行四边形沿任一直线对折都不能完全重合.\n【详解】原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "22-23四年级下·湖南湘西·期末"
+    },
+    {
+        "id":  "322",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "圆，长方形，正方形，正三角形都是轴对称图形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】这些图形都能找到至少一条对称轴并可对折重合.\n【详解】原说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "22-23四年级下·湖南永州·期末"
+    },
+    {
+        "id":  "323",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "图形![](images/2026-03-14%20141313.jpg)有3条对称轴。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】根据该图形的轴对称特征，可找到3条能使两侧重合的折线.\n【详解】该图形有3条对称轴。原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "22-23四年级下·湖南常德·期末"
+    },
+    {
+        "id":  "324",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "平移后的图形与原图形相比，大小，形状，位置都不变。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平移后大小，形状不变，但位置会改变.\n【详解】原说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "22-23四年级下·湖南常德·期末"
+    },
+    {
+        "id":  "325",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "苹果落地不属于平移运动。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】苹果下落沿直线方向移动，符合平移特征.\n【详解】苹果落地属于平移运动，原说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "22-23四年级下·湖南怀化·期末"
+    },
+    {
+        "id":  "326",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "图形平移只改变图形的位置，其大小和形状都不会发生变化。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】平移定义决定其仅改变位置.\n【详解】原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "22-23四年级下·湖南娄底·期末"
+    },
+    {
+        "id":  "327",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "直角三角形一定是轴对称图形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】只有等腰直角三角形是轴对称图形，普通直角三角形不是.\n【详解】原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的分类",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "22-23四年级下·山东日照·期末"
+    },
+    {
+        "id":  "328",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "平行四边形，长方形，等边三角形都是轴对称图形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平行四边形不是轴对称图形.\n【详解】原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "轴对称的认识及辨认",
+                                 "长方形的概念及特点",
+                                 "平行四边形的概念及特点"
+                             ],
+        "source":  "22-23五年级上·河南周口·期末"
+    },
+    {
+        "id":  "329",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "平移不改变图形的大小，位置和形状，只改变图形的方向。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平移不改变大小，形状，方向，只改变位置.\n【详解】原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "23-24四年级下·全国·单元测试"
+    },
+    {
+        "id":  "330",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "图形在旋转和平移时，形状和大小都不发生变化。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】平移与旋转都保持图形形状和大小不变.\n【详解】原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "旋转与旋转现象",
+                                 "平移与平移现象"
+                             ],
+        "source":  "23-24五年级上·河南周口·期末"
+    },
+    {
+        "id":  "331",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "长方形，等腰三角形，等腰梯形都是轴对称图形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】这三类图形都能沿某条直线对折后两边重合.\n【详解】原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "轴对称的认识及辨认",
+                                 "长方形的概念及特点"
+                             ],
+        "source":  "23-24五年级上·重庆垫江·期末"
+    },
+    {
+        "id":  "332",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "三角形有3条对称轴。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】只有等边三角形有3条对称轴;等腰三角形1条，一般三角形0条.\n【详解】原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "三角形的分类",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "10-11五年级下·云南楚雄·期中"
+    },
+    {
+        "id":  "333",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "从![](images/1ca20b72147901e43ccea4ae34bd82a1b4edebe770afde36fa6362d29f1fb2d8.jpg)上剪下来的是![](images/581e44980ac963e33afced4afb60a6bdc76b4be0f51b48f07ba3bf3f6d29cc7a.jpg)。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】根据轴对称剪纸规律，先确定折痕与半图形，再还原完整图形判断.\n【详解】题干所示剪下图形与还原结果一致，原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "轴对称的剪纸问题"
+                             ],
+        "source":  "23-24五年级上·河南周口·期中"
+    },
+    {
+        "id":  "334",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "数字\u00278\u0027的对称轴有2条，字母\u0027W\u0027的对称轴有一条。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】根据轴对称图形定义，若沿某直线对折后两侧完全重合，则该直线为对称轴.\n【详解】数字\u00278\u0027有2条对称轴，字母\u0027W\u0027有1条对称轴，原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "23-24五年级上·河南驻马店·期中"
+    },
+    {
+        "id":  "335",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "\u0027画\u0027和\u0027图\u0027这两个字都是轴对称图形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】判断汉字是否为轴对称图形，关键看是否存在一条对称轴使对折后两边重合.\n【详解】\u0027画\u0027是轴对称图形,\u0027图\u0027不是轴对称图形，原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "19-20五年级上·河北邯郸·期中"
+    },
+    {
+        "id":  "336",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "字母A的对称轴只有一条。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】字母A沿竖直中线对折可重合.\n【详解】字母A只有1条对称轴，原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "19-20五年级上·江西抚州·期中"
+    },
+    {
+        "id":  "337",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "平移会改变图形的位置，大小和形状。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平移只改变位置，不改变形状，大小和方向.\n【详解】原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "22-23五年级上·四川宜宾·期中"
+    },
+    {
+        "id":  "338",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "字母\u0027O\u0027是轴对称图形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】字母O可沿多条直线对折后重合.\n【详解】字母\u0027O\u0027是轴对称图形，原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "22-23五年级上·四川宜宾·期中"
+    },
+    {
+        "id":  "339",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "等腰三角形是轴对称图形，并且只有1条对称轴。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】等腰三角形沿底边高所在直线对折后两边重合.\n【详解】等腰三角形是轴对称图形，且只有1条对称轴。![](images/095a0dee69afe994aecf2444f8547595999abee36647c6b18efd1b768963ba4a.jpg) 原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认",
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "20-21五年级上·河南郑州·期中"
+    },
+    {
+        "id":  "340",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "平移会改变图形的位置和大小。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平移后图形大小不变，只改变位置.\n【详解】原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "22-23四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "341",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "长方形有四条对称轴，三角形有三条对称轴。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】长方形只有2条对称轴;三角形中只有等边三角形有3条，其他不一定.\n【详解】原题说法错误。![](images/b1da23a6476cec695db561e124e02439.png) 故答案为: ×",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "22-23四年级下·山东临沂·期末"
+    },
+    {
+        "id":  "342",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "一个图形经过平移后得到的图形与原图形大小相等，形状相同。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】平移不改变图形形状和大小.\n【详解】原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "22-23四年级下·山东菏泽·期末"
+    },
+    {
+        "id":  "343",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "对称轴两边的对称点的连线和对称轴互相垂直。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】轴对称图形性质：对称点到对称轴距离相等，连线与对称轴垂直.\n【详解】原题说法正确。![](images/98bf1576773eee99d64829ea167fcb923683a7ffad01136827f75a421f3a39f8.jpg) 故答案为:√",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "22-23四年级下·山东菏泽·期末"
+    },
+    {
+        "id":  "344",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "一个图形经过平移后，形状发生了变化。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平移只改变位置，不改变形状与大小.\n【详解】原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "21-22五年级上·吉林四平·期末"
+    },
+    {
+        "id":  "345",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "字母\u0027Z\u0027可以看作是轴对称图形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】字母Z沿任一直线对折后通常不能完全重合.\n【详解】字母\u0027Z\u0027不是轴对称图形，原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "21-22五年级上·吉林辽源·期末"
+    },
+    {
+        "id":  "346",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "一个轴对称图形，对称轴两边的图形通过平移可以完全重合。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】轴对称图形应通过沿对称轴对折重合，不是通过平移重合.\n【详解】原题说法错误。![](images/1b46ea92a98b1f96de15ae6dd9712fc4.png) 故答案为: ×",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "21-22五年级上·吉林白山·期末"
+    },
+    {
+        "id":  "347",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "在等边三角形，平行四边形，正方形，圆中，对称轴最多的图形是圆。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】等边三角形3条，平行四边形0条，正方形4条，圆无数条.\n【详解】圆对称轴最多，原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "20-21四年级下·山东临沂·期末"
+    },
+    {
+        "id":  "348",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "正方形，圆，平行四边形和等腰梯形都是轴对称图形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平行四边形一般不是轴对称图形.\n【详解】原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "22-23六年级下·湖北十堰·期末"
+    },
+    {
+        "id":  "349",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "一个图形经过平移后本身的形状改变了。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平移不改变图形形状和大小.\n【详解】原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "20-21四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "350",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "三角形只有一条对称轴。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】等边三角形有3条，等腰三角形有1条，一般三角形无对称轴.\n【详解】不是所有三角形都只有1条对称轴。故答案为: ×",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "22-23五年级·全国·假期作业"
+    },
+    {
+        "id":  "351",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "长方形有两条对称轴。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】长方形沿两组对边中点连线可对折重合.\n【详解】长方形有2条对称轴。![](images/5774e05a5e2250aff223c6bca1c44f2f6088acecddbef4241f4f132f42c48e5f.jpg) 原说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "22-23五年级·全国·假期作业"
+    },
+    {
+        "id":  "352",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "用传送带传送物品，推拉抽屉都是平移现象。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】两者都沿直线方向整体移动.\n【详解】都属于平移现象，原说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "22-23四年级·全国·假期作业"
+    },
+    {
+        "id":  "353",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "等腰直角三角形是轴对称图形，它只有一条对称轴。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】等腰直角三角形沿斜边中线(垂直平分线)对折可重合.\n【详解】它只有1条对称轴。![](images/2206705fd80ce658e8d83ac05208e72b1b70d3855a810fdcf079e3481573317c.jpg) 原说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "20-21四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "354",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "长方形，圆，平行四边形都是轴对称图形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平行四边形不是轴对称图形.\n【详解】原题说法错误。![](images/96b28600ccf5072f72898a33be93877cdd126efbf9b186a6a591636b1d196110.jpg) 故答案为: ×",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "21-22六年级下·重庆璧山·期末"
+    },
+    {
+        "id":  "355",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "在等边三角形，平行四边形，正方形，圆中，对称轴最多的图形是圆，对称轴最少的图形是平行四边形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】从给出的数量关系看：圆最多，平行四边形最少是正确结论.\n【详解】原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "20-21四年级下·山东菏泽·期末"
+    },
+    {
+        "id":  "356",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "平行四边形有2条对称轴，等腰三角形有1条对称轴，直角梯形没有对称轴。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平行四边形一般没有对称轴;等腰三角形有1条;直角梯形通常无对称轴.\n【详解】原题首句错误，故整体错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "22-23五年级上·安徽黄山·期末"
+    },
+    {
+        "id":  "357",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "沿着平行四边形的任何一条对角线剪开，可以得到两个完全相同的三角形，所以平行四边形对角线所在的直线是它的对称轴。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】\u0027剪开得到全等三角形\u0027不等于\u0027沿该线对折能重合\u0027.\n【详解】平行四边形沿对角线对折不能重合，不是其对称轴。故答案为: ×",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "2023·全国·小升初模拟"
+    },
+    {
+        "id":  "358",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "周一早上学校升国旗时国旗的运动是平移现象。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】国旗沿旗杆上下做直线平行移动.\n【详解】属于平移现象，原说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "20-21三年级上·江苏·单元测试"
+    },
+    {
+        "id":  "359",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "圆的对称轴只有4条。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】圆的任意直径所在直线都是对称轴.\n【详解】圆有无数条对称轴，不是4条。故答案为: ×",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "20-21四年级下·湖南常德·期末"
+    },
+    {
+        "id":  "360",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "长方形和正方形的对角线所在的直线都不是图形的对称轴。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】长方形对角线不是其对称轴;正方形两条对角线是其对称轴.\n【详解】原题\u0027都不是\u0027错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "正方形的概念及特点",
+                                 "长方形的概念及特点"
+                             ],
+        "source":  "21-22五年级上·重庆彭水·期末"
+    },
+    {
+        "id":  "361",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "正方形有两条对称轴。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】正方形有4条对称轴(两条中线，两条对角线).\n【详解】原题说法错误。![](images/04c58a5dce09aa519f8a9dc772c961ad38c0a5093cc66ee4631b8ed286260a4d.jpg) 故答案为: ×",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "18-19四年级下·江苏盐城·期中"
+    },
+    {
+        "id":  "362",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "物体平移时大小，形状都不发生变化，只是位置发生变化。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】平移特征：大小，形状，方向不变，位置变.\n【详解】原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "21-22五年级上·河南商丘·期末"
+    },
+    {
+        "id":  "363",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "物体平移后，形状和大小不变，改变的是位置。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】平移不改变图形大小和形状，只改变位置.\n【详解】原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "21-22四年级下·陕西安康·期末"
+    },
+    {
+        "id":  "364",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "不同的轴对称图形，对称轴的条数可能不一样。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】不同图形对称轴条数不同，如长方形2条，正方形4条.\n【详解】原题说法正确。![](images/2eb4f9844769ce0f1429d3819c49a1ae.png) 故答案为:√",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "21-22四年级下·湖北咸宁·期末"
+    },
+    {
+        "id":  "365",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "平移不会改变图形的位置和大小。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平移会改变位置，但不改变大小.\n【详解】原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "作平移后的图形",
+                                 "平移与平移现象"
+                             ],
+        "source":  "21-22四年级下·湖北咸宁·期末"
+    },
+    {
+        "id":  "366",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "将5厘米长的线段向左平移3米，平移后的线段长8厘米。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平移不改变线段长度.\n【详解】平移后仍是5厘米，不是8厘米。故答案为: ×",
+        "knowledge_points":  [
+                                 "作平移后的图形",
+                                 "平移与平移现象"
+                             ],
+        "source":  "21-22四年级下·湖北孝感·期末"
+    },
+    {
+        "id":  "367",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "不同的轴对称图形，对称轴的条数一定不一样。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】不同图形对称轴条数可能相同，如等腰三角形与等腰梯形都可为1条.\n【详解】题干中的\u0027一定不一样\u0027错误。![](images/1dbbbd27e0b96ed316b69bad8a4dae2a2d868748c9902c85126d669d82ee492a.jpg) ![](images/9a3157d70dd256854553100d97660a520293418258fe992b968798ecfd48e460.jpg) 故答案为: ×",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "21-22四年级下·湖北黄冈·期末"
+    },
+    {
+        "id":  "368",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "一个图形经过平移后，它的大小和形状都不发生改变。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】平移前后图形大小，形状不变.\n【详解】原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "17-18四年级下·全国·单元测试"
+    },
+    {
+        "id":  "369",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "升国旗是平移运动。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】国旗沿旗杆方向上下平行移动，属于平移.\n【详解】原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "21-22四年级下·四川南充·期末"
+    },
+    {
+        "id":  "370",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "升国旗是平移运动。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】国旗升降时沿着旗杆方向作直线平行移动，形状和大小不变.\n【详解】升国旗属于平移运动，原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "21-22四年级下·四川南充·期末"
+    },
+    {
+        "id":  "371",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "把一个图形先平移再旋转后，形状不变，大小发生变化。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平移和旋转都不改变图形的形状和大小，只改变位置(旋转还会改变方向).\n【详解】原题中\u0027大小发生变化\u0027错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "旋转与旋转现象",
+                                 "平移与平移现象"
+                             ],
+        "source":  "21-22五年级上·重庆綦江·期末"
+    },
+    {
+        "id":  "372",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "平移不改变图形的大小和形状，只改变图形的位置。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】平移的特征是大小，形状，方向不变，位置改变.\n【详解】原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "21-22四年级下·湖南湘潭·期末"
+    },
+    {
+        "id":  "373",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "长方形和正方形都有4条对称轴。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】长方形有2条对称轴，正方形有4条对称轴.\n【详解】原题把两者都说成4条错误。![](images/47bea20904b232407855f4d8d6194fad7fe54236c57ffc822cc7c921e4ad9746.jpg) ![](images/d85165e47f49148ce47a1f6489adbf41a9772c34e503c631ca0aa174cb7e6144.jpg) 故答案为: ×",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "正方形的概念及特点",
+                                 "长方形的概念及特点"
+                             ],
+        "source":  "2018·江苏苏州·小升初真题"
+    },
+    {
+        "id":  "374",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "长方形，正方形，等腰三角形都是轴对称图形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】这三类图形都可找到至少一条对称轴并对折重合.\n【详解】原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "21-22四年级下·湖南怀化·期末"
+    },
+    {
+        "id":  "375",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.94,
+        "question":  "平移后的图形，形状，大小不变，位置发生了变化。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】平移本质是同方向同距离移动，大小形状不变，仅位置变.\n【详解】原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "21-22四年级下·湖南怀化·期末"
+    },
+    {
+        "id":  "376",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "如果两个图形成轴对称，那么其中一个图形可以由另一个图形平移得到。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】轴对称与平移是不同变换;轴对称一般不能仅靠平移得到.\n【详解】原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认",
+                                 "平移与平移现象"
+                             ],
+        "source":  "21-22四年级下·河南许昌·期末"
+    },
+    {
+        "id":  "377",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "等腰三角形，等腰梯形，扇形，平行四边形都只有一条对称轴。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】前三者都可有1条对称轴，但平行四边形一般不是轴对称图形.\n【详解】原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "2021·河南驻马店·小升初真题"
+    },
+    {
+        "id":  "378",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "数字\u00270,4,8\u0027都是轴对称图形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】数字0和8是轴对称图形,4通常不是.\n【详解】原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "21-22五年级下·广西贵港·期末"
+    },
+    {
+        "id":  "379",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "在地面上向前推桌子属于平移现象。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】桌子整体沿直线方向移动，方向不变.\n【详解】属于平移现象，原说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "旋转与旋转现象",
+                                 "平移与平移现象"
+                             ],
+        "source":  "21-22四年级下·河南周口·期末"
+    },
+    {
+        "id":  "380",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "平移后的图形和原图形相比，位置变了，形状和大小都没变。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】平移只改变位置，不改变形状和大小.\n【详解】原说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "21-22四年级下·四川凉山·期末"
+    },
+    {
+        "id":  "381",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "长方形和正方形一样，都是轴对称图形，都有四条对称轴。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】长方形是轴对称图形但只有2条对称轴;正方形有4条.\n【详解】\u0027都4条\u0027错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "21-22四年级下·山东临沂·期末"
+    },
+    {
+        "id":  "382",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "直角梯形，正三角形都是轴对称图形，而且正三角形有三条对称轴。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】正三角形是轴对称图形且有3条对称轴;直角梯形一般不是轴对称图形.\n【详解】因前半句错误，整体为错。故答案为: ×",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "21-22四年级下·浙江宁波·期末"
+    },
+    {
+        "id":  "383",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "有4条对称轴的图形一定是正方形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】有4条对称轴的不一定只有正方形，还可能是其他组合图形.\n【详解】原题说法错误。![](images/3cf885d791d7672632b1ebad37b57160ce5c32da4b84e7b9b79a13bf3df5a6e8.jpg) 故答案为: ×",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "21-22四年级下·山东菏泽·期末"
+    },
+    {
+        "id":  "384",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "平移后的图形，形状和大小与平移前的图形完全一样。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】平移保形保大小，仅改位置.\n【详解】原说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "21-22四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "385",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "长方形，正方形，等腰三角形和圆都是轴对称图形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】这些图形都存在对称轴，沿对称轴可对折重合.\n【详解】原说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "2022·甘肃庆阳·小升初真题"
+    },
+    {
+        "id":  "386",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "等腰三角形，正方形，平行四边形都是轴对称图形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平行四边形通常不是轴对称图形.\n【详解】原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "2022·重庆忠县·小升初真题"
+    },
+    {
+        "id":  "387",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "图形平移后，大小不变，位置不变，方向变了。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平移后位置改变，大小和方向不变.\n【详解】题干三项中\u0027位置不变，方向变了\u0027均错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "21-22四年级下·广东阳江·期末"
+    },
+    {
+        "id":  "388",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "长方形有2条对称轴，正方形有4条对称轴。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】长方形沿两组对边中点连线对称;正方形在此基础上还有两条对角线对称.\n【详解】原说法正确。![](images/9d3e144a19ca6ae5b11cc78d7f27be6eec04c805c4d50697c7c220d4b6a2c3e2.jpg) 故答案为:√",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "21-22五年级下·河北保定·期末"
+    },
+    {
+        "id":  "389",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "圆，正方形，长方形，平行四边形都是轴对称图形，其中圆的对称轴最多。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】圆，正方形，长方形是轴对称图形，但平行四边形不是.\n【详解】由于前提错误，整句错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认",
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "21-22五年级下·江西宜春·期末"
+    },
+    {
+        "id":  "390",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "算盘上算珠的运动是平移现象。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】算珠沿珠杆方向做直线移动.\n【详解】属于平移现象，原说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "21-22四年级下·全国·假期作业"
+    },
+    {
+        "id":  "391",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "汽车在笔直的公路上行驶是平移。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】车体在笔直公路上整体沿直线移动.\n【详解】属于平移，原说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "21-22四年级下·全国·假期作业"
+    },
+    {
+        "id":  "392",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "一条2厘米长的小纸船放在水面上，若船头向前平移80厘米，则船尾向前平移了78厘米。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平移时物体上所有点移动距离相同.\n【详解】船尾也应平移80厘米，不是78厘米。故答案为: ×",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "21-22四年级下·全国·假期作业"
+    },
+    {
+        "id":  "393",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "平移改变图形的大小。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平移不改变图形大小.\n【详解】原说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "21-22四年级下·全国·假期作业"
+    },
+    {
+        "id":  "394",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "等边三角形是轴对称图形，有1条对称轴。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】等边三角形有3条对称轴.\n【详解】原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "21-22四年级下·广东东莞·期末"
+    },
+    {
+        "id":  "395",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "长方形的对称轴比等边三角形的对称轴多。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】长方形2条，等边三角形3条.\n【详解】长方形更少，不是更多。故答案为: ×",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "18-19四年级下·广东肇庆·期末"
+    },
+    {
+        "id":  "396",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "图形平移后的形状和方向都保持不变。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】平移不改变形状，大小和方向，仅改变位置.\n【详解】原说法正确。![](images/1d9a7fedfb8eb2009f13daab735879816a4e661459bfe39e8661ecb0b50ceecb.jpg) 故答案为:√",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "20-21四年级下·山东日照·期末"
+    },
+    {
+        "id":  "397",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "王明的作业评价是\u0027A\u0027,字母\u0027A\u0027可以看作是轴对称图形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】字母A沿竖直中线对折可重合.\n【详解】可看作轴对称图形。![](images/3e4d05e46d26b20f8d1d409d5c77adc052a5835d74d038d7c934ec80fdad2235.jpg) 故答案为:√",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "21-22三年级上·贵州毕节·期末"
+    },
+    {
+        "id":  "398",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "长方形，正方形，平行四边形，圆，等腰梯形都是轴对称图形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平行四边形一般不是轴对称图形.\n【详解】因此不是\u0027都\u0027,原题错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "21-22六年级下·贵州遵义·期末"
+    },
+    {
+        "id":  "399",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "求下图的周长。(单位:cm)![](images/170489dbed16f975993769f873479d2d1cc7723b553a8d41d5ab193e198580bf.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "18cm",
+        "solution":  "【分析】用平移方法把图形转化成长5cm,宽4cm的长方形，再算周长.\n【详解】如图:![](images/1e1de605a08578abdf7ea337b8599129bfe3cc3a0f3693b619b69831640f9a80.jpg)\n(5+4)×2=9×2=18(cm)。",
+        "knowledge_points":  [
+                                 "利用平移法巧算周长",
+                                 "含多边形的组合图形的周长",
+                                 "长方形的周长"
+                             ],
+        "source":  "24-25四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "400",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "计算下面图形阴影部分的面积。![](images/c3cec12d9d13aaf9cc74bbf4e375a9eb31511d60df92ed3cc29531dcca010910.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "50平方厘米",
+        "solution":  "【分析】将上方半圆对应部分平移到下方后，阴影可拼成长10厘米，宽5厘米的长方形.\n【详解】10×5=50(平方厘米),所以阴影部分面积是50平方厘米。",
+        "knowledge_points":  [
+                                 "利用平移法巧算面积",
+                                 "求组合图形中阴影部分的面积",
+                                 "长方形的面积"
+                             ],
+        "source":  "24-25四年级下·广西河池·期末"
+    },
+    {
+        "id":  "401",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面现象属于平移的是()。",
+        "options":  [
+                        "A. 升国旗",
+                        "B. 风车转动",
+                        "C. 钟表指针的转动"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】平移是指在平面内，将一个图形沿着某个方向移动一定的距离，这样的运动叫做图形的平移。旋转是指在平面内，将一个图形绕一点按某个方向转动一个角度，这样的运动叫做图形的旋转.\n【详解】A。升国旗属于平移;B。风车转动属于旋转;C。钟表指针的转动属于旋转。故答案为:A",
+        "knowledge_points":  [
+                                 "平移与平移现象，旋转与旋转现象"
+                             ],
+        "source":  "24-25四年级下·河南安阳·期末"
+    },
+    {
+        "id":  "402",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列图形中，对称轴最多的是()。",
+        "options":  [
+                        "A. 等腰梯形",
+                        "B. 正方形",
+                        "C. 长方形",
+                        "D. 等边三角形"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】把一个图形沿着一条直线对折，对折后直线两边的图形能完全重合，这样的图形就是轴对称图形，这条直线就是这个图形的对称轴。据此判断选项四个图形各有几条对称轴，再选择.\n【详解】A。等腰梯形有1条对称轴。B。正方形有4条对称轴。C。长方形有2条对称轴。D。等边三角形有3条对称轴。因此正方形对称轴最多。故答案为:B",
+        "knowledge_points":  [
+                                 "直角梯形和等腰梯形的概念及特点，对称轴的画法及数量，等腰三角形和等边三角形的认识及特征，正方形的概念及特点"
+                             ],
+        "source":  "24-25四年级下·浙江宁波·期末"
+    },
+    {
+        "id":  "403",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "一个图形平移后()是会改变的。",
+        "options":  [
+                        "A. 大小",
+                        "B. 位置",
+                        "C. 形状",
+                        "D. 无法确定"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】平移是指在平面内，将一个图形沿某个方向移动一定的距离，不改变图形的形状和大小，只改变其位置.\n【详解】一个图形平移后位置是会改变的。故答案为:B",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "24-25四年级下·福建福州·期末"
+    },
+    {
+        "id":  "404",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面的图形中，对称轴最多的是()。",
+        "options":  [
+                        "A. 正方形",
+                        "B. 等边三角形",
+                        "C. 长方形",
+                        "D. 圆"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】把一个平面图形沿一条直线对折，折痕两边的图形能够完全重合，这样的图形叫做轴对称图形，折痕所在的直线是轴对称图形的对称轴.\n【详解】![](images/af9afe1e0a681d8ca161aa6dc7ba72df.png)A。正方形是轴对称图形，有4条对称轴,B。等边三角形是轴对称图形，有3条对称轴,C。长方形是轴对称图形，有2条对称轴，如图:D。圆是轴对称图形，有无数条对称轴，如图：综上所述，对称轴最多的是圆。故答案为:D",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "24-25四年级下·江西萍乡·期末"
+    },
+    {
+        "id":  "405",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面说法中正确的是()。",
+        "options":  [
+                        "A. 大于0.5而小于0.6的小数只有9个。",
+                        "B. 一个三角形至少有两个锐角。",
+                        "C. 因为0.6与0.60大小相同，所以计数单位也相同。",
+                        "D. 汉字\"中，目，口，十，丈\"都是轴对称图形。"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】大于0.5而小于0.6的两位小数有0.51,0.52,0.53,0.54,0.55,0.56,0.57,0.58,0.59，共9个;但是还有三位小数，四位小数等，因此有无数个。锐角三角形三个角都是锐角，钝角三角形有一个钝角，两个锐角，直角三角形有一个直角，两个锐角，所以三角形至少有两个锐角。根据小数的性质，小数末尾增加或者去掉几个0，小数的大小不变;一位小数的计数单位是0.1，两位小数的计数单位是0.01。一个图形沿着一条直线折叠，直线两侧的部分能够完全重合，这样的图形就是轴对称图形.\n【详解】A。大于0.5而小于0.6的小数有无数个，说法错误。B。一个三角形至少有两个锐角，说法正确。C。0.6与0.60大小相同，但计数单位不同,0.6的计数单位是0.1,0.60的计数单位是0.01，说法错误。D。汉字\"中，目，口，十\"都是轴对称图形，\"丈\"不是轴对称图形，说法错误。故答案为:B",
+        "knowledge_points":  [
+                                 "多位小数的大小比较，三角形的分类，小数的性质，轴对称的认识及辨认"
+                             ],
+        "source":  "24-25四年级下·内蒙古鄂尔多斯·期末"
+    },
+    {
+        "id":  "406",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列说法正确的是()。",
+        "options":  [
+                        "A. 一个三位小数，保留两位小数取近似数是6.30，这个三位小数最大是6.299。",
+                        "B. 等边三角形有3条对称轴。",
+                        "C. 40.0800化简后是40.8。"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】6.30可能是\"四舍\"得来的，也可能是\"五入\"得来的近似数，通过\"五入\"得到的6.30，最小是6.295;通过\"四舍\"得到的6.30，最大是6.304;三条边相等的三角形是等边三角形，等边三角形有3条对称轴;小数的末尾填上\"0\"或去掉\"0\"，小数的大小不变,40.0800化简后是40.08.\n【详解】A。一个三位小数，保留两位小数取近似数是6.30，这个三位小数最大是6.304，原说法错误;B。等边三角形有3条对称轴，原说法正确;C。40.0800化简后是40.08，原说法错误。故答案为:B",
+        "knowledge_points":  [
+                                 "小数的近似数，对称轴的画法及数量，等腰三角形和等边三角形的认识及特征，小数的性质"
+                             ],
+        "source":  "24-25四年级下·四川德阳·期末"
+    },
+    {
+        "id":  "407",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "三角形ABC进行平移运动，其平移过程正确的是()。![](images/bc2c565c0ee65bac920168758b19cb6360d5876104e0299415621ccedc3dc9ac.jpg)",
+        "options":  [
+                        "A. 先向右平移2格，再向下平移2格",
+                        "B. 先向右平移4格，再向下平移2格",
+                        "C. 先向下平移2格，再向左平移2格",
+                        "D. 先向下平移2格，再向左平移4格"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】要想知道一个图形平移的方向和距离，只需要找到图形中的一个关键点，数出这个关键点平移的方向和距离即可。据此解答.\n【详解】由题意得，选取三角形最上面的顶点A点作为关键点，数一数可知，三角形ABC先向右平移了4格，再向下平移了2格。也可能先向下平移了2格，再向右平移了4格。故答案为:B",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "24-25四年级下·浙江杭州·期末"
+    },
+    {
+        "id":  "408",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面几句话中，正确的是()。",
+        "options":  [
+                        "A. 边长4厘米的正方形，它的周长和面积相等。",
+                        "B. 100张A4纸叠起来的高度是1厘米,1亿张A4纸叠起来的高度大约是1000米。",
+                        "C. 长方形，正方形，平行四边形和等腰梯形都是轴对称图形。",
+                        "D. □。□≈2，这个数最小可能是1.5，最大可能是2.4。"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】封闭图形一周的长度叫周长，面积是指物体表面或封闭图形所占空间的大小，两个不能比较;先将1亿改写成100000000，除以100求出1亿张里面有多少个100张，再乘1即可求出叠起来的高度大约是多少厘米，根据1米=100厘米，换算成米为单位即可;平面内，一个图形沿一条直线折叠，直线两旁的部分能够完全重合，这个图形就是轴对称图形;根据小数的近似数，一个一位小数保留整数部分是2，要使这个小数最小，则十分位上的数大于或等于5向个位进1，要使这个小数最大，则十分位上的数小于5需要舍去.\n【详解】A。周长和面积是两个不同的概念，无法比较，选项说法错误;B。1亿=100000000,100000000÷100×1=1000000(厘米),1000000厘米=10000米,1亿张A4纸叠起来的高度大约是10000米，选项说法错误;C。长方形，正方形和等腰梯形是轴对称图形，平行四边形不是轴对称图形，选项说法错误;D。1.5≈2,2.4≈2，所以这个数最小可能是1.5，最大可能是2.4，选项说法正确。故答案为:D",
+        "knowledge_points":  [
+                                 "小数的近似数",
+                                 "一亿有多大",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "24-25四年级下·浙江宁波·期末"
+    },
+    {
+        "id":  "409",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列图形中，对称轴最多的是()。",
+        "options":  [
+                        "A. 等边三角形",
+                        "B. 正方形",
+                        "C. 圆"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】一个图形沿一条直线折叠，直线两旁的部分能够完全重合，这样的图形叫做轴对称图形;折叠的这条直线叫做这个图形的对称轴，据此对每个选项进行分析.\n【详解】A。等边三角形有3条对称轴;B。正方形有4条对称轴;C。圆有无数条对称轴;对称轴最多的是圆。故答案为:C",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认",
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "24-25四年级下·河南洛阳·期末"
+    },
+    {
+        "id":  "410",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "图形A向右平移4格得到图形B,图形B再向下平移3格得到图形C,图形C在图形A的()方向。",
+        "options":  [
+                        "A. 左下方",
+                        "B. 右上方",
+                        "C. 右下方"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】根据平移的方向和距离来判断图形的位置。平移：作图时要先找到图形的关键点，分别把这几个关键点按照平移的方向和距离确定对应点后，再顺次连接对应点即可得到平移后的图形。假设图形A是三角形，依次画出图形B和图形C,最后观察位置即可求解.\n【详解】如图:![](images/91760cc08d5887fee06b0d33f6be9b14b28e4b8989be80297a7a91740b84b918.jpg)根据图示可知，图形A向右平移4格得到图形B,图形B再向下平移3格得到图形C,图形C在图形A的右下方方向。故答案为:C",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "24-25四年级下·吉林·期末"
+    },
+    {
+        "id":  "411",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列说法正确的是()。",
+        "options":  [
+                        "A. 用2根5厘米和2根3厘米的小棒可以围成两种三角形(每条边只用一根小棒)",
+                        "B. 三角形有三条高，平行四边形有两条高",
+                        "C. 方格图中点A的位置用数对表示是(4,5),向下平移了3格后，点A的位置用数对表示是(1,5)"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】A。根据三角形三边关系：三角形任意两边之和大于第三边。在实际判断时，通常只需判断较短两边之和是否大于第三边即可。B。在平行四边形中，从一条边上的任意一点向对边作垂线，这点与垂足间的距离叫做以这条边为底的平行四边形的高，平行四边形有无数条高;经过三角形的顶点向对边作垂线，顶点和垂足之间的线段就是三角形的一条高，三角形有三条高。C。数对表示位置时，前一个数字表示列，后一个数字表示行;点向下平移，行数减少.\n【详解】A。3厘米+5厘米\u003e5厘米,3厘米+3厘米\u003e5厘米，可以用2根5厘米与1根3厘米，或1根5厘米与2根3厘米的小棒围成三角形，即用2根5厘米和2根3厘米的小棒可以围成两种三角形，选项说法正确。B。根据分析可知，三角形有三条高，平行四边形有无数条高，选项说法错误。C。方格图中A点的数对是(4,5),向下平移了3格后数对是(4,2),选项说法错误。故答案为:A",
+        "knowledge_points":  [
+                                 "用数对表示位置",
+                                 "三角形三边关系",
+                                 "三角形的高及画法",
+                                 "平行四边形的高及画法"
+                             ],
+        "source":  "24-25四年级下·河南新乡·期末"
+    },
+    {
+        "id":  "412",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列说法中，正确的有()。",
+        "options":  [
+                        "A. 直角三角形的直角边就是它的高，所以直角三角形有2条高。",
+                        "B. 电动伸缩门的设计运用了三角形的稳定性。",
+                        "C. 有一个角是60°的等腰三角形一定是等边三角形。",
+                        "D. 正方形，长方形，三角形和平行四边形都是轴对称图形。"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】(1)从三角形任一顶点向它的对边或者对边的延长线作垂线，从顶点到垂足间的线段叫做三角形的高。每个三角形都有三个底和对应的高。(2)三角形具有稳定性，电动伸缩门运用的是平行四边形的不稳定性。(3)等腰三角形的两个底角相等，三角形内角和为180°,如果一个角是60度，那么另外两个角也是60度。(4)如果一个图形沿着一条直线对折后两部分完全重合，这样的图形叫做轴对称图形.\n【详解】A。直角三角形也有三条高,A错误;B。电动伸缩门运用的是平行四边形的不稳定性，三角形具有稳定性，所以B错误。C。当顶角为60度时，底角为(180°-60°)÷2=60°;当底角为60度时，顶角为180°-60°*2=60°,三个角都是60度，一定是等边三角形。所以C正确。D。三角形中，只有等腰三角形(含等边三角形)是轴对称图形;平行四边形不是轴对称图形，正方形，长方形是轴对称图形，所以D错误。故答案为:C",
+        "knowledge_points":  [
+                                 "三角形的稳定性及应用",
+                                 "三角形的高及画法",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "24-25四年级下·辽宁鞍山·期末"
+    },
+    {
+        "id":  "413",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列图形中，对称轴最多的是()。",
+        "options":  [
+                        "A. 正方形",
+                        "B. 圆",
+                        "C. 长方形"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】一个图形沿一条直线对折，直线两旁的图形完全重合，这样的图形叫做轴对称图形，折痕所在的直线就是对称轴。由题意得，分别数出选项中的图形的对称轴数量，然后找出对称轴数量最多的图形即可.\n【详解】A。正方形有4条对称轴，如图:![](images/beb05c9ee92d7cf67d4bd5641832334e6fc77a08c98f7e38c50ad2ca2c0922de.jpg)\nB。圆有无数条对称轴，如图:![](images/3d926d606c4613a0af83a505495dcbc07efebf10933d1e5d4e96ada3295128ef.jpg)\nC。长方形有2条对称轴，如图:![](images/40b76c68b2daef7dca6fd80f63b656b3705e08836e8b1ed9455e2ae06672238b.jpg)\n对比可知，圆的对称轴数量最多。故答案为:B",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "17-18四年级下·山东菏泽·期末"
+    },
+    {
+        "id":  "414",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面图形中,()一定是轴对称图形。",
+        "options":  [
+                        "A. 锐角三角形",
+                        "B. 平行四边形",
+                        "C. 等腰梯形"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】根据轴对称图形的意义：如果一个图形沿着一条直线对折后两部分完全重合，这样的图形叫轴对称图形，这条直线叫对称轴;依次进行判断即可.\n【详解】A。锐角三角形不一定是轴对称图形，只有三角形是等腰三角形或等边三角形时，才是轴对称图形，不符合题意;B。平行四边形没有对称轴，不是轴对称图形，不符合题意;C。过等腰梯形底的中点的垂线是等腰梯形的对称轴，所以是轴对称图形，符合题意。故答案为:C",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "24-25四年级下·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "415",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面的图形,()不是轴对称图形。",
+        "options":  [
+                        "A. 平行四边形",
+                        "B. 长方形",
+                        "C. 等腰三角形",
+                        "D. 圆"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】如果一个图形沿着一条直线对折后两部分完全重合，这样的图形叫轴对称图形，这条线叫对称轴，由此找出图形的对称轴即可解答.\n【详解】A。平行四边形没有对称轴，不是轴对称图形;B。长方形有两条对称轴，是轴对称图形;C。等腰三角形有一条对称轴，是轴对称图形;D。圆有无数条对称轴，是轴对称图形。故答案为:A",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "23-24四年级下·河南安阳·期末"
+    },
+    {
+        "id":  "416",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列说法错误的是()。",
+        "options":  [
+                        "A. 正方形，长方形，等腰梯形都是轴对称图形",
+                        "B. 三角形中两个内角和等于第三个角时，这个三角形一定是直角三角形",
+                        "C. 把348缩小到原来的1÷1000与把0.0348扩大到它的10倍的结果相等",
+                        "D. 六边形的内角和是180°*6"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】平面内，一个图形沿一条直线折叠，直线两旁的部分能够完全重合，这个图形就是轴对称图形。三角形的内角和是180°。两个内角度数和等于第三个角，说明内角和等于2个第三个角的度数。用180°除以2，可以算出第三个角的度数，据此判断是不是直角三角形。把一个数缩小到原来的1÷1000，也就相当于把这个数的小数点向左移动三位。把一个数扩大到它的10倍，也就相当于把这个数的小数点向右移动一位。多边形的内角和=(边数-2)×180°.\n【详解】A。正方形，长方形，等腰梯形对折后能完全重合，都是轴对称图形，原题干说法正确。B。180°/2=90°,这个角是直角。三角形中两个内角和等于第三个角时，这个三角形一定是直角三角形，原题干说法正确。C。把348的小数点向左移动三位是0.348，把0.0348小数点向右移动一位是0.348，结果相等。原题干说法正确。D。六边形的内角和是(6-2)×180°=4×180°=720°,原题干说法错误。故答案为:D",
+        "knowledge_points":  [
+                                 "小数点向左移动引起小数大小变化的规律",
+                                 "小数点向右移动引起小数大小变化的规律",
+                                 "多边形的内角和",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "23-24四年级下·浙江嘉兴·期末"
+    },
+    {
+        "id":  "417",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面的图形中,()不是轴对称图形。",
+        "options":  [
+                        "A. 长方形",
+                        "B. 平行四边形",
+                        "C. 等边三角形",
+                        "D. 等腰梯形"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据题意，轴对称图形是指可以沿一条直线(对称轴)对折后，两边完全重合的图形。以此分析各选项进行判断选择即可.\n【详解】根据分析可知:A。长方形：有两条对称轴(水平中线和垂直中线),是轴对称图形。B。平行四边形：在一般情况下(非矩形，菱形等特殊类型),没有对称轴，因为沿任何直线对折，两边都无法完全重合。不是轴对称图形。C。等边三角形：有三条对称轴(每条高所在的直线),是轴对称图形。D。等腰梯形：有一条对称轴(通过上底和下底中点的垂直线),是轴对称图形。图形中，平行四边形不是轴对称图形。故答案为:B",
+        "knowledge_points":  [
+                                 "直角梯形和等腰梯形的概念及特点",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "轴对称的认识及辨认",
+                                 "平行四边形的概念及特点"
+                             ],
+        "source":  "24-25四年级下·黑龙江佳木斯·期末"
+    },
+    {
+        "id":  "418",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "长方形，正方形，等边三角形和等腰梯形都是轴对称图形，它们中对称轴最多的图形是()。",
+        "options":  [
+                        "A. 等腰梯形",
+                        "B. 等边三角形",
+                        "C. 长方形",
+                        "D. 正方形"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】一个图形沿一条直线对折，直线两旁的图形完全重合，这样的图形叫做轴对称图形，折痕所在的直线就是对称轴;据此选择即可.\n【详解】A。等腰梯形有1条对称轴，如图:![](images/028987500db3c52d45c97e6dae2f36aefbeb3d532d7870373e03a8c8369cb067.jpg)\nB。等边三角形有3条对称轴，如图:![](images/798dca95cacccdb1a3b1dbba16752703f1b19e782a455b687b4db35c23bba8cd.jpg)\nC。长方形有2条对称轴，如图:![](images/90ba7727677642b95b8a1ece6c380ef1483f68b7f53606fd4566c1a8490b300d.jpg)\nD。正方形有4条对称轴，如图:![](images/a1e960e42348dd24357d2d9bde47ed255904e6c5b6ac1ebd3a5b7a38c1d51d74.jpg)\n1\u003c2\u003c3\u003c4，对称轴最多的图形是正方形。故答案为:D",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "正方形的概念及特点",
+                                 "长方形的概念及特点"
+                             ],
+        "source":  "24-25四年级下·湖北武汉·期末"
+    },
+    {
+        "id":  "419",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面说法正确的有()个。",
+        "options":  [
+                        "A. 1",
+                        "B. 2",
+                        "C. 3"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】(1)计算68×24的积后添上1个0，相当于原积乘10;而680×24=(68×10)×24=68×24×10，与前者相等;(2)任何三角形都有3条高，钝角三角形有3条高;(3)等腰三角形沿底边上的高所在的直线可重合;等腰梯形沿上下底中点连线所在直线对折可重合，均为轴对称图形;(4)三根小棒围三角形需要满足\"任意两边之和大于第三边\"。若两根等长小棒长度为1，第三根为3，两边之和小于第三边，无法构成三角形.\n【详解】由分析可知,(1)在68×24的积的后面添上1个0，就和680×24的积相等。说法正确。(2)一个钝角三角形有3条高，原说法错误。(3)等腰三角形和等腰梯形都是轴对称图形。说法正确。(4)用三根小棒围三角形，其中只要有两根长度相等，不一定能围成三角形。原说法错误。说法正确的有2个。故答案为:B",
+        "knowledge_points":  [
+                                 "三角形三边关系",
+                                 "两位数乘两位数的进位乘法",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "24-25四年级下·河南洛阳·期末"
+    },
+    {
+        "id":  "420",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面说法错误的有()个。",
+        "options":  [
+                        "A. 2",
+                        "B. 3",
+                        "C. 4"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】分析各个说法，统计出说法错误的有几个即可。①根据小数的性质：小数的末尾添上0或者去掉0，小数的大小不变，但是小数的计数单位会发生改变。②锐角三角形是三个角都是锐角的三角形，已知三角形有一个角是60°时，其他的角可能是钝角，直角，那么三角形可能是钝角三角形，直角三角形，不一定就是锐角三角形。③根据轴对称图形的定义：如果一个平面图形能够沿一条直线对折，且对折后的两部分能够完全重合，那么这个图形称为轴对称图形。这条直线称为对称轴。正方形，长方形，等腰三角形和等腰梯形都是轴对称图形，但是平行四边形不一定是轴对称图形。④0乘任何数都得0;0除以任何非0的数，也得0;0加上任何非0的数，得这个数本身;任何非0的数减去0，得到这个数本身。⑤根据计算小数加减法的计算方法：把相同的数位对齐，即把小数点对齐;按照整数加减法计算法则计算;将结果补上小数点;若小数的末尾有0，一般可以去掉.\n【详解】①0.06=0.060,0.06表示小数精确到百分位，计数单位是0.01;0.060表示小数精确到千分位，计数单位是0.001。因此0.06和0.060的大小相等，但计数单位不同。该说法正确。②三角形有一个角是60°,则这个三角形不一定是锐角三角形，因此原题说法错误。③正方形，长方形，等腰三角形和等腰梯形都是轴对称图形，但是平行四边形不一定是轴对称图形。因此原题说法错误。④0只有进行乘除法的运算时，结果是0。进行加减法运算时，结果不是0。因此原题说法错误。⑤用竖式计算小数加减法和整数加减法的方法相同，都是相同数位对齐，然后相加减。该说法正确。因此说法错误的有②,③,④,共3个。故答案为:B",
+        "knowledge_points":  [
+                                 "小数的数位和计数单位的认识",
+                                 "三角形的分类",
+                                 "小数的性质",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "24-25四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "421",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列说法中，正确的是()。",
+        "options":  [
+                        "A. 小数3.598精确到十分位约是3.60",
+                        "B. 8和8.0大小相等，计数单位相同。",
+                        "C. 钝角三角形中，钝角所对的那条边不一定是最长的。",
+                        "D. 等腰梯形，等边三角形，平行四边形都是轴对称图形。"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】A。用\"四舍五入\"法求近似数，看需要保留的下一位，是0~4舍去，是5~9向前一位进一;B。在小数的末尾添上0或者去掉0，小数的大小不变，整数的计数单位是1，一位小数的计数单位是0.1，两位小数的计数单位是0.01,8=8.0，大小相等，计数单位不同;C。在三角形中，角度越大，所对的那条边越长，钝角三角形中，钝角所对的那条边一定是最长的;D。平行四边形对折后两侧的图形不能完全重合，不是轴对称图形.\n【详解】A。3.598≈3.6，原题干说法正确;B。8=8.0，大小相等，计数单位不同，原题干说法错误;C。钝角三角形中，钝角所对的那条边一定是最长的，原题干说法错误;D。平行四边形不是轴对称图形，原题干说法错误。故答案为:A",
+        "knowledge_points":  [
+                                 "小数的近似数",
+                                 "小数的数位和计数单位的认识",
+                                 "小数的性质",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "24-25四年级下·浙江·期末"
+    },
+    {
+        "id":  "422",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列图形中，对称轴最多的是()。",
+        "options":  [
+                        "A. 等边三角形",
+                        "B. 长方形",
+                        "C. 正方形",
+                        "D. 平行四边形"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】将图形沿着一条直线对折，如果直线两侧的部分能够完全重合，折痕所在的直线叫做它的对称轴，根据对称轴定义找出各图形对称轴的数量.\n【详解】根据分析可知:\nA。等边三角形的对称轴：每条高所在的直线，共有3条对称轴，如图:![](images/ca95e3236e7cd1a785df1a97bf9482cedd225a850843c4ee05042ce716eb7727.jpg)B。长方形的对称轴：每组对边中点连线所在的直线，共有2条对称轴，如图:![](images/f16a2c891e961762be7934918b187e3e5d79515c6d03213446b71d994ac92e46.jpg)C。正方形的对称轴有4条，如图:![](images/b01e6256d2cd97b3abee900d051a610638c0b8e787c6c4dacde9dac1d17aa03c.jpg)D。平行四边形不是轴对称图形，没有对称轴。对称轴最多的是正方形。故答案为:C",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "正方形的概念及特点",
+                                 "长方形的概念及特点"
+                             ],
+        "source":  "24-25四年级下·江西吉安·期末"
+    },
+    {
+        "id":  "423",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面的说法，错误的是()。",
+        "options":  [
+                        "A. 三角形具有稳定性，且有3条高",
+                        "B. 5.4和5.40大小相等，计数单位不相同",
+                        "C. 25×52可以用25×4×13进行简算",
+                        "D. 等腰三角形和等边三角形都有2条对称轴"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】三角形的特性是具有稳定性，三角形的高是从顶点到对边的垂直距离，三角形有3条边，所以有3条高;小数的末尾添上0或去掉0不改变大小，但是会改变计数单位，一位小数的计数单位是0.1，两位小数的计数单位是0.01;把52拆成4×13，利用25×4=100简化计算;等腰三角形两边相等，有1条对称轴，等边三角形3边相等，有3条对称轴.\n【详解】A。三角形的特性是具有稳定性，且每一条边都有对应一条高，选项说法正确;B。5.4=5.40,5.4有一位小数，计数单位是0.1,5.40有两位小数，计数单位是0.01，计数单位不同，选项说法正确;C。25×52=25×4×13=100×13=1300，选项说法正确;D。等腰三角形两边相等，有1条对称轴，等边三角形3边相等，有3条对称轴，选项说法错误。故答案为:D",
+        "knowledge_points":  [
+                                 "三角形的稳定性及应用",
+                                 "小数的数位和计数单位的认识",
+                                 "对称轴的画法及数量",
+                                 "小数的性质"
+                             ],
+        "source":  "24-25四年级下·湖南株洲·期末"
+    },
+    {
+        "id":  "424",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列图形中，不是轴对称图形的是()。",
+        "options":  [
+                        "A. 等腰梯形",
+                        "B. 平行四边形",
+                        "C. 等边三角形"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】一个图形沿一条直线对折，直线两旁的图形完全重合，这样的图形叫做轴对称图形，折痕所在的直线就是对称轴;两条腰长度相等的梯形叫作等腰梯形。两组对边分别平行的四边形是平行四边形。三条边都相等的三角形叫作等边三角形。据此解答.\n【详解】A。如图，等腰梯形关于直线对称，它是轴对称图形。![](images/070d96fb83d42cb3766f3f52d7abebaadaf8bd0f39df87eca6d494e44a40e5eb.jpg)B。如图，无法找到直线使得平行四边形关于直线对称，它不是轴对称图形。![](images/a9a97047c2d52d81864aea9f1bc1cf4b4f7426a2cfbc619dd3d328c8136eb162.jpg)C。如图，等边三角形关于直线对称，它是轴对称图形。![](images/23f9f65d9746e7dff782a108cac8733d8dfcf5dc4eb789c2c8d7417fa8186381.jpg)故答案为:B",
+        "knowledge_points":  [
+                                 "直角梯形和等腰梯形的概念及特点",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "轴对称的认识及辨认",
+                                 "平行四边形的概念及特点"
+                             ],
+        "source":  "24-25四年级下·湖北黄冈·期末"
+    },
+    {
+        "id":  "425",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面图形中，不是轴对称图形的是()。",
+        "options":  [
+                        "A. 长方形",
+                        "B. 平行四边形",
+                        "C. 正方形",
+                        "D. 等边三角形"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】一个图形沿一条直线对折后，直线两旁的部分能够完全重合，这样的图形叫做轴对称图形，这条直线叫做对称轴。判断是不是轴对称图形的关键是看能否找出对称轴，轴对称图形沿对称轴对折后两部分能完全重合.\n【详解】A。长方形有2条对称轴，是轴对称图形。B。平行四边形找不到一条对称轴，不是轴对称图形。C。正方形有4条对称轴，是轴对称图形。D。等边三角形有3条对称轴，是轴对称图形。故答案为:B",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "24-25四年级下·河北沧州·期末"
+    },
+    {
+        "id":  "426",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列图形中，对称轴最多的是()。",
+        "options":  [
+                        "A. 等边三角形",
+                        "B. 长方形",
+                        "C. 圆"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】如果一个图形沿一条直线折叠，直线两旁的部分能够互相重合，这个图形叫做轴对称图形，这条直线叫做对称轴。据此判断即可.\n【详解】A。等边三角形有3条对称轴，如图:![](images/83cb1ae973d12fdf3c9a0307ae41e0fe098a1c7ec01a96b0b7fba108e6731f81.jpg)B。长方形有2条对称轴，如图:![](images/5c4494ed29c75f0a31b799d2cf67d440b62d4de7bf33c126d7712e53d0e24e60.jpg)C。圆有无数条对称轴，如图:![](images/6b60381c518544a2dd0c264d3154c0bbc74fd88342c3f69bb955113bc53336aa.jpg)故答案为:C",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "24-25四年级下·湖南岳阳·期末"
+    },
+    {
+        "id":  "427",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "等腰三角形，直角三角形，长方形，平行四边形这些图形中，一定是轴对称图形的有()个。",
+        "options":  [
+                        "A. 2",
+                        "B. 3",
+                        "C. 4"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】一个图形沿一条直线对折，直线两旁的图形完全重合，这样的图形叫做轴对称图形，折痕所在的直线就是对称轴。有两条边相等的三角形叫作等腰三角形。有一个角是直角的三角形是直角三角形。两组对边分别平行，四个角都是直角的四边形叫做长方形。两组对边分别平行的四边形叫作平行四边形。据此解答.\n【详解】根据题意作图如下：等腰三角形:![](images/167634b209040764d978910ec18d33886927321994d2a0b4424c6110d9d5d111.jpg)直角三角形:![](images/5a92b084337697b6e3ae5161aa53d64ff316cee418fb1c4806883cad0e76a96c.jpg)长方形:![](images/0408545e0beff371f3a0300f25c03993eb7e87d95324e9d8a19fc2b2bfab30bf.jpg)平行四边形:![](images/e817efab0ef1c0ab46c10ef665dd98208e5247b492669df16472a837ef52b71e.jpg)由图可知，等腰三角形和长方形关于直线对称，它们都是轴对称图形;直角三角形和平行四边形对折后都无法重合，所以它们都不是轴对称图形。所以，一定是轴对称图形的有2个。故答案为:A",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "轴对称的认识及辨认",
+                                 "长方形的概念及特点",
+                                 "平行四边形的概念及特点"
+                             ],
+        "source":  "24-25四年级下·湖北武汉·期末"
+    },
+    {
+        "id":  "428",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "平行四边形，长方形，圆，等腰梯形和等边三角形都是轴对称图形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】轴对称图形是指在平面内沿一条直线折叠，直线两旁的部分能够完全重合的图形。判断是不是轴对称图形的关键是找出对称轴，图形两部分沿对称轴折叠后能完全重合.\n【详解】长方形，圆，等腰梯形和等边三角形都是轴对称图形，平行四边形不是轴对称图形。故答案为: ×",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "2022六年级下·全国·专题练习"
+    },
+    {
+        "id":  "429",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "正方形有4条对称轴;三角形有3条对称轴。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】正方形有4条对称轴，长方形有两条对称轴，等腰三角形有1条对称轴，等边三角形有3条对称轴，圆有无数条对称轴。本题主要考查学生对常见图形对称轴数量的掌握.\n【详解】根据分析可知，正方形有4条对称轴，等边三角形有3条对称轴，不是所有的三角形都有3条对称轴，原说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "20-21四年级下·山东临沂·期末"
+    },
+    {
+        "id":  "430",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "正方形纸片对折后所得到的图形不是轴对称图形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】一个图形沿一条直线对折，直线两旁的图形完全重合，这样的图形叫做轴对称图形，折痕所在的直线就是对称轴。关键是熟悉轴对称图形的特点，正方形是轴对称图形，有4条对称轴.\n【详解】如图，正方形纸片对折后所得到的图形是轴对称图形.\n![](images/40a918f153a48c6a458e09b1264c76b0b43e6b7f11fa7d3a31fc9feee2888ce8.jpg)\n故答案为: ×",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "20-21六年级下·四川宜宾·期末"
+    },
+    {
+        "id":  "431",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "平移既能改变图形的位置，又能改变图形的形状。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】本题考查平移的基本特征。平移是指在平面内，将一个图形上的所有点都按照某个直线方向做相同距离的移动。平移只改变图形的位置，不改变图形的形状和大小.\n【详解】平移是图形沿直线移动一定距离的运动，平移前后位置发生变化而形状，大小不变。因此平移能改变图形的位置，但不能改变图形的形状。故答案为: ×",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "20-21四年级下·黑龙江牡丹江·期末"
+    },
+    {
+        "id":  "432",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "正方形有4条对称轴，圆有2条对称轴。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】根据轴对称图形的定义及性质求解：如果一个图形沿一条直线折叠，直线两旁的部分能够互相重合，这个图形叫做轴对称图形，这条直线叫做对称轴。本题考查轴对称图形的知识，要求掌握好轴对称的概念;判断是不是轴对称图形的关键是寻找对称轴，图形折叠后直线两旁的部分能够互相重合.\n【详解】正方形有4条对称轴，圆有无数条对称轴，故原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "20-21四年级下·黑龙江七台河·期末"
+    },
+    {
+        "id":  "433",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "B和E这两个字母都是轴对称图形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】平面内，一个图形沿一条直线折叠，直线两旁的部分能够完全重合的图形叫做轴对称图形，这条直线叫做对称轴，据此判断。本题考查学生对轴对称图形的认识与判断，可以实际画出对称轴判断.\n【详解】如图所示:![](images/2026-03-14%20144532.jpg)B和E这两个字母都是轴对称图形，原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "20-21三年级下·辽宁营口·期末"
+    },
+    {
+        "id":  "434",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "等腰三角形所在底边上的高就是它的对称轴。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】依据轴对称图形的意义，即在同一个平面内，一个图形沿某条直线对折，对折后的两部分都能完全重合，则这个图形就是轴对称图形，这条直线就是其对称轴，从而可以画出它们的对称轴。此题主要考查如何确定轴对称图形的对称轴条数及位置.\n【详解】由轴对称图形的意义可知：等腰三角形是轴对称图形，它底边上的高所在的直线就是它的对称轴，高是线段，不能说高就是对称轴。故答案为: ×",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "20-21六年级下·辽宁朝阳·期末"
+    },
+    {
+        "id":  "435",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "所有的圆都是轴对称图形，所有的直角三角形都不是轴对称图形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】根据轴对称图形的意义：如果一个图形沿着一条直线对折后两部分完全重合，这样的图形叫做轴对称图形，这条直线叫做对称轴;据此判断即可。此题主要考查轴对称图形的定义及应用。判断一个图形是否是轴对称图形，关键是找它的对称轴，要想象沿着这条线翻折能不能重叠.\n【详解】根据分析可得：所有的圆都是轴对称图形，等腰直角三角形是轴对称图形;故原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "20-21四年级下·四川内江·期末"
+    },
+    {
+        "id":  "436",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "三角形不一定是轴对称图形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】根据轴对称图形的意义：如果一个图形沿着一条直线对折后两部分完全重合，这样的图形叫做轴对称图形，这条直线叫做对称轴;据此进行判断即可。此题考查了轴对称图形的意义，判断轴对称图形的关键是寻找对称轴，看图形对折后两部分是否完全重合.\n【详解】等腰三角形和等边三角形都是轴对称图形，其它的三角形不是轴对称图形;所以原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "20-21四年级下·河南濮阳·期末"
+    },
+    {
+        "id":  "437",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "一个图形经过平移后，与原图形成轴对称。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】一个图形经过平移后，与原图可能形成轴对称，也可能不形成轴对称，举例解答即可。判断一个图案是否是轴对称图形的关键是看在这个图形中能否找到一条直线，使图形沿着这条直线对折后能够完全重合。解决本题时可亲自画图试一试，即可得出结论.\n【详解】如图所示:\n![](images/6b038a5063597c4acf66faca366d6ade9c2e82bc7d88ffeeae848aac039969ad.jpg)\n图A向右平移5格得到图B,图A和图B形成轴对称。图A先向右平移1格，再向下平移4格得到图C,图A和图C不形成轴对称。原说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平移与平移现象",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "18-19四年级下·江西抚州·期末"
+    },
+    {
+        "id":  "438",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "四边形都是对称图形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】根据轴对称图形的定义可得：一个图形沿某条直线对折后直线两旁的部分能够完全重合，是轴对称图形，否则就不是轴对称图形。此题主要考查轴对称图形的定义，并不是所有的四边形都是轴对称图形.\n【详解】例如正方形就是轴对称图形;平行四边形就不是轴对称图形，所以原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "18-19三年级下·全国·期中"
+    },
+    {
+        "id":  "439",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "利用平移，对称可以设计出许多漂亮图案。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】举出利用平移，对称设计的漂亮图案即可判断正确。本题考查对平移和对称的理解及在实际当中的运用.\n【详解】下面图形是利用平移，对称设计出的漂亮图案，所以判断正确。![](images/e821d7385972e9077311fffe047f324a2fdfbdeed78197d21fa39dc8d74d16a7.jpg)故答案为:√",
+        "knowledge_points":  [
+                                 "运用平移",
+                                 "对称设计图案"
+                             ],
+        "source":  "15-16五年级上·全国·单元测试"
+    },
+    {
+        "id":  "440",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "风车的转动是旋转现象，缆车的运行是平移现象。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】判断生活中的现象，是否是平移，要根据平移的定义进行选择，图形平移前后的形状和大小没有变化，只是位置发生变化;旋转现象必须具备有旋转中心，旋转方向，旋转角，根据旋转的性质，进行选择。此题是考查对平移与旋转的理解及在实际当中的运用.\n【详解】由旋转和平移的定义可知：风车的转动是旋转现象，缆车的运行是平移现象。所以判断正确。故答案为:√",
+        "knowledge_points":  [
+                                 "旋转与旋转现象",
+                                 "平移与平移现象"
+                             ],
+        "source":  "21-22三年级上·山东枣庄·期末"
+    },
+    {
+        "id":  "441",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "等腰梯形和平行四边形都是轴对称图形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】如果一个图形沿着一条直线对折后两部分完全重合，这样的图形叫做轴对称图形，由此进行判断即可。此题主要考查了对轴对称图形的了解掌握.\n【详解】等腰梯形沿上下底边的中垂线对折后两边能完全重合，所以等腰梯形为轴对称图形;而平行四边形无论怎样对折两边都不能完全重合，所以平行四边形不是轴对称图形，故此说法不正确。故答案为: ×",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "21-22四年级上·湖北襄阳·期末"
+    },
+    {
+        "id":  "442",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "把一个图形平移后，图形的大小不变。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】图形的平移只是改变图形的位置，图形的形状和大小不改变。明确平移的特点是解答本题的关键.\n【详解】把一个图形平移后，图形的大小不变，说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "18-19五年级上·辽宁·单元测试"
+    },
+    {
+        "id":  "443",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "长方形的对称轴有4条，平行四边形没有对称轴。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】一个图形沿一条直线对折，直线两旁的图形完全重合，这样的图形叫做轴对称图形，折痕所在的直线就是对称轴。明确轴对称图形的特点是解答本题的关键.\n【详解】长方形分别沿长和宽的中线所在的直线对折，对折后两边完全重合，所以长方形是轴对称图形，它有2条对称轴;平行四边形没有对称轴，原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认",
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "18-19五年级上·河南商丘·期末"
+    },
+    {
+        "id":  "444",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "沿直的铁轨运行的高铁是平移。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】在平面内，把一个图形整体沿某条直线方向平行移动的过程，称为平移。决定平移后图形位置的要素：一是平移的方向，二是平移的距离.\n【详解】沿直的铁轨运行的高铁是平移，说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "18-19五年级上·河南商丘·期末"
+    },
+    {
+        "id":  "445",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "等边三角形的对称轴的条数比长方形的条数多。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】根据轴对称图形的意义：如果一个图形沿着一条直线对折后两部分完全重合，这样的图形叫做轴对称图形，这条直线叫做对称轴;据此判断即可。本题考查轴对称图形的意义，明确它的意义是解题的关键.\n【详解】等边三角形有3条对称轴，长方形有2条对称轴，所以等边三角形的对称轴的条数比长方形的条数多。故答案为:√",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "22-23五年级上·全国·假期作业"
+    },
+    {
+        "id":  "446",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "四条边都相等的平行四边形是轴对称图形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】根据轴对称图形的意义：如果一个图形沿一条直线对折后两部分完全重合，这样的图形叫做轴对称图形，这条直线叫做对称轴;据此判断即可。正确理解轴对称的意义及菱形的特征，是解答此题的关键.\n【详解】四条边都相等的平行四边形是菱形，菱形属于轴对称图形，所以原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "21-22四年级下·全国·单元测试"
+    },
+    {
+        "id":  "447",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "圆和扇形都有无数条对称轴。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】将图形沿着一条直线对折，如果直线两侧的部分能够完全重合，折痕所在的直线叫做它的对称轴，根据对称轴定义找出图形的对称轴即可。准确根据对称轴的意义判断出对称轴的数量是解答题目的关键.\n【详解】圆形的对称轴：每条直径所在的直线，则圆形有无数条对称轴;扇形的对称轴：圆心和弧中点连线所在的直线，则扇形只有1条对称轴。故答案为: ×",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "20-21六年级上·山西忻州·期末"
+    },
+    {
+        "id":  "448",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "轴对称图形，对称轴两边的图形形状，大小完全一样。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】根据轴对称图形的意义：如果一个图形沿着一条直线对折后两部分完全重合，这样的图形叫做轴对称图形，这条直线叫做对称轴;据此判断即可。此题考查的是轴对称图形的特点，掌握判断轴对称图形的关键是寻找对称轴，图形两部分折叠后可完全重合是解题关键.\n【详解】轴对称图形是特殊的平面图形，对称轴两边大小要相等，形状相同，完全一样。故答案为:√",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "20-21五年级上·山西忻州·期中"
+    },
+    {
+        "id":  "449",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "平行四边形和圆都是轴对称图形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】本题考查轴对称图形的判断。轴对称图形是指沿一条直线对折后，两侧能够完全重合的图形。圆是轴对称图形，有无数条对称轴。但平行四边形中，只有特殊的平行四边形(如长方形，正方形，菱形)是轴对称图形，一般的平行四边形不是轴对称图形.\n【详解】平行四边形中，长方形，正方形，菱形是轴对称图形，一般的平行四边形不是轴对称图形。故答案为: ×",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "21-22五年级上·山东枣庄·期中"
+    },
+    {
+        "id":  "450",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "圆，正方形，梯形都是轴对称图形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】一个图形沿一条直线对折，直线两旁的图形完全重合，这样的图形叫做轴对称图形，据此解答即可。明确轴对称图形的特点是解答本题的关键.\n【详解】圆，正方形都是轴对称图形，梯形中只有等腰梯形是轴对称图形，一般梯形不是轴对称图形，原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "21-22五年级上·山东滨州·期中"
+    },
+    {
+        "id":  "451",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "有4条对称轴的图形是正方形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】一个图形沿一条直线对折，直线两旁的部分能够完全重合，这个图形就是轴对称图形，这条直线就是它的对称轴;据此并结合对轴对称图形的认识，进行解答即可。此题考查了轴对称图形的辨识，能够根据给出的对称轴的条数，例举出符合条件的图形.\n【详解】根据轴对称图形的定义可知，正方形有4条对称轴，但是有4条对称轴的图形不一定是正方形，如图，这个组合图形的对称轴就有4条，原题说法错误.\n![](images/4ebf9853c040263906037b11f5be641d50ded00beaba2f2939afcf2900aa4b77.jpg)\n故答案为: ×",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "21-22六年级上·全国·课后作业"
+    },
+    {
+        "id":  "452",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "对称图形对称轴两边，图形一边平移就一定能与另一边重合。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】根据轴对称图形的意义：如果一个图形沿着一条直线对折后两部分完全重合，这样的图形叫做轴对称图形，这条直线叫做对称轴;平移不一定能与另一边重合，据此判断即可。判断轴对称图形的关键是寻找对称轴，图形两部分折叠后可完全重合.\n【详解】对称图形对称轴两边，图形一边平移就一定能与另一边重合，此说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平移与平移现象",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "19-20三年级上·河北石家庄·期中"
+    },
+    {
+        "id":  "453",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "直角梯形是轴对称图形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】判断是不是轴对称图形的关键是看能否找出对称轴，图形沿对称轴对折后两部分能完全重合。明确轴对称图形的特点是解答本题的关键.\n【详解】直角梯形找不到一条对折后使两部分能完全重合的对称轴，它不是轴对称图形，所以判断错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "20-21四年级上·湖南怀化·期末"
+    },
+    {
+        "id":  "454",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "平移只改变图形的位置，不改变图形的面积。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】平移，是指在平面内，将一个图形上的所有点都按照某个直线方向做相同距离的移动，这样的图形运动叫做图形的平移运动，简称平移。平移不改变图形的形状和大小，只改变物体的位置。本题考查的是平移的概念，属于基础题.\n【详解】平移只会改变图形的位置，不改变图形的大小，说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "20-21四年级下·河南周口·期末"
+    },
+    {
+        "id":  "455",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "小朋友玩滑滑梯的运动是旋转现象。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平移：在平面内，将一个图形上的所有点都按照某个方向作相同距离移动的图形运动。平移后图形的位置改变，形状，大小，方向不变。旋转：在平面内，将一个图形绕一点按某个方向转动一个角度，这样的运动叫作图形的旋转。旋转前后图形的位置和方向改变，形状，大小不变;据此解答即可。此题考查了平移与旋转的意义及在实际当中的运用.\n【详解】小朋友玩滑滑梯的运动是平移现象，小朋友身上的所有点都按照某个方向作相同距离移动，而不是旋转，原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平移与平移现象",
+                                 "旋转与旋转现象"
+                             ],
+        "source":  "20-21三年级下·河南商丘·期末"
+    },
+    {
+        "id":  "456",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "等边三角形，正方形都是轴对称图形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】沿一条直线对折，直线两边的部分能够完全重合的图形叫做轴对称图形，据此解答即可。本题考查轴对称，解答本题的关键是掌握沿一条直线对折，直线两边的部分能够完全重合的图形叫做轴对称图形.\n【详解】等边三角形，正方形都是轴对称图形，原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "20-21四年级下·河南商丘·期末"
+    },
+    {
+        "id":  "457",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "用对折的方法，能找出轴对称图形的对称轴。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】将这个图形沿一条直线对折后，看两边的图形能不能完全重合。能完全重合的就是轴对称图形，这条直线就轴对称图形的对称轴;如果无论沿着哪条直线对折，两边的图形都不能完全重合，则不是轴对称图形。据此判断即可。本题考查根据轴对称的定义找对称轴的方法，需熟练掌握.\n【详解】用对折的方法，图形沿着一条直线对折后，折痕所在的直线就是这个图形的对称轴。故答案为:√",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "20-21三年级下·河南商丘·期末"
+    },
+    {
+        "id":  "458",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "在圆，正方形，等边三角形，长方形中，正方形的对称轴条数最多。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】把一个图形沿着某一条直线折叠，如果它能够与另一个图形重合，那么就说这两个图形关于这条直线(成轴)对称，这条直线就是它的对称轴，这个图形就是轴对称图形，据此判断各个图形对称轴的条数即可解题。熟记常见图形轴对称的条数是解答此题的关键.\n【详解】圆有无数条对称轴，正方形有4条对称轴，等边三角形有3条对称轴，长方形有2条对称轴，圆的对称轴最多，所以原题干说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "20-21四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "459",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "所有的平行四边形都是轴对称图形，它有2条或4条对称轴。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】一个图形沿一条直线对折，直线两旁的图形完全重合，这样的图形叫做轴对称图形，折痕所在的直线就是对称轴。据此可知，平行四边形不是轴对称图形，没有对称轴。本题考查轴对称的认识。判断一个图形是否是轴对称图形，关键是找它的对称轴，要想象沿着这条线翻折能不能重叠.\n【详解】根据分析可知，所有的平行四边形都不是轴对称图形，它没有对称轴。故答案为: ×",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "19-20四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "460",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "数字\"0\"和\"8\"都是轴对称图形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】根据对轴对称图形的概念：图形能沿着一条直线对折后对称就是轴对称图形，这条直线为对称轴，据此判断\"0\"和\"8\"是不是轴对称图形。本题考查的是轴对称图形的判断，关键要找到对称轴.\n【详解】\"0\"和\"8\"能找到对称轴，是轴对称图形。故题干说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "20-21三年级下·河南鹤壁·期末"
+    },
+    {
+        "id":  "460",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "数字\"0\"和\"8\"都是轴对称图形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】根据对轴对称图形的概念：图形能沿着一条直线对折后对称就是轴对称图形，这条直线为对称轴，据此判断\"0\"和\"8\"是不是轴对称图形。本题考查的是轴对称图形的判断，关键要找到对称轴.\n【详解】\"0\"和\"8\"能找到对称轴，是轴对称图形。故题干说法正确。",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "20-21三年级下·河南鹤壁·期末"
+    },
+    {
+        "id":  "461",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "平行四边形和正方形都是轴对称图形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】如下图，平行四边形找不到一条对称轴，不是轴对称图形，正方形有4条对称轴，是轴对称图形。![](images/09d6e3090bded075b43e492412defd52885aa90b7be89b44eb313d267a952755.jpg)![](images/95c0c7df841e6537a28f389120973129562805ffb2809c98f2ebcdfcd985e077.jpg)本题主要考查学生对轴对称图形判定方法的掌握和灵活运用.\n【详解】根据分析可知，平行四边形不是轴对称图形，正方形是轴对称图形，所以判断错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "19-20四年级下·河南洛阳·期末"
+    },
+    {
+        "id":  "462",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "所有的三角形都不是轴对称图形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】一个图形沿一条直线折叠，直线两旁的部分能够完全重合，这样的图形叫做轴对称图形;依此分析即可。熟练掌握轴对称图形的特点是解答此题的关键.\n【详解】等腰三角形的两条腰相等，两个底角也相等;等边三角形的特点是三条边都相等，三个角也相等。因此等腰三角形和等边三角形是轴对称图形。故答案为: ×",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "18-19三年级下·山东青岛·单元测试"
+    },
+    {
+        "id":  "463",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "圆形和等腰三角形都是轴对称图形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】一个图形沿一条直线对折，直线两旁的图形完全重合，这样的图形叫做轴对称图形，折痕所在的直线就是对称轴。判断是不是轴对称图形，关键看能否找到对称轴.\n【详解】圆形和等腰三角形都能找到对称轴，所以圆形和等腰三角形都是轴对称图形。故题干说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "20-21四年级下·河南开封·期末"
+    },
+    {
+        "id":  "464",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "图形平移之后，与原来图形相比形状，大小都没有改变。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】根据平移的特征：物体或图形平移后，它们的形状，大小，方向都不改变，只是位置发生了变化。即可判断。本题考查的是对平移的理解，平移只改变位置，不会改变图形的形状和大小.\n【详解】图形平移之后，与原来图形相比形状，大小都没有改变。故答案为:√",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "20-21四年级下·河南开封·期末"
+    },
+    {
+        "id":  "465",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "等边三角形有两条对称轴。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】一个图形沿一条直线对折，直线两旁的图形完全重合，这样的图形叫做轴对称图形，折痕所在的直线就是对称轴。据此判断即可。本题主要考查了轴对称图形的对称轴条数，应熟记常见图形的对称轴条数，例如长方形有2条对称轴，正方形有4条对称轴，等腰三角形有1条对称轴等.\n【详解】如图所示，等边三角形应有三条对称轴。![](images/19952d7de8c7c07de8fe89e95b7abc79c6a2fa028b45ca25f5deab34a45f264b.jpg)故答案为: ×",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "19-20四年级下·内蒙古呼和浩特·期末"
+    },
+    {
+        "id":  "466",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "\"凹\"是轴对称图形，只有1条对称轴。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】一个图形沿一条直线对折，直线两旁的部分能够完全重合，那么这个图形就是轴对称图形，这条直线就是这个图形的一条对称轴。据此解题即可。此题考查了利用轴对称图形的定义判断轴对称图形的对称轴的条数的灵活应用.\n【详解】如图:![](images/cd54afea4d71c062942266f7ef5fbef69dc769a2c7d5ef8ea8b3ddf5a85fc6d8.jpg)\"凹\"是轴对称图形，只有一条对称轴。故答案为:√",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "20-21四年级下·内蒙古呼伦贝尔·期末"
+    },
+    {
+        "id":  "467",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "平移和旋转改变图形的位置。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】平移和旋转都是物体或图形的位置发生变化而形状，大小不变。区别在于，平移时物体沿直线运动，本身方向不发生改变;旋转是物体绕着某一点或轴运动，本身方向发生了变化。据此判断即可。本题考查平移和旋转的特点，应注意区分.\n【详解】根据分析可知，平移和旋转均改变图形的位置。故答案为:√",
+        "knowledge_points":  [
+                                 "平移与平移现象",
+                                 "旋转与旋转现象"
+                             ],
+        "source":  "20-21四年级下·广西防城港·期末"
+    },
+    {
+        "id":  "468",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "图形在平移过程中，位置发生改变，形状和大小不变。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】物体或图形平移后，它们的形状，大小，方向都不改变，只是位置发生了变化。决定平移后图形位置的要素：一是平移的方向，二是平移的距离.\n【详解】图形在平移过程中，位置发生改变，形状和大小不变，说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "19-20五年级下·湖北武汉·期末"
+    },
+    {
+        "id":  "469",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "经过平移后的图形，它的形状大小都发生了变化。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】本题考查平移的基本特征。平移是指在平面内，将一个图形上的所有点都按照某个直线方向做相同距离的移动。平移不改变图形的形状和大小，只改变图形的位置.\n【详解】图形平移前与平移后，图形的形状，大小不变，只是位置发生了改变。故答案为: ×",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "20-21四年级下·新疆省直辖县级单位·期末"
+    },
+    {
+        "id":  "470",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "长方形只有一条对称轴。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】依据轴对称图形的概念，即在平面内，如果一个图形沿一条直线对折，直线两旁的部分能够完全重合，这样的图形叫做轴对称图形，据此即可进行判断。解答此题的主要依据是：轴对称图形的定义及其对称轴的条数.\n【详解】如图:![](images/66232ef039f2496ef25b24d4c20b5a0bde61bc14a75250f06b7f04e7ce61f3b3.jpg)长方形有2条对称轴，原说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "18-19四年级下·新疆省直辖县级单位·期末"
+    },
+    {
+        "id":  "471",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "一个正方形沿着对称轴对折后得到的一定是长方形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】正方形有4条对称轴，沿着对称轴对折后得到的可能是长方形也可能是三角形，据此判断。此题考查了对称图形，注意思考问题要全面.\n【详解】当沿着正方形的对角线对折后得到的是一个三角形，原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "20-21六年级下·山西朔州·期末"
+    },
+    {
+        "id":  "472",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "圆有10条对称轴。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】如果一个图形沿一条直线对折，直线两侧的部分能够完全重合，这个图形就是轴对称图形，这条直线就是它的对称轴。据此解答。因为圆形自身的特点决定了它有无数条对称轴，可具体在纸上画一画，就不难得出答案了.\n【详解】结合圆形的具体特征可知：圆有无数条对称轴，原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "圆的概念及特点"
+                             ],
+        "source":  "19-20二年级下·甘肃天水·期末"
+    },
+    {
+        "id":  "473",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "汉字\"由\"是轴对称图形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】一个图形沿一条直线对折，直线两旁的图形完全重合，这样的图形叫做轴对称图形，折痕所在的直线就是对称轴;据此进行判断即可。判断一个图形是否是轴对称图形，关键是找它的对称轴，要想象沿着这条线翻折能不能重叠.\n【详解】根据轴对称图形的定义可知，汉字\"由\"是轴对称图形。故答案为:√",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "20-21四年级下·全国·假期作业"
+    },
+    {
+        "id":  "474",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "平移可以改变图形的大小和方向。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】本题考查平移的基本特征。平移是指在平面内，将一个图形上的所有点都按照某个直线方向做相同距离的移动。平移只改变图形的位置，不改变图形的形状，大小和方向.\n【详解】平移，是指在同一平面内，将一个图形上的所有点都按照某个直线方向做相同距离的移动，这样的图形运动叫做图形的平移运动，简称平移。平移不改变图形的形状和大小。故答案为: ×",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "18-19四年级下·云南临沧·期末"
+    },
+    {
+        "id":  "475",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "轴对称图形的对称轴至少有一条。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】一个图形沿一条直线折叠，直线两旁的部分能够完全重合，这样的图形叫做轴对称图形;折叠的这条直线叫做这个图形的对称轴，此题依此判断即可。熟练掌握对称轴的画法是解答此题的关键.\n【详解】轴对称图形的对称轴至少有一条，如长方形是轴对称图形，它的对称轴有两条。故答案为:√",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "18-19四年级下·云南临沧·期末"
+    },
+    {
+        "id":  "476",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "正方形，长方形和梯形都是轴对称图形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】一个图形沿一条直线折叠，直线两旁的部分能够完全重合，这样的图形叫做轴对称图形此题依此判断即可。熟练掌握轴对称图形的特点是解答此题的关键.\n【详解】正方形，长方形沿一条直线折叠，直线两旁的部分都能够完全重合，因此正方形和长方形是轴对称图形;梯形沿一条直线折叠，直线两旁的部分不一定能够完全重合，因此梯形不一定是轴对称图形。故答案为: ×",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "19-20四年级下·安徽滁州·期末"
+    },
+    {
+        "id":  "477",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "长方形，平行四边形，正方形都是轴对称图形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】一个图形沿一条直线折叠，直线两旁的部分能够完全重合，这样的图形叫做轴对称图形。熟练掌握轴对称图形的特点是解答此题的关键.\n【详解】长方形，正方形沿一条直线折叠，直线两旁的部分都能够完全重合，因此长方形和正方形都是轴对称图形。平行四边形沿一条直线折叠，直线两旁的部分不一定能够完全重合，因此平行四边形不是轴对称图形。故答案为: ×",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "19-20四年级下·江西宜春·期末"
+    },
+    {
+        "id":  "478",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "正方形的对角线是它的对称轴。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】一个图形沿一条直线折叠，直线两旁的部分能够完全重合，这样的图形叫做轴对称图形。折叠的直线就是轴对称图形的对称轴。熟练掌握对称轴是解答此题的关键.\n【详解】将正方形沿对角线对折，对角线两旁的部分能够完全重合，所以对角线所在的直线就是它的对称轴，但是正方形的对角线是线段，所以不能说对角线是对称轴。故答案为: ×",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "15-16四年级下·甘肃武威·期末"
+    },
+    {
+        "id":  "479",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "等边三角形，长方形和正方形都是轴对称图形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】一个图形沿一条直线对折，直线两旁的图形完全重合，这样的图形叫做轴对称图形，折痕所在的直线就是对称轴;据此进行判断即可。判断一个图案是否是轴对称图形的关键是看在这个图形中能否找到一条直线，使图形沿着这条直线对折后能够完全重合.\n【详解】等边三角形是轴对称图形，有3条对称轴。长方形是轴对称图形，有2条对称轴。正方形是轴对称图形，有4条对称轴。故答案为:√",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "18-19四年级下·甘肃兰州·期末"
+    },
+    {
+        "id":  "480",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "平移和轴对称只改变图形的位置，不改变图形的形状和大小。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】本题考查平移和轴对称的基本特征。平移只改变图形的位置，不改变图形的形状，大小和方向。轴对称(翻折)也只改变图形的位置和方向(相当于镜像),但不改变图形的形状和大小。因此，两者都不改变图形的形状和大小.\n【详解】平移和轴对称都不改变图形的形状和大小，只改变图形的位置，所以判断正确。故答案为:√",
+        "knowledge_points":  [
+                                 "平移与平移现象",
+                                 "旋转与旋转现象"
+                             ],
+        "source":  "19-20四年级下·黑龙江齐齐哈尔·期末"
+    },
+    {
+        "id":  "481",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "一个图形平移后，形状和位置都不变。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】物体或图形沿着某个方向移动了一定距离叫做平移。特点：大小，形状，方向不变，位置变化。熟悉平移的特征是解答此题的关键.\n【详解】一个图形平移后，形状不变，位置改变。故答案为: ×",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "20-21四年级下·全国·单元测试"
+    },
+    {
+        "id":  "482",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "红领巾是一个等腰三角形，它只有1条对称轴。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】红领巾的两条边相等，是等腰三角形。等腰三角形顶点到底边中点所在的直线就是对称轴。如果一个图形沿着一条直线对折后两部分完全重合，这样的图形叫做轴对称图形，这条直线叫做对称轴.\n【详解】红领巾是一个等腰三角形，它只有1条对称轴，所以判断正确。故答案为:√",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "19-20五年级下·全国·单元测试"
+    },
+    {
+        "id":  "483",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "长方形，圆，正方形，平行四边形和三角形都是轴对称图形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】如果一个平面图形沿着一条直线折叠后，直线两旁的部分能够互相重合，那么这个图形叫做轴对称图形，这条直线叫对称轴;判断一个图形是否是轴对称图形，关键是找它的对称轴，要想象沿着这条线翻折能不能重叠。解答此题的依据是：轴对称图形的意义及特征.\n【详解】长方形，圆，正方形是轴对称图形，平行四边形不是轴对称图形，等腰三角形，等边三角形是轴对称图形，一般的三角形不是轴对称图形。故答案为: ×",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "20-21四年级下·全国·单元测试"
+    },
+    {
+        "id":  "484",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "镜子外的人举起左手，镜子里的影子也一定举起左手。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】根据镜面对称的特征，镜中的景物与实际景物上下前后方向一致，左右方向相反，大小不变，且关于镜对称，镜子外的人举起左手，镜子里的影子应举起右手。此题主要明白镜面对称的特点是：上下前后方向一致，左右方向相反.\n【详解】镜子外的人举起左手，镜子里的影子也一定举起左手是错误的，镜子里的影子应举起右手。故答案为: ×",
+        "knowledge_points":  [
+                                 "镜面对称"
+                             ],
+        "source":  "20-21四年级下·全国·单元测试"
+    },
+    {
+        "id":  "485",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "正六边形一定有6条对称轴。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】依据轴对称图形的意义，即在平面内，如果一个图形沿一条直线对折，对折后的两部分都能完全重合，这样的图形叫做轴对称图形，这条直线就是其对称轴，据此即可进行判断。解答此题的主要依据是：轴对称图形的意义及其对称轴的条数，解答时要把所有的对称轴都考虑到.\n【详解】因为正六边形沿三组对应边的中点所在的直线和三条对角线对折，对折后的两部分都能完全重合，则正六边形是轴对称图形，三组对应边的中点所在的直线和三条对角线所在的直线即是它的对称轴，所以正六边形有6条对称轴，原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "20-21五年级下·全国·单元测试"
+    },
+    {
+        "id":  "486",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "平移可以改变图形的大小，但不能改变图形的形状。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平移和旋转都是物体或图形的位置发生变化而形状，大小不变。区别在于，平移时物体沿直线运动，本身方向不发生改变;旋转是物体绕着某一点或轴运动，本身方向发生了变化。熟练掌握平移的定义和特征是解答本题的关键.\n【详解】根据分析可知，平移可以改变图形的位置，但图形的形状和大小不变;所以判断错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "19-20四年级下·重庆秀山·期末"
+    },
+    {
+        "id":  "487",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "圆有无数条对称轴，正方形有4条对称轴，平行四边形有2条对称轴。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】一个图形沿一条直线对折，直线两旁的图形完全重合，这样的图形叫做轴对称图形，折痕所在的直线就是对称轴。据此可知，圆有无数条对称轴，正方形有4条对称轴，长方形有2条对称轴，平行四边形没有对称轴。据此判断即可。本题考查对称轴的认识，应熟记常见图形对称轴数量.\n【详解】圆有无数条对称轴，正方形有4条对称轴，平行四边形没有对称轴。故答案为: ×",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "19-20四年级下·重庆巴南·期末"
+    },
+    {
+        "id":  "488",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "等腰三角形是轴对称图形，它有2条对称轴。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】等腰三角形包括两边相等的等腰三角形和三边相等的等边三角形，两边相等的等腰三角形的对称轴是底边上的高所在的直线，等边三角形的对称轴是三边上的高所在的直线。明确等边三角形是特殊的等腰三角形是解答此题的关键.\n【详解】根据分析可知：等腰三角形的对称轴可以有1条或3条;等腰三角形是轴对称图形，它有2条对称轴，所以判断错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "19-20四年级下·天津宝坻·期末"
+    },
+    {
+        "id":  "489",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "字母T,W,N都是轴对称图形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】一个图形沿一条直线对折，直线两旁的图形完全重合，这样的图形叫做轴对称图形，折痕所在的直线就是对称轴;据此进行判断即可。判断一个图案是否是轴对称图形的关键是看在这个图形中能否找到一条直线，使图形沿着这条直线对折后能够完全重合.\n【详解】字母T,W是轴对称图形,N不是轴对称图形。故答案为: ×",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "19-20三年级下·宁夏银川·期末"
+    },
+    {
+        "id":  "490",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "树叶都是轴对称图形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】一个图形沿一条直线对折后，直线两旁的部分能够完全重合，这样的图形叫做轴对称图形，据此进行判断即可。判断是不是轴对称图形的关键是找出对称轴，看图形沿对称轴对折后两部分能否完全重合.\n【详解】根据轴对称图形的意义可知，并不是所有的树叶都是轴对称图形，所以判断错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "20-21三年级上·河北承德·期中"
+    },
+    {
+        "id":  "491",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "一个图形先向上平移5格，再向下平移6格，得到的图形就相当于向下平移了1格。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】在平面内，将一个图形上的所有点都按照某个直线方向做相同距离的移动，这样的图形运动叫做图形的平移运动，简称平移。平移不改变图形的形状和大小.\n【详解】一个图形先向上平移5格，再向下平移6格，得到的图形就相当于向下平移了1格。原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "20-21五年级上·全国·期中"
+    },
+    {
+        "id":  "492",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "小青坐电梯从一楼到十二楼，电梯的运动是平移运动。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】把一个图形整体沿某个方向移动一定的距离，图形的这种运动叫做平移，据此解答。本题考查了平移现象在生活中的实际实例，数学联系生活，要学以致用.\n【详解】根据分析可得：小青坐电梯从一楼到十二楼，电梯的运动是平移运动。原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "19-20四年级下·甘肃庆阳·期末"
+    },
+    {
+        "id":  "493",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "长方形，正方形和等边三角形一定是轴对称图形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】轴对称的定义：一个图形沿一条直线对折，直线两旁的图形完全重合，这样的图形叫做轴对称图形，折痕所在的直线就是对称轴;常见的轴对称图形及对称轴数量如：圆有无数条对称轴，正方形有4条对称轴，长方形有2条对称轴，等腰梯形有1条对称轴，等边三角形有3条对称轴，等腰三角形有1条对称轴等。本题考查了轴对称图形的辨别，关键是掌握轴对称图形的定义和常用图形那些是轴对称图形.\n【详解】根据分析可得：长方形，正方形和等边三角形一定是轴对称图形。原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "19-20四年级下·甘肃天水·期末"
+    },
+    {
+        "id":  "494",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "等边三角形和长方形的对称轴条数相等。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】轴对称图形定义为平面内，一个图形沿一条直线折叠，直线两旁的部分能够完全重合的图形，直线叫做对称轴，分别确定等边三角形和长方形对称轴的条数，再做判断。本题考查了对称轴的数量，关键是明确轴对称和对称轴，熟悉平面图形的特征.\n【详解】等边三角形有3条对称轴，长方形有2条对称轴，对称轴条数不相等，所以原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "19-20六年级下·山东菏泽·期末"
+    },
+    {
+        "id":  "495",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "圆一定是轴对称图形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】根据轴对称图形的意义：如果一个图形沿着一条直线对折后两部分完全重合，这样的图形叫做轴对称图形，这条直线叫做对称轴;据此解答。本题考查了轴对称图形的辨认，掌握轴对称图形的意义是解题的关键.\n【详解】根据分析可得：圆是轴对称图形，圆的任意一条直径都是它的对称轴，所以圆一定是轴对称图形。故答案为:√",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "19-20四年级·广西河池·期末"
+    },
+    {
+        "id":  "496",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "等边三角形只有一条对称轴。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】一个图形沿一条直线对折，直线两旁的图形完全重合，这样的图形叫做轴对称图形，折痕所在的直线就是对称轴;据此进行判断即可。本题考查对称轴的定义，灵活运用定义判断一个图形有几条对称轴.\n【详解】根据对称轴的定义可知，等边三角形有三条对称轴。故答案为: ×",
+        "knowledge_points":  [
+                                 "对称轴的画法及数量",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "19-20四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "497",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "一个图形不论是向左平移还是向下平移，它的形状和大小都不变。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】平移是指在平面内，将一个图形上的所有点都按照某个方向作相同距离的移动，平移不改变图形的形状和大小，只是改变位置;解答判断即可。本题主要考查平移的意义和特征.\n【详解】一个图形无论怎样平移，都只是位置发生了变化，它的形状，大小都不会发生变化;所以，一个图形不论是向左平移还是向下平移，它的形状和大小都不变，故此说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "平移与平移现象"
+                             ],
+        "source":  "19-20四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "498",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "平行四边形一定是轴对称图形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】一个图形沿一条直线对折后，直线两旁的部分能够完全重合，这样的图形叫做轴对称图形，据此进行判断即可。此题考查了轴对称图形的判断方法，及学生思维的严密性.\n【详解】对于一般的平行四边形，无论沿哪一条直线对折，平行四边形的两边都不会完全重合，所以平行四边形一定是轴对称图形，故此说法不正确。故答案为: ×",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认",
+                                 "平行四边形的概念及特点"
+                             ],
+        "source":  "19-20五年级下·河北·单元测试"
+    },
+    {
+        "id":  "499",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "计算图中阴影部分的面积。(单位：米)![](images/d7e4209d5a5e3d8d629375f25203fed3a64fe3892692c393f890e8e7ee3d76db.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "84平方米",
+        "solution":  "【分析】观察图形，图中的空白部分是十字形道路，我们可以把横向和纵向的空白道路分别平移到边缘，这样阴影部分就可以拼成一个新的长方形;原来长方形的长是16米，由于纵向道路宽2米，平移后新长方形的长为16-2=14米，原来长方形的宽是8米，由于横向道路宽2米，平移后新长方形的宽为8-2=6米;最后根据\"长方形面积=长×宽\"计算出阴影部分的面积.\n【详解】(16-2)×(8-2)\n=14×6\n=84(平方米)\n所以阴影部分的面积是84平方米。",
+        "knowledge_points":  [
+                                 "利用平移法巧算周长",
+                                 "利用平移巧算周长与面积",
+                                 "长方形的面积"
+                             ],
+        "source":  "2024·辽宁鞍山·小升初真题"
+    },
+    {
+        "id":  "500",
+        "unit":  "第七单元图形的运动",
+        "question_type":  "计算题",
+        "difficulty":  0.85,
+        "question":  "求出下面图形的周长。(单位：厘米)![](images/cd0b171dddacedd8d8301f42c65b4e90bca25071a520a9d1e659b6421ddf3892.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "28厘米",
+        "solution":  "【分析】如图所示，按箭头所示将线段进行平移，此时图形可以变成一个边长是7厘米的正方形，正方形周长=边长×4，据此解题.\n![](images/46b1ce64ad55c7d2257867ed0fc29bc14297414953a2e83bf4609ce417cfd521.jpg)\n【详解】7×4=28(厘米)，图形的周长是28厘米。",
+        "knowledge_points":  [
+                                 "利用平移法巧算周长",
+                                 "利用平移巧算周长与面积",
+                                 "正方形的周长"
+                             ],
+        "source":  "23-24四年级下·全国·单元测试"
+    }
+]
diff --git "a/\345\233\233\345\271\264\347\272\247\351\242\230\351\233\206/\345\233\233\345\271\264\347\272\247\344\270\213/\345\233\233\345\271\264\347\272\247\344\270\213\347\254\254\344\270\211\345\215\225\345\205\203\350\277\220\347\256\227\345\276\213.json" "b/\345\233\233\345\271\264\347\272\247\351\242\230\351\233\206/\345\233\233\345\271\264\347\272\247\344\270\213/\345\233\233\345\271\264\347\272\247\344\270\213\347\254\254\344\270\211\345\215\225\345\205\203\350\277\220\347\256\227\345\276\213.json"
new file mode 100644
index 0000000000000000000000000000000000000000..fcfd98c7729f4be1b3699ba560f585a53f3a92b8
--- /dev/null
+++ "b/\345\233\233\345\271\264\347\272\247\351\242\230\351\233\206/\345\233\233\345\271\264\347\272\247\344\270\213/\345\233\233\345\271\264\347\272\247\344\270\213\347\254\254\344\270\211\345\215\225\345\205\203\350\277\220\347\256\227\345\276\213.json"
@@ -0,0 +1,8276 @@
+[
+    {
+        "id":  "1",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在()里填上\"\u003e\"\"\u003c\"或\"=\".\n48×48()96×24;180÷12()180/15;9999()208708",
+        "options":  [
+
+                    ],
+        "answer":  "=;\u003e;\u003c",
+        "solution":  "【分析】比较48×48与96×24的大小，因为96=48×2，24=48÷2，所以96×24=(48×2)×(48÷2)，根据乘法结合律可判断它们的大小。比较180÷12与180÷15的大小，在除法运算中，当被除数相同时，除数越小，商越大。比较9999与208708的大小，先看位数，位数多的数大;9999是四位数，208708是六位数，所以208708大。【详解】96×24=(48×2)×(48÷2)=48×48，所以48×48=96×24。因为12\u003c15，且两个式子的被除数都是180，因此180÷12\u003e180÷15。六位数大于四位数，因此9999\u003c208708。",
+        "knowledge_points":  [
+                                 "整数乘法结合律，除数是两位数的笔算除法，两位数乘两位数的进位乘法，两位数除以一位数，首位能除尽"
+                             ],
+        "source":  "24-25四年级上·河北衡水·期末"
+    },
+    {
+        "id":  "2",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\"\u003e\"\"\u003c\"或\u0027=\u0027.\n439999()44万;1周角()2直角;39×40()4×390;999×202()200000",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003e;=;\u003e",
+        "solution":  "【分析】比较大小时，先来看位数，位数多的数就大，位数相同比较最高位上的数，最高位上的数大的数就大，依次比下去即可;比较439999与44万的大小：因为1万=10000，所以44万=440000，然后比较439999与440000，再判断大小。比较1周角与2直角的大小:1周角=360,1直角=90。先计算2直角的度数，即2乘以90等于180，再比较360与180，判断大小。比较39×40与4×390的大小：根据积不变的规律，一个因数扩大几倍，另一个因数缩小相同的倍数，积不变。在39×40中，将39扩大10倍变为390，将40缩小10倍变为4，积不变。再判断大小。比较999×202与200000的大小：首先计算999×202的积。可以将999看作(1000-1),利用乘法分配律进行计算，再判断大小。【详解】1万=10000,44万=440000，都是六位数，十万位上都是4，来比较万位上的数字,3\u003c4，再判断大小。因此,439999\u003c44万。1周角=360,1直角=90,2个直角是180,360\u003e180，因此,1周角\u003e2直角。根据积不变的规律,39×40=4×390.999×202=(1000-1)×202=1000×202-1×202=202000-202=201798都是六位数，十万位上都是2，万位上都是0，来比较千位上的数字,1\u003e0,201798\u003e200000，所以999×202\u003e200000。",
+        "knowledge_points":  [
+                                 "大数的比较，整数乘法分配律，积的变化规律(整数乘法),亿以内数的改写"
+                             ],
+        "source":  "24-25四年级上·云南曲靖·期末"
+    },
+    {
+        "id":  "3",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "(a+b)×c=ac+()这里运用的是()律。a×b=b×()这里运用的是()律。",
+        "options":  [
+
+                    ],
+        "answer":  "bc;乘法分配;a;乘法交换",
+        "solution":  "【分析】本题考查乘法运算定律。乘法分配律：两个数的和与一个数相乘，可以先把它们与这个数分别相乘，再相加。用字母表示为(a+b)×c=a×c+b×c。乘法交换律：交换两个因数的位置，积不变。用字母表示为a×b=b×a。【详解】两个数的和与一个数相乘，可以把这两个数分别与这个数相乘，再相加。用字母表示为(a+b)×c=ac+bc,运用的是乘法分配律。两个数相乘，交换两个乘数的位置，积不变。用字母表示为a×b=b×a,运用的是乘法交换律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律，整数乘法交换律"
+                             ],
+        "source":  "25-26五年级上·新疆乌鲁木齐·期中"
+    },
+    {
+        "id":  "4",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "12×201=12×200+12是运用了()律;400-65-135=400-(65+135)是运用了();20×168×5=168×(20×5)既运用了()律，又运用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法分配;减法的运算性质;乘法交换;乘法结合",
+        "solution":  "【分析】乘法分配律是指两个数的和与一个数相乘，可以先把它们分别与这个数相乘，再相加，将201拆分为200+1，符合乘法分配律的形式，故运用了乘法分配律;减法的运算性质是指从一个数里连续减去两个数，可以减去这两个数的和,400连续减去65和135，先计算65+135，则运用了减法的运算性质;乘法交换律是指两个数相乘，交换因数的位置，它们的积不变,20与168交换了位置，所以运用了乘法交换律;乘法结合律是指三个数相乘，先把前两个数相乘，再和另外一个数相乘，或先把后两个数相乘，再和另外一个数相乘，积不变，先计算20×5，则运用了乘法结合律。【详解】由分析可知12×201=12×200+12是运用了乘法分配律;400-65-135=400-(65+135)是运用了减法的运算性质;20×168×5=168×(20×5)既运用了乘法交换律，又运用了乘法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律，整数乘法结合律，整数乘法交换律，整数减法的性质"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "5",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "(a×b)×c=();a×(b×c)表示的运算定律是();乘法分配律用字母表示是()。",
+        "options":  [
+
+                    ],
+        "answer":  "a×(b×c);乘法结合律;(a+b)×c=a×c+b×c",
+        "solution":  "【分析】乘法结合律：三个数相乘，先把前两个数相乘或者先把后两个数相乘，积不变，用字母表示为(a×b)×c=a×(b×c)。乘法分配律：两个数的和与第三个数相乘，可以把这两个数与第三个数分别相乘再相加，用字母表示为(a+b)×c=a×c+b×c或a×(b+c)=a×b+a×c。【详解】(a×b)×c=a×(b×c);a×(b×c)表示的运算定律是乘法结合律;乘法分配律用字母表示是(a+b)×c=a×c+b×c。",
+        "knowledge_points":  [
+                                 "整数乘法结合律，整数乘法分配律"
+                             ],
+        "source":  "25-26四年级上·广东揭阳·月考"
+    },
+    {
+        "id":  "6",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在计算297+298+299+300+301+302+303时，怎么算最快呢。",
+        "options":  [
+
+                    ],
+        "answer":  "将首尾两两相加，再加上中间数",
+        "solution":  "【分析】观察算式中的数字，发现第一个数297和最后一个数303，第二个数298和倒数第二个数302，第三个数299和倒数第三个数301，它们的和均为600。因此可以将它们首尾两两相加，再加上中间的数，快速算出结果。【详解】根据分析可知：将它们首尾两两相加，再加上中间的数，快速算出结果。297+298+299+300+301+302+303=(297+303)+(298+302)+(299+301)+300=600+600+600+300=600×3+300=1800+300=2100",
+        "knowledge_points":  [
+                                 "整数加法结合律，整数加法交换律，整十，整百，整千数与一位数的乘法，两，三位数的连续进位加法"
+                             ],
+        "source":  "25-26四年级上·河南南阳·期中"
+    },
+    {
+        "id":  "7",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算器上的数字键\"3\"坏了，怎样用这个计算器计算。请用综合算式表示出468×63=()计算的思考过程。()",
+        "options":  [
+
+                    ],
+        "answer":  "468×63=468×7×9",
+        "solution":  "【分析】根据题意，乘法结合律指三个数相乘时，先乘前两个数或先乘后两个数，结果相同。乘法结合律:(a×b)×c=a×(b×c)或a×b×c=a×(b×c)把这63写成两个数相乘的形式，然后再相乘，只要不出现数字3即可，积不变。据此答题即可。【详解】根据乘法结合律，把这63写成两个数相乘的形式，如7×9=63，然后再与468相乘，即468×63=468×7×9。(答案不唯一)",
+        "knowledge_points":  [
+                                 "计算器的初步认识与使用，整数乘法结合律"
+                             ],
+        "source":  "25-26四年级上·河南南阳·期中"
+    },
+    {
+        "id":  "8",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "1+2+3+4+5+95+96+97+98+99=()",
+        "options":  [
+
+                    ],
+        "answer":  "500",
+        "solution":  "【分析】根据题意，仔细观察题目可知,1+99=100,2+98=100,3+97=100,4+96=100,5+95=100，再把5个100相加，列式计算即可。【详解】根据分析:1+2+3+4+5+95+96+97+98+99=(1+99)+(2+98)+(3+97)+(4+96)+(5+95)=100+100+100+100+100=500即1+2+3+4+5+95+96+97+98+99=500",
+        "knowledge_points":  [
+                                 "整数加法结合律，整数加法交换律"
+                             ],
+        "source":  "25-26四年级上·河南南阳·期中"
+    },
+    {
+        "id":  "9",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "下面的算式分别运用了什么运算律。(填序号)\n(1)57+98=98+57。() (2)47+36+53=36+(47+53)。( ) (3)a+(20+34)=(a+20)+34。()(4)A+68=68+A。() (5)81+67+19+33=(81+19)+(67+33)。()",
+        "options":  [
+                        "①加法交换律",
+                        "②加法结合律",
+                        "③加法交换律和加法结合律"
+                    ],
+        "answer":  "①;③;②;①;③",
+        "solution":  "【分析】分析每个算式，根据加法交换律和结合律的定义判断运用的运算定律。(1)57+98=98+57中，交换了57和98的位置，符合加法交换律。(2)47+36+53=36+(47+53)中，先交换47和36的位置，再把47和53结合，运用了加法交换律和加法结合律。(3)a+(20+34)=(a+20)+34中，只改变了运算顺序，符合加法结合律。(4)A+68=68+A中，交换了A和68的位置，符合加法交换律。(5)81+67+19+33=(81+19)+(67+33)中，先交换19和67的位置，再分别结合，运用了加法交换律和加法结合律。【详解】由分析可知：(1)填①; (2)填③; (3)填②; (4)填①; (5)填③。",
+        "knowledge_points":  [
+                                 "整数加法结合律，整数加法交换律"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "10",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在()里填上\"\u003e\"\"\u003c\"或\u0027=\u0027.\n175-15-45()175-45-15;316-(195-116)()316-116-195;589-(389+48)()589-389+48;186-(33+67)()186-33-67",
+        "options":  [
+
+                    ],
+        "answer":  "=;\u003e;\u003c;=",
+        "solution":  "【分析】根据四则混合运算顺序，分别计算左右两边的式子，再比较得数大小。【详解】175-15-45=160-45=115175-45-15=130-15=115115=115，所以,175-15-45=175-45-15;316-(195-116)=316-79=237;316-116-195=200-195=5;237\u003e5，所以,316-(195-116)\u003e316-116-195;589-(389+48)=589-437=152;589-389+48=200+48=248;152\u003c248，所以,589-(389+48)\u003c589-389+48;186-(33+67)=186-100=86;186-33-67=153-67=86;86=86，所以,186-(33+67)=186-33-67。",
+        "knowledge_points":  [
+                                 "整数减法的性质，带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "11",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "10-9+8-7+6-5+4-3+2-1=()",
+        "options":  [
+
+                    ],
+        "answer":  "5",
+        "solution":  "【分析】观察算式，从左向右，依次每两项的计算结果都是1，应用加法的结合律，可知整个算式能看成五个减法算式的和，结果就是五个1的和，据此解答。【详解】10-9+8-7+6-5+4-3+2-1=(10-9)+(8-7)+(6-5)+(4-3)+(2-1)=1+1+1+1+1=5，所以10-9+8-7+6-5+4-3+2-1=5。",
+        "knowledge_points":  [
+                                 "整数加法结合律，加减混合运算(10以内)"
+                             ],
+        "source":  "2025·内蒙古通辽·小升初真题"
+    },
+    {
+        "id":  "12",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "3454+1634+366+46=() ○ () ○() ○()。",
+        "options":  [
+
+                    ],
+        "answer":  "3454;46;1634;366",
+        "solution":  "【分析】根据加法交换律，把式子变成3454+46+1634+366，再根据加法结合律把式子变成(3454+46)+(1634+366)，再进行简便计算。【详解】3454+1634+366+46=3454+46+1634+366=(3454+46)+(1634+366)=3500+2000=5500。",
+        "knowledge_points":  [
+                                 "整数加法结合律，整数加法交换律"
+                             ],
+        "source":  "24-25四年级上·山东菏泽·期中"
+    },
+    {
+        "id":  "13",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号中填上\"\u003e\"\"\u003c\"或\u0027=\u0027.\n5亿()50000001;496870()498706\n110×28()280×11;360×25()360×24",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;\u003c;=;\u003e",
+        "solution":  "【分析】把5亿化成整数后再进行比较。比较两个数的大小，首先比位数，位数多的数大;位数相同，再从高位到低位依次比较。根据乘法交换律和积的变化规律进行解答。【详解】5亿=500000000，500000000\u003e50000001，所以5亿\u003e50000001;496870\u003c498706;110×28=(110÷10)×(28×10)=11×280，11×280=280×11，所以110×28=280×11;25\u003e24，所以360×25\u003e360×24。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法),整数乘法交换律，大数的比较"
+                             ],
+        "source":  "24-25四年级上·吉林白城·期中"
+    },
+    {
+        "id":  "14",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "用简便方法计算245+649+55=649+245+55=649+(245+55)=949的依据是加法()律和加法()律。",
+        "options":  [
+
+                    ],
+        "answer":  "交换;结合",
+        "solution":  "【分析】加法交换律:a+b=b+a,加法结合律:a+b+c=a+(b+c),245+649+55=649+245+55运用了加法交换律，交换加数649和245的位置,649+245+55=649+(245+55)运用了加法结合律，据此解答。【详解】根据分析可知，用简便方法计算245+649+55=649+245+55=649+(245+55)=949的依据是加法交换律和加法结合律。",
+        "knowledge_points":  [
+                                 "整数加法结合律，整数加法交换律"
+                             ],
+        "source":  "24-25四年级上·重庆黔江·期中"
+    },
+    {
+        "id":  "15",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "三个数相乘，先乘(),或者先乘后两个数，积不变。这叫做()。用字母表示是()。",
+        "options":  [
+
+                    ],
+        "answer":  "前两个数;乘法结合律;(a×b)×c=a×(b×c)",
+        "solution":  "【分析】根据乘法结合律指三个数相乘时，先乘前两个数或先乘后两个数，结果相同。用字母表示是(a×b)×c=a×(b×c)以此答题即可。【详解】根据分析可知：三个数相乘，先乘前两个数，或者先乘后两个数，积不变。这叫做乘法结合律。用字母表示是(a×b)×c=a×(b×c)。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·广东汕头·期末"
+    },
+    {
+        "id":  "16",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "萌萌用计算器计算\"125×24\"时，发现按键\"4\"坏了。如果还用这个计算器，那么他可以怎样计算。请写出算式:()。",
+        "options":  [
+
+                    ],
+        "answer":  "125×8×3",
+        "solution":  "【分析】根据乘法结合律:(a×b)×c=a×(b×c)或a×b×c=a×b(b×c)),把24改写成8×3或12×2，用计算器计算即可。【详解】根据分析可知:125×24=125×(8×3)=125×8×3=1000×3=3000 萌萌用计算器计算\"125×24\"时，发现按键\"4\"坏了。如果还用这个计算器，那么他可以怎样计算。请写出算式:125×8×3。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·河北邢台·期末"
+    },
+    {
+        "id":  "17",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "萌萌用计算器计算\"125×24\"时，发现按键\"4\"坏了。如果还用这个计算器，那么他可以怎样计算。请写出算式()。",
+        "options":  [
+
+                    ],
+        "answer":  "125×3×8",
+        "solution":  "【分析】按键\"4\"坏了，就不能直接按出24。我们可以把24拆成两个数的乘积或和，比如24=3×8，或者24=20+4，但4坏了所以不能用4。可以用125×24=125×3×8，因为3和8的键都是好的。【详解】把24拆成8×3:125×24=125×3×8。如果还用这个计算器，那么他可以怎样计算。请写出算式:125×3×8。(答案不唯一)",
+        "knowledge_points":  [
+                                 "整数乘法结合律，计算器的初步认识与使用"
+                             ],
+        "source":  "24-25四年级下·河北邢台·期末"
+    },
+    {
+        "id":  "18",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "两个数的和与一个数相乘，可以先把它们与这个数分别(),再(),这叫做，用字母表示是。",
+        "options":  [
+
+                    ],
+        "answer":  "相乘;相加;乘法分配律;(a+b)×c=a×c+b×c",
+        "solution":  "【详解】两个数的和与一个数相乘，可以先把它们与这个数分别相乘，再相加，这叫做乘法分配律，用字母表示是:(a+b)×c=a×c+b×c。例如:(100+5)×24=100×24+5×24=2400+120=2520",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·安徽淮北·期末"
+    },
+    {
+        "id":  "19",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "已知A×13+87×A=3500,A=();已知250-B×6=130,B=( )。",
+        "options":  [
+
+                    ],
+        "answer":  "35;20",
+        "solution":  "【分析】A×13+87×A利用乘法分配律a×c+b×c=(a+b)×c,变为A×(13+87)=3500，根据因数=积/另一个因数求A;已知250-B×6=130，把B×6看成一个整体，根据减数=被减数-差，可知B×6=250-130=120，再根据因数=积/另一个因数，即可求出B。【详解】A×13+87×A=3500 A×(13+87)=3500 A×100=3500 A=3500÷100=35250-B×6=130 B×6=250-130=120 B=120÷6=20 即已知A×13+87×A=3500,A=35;已知250-B×6=130,B=20。",
+        "knowledge_points":  [
+                                 "整数乘法分配律，乘，除法的意义和各部分间的关系，加，减法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·湖北十堰·期末"
+    },
+    {
+        "id":  "20",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "婷婷在计算(45+A)×20时漏掉了括号，这样所得的结果与正确的结果相差()。",
+        "options":  [
+
+                    ],
+        "answer":  "855",
+        "solution":  "【分析】漏掉小括号后，式子为45+A×20，而带小括号的式子，根据乘法分配律去掉小括号后，把20分别与45,A相乘，再把两个积相加，两个式子的差别在于，加号左边的数一个是45，另一个是45×20，所以用45与20的积，减45即可解答。【详解】(45+A)×20=45×20+20×A=900+20A(45+A)×20漏掉括号后，式子为45+A×20=45+20A;45×20-45=900-45=855婷婷在计算(45+A)×20时漏掉了括号，这样所得的结果与正确的结果相差855。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算，整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·山东日照·期末"
+    },
+    {
+        "id":  "21",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\"\u003e\"\"\u003c\"或\"=\".\na+30()30+a;125×102()125×100+2;985-327-27()985-(327-27);3200-758+142()3200-(758+142)",
+        "options":  [
+
+                    ],
+        "answer":  "=;\u003e;\u003c;\u003e",
+        "solution":  "【分析】根据加法交换律:a+b=b+a，据此比较即可;先将102写成100+2的形式，再根据乘法分配律:(a+b)×c=a×c+b×c,将左边算式简便计算后比较即可;根据减法的性质a-b-c=a-(b+c),将左边算式写成有小括号的形式后比较即可;根据减法的性质，去掉右边算式的小括号后比较即可。【详解】a+30=30+a;125×102=125×(100+2)=125×100+125×2,125×100+125×2和125×100+2前面都是125×100,125×2×2，所以125×100+125×2\u003e125×100+2,125×102\u003e125×100+2;985-327-27=985-(327+27),985-(327+27)和985-(327-27)被减数都是985，减数327+27\u003e327-27，减数越大差越小，所以985-(327+27)\u003c985-(327-27),985-327-27\u003c985-(327-27);3200-(758+142)=3200-758-142,3200-758-142和3200-758+142前面都是3200-758，减去142肯定比加142小，所以3200-758-142\u003c3200-758+142,3200-758+142\u003e3200-(758+142)。a+30=30+a;125×102\u003e125×100+2;985-327-27\u003c985-(327-27);3200-758+142\u003e3200-(758+142)。",
+        "knowledge_points":  [
+                                 "整数乘法分配律，整数减法的性质，整数加法交换律"
+                             ],
+        "source":  "24-25四年级下·四川南充·期末"
+    },
+    {
+        "id":  "22",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "4×8×25×125=(4×25)×(8×125))运用了()律和()律。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法交换;乘法结合",
+        "solution":  "【分析】乘法交换律：两个数相乘，交换两个乘数的位置，积不变。用字母表示为:a×b=b×a。乘法结合律：三个数相乘，先把前两个数相乘或者先把后两个数相乘，积不变。用字母表示为:a×b×c=a×(b×c))。由题意得，仔细观察算式4×8×25×125=(4×25)×(8×125))可知，交换乘数\"8\"和\"25\"的位置，先算4×25和8×125，整个过程中运用了乘法交换律和乘法结合律。【详解】4×8×25×125=(4×25)×(8×125)运用了乘法交换律和乘法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法交换律，整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·浙江杭州·期末"
+    },
+    {
+        "id":  "23",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算.\n200-198+196-194+192-190+。+8-6+4-2",
+        "options":  [
+
+                    ],
+        "answer":  "100",
+        "solution":  "【分析】这是一道加减混合运算。观察可知，相邻两个数分为一组后，每组结果都相等，可利用分组法进行简便计算。【详解】200-198=2，196-194=2，192-190=2，…，8-6=2，4-2=2。从2到200一共有100个偶数，每两个数一组，共有50组。每组结果都是2，所以总和为2×50=100。",
+        "knowledge_points":  [
+                                 "两，三位数的加减混合运算"
+                             ],
+        "source":  "2024四年级·全国·竞赛"
+    },
+    {
+        "id":  "24",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "我们在计算15×24时，先算5乘24，再算10乘24，最后再相加，即(5+10)×24=5×24+10×24。笔算乘法的竖式计算，体现了乘法()律。用文字表示为\"两个数的()与一个数相乘，可以先把它们与这个数()相乘，再相加.\"用字母表示为()。",
+        "options":  [
+
+                    ],
+        "answer":  "分配;和;分别;(a+b)×c=a×c+b×c",
+        "solution":  "【分析】两个数的和与一个数相乘，可以先把它们分别与这个数相乘，再相加。这就是乘法分配律，据此解答。【详解】(5+10)×24=5×24+10×24笔算乘法的竖式计算，体现了乘法分配律。用文字表示为\"两个数的和与一个数相乘，可以先把它们分别与这个数相乘，再相加.\"用字母表示为(a+b)×c=a×c+b×c。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·河南省直辖县级单位·期末"
+    },
+    {
+        "id":  "25",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "刘阿姨到洛阳买了12盒牡丹花茶，付钱时是这样算的:10×24=240(元),2×24=48(元),240+48=288(元),这种盒装的牡丹花茶，每盒()元，它在计算过程中运用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "24;乘法分配律",
+        "solution":  "【分析】根据题意，将12盒分成10盒与2盒，再与牡丹花茶的单价相乘。算出10盒与2盒各是多少元。再相加，就是12盒一共多少元。根据单价×数量=总价，也可以直接用每盒牡丹花茶的价钱乘12，就是一共多少元。这里符合乘法分配律。两个数的和与一个数相乘，可以先把它们分别与这个数相乘，再相加。用字母表示是(a+b)×c=a×c+b×c。根据算式可以推算出每盒牡丹花茶需要花费的钱。【详解】根据分析，这种盒装的牡丹花茶，每盒24元。商品总价为:10×24+2×24=(10+2)×24=12×24=288(元)所以，它在计算过程中运用了乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律，经济问题"
+                             ],
+        "source":  "24-25四年级下·河南洛阳·期末"
+    },
+    {
+        "id":  "26",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "计算87×101时可以应用()律来简便计算，结果是()。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法分配律;8787",
+        "solution":  "【分析】计算87×101时可以先把101改写成100+1，然后运用乘法的分配律进行计算，会比较简便。【详解】由分析可得:87×101=87×(100+1)=87×100+87×1=8700+87=8787计算87×101时可以应用乘法分配律来简便计算，结果是8787。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·河南许昌·期末"
+    },
+    {
+        "id":  "27",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "判断325+231=556的计算是否正确，可用()+()来验算，这种验算方法运用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "231;325;加法交换",
+        "solution":  "【分析】加法交换律是指两个数相加，交换加数的位置，和不变，用字母表示为a+b=b+a。【详解】根据加法交换律a+b=b+a，那么325+231=231+325。所以判断325+231=556的计算是否正确，可用231+325来验算，这种验算方法运用了加法交换律。",
+        "knowledge_points":  [
+                                 "整数加法交换律"
+                             ],
+        "source":  "24-25四年级下·河南焦作·期末"
+    },
+    {
+        "id":  "28",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "18+25+75=18+(25+75)这是根据()律进行简便计算。",
+        "options":  [
+
+                    ],
+        "answer":  "加法结合",
+        "solution":  "【分析】加法结合律是指三个数相加，先把前两个数相加，或者先把后两个数相加，和不变。原式18+25+75中，将25和75先相加(即25+75=100),再与18相加，得到18+100=118;符合加法结合律，据此解答。【详解】由分析可知,18+25+75=18+(25+75)这是根据加法结合律进行简便计算。",
+        "knowledge_points":  [
+                                 "整数加法结合律"
+                             ],
+        "source":  "24-25四年级下·河南洛阳·期末"
+    },
+    {
+        "id":  "29",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\"\u003e\"\"\u003c\"或\u0027=\u0027.\n240×3()24×40;400个万()308万;720÷(9×5)()720÷9×5;231×101()231×100+1",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003e;\u003c;\u003e",
+        "solution":  "【分析】比较240×3和24×40的大小，计算出左右两边式子的结果，再比较大小。比较400个万和308万的大小,400个万是400万，因为400万\u003e308万，所以400个万\u003e308万。比较720÷((9×5))和720÷9×5的大小，计算出左右两边式子的结果，再比较大小。720÷(9×5):先算括号里的乘法，再算括号外除法;720÷9×5，按照从左到右的顺序计算，计算出结果进行比较即可。比较231×101和231×100+1的大小，计算出左右两边式子的结果，再比较大小。【详解】240×3=720;24×40=960，因为720\u003c960，所以240×3\u003c24×40。因为400万\u003e308万，所以400个万\u003e308万。720÷(9×5)=720÷45=16720÷9×5=80×5=400 因为16\u003c400，所以720÷((9×5))\u003c720÷9×5.231×101=231×(100+1)=231×100+231×1=23100+231=23331231×100+1=23100+1=23101因为23331\u003e23101，所以231×101\u003e231×100+1",
+        "knowledge_points":  [
+                                 "整数乘法分配律，带有小括号的混合运算，大数的比较"
+                             ],
+        "source":  "24-25四年级下·河南洛阳·期末"
+    },
+    {
+        "id":  "30",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "7200÷15/4=7200÷ (□*□);348×23+23×52=(□+ □)×23。",
+        "options":  [
+
+                    ],
+        "answer":  "15;4;348;52",
+        "solution":  "【分析】根据除法的性质，连续除以两个数等于除以这两个数的积。根据乘法分配律，两个数的和乘第三个数，等于这两个数分别乘第三个数，再把它们的积相加;据此解答。【详解】7200÷15÷4=7200÷(15×4)=7200÷60=120。348×23+23×52=(348+52)×23=400×23=9200。所以，7200÷15÷4=7200÷(15×4);348×23+23×52=(348+52)×23。",
+        "knowledge_points":  [
+                                 "整数除法的性质，整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·四川乐山·期末"
+    },
+    {
+        "id":  "31",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "已知A×24+76×A=4800,A=()",
+        "options":  [
+
+                    ],
+        "answer":  "48",
+        "solution":  "【分析】本题符合乘法分配律的特点，一个数分别与两个数相乘再相加，等于这个数乘这两个数的和;再根据积除以一个因数等于另一个因数算出A的值。【详解】A×24+76×A=A×(24+76)=A×100。A×100=4800，所以A=4800÷100=48。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·山东日照·期末"
+    },
+    {
+        "id":  "32",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "125×25*□,在方框里填()能使计算简便(□里只填一个数),计算结果是()。",
+        "options":  [
+
+                    ],
+        "answer":  "32;100000",
+        "solution":  "【分析】因为125×8=1000,25×4=100，所以要使得计算简便，可以在方框里填4×8，也就是32。根据乘法交换律和乘法结合律进行简算。【详解】根据分析，在方框里填32，算式为:125×25×32将32看作4×8利用乘法交换律和乘法结合律，将算式变成:125×25×32=125×25×4×8=(125×8)×(25×4)=1000×100=100000所以计算结果为100000。(答案不唯一)",
+        "knowledge_points":  [
+                                 "整数乘法交换律，整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·河南洛阳·期末"
+    },
+    {
+        "id":  "33",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明到游泳馆游泳，游泳池长50米，他每次都游12个来回。小明每次要游()米。",
+        "options":  [
+
+                    ],
+        "answer":  "1200",
+        "solution":  "【分析】1个来回是游了2个游泳池长，先用12乘2求出小明一共游了几个游泳池的长，再乘50即可求解。【详解】12×2×50=12×(2×50)=12×100=1200(米)。所以小明每次要游1200米。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·四川凉山·期末"
+    },
+    {
+        "id":  "34",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "小明将17×(□+3)算成17*□+3，他算出的结果与正确答案相差()。",
+        "options":  [
+
+                    ],
+        "answer":  "48",
+        "solution":  "【分析】根据乘法分配律求出17×(□+3)的结果，再减去17×□+3，然后再进一步解答。【详解】17×(□+3)-(17×□+3)=17×□+17×3-17×□-3=(17×□-17×□)+(17×3-3)=51-3=48。小明将17×(□+3)算成17×□+3，他算出的结果与正确答案相差48。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·四川凉山·期末"
+    },
+    {
+        "id":  "35",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "7×42+13×42=(7+13)×42这里运用了()律;如果■+▲=100，那么68*■+68*▲=()。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法分配律;6800",
+        "solution":  "【分析】7与13分别乘42，再将积相加，可以逆用乘法分配律，先计算7与13的和，再把这个和与42相乘。给68分别乘■与▲，再把积相加，也可以逆用乘法分配律，先计算■与▲的和，再把这个和与68相乘。【详解】7×42+13×42=(7+13)×42，这里运用了乘法分配律;68×■+68×▲=68×(■+▲)=68×100=6800，所以如果■+▲=100，那么68×■+68×▲=6800。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·福建漳州·期末"
+    },
+    {
+        "id":  "36",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "125×88=125×(80+8)=125×80+125×8，这是运用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法分配律",
+        "solution":  "【分析】计算125×88时，可以先把88拆成80+8，算式变成125×(80+8)，然后根据乘法分配律把算式变成125×80+125×8，据此解答。【详解】125×88=125×(80+8)=125×80+125×8=10000+1000=11000，这是运用了乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "25-26五年级上·黑龙江佳木斯·开学考试"
+    },
+    {
+        "id":  "37",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算:25×(40+4)时，应用()律可使计算简便，结果是()。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法分配;1100",
+        "solution":  "【分析】在计算25×(40+4)时，观察到括号内40和4相加，可以运用乘法分配律将原式展开为25×40+25×4，25×40=1000，25×4=100，这样计算更简便。【详解】25×(40+4)=25×40+25×4=1000+100=1100，所以计算25×(40+4)时，应用乘法分配律可使计算简便，结果是1100。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·江西萍乡·期末"
+    },
+    {
+        "id":  "38",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果a+b=40，那么5×a+5×b=( )。",
+        "options":  [
+
+                    ],
+        "answer":  "200",
+        "solution":  "【分析】根据乘法分配律a×b+a×c=a×(b+c)，将5×a+5×b转化为5×(a+b)，再代入已知条件a+b=40计算即可。【详解】5×a+5×b=5×(a+b)=5×40=200。如果a+b=40，那么5×a+5×b=200。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·江西萍乡·期末"
+    },
+    {
+        "id":  "39",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果A×B=8，那么125×A×9×B=()",
+        "options":  [
+
+                    ],
+        "answer":  "9000",
+        "solution":  "【分析】根据乘法结合律，125×A×9×B=125×9×(A×B)，再把A×B=8代入算式;再根据乘法交换律，把算式变成125×8×9，列式计算即可。【详解】125×A×9×B=125×9×(A×B)=125×9×8=125×8×9=1000×9=9000。所以如果A×B=8，那么125×A×9×B=9000。",
+        "knowledge_points":  [
+                                 "整数乘法交换律，整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·江西九江·期末"
+    },
+    {
+        "id":  "40",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "125×32=125×()×(),运用()律。",
+        "options":  [
+
+                    ],
+        "answer":  "8;4;乘法结合",
+        "solution":  "【分析】可将32看成8×4，这样125×32就可以变成125×8×4，先计算125×8，这是运用了乘法结合律;整数乘法结合律：三个数相乘，先乘前两个数，或者先乘后两个数，积不变，这叫做乘法结合律，用字母表示:(a×b)×c=a×(b×c);据此解答.\n【详解】根据分析:125×32=125×8×4，运用了乘法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·福建福州·期末"
+    },
+    {
+        "id":  "41",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用计算器计算\"35×32\"时，发现按键\"3\"坏了，如果还要继续使用这个计算器，你会怎样计算。请列出算式()。",
+        "options":  [
+
+                    ],
+        "answer":  "5×7×4×8",
+        "solution":  "【分析】因为计算器上\"3\"键坏了，不能直接用35和32相乘，所以考虑把35和32分别分解成其他数相乘的形式，再利用乘法结合律来计算它们的乘积.\n【详解】35可以写成5和7相乘，即35=5×7，同样,32可以写成4和8相乘，也就是32=4×8，所以原来的35×32，就可以写成5×7×4×8。(答案不唯一)",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·江西上饶·期末"
+    },
+    {
+        "id":  "42",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "小红家每天要买一瓶3元的牛奶和一袋2元的豆浆，她家一星期买牛奶和豆浆要花()元。",
+        "options":  [
+
+                    ],
+        "answer":  "35",
+        "solution":  "【分析】分别计算出一个星期买牛奶的钱和一个星期买豆浆的钱，最后再相加即可，再根据乘法分配律a×b+a×c=a×(b+c)进行简便计算。【详解】3×7+2×7=(3+2)×7=5×7=35(元)。她家一星期买牛奶和豆浆要花35元。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·江西赣州·期末"
+    },
+    {
+        "id":  "43",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "小红在用计算器计算9876×49时发现键4坏了，于是她这样计算:9876×50-9876，她这样算的依据是()。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法分配律",
+        "solution":  "【分析】根据乘法分配律，两个数的差与一个数相乘，可以先把它们分别与这个数相乘，再相减;49=50-1，所以9876×49=9876×(50-1),利用乘法分配律为9876×50-9876，据此解题.\n【详解】小红在用计算器计算9876×49时发现键4坏了，于是她这样计算:9876×50-9876，她这样算的依据是乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·江西赣州·期末"
+    },
+    {
+        "id":  "44",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算25×(40+4)时，应用()律可使计算简便，结果是()。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法分配;1100",
+        "solution":  "【分析】根据题意，在计算25×(40+4)时，观察算式结构，发现括号内是两个数相加，且外部的因数25与括号内的两个加数40和4相乘时都能得到整百数，符合乘法分配律的应用条件。应用乘法分配律将原式展开为25×40+25×4，分别计算后相加，可使计算简便.\n【详解】根据分析可知:25×(40+4)=25×40+25×4=1000+100=1100 计算25×(40+4)时，应用乘法分配律可使计算简便，结果是1100。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·江西萍乡·期末"
+    },
+    {
+        "id":  "45",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "如果a+b=40，那么5×a+5×b=()",
+        "options":  [
+
+                    ],
+        "answer":  "200",
+        "solution":  "【分析】根据乘法分配律a×b+a×c=a×(b+c),将5×a+5×b转化为5×(a+b),再代入已知条件a+b=40计算即可.\n【详解】5×a+5×b=5×(a+b)=5×40=200如果a+b=40，那么5×a+5×b=200。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·江西萍乡·期末"
+    },
+    {
+        "id":  "46",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "5×75×20=5×()×75，这是应用了()。",
+        "options":  [
+
+                    ],
+        "answer":  "20;乘法交换律",
+        "solution":  "【分析】乘法交换律是指两个数相乘，交换因数的位置，它们的积不变,a×b=b×a。据此解答.\n【详解】我们可以先交换75和20的位置，得到5×20×75。所以5×75×20=5×20×75，这是应用了乘法交换律。",
+        "knowledge_points":  [
+                                 "整数乘法交换律"
+                             ],
+        "source":  "24-25四年级下·江西赣州·期末"
+    },
+    {
+        "id":  "47",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果○-△=10，那么47*○-47*△=(),如果☆*○=50，那么☆*(○*8)=()",
+        "options":  [
+
+                    ],
+        "answer":  "470;400",
+        "solution":  "【分析】根据乘法分配律的逆运算，可将47×○-47×△变为47×(○-△)，然后代入数据进行计算。根据乘法结合律，可将☆×(○×8)变为(☆×○)×8，然后代入数据进行计算。【详解】如果○-△=10，那么47×○-47×△=47×(○-△)=47×10=470;如果☆×○=50，那么☆×(○×8)=(☆×○)×8=50×8=400。",
+        "knowledge_points":  [
+                                 "整数乘法分配律，整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·陕西渭南·期末"
+    },
+    {
+        "id":  "48",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "比较算式的大小:12×99+12()(12+1)×99",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c",
+        "solution":  "【分析】利用乘法分配律:(a+b)×c=a×c+b×c,右边的算式(12+1)×99=12×99+99，此算式跟左边的算式对比，前半部分一样为12×99，后半部分99比12大，所以12×99+12\u003c(12+1)×99.\n【详解】根据分析,(12+1)×99=12×99+99，所以12×99+12\u003c(12+1)×99。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·浙江杭州·期末"
+    },
+    {
+        "id":  "49",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "聪聪在计算40×(△+5)时，由于粗心把算式写成了40*△+5，这样算出的结果与正确结果相差()。",
+        "options":  [
+
+                    ],
+        "answer":  "195",
+        "solution":  "【分析】乘法分配律：两个数的和与一个数相乘，可以先把它们分别与这个数相乘，再相加。用字母表示是(a+b)×c=a×c+b×c。将40×(△+5)根据乘法分配律变成40*△+40×5，再与40*△+5比较。发现大家都有40*△,相差的是40×5和5之间的差。据此解答.\n【详解】40×(△+5)=40*△+40×5=40*△+200 与40*△+5，相差的是200-5=195 所以，这样算出的结果与正确结果相差195。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·贵州铜仁·期末"
+    },
+    {
+        "id":  "50",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "小丽计算25×24时，将\"24\"看成了\"4\"计算出了100，现在需要将100 ○ □ 就可以得到正确的结果。(在圆圈内填合适的运算符号，在方框内填合适的数或式子)",
+        "options":  [
+
+                    ],
+        "answer":  "×;6",
+        "solution":  "【分析】首先分析正确计算与错误计算的差异：正确算式是25×24，小丽错算成25×4=100.24=4×6，因此25×24=25×(4×6)=(25×4)×6。小丽已经算出25×4=100，所以只需将100×6，就能得到正确结果.\n【详解】由分析知:25×24=25×(4×6)=(25×4)×6 所以小丽计算25×24时，将\"24\"看成了\"4\"计算出了100，现在需要将100×6就可以得到正确的结果。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·河南新乡·期末"
+    },
+    {
+        "id":  "51",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "用计算器计算\"125×16\"时，发现按键\"6和8\"坏了，聪明的小明用\"125×()×()\"解决了问题。",
+        "options":  [
+
+                    ],
+        "answer":  "4;4",
+        "solution":  "【分析】计算器的按键\"6\"坏了，则无法按出16，可以将16写成8×2或4×4，但是\"8\"也坏了，所以只能写成4×4，则算式变为125×4×4;据此解答即可。【详解】根据分析可知：用计算器计算\"125×16\"时，发现按键\"6和8\"坏了，聪明的小明用\"125×4×4\"解决了问题。",
+        "knowledge_points":  [
+                                 "整数乘法结合律，计算器的初步认识与使用"
+                             ],
+        "source":  "24-25四年级下·江西南昌·期末"
+    },
+    {
+        "id":  "52",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "计算25×(4+8)时，可以用()律简便计算，结果是()。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法分配律;300",
+        "solution":  "【分析】仔细观察算式25×(4+8)可知，可以利用乘法分配律a×b+a×c=a×(b+c)),将算式转化为25×4+25×8计算起来比较简便.\n【详解】25×(4+8)=25×4+25×8=100+200=300所以，计算25×(4+8)时，可以用乘法分配律简便计算，结果是300。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·江西南昌·期末"
+    },
+    {
+        "id":  "53",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "简便计算25×44=25×4×11运用了()这一运算定律。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法结合律",
+        "solution":  "【分析】根据题意，观察发现25×44=25×4×11是将44拆成了(4×11),再运用乘法结合律:(a×b)×c=a×(b×c))或a×b×c=a×(b×c);【详解】根据分析可得:25×44=25×(4×11)=25×4×11=(25×4)×11=1100，所以25×44=25×4×11，这里运用了乘法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·河南新乡·期末"
+    },
+    {
+        "id":  "54",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小丽在计算25×(4+30)时，算成了25×4+30，所得的结果比正确结果少()。",
+        "options":  [
+
+                    ],
+        "answer":  "720",
+        "solution":  "【分析】根据乘法分配律a×(b+c)=a×b+a×c，将25×(4+30)转化成25×4+25×30。再按正常运算顺序计算结果，25×4+30先算乘法，再算加法。将正确结果减去错误结果即可求出两者差值。【详解】25×(4+30)=25×4+25×30=100+750=850。25×4+30=100+30=130。差值：850-130=720，所以所得的结果比正确结果少720。",
+        "knowledge_points":  [
+                                 "乘加，乘减混合运算，带有小括号的混合运算，整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·贵州遵义·期末"
+    },
+    {
+        "id":  "55",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小马虎在计算\"35×(8-A)\"时错算成\"35×8-A\"，他算得的结果比正确结果大了102，那么A是(),这道题的正确答案是()。",
+        "options":  [
+
+                    ],
+        "answer":  "3;175",
+        "solution":  "【分析】根据乘法分配律(a-b)×c=a×c-b×c,将35×(8-A)写成35×8-35×A,错算成了35×8-A,少减了34个A,是102，所以用102÷34即可计算出A=3，然后带入算式即可计算出结果.\n【详解】35×(8-A)=35×8-35×A (35-1)×A=34×A=102 A=102÷34=335×(8-3)=35×5=175 小马虎在计算\"35×(8-A)\"时错算成\"35×8-A\"，他算得的结果比正确结果大了102，那么A是3，这道题的正确答案是175。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·浙江台州·期末"
+    },
+    {
+        "id":  "56",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "①现有\"6,2,3,4\"四个数，经过怎样的运算才能得到24呢。请你写出：综合算式:()=24。②已知M-N=1，求M×79-N×79=();",
+        "options":  [
+
+                    ],
+        "answer":  "6×4×(3-2);79",
+        "solution":  "【分析】①题目要求用6、2、3、4四个数通过运算得到24。可通过尝试不同的运算组合，结合四则运算和括号调整顺序，找到符合条件的算式。例如：6×4=24，剩余的两个数3和2可以通过运算得到1，所以可列式为6×4×(3-2)=24。②已知M-N=1，利用乘法分配律的逆运算，将M×79-N×79转化为79×(M-N)，代入已知条件计算即可。【详解】①综合算式：6×4×(3-2)=24。(答案不唯一)②M×79-N×79=79×(M-N)=79×1=79。",
+        "knowledge_points":  [
+                                 "整数乘法分配律，带有小括号的混合运算,24点"
+                             ],
+        "source":  "24-25四年级下·浙江台州·期末"
+    },
+    {
+        "id":  "57",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小津用计算器计算456×78时，发现按键\"7\"坏了，请帮他把算式补充完整:456×80-456×()。",
+        "options":  [
+
+                    ],
+        "answer":  "2",
+        "solution":  "【分析】乘法分配律，两个数的和与一个数相乘，可以先把它们与这个数分别相乘，再相加。首先78可以看作80减去2，然后根据乘法分配律进行计算即可.\n【详解】456×78=456×(80-2)=456×80-456×2=36480-912=35568因此456×78=456×80-456×2",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·河南洛阳·期末"
+    },
+    {
+        "id":  "58",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "乘法分配律用字母表示为:();乘法结合律用字母表示为:()。",
+        "options":  [
+
+                    ],
+        "answer":  "a×(b+c)=a×b+a×c;a×b×c=a×(b×c)",
+        "solution":  "【分析】(1)根据乘法分配律的定义：两个数的和与一个数相乘，可以先把它们与这个数分别相乘，再相加。因此用字母表示为:a×(b+c)=a×b+a×c。据此解答即可。(2)根据乘法结合律的定义：三个数相乘，先乘前两个数，或者先乘后两个数，积不变。因此用字母表示为:a×b×c=a×(b×c)。据此解答即可.\n【详解】乘法分配律用字母表示为:a×(b+c)=a×b+a×c;乘法结合律用字母表示为:a×b×c=a×(b×c)。",
+        "knowledge_points":  [
+                                 "整数乘法分配律，整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·河南洛阳·期末"
+    },
+    {
+        "id":  "59",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果A+B=30，那么(A+10)+(B-5)=()。如果C+D=50,C-D=9，那么C×20-D×20=()。",
+        "options":  [
+
+                    ],
+        "answer":  "35;180",
+        "solution":  "【分析】(A+10)+(B-5)可以写成A+10+B-5，再利用加法交换律和结合律变成A+B+10-5，然后代入A+B=30计算即可。C×20-D×20根据乘法分配律可以写成(C-D)×20，再代入C-D=9计算即可。【详解】如果A+B=30，那么(A+10)+(B-5)=A+B+10-5=30+10-5=35。因为C-D=9，所以C×20-D×20=(C-D)×20=9×20=180。",
+        "knowledge_points":  [
+                                 "整数加法交换律，整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·浙江嘉兴·期末"
+    },
+    {
+        "id":  "60",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果 △ +○=40，那么12*△+12*○=()。",
+        "options":  [
+
+                    ],
+        "answer":  "480",
+        "solution":  "【分析】乘法分配律:a×c+b×c=(a+b)×c,所以12*△+12*○=12×(△+○),△+○=40，所以用12×40即可.\n【详解】12*△+12*○=12×(△+○)=12×40=480，如果 △ +○=40，那么12*△+12*○=480。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·河南开封·期末"
+    },
+    {
+        "id":  "61",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "125×88=125×8×11，运用了()律,125×88=125×(80+8)=125×80+125×8，运用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法结合律;乘法分配律",
+        "solution":  "【分析】乘法结合律是指三个数相乘，先把前两个数相乘，再和另外一个数相乘，或先把后两个数相乘，再和另外一个数相乘，积不变。乘法分配律是指两个数的和与一个数相乘，可以先把它们分别与这个数相乘，再相加。在125×88=125×8×11中，是把88拆分成8×11，然后先计算125×8，这符合乘法结合律的特征，所以运用了乘法结合律。在125×88=125×(80+8)=125×80+125×8中，是把88拆分成80+8，然后分别与125相乘再相加，这符合乘法分配律的特征，所以运用了乘法分配律.\n【详解】根据分析可知,125×88=125×8×11运用了乘法结合律,125×88=125×(80+8)=125×80+125×8，运用了乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律，整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·河南开封·期末"
+    },
+    {
+        "id":  "62",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果□-△=16，那么5×□-△×5=();如果○÷25÷☆=○÷100，那么☆=()。",
+        "options":  [
+
+                    ],
+        "answer":  "80;4",
+        "solution":  "【分析】第一个问题，利用乘法分配律将5×□-5×△转化为5×(□-△)，代入已知条件□-△=16即可求解。第二个问题应用除法的性质，将连续除法转换为除以两个数的积，通过等式关系求出☆的值。【详解】5×□-5×△=5×(□-△)=5×16=80。○÷25÷☆=○÷(25×☆)=○÷100，所以25×☆=100，☆=4。",
+        "knowledge_points":  [
+                                 "整数除法的性质，整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·浙江台州·期末"
+    },
+    {
+        "id":  "63",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "如果M-N=8，那么125×M-N×125=()。",
+        "options":  [
+
+                    ],
+        "answer":  "1000",
+        "solution":  "【分析】根据乘法分配律a×b-a×c=a×(b-c)，125×M-N×125可以变成125×(M-N)，再将M-N=8代入计算即可。【详解】125×M-N×125=125×(M-N)=125×8=1000，所以如果M-N=8，那么125×M-N×125=1000。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·江西赣州·期末"
+    },
+    {
+        "id":  "64",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "1+3+5+7+…+15+17+19=()",
+        "options":  [
+
+                    ],
+        "answer":  "100",
+        "solution":  "【分析】本题根据加法交换律和结合律进行简算。1+3+5+7+。+15+17+19发现的规律是:1+19=20,3+17=20,5+15=20，以此类推,1,3,5。19一共有10个数，两两相加得出5个和，因为把式子的两头的两个数相加，和都是20，一共有5个20，求5个20是多少用乘法，据此解答.\n【详解】1+3+5+7+。+15+17+19=(1+19)+(3+17)+(5+15)+(7+13)+(9+11)=20+20+20+20+20=5×20=100 所以1+3+5+7+。+15+17+19=100",
+        "knowledge_points":  [
+                                 "整数加法结合律，整数加法交换律"
+                             ],
+        "source":  "18-19六年级上·河南南阳·期末"
+    },
+    {
+        "id":  "65",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据运算律填一填。(1)355+128+45=(□ + □)+□ (2)17×125×8=□*(□*□)",
+        "options":  [
+
+                    ],
+        "answer":  "355;45;128;17;125;8",
+        "solution":  "【分析】(1)计算时先利用加法交换律交换128与45的位置变为355+45+128，再利用加法结合律变为(355+45)+128，然后先算小括号内的加法，再算小括号外的加法;(2)计算时利用乘法结合律变为17×((125×8)),然后先算小括号内的乘法，再算小括号外的乘法.\n【详解】(1)355+128+45=355+45+128=(355+45)+128=400+128=528 即355+128+45=(355+45)+128;(2)17×125×8=17×(125×8)=17×1000=17000 即17×125×8=17×(125×8)。",
+        "knowledge_points":  [
+                                 "整数加法交换律，整数加法结合律，整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·陕西安康·期末"
+    },
+    {
+        "id":  "66",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用计算器计算\"879×49\"时，发现数字键\"4\"坏了。如果还用这个计算器，你会怎样计算。请写出算式:()。",
+        "options":  [
+
+                    ],
+        "answer":  "879×50-879",
+        "solution":  "【分析】根据乘法分配律：两个数的和与一个数相乘，可以把它们与这个数分别相乘，再相加;把49看成(50-1),然后根据乘法分配律列式即可.\n【详解】879×49=879×(50-1)=879×50-879×1=879×50-879 (答案不唯一)用计算器计算\"879×49\"时，发现数字键\"4\"坏了。如果还用这个计算器，你会怎样计算。请写出算式:879×50-879。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·贵州黔南·期末"
+    },
+    {
+        "id":  "67",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "算式457+236+□,在□中填上()就能使计算简便(只填一个数),计算结果是()。",
+        "options":  [
+
+                    ],
+        "answer":  "64;757",
+        "solution":  "【分析】根据加法结合律，将后两个数236与□结合凑成整百数，可使计算简便。236需加64得到300，再与457相加，结果为757。合理即可.\n【详解】457+236+64=457+(236+64)=457+300=757 算式457+236+□,在 □ 中填上64就能使计算简便(只填一个数),计算结果是757。(答案不唯一)",
+        "knowledge_points":  [
+                                 "整数加法结合律"
+                             ],
+        "source":  "24-25四年级下·内蒙古鄂尔多斯·期末"
+    },
+    {
+        "id":  "68",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "56×4×25=56×(4×25),运用了乘法()律。",
+        "options":  [
+
+                    ],
+        "answer":  "结合",
+        "solution":  "【分析】三个数相乘，可以先把前两个数相乘再乘第三个数，或者先把后两个数相乘再乘第一个数，它们的积不变，这叫做乘法结合律;这里把4与25结合先计算，所以运用了乘法结合律.\n【详解】由分析可知:56×4×25=56×(4×25),运用了乘法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·新疆克孜勒苏·期末"
+    },
+    {
+        "id":  "69",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "125×24×8=24×(125×8)是运用了(        )律和(        )律。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法交换;乘法结合",
+        "solution":  "【分析】125×24×8先根据乘法交换律a×b=b×a,变成24×125×8，再根据乘法结合律(a×b)×c=a×(b×c),变成24×(125×8)使得计算简便.\n【详解】125×24×8=24×125×8=24×(125×8)=24×1000=24000 根据分析,125×24×8=24×(125×8)是运用了乘法交换律和乘法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法结合律，整数乘法交换律"
+                             ],
+        "source":  "24-25四年级下·内蒙古呼伦贝尔·期末"
+    },
+    {
+        "id":  "70",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "已知M-N=7，那么M×125-125×N+125=(),10-M+N=()。",
+        "options":  [
+
+                    ],
+        "answer":  "1000;3",
+        "solution":  "【分析】已知M-N=7，利用乘法分配律，将M×125-125×N+125，变为125×(M-N)+125，再将所求表达式中的M-N替换为7，从而简化计算;将原式10-M+N,变为10-(M-N),再将所求表达式中的M-N替换为7，从而简化计算.\n【详解】M×125-125×N+125\n=125×(M-N)+125\n=125×7+125\n=875+125\n=1000\n10-M+N\n=10-(M-N)\n=10-7\n=3",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·浙江宁波·期中"
+    },
+    {
+        "id":  "71",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "125×13×8=13×(125×8)是运用了()律和()律。乘法分配律用字母表示为:()。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法交换;乘法结合;(a+b)×c=a×c+b×c",
+        "solution":  "【分析】乘法交换律：两个数相乘，交换两个乘数的位置，积不变。用字母表示为:a×b=b×a。乘法结合律：三个数相乘，先把前两个数相乘或者先把后两个数相乘，积不变。用字母表示为:a×b×c=a×(b×c)。乘法分配律：两个数的和与第三个数相乘，可以把这两个数与第三个数分别相乘再相加。用字母表示为:(a+b)×c=a×c+b×c由题意得，在算式125×13×8=13×(125×8))中，乘数的位置改变了，而且先把后两个乘数相乘，所以整个算式运用了乘法交换律;乘法分配律用算式表示为:(a+b)×c=a×c+b×c.\n【详解】125×13×8=13×(125×8)是运用了乘法交换律和乘法结合律。乘法分配律用字母表示为:(a+b)×c=a×c+b×c。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数乘法结合律",
+                                 "整数乘法交换律"
+                             ],
+        "source":  "24-25四年级下·江西赣州·期中"
+    },
+    {
+        "id":  "72",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据运算定律在括号里填上合适的数或字母.\nF+48=()+()\n125×(8*□)=(125*□)×17",
+        "options":  [
+
+                    ],
+        "answer":  "48;F;17;8",
+        "solution":  "【分析】根据加法交换律:a+b=b+a,所以F+48=48+F;根据乘法结合律:(a×b)×c=a×(b×c)(a×b),所以125×(8×17)=(125×8)×17.\n【详解】根据分析:F+48=48+F;125×(8×17)=(125×8)×17。",
+        "knowledge_points":  [
+                                 "整数乘法结合律",
+                                 "整数加法交换律"
+                             ],
+        "source":  "24-25四年级下·云南玉溪·期末"
+    },
+    {
+        "id":  "73",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果△+○=40，那么25*△+25*○=()。",
+        "options":  [
+
+                    ],
+        "answer":  "1000",
+        "solution":  "【分析】计算25*△+25*○时，应用乘法分配律(a+b)×c=a×c+b×c的逆运算将算式变成(△+○)×25，然后把△+○=40代入计算.\n【详解】25*△+25*○\n=25×(△+○)\n=25×40\n=1000\n所以25*△+25*○=1000。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·云南玉溪·期末"
+    },
+    {
+        "id":  "74",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填\"\u003e\"\"\u003c\"或\"=\"\n68×100()68×99+1;24×15()12×30;0个千万()1亿;7499000()750万",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;=;\u003c;\u003c",
+        "solution":  "【分析】整数大小的比较方法，首先看数位，若不同，数位多的整数大;若相同，从高位看起，相同数位上的数大的那个整数就大。(1)把100拆分成((99+1),算式68×100化成68×(99+1),根据乘法分配律可知,68×68×(99+1)=68×99+68×1，据此比较68×99+68×1和68×99+1的大小即可。(2)计算出算式的结果，再比较大小。(3)先算出0个千万的结果，再比较大小。(4)将大数统一单位后，再比较大小.\n【详解】(1)68×100\n=68×(99+1)\n=68×99+68×1\n68×99+68×1\u003e68×99+1，所以68×100\u003e68×99+1。(2)24×15=360,12×30=360,360=360，所以24×15=12×30。(3)0个千万是0,0\u003c1亿，所以0个千万\u003c1亿。(4)750万=7500000,7499000\u003c7500000，所以7499000\u003c750万。",
+        "knowledge_points":  [
+                                 "亿以内数的改写",
+                                 "整数乘法分配律",
+                                 "大数的比较"
+                             ],
+        "source":  "24-25四年级下·河南新乡·期末"
+    },
+    {
+        "id":  "75",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果a+b=100，那么500+a+b=(),500-a-b=()。",
+        "options":  [
+
+                    ],
+        "answer":  "600;400",
+        "solution":  "【分析】加法结合律：三个数相加，先把前两个数相加或者先把后两个数相加，和不变。用字母表示为:a+b+c=a+(b+c);减法的性质：一个数连续减去两个数，等于这个数减去两个数的和。用字母表示为:a-b-c=a-(b+c)。由题意得，可以利用加法结合律将算式500+a+b转化为500+(a+b),然后将a+b的值代入算出500+a+b的值。利用减法的性质将算式500-a-b转化为500-(a+b),然后将a+b的值代入算出500-a-b的值.\n【详解】500+a+b\n=500+(a+b)\n=500+100\n=600\n500-a-b\n=500-(a+b)\n=500-100\n=400\n如果a+b=100，那么500+a+b=600,500-a-b=400。",
+        "knowledge_points":  [
+                                 "整数减法的性质",
+                                 "整数加法结合律"
+                             ],
+        "source":  "24-25四年级下·浙江温州·期末"
+    },
+    {
+        "id":  "76",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "丽丽在计算(48+52)×a时，漏看了小括号，算出的结果比正确答案少了624，检查时发现了错误，又重新计算。她算出的正确结果应该是()。",
+        "options":  [
+
+                    ],
+        "answer":  "1400",
+        "solution":  "【分析】在计算(48+52)×a时，忘记了小括号，丽丽计算的算式是48+52×a,根据乘法分配律，将算式(48+52)×a转换成48×a+52×a,用正确算式(48×a+52×a)减去(48+52×a)的差为624，可以得到48×a-48=624，用624加48再除以48即可计算出a是多少，最后代入算式(48+52)×a,先计算括号里加法，再计算括号外的乘法即可.\n【详解】(48+52)×a=48×a+52×a\n48×a+52×a-(48+52×a)\n=48×a+52×a-48-52×a\n=48×a-48\n48×a-48=624\na代表的数:\n(624+48)÷48\n=672÷48\n=14\n(48+52)×14\n=100×14\n=1400\n所以正确结果是1400。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·浙江·期末"
+    },
+    {
+        "id":  "77",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小军计算6×(□+5)错算成6*□+5，他得到的结果与正确答案相差( )。",
+        "options":  [
+
+                    ],
+        "answer":  "25",
+        "solution":  "【分析】根据题意，乘法分配律是指两个数的和同一个数相乘，等于把两个加数分别同这个数相乘，再把两个积加起来，结果不变。用字母表示为:(a+b)×c=a×c+b×c。根据乘法分配律的定义,6×(□+5)的正确答案应该是6*□+5×6。则他的计算结果与正确答案相差6*□+5×6-(6*□+5)。以此答题即可.\n【详解】根据分析可知:\n6×(□+5)=6*□+5×6\n6*□+5×6-(6*□+5)\n=6*□+5×6-6*□-5\n=30-5\n=25\n小军计算6×(□+5)错算成6*□+5，他得到的结果与正确答案相差25。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "78",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小丹在用计算器计算1258×49时，发现数字键\u00274\u0027坏了。如果还是使用这个计算器，要算出正确结果，你能帮小丹想到什么办法。你想到的办法是()。",
+        "options":  [
+
+                    ],
+        "answer":  "把49拆成7×7",
+        "solution":  "【分析】三个数相乘，先乘前两个数，或者先乘后两个数，积不变。依据乘法结合律和表内乘法口诀，可以把\"49\"看成\"7×7\"。据此解答.\n【详解】1258×49\n=1258×(7×7)\n=1258×7×7\n=8806×7\n=61642\n综上可知，数字键\"4\"坏了，我们可以将1258×49转化为1258×7×7，计算结果也不会发生改变。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "23-24四年级下·河北保定·期末"
+    },
+    {
+        "id":  "79",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "(b+20)×3=( )×( )+( )×( ),运用的运算定律是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "b;3;20;3;乘法分配律",
+        "solution":  "【分析】根据乘法分配律a×(b+c)=a×b+a×c，据此解答即可。【详解】根据分析，(b+20)×3=b×3+20×3，运用的运算定律是乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·吉林松原·期末"
+    },
+    {
+        "id":  "80",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "(a+b)+c=a+(b+c)所表示的运算定律是()律。",
+        "options":  [
+
+                    ],
+        "answer":  "加法结合",
+        "solution":  "【分析】三个数相加，先把前两个数相加，或者先把后两个数相加，和不变，字母表示为:(a+b)+c=a+(b+c);据此即可解答.\n【详解】根据分析可知,(a+b)+c=a+(b+c)所表示的运算定律是加法结合律。",
+        "knowledge_points":  [
+                                 "整数加法结合律"
+                             ],
+        "source":  "24-25四年级下·广东潮州·期末"
+    },
+    {
+        "id":  "81",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "彤彤在计算24×18+12时，错算成24×(18+12),结果与正确得数相差(),正确的得数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "276;444",
+        "solution":  "【分析】根据乘法分配律，把算式24×(18+12)化成24×18+24×12，再用错误的算式减去正确的算式，即可求出结果与正确得数相差多少，再计算算式24×18+12，计算出结果，即可解答.\n【详解】24×(18+12)-(24×18+12)\n=24×18+24×12-24×18-12\n=24×12-12\n=288-12\n=276\n24×18+12\n=432+12\n=444\n彤彤在计算24×18+12时，错算成24×(18+12),结果与正确得数相差276，正确的得数是444",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·广东潮州·期末"
+    },
+    {
+        "id":  "82",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在计算25×88时，小旭是这样想的:88×25=22×(4×25)=22×100=2200，这里运用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法结合",
+        "solution":  "【分析】根据乘法结合律：三个数相乘，先乘前两个数，或者先乘后两个数，积不变。小旭将88看成22×4，所以88×25=(22×4)×25=22×(4×25)),运用了乘法结合律，即可解题.\n【详解】由分析可知:\n88×25\n=(22×4)×25\n=22×(4×25)\n=22×100\n=2200\n所以计算25×88时，小旭是这样想的:88×25=22×(4×25)=22×100=2200，这里运用了乘法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "83",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "185+79+15+21=(185+15)+(79+21)运用的运算律是()和()。",
+        "options":  [
+
+                    ],
+        "answer":  "加法交换律;加法结合律",
+        "solution":  "【分析】加法交换律：两个数相加，交换两个加数的位置，和不变。用字母表示:a+b=b+a;;加法结合律：三个数相加，先把前两个数相加，或者先把后两个数相加，和不变。用字母表示为:(a+b)+c=a+(b+c),据此解答.\n【详解】由分析可知:185+79+15+21=(185+15)+(79+21)运用的运算律是加法交换律和加法结合律。",
+        "knowledge_points":  [
+                                 "整数加法结合律",
+                                 "整数加法交换律"
+                             ],
+        "source":  "24-25四年级下·海南省直辖县级单位·期末"
+    },
+    {
+        "id":  "84",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "某工厂生产的肥皂每块重125克,5块装一盒，买这样的8盒肥皂重()克，合()千克。",
+        "options":  [
+
+                    ],
+        "answer":  "5000;5",
+        "solution":  "【分析】由题意得，某工厂生产的肥皂每块重125克,5块装一盒，可以先用125乘5算出每盒肥皂重多少克，然后再乘上8即可算出买这样的8盒肥皂一共重多少克。最后根据1000克=1千克来转化单位即可。计算时，利用乘法交换律可使计算简便.\n【详解】125×5×8\n=125×8×5\n=1000×5\n=5000(克)\n1000克=1千克，所以5000克=5千克.\n某工厂生产的肥皂每块重125克,5块装一盒，买这样的8盒肥皂重5000克，合5千克。",
+        "knowledge_points":  [
+                                 "整数乘法交换律",
+                                 "克，千克之间的换算与比较"
+                             ],
+        "source":  "24-25四年级下·湖南岳阳·期末"
+    },
+    {
+        "id":  "85",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "(36×25)×4=36×(25×4)运用了()(填运算律)。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法结合律",
+        "solution":  "【分析】根据乘法结合律:a×b×c=a×(b×c))可知,(36×25)×4=36×(25×4)运用了乘法结合律(填运算律).\n【详解】根据分析可知,(36×25)×4=36×(25×4)运用了乘法结合律(填运算律)。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·广东佛山·期末"
+    },
+    {
+        "id":  "86",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "7×425+13×425=(7+13)×425，这里运用了( )律;如果□+△=100，那么54*□+54*△=( )。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法分配;5400",
+        "solution":  "【分析】乘法分配律是指两个数的和与一个数相乘，可以先把它们分别与这个数相乘，再相加，根据乘法分配律:(a+b)×c=a×c+b×c的逆运用，将算式54*□+54*△写成54×(□+△),据此代入数字计算即可.\n【详解】7×425+13×425\n=(7+13)×425\n=20×425\n=8500\n54*□+54*△\n=54×(□+△)\n=54×100\n=5400\n7×425+13×425=(7+13)×425，这里运用了乘法分配律;如果□+△=100，那么54*□+54*△=5400。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·云南红河·期末"
+    },
+    {
+        "id":  "87",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "24+42+76+58用简便计算可以写成()+()。",
+        "options":  [
+
+                    ],
+        "answer":  "24+76;42+58",
+        "solution":  "【分析】加法结合律：三个数相加，可以先把前两个数相加，或者先把后两个数相加，和不变。加法交换律：两个数相加，交换它们的位置，和不变。据此解答即可。【详解】24+42+76+58先根据加法交换律变成24+76+42+58，再根据加法结合律变成(24+76)+(42+58)，使计算简便。",
+        "knowledge_points":  [
+                                 "整数加法结合律",
+                                 "整数加法交换律"
+                             ],
+        "source":  "24-25四年级下·广东佛山·期末"
+    },
+    {
+        "id":  "88",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "37×415-6×415+9×415=(□-□+□)×415。",
+        "options":  [
+
+                    ],
+        "answer":  "37;6;9",
+        "solution":  "【分析】乘法分配律：两个数的和与第三个数相乘，可以把这两个数与第三个数分别相乘再相加。用字母表示为:(a+b)×c=a×c+b×c。据此解答.\n【详解】由乘法分配律可知,37×415-6×415+9×415=(37-6+9)×415。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·吉林四平·期末"
+    },
+    {
+        "id":  "89",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在计算125×47×8=47×(125×8)时，这是运用了()律和()律。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法交换;乘法结合",
+        "solution":  "【分析】乘法交换律:a×b=b×a,将125和47交换了位置，所以运用了乘法交换律;乘法结合律:(a×b)×c=a×(b×c),将125×8结合在一起，所以是运用了乘法结合律.\n【详解】根据分析，在计算125×47×8=47×(125×8)时，这是运用了乘法交换律和乘法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法结合律",
+                                 "整数乘法交换律"
+                             ],
+        "source":  "24-25四年级下·湖北孝感·期末"
+    },
+    {
+        "id":  "90",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "125×32×25=(125*□)×(25*□),这里运用了()律和()律。",
+        "options":  [
+
+                    ],
+        "answer":  "8;4;乘法结合;乘法交换",
+        "solution":  "【分析】根据题意，计算125×32×25时，把32改写成8×4，再根据乘法交换律:a×b=b×a和乘法结合律:(a×b)×c=a×(b×c)或a×b×c=a×(b×c),变算式为:(125×8)×(25×4),再进行计算.\n【详解】根据分析可知:125×32×25=(125×8)×(25×4),这里运用了乘法结合律和乘法交换律。",
+        "knowledge_points":  [
+                                 "整数乘法结合律",
+                                 "整数乘法交换律"
+                             ],
+        "source":  "24-25四年级下·黑龙江佳木斯·期末"
+    },
+    {
+        "id":  "91",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "25×(19*□)=19×(□ ○ 4)3500÷14=3500÷□/2",
+        "options":  [
+
+                    ],
+        "answer":  "4;25;*;7",
+        "solution":  "【分析】乘法结合律：三个数相乘，先把前两个数相乘，或先把后两个数相乘，积不变;乘法交换律：两个数相乘，交换因数的位置，它们的积不变。第一个算式先利用乘法结合律去掉小括号，再利用乘法交换律交换19与25的位置，再利用乘法结合律先算25×4即可;除法的性质：一个数连续除以两个数，可以用这个数除以后面两个数的积。第二个算式中14=7×2，先把算式写成3500÷(7×2)),再利用除法的性质变为3500÷7/2即可.\n【详解】25×25×(19×4)=19×(25×4),3500÷14=3500÷7/2",
+        "knowledge_points":  [
+                                 "整数除法的性质",
+                                 "整数乘法结合律",
+                                 "整数乘法交换律"
+                             ],
+        "source":  "24-25四年级下·辽宁鞍山·期末"
+    },
+    {
+        "id":  "92",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "用简便方法计算44×25时,44×25=11×(4×25)这里运用了()律,44×25=40×25+4×25这里运用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法结合;乘法分配",
+        "solution":  "【分析】观察发现44×25=11×(4×25)是将44拆为了(11×4),再运用乘法结合律:(a×b)×c=a×(b×c)或a×b×c=a×(b×c);44×25=40×25+4×25是将44拆为了(40+4),再运用乘法分配律:(a+b)×c=a×c+b×c;据此解答.\n【详解】根据分析:44×25\n=(11×4)×25\n=11×(4×25)\n=11×100=1100\n所以44×25=11×(4×25)这里运用了乘法结合律.\n44×25\n=(40+4)×25\n=40×25+4×25\n=1000+100\n=1100\n所以44×25=40×25+4×25这里运用了乘法分配律;",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·海南三亚·期末"
+    },
+    {
+        "id":  "93",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "根据乘法分配律，填空:45×99+45=45×(□+□)",
+        "options":  [
+
+                    ],
+        "answer":  "99;1",
+        "solution":  "【分析】两个数的和与一个数相乘，可以用这两个数分别和这个数相乘，再把它们的积相加，这叫乘法分配律，用字母表示为:(a+b)×c=a×c+b×c.\n【详解】45×99+45\n=45×(99+1)\n=45×100\n=4500\n根据乘法分配律，填空:45×99+45=45×(99+1)。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·黑龙江佳木斯·期末"
+    },
+    {
+        "id":  "94",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果☆+□=12,☆*□=32，那么25*☆+25*□=( ),480÷☆/□=( )。",
+        "options":  [
+
+                    ],
+        "answer":  "300;15",
+        "solution":  "【分析】计算25*☆+25*□时，应用乘法分配律，变成25×(☆+□),然后把☆+□=12代入计算;计算480÷☆/□时，应用除法的性质，变成480÷(☆*□),然后把☆*□=32代入计算.\n【详解】25*☆+25*□\n=25×(☆+□)\n=25×12\n=300\n480÷☆/□\n=480÷(☆*□)\n=480÷32\n=15\n因此，如果☆+□=12,☆*□=32，那么25*☆+25*□=300,480÷☆/□=15。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数除法的性质"
+                             ],
+        "source":  "24-25四年级下·浙江杭州·期末"
+    },
+    {
+        "id":  "95",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "(4+40)×25=4×25+()×25，应用了()律，用字母表示是()。",
+        "options":  [
+
+                    ],
+        "answer":  "40;乘法分配;(a+b)×c=a×c+b×c",
+        "solution":  "【分析】乘法分配律：两个数的和与一个数相乘，可以先把它们与这个数分别相乘，再相加，字母表示为(a+b)×c=a×c+b×c;据此即可解答.\n【详解】(4+40)×25=4×25+40×25，应用了乘法分配律，用字母表示是:(a+b)×c=a×c+b×c。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·河南濮阳·期末"
+    },
+    {
+        "id":  "96",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "用自己喜欢的方式表示运算律.\n加法交换律:();乘法结合律:()。",
+        "options":  [
+
+                    ],
+        "answer":  "a+b=b+a;(a×b)×c=a×(b×c)",
+        "solution":  "【分析】加法交换律：两个数相加，交换两个加数的位置，和不变。用字母表示为:a+b=b+a;乘法结合律：三个数相乘，先把前两个数相乘或者先把后两个数相乘，积不变。用字母表示为:(a×b)×c=a×(b×c)。据此解答.\n【详解】加法交换律:a+b=b+a;乘法结合律:(a×b)×c=a×(b×c)。(答案不唯一)",
+        "knowledge_points":  [
+                                 "整数加法交换律",
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·陕西西安·期末"
+    },
+    {
+        "id":  "97",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在□内填适当的数或字母.\n6×32+4×32=(□+□)*□.\n(m-40)×2=□*□-□*□。",
+        "options":  [
+
+                    ],
+        "answer":  "6;4;32;m;2;40;2",
+        "solution":  "【分析】乘法分配律:a×(b+c)=a×b+a×c;根据乘法分配律进行填空即可解答.\n【详解】6×32+4×32=(6+4)×32。(m-40)×2=m×2-40×2。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·河南信阳·期末"
+    },
+    {
+        "id":  "98",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "4×a+5×a=(□+□)○a，运用了乘法()律;25×4×□=(25×□)×13，运用了乘法()律。",
+        "options":  [
+
+                    ],
+        "answer":  "4;5;×;分配;13;4;结合",
+        "solution":  "【分析】根据乘法分配律a×b+a×c=a×(b+c)可得，4×a+5×a满足乘法分配律的格式要求，所以4×a+5×a=(4+5)×a。计算25×4×13时，可以把25与4结合起来先算，变为(25×4)×13，运用了乘法结合律。【详解】根据分析可得：4×a+5×a=(4+5)×a，运用了乘法分配律;25×4×13=(25×4)×13，运用了乘法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法结合律",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·辽宁盘锦·期末"
+    },
+    {
+        "id":  "99",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用简便方法计算44×25时，明明把算式改写成11×(4×25),运用的运算律是( );聪聪把算式改写成40×25+4×25运用的运算律是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法结合律;乘法分配律",
+        "solution":  "【分析】根据乘法结合律：三个数相乘，先把前两个数相乘，再和另外一个数相乘，或先把后两个数相乘，再和另外一个数相乘，积不变;(a×b)×c=a×(b×c);乘法分配律：是指两个数的和与一个数相乘，可以先把它们分别与这个数相乘，再相加;(a+b)×c=a×c+b×c。明明把44拆分成11×4，算式变成11×4×25=11×(4×25);聪聪把44拆分成40+4，算式变成(40+4)×25=40×25+4×25，据此即可解答.\n【详解】用简便方法计算44×25时，明明把算式改写成11×(4×25),运用的运算律是乘法结合律;聪聪把算式改写成40×25+4×25运用的运算律是乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法结合律",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "100",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "计算25×48时，莹莹和东东出现不同计算方法，莹莹运用了()律，东东运用了()律.\n![](images/2026-03-07%20134950.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "乘法结合;乘法分配",
+        "solution":  "【分析】莹莹的方法：莹莹把48拆分成4×12，然后先计算25×4，即25×(4×12)=(25×4)×12，这运用了乘法结合律(a×b)×c=a×(b×c),利用25与4相乘得整百数的特性来简化计算。东东的方法：东东将48拆分为40+8，然后按照25×(40+8)=25×40+25×8的形式进行计算，这符合乘法分配律a×(b+c)=a×b+a×c,通过这种方式将复杂的乘法运算转化为较简单的乘法和加法运算.\n【详解】由分析得，计算25×48时，莹莹和东东出现不同计算方法，莹莹运用了乘法结合律，东东运用了乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法结合律",
+                                 "整数乘法交换律"
+                             ],
+        "source":  "24-25四年级下·广东广州·期末"
+    },
+    {
+        "id":  "101",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "为给市民提供温馨舒适的阅读环境，某城市书房买来70套桌椅，每张桌子125元，每把椅子75元，买这些桌椅一共用了()元。",
+        "options":  [
+
+                    ],
+        "answer":  "14000",
+        "solution":  "【分析】用每张桌子的钱数乘数量计算出桌子需要花的总钱数，再用每把椅子的钱数乘数量计算出椅子需要花的总钱数，然后再加一起即等于桌椅总共要花的钱，计算时可以利用乘法分配律进行简算.\n【详解】125×70+75×70\n=(125+75)×70\n=200×70\n=14000(元)为给市民提供温馨舒适的阅读环境，某城市书房买来70套桌椅，每张桌子125元，每把椅子75元，买这些桌椅一共用了14000元。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "102",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "根据运算定律，在方框内填上适当的数.\n(57+34)+66=57+(□+□);(29×125)*□=29×(□*8)",
+        "options":  [
+
+                    ],
+        "answer":  "34;66;8;125",
+        "solution":  "【分析】第一个算式根据加法的结合律即(a+b)+c=a+(b+c),即可得到答案;第二个算式根据乘法的结合律a×b×c=a×(b×c)即可得到答案.\n【详解】(57+34)+66=57+(34+66)\n(29×125)×8=29×(125×8)",
+        "knowledge_points":  [
+                                 "整数加法结合律",
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·安徽芜湖·期末"
+    },
+    {
+        "id":  "103",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "15×4=();15×8=15×4×()=()",
+        "options":  [
+
+                    ],
+        "answer":  "60;2;120",
+        "solution":  "【分析】根据题意，乘法结合律指三个数相乘时，先乘前两个数或先乘后两个数，结果相同。乘法结合律:(a×b)×c=a×(b×c)或a×b×c=a×(b×c),把8改写成4×2，再进行计算即可.\n【详解】根据分析可知:\n15×4=60\n15×8=15×4×2=60×2=120",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·河南洛阳·期末"
+    },
+    {
+        "id":  "104",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "a+b=125,c-a-b=325，则c=();a-b=125，那么8×a-8×b=()。",
+        "options":  [
+
+                    ],
+        "answer":  "450;1000",
+        "solution":  "【分析】(1)减法的性质：一个数连续减去两个数，等于这个数减去两个数的和。用字母表示为:a-b-c=a-(b+c)。由题意得，利用减法的性质将算式c-a-b=325转化为c-(a+b)=325，然后将a+b的值代入即可算出c的值。(2)乘法分配律：两个数的和与第三个数相乘，可以把这两个数与第三个数分别相乘再相加。用字母表示为:(a+b)*=a×c+b×c。由题意得，利用乘法分配律的逆运算将算式8×a-8×b转化为8×(a-b),然后将a-b的值代入即可算出8×a-8×b的值.\n【详解】(1)c-a-b=325，即c-(a+b)=325。a+b=125，所以c=325+125=450。(2)8×a-8×b=8×(a-b)=8×125=1000。a+b=125,c-a-b=325，则c=450;a-b=125，那么8×a-8×b=1000",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数减法的性质"
+                             ],
+        "source":  "24-25四年级下·山东济南·期末"
+    },
+    {
+        "id":  "105",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在()里填上\"\u003e\"\u003c\"或\"=\".\n86500÷4/25()86500÷4×25;234×25×5()234×(25+5);508-35+93()508+93-35;167-98()167-100-2",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003e;=;\u003e",
+        "solution":  "【分析】(1)根据除法的性质:a/b/c=a/(b×c),先将左边算式的值计算出来，再按照运算顺序，计算出右边算式的值，进行大小比较即可;(2)按照运算顺序，从左至右计算出左边算式的值，再按照运算顺序，先计算小括号内的加法，再计算小括号外的乘法，得出右边算式的值，进行大小比较即可;(3)根据带符号搬家，交换35和93的值，再进行大小比较即可;(4)先计算出左边算式的值，再根据减法的性质:a-b-c=a-(b+c),把100和2结合起来，计算出右边算式的值，进行大小比较即可.\n【详解】86500÷4/25\n86500÷(4×25)\n=86500÷100\n=865又86500÷4×25\n=21625×25\n=540625\n865\u003c540625\n因此:86500÷4/25\u003c86500÷4×25\n234×25×5\n=5850×5\n=29250\n234×(25+5)\n=234×30\n=7020\n29250\u003e7020\n因此:234×25×5\u003e234×(25+5)\n508-35+93\n=508+93-35\n=601-35\n=566\n508+93-35\n=601-35\n=566\n566=566\n因此,508-35+93=508+93-35\n167-98=69\n167-100-2\n=167-(100+2)\n=167-102\n=65\n69\u003e65\n因此,167-98\u003e167-100-2",
+        "knowledge_points":  [
+                                 "整数除法的性质",
+                                 "整数减法的性质",
+                                 "整数加法交换律",
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "24-25四年级下·河南信阳·期末"
+    },
+    {
+        "id":  "106",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在方框内填上适当的数，并在括号里填一填运用了什么运算律.\n(1)(144+73)+27=144+(73+□),运用了().\n(2)125×74*□=□*8×74，运用了().\n(3)28×168+168×72=(□+□)×168，运用了()。",
+        "options":  [
+
+                    ],
+        "answer":  "27;加法结合律;8;125;乘法交换律;28;72;乘法分配律",
+        "solution":  "【分析】(1)加法结合律：三个数相加，先把前两个数相加，或者先把后两个数相加，和不变;(2)乘法交换律：两个数相乘，交换两个因数的位置，积不变;(3)乘法分配律：两个数与一个数分别相乘，可以先算这两个数的和，再与这个数相乘;据此解答.\n【详解】(1)(144+73)+27=144+(73+27),运用了加法结合律。(2)125×74×8=125×8×74，运用了乘法交换律。(3)28×168+168×72=(28+72)×168，运用了乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数乘法交换律",
+                                 "整数加法结合律"
+                             ],
+        "source":  "24-25四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "107",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "聪聪在用计算器计算49×567时，发现按键\"4\"坏了。请你帮聪聪想一想该如何用计算器计算，请把想法写到括号里()",
+        "options":  [
+
+                    ],
+        "answer":  "50×567-567",
+        "solution":  "【分析】由于计算器的按键\"4\"损坏，无法直接输入49，因此需要将49转化为其他数的乘积形式。观察到49=7×7，因此原式49×567可转化为7×7×567进行计算，避免使用数字4;也可以把49写成50-1，即把原式转化为(50-1)×567，根据乘法分配律(a-b)×c=a×c-b×c可知,(50-1)×567=50×567-567，在计算器上依次输入50,*,567,-,567，即可得到正确结果，且过程中无需使用按键\"4\".\n【详解】根据分析可知：聪聪在用计算器计算49×567时，发现按键\"4\"坏了。聪聪可以用计算器这样计算:((50×567-567)。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "计算器的初步认识与使用"
+                             ],
+        "source":  "24-25四年级下·河南信阳·期末"
+    },
+    {
+        "id":  "108",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "已知A×B=200，那么(A×10)×B=(),(A×20)×(B/20)=(),800/A/B=()。",
+        "options":  [
+
+                    ],
+        "answer":  "2000;200;4",
+        "solution":  "【分析】根据积的变化规律：两个因数相乘(0除外),如果一个因数不变，一个因数扩大到原来的几倍或缩小为原来的几分之一，积也扩大到原来的几倍或缩小为原来的几分之一;商的变化规律：被除数和除数同时乘或除以相同的数(0除外),商不变;除法的性质，一个数连续除以两个数，等于除以两个数的乘积，据此解答即可.\n【详解】(A×10)×B\n=200×10\n=2000\n(A×20)×(B/20)=200\n800/A/B\n=800÷(A×B)\n=800÷200\n=4\n已知A×B=200，那么(A×10)×B=2000,(A×20)×(B/20)=200,800/A/B=4。",
+        "knowledge_points":  [
+                                 "整数除法的性质",
+                                 "积的变化规律(整数乘法)",
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "24-25四年级下·江苏南京·期末"
+    },
+    {
+        "id":  "109",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算\"200×45\"时，第二个因数看成54，结果会比正确的积多()。",
+        "options":  [
+
+                    ],
+        "answer":  "1800",
+        "solution":  "【分析】用(200×54)减去(200×45),计算出结果比正确的积多多少;计算时可以运用整数乘法分配律：两个数的差与一个数相乘，可以用被减数和减数分别与这个数相乘，再相减，用字母表示:(a-b)×c=a×c-b×c,乘法分配律可以逆运用，用字母表示:a×c-b×c=(a-b)×c;据此解答.\n【详解】根据分析:(200×54)-(200×45)=(54-45)×200=9×200=1800，所以结果会比正确的积多1800。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·浙江宁波·期末"
+    },
+    {
+        "id":  "110",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "125×25×32=(125*□)×(□*25),这里运用了()律和()律。(方框内也要填上答案)",
+        "options":  [
+
+                    ],
+        "answer":  "8;4;乘法交换;乘法结合",
+        "solution":  "【分析】125×25×32先根据乘法交换律:a×b=b×a,将算式125×25×32变成125×32×25，再将32写成8×4的形式，即125×(8×4)×25，然后根据乘法结合律:(a×b)×c=a×(b×c),将算式125×(8×4)×25变成(125×8)×(4×25),最后按照运算顺序计算即可.\n【详解】125×25×32\n=125×32×25\n=125×(8×4)×25\n=(125×8)×(4×25)\n=1000×100\n=100000\n125×25×32=(125×8)×(4×25),这里运用了乘法交换律和乘法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法结合律",
+                                 "整数乘法交换律"
+                             ],
+        "source":  "24-25四年级下·河北沧州·期末"
+    },
+    {
+        "id":  "111",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算88×125时,①88×125=80×125+8×125，运用了()律;②88×125=11×(8×125),运用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法分配;乘法结合",
+        "solution":  "【分析】乘法分配律：两个数的和与第三个数相乘，可以把这两个数与第三个数分别相乘再相加。用字母表示为:(a+b)×c=a×c+b×c;乘法结合律：三个数相乘，先把前两个数相乘或者先把后两个数相乘，积不变。用字母表示为:a×b×c=a×(b×c)。由题意得，计算88×125时，可以先把88转化为80+8，然后利用乘法分配律使计算简便;也可以把88转化为11×8，然后利用乘法结合律将原式转化为11×(8×125)可使计算简便.\n【详解】88×125\n=(80+8)×125\n=80×125+8×125\n88×125\n=(11×8)×125\n=11×(8×125)\n计算88×125时,①88×125=80×125+8×125，运用了乘法分配律②88×125=11×(8×125),运用了乘法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·湖南株洲·期末"
+    },
+    {
+        "id":  "112",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "太极拳是国家级非物质文化遗产之一。在一场大型的太极拳表演活动中，男生有24排，女生有26排，每排站15人。一共有()人参加表演。",
+        "options":  [
+
+                    ],
+        "answer":  "750",
+        "solution":  "【分析】分别用男生，女生的排数乘上每排的人数，求出男生，女生的人数，再相加求和即可求出参加表演的总人数.\n【详解】24×15+26×15\n=(24+26)×15\n=50×15\n=750(人)\n一共有750人参加表演。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "乘加",
+                                 "乘减混合运算"
+                             ],
+        "source":  "24-25四年级下·湖北省直辖县级单位·期末"
+    },
+    {
+        "id":  "113",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "我们如果口算24×25，可以用()律进行简便计算，如果列竖式笔算，在计算过程中运用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法结合律;乘法分配律",
+        "solution":  "【分析】本题需要明确两种计算24×25的方法分别运用了什么运算律。乘法结合律是三个数相乘，先把前两个数相乘，再乘第三个数;或先把后两个数相乘，再乘第一个数，积不变。乘法分配律是两个数的和与一个数相乘，可以先把两个加数分别与这个数相乘，再将两个积相加，结果不变。对于口算，要找到合适的变形方式使计算简便;对于列竖式笔算，要理解其计算过程所依据的原理。【详解】口算24×25时，把24拆分成6×4，则24×25=6×4×25。根据乘法结合律，这里先计算4×25=100，再计算6×100=600，所以口算24×25运用了乘法结合律。列竖式计算24×25时，把25拆分成20+5，那么24×25=24×(20+5)=24×20+24×5，所以列竖式笔算运用了乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·湖北武汉·期末"
+    },
+    {
+        "id":  "114",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在列竖式计算125×18时，其实就是把()个125和()个125相加，运用的是()律。",
+        "options":  [
+
+                    ],
+        "answer":  "8;10;乘法分配律",
+        "solution":  "【分析】在列竖式计算125×18时，将18分解为10和8，分别用125乘10和125乘8，再将结果相加。这符合乘法分配律的应用，即125×(10+8)=125×10+125×8。【详解】计算125×18时，先用18个位上的8乘125，得到125×8=1000，表示8个125相加;再用18十位上的1乘125，表示1个十，也就是10个125相加，即125×10=1250。最后将两部分积相加：1000+1250=2250。因此，在列竖式计算125×18时，其实就是把8个125和10个125相加，运用的是乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "24-25四年级下·湖北武汉·期末"
+    },
+    {
+        "id":  "115",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "算式125×88，运用乘法分配律计算的算式是()，运用乘法结合律计算的算式是()。",
+        "options":  [
+
+                    ],
+        "answer":  "125×(80+8)或125×(8+80);125×8×11或(125×8)×11",
+        "solution":  "【分析】由题意得，简算125×88时，可以把88转化为80+8，然后利用乘法分配律使计算简便;也可以把88转化为8×11，然后利用乘法结合律使计算简便.\n【详解】125×88\n=125×(80+8)\n=125×80+125×8\n125×88\n=125×(8×11)\n=(125×8)×11\n算式125×88，运用乘法分配律计算的算式是125×(80+8)或125×(8+80),运用乘法结合律计算的算式是125×8×11或(125×8)×11。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·湖南邵阳·期末"
+    },
+    {
+        "id":  "116",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果x×y=24，那么x×y×5=();如果m/n=30，那么m/(n×5)=()。",
+        "options":  [
+
+                    ],
+        "answer":  "120;6",
+        "solution":  "【分析】已知x×y=24，根据乘法结合律,x×y×5=( x×y)×5，将x×y=24代入即可算出;已知m/n=30，当除数n扩大5倍时，商缩小到原来的五分之一，用30除以5即可.\n【详解】x×y×5\n=(x×y)×5\n=24×5\n=120\n已知m/n=30，当除数n乘以5时，根据除法运算性质:\nm/(n×5)\n=(m/n)÷5\n=30÷5\n=6\n即如果x×y=24，那么x×y×5=120;如果m/n=30，那么m/(n×5)=6。",
+        "knowledge_points":  [
+                                 "商不变的规律及应用",
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·河南许昌·期末"
+    },
+    {
+        "id":  "117",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "我们已经掌握了加法，乘法的运算律，也学会了探究运算定律的一般方法。请用学过的方法试着研究下面的规律:\n(42+35)÷7=11,42÷7+35÷7=11→(42+35)÷7=42÷7+35÷7\n(36+54)÷9=10,36÷9+54÷9=10→(36+54)÷9=36÷9+54÷9\n你还能写出一组这样等式吗。你发现了什么。\n举例:()我发现:()",
+        "options":  [
+
+                    ],
+        "answer":  "(24+16)÷4=10;两个数的和除以一个数，等于这两个数分别除以这个数，再把所得的商相加",
+        "solution":  "【分析】根据所给算是计算出四个算是的结果，再根据结果及算式的特点找出规律，并根据规律写出一组算式即可。我发现：两个数的和除以一个数(0除外),等于这两个数分别除以这个数，再把它们的商相加。用字母表示:(a+b)/c=a/c+b/c。以此答题即可.\n【详解】根据分析可知:\n举例:(24+16)÷4=10.\n我发现：两个数的和除以一个数(0除外),等于这两个数分别除以这个数，再把它们的商相加。用字母表示:(a+b)/c=a/c+b/c。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·河北保定·期末"
+    },
+    {
+        "id":  "118",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "11×135+135×9=(11+9)×135这里运用了()律;如果☆-△=100，那48*☆-48*△=()。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法分配;4800",
+        "solution":  "【分析】乘法分配律：两个数的和与一个数相乘，可以先把它们与这个数分别相乘，再相加。根据乘法分配律的逆运算，可把原式变为(☆-△)×48，再按照运算顺序计算即可。【详解】11×135+135×9=(11+9)×135，这里运用了乘法分配律;如果☆-△=100，那么48×☆-48×△=(☆-△)×48=100×48=4800。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·河北保定·期末"
+    },
+    {
+        "id":  "119",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "已知a-b=50,a×c-b×c=1000，则c=()。",
+        "options":  [
+
+                    ],
+        "answer":  "20",
+        "solution":  "【分析】运用乘法分配律进行简便运算,a×c-b×c=(a-b)×c,已知a-b=50,a×c-b×c=1000，代入得50×c=1000，根据因数=积/另一个因数，求出Co\n【详解】已知a-b=50,a×c-b×c=1000,a×c-b×c=(a-b)×c所以c=1000÷50=20。",
+        "knowledge_points":  [
+                                 "用字母表示运算定律及计算公式",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级上·湖南邵阳·期末"
+    },
+    {
+        "id":  "120",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算125×88时，可以用125×()×()进行简便计算，这是运用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "8;11;乘法结合",
+        "solution":  "【分析】乘法结合律的特点是三个数相乘，先把前两个数相乘，或先把后两个数相乘，积不变;计算125×88时，观察到125与8相乘可得1000，便于简便计算。因此将88拆分为8×11，变为125×(8×11),再利用乘法结合律去掉小括号变为125×8×11，然后再从左至右依次计算乘法.\n【详解】125×88=125×(8×11)=125×8×11=1000×11=11000计算125×88时，可以用125×8×11进行简便计算，这是运用了乘法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·湖北黄石·期末"
+    },
+    {
+        "id":  "121",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "当☆=()时(填\"82\"，\"153\"，\"48\"),算式653-218-☆就可以转换成653-(218+☆)进行简便运算，最后计算结果是()。",
+        "options":  [
+
+                    ],
+        "answer":  "82;353",
+        "solution":  "【分析】根据减法的运算性质a-b-c=a-(b+c),把算式653-218-☆转换成653-(218+☆)进行简便运算，说明218+☆和是整百数，使653减整百数更方便计算。据此选择，并计算即可.\n【详解】218+82=300，整百数,653-300=353;218+153=371，不是整百数;218+48=266，不是整百数;当☆=(82)时(填\"82\"，\"153\"，\"48\"),算式653-218-☆就可以转换成653-(218+☆)进行简便运算，最结果是(353)。",
+        "knowledge_points":  [
+                                 "整数减法的性质"
+                             ],
+        "source":  "24-25四年级下·湖北武汉·期末"
+    },
+    {
+        "id":  "122",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "已知◎+☆=100，那么◎*26+☆*26=(      )。若◎*☆=100，那么(☆*18)*◎=(      )。",
+        "options":  [
+
+                    ],
+        "answer":  "2600;1800",
+        "solution":  "【分析】根据乘法分配律:(a+b)×c=a×c+b×c,已知◎+☆=100，那么◎*26+☆*26=(◎+☆)×26，代入数据，计算即可。根据乘法结合律:(a×b)×c=a×(b×c)或a×b×c=a×(b×c),乘法交换律,a×b=6×a,◎*☆=100，那么(☆*18)*◎=◎*☆*18，代入数据，计算即可.\n【详解】根据分析可知:◎+☆=100 ◎*26+☆*26=(◎+☆)×26 =100×26 =2600 ◎*☆=100\n(☆*18)*◎=◎*☆*18=100×18=1800已知◎+☆=100，那么◎*26+☆*26=2600。若◎*☆=100，那么(☆*18)*◎=1800。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数乘法结合律",
+                                 "整数乘法交换律"
+                             ],
+        "source":  "24-25四年级下·湖北随州·期末"
+    },
+    {
+        "id":  "123",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算25×12×4时，为了计算简便，可以先算()×(),这是应用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "25;4;乘法交换",
+        "solution":  "【分析】乘法交换律:a×b=b×a;据此可知，观察算式25×12×4,25×4的积是整百数，所以可以将原式转化成25×4×12再计算会比较简便.\n【详解】25×12×4=25×4×12=100×12=1200计算25×12×4时，为了计算简便，可以先算25×4，这是应用了乘法交换律。",
+        "knowledge_points":  [
+                                 "整数乘法交换律"
+                             ],
+        "source":  "24-25四年级下·湖北孝感·期末"
+    },
+    {
+        "id":  "124",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "如果a×b=64,c×a=146，那么a×(b+c)=()。",
+        "options":  [
+
+                    ],
+        "answer":  "210",
+        "solution":  "【分析】根据乘法分配律,a×(b+c)= a×b+ a×c,再把a×b=64,c×a=146代入计算即可解答.\n【详解】a×(b+c)= a×b+ a×c=64+146=210",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·湖北武汉·期末"
+    },
+    {
+        "id":  "125",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "明明运用加法交换律写了一个等式:□6+5□=□4+□6，如果这两个加数的和是100，那么这两个加数分别是()和()。",
+        "options":  [
+
+                    ],
+        "answer":  "46;54",
+        "solution":  "【分析】加法交换律：两个数相加，交换加数的位置，和不变。因为□6+5□=□4+□6运用了加法交换律，所以5□个位上的数字和□4个位上的数字相同，即5□个位上是4，这个数是54。已知两个加数的和是100，其中一个加数是54，那么另一个加数是100-54=46，据此解答即可.\n【详解】由分析可知,5□个位上是4，这个数是54.100-54=46。所以这两个加数分别是54和46。",
+        "knowledge_points":  [
+                                 "整数加法交换律"
+                             ],
+        "source":  "24-25四年级下·湖北咸宁·期末"
+    },
+    {
+        "id":  "126",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小乐计算36+24*□时弄错了运算顺序，结果算出来的答案比正确答案多72,□代表的数是(),正确的结果是()。",
+        "options":  [
+
+                    ],
+        "answer":  "3;108",
+        "solution":  "【分析】由题意得，小乐计算36+24*□时弄错了运算顺序，即小乐计算的算式为:(36+24)*□。可以利用乘法分配律将算式(36+24)*□转化为36*□+24*□。小乐算出来的答案比正确答案多72，即36*□-36=72，据此可以将□的值算出来。最后把□的值代入算式36+24*□即可算出正确的结果.\n【详解】(36+24)*□=36*□+24*□=36*□+24*□-(36+24*□)=36*□+24*□-36-24*□=36*□+24*□-24*□-36\n=36*□-3636*□-36=72，那么36*□=36+72=108。所以□=108÷36=3.36+24*□=36+24×3=36+72=108.\n小乐计算36+24*□时弄错了运算顺序，结果算出来的答案比正确答案多72,□代表的数是3，正确的结果是108。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·湖北黄冈·期末"
+    },
+    {
+        "id":  "127",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "88×125可以用以下两种方法进行简便计算.\n方法一:88×125=11×8×125=11×(8×125)=11×1000=11000\n方法二:88×125=(80+8)×125=80×125+8×125=10000+1000=11000\n这两种方法，方法一是运用了()律进行简便运算的，方法二是运用了()律进行简便运算的。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法结合;乘法分配",
+        "solution":  "【分析】方法一：计算88×125时，先把88看成11×8，因为8×125的积是1000，所以先计算8×125的积，再用11乘所得的积，这是运用了乘法结合律，先把前两个数相乘或者先把后两个数相乘，积不变。方法二：计算88×125时，先把88看成80+8，分别计算80×125和8×125，再把积相加。这是运用了乘法分配律，即两个数的和与一个数相乘，可以先把这两个数分别与这个数相乘，再把积相加.\n【详解】方法一:88×125=11×8×125=11×(8×125)=11×1000=11000 方法二:88×125=(80+8)×125=80×125+8×125=10000+1000=11000 方法一是运用了(乘法结合)律进行简便运算的，方法二是运用了(乘法分配)律进行简便运算的。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·河北保定·期末"
+    },
+    {
+        "id":  "128",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "58+164+42+136=(58+42)+(164+136),这里运用了()律;如果▲+□=220，那么560-▲-□=()。",
+        "options":  [
+
+                    ],
+        "answer":  "加法交换律和加法结合;340",
+        "solution":  "【分析】加法交换律：两个数相加，交换两个加数的位置，和不变。用字母表示为:a+b=b+a。加法结合律：三个数相加，先把前两个数相加或者先把后两个数相加，和不变。用字母表示为:a+b+c=a+(b+c);减法的性质：一个数连续减去两个数，等于这个数减去两个数的和。用字母表示为:a-b-c=a-(b+c)。(1)在算式58+164+42+136=(58+42)+(164+136)中，加数的位置改变了，而且先把前两个数和后两个数相加，它运用了加法交换律和加法结合律。(2)利用减法的性质将算式560-▲-□转化为560-(▲+□),然后将▲+□=220代入算式即可解答.\n【详解】(1)由分析可知，算式58+164+42+136=(58+42)+(164+136)运用了加法交换律和加法结合律。(2)560-▲-□=560-(▲+□)=560-220=340,58+164+42+136=(58+42)+(164+136),这里运用了加法交换律和加法结合律;如果▲+□=220，那么560-▲-□=340。",
+        "knowledge_points":  [
+                                 "整数减法的性质",
+                                 "整数加法结合律",
+                                 "整数加法交换律"
+                             ],
+        "source":  "24-25四年级下·湖南张家界·期末"
+    },
+    {
+        "id":  "129",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小马在计算45×(□+3)时，错误地算成45*□+3，他计算的结果比正确结果少(      )。",
+        "options":  [
+
+                    ],
+        "answer":  "132",
+        "solution":  "【分析】根据题意，乘法分配律：两个数的和与一个数相乘，可以先把它们与这个数分别相乘，再相加。根据乘法分配律把45×(□+3)展开，再与45*□+3相减，可得他得到的结果比正确结果少多少。以此答题即可.\n【详解】根据分析可知:45×(□+3)=45*□+45×345*□+45×3-(45*□+3) =45*□+45×3-45*□-3\n=45×3-3=135-3=132\n小马在计算45×(□+3)时，错误地算成45*□+3，他计算的结果比正确结果少132。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·湖南张家界·期末"
+    },
+    {
+        "id":  "130",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果330÷○/☆=33，那么○*☆=(      )。",
+        "options":  [
+
+                    ],
+        "answer":  "10",
+        "solution":  "【分析】已知330÷○/☆=33，根据除法的性质a/b/c=a/(b×c)把原式变成330÷(○*☆)=33，然后将○*☆看作一个整体，根据\"除数=被除数/商\"，即可求出○*☆的值.\n【详解】330÷○/☆=33330÷(○*☆)=33 ○*☆=330÷33 ○*☆=10 如果330÷○/☆=33，那么○*☆=10。",
+        "knowledge_points":  [
+                                 "整数除法的性质",
+                                 "乘",
+                                 "除法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·湖南益阳·期末"
+    },
+    {
+        "id":  "131",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "125×(8*□)=(125*□)×38。",
+        "options":  [
+
+                    ],
+        "answer":  "38;8",
+        "solution":  "【分析】通过观察等式可知，算式是3个数进行相乘，利用乘法结合律，先计算125乘8，再乘38，据此解答.\n【详解】根据分析可知,125×(8×38)=(125×8)×38。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·湖南益阳·期末"
+    },
+    {
+        "id":  "132",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "下面每一个汉字表示一个数，计算一下\"邦\"代表的数是( )。德=100000，礼=125，仪=8，名=4，礼×仪×名×邦=德。",
+        "options":  [
+
+                    ],
+        "answer":  "25",
+        "solution":  "【分析】根据题意，已知礼=125，仪=8，名=4，德=100000，且等式为礼×仪×名×邦=德。将已知数值代入等式后，根据除法的性质:a/b/c=a/(b×c),通过除法运算，即可求出\"邦\"的值.\n【详解】根据分析可知：礼×仪×名×邦=德 礼=125，仪=8，名=4，德=100000100000÷125/8÷4\n=100000÷(125×8×4)=100000÷(1000×4)=100000÷4000=25下面每一个汉字表示一个数，计算一下\"邦\"代表的数是25。",
+        "knowledge_points":  [
+                                 "整数除法的性质"
+                             ],
+        "source":  "24-25四年级下·云南玉溪·期末"
+    },
+    {
+        "id":  "133",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算器上数字键\"3\"坏了，计算420÷35可以用420÷()/()代替。",
+        "options":  [
+
+                    ],
+        "answer":  "75",
+        "solution":  "【分析】根据题意可知,35可以写成5×7，这样就不会使用数字键\"3\"，因此可根据整数除法的性质进行简算，依此解答.\n【详解】由分析得,420÷35=420÷(7×5)=420÷7/5所以，计算器上数字键\"3\"坏了，计算420÷35可以用420÷7/5代替。",
+        "knowledge_points":  [
+                                 "计算器的初步认识与使用",
+                                 "整数除法的性质"
+                             ],
+        "source":  "24-25四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "134",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "(25+8)×4=25×4+8×4这里运用的运算定律是()。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法分配律",
+        "solution":  "【分析】乘法分配律为:a×b+a×c=a×(b+c),据此解答即可.\n【详解】根据分析可得:(25+8)×4=25×4+8×4，运用的运算定律为：乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·吉林松原·期末"
+    },
+    {
+        "id":  "135",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "(25×5)×2=25×((5×2),这里运用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法结合",
+        "solution":  "【分析】乘法结合律:a×b×c=a×(b×c),算式给(5×2)加上了括号，使算式先计算5×2，运用了乘法的结合律.\n【详解】(25×5)×2=25×(5×2)运用了乘法的结合律。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·宁夏固原·期末"
+    },
+    {
+        "id":  "136",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在计算a/(b×c)时，小华错算成a/b×c,计算结果是正确结果的36倍，则c=()。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】小华看错了运算顺序，但计算过程是正确的。通过对比正确的算式a/(b×c)和错误算式a/b×c,找出两者结果的关系，从而求出c的值。a/(b×c)=a/b/c,小华错算成了a/b×c,计算结果是正确结果的36倍，相当于c×c=36。据此解答.\n【详解】c×c=36\n6×6=36\n所以c=6",
+        "knowledge_points":  [
+                                 "整数除法的性质",
+                                 "乘",
+                                 "除法的意义和各部分间的关系",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·湖南怀化·期末"
+    },
+    {
+        "id":  "137",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小强和芳芳放学同时从学校出发，沿同一条路去科技馆，小强的速度是70米/分，芳芳的速度是60米/分。30分钟后小强到了科技馆，这时芳芳离科技馆还有()米。",
+        "options":  [
+
+                    ],
+        "answer":  "300",
+        "solution":  "【分析】小强的速度是70米/分,30分钟后小强到了科技馆，根据路程=速度×时间，即可求出学校到科技馆的路程;又已知芳芳的速度是60米/分，求30分钟后小强到了科技馆，这时芳芳离科技馆还有多少米，则根据路程=速度×时间，先求出芳芳30分钟行走的路程，再与求出的学校与科技馆的路程作差，即可解答。计算时，可以根据乘法分配律进行简算.\n【详解】70×30-60×30=(70-60)×30=10×30=300(米)即小强和芳芳放学同时从学校出发，沿同一条路去科技馆，小强的速度是70米/分，芳芳的速度是60米/分。30分钟后小强到了科技馆，这时芳芳离科技馆还有300米。",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·山东临沂·期末"
+    },
+    {
+        "id":  "138",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果★+▲=25，那么12*★+12*▲=()。",
+        "options":  [
+
+                    ],
+        "answer":  "300",
+        "solution":  "【分析】根据乘法分配律:(a+b)×c=a×c+b×c,将算式12*★+12*▲变成12×(★+▲),已知★+▲=25，据此代入数字即可计算出结果.\n【详解】12*★+12*▲\n=12×(★+▲)\n=12×25\n=300\n如果★+▲=25，那么12*★+12*▲=300。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·浙江温州·期末"
+    },
+    {
+        "id":  "139",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "观察下边的竖式，明明在计算的过程中运用到的运算律是().\n![](images/2026-03-07%20150949.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "乘法分配律",
+        "solution":  "【分析】乘法分配律是指两个数的和同一个数相乘，等于把两个加数分别同这个数相乘，再把两个积加起来，结果不变，本题是把12看成(2+10)。先用12个位上的2×25=50，再用12十位上的1，也就是10×25=250，再把50和250相加.\n【详解】根据分析，明明在计算的过程中运用到的运算律是乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·山东菏泽·期末"
+    },
+    {
+        "id":  "140",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "99×9+9=(99 ∘ 1) ∘ 9;510-□-180=510-(220+□ )",
+        "options":  [
+
+                    ],
+        "answer":  "+;*;220;180",
+        "solution":  "【分析】99×9+9先将9写成9×1的形式，即99×9+9×1，然后根据乘法分配律:(a+b)×c=a×c+b×c,将算式99×9+9×1变成(99+1)×9;根据减法的性质:a-b-c=a-(b+c),右边的算式括号里其中一个加数是220，且左边的算式中没有，则左边第一个减去的是220，据此填空即可.\n【详解】99×9+9\n=99×9+9×1\n=(99+1)×9\n=100×9\n=900\n510-220-180\n=510-(220+180)\n=510-400\n=110\n99×9+9=(99+1)×9;510-220-180=510-(220+180)。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数减法的性质"
+                             ],
+        "source":  "24-25四年级下·山东菏泽·期末"
+    },
+    {
+        "id":  "141",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "下图展示了奇奇和妙妙对\"125×88\"的计算过程，奇奇运用了()律，妙妙运用了()律.\n125×88\n=125×(80+8)\n=125×80+125×8\n=10000+1000\n=11000\n奇奇\n88×125\n=11×8×125\n=11×(8×125)\n=11×1000\n=11000\n妙妙",
+        "options":  [
+
+                    ],
+        "answer":  "乘法分配;乘法结合",
+        "solution":  "【分析】乘法分配律是指两个数的和与一个数相乘，可以先把它们分别与这个数相乘，再相加，奇奇先将88写成80+8的形式，即125×(80+8),然后根据乘法分配律:(a+b)×c=a×c+b×c,把算式125×(80+8),变成125×80+125×8，最后按照运算顺序计算即可;乘法结合律是指三个数相乘，先把前两个数相乘，再和另外一个数相乘，或先把后两个数相乘，再和另外一个数相乘，积不变，妙妙先将88写成11×8的形式，即11×8×125，然后根据乘法结合律:a×b×c=a×(b×c),将算式11×8×125变成11×(8×125),最后按照运算顺序计算即可.\n【详解】下图展示了奇奇和妙妙对\"125×88\"的计算过程，奇奇运用了乘法分配律，妙妙运用了乘法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·山东菏泽·期末"
+    },
+    {
+        "id":  "142",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算125×88时，可以用125×(□*□)进行简便计算，结果是()。",
+        "options":  [
+
+                    ],
+        "answer":  "8;11;11000",
+        "solution":  "【分析】计算125×88时，观察到125与8相乘可得1000，便于简便计算。因此将88拆分为8×11，利用乘法结合律a×b×c=a×(b×c)先计算125×8，再乘11，从而快速得出结果.\n【详解】125×88\n= 125×(8×11)\n=125×8×11\n=1000×11\n=11000\n即125×88的简便计算过程为125×(8×11),结果为11000。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·湖北黄石·期末"
+    },
+    {
+        "id":  "143",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据加法的交换律或结合律填空.\n( )+65=65+20;(25+68)+32=25+( + )",
+        "options":  [
+
+                    ],
+        "answer":  "20;68;32",
+        "solution":  "【分析】加法交换律a+b=b+c和结合律(a+b)+c=a+(b+c)。第一空根据加法交换律，等式两边加数位置互换;第二空根据加法结合律，调整加法运算顺序.\n【详解】根据加法交换律，等式左边为( )+65，右边为65+20。交换律表明两数相加，交换位置和不变，因此括号内应填20，即等式变为20+65=65+20。原式为(25+68)+32=25+( + )。根据加法结合律，可将前两个数的和与第三个数相加，转化为第一个数与后两个数的和相加。因此,68和32结合相加，等式变为25+(68+32)。",
+        "knowledge_points":  [
+                                 "整数加法结合律",
+                                 "整数加法交换律"
+                             ],
+        "source":  "24-25四年级下·湖北黄石·期末"
+    },
+    {
+        "id":  "144",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果口算450×18，可以先口算450×20，再减去()个450，最后结果是()。",
+        "options":  [
+
+                    ],
+        "answer":  "2;8100",
+        "solution":  "【分析】18比较接近20，可以看成20-2。根据乘法分配律可得450×18=450×(20-2)=450×20-450×2450×18=450*,表示20个450减去2个450，最后求出结果即可;据此解答.\n【详解】450×18\n=450×(20-2)\n=450×20-450×2\n=9000-900\n=8100\n即如果口算450×18，可以先口算450×20，再减去2个450，最后结果是8100。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·河北邯郸·期末"
+    },
+    {
+        "id":  "145",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "用简便方法计算37×25×4，先算(),这是运用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "25×4;乘法结合",
+        "solution":  "【分析】乘法结合律：三个数相乘，可以先把前两个数相乘，或者先把后两个数相乘，积不变。用字母表示是(a×b)×c=a×(b×c).\n【详解】根据分析，用简便方法计算37×25×4，先算25×4，这是运用了乘法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·湖南湘潭·期末"
+    },
+    {
+        "id":  "146",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小美用计算器计算7255×63时，发现计算器的按键\"3\"坏了。如果要用这个计算器算出正确结果，请你帮她想出一个可以替代的算式()。",
+        "options":  [
+
+                    ],
+        "answer":  "7255×7×9",
+        "solution":  "【分析】乘法结合律：三个数相乘，先把前两个数相乘，再和另外一个数相乘，或先把后两个数相乘，再和另外一个数相乘，积不变;(a×b)×c=a×(b×c);把63拆分成7×9，算式变成7255×7×9，据此计算即可.\n【详解】小美用计算器计算7255×63时，发现计算器的按键\"3\"坏了。如果要用这个计算器算出正确结果，请你帮她想出一个可以替代的算式7255×7×9。(答案不唯一)",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·江西新余·期末"
+    },
+    {
+        "id":  "147",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "毛毛把6×(m+5)错写成了6×m+5，他的计算结果与正确答案相差( )。",
+        "options":  [
+
+                    ],
+        "answer":  "25",
+        "solution":  "【分析】根据题意，乘法分配律是指两个数的和同一个数相乘，等于把两个加数分别同这个数相乘，再把两个积加起来，结果不变。用字母表示为:(a+b)×c=a×c+b×c。根据乘法分配律的定义,6×(m+5)的正确答案应该是6×m+5×6。则他的计算结果与正确答案相差6×m+5×6-(6×m+5)。以此答题即可.\n【详解】根据分析可知:\n6×(m+5)=6×m+5×6\n6×m+5×6-(6×m+5)\n=6×m+5×6-6×m-5\n=30-5\n=25\n毛毛把6×(m+5)错写成了6×m+5，他的计算结果与正确答案相差25。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·江西新余·期末"
+    },
+    {
+        "id":  "148",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小迷糊把60×(□+6)错算成60*□+6。他得到的结果与正确结果相差()。",
+        "options":  [
+
+                    ],
+        "answer":  "354",
+        "solution":  "【分析】根据乘法分配律:(a+b)×c=a×c+b×c,将正确算式展开后与错误算式进行比较，计算两者的差值.\n【详解】根据分析可知:\n60×(□+6)=60*□+60×6=60*□+360\n(60*□+360)-(60*□+6)=360-6=354\n小迷糊把60×(□+6)错算成60*□+6。他得到的结果与正确结果相差354。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·重庆南岸·期末"
+    },
+    {
+        "id":  "149",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小思用下面的方法计算54×101，她这样计算可以吗。\n54×101\n=54×100+54×1\n=5400+54\n=5454\n()(选\"可以\"或\"不可以\")这样算。请说明理由:()。",
+        "options":  [
+
+                    ],
+        "answer":  "可以;利用乘法分配律，算式变为54×(100+1)，这样计算更简便",
+        "solution":  "【分析】乘法分配律：是指两个数的和与一个数相乘，可以先把它们分别与这个数相乘，再相加;a×c+b×c=(a+b)×c。小思把101看成(100+1),然后利用乘法分配律计算.\n【详解】54×101\n=54×(100+1)\n=54×100+54×1\n=5400+54×1\n=5400+54\n=5454\n可以这样算。理由：利用乘法分配律，算式变为54×(100+1),这样计算更简便。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·广西柳州·期末"
+    },
+    {
+        "id":  "150",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "28×9+72×9=(28 ∘ 72)∘ 9;1001-□-540=1001-(460+□)",
+        "options":  [
+
+                    ],
+        "answer":  "+;×;460;540",
+        "solution":  "【分析】根据乘法分配律，两个数的和与另一个数相乘，可以先把它们与这个数分别相乘，再相加，乘法分配律可互逆运算;根据减法的性质，一个数连续减去两个数，可以减去这两个数的和.\n【详解】28×9+72×9\n=(28+72)×9\n1001-460-540\n=1001-(460+540)",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数减法的性质"
+                             ],
+        "source":  "24-25四年级下·山东菏泽·期末"
+    },
+    {
+        "id":  "151",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "粗心的小星把(6+□)×8算成了6+□×8，他算出的结果与正确结果相差()。",
+        "options":  [
+
+                    ],
+        "answer":  "42",
+        "solution":  "【分析】由题意可知：这道题符合乘法分配律的特点，一个数乘两个数的和等于分别与这两个数相乘，再相加。小星计算的时候前面的6没有和8相乘，这样算出差值即可.\n【详解】由分析可得:(6+□)×8=6×8+□×8，小星把它算成了6+□×8;两个算式的差值:6×8-6=48-6=42。粗心的小星把(6+□)×8算成了6+□×8，他算出的结果与正确结果相差42。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·云南曲靖·期末"
+    },
+    {
+        "id":  "152",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "\"五一\"假期期间，商场的冰箱专卖店进行促销大酬宾活动，一台冰箱的原价是5688元，活动期间，降价588元，再返212元现金。购买这台冰箱实际要花()元。",
+        "options":  [
+
+                    ],
+        "answer":  "4888",
+        "solution":  "【分析】用原价减去降价的588，再减去返现的212，即可求出购买这台冰箱实际价格;计算过程可以利用减法的性质进行简便计算.\n【详解】5688-588-212\n=5688-(588+212)\n=5688-800\n=4888(元)\n所以购买这台冰箱实际要花4888元。",
+        "knowledge_points":  [
+                                 "整数减法的性质"
+                             ],
+        "source":  "24-25四年级下·四川乐山·期末"
+    },
+    {
+        "id":  "153",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算88×125时，琪琪写成88×125=11×(8×125),是运用()律计算的，皮皮写成88×125=80×125+8×125，是运用()律计算的。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法结合律;乘法分配律",
+        "solution":  "【分析】乘法结合律：三个数相乘，先把前两个数相乘，再和另外一个数相乘，或先把后两个数相乘，再和另外一个数相乘，积不变;(a×b)×c=a×(b×c);由此可知，琪琪计算时，把88拆分成11×8，再利用乘法结合律计算;乘法分配律：是指两个数的和与一个数相乘，可以先把它们分别与这个数相乘，再相加;a×c+b×c=(a+b)×c。由此可知，皮皮计算时，把88拆分成(80+8),再利用乘法分配律计算;据此解答即可.\n【详解】由分析可知，计算88×125时，琪琪写成88×125=11×(8×125),是运用乘法结合律计算的，皮皮写成88×125=80×125+8×125，是运用乘法分配律计算的。",
+        "knowledge_points":  [
+                                 "整数乘法结合律",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·四川乐山·期末"
+    },
+    {
+        "id":  "154",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "临川四梦大剧院共有35排座位，每排40个。用乘法分配律计算总座位数是:\n35×40=()×()+()×()=()。",
+        "options":  [
+
+                    ],
+        "answer":  "30;40;5;40;1400",
+        "solution":  "【分析】乘法分配律:(a+b)×c=a×c+b×c,根据乘法分配律,35乘40，可以把35分成30和5，然后用30和5分别与40相乘，再把乘得的积相加即可解答.\n【详解】35×40=(30+5)×40=30×40+5×40=1200+200=1400",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·江西抚州·期末"
+    },
+    {
+        "id":  "155",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在数学课上，小宇和小萱分别运用不同的乘法运算定律计算125×24。\n小宇是这样计算的:125×24=125×(8×3)=(125×8)×3=1000×3=3000;小萱则采用另一种方法:125×24=125×(20+4)=125×20+125×4=2500+500=3000。\n说一说：小宇应用的运算律是(),小萱应用的运算律是()",
+        "options":  [
+
+                    ],
+        "answer":  "乘法结合律;乘法分配律",
+        "solution":  "【分析】乘法的运算律：乘法交换律是交换两个因数的位置，积不变;乘法结合律是三个数相乘，先把前两个数相乘和先把后两个数相乘，结果不变;乘法分配律是两个数的和与第三个数相乘，等于这两个数分别与第三个数相乘的积再相加;据此解答.\n【详解】小宇:125×24=125×(8×3)=(125×8)×3=1000×3=3000;把24拆成8×3，符合乘法结合律，所以小宇应用的运算律是乘法结合律。小萱:125×24=125×(20+4)=125×20+125×4=2500+500=3000，把24拆成20+4，用125分别乘20和4，符合乘法分配律，所以小萱应用的运算律是乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·陕西西安·期末"
+    },
+    {
+        "id":  "156",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果36×A+36×B运用了乘法分配律进行了简便计算后，结果是3600，则A可以是(),B可以是()。",
+        "options":  [
+
+                    ],
+        "answer":  "50;50",
+        "solution":  "【分析】根据乘法分配律将36×A+36×B改写为(A+B)×36，又已知36×A+36×B运用了乘法分配律进行了简便计算后的结果是3600，所以(A+B)×36=3600，根据积/一个因数=另一个因数，可知(A+B)=3600÷36，从而可以得出A+B=100;据此再进一步解答即可.\n【详解】因为36×A+36×B=(A+B)×36=3600所以(A+B)=3600÷36=100所以A可以是50,B可以是50。(答案不唯一)",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "乘",
+                                 "除法的意义和各部分间的关系",
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "24-25四年级下·河南许昌·期末"
+    },
+    {
+        "id":  "157",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "125×29×()=29×(()×8)。",
+        "options":  [
+
+                    ],
+        "answer":  "8;125",
+        "solution":  "【分析】根据乘法结合律，三个数相乘，先把前两个数相乘，再和另外一个数相乘，或先把后两个数相乘，再和另外一个数相乘，积不变，据此解答即可.\n【详解】由分析可知:125×29×8=29×(125×8),因为125×29×8交换125与29的位置，再利用乘法结合律把125与8结合计算，变式为29×(125×8)",
+        "knowledge_points":  [
+                                 "整数乘法结合律",
+                                 "整数乘法交换律"
+                             ],
+        "source":  "24-25四年级下·广东汕头·期末"
+    },
+    {
+        "id":  "158",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "淘气在做作业时，由于马虎，把60×(c+4)错算成了60c+4，那么他所得到的结果与正确的结果相差()。",
+        "options":  [
+
+                    ],
+        "answer":  "236",
+        "solution":  "【分析】乘法分配律是指两个数的和同一个数相乘，等于把两个加数分别同这个数相乘，再把两个积加起来，结果不变。根据乘法分配律的定义可知,60×(c+4)的正确结果应该是60×c+60×4，再用正确结果减去得到的结果解答即可.\n【详解】60×(c+4)\n=60×c+60×4\n=60c+240\n60c+240-(60c+4)\n=60c+240-60c-4\n=240-4\n=236\n即淘气在做作业时，由于马虎，把60×(c+4)错算成了60c+4，那么他所得到的结果与正确的结果相差236。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·湖北十堰·期末"
+    },
+    {
+        "id":  "159",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "聪聪和强强从学校操场的同一点出发背对背而行，聪聪每分钟走45米，强强每分钟走55米,2分钟后二人相遇，操场全长()米。",
+        "options":  [
+
+                    ],
+        "answer":  "200",
+        "solution":  "【分析】根据路程=速度×时间，可以先算出每人行驶的路程再相加，列式为45×2+55×2;也可以先算出两个人的速度和，因为他们行驶的时间相同，再乘相遇时间，列式为(45+55)×2，这样计算更简便.\n【详解】45×2+55×2\n=(45+55)×2\n=100×2\n=200(米)\n即聪聪和强强从学校操场的同一点出发背对背而行，聪聪每分钟走45米，强强每分钟走55米,2分钟后二人相遇，操场全长200米。",
+        "knowledge_points":  [
+                                 "相遇问题",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·北京通州·期末"
+    },
+    {
+        "id":  "160",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "哪吒和敖丙合力对抗天劫，若两人法力之和为100，则67*哪吒法力+67*敖丙法力=(),运用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "6700;乘法分配律",
+        "solution":  "【分析】乘法分配律是指两个数的和与一个数相乘，可以先把它们分别与这个数相乘，再相加。用字母表示为(a+b)×c=a×c+b×c。我们可以把哪吒法力和敖丙法力看作a和b,67看作c,利用乘法分配律来解题.\n【详解】根据乘法分配律(a+b)×c=a×c+b×c,在这里a是哪吒法力,b是敖丙法力,c=67。已知哪吒法力+敖丙法力=100，那么67*哪吒法力+67*敖丙法力=67×(哪吒法力+敖丙法力)。因为哪吒法力+敖丙法力=100，所以67×(哪吒法力+敖丙法力)=67×100=6700。所以,67*哪吒法力+67*敖丙法力=6700，运用了乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·江西赣州·期末"
+    },
+    {
+        "id":  "161",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "已知两个数的乘积是810，如果一个数不变，另一个数拆成10×9，那么这三个数的积是()。",
+        "options":  [
+
+                    ],
+        "answer":  "810",
+        "solution":  "【分析】原来是两个数相乘，把另一个数拆成10×9之后变成了三个数相乘，根据乘法结合律，可知积不变。乘法结合律：三个数相乘，先乘前两个数，或者先乘后两个数，积不变.\n【详解】例如:9×90\n=9×10×9\n=90×9\n=810\n已知两个数的乘积是810，如果一个数不变，另一个数拆成10×9，那么这三个数的积是(810)。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·湖北省直辖县级单位·期中"
+    },
+    {
+        "id":  "162",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "如果a×b=30，那么(a×3)×(b×4)=()。如果a-b=8，那么125×a-125×b=()。",
+        "options":  [
+
+                    ],
+        "answer":  "360;1000",
+        "solution":  "【分析】本题主要涉及乘法分配律和乘法结合律的运用.\n乘法分配律是指两个数的差与一个数相乘，可以先把它们分别与这个数相乘，再相减;乘法结合律是指三个数相乘，先把前两个数相乘，再和另外一个数相乘，或先把后两个数相乘，再和另外一个数相乘，积不变。我们可以根据这两个运算律来计算题目中的式子.\n【详解】(a×3)×(b×4)\n=a×3×b×4\n=(a×b)×(3×4)\n=30×12\n=360\n所以，如果a×b=30，那么(a×3)×(b×4)=360\n125×a-125×b\n=125×(a-b)\n=125×8\n=1000\n所以如果a-b=8，那么125×a-125×b=1000",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·湖北省直辖县级单位·期中"
+    },
+    {
+        "id":  "163",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果☆*△=60，那么(☆*4)×(△*5)=()。如果A+B=8，那么125×A+125×B=()。",
+        "options":  [
+
+                    ],
+        "answer":  "1200;1000",
+        "solution":  "【分析】(1)利用乘法交换律a×b=b×a和乘法结合律(a×b)×c=a×(b×c),将(☆*4)×(△*5)变形为☆*△*4×5，再代入已知条件进行计算即可解答;(2)应用乘法分配律的逆运算a×c+b×c=(a+b)×c,将125×A+125×B转化为125×(A+B),直接代入已知值计算即可解答.\n【详解】(☆*4)×(△*5)\n=☆*△*4×5\n=60×4×5\n=240×5\n=1200\n125×A+125×B\n=125×(A+B)\n=125×8\n=1000\n如果☆*△=60，那么(☆*4)×(△*5)=1200。如果A+B=8，那么125×A+125×B=1000。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数乘法结合律",
+                                 "整数乘法交换律"
+                             ],
+        "source":  "24-25四年级下·湖北十堰·期中"
+    },
+    {
+        "id":  "164",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算125×(8×17)时，可以先算(),这样计算是应用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "125×8;乘法结合",
+        "solution":  "【分析】观察算式125×(8×17),125×8=1000，可以利用乘法结合律，先计算125×8，再用积乘17.\n【详解】125×(8×17)\n=125×8×17\n=1000×17\n=17000\n计算125×(8×17)时，可以先算125×8，这样计算是应用了乘法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·湖南长沙·期中"
+    },
+    {
+        "id":  "165",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "计算((23×125)×8×8时，为了计算简便，可以先算(),这样计算是根据()。",
+        "options":  [
+
+                    ],
+        "answer":  "125×8;乘法结合律",
+        "solution":  "【分析】乘法结合律：三个数相乘，先乘前两个数，或者先乘后两个数，积不变;字母表示为:(a×b)×c=a×(b×c);因为125×8=1000，则此题可利用乘法结合律计算较简便.\n【详解】计算(23×125)×8时，为了计算简便，可以先算125×8，这样计算是根据乘法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·湖北黄冈·期中"
+    },
+    {
+        "id":  "166",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "根据运算律填空125×88=125×()×8，这是应用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "11;乘法结合",
+        "solution":  "【分析】乘法结合律:a×(b×c)=(a×b)×c,88=a×11×8,125×88=125×(11×8)=125×11×8，运用的是乘法结合律，据此解题.\n【详解】根据运算律填空125×88=125×11×8，这是应用了乘法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·湖南长沙·期中"
+    },
+    {
+        "id":  "167",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "用字母表示乘法分配律:a×(b+c)=()。",
+        "options":  [
+
+                    ],
+        "answer":  "a×b+a×c",
+        "solution":  "【分析】乘法分配律：两个数的和与一个数相乘，可以先把它们与这个数分别相乘，再相加。字母表示为:a×(b+c)=a×b+a×c;据此即可解答.\n【详解】用字母表示乘法分配律:a×(b+c)=a×b+a×c",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·湖南长沙·期中"
+    },
+    {
+        "id":  "168",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "根据运算律填空.\n32×18+32×12=32×(()+12)",
+        "options":  [
+
+                    ],
+        "answer":  "18",
+        "solution":  "【分析】乘法分配律：两个数的和与第三个数相乘，可以把这两个数与第三个数分别相乘再相加。用字母表示为:(a+b)×c=a×c+b×c。据此解答.\n【详解】由乘法分配律可知,32×18+32×12=32×(18+12)。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·湖南长沙·期中"
+    },
+    {
+        "id":  "169",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "68×99=68×(100-1)=68×100-68，这是应用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法分配律",
+        "solution":  "【分析】乘法分配律是指两个数的和与一个数相乘，可以先把它们分别与这个数相乘，再相加，用字母表示为:(a+b)×c=a×c+b×c。原式将99拆分为(100-1),并根据乘法分配律计算.\n【详解】68×99\n=68×(100-1)\n=68×100-68\n=6800-68\n=6732\n68×99=68×(100-1)=68×100-68，这是应用了乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·湖南长沙·期中"
+    },
+    {
+        "id":  "170",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "小丽在计算(□+24)×6时，漏看了括号，算成□+24×6，结果相差( )。",
+        "options":  [
+
+                    ],
+        "answer":  "5□",
+        "solution":  "【分析】根据乘法分配律(a+b)×c= a×c+b×c,把(□+24)×6变式为□*6+24×6，再与算式□+24×6比较即可.\n【详解】(□+24)×6\n=□*6+24×6\n□*6+24×6-(□+24×6)\n=□*6+24×6-□-24×6\n=□*6-□\n=5□\n所以小丽在计算(□+24)×6时，漏看了括号，算成□+24×6，结果相差5□。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·湖南长沙·期中"
+    },
+    {
+        "id":  "171",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据运算定律填空.\n36×75=(□*□)×9=2700;125×(80-8)=125*□-125*□=9000",
+        "options":  [
+
+                    ],
+        "answer":  "4;75;80;8",
+        "solution":  "【分析】(1)把36拆分成4×9，算式变成4×9×75，根据乘法交换律，交换9和75的位置，再利用乘法结合律计算即可;(2)乘法分配律：是指两个数的和与一个数相乘，可以先把它们分别与这个数相乘，再相加;(a+b)×c=a×c+b×c。据此解答即可.\n【详解】36×75\n=4×9×75\n=4×75×9\n=300×9\n=2700\n125×(80-8)\n=125×80-125×8\n=10000-1000\n=9000\n36×75=(4×75)×9=2700\n125×(80-8)=125×80-125×8=9000",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数乘法结合律",
+                                 "整数乘法交换律"
+                             ],
+        "source":  "24-25四年级下·湖南长沙·期中"
+    },
+    {
+        "id":  "172",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据运算定律填空.\n48×25=(□*□)×12=1200;89×101-89=89×(□-□)=8900",
+        "options":  [
+
+                    ],
+        "answer":  "4;25;101;1",
+        "solution":  "【分析】(1)根据乘法结合律a×b×c=a×(b×c)和乘法交换律a×b=b×a,将48看成4×12，先算4×25，再用积乘12。(2)根据乘法分配律a×b+a×c=a×(b+c),先计算101-1，再用89乘这个差.\n【详解】48×25=(4×25)×12=1200,89×101-89=89×(101-1)=8900",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数乘法结合律",
+                                 "整数乘法交换律"
+                             ],
+        "source":  "24-25四年级下·湖南岳阳·期中"
+    },
+    {
+        "id":  "173",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算(13×125)×8时，可以先算(),运用的运算定律是()。",
+        "options":  [
+
+                    ],
+        "answer":  "125×8;乘法结合律",
+        "solution":  "【分析】根据题意，乘法结合律指三个数相乘时，先乘前两个数或先乘后两个数，结果相同。用字母表示为:(a×b)×c=a×(b×c)或a×b×c=a×(b×c),在计算(13×125)×8时，观察到125与8相乘可得整千数，因此利用乘法结合律改变运算顺序，先计算125×8，使计算简便。以此答题即可.\n【详解】根据分析可知:\n(13×125)×8\n=13×(125×8)\n=13×1000\n=13000\n计算(。13×125))×8时，可以先算125×8，运用的运算定律是乘法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·四川德阳·期中"
+    },
+    {
+        "id":  "174",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一本相册有30页，每页可以插5张照片。小红家有900张照片，几本相册才够用。\n(1)算式900÷((30×5))中,30×5表示().\n(2)算式900÷5/30中,900÷5表示()。",
+        "options":  [
+
+                    ],
+        "answer":  "一本相册能插多少张照片;需要多少页",
+        "solution":  "【分析】(1)算式900÷(30×5)中,30表示一本相册的页数,5表示每页插照片数量。用一本相册的页数乘每页插照片数量，求出一本相册插照片数量。即30×5表示一本相册能插多少张照片。(2)算式900÷5/30中,900表示照片总数量,5表示每页插照片数量。用照片总数量除以每页插照片数量，求出需要的页数。即900÷5表示需要多少页.\n【详解】(1)算式900÷(30×5)中,30×5表示一本相册能插多少张照片。(2)算式900÷5/30中,900÷5表示需要多少页。",
+        "knowledge_points":  [
+                                 "整数除法的性质"
+                             ],
+        "source":  "24-25四年级下·内蒙古赤峰·期中"
+    },
+    {
+        "id":  "175",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "a×(b+c)=a*□+□*c,这是的运用。",
+        "options":  [
+
+                    ],
+        "answer":  "b;a;乘法分配律",
+        "solution":  "【分析】乘法分配律指一个数乘两数之和，等于这个数分别乘这两个数再相加。用字母表示为:(a+b)×c=a×c+b×c,以此答题即可.\n【详解】根据分析可知:a×(b+c)=a×b+a×c,这是乘法分配律的运用。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·内蒙古赤峰·期中"
+    },
+    {
+        "id":  "176",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "b×47+b×53=□*□(□+□)。",
+        "options":  [
+
+                    ],
+        "answer":  "b;47;53",
+        "solution":  "【分析】观察b×47+b×53发现，两个乘法中都有相同的因数b,可以利用乘法分配律a×c+b×c=(a+b)×c进行简算.\n【详解】由乘法分配律可知:b×47+b×53=b×(47+53)。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·新疆乌鲁木齐·期中"
+    },
+    {
+        "id":  "177",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "12×32=32×12，运用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法交换",
+        "solution":  "【分析】根据乘法交换律，两个数相乘，交换因数的位置，积不变。表示为:a×b=b×a,因此,12×32=32×12，运用了乘法交换律.\n【详解】根据分析可知:12×32=32×12，运用了乘法交换律。",
+        "knowledge_points":  [
+                                 "整数乘法交换律"
+                             ],
+        "source":  "24-25四年级下·云南玉溪·期中"
+    },
+    {
+        "id":  "178",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "简便计算26×99，正确的做法是(),计算中用到的运算定律是()。",
+        "options":  [
+
+                    ],
+        "answer":  "26×100-26;乘法分配律",
+        "solution":  "【分析】乘法分配律指一个数乘两数之和，等于这个数分别乘这两个数再相加。乘法分配律:(a+b)×c=a×c+b×c,观察算式26×99，发现99接近100，因此将99转化100-1，利用乘法分配律展开，即26×26×(100-1)=26×100-26×1，简化计算过程.\n【详解】根据分析可知:\n26×99\n=26×(100-1)\n=26×100-26×1\n=2600-26\n=2574\n简便计算26×99，正确的做法是26×100--26×1，计算中用到的运算定律是乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·新疆乌鲁木齐·期末"
+    },
+    {
+        "id":  "179",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "计算186×37=37×186时，应用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法交换",
+        "solution":  "【分析】乘法交换律：两个数相乘，交换两个乘数的位置，积不变。用字母表示为:a×b=b×a.\n【详解】计算186×37=37×186时，应用了乘法交换律。",
+        "knowledge_points":  [
+                                 "整数乘法交换律"
+                             ],
+        "source":  "24-25四年级下·重庆渝中·期中"
+    },
+    {
+        "id":  "180",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用计算器计算56×49时，发现\"4\"键坏了，如果还用这个计算器，你会怎样算。请写出算式()。",
+        "options":  [
+
+                    ],
+        "answer":  "56×7×7=2744(答案不唯一)",
+        "solution":  "【分析】由题意得，用计算器计算56×49时，发现\"4\"键坏了，要想计算出56×49的结果，可以把56×49转化为其它与其相等但不含数字\"4\"的算式。可以把49转化为7×7，然后利用乘法结合律将原式转化为56×7×7。据此解答.\n【详解】56×49\n=56×(7×7)\n=56×7×7\n=392×7\n=2744\n故算式56×7×7可以计算出算式56×49的结果。",
+        "knowledge_points":  [
+                                 "计算器的初步认识与使用",
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·内蒙古鄂尔多斯·期中"
+    },
+    {
+        "id":  "181",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "黄老师为参加趣味数学比赛获奖的小朋友采购了45个标价为9元的魔方，后来又采购了9副标价为15元的七巧板，黄老师一共花了()元。",
+        "options":  [
+
+                    ],
+        "answer":  "540",
+        "solution":  "【分析】根据题意，可以魔方和七巧板各自的标价乘购买的数量，就是魔方和七巧板各花了多少元。再相加就是一共花了多少元。在计算时，可以运用乘法分配律，把45和15先相加，使得计算简便.\n【详解】45×9+9×15\n=(45+15)×9\n=60×9\n=540(元)\n所以，黄老师一共花了540元。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·重庆大渡口·期中"
+    },
+    {
+        "id":  "182",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "马小虎在计算40×(△+5)错算成40*△+5，他得到的结果与正确结果相差()。",
+        "options":  [
+
+                    ],
+        "answer":  "195",
+        "solution":  "【分析】根据乘法分配律可知，正确算式40×(△+5)展开后为40*△+200，错误算式为40*△+5。两个算式有相同的40*△,两者的差由200和5的差决定，即200-5=195.\n【详解】正确计算:\n40×(△+5)\n=40*△+40×5\n=40△+200\n错误计算:\n40*△+5\n正确结果与错误结果的差为:\n200-5=195\n因此，得到的结果与正确结果相差195。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·重庆南川·期中"
+    },
+    {
+        "id":  "183",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "学校开展\"点燃读书激情，书香溢满校园\"阅读活动，新购进一批图书，同时新增设了12个相同的书架，每个书架有8层，每层放了125本，新增设的书架上一共放了()本书。",
+        "options":  [
+
+                    ],
+        "answer":  "12000",
+        "solution":  "【分析】已知新增设了12个相同的书架，每个书架有8层，用每个书架的层数乘书架的个数，求出一共有多少层，再乘每层放图书的本数，即是新增设的书架上一共放图书的总本数。计算时可根据乘法交换律a×b=b×a进行简算.\n【详解】8×12×125\n=8×125×12\n=1000×12\n=12000(本)\n新增设的书架上一共放了12000本书。",
+        "knowledge_points":  [
+                                 "整数乘法交换律"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期中"
+    },
+    {
+        "id":  "184",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小马虎在计算(□+15)×4时，漏看了括号，算成□+15×4，结果相差()。",
+        "options":  [
+
+                    ],
+        "answer":  "3□",
+        "solution":  "【分析】根据乘法分配律(a+b)×c=a×c+b×c,把(□+15)×4变式为□*4+15×4，再与算式□+15×4比较即可.\n【详解】(□+15)×4=□*4+15×4,□*4+15×4比□+15×4多了□*4-□=3□,即小马虎在计算(□+15)×4时，漏看了括号，算成□+15×4，结果相差3□。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·湖南岳阳·期中"
+    },
+    {
+        "id":  "185",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用简便方法计算:125×32=125×()×(),结果是()。",
+        "options":  [
+
+                    ],
+        "answer":  "8;4;4000",
+        "solution":  "【分析】乘法结合律：三个数相乘，先把前两个数相乘，再与第三个数相乘，或者是先把后两个数相乘，再与第个数相乘，积不变，用字母表示为(a×b)×c=a×(b×c)。用简便方法计算125×32，先把32分解成(8×4)),再根据乘法结合律计算.\n【详解】125×32\n=125×(8×4)\n=125×8×4\n=1000×4\n=4000\n用简便方法计算:125×32=125×8×4，结果是4000。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·湖南岳阳·期中"
+    },
+    {
+        "id":  "186",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "67×125×8=67×(125×8),运用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法结合",
+        "solution":  "【分析】乘法结合律：三个数相乘，可以先把前两个数相乘，或者先把后两个数相乘，积不变。用字母表示是(a×b)×c=a×(b×c)。据此解答.\n【详解】根据分析,67×125×8=67×(125×8),运用了乘法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·辽宁鞍山·期中"
+    },
+    {
+        "id":  "187",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "99×125×8=99×(125×8)运用了()律。计算618+324+76时，为了计算简便，可以先算(),这是根据()律。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法结合;324+76;加法结合",
+        "solution":  "【分析】乘法结合律是指三个数相乘，先把前两个数相乘，再和另外一个数相乘，或先把后两个数相乘，再和另外一个数相乘，积不变，计算99×125×8，可以先计算125×8的结果，再与99相乘，据此填空即可;根据加法结合律：先把前两个数相加，再加第三个数，或者先把后两个数相加，再加第一个数，和相等;618+324+76中，可以先计算324+76，得400，计算更简便.\n【详解】99×125×8=99×(125×8)运用了乘法结合律。计算618+324+76时，为了计算简便，可以先算324+76，这是根据加法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法结合律",
+                                 "整数加法结合律"
+                             ],
+        "source":  "24-25四年级下·河北邢台·期中"
+    },
+    {
+        "id":  "188",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "把\"635×24+65×24\"，改写成乘法分配律形式是(),结果是()。",
+        "options":  [
+
+                    ],
+        "answer":  "(35+65)×24;2400",
+        "solution":  "【分析】乘法分配律：两个数的和与一个数相乘，可以先把两个加数分别与这个数相乘，再将两个积相加，结果不变，用字母表示为a×c+b×c=(a+b)×c依此计算.\n【详解】35×24+65×24\n=(35+65)×24\n=100×24\n=2400\n所以把6635×24+65×24，改写成乘法分配律形式是(35+65)×24，结果是2400。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·湖南岳阳·期中"
+    },
+    {
+        "id":  "189",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "已知1+2+3+4=10，那么3+6+9+12的结果是多少。有人这样思考3+6+9+12=3×(1+2+3+4)=3×10=30。已知1+2+3+。+10=55，那么8+16+24+。+80=()。",
+        "options":  [
+
+                    ],
+        "answer":  "440",
+        "solution":  "【分析】观察题目中的例子，发现新数列是原数列每个数乘相同倍数后的和，因此可以利用原数列的和乘倍数得到结果。8+16+24+。+80每个数是原数列对应项的8倍，故总和为原数列和55乘以8。两个数的和与一个数相乘，可以先把它们分别与这个数相乘，再相加。也可以是多个数的和与一个数相乘，可以先把它们分别与这个数相乘，再相加。符合乘法分配律的形式.\n【详解】根据分析可得:\n8+16+24+。+80\n=8×(1+2+3+。+10)\n=8×55\n=440\n那么8+16+24+。+80=440",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·福建福州·期中"
+    },
+    {
+        "id":  "190",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "杨老师为学校采购了65个标价为18元的皮球，后来又采购了18个标价为35元的排球，杨老师一共要付()元。",
+        "options":  [
+
+                    ],
+        "answer":  "1800",
+        "solution":  "【分析】根据单价×数量=总价，代入数据分别求出65个皮球和18个排球的价钱，再相加即可;计算过程可以利用乘法分配律进行简便计算.\n【详解】65×18+18×35\n=(65+35)×18\n=100×18\n=1800(元)\n所以杨老师一共要付1800元。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·福建莆田·期中"
+    },
+    {
+        "id":  "191",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "笑笑把125×(□+5)错写成125*□+5，她得到的结果与正确答案相差()。",
+        "options":  [
+
+                    ],
+        "answer":  "620",
+        "solution":  "【分析】根据乘法分配律将算式125×(□+5)去掉括号后再与算式125*□+5比较，求出相差的值。乘法分配律：两个数的和与一个数相乘，可以先把两个加数分别与这个数相乘，再将两个积相加，结果不变,a×c+b×c=(a+b)×c.\n【详解】125×(□+5)=125*□+125×5\n125×5-5\n=625-5\n=620\n所以，她得到的结果与正确答案相差620。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·安徽宣城·期中"
+    },
+    {
+        "id":  "192",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "计算89×101时，可以简化为89×(100+1),这是运用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法分配律",
+        "solution":  "【分析】乘法分配律是指两个数的和与一个数相乘，可以先把它们分别与这个数相乘，再把积加起来,a×(b+c)=a×b+a×c.\n【详解】将101拆分成100+1，那么89×101就可以写成89×(100+1),在这里a=89,b=100,c=1，所以89×(100+1)=89×100+89×1=89×100+89。所以，计算89×101时，可以简化为89×(100+1),这是运用了乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·湖南岳阳·期中"
+    },
+    {
+        "id":  "193",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果42×A+42×B的计算结果是4200，那么A+B的和是()。",
+        "options":  [
+
+                    ],
+        "answer":  "100",
+        "solution":  "【分析】根据乘法分配律：是指两个数的和与一个数相乘，可以先把它们分别与这个数相乘，再相加;a×c+b×c=(a+b)×c。据此解答即可.\n【详解】42×A+42×B=42×(A+B),4200÷42=100，如果42×A+42×B的计算结果是4200，那么A+B的和是100。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·河南许昌·期末"
+    },
+    {
+        "id":  "194",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据运算律在横线上填上合适的数或字母.\n25×125×8×4=(□*□)×(□*□)\n78×101-78=□*(□-□)",
+        "options":  [
+
+                    ],
+        "answer":  "25;4;125;8;78;101;1",
+        "solution":  "【分析】根据题意，乘法结合律指三个数相乘时，先乘前两个数或先乘后两个数，结果相同。乘法结合律:(a×b)×c=a×(b×c)或a×b×c=a×(b×c)。乘法分配律指一个数乘以两数之和，等于这个数分别乘以这两个数再相加。乘法分配律:(a+b)×c=a×c+b×c,以此答题即可.\n【详解】根据分析可知:\n25×125×8×4=(25×4)×(125×8)\n78×101-78=78×(101-1)",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期中"
+    },
+    {
+        "id":  "195",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算125×88时，一般会把88拆分，这样可以简便计算。若运用乘法结合律，则这道算式可以写成();若运用乘法分配律算式可以改写成()。",
+        "options":  [
+
+                    ],
+        "answer":  "125×(8×11);125×(80+8)",
+        "solution":  "【分析】根据题意，乘法结合律指三个数相乘时，先乘前两个数或先乘后两个数，结果相同。乘法结合律:(a×b)×c=a×(b×c)或a×b×c=a×(b×c);乘法分配律指一个数乘以两数之和，等于这个数分别乘以这两个数再相加。乘法分配律:(a+b)×c=a×c+b×c;计算125×88时，通过拆分88简化运算。乘法结合律需要将数拆分为多个因数相乘并重新组合，而乘法分配律需拆分为两个数的和，分别相乘再相加。选择拆分方式时，优先考虑与125乘能得整千，整百的数，如8和80，便于快速计算。以此答题即可.\n【详解】根据分析可知:\n125×88\n=125×(8×11)\n=125×8×11\n=1000×11\n=11000\n125×88\n=125×(80+8)\n=125×80+125×8\n=10000+1000\n=11000\n计算125×88时，一般会把88拆分，这样可以简便计算。若运用乘法结合律，则这道算式可以写成125×(8×11);若运用乘法分配律算式可以改写成125×(80+8)。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·山东济南·期中"
+    },
+    {
+        "id":  "196",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "张强由于粗心把25×(△+4)错算成25*△+4，他得到的结果与正确结果相差()。",
+        "options":  [
+
+                    ],
+        "answer":  "96",
+        "solution":  "【分析】正确算式25×(△+4)根据乘法分配律展开为25*△+25×4，而错误算式为25*△+4。两者的差为25×4-4，计算即可得出结果.\n【详解】正确算式:25×(△+4)=25*△+25×4=25*△+100\n错误算式:25*△+4\n相差结果:\n(25*△+100)-(25*△+4)=100-4=96\n因此，他得到的结果与正确结果相差96。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·山东济南·期中"
+    },
+    {
+        "id":  "197",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "王明在用计算器计算\"236×45\u0027\"时，发现按键\"4\"坏了。如果还是用这个计算器，他可以怎么计算。请写出算式()。",
+        "options":  [
+
+                    ],
+        "answer":  "236×5×9",
+        "solution":  "【分析】本题可以考虑45拆成2个不含数字4的因数相乘，即45=5×9，然后再根据乘法结合律进行简算，乘法结合律的特点是三个数相乘，先把前两个数相乘，或先把后两个数相乘，积不变，用字母表示为:a×c×b=a×(c×b)。据此即可解答.\n【详解】236×45\n=236×(5×9)\n=236×5×9\n=1180×9\n=10620\n王明在用计算器计算\"236×45\"时，发现按键\"4\"坏了。如果还是用这个计算器，他可以怎么计算。请写出算式:236×5×9。(答案不唯一)",
+        "knowledge_points":  [
+                                 "计算器的初步认识与使用",
+                                 "整数乘法结合律"
+                             ],
+        "source":  "23-24四年级下·浙江舟山·期末"
+    },
+    {
+        "id":  "198",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算(26×25)×4时，为了计算简便，可以先算(),运用的是()律。",
+        "options":  [
+
+                    ],
+        "answer":  "25×4;乘法结合",
+        "solution":  "【分析】乘法结合律：三个数相乘，先乘前两个数，或者先乘后两个数，积不变。字母表示为:(a×b)×c=a×(b×c),据此解答.\n【详解】(26×25)×4\n=26×(25×4)\n=26×100\n=2600\n计算(26×25)×4时，为了计算简便，可以先算25×4，运用的是乘法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·四川遂宁·期中"
+    },
+    {
+        "id":  "199",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "用自己喜欢的方式表示加法交换律:()。",
+        "options":  [
+
+                    ],
+        "answer":  "a+b=b+a",
+        "solution":  "【分析】加法交换律：两个数相加，交换加数的位置，和不变;【详解】用自己喜欢的方式表示加法交换律:a+b=b+a或甲+乙=乙+甲。(答案不唯一)。",
+        "knowledge_points":  [
+                                 "整数加法交换律"
+                             ],
+        "source":  "24-25四年级下·海南省直辖县级单位·期中"
+    },
+    {
+        "id":  "200",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "明明在计算34×(☆-10)时，不小心算成了34*☆-10，他的结果比正确结果少()。",
+        "options":  [
+
+                    ],
+        "answer":  "330",
+        "solution":  "【分析】明明在计算时看错了运算顺序，但计算的过程是正确的，可以结合\"看错\"与\"正确\"算式的关系，先用乘法分配律将原来的算式去掉括号，再和看错了的算式比较，来确定结果的差值.\n【详解】34×(☆-10)=34×☆-34×10\n34×10-10\n=340-10\n=330\n所以，他的结果比正确结果少330。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·内蒙古鄂尔多斯·期末"
+    },
+    {
+        "id":  "201",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "用计算器计算\"79×94328\"时发现\"7\"键坏了。如果还用这个计算器且运用乘法分配律的知识，你会怎么计算。请写出算式:()。",
+        "options":  [
+
+                    ],
+        "answer":  "(80-1)×94328",
+        "solution":  "【分析】用乘法分配律计算79×94328时，把79看作80-1，即(80-1)×94328，据此解答即可.\n【详解】用计算器计算\"79×94328\"时发现\"7\"键坏了。如果还用这个计算器且运用乘法分配律的知识，你会怎么计算。请写出算式:(80-1)×94328。(答案不唯一)",
+        "knowledge_points":  [
+                                 "计算器的初步认识与使用",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·浙江杭州·期中"
+    },
+    {
+        "id":  "202",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "3×4×25=3×(4×25)运用了乘法()律。",
+        "options":  [
+
+                    ],
+        "answer":  "结合",
+        "solution":  "【分析】乘法结合律是指三个数相乘，先把前两个数相乘，再和另外一个数相乘，或先把后两个数相乘，再和另外一个数相乘，积不变;据此解答即可.\n【详解】根据分析可得:3×4×25=3×(4×25)运用了乘法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·浙江绍兴·期中"
+    },
+    {
+        "id":  "203",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "25×5×2=25×(5×2)这里运用的运算律是(),用字母表示是()。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法结合律;(a×b)×c=a×(b×c)",
+        "solution":  "【分析】乘法结合律：三个数相乘，可以先把前两个数相乘，或者先把后两个数相乘，积不变。用字母表示是(a×b)×c=a×(b×c).\n【详解】根据分析可知,25×5×2=25×(5×2)这里运用了乘法结合律，用字母表示是(a×b)×c=a×(b×c)。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·云南昆明·期中"
+    },
+    {
+        "id":  "204",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "如果A+B=40，那么78×A+78×B=()。",
+        "options":  [
+
+                    ],
+        "answer":  "3120",
+        "solution":  "【分析】乘法分配律是指两个数的和与一个数相乘，可以先把它们分别与这个数相乘，再相加。用字母表示为(a+b)×c=a×c+b×c。所以78×A+78×B=78×(A+B).\n【详解】已知A+B=40，将其代入78×(A+B)可得:78×40=3120。所以,78×A+78×B=3120。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·湖北黄石·期中"
+    },
+    {
+        "id":  "205",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "小军在计算\"864÷36\"时，发现数字键\"3\"坏了。他按\"864÷4/9\"就能得出答案。还可以按()或()。",
+        "options":  [
+
+                    ],
+        "answer":  "864÷2/18;864÷6/6",
+        "solution":  "【分析】这是一道关于运算方法灵活运用的问题，关键在于将除数36进行合理拆分，利用除法的性质，把一个数除以两个数的积转化为连续除以这两个数，从而在数字键\"3\"损坏的情况下得出结果。分析已知拆分方式：已知36=4×9，所以864÷36=864÷(4×9),根据除法的性质，一个数除以两个数的积等于这个数连续除以这两个数，即864÷(4×9)=864÷4/9.\n【详解】因为36=2×18，所以864÷36=864÷(2×18)=864÷2/18;又因为36=6×6，所以864÷36=864÷(6×6)=864÷6/6。所以，小军在计算\"864÷36\"时，发现数字键\"3\"坏了。他按\"864÷4/9\"就能得出答案。还可以按864÷2/18或864÷6/6。",
+        "knowledge_points":  [
+                                 "整数除法的性质"
+                             ],
+        "source":  "24-25四年级下·湖北黄石·期中"
+    },
+    {
+        "id":  "206",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "(28×5)×2=28×(5×2)这是运用了()律,(8+4)×25=8×25+4×25这是运用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法结合;乘法分配",
+        "solution":  "【分析】乘法结合律是三个数相乘可以先算前两个数相乘，也可以先算后两个数相乘，积不变，表示为(a×b)×c=a×(b×c);乘法分配律是两个数的和与一个数相乘，等于把两个数分别同这个数相乘，再把两个积相加，用字母表示:a×(b+c)=a×b+a×c.\n【详解】(28×5)×2=28×(5×2)这是运用了乘法结合律,(8+4)×25=8×25+4×25这是运用了乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·湖北武汉·期中"
+    },
+    {
+        "id":  "207",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "新能源汽车已经走进我们的生活，逐渐为大家所青睐。一处新能源汽车共享充电站有16个充电桩，一个充电桩最多可同时给8辆车充电，如果每辆车都充电1小时，那么25小时最多可给()辆车充电。",
+        "options":  [
+
+                    ],
+        "answer":  "3200",
+        "solution":  "【分析】根据题意，已知一处新能源汽车共享充电站有16个充电桩，一个充电桩最多可同时给8辆车充电，先用16乘8，求出16个充电桩最多能充电的车辆数;再乘25，就是25小时最多可给多少辆车充电;列出算式。根据乘法交换律:a×b=b×a和乘法结合律:(a×b)×c=a×(b×c)或a×b×c=a×(b×c),变算式为:4×8×(4×25),再进行计算.\n【详解】根据分析可知:16×8×25=4×4×8×25=4×8×(4×25)=32×100=3200(辆)。新能源汽车已经走进我们的生活，逐渐为大家所青睐。一处新能源汽车共享充电站有16个充电桩，一个充电桩最多可同时给8辆车充电，如果每辆车都充电1小时，那么25小时最多可给3200辆车充电。",
+        "knowledge_points":  [
+                                 "整数乘法结合律",
+                                 "整数乘法交换律"
+                             ],
+        "source":  "24-25四年级下·湖北荆州·期中"
+    },
+    {
+        "id":  "208",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "a×8+8×17=□*(□+□)。",
+        "options":  [
+
+                    ],
+        "answer":  "8;a;17",
+        "solution":  "【分析】乘法分配律是指两个数的和同一个数相乘，等于把两个加数分别同这个数相乘，再把两个积加起来，结果不变。据此计算a×8+8×17，先计算a+17，再用和乘8.\n【详解】a×8+8×17=8×(a+17)。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·河南焦作·期中"
+    },
+    {
+        "id":  "209",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算:13×97+8×14×125=()。",
+        "options":  [
+
+                    ],
+        "answer":  "15261",
+        "solution":  "【分析】观察算式后发现，可以将97写成100-3，然后13×97可以运用乘法分配律进行简便计算;将14和125交换位置，运用乘法分配律8×14×125可以进行简便计算。由此即可解决.\n【详解】13×97+8×14×125=13×(100-3)+8×125×14=13×100-13×3+1000×14=1300-39+14000=1261+14000=15261。故13×97+8×14×125=15261。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数乘法交换律"
+                             ],
+        "source":  "2023四年级·全国·竞赛"
+    },
+    {
+        "id":  "210",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "下图是我们上学期学过的三位数乘两位数的笔算。其实，笔算过程所运用的运算定律就是()。根据笔算过程，用横式表达笔算乘法的算理并填空:\n![](images/41f7e70a91a3d28ca1b247f872b7b5c22079f6179e6b04353f04a5a00554aee6.jpg)\n145×12\n=145×(□+□)\n=(□*□)+(□*□)\n=□+□\n=□",
+        "options":  [
+
+                    ],
+        "answer":  "乘法分配律;2;10;145;2;145;10;290;1450;1740",
+        "solution":  "【分析】乘法分配律指一个数乘两数之和，等于这个数分别乘这两个数再相加。乘法分配律:(a+b)×c=a×c+b×c,题目中三位数乘两位数的笔算,145×12，把12改写成2+10的算式，先用145×2=290，再用145×10=1450，最后算290+1450=1740。以此答题即可.\n【详解】根据分析可知：下图是我们上学期学过的三位数乘两位数的笔算。其实，笔算过程所运用的运算定律就是乘法分配律。根据笔算过程，用横式表达笔算乘法的算理并填空:145×12=145×(2+10)=(2×145)+(10×145)=290+1450=1740。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "三位数与两位数的乘法"
+                             ],
+        "source":  "24-25四年级下·云南昭通·期中"
+    },
+    {
+        "id":  "211",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "125×(8×24)=(125×8)×24，这个算式运用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法结合",
+        "solution":  "【分析】乘法结合律：三个数相乘，先乘前两个数，或者先乘后两个数，积不变.\n【详解】125×(8×24)=(125×8)×24，这个算式运用了乘法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·河北张家口·期中"
+    },
+    {
+        "id":  "212",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "25×5×2=25×(5×2)这里运用的运算律是(),用字母表示是()。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法结合律;(a×b)×c=a×(b×c)",
+        "solution":  "【分析】乘法结合律：三个数相乘，可以先把前两个数相乘，或者先把后两个数相乘，积不变。用字母表示是(a×b)×c=a×(b×c).\n【详解】根据分析,25×5×2=25×(5×2)这里运用了乘法结合律，用字母表示是(a×b)×c=a×(b×c)。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·云南保山·期中"
+    },
+    {
+        "id":  "213",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "根据运算律填空.\n64+51+36=64+□+51;26×25×8=26×(□□□)",
+        "options":  [
+
+                    ],
+        "answer":  "36;25;*;8",
+        "solution":  "【分析】加法交换律的特点是两个数相加，交换加数的位置，和不变，用字母表示为:a+b=b+a。乘法结合律的特点是三个数相乘，先把前两个数相乘，或先把后两个数相乘，积不变，用字母表示为:a×c×b=a×(c×b);依此填空.\n【详解】64+51+36=64+36+51,26×25×8=26×(25×8)。",
+        "knowledge_points":  [
+                                 "整数加法交换律",
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·新疆喀什·期中"
+    },
+    {
+        "id":  "214",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "53×99+53=□*(□+□)这是根据()律进行简便计算的。",
+        "options":  [
+
+                    ],
+        "answer":  "53;99;1;乘法分配",
+        "solution":  "【分析】此题可以逆用乘法分配律，先计算99与1的和，再用这个和乘53即可，据此解答.\n【详解】53×99+53=53×(99+1)这是根据乘法分配律进行简便计算的。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·新疆阿克苏·期中"
+    },
+    {
+        "id":  "215",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "49×□+□×51=68×(□+□),这里运用了()，用字母可表示为()。",
+        "options":  [
+
+                    ],
+        "answer":  "68;68;49;51;乘法分配律;a×b+a×c=a×(b+c)",
+        "solution":  "【分析】根据题意，49×□+□×51=68×(□+□)，这是一个数分别与两个数相乘，等于这个数与那两个数的和相乘，是乘法分配律，据此填空。\n【详解】49×68+68×51=68×(49+51)，这里运用了乘法分配律，用字母表示是a×b+a×c=a×(b+c)。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·贵州安顺·期中"
+    },
+    {
+        "id":  "216",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "25×(4×8)=(25×4)×8，这里运用了()律;125×(8+4)=125×8+125×4，这里运用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法结合;乘法分配",
+        "solution":  "【分析】乘法结合律：三个数相乘，先把前两个数相乘，或者先把后两个数相乘，积不变，用字母表示为:(a×b)×c=a×(b×c);乘法分配律：两个数的和与第三个数相乘，可以把这两个数与第三个数分别相乘，再相加，用字母表示为:a×(b+c)=a×b+a×c.\n【详解】25×(4×8)=(25×4)×8，这里运用了乘法结合律;125×(8+4)=125×8+125×4，这里运用了乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·吉林松原·期中"
+    },
+    {
+        "id":  "217",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用计算器计算223×19时发现键\"9\"坏了，如果还用这个计算器，你会怎样计算，写出算式(),正确的结果是()。",
+        "options":  [
+
+                    ],
+        "answer":  "223×20-223;4237",
+        "solution":  "【分析】根据题意可知，用计算器计算223×19时，可将19写成20-1，然后再根据乘法分配律的特点(a-b)×c=a×c-b×c,变式为223×(20-1)=223×20-223进行简算，依此解答.\n【详解】223×19=223×(20-1)=223×20-223=4460-223=4237。即写出算式(223×20-223),正确结果为4237。(算式不唯一)",
+        "knowledge_points":  [
+                                 "计算器的初步认识与使用",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·贵州铜仁·期中"
+    },
+    {
+        "id":  "218",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "计算(40+4)×25=40×25+4×25时运用了()律;计算13×8×125=13×(8×125)时运用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法分配;乘法结合",
+        "solution":  "【分析】乘法分配律：两个数的和与一个数相乘，可以先把它们与这个数分别相乘，再相加;字母表示为:(a+b)×c=a×c+b×c。乘法结合律：三个数相乘，先乘前两个数，或者先乘后两个数，积不变;字母表示为:(a×b)×c=a×(b×c).\n【详解】计算(40+4)×25=40×25+4×25时运用了乘法分配律;计算13×8×125=13×(8×125)时运用了乘法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·贵州铜仁·期中"
+    },
+    {
+        "id":  "219",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "下面的竖式在计算过程中运用了()律，列出相应的算式:().\n![](images/b51d153e491d4ad3709638c1a465e69df6df31cd7fa4c7cdbc6f8db7a6c4eab9.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "乘法分配;25×2+25×10=300",
+        "solution":  "【分析】根据题意，两位数乘两位数，先用两位数个位上的数去乘两位数，再用十位上的数去乘两位数，最后相加。25×12中，先算25×2=50，再算25×10=250，最后50+250=300。根据乘法分配律:(a+b)×c=a×c+b×c,列式为25×2+25×10=300.\n【详解】根据分析可知：该竖式在计算过程中运用了乘法分配律，列出相应算式为25×2+25×10=300。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·山东临沂·期中"
+    },
+    {
+        "id":  "220",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "算式\"97×a+97×b\"可以利用乘法分配律进行简便计算，结果是9700。那么a是(),b是()。",
+        "options":  [
+
+                    ],
+        "answer":  "3;97",
+        "solution":  "【分析】乘法分配律是指两个数的和与一个数相乘，可以先把它们分别与这个数相乘，再相加。根据(a+b)×c=a×c+b×c,可把算式看成(a+b)×97=9700，据此求解.\n【详解】9700÷97=100，所以a+b=100。取a=3，b=97时，97×a+97×b=9700。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·浙江杭州·期末"
+    },
+    {
+        "id":  "221",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "李阿姨在体育用品专卖店购进了60套运动服，其中上衣75元，裤子45元，一共花了多少钱。小明列出的算式为75×60+45×60，解决这个问题还可以列式为()。(只列综合算式不计算)",
+        "options":  [
+
+                    ],
+        "answer":  "(75+45)×60",
+        "solution":  "【分析】乘法分配律:(a+b)×c=a×c+b×c,可知75×60+45×60=(75+45)×60.\n【详解】根据分析可知:75×60+45×60=(75+45)×60=120×60=7200(元)。解决这个问题还可以列式为(75+45)×60。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·内蒙古鄂尔多斯·期末"
+    },
+    {
+        "id":  "222",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "20-19+18-17+。+2-1=()",
+        "options":  [
+
+                    ],
+        "answer":  "10",
+        "solution":  "【分析】观察算式可将其写成(20-19)+(18-17)+。+(2-1),每一组都等于1，共10组，因此结果为10.\n【详解】20-19+18-17+。+2-1=(20-19)+(18-17)+。+(2-1)=1+1+。+1=10×1=10。",
+        "knowledge_points":  [
+                                 "整数加法结合律"
+                             ],
+        "source":  "23-24四年级下·重庆丰都·期末"
+    },
+    {
+        "id":  "223",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "减法的性质用字母表示为()。",
+        "options":  [
+
+                    ],
+        "answer":  "a-b-c=a-(b+c)",
+        "solution":  "【分析】减法的性质：在减法算式里，连续减去两个数等于减去两个数的和.\n【详解】由分析可知，减法的性质用字母表示为a-b-c=a-(b+c)。",
+        "knowledge_points":  [
+                                 "整数减法的性质"
+                             ],
+        "source":  "23-24四年级下·重庆丰都·期末"
+    },
+    {
+        "id":  "224",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "计算12×101时，丽丽用简便方法12×100+12计算，这是运用了乘法()律。",
+        "options":  [
+
+                    ],
+        "answer":  "分配",
+        "solution":  "【分析】12×101可看作12×(100+1),利用乘法分配律(a+b)×c=a×c+b×c.\n【详解】12×101=12×(100+1)=12×100+12×1=12×100+12=1212。即运用了乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·浙江绍兴·期末"
+    },
+    {
+        "id":  "225",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "265×95+265×5=265×(95+5),计算的依据是()。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法分配律",
+        "solution":  "【分析】该式是两个积有相同因数265，可根据乘法分配律a×c+b×c=(a+b)×c进行变形.\n【详解】265×95+265×5=265×(95+5)=265×100=26500，因此依据是乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·重庆云阳·期末"
+    },
+    {
+        "id":  "226",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用计算器计算\"236×49\"时，发现键\"4\"坏了。如果还用这个计算器，你会怎样计算。请写出算式:()。",
+        "options":  [
+
+                    ],
+        "answer":  "236×50-236",
+        "solution":  "【分析】把49转化为50-1，再利用乘法分配律a×(b-c)=a×b-a×c.\n【详解】根据分析可知，把49转化为50-1，算式变为236×50-236。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数乘法结合律"
+                             ],
+        "source":  "23-24四年级下·重庆秀山·期末"
+    },
+    {
+        "id":  "227",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "灵灵将30×(□+5)错算成了30*□+5，请你算一算，正确的答案与错误的答案相差()。",
+        "options":  [
+
+                    ],
+        "answer":  "145",
+        "solution":  "【分析】30×(□+5)=30*□+30×5，错误算式是30*□+5，相差(30×5-5).\n【详解】30×5-5=150-5=145，所以正确答案与错误答案相差145。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "23-24四年级下·重庆·期末"
+    },
+    {
+        "id":  "228",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "如果40×16+A×40=800，那么A=()",
+        "options":  [
+
+                    ],
+        "answer":  "4",
+        "solution":  "【分析】根据乘法分配律可提取40，得40×(16+A)=800.\n【详解】40×(16+A)=800,16+A=800÷40=20，所以A=20-16=4。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·重庆开州·期末"
+    },
+    {
+        "id":  "229",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "算式228+167+□,在□中填上()就能使计算简便(只填一个数),计算结果是()。",
+        "options":  [
+
+                    ],
+        "answer":  "33;428",
+        "solution":  "【分析】根据加法结合律，把167与□凑整，取33可得167+33=200，便于计算.\n【详解】228+167+33=228+(167+33)=228+200=428。",
+        "knowledge_points":  [
+                                 "整数加法结合律"
+                             ],
+        "source":  "23-24四年级下·福建莆田·期末"
+    },
+    {
+        "id":  "230",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算器上数字键\"6\"坏了，如果想用这台计算器计算\"135×66\"的积，你可以在计算器上输入算式:(),理由是:()。",
+        "options":  [
+
+                    ],
+        "answer":  "135×22×3;66看作22×3，利用乘法结合律计算135×66=135×22×3",
+        "solution":  "【分析】将66分解为22×3，利用乘法结合律把135×66转化为135×22×3.\n【详解】135×66=135×(22×3)=135×22×3=2970×3=8910。可输入135×22×3。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "23-24四年级下·福建莆田·期末"
+    },
+    {
+        "id":  "231",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "9×25×3×4=9×3×(4×25)计算过程中既运用了()律，又运用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法交换;乘法结合",
+        "solution":  "【分析】先交换25和3的位置用乘法交换律，再把4和25结合用乘法结合律.\n【详解】9×25×3×4=9×3×25×4=9×3×(25×4)=9×3×100=2700，所以既运用了乘法交换律，又运用了乘法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法结合律",
+                                 "整数乘法交换律"
+                             ],
+        "source":  "23-24四年级下·重庆秀山·期末"
+    },
+    {
+        "id":  "232",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "张乐在学习运算律时遇到的问题.\n(1)在计算36×48+36×52时，他是这样想的:36×48+36×52=36×(48+52),他依据的运算律是().\n(2)在计算(24×125)×8时，为了使计算简便，可以先算(),他这样计算是运用了()律.\n(3)在计算868-77-□,他想使这个算式可以进行简便计算,□里可以填().\n(4)在计算22×25时，他想把22拆分后进行简便计算，运用乘法分配律可以把22拆分成();运用乘法结合律可以把22拆分成()。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法分配律;125×8;乘法结合;23;20+2;11×2",
+        "solution":  "【分析】(1)36×48+36×52=36×(48+52)符合乘法分配律a×c+b×c=(a+b)×c.\n(2)(24×125)×8中先算125×8更简便，运用乘法结合律a×b×c=a×(b×c).\n(3)868-77-□中若77与□凑整百可简算，取23可得868-(77+23).\n(4)22可拆成20+2用于分配律，也可拆成11×2用于结合律.\n【详解】(1)依据乘法分配律.\n(2)可以先算125×8，运用了乘法结合律.\n(3)□里可填23。(答案不唯一)\n(4)用乘法分配律可拆成20+2;用乘法结合律可拆成11×2。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数乘法结合律",
+                                 "整数减法的性质"
+                             ],
+        "source":  "24-25四年级下·山西忻州·期中"
+    },
+    {
+        "id":  "233",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小糊涂在计算16×(5+n)时，错算成了16×5+n,他算出的结果和正确的结果相差()。",
+        "options":  [
+
+                    ],
+        "answer":  "15n",
+        "solution":  "【分析】正确算式16×(5+n)=16×5+16n,错误算式是16×5+n,两者相差(16n-n)=15n.\n【详解】16×(5+n)-(16×5+n)=16×5+16n-16×5-n=15n。所以结果相差15n。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·山西忻州·期中"
+    },
+    {
+        "id":  "234",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算125×16时，可以转化成125×()×()。",
+        "options":  [
+
+                    ],
+        "answer":  "8;2",
+        "solution":  "【分析】计算125×16时，先将16写成8×2的形式，即125×(8×2),再根据乘法结合律:(a×b)×c=a×(b×c),将算式125×(8×2)变成125×8×2，据此即可简便计算.\n【详解】125×16=125×(8×2)=125×8×2=1000×2=2000。计算125×16时，可以转化成125×8×2。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·四川乐山·期中"
+    },
+    {
+        "id":  "235",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据运算定律在横线上填数.\n(1)125×(8×□)=(125×□)×13\n(2)4800÷25/4=4800÷(□×□)",
+        "options":  [
+
+                    ],
+        "answer":  "13;8;25;4",
+        "solution":  "【分析】(1)从算式可以看出，运用乘法结合律，三个数相乘，先把8与13相乘再与125相乘，或先把125与8相乘，再与13相乘，它们的积相等。(2)运用除法的性质,4800除以25，再除以4，等于4800除以25与4的积。据此解答.\n【详解】(1)125×(8×13)=(125×8)×13;(2)4800÷25/4=4800÷(25×4)。",
+        "knowledge_points":  [
+                                 "整数除法的性质",
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·河南信阳·期中"
+    },
+    {
+        "id":  "236",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "两个数相乘，交换两个()的位置，积(),这叫作乘法交换律;用字母来表示为a×b=()",
+        "options":  [
+
+                    ],
+        "answer":  "因数;不变;b×a",
+        "solution":  "【详解】如:2×3=3×2,10×5=5×10。两个数相乘，交换两个因数的位置，积不变，这叫作乘法交换律;用字母来表示为a×b=b×a。",
+        "knowledge_points":  [
+                                 "整数乘法交换律"
+                             ],
+        "source":  "24-25四年级下·广东江门·期中"
+    },
+    {
+        "id":  "237",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里里填上合适的数，使算式可以简便计算.\n2000÷25/();12×()+23×88;648+54-()",
+        "options":  [
+
+                    ],
+        "answer":  "4;23;48",
+        "solution":  "【分析】根据除法的性质:a/b/c=a/(b×c),只需要找出25乘哪个数的积是几百或几千，即可简便计算;根据乘法分配律:(a+b)×c=a×c+b×c,12+88=100，则括号里的数是23;同级运算中，可以交换两个计算顺序，改为先计算减法再计算加法，只需要找出648减去哪个数可以得到整十数或整百数，即可简便计算.\n【详解】2000÷25/4=2000÷(25×4)=2000÷100=20;12×23+23×88=(12+88)×23=100×23=2300;648+54-48=648-48+54=600+54=654。故可填4,23,48。",
+        "knowledge_points":  [
+                                 "整数加法交换律",
+                                 "整数乘法分配律",
+                                 "整数除法的性质"
+                             ],
+        "source":  "24-25四年级下·江西抚州·期中"
+    },
+    {
+        "id":  "238",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算(27×50)×4时，为了计算简便，可以先算(),这样计算是根据()律。",
+        "options":  [
+
+                    ],
+        "answer":  "50×4;乘法结合律",
+        "solution":  "【分析】计算(27×50)×4时，可以把50和4组合相乘简便，再乘27，因为乘法结合律是前两个数相乘再乘第三个数，等于后两个数相乘再乘第一个数，这道题符合乘法结合律.\n【详解】(27×50)×4=27×(50×4)=27×200=5400。即可以先算50×4，这样计算是根据乘法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·江西抚州·期中"
+    },
+    {
+        "id":  "239",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "629+(b+271)=(629+271)+b运用的运算律是()和()。",
+        "options":  [
+
+                    ],
+        "answer":  "加法交换律;加法结合律",
+        "solution":  "【分析】加法交换律：交换两个加数的位置，和不变。加法结合律：先把前两个数相加，或者把后两个数相加，和不变。算式中先交换b和271的位置，再利用加法结合律计算.\n【详解】629+(b+271)=(629+271)+b运用的运算律是加法交换律和加法结合律。",
+        "knowledge_points":  [
+                                 "整数加法交换律",
+                                 "整数加法结合律"
+                             ],
+        "source":  "24-25四年级下·陕西渭南·期中"
+    },
+    {
+        "id":  "240",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "三个数相加，先把()相加，或者先把()相加，和(),这叫作加法结合律。用字母表示为()。",
+        "options":  [
+
+                    ],
+        "answer":  "前两个数;后两个数;不变;(a+b)+c=a+(b+c)",
+        "solution":  "【分析】本题考查加法结合律的定义和字母表示。加法结合律是指三个数相加，先把前两个数相加，或者先把后两个数相加，和不变。用字母表示为(a+b)+c=a+(b+c).\n【详解】三个数相加，先把前两个数相加，或者先把后两个数相加，和不变，这叫作加法结合律。用字母表示为(a+b)+c=a+(b+c)。",
+        "knowledge_points":  [
+                                 "整数加法结合律"
+                             ],
+        "source":  "24-25四年级下·河南新乡·期中"
+    },
+    {
+        "id":  "241",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用含有字母的式子表示运算定律.\n(1)加法结合律:();(2)乘法分配律:();",
+        "options":  [
+
+                    ],
+        "answer":  "(a+b)+c=a+(b+c);(a+b)×c=a×c+b×c",
+        "solution":  "【分析】(1)加法结合律是指三个数相加，先把前两个数相加，或者先把后两个数相加，和不变。(2)乘法分配律是指两个数的和与一个数相乘，可以先把它们分别与这个数相乘，再相加.\n【详解】(1)加法结合律可写为(a+b)+c=a+(b+c);(2)乘法分配律可写为(a+b)×c=a×c+b×c。",
+        "knowledge_points":  [
+                                 "整数加法结合律",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·广东汕头·期中"
+    },
+    {
+        "id":  "242",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "学校文艺汇演，班级里采购了5包标价为16元的手花和24个标价为5元的头饰，一共用了()元。",
+        "options":  [
+
+                    ],
+        "answer":  "200",
+        "solution":  "【分析】已知采购了5包手花，每包标价16元，那么购买手花的花费就是包数乘每包的价格;采购了24个头饰，每个标价5元，所以购买头饰的花费也是个数乘每个的价格;最后把购买手花的费用和购买头饰的费用加起来就是总费用。计算时，用乘法分配律进行简算.\n【详解】5×16+24×5=5×(16+24)=5×40=200(元)。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·广东珠海·期中"
+    },
+    {
+        "id":  "243",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "计算26×125×8时，可以先算()，再算()，这样计算采用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "125×8=1000;26×1000=26000;乘法结合",
+        "solution":  "【分析】乘法结合律是指三个数相乘，先把前两个数相乘，再乘第三个数，或先把后两个数相乘，再乘第一个数，积不变。26×125×8可变为26×(125×8).\n【详解】先算125×8=1000，再算26×1000=26000，这样计算采用了乘法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·河南新乡·期中"
+    },
+    {
+        "id":  "244",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "已知a+b=10,47×a+47×b=()。",
+        "options":  [
+
+                    ],
+        "answer":  "470",
+        "solution":  "【分析】根据乘法分配律a×b+a×c=a×(b+c)可知,47×a+47×b=47×(a+b),再根据a+b=10即可解决.\n【详解】47×a+47×b=47×(a+b)=47×10=470。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·内蒙古鄂尔多斯·期中"
+    },
+    {
+        "id":  "245",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用计算器计算65×49时，发现\"4\"键坏了，如果还用这个计算器，你会怎样算。请写出算式:()。",
+        "options":  [
+
+                    ],
+        "answer":  "65×7×7",
+        "solution":  "【分析】计算器上的数字键\"4\"坏了，把49分解成用不到数字4的算式，例如49=7×7,49=10+39,49=50-1等形式，再用计算器计算.\n【详解】65×49=65×(7×7)=65×7×7=455×7=3185。可写算式65×7×7=3185。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·内蒙古鄂尔多斯·期中"
+    },
+    {
+        "id":  "246",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "4×18×25=18×(),运用了()律和()律。",
+        "options":  [
+
+                    ],
+        "answer":  "4×25或25×4;乘法交换;乘法结合",
+        "solution":  "【分析】观察4×18×25，先把18与4交换位置，再把4与25结合，可实现简便计算.\n【详解】4×18×25=18×(4×25)。由于交换了18和4的位置，运用了乘法交换律;又把4和25先结合，运用了乘法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法交换律",
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·内蒙古鄂尔多斯·期中"
+    },
+    {
+        "id":  "247",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在横线上填上适当的数，并在括号里填一填运用了什么运算律.\n(56+72)+28=56+(72+□),运用了().\n125×60×□=125×□×60，运用了().\n28×155+155×72=(□+□)×155，运用了()。",
+        "options":  [
+
+                    ],
+        "answer":  "28;加法结合律;8;乘法交换律;28;72;乘法分配律",
+        "solution":  "【分析】加法结合律:(a+b)+c=a+(b+c);乘法交换律:a×b=b×a;乘法分配律:a×c+b×c=(a+b)×c.\n【详解】(56+72)+28=56+(72+28),运用了加法结合律。125×60×8=125×8×60，运用了乘法交换律。28×155+155×72=(28+72)×155，运用了乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数乘法交换律",
+                                 "整数加法结合律"
+                             ],
+        "source":  "24-25四年级下·江西赣州·期中"
+    },
+    {
+        "id":  "248",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "用简便方法计算47×25×4=47×(□*□),这里运用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "25;4;乘法结合",
+        "solution":  "【分析】乘法结合律：三个数相乘，可以先把前两个数相乘，或者先把后两个数相乘，积不变。47×25×4可先算25×4使计算简便.\n【详解】47×25×4=47×(25×4)=47×100=4700。所以运用了乘法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·安徽黄山·期中"
+    },
+    {
+        "id":  "249",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明在计算\"36-(a+8)\"时错算成\"36-a+8\"，这样算出的结果与正确的结果相差()。",
+        "options":  [
+
+                    ],
+        "answer":  "16",
+        "solution":  "【分析】根据减法的性质：一个数连续减去两个数等于减去这两个数的和，所以36-(a+8)=36-a-8。与36-a+8比较，相差8+8=16.\n【详解】36-a+8-[36-(a+8)]=36-a+8-[36-a-8]=8+8=16。",
+        "knowledge_points":  [
+                                 "整数减法的性质"
+                             ],
+        "source":  "24-25四年级下·广西河池·期中"
+    },
+    {
+        "id":  "250",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "用简便方法计算45×102-45×2=45×(),运用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "102-2;乘法分配",
+        "solution":  "【分析】式子45×102-45×2中前后两项都有因数45，符合乘法分配律的减法形式a×c-b×c=(a-b)×c.\n【详解】45×102-45×2=45×(102-2)。运用了乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·湖南永州·期中"
+    },
+    {
+        "id":  "251",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算253+75+108+47+92时，可以使用()律和()律使运算简便。",
+        "options":  [
+
+                    ],
+        "answer":  "加法交换律;加法结合律",
+        "solution":  "【分析】加法交换律：两个数相加，交换两个加数的位置，和不变;加法结合律：三个数相加，先把前两个数相加或者先把后两个数相加，和不变.\n【详解】253+75+108+47+92=253+47+108+92+75=(253+47)+(108+92)+75，故可使用加法交换律和加法结合律。",
+        "knowledge_points":  [
+                                 "整数加法结合律",
+                                 "整数加法交换律"
+                             ],
+        "source":  "24-25四年级下·江西九江·期中"
+    },
+    {
+        "id":  "252",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算72×125×8时，为了使计算简便，可以先算(),这是根据()律;99×a+a=()×a,运用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "125×8;乘法结合律;99+1;乘法分配律",
+        "solution":  "【分析】(1)72×125×8中可先算125×8=1000，应用乘法结合律。(2)99×a+a可提取公因数a,变成(99+1)×a,应用乘法分配律.\n【详解】计算72×125×8时可先算125×8，这是根据乘法结合律;99×a+a=(99+1)×a,运用了乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·江西赣州·期中"
+    },
+    {
+        "id":  "253",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "用简便方法计算\"692+376+24\"，要先算(),这是根据()律，用字母表示这一运算定律为。",
+        "options":  [
+
+                    ],
+        "answer":  "376+24;加法结合律;(a+b)+c=a+(b+c)",
+        "solution":  "【分析】加法结合律：三个数相加，先把前两个数相加或先把后两个数相加，和不变。根据数字特点先算376+24更简便.\n【详解】692+376+24中，先算376+24，这是根据加法结合律，字母表示为(a+b)+c=a+(b+c)。",
+        "knowledge_points":  [
+                                 "整数加法结合律"
+                             ],
+        "source":  "24-25四年级下·辽宁鞍山·期中"
+    },
+    {
+        "id":  "254",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "加法，减法，乘法，除法统称为();一个数加上0还得();用字母表示乘法分配律()。",
+        "options":  [
+
+                    ],
+        "answer":  "四则运算;这个数;(a+b)×c=a×c+b×c",
+        "solution":  "【分析】加法，减法，乘法，除法统称为四则运算;一个数加上0还得这个数本身;乘法分配律是(a+b)×c=a×c+b×c.\n【详解】加法，减法，乘法，除法统称为四则运算;一个数加上0还得这个数;用字母表示乘法分配律为(a+b)×c=a×c+b×c。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "乘，除法的意义和各部分间的关系",
+                                 "加，减法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·辽宁鞍山·期中"
+    },
+    {
+        "id":  "255",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用简便方法计算88×125，可以写成(),结果是()。",
+        "options":  [
+
+                    ],
+        "answer":  "(80+8)×125或11×(8×125);11000",
+        "solution":  "【分析】88×125可用两种思路简算：一是把88拆成80+8，用乘法分配律;二是把88拆成11×8，用乘法结合律.\n【详解】88×125=(80+8)×125=80×125+8×125=10000+1000=11000;也可写成11×(8×125)=11×1000=11000。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·江西九江·期中"
+    },
+    {
+        "id":  "256",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "25×(40+8)=25×()+25×(),这是运用了()。",
+        "options":  [
+
+                    ],
+        "answer":  "40;8;乘法分配律",
+        "solution":  "【分析】两个数的和与一个数相乘，可以用这两个数分别和这个数相乘，再把它们的积相加，这叫乘法分配律.\n【详解】25×(40+8)=25×40+25×8，这是运用了乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·河南濮阳·期中"
+    },
+    {
+        "id":  "257",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "乘法交换律用字母表示是(),乘法结合律用字母表示是(),乘法分配律用字母表示是()。",
+        "options":  [
+
+                    ],
+        "answer":  "a×b=b×a;(a×b)×c=a×(b×c);(a+b)×c=a×c+b×c",
+        "solution":  "【分析】乘法交换律：交换因数位置，积不变;乘法结合律：改变结合顺序，积不变;乘法分配律：一个数乘和等于分别相乘再相加.\n【详解】乘法交换律是a×b=b×a,乘法结合律是(a×b)×c=a×(b×c),乘法分配律是(a+b)×c=a×c+b×c。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数乘法结合律",
+                                 "整数乘法交换律"
+                             ],
+        "source":  "24-25四年级下·安徽合肥·期中"
+    },
+    {
+        "id":  "258",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "用\"交换位置再加一遍\"的方法验算加法时，运用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "加法交换律",
+        "solution":  "【分析】两个数相加，交换加数的位置，和不变，这叫做加法交换律.\n【详解】例如123+684=807,684+123=807，因此\"交换位置再加一遍\"运用了加法交换律。",
+        "knowledge_points":  [
+                                 "整数加法交换律"
+                             ],
+        "source":  "24-25四年级下·安徽合肥·期中"
+    },
+    {
+        "id":  "259",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "5×4×25×2=(5×2)×(25×4)是运用了()和()。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法交换律;乘法结合律",
+        "solution":  "【分析】先交换2与4,25的位置，再把5与2,25与4分别结合，既用了乘法交换律，又用了乘法结合律.\n【详解】5×4×25×2=5×2×25×4=(5×2)×(25×4)=10×100=1000。",
+        "knowledge_points":  [
+                                 "整数乘法交换律",
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·安徽淮北·期中"
+    },
+    {
+        "id":  "260",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "乒乓球是我国的国球。某乒乓球厂生产了5800个乒乓球，每25个装一袋，每4个装一箱，一共可以装()箱。",
+        "options":  [
+
+                    ],
+        "answer":  "58",
+        "solution":  "【分析】5800个乒乓球每25个装一袋，用5800÷25求袋数，再除以4求箱数。可利用除法性质a/b/c=a/(b×c)简算.\n【详解】5800÷25/4=5800÷(25×4)=5800÷100=58(箱)。",
+        "knowledge_points":  [
+                                 "整数除法的性质"
+                             ],
+        "source":  "24-25四年级下·福建厦门·期中"
+    },
+    {
+        "id":  "261",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小红和小兰分别用不同的乘法运算定律计算25×16.\n小红是这样做的:25×16=25×(10+6)=25×10+25×6;小兰认为自己的方法更简单，她的想法是:25×16=25×4×4=(25×4)×4.\n说一说：小红应用的运算律是(),小兰应用的运算律是()。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法分配律;乘法结合律",
+        "solution":  "【分析】小红把16拆成10+6并分别与25相乘后相加，运用的是乘法分配律;小兰把16拆成4×4并改变乘法结合顺序，运用的是乘法结合律.\n【详解】根据分析，小红应用的是乘法分配律，小兰应用的是乘法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·贵州六盘水·期中"
+    },
+    {
+        "id":  "262",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在方框内填适当的数或运算符号。\n125×32×25=(125×□)×(□×25)\n267-39-61=267-(39□61)\n4000÷25/8=4000÷(□□□)",
+        "options":  [
+
+                    ],
+        "answer":  "8;4;+;25;×;8",
+        "solution":  "【分析】乘法结合律：三个数相乘可改变结合顺序;减法性质：连续减去两个数等于减去它们的和;除法性质：连续除以两个数等于除以它们的积.\n【详解】125×32×25=(125×8)×(4×25);267-39-61=267-(39+61);4000÷25/8=4000÷(25×8)。",
+        "knowledge_points":  [
+                                 "整数除法的性质",
+                                 "整数乘法结合律",
+                                 "整数减法的性质"
+                             ],
+        "source":  "24-25四年级下·福建漳州·期中"
+    },
+    {
+        "id":  "263",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "明明在计算70×(A+8)时，把算式抄成了70×A+8，这样两道计算结果相差()。",
+        "options":  [
+
+                    ],
+        "answer":  "552",
+        "solution":  "【分析】根据乘法分配律,70×(A+8)=70×A+70×8。与错误算式70×A+8相比，差值为70×8-8.\n【详解】70×A+70×8-(70×A+8)=70×8-8=560-8=552。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·福建三明·期中"
+    },
+    {
+        "id":  "264",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "25×8×4×125=(25×4)×(8×125),应用了乘法()律和乘法()律。",
+        "options":  [
+
+                    ],
+        "answer":  "交换;结合",
+        "solution":  "【分析】先交换8和4的位置，再把25与4,8与125分别结合计算.\n【详解】25×8×4×125=(25×4)×(8×125),应用了乘法交换律和乘法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法结合律",
+                                 "整数乘法交换律"
+                             ],
+        "source":  "24-25四年级下·福建三明·期中"
+    },
+    {
+        "id":  "265",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果○-△=50，那么8×○-8×△=();如果A×B=40，那么A×5×B=()。",
+        "options":  [
+
+                    ],
+        "answer":  "400;200",
+        "solution":  "【分析】8×○-8×△可用乘法分配律写成8×(○-△);A×5×B可用乘法交换律写成A×B×5.\n【详解】8×○-8×△=8×(○-△)=8×50=400;A×5×B=A×B×5=40×5=200。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数乘法交换律"
+                             ],
+        "source":  "24-25四年级下·福建福州·期中"
+    },
+    {
+        "id":  "266",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明计算(125+40)×8时，他按125×8+40来算，那么算式的结果比正确的()了()。(填：多或少)",
+        "options":  [
+
+                    ],
+        "answer":  "少;280",
+        "solution":  "【分析】正确计算应是(125+40)×8=125×8+40×8，与125×8+40相比，差在40×8与40之间.\n【详解】40×8-40=320-40=280。所以错误结果比正确结果少280。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·福建福州·期中"
+    },
+    {
+        "id":  "267",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算下面各题，你有什么发现。\n240÷8+160÷8=50;630÷7-210÷7=();(240+160)÷8=();(630-210)÷7=60\n发现：求两个数除以同一个数(0除外)的商的和或差，可以先求这两个数的(),再()这个相同的除数。",
+        "options":  [
+
+                    ],
+        "answer":  "60;50;和或差;除以",
+        "solution":  "【分析】两个数除以同一个数(0除外)的商的和或差，等于这两个数的和或差再除以原来的除数:a/c+b/c=(a+b)/c,a/c-b/c=(a-b)/c.\n【详解】630÷7-210÷7=90-30=60;(240+160)÷8=400÷8=50。可得发现：先求两个数的和或差，再除以这个相同的除数。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·福建龙岩·期中"
+    },
+    {
+        "id":  "268",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在计算27×4×25时，可以先把()和()相乘，得到(),再与()相乘。这是运用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "4;25;100;27;乘法结合律",
+        "solution":  "【分析】27×4×25可根据乘法结合律先算4×25，再与27相乘，使计算简便.\n【详解】先把4和25相乘得100，再与27相乘。这是运用了乘法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·湖北十堰·期中"
+    },
+    {
+        "id":  "269",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "5个工人2天加工360个零件，平均每个工人每天加工()个零件。",
+        "options":  [
+
+                    ],
+        "answer":  "36",
+        "solution":  "【分析】先求每个工人2天加工的零件数:360÷5，再除以2求每天加工数。可用除法性质a/b/c=a/(b×c).\n【详解】360÷5/2=360÷(5×2)=360÷10=36(个)。",
+        "knowledge_points":  [
+                                 "多位数与一位数的连除",
+                                 "整数除法的性质",
+                                 "有具体量的工程问题"
+                             ],
+        "source":  "24-25四年级下·湖北孝感·期中"
+    },
+    {
+        "id":  "270",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "算式102×44=100×44+2×44，这是应用了()律;算式9×125×8=9×(125×8),这是应用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法分配律;乘法结合律",
+        "solution":  "【分析】102×44可把102看作100+2，再用乘法分配律(a+b)×c=a×c+b×c;9×125×8先算125×8更简便，用乘法结合律(a×b)×c=a×(b×c).\n【详解】102×44=(100+2)×44=100×44+2×44;9×125×8=9×(125×8)。分别应用了乘法分配律和乘法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法结合律",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·河北唐山·期中"
+    },
+    {
+        "id":  "271",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "64×64+36×64=(□+□)×64，这道题运用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "64;36;乘法分配律",
+        "solution":  "【分析】64×64+36×64两项有共同因数64，可提取公因数，符合乘法分配律a×c+b×c=(a+b)×c.\n【详解】64×64+36×64=(64+36)×64，运用了乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·河北唐山·期中"
+    },
+    {
+        "id":  "272",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "明明在计算□-(165+22)时，漏看了小括号，算成了□-165+22。他计算的结果与实际结果相比,()(填\"变大了\"\"变小了\"或\"不变\")。",
+        "options":  [
+
+                    ],
+        "answer":  "变大了",
+        "solution":  "【分析】□-(165+22)=□-165-22，误算成□-165+22，相当于把\"减22\"变成\"加22\"，结果会增大.\n【详解】误算结果与正确结果相差44，因此计算结果变大了。",
+        "knowledge_points":  [
+                                 "整数减法的性质"
+                             ],
+        "source":  "24-25四年级下·河南三门峡·期中"
+    },
+    {
+        "id":  "273",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "丁丁和小宇利用周末发放\"防溺水\"宣传单，丁丁每分发放16份，小宇每分发放21份，经过40分正好发放完。丁丁比小宇少发放()份\"防溺水\"宣传单。",
+        "options":  [
+
+                    ],
+        "answer":  "200",
+        "solution":  "【分析】分别求两人40分钟发放量，再求差:21×40-16×40，可用乘法分配律简算.\n【详解】21×40-16×40=(21-16)×40=5×40=200(份)。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·河南新乡·期中"
+    },
+    {
+        "id":  "274",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算88×125时，聪聪这样做:88×125=(80+8)×125=80×125+8×125，他运用的是()律;明明这样做:88×125=(11×8)×125=11×(8×125),他运用的是()律。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法分配律;乘法结合律",
+        "solution":  "【分析】把88拆成80+8并分别乘125，是乘法分配律;把88拆成11×8并调整运算顺序，是乘法结合律.\n【详解】聪聪运用了乘法分配律;明明运用了乘法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法结合律",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·河南漯河·期中"
+    },
+    {
+        "id":  "275",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "天天在计算25×34时，不小心把\"34\"写成了\"43\"，她的计算结果和正确的得数相差()。",
+        "options":  [
+
+                    ],
+        "answer":  "225",
+        "solution":  "【分析】错误与正确之差为25×43-25×34=25×(43-34),用乘法分配律简算.\n【详解】25×(43-34)=25×9=225，所以相差225。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·湖南怀化·期中"
+    },
+    {
+        "id":  "276",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "为了使(128+147)+△的计算简便，聪聪将其改写成128+(147+△)。△可能是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "53",
+        "solution":  "【分析】运用加法结合律后，为便于计算应让147+△凑整百,147+53=200最简便.\n【详解】(128+147)+△=128+(147+53)=128+200=328，所以△可填53。(答案不唯一)",
+        "knowledge_points":  [
+                                 "整数加法结合律"
+                             ],
+        "source":  "24-25四年级下·甘肃天水·期中"
+    },
+    {
+        "id":  "277",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据运算定律在括号里填上合适的数.\n25×25×4=25×();125×(8+4)=125×()+125×()。",
+        "options":  [
+
+                    ],
+        "answer":  "25×4;8;4",
+        "solution":  "【分析】25×25×4可按乘法结合律写成25×(25×4);125×(8+4)按乘法分配律展开为125×8+125×4.\n【详解】应填25×4;8;4。",
+        "knowledge_points":  [
+                                 "整数乘法结合律",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·甘肃天水·期中"
+    },
+    {
+        "id":  "278",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上合适的运算符号，使等号两边相等.\n528-(91-73)=528-91()73;4+4()4()4=7",
+        "options":  [
+
+                    ],
+        "answer":  "+;-;/",
+        "solution":  "【分析】减号前去括号要变号，所以528-(91-73)=528-91+73。后式要得7，可先得8再减1，即4+4-4÷4.\n【详解】应填\"+;-;/\"。",
+        "knowledge_points":  [
+                                 "无括号的运算顺序",
+                                 "整数减法的性质"
+                             ],
+        "source":  "24-25四年级下·河南南阳·期中"
+    },
+    {
+        "id":  "279",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "97×(99+1)=97×()+97×(),这里用的是()律。",
+        "options":  [
+
+                    ],
+        "answer":  "99;1;乘法分配律",
+        "solution":  "【分析】符合乘法分配律a×(b+c)=a×b+a×c.\n【详解】97×(99+1)=97×99+97×1，这里用的是乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·河北廊坊·期中"
+    },
+    {
+        "id":  "280",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在25×36+64*□中的□填(),这个算式就能运用乘法分配律进行简便计算了。",
+        "options":  [
+
+                    ],
+        "answer":  "25",
+        "solution":  "【分析】要用乘法分配律，需两项有相同因数，令64*□中的□为25，可得25×36+64×25=(36+64)×25.\n【详解】□应填25。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·河南信阳·期中"
+    },
+    {
+        "id":  "281",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算125×88时，笑笑和淘气用不同的方式计算，他们分别运用了什么运算律。![](images/2026-03-07%20174249.jpg)\n笑笑运用了()律;淘气运用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法分配律;乘法结合律",
+        "solution":  "【分析】笑笑把88拆成80+8后分别与125相乘，是乘法分配律;淘气把88拆成8×11并先算125×8，是乘法结合律.\n【详解】笑笑运用了乘法分配律;淘气运用了乘法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法结合律",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·陕西延安·期中"
+    },
+    {
+        "id":  "282",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "明明在计算24×205时，用24×(200+5)=24×200+24×5，他运用了()律，请用该方法计算:99×125=()×125=()",
+        "options":  [
+
+                    ],
+        "answer":  "乘法分配律;100-1;100×125-1×125",
+        "solution":  "【分析】24×(200+5)=24×200+24×5体现乘法分配律。99可看成100-1，再用分配律.\n【详解】99×125=(100-1)×125=100×125-1×125。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·河南三门峡·期中"
+    },
+    {
+        "id":  "283",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "超市运来600千克面粉，上午卖出126千克，下午卖出74千克，还剩多少千克面粉。莉莉列式为600-(126+74),她先求(),再求()。",
+        "options":  [
+
+                    ],
+        "answer":  "一共卖出多少千克;还剩多少千克面粉",
+        "solution":  "【分析】先把上午和下午卖出的量相加，得到一共卖出量;再用总量减去卖出量，得到剩余量.\n【详解】600-(126+74)中，先求一共卖出多少千克，再求还剩多少千克面粉。",
+        "knowledge_points":  [
+                                 "整数减法的性质"
+                             ],
+        "source":  "24-25四年级下·河南开封·期中"
+    },
+    {
+        "id":  "284",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在方框内填适当的数.\n□*18+15*□=15×(□+82)\n25×44=25*□*□\n400÷25/4=400÷(□*□)",
+        "options":  [
+
+                    ],
+        "answer":  "15;82;18;4;11;25;4",
+        "solution":  "【分析】第一式按乘法分配律配出同因数15;第二式把44分解为4×11并用结合律;第三式用除法性质a/b/c=a/(b×c).\n【详解】可填:15,82,18;4,11;25,4。",
+        "knowledge_points":  [
+                                 "整数乘法结合律",
+                                 "整数乘法分配律",
+                                 "整数除法的性质"
+                             ],
+        "source":  "24-25四年级下·广东东莞·期中"
+    },
+    {
+        "id":  "285",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "计算25×(4+8)时，先算()法，再算()法，可以应用()律使计算简便。",
+        "options":  [
+
+                    ],
+        "answer":  "加;乘;乘法分配律",
+        "solution":  "【分析】有小括号先算括号内加法，再算括号外乘法;也可用乘法分配律化为25×4+25×8简算.\n【详解】先算加法，再算乘法，可以应用乘法分配律使计算简便。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·甘肃天水·期中"
+    },
+    {
+        "id":  "286",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "游戏公司举办活动，为玩家准备了三种虚拟道具，分别是魔法药剂，能量水晶和传送卷轴，一共875个。其中魔法药剂有234个，能量水晶有366个，那么传送卷轴有()个。",
+        "options":  [
+
+                    ],
+        "answer":  "275",
+        "solution":  "【分析】用总数减去前两种道具数量，可求第三种数量.\n【详解】875-234-366=875-(234+366)=875-600=275(个)。",
+        "knowledge_points":  [
+                                 "整数减法的性质"
+                             ],
+        "source":  "24-25四年级下·陕西西安·期中"
+    },
+    {
+        "id":  "287",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "125×16×8=(125×8)×16运用了()律和()律。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法交换律;乘法结合律",
+        "solution":  "【分析】先交换16与8的位置，再把125与8先结合计算.\n【详解】125×16×8=(125×8)×16，运用了乘法交换律和乘法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法交换律",
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·陕西西安·期中"
+    },
+    {
+        "id":  "288",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "25×7×4=25×4×7运用了()律,125×(4+8)=125×4+125×8运用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法交换律;乘法分配律",
+        "solution":  "【分析】25×7×4中交换了7和4的位置，用乘法交换律;125×(4+8)展开为两项乘积和，用乘法分配律.\n【详解】第一空填乘法交换律，第二空填乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法交换律",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·河南信阳·期中"
+    },
+    {
+        "id":  "289",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果△+□=36，那么△*5+□*5=()。",
+        "options":  [
+
+                    ],
+        "answer":  "180",
+        "solution":  "【分析】△*5+□*5可写为(△+□)×5，再代入△+□=36.\n【详解】(△+□)×5=36×5=180。",
+        "knowledge_points":  [
+                                 "两，三位数与一位数连续进位的乘法",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·河南新乡·期中"
+    },
+    {
+        "id":  "290",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小马虎在计算9×(25+a)时，算成了9×25+a,结果比正确答案少56，正确答案是()。",
+        "options":  [
+
+                    ],
+        "answer":  "288",
+        "solution":  "【分析】正确式9×(25+a)=9×25+9a,错误式为9×25+a,二者相差8a。已知少56，则8a=56,a=7.\n【详解】正确答案=9×(25+7)=9×32=288。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·浙江杭州·期中"
+    },
+    {
+        "id":  "291",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\"\u003e\"\"\u003c\"或\"=\".\n98×5×8()98×(5×8);135÷(9×8)()135÷9×8",
+        "options":  [
+
+                    ],
+        "answer":  "=;\u003c",
+        "solution":  "【分析】98×5×8与98×(5×8)由乘法结合律相等;135÷(9×8)=135÷72，而135÷9×8=15×8=120，前者小于后者.\n【详解】应填\"=;\u003c\"。",
+        "knowledge_points":  [
+                                 "因数中间或末尾是0的乘法",
+                                 "带有小括号的混合运算",
+                                 "整数除法的性质"
+                             ],
+        "source":  "24-25四年级下·浙江杭州·期中"
+    },
+    {
+        "id":  "292",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.4,
+        "question":  "算式1×2×3×4×5=120的积的末尾有1个\"0\";算式1×2×3×4×5×…×20的积的末尾有()个连续的\"0\"。",
+        "options":  [
+
+                    ],
+        "answer":  "4",
+        "solution":  "【分析】积末尾0来自因数2和5的配对。1到20中2的个数充足，只需数5的个数:5,10,15,20共4个5.\n【详解】因此1×2×3×4×5×…×20的积末尾有4个连续的0。",
+        "knowledge_points":  [
+                                 "整数乘法交换律",
+                                 "整数乘法结合律",
+                                 "算式的规律(整数)"
+                             ],
+        "source":  "2025四年级下·全国·专题练习"
+    },
+    {
+        "id":  "293",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "李叔叔和同事骑行四天去相距435千米的某地游玩，第一天骑行了109千米，第二天骑行了135千米，第三天骑行了91千米，按照这几天的行程，第四天要骑行()千米才能按时到达。",
+        "options":  [
+
+                    ],
+        "answer":  "100",
+        "solution":  "【分析】第四天路程=总路程-前三天路程，可利用减法性质简算.\n【详解】435-109-135-91=435-135-109-91=300-(109+91)=300-200=100(千米)。",
+        "knowledge_points":  [
+                                 "两，三位数的不退位减法",
+                                 "整数减法的性质"
+                             ],
+        "source":  "24-25四年级下·甘肃庆阳·期中"
+    },
+    {
+        "id":  "294",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果m-n=4，那么50×m-50×n=();如果38×☆+62×☆=900，那么☆代表的数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "200;9",
+        "solution":  "【分析】50×m-50×n可按分配律写成50×(m-n);38×☆+62×☆可提取☆得(38+62)×☆.\n【详解】50×(m-n)=50×4=200;(38+62)×☆=900,100×☆=900,☆=9。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·甘肃庆阳·期中"
+    },
+    {
+        "id":  "295",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "请你用最简洁的方法表示加法结合律:( )。",
+        "options":  [
+
+                    ],
+        "answer":  "(a+b)+c=a+(b+c)",
+        "solution":  "【分析】三个数相加，先加前两个或先加后两个，和不变.\n【详解】加法结合律可表示为(a+b)+c=a+(b+c)。",
+        "knowledge_points":  [
+                                 "整数加法结合律"
+                             ],
+        "source":  "24-25四年级下·山西晋中·期中"
+    },
+    {
+        "id":  "296",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在横线上填入合适的数，使计算变得简便，在括号里填上运算律.\n45*□+□*99=99×(□+□),运用了().\na+56+44=a+(□+□),运用了()。",
+        "options":  [
+
+                    ],
+        "answer":  "99;55;45;55;乘法分配律;56;44;加法结合律",
+        "solution":  "【分析】第一式要凑共同因数99，写成99×(45+55),用乘法分配律;第二式把56和44先结合，用加法结合律.\n【详解】45×99+55×99=99×(45+55),运用了乘法分配律;a+56+44=a+(56+44),运用了加法结合律。",
+        "knowledge_points":  [
+                                 "整数加法结合律",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·广西柳州·期中"
+    },
+    {
+        "id":  "297",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在做门票收入的预算时，小刚的计算方法是275×64+275×36，小云的计算方法是275×(64+36),()的计算方法更简便，利用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "小云;乘法分配律",
+        "solution":  "【分析】275×(64+36)先算括号内得100，再乘275更简便，且符合a×(b+c)=a×b+a×c.\n【详解】更简便的是小云的方法，利用了乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·广西玉林·期中"
+    },
+    {
+        "id":  "298",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "要使125*□=125×10+125×8成立，那么□内的数应该是(),这是运用了()律;如果a×(b×10)=150，那么a×b=()。",
+        "options":  [
+
+                    ],
+        "answer":  "18;乘法分配律;15",
+        "solution":  "【分析】125×10+125×8=125×(10+8)=125×18，运用乘法分配律。a×(b×10)=(a×b)×10=150，所以a×b=150÷10.\n【详解】□应填18，这是乘法分配律;a×b=15。",
+        "knowledge_points":  [
+                                 "乘，除法的意义和各部分间的关系",
+                                 "整数乘法结合律",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·河南南阳·期中"
+    },
+    {
+        "id":  "299",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "丽丽说：通过学习，我知道了a×b×c=a×(b×c),那么我猜想,a/b/c=a/(b/c)(b,c均不为0).\n(1)丽丽的猜想是()的。(填\"正确\"或\"错误\")\n(2)我能举例子证明:()。",
+        "options":  [
+
+                    ],
+        "answer":  "错误;600÷6/2=50,600÷(6/2)=200",
+        "solution":  "【分析】除法性质是a/b/c=a/(b×c),而不是a/(b/c),所以猜想错误。可用具体数值对比验证.\n【详解】(1)猜想错误.\n(2)例如:600÷6/2=50，而600÷(6/2)=200，二者不相等。",
+        "knowledge_points":  [
+                                 "整数乘法结合律",
+                                 "整数除法的性质"
+                             ],
+        "source":  "24-25四年级下·河南南阳·期中"
+    },
+    {
+        "id":  "300",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "刘老师从家骑共享单车去学校，平均每分钟行199米，用了24分钟到学校。要计算刘老师家距离学校多少米，可以把\"199\"凑成整百数进行简便计算，列式为()×()-(),这里运用了()律，结果是()米。",
+        "options":  [
+
+                    ],
+        "answer":  "200;24;24;乘法分配律;4776",
+        "solution":  "【分析】路程=速度×时间，即199×24。把199看作200-1，可用乘法分配律(a-b)×c=a×c-b×c简算.\n【详解】199×24=(200-1)×24=200×24-24=4800-24=4776(米)。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "基础行程问题"
+                             ],
+        "source":  "24-25四年级下·河南南阳·期中"
+    },
+    {
+        "id":  "301",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "方方在计算(6+□)×15时把小括号去掉了，算出的结果与正确的结果相差()。",
+        "options":  [
+
+                    ],
+        "answer":  "84",
+        "solution":  "【分析】(6+□)×15把小括号去掉后是6+□×15;而原式可按乘法分配律写成6×15+□×15，比较两式差值即可.\n【详解】(6+□)×15=6×15+□×15，与6+□×15相比，都有□×15，相差6×15-6=90-6=84，所以相差84。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·河南南阳·期中"
+    },
+    {
+        "id":  "302",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在计算8×(125×13)=(8×125)×13时，运用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法结合律",
+        "solution":  "【分析】三个数相乘，先乘前两个或先乘后两个，积不变，这是乘法结合律.\n【详解】8×(125×13)=(8×125)×13=1000×13=13000，运用了乘法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·河南南阳·期中"
+    },
+    {
+        "id":  "303",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果△+☆=50，那么16×△+☆×16=()。",
+        "options":  [
+
+                    ],
+        "answer":  "800",
+        "solution":  "【分析】由乘法分配律,16×△+☆×16=(△+☆)×16，再代入△+☆=50.\n【详解】16×△+☆×16=(△+☆)×16=50×16=800。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·河南南阳·期中"
+    },
+    {
+        "id":  "304",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "军军和丁丁在计算27×(□+35)时，都算错了。(1)军军把算式看成了27×□+35，他的计算结果与正确答案相差()。(2)丁丁把算式看成了27×35+□,他算出的结果比正确结果少了78，则正确结果是()。",
+        "options":  [
+
+                    ],
+        "answer":  "910;1026",
+        "solution":  "【分析】原式27×(□+35)=27×□+27×35。分别与错误算式比较即可.\n【详解】(1)27×(□+35)=27×□+945，与27×□+35相差945-35=910.\n(2)27×□+27×35与27×35+□相差26个□,已知差78，得□=78÷26=3。原式=27×(3+35)=27×38=1026。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·湖南怀化·期中"
+    },
+    {
+        "id":  "305",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果a/25/b=a/75(a和b均不为0),那么b=()。",
+        "options":  [
+
+                    ],
+        "answer":  "3",
+        "solution":  "【分析】连续除以两个数等于除以这两个数的积:a/25/b=a/(25×b)。与a/75对应可得25×b=75.\n【详解】b=75÷25=3。",
+        "knowledge_points":  [
+                                 "整数除法的性质",
+                                 "乘，除法的意义和各部分间的关系",
+                                 "含有字母式子的化简与求值"
+                             ],
+        "source":  "24-25四年级下·河北邢台·期中"
+    },
+    {
+        "id":  "306",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据运算律填一填。(263+□)+58=□+(42+□);25×75×4=75×(□*□);(8+10)×125=□*□+□*□。",
+        "options":  [
+
+                    ],
+        "answer":  "42;263;58;25;4;8;125;10;125",
+        "solution":  "【分析】分别运用加法结合律，乘法交换律与结合律，乘法分配律.\n【详解】(263+42)+58=263+(42+58);25×75×4=75×(25×4);(8+10)×125=8×125+10×125。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数乘法结合律",
+                                 "整数乘法交换律",
+                                 "整数加法结合律"
+                             ],
+        "source":  "24-25四年级下·河北保定·期中"
+    },
+    {
+        "id":  "307",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "计算256+344后，可以交换两个数的位置验算，运用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "加法交换律",
+        "solution":  "【分析】交换两个加数的位置，和不变，是加法交换律.\n【详解】计算256+344后可用344+256验算，运用了加法交换律。",
+        "knowledge_points":  [
+                                 "整数加法交换律"
+                             ],
+        "source":  "24-25四年级下·河北保定·期中"
+    },
+    {
+        "id":  "308",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "计算23×4×25时，可以先算(),这是应用了()。",
+        "options":  [
+
+                    ],
+        "answer":  "4×25;乘法结合律",
+        "solution":  "【分析】三个数连乘可先算后两个，利用乘法结合律使计算简便.\n【详解】23×4×25=23×(4×25),可先算4×25，应用了乘法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·河北唐山·期中"
+    },
+    {
+        "id":  "309",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "文文的计算器上的数字键\"9\"坏了，如果想用这个计算器计算出39×13的得数，可以将原来的算式变成()。",
+        "options":  [
+
+                    ],
+        "answer":  "3×13×13",
+        "solution":  "【分析】为避开数字键9，把39分解为3×13，再用乘法结合律改写.\n【详解】39×13=(3×13)×13=3×13×13。",
+        "knowledge_points":  [
+                                 "计算器的初步认识与使用",
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·山东临沂·期中"
+    },
+    {
+        "id":  "310",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "文文要看一本254页的作文书，她第一天看了87页，第二天看了113页，还剩()页没看。",
+        "options":  [
+
+                    ],
+        "answer":  "54",
+        "solution":  "【分析】总页数减去两天阅读页数，计算时可用减法性质先合并减数.\n【详解】254-87-113=254-(87+113)=254-200=54(页)。",
+        "knowledge_points":  [
+                                 "整数减法的性质"
+                             ],
+        "source":  "24-25四年级下·河北邯郸·期中"
+    },
+    {
+        "id":  "311",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果a×c=60,b×c=20，那么(a-b)×c=()。如果1500/a/b=30，那么a×b=()。",
+        "options":  [
+
+                    ],
+        "answer":  "40;50",
+        "solution":  "【分析】(a-b)×c可用乘法分配律写成a×c-b×c;1500/a/b可用除法性质写成1500÷(a×b).\n【详解】(a-b)×c=60-20=40.1500÷(a×b)=30，所以a×b=1500÷30=50。",
+        "knowledge_points":  [
+                                 "整数除法的性质",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·河南南阳·期中"
+    },
+    {
+        "id":  "312",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "花花在计算60×(☆+2)时，把算式抄成了60*☆+2，这样计算出的得数与正确得数相比变()(填\"大\"或\"小\"),两个算式的结果相差()。",
+        "options":  [
+
+                    ],
+        "answer":  "小;118",
+        "solution":  "【分析】正确式60×(☆+2)=60*☆+120，错误式是60*☆+2，因此错误结果更小，差120-2.\n【详解】差值=120-2=118，所以变小，相差118。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·河北唐山·期中"
+    },
+    {
+        "id":  "313",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算17×4×25时，为了计算简便，可以先算()×(),这是根据()律。",
+        "options":  [
+
+                    ],
+        "answer":  "4;25;乘法结合律",
+        "solution":  "【分析】4×25可凑整成100，先算更简便，依据乘法结合律.\n【详解】17×4×25=17×(4×25)=17×100=1700。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·河北唐山·期中"
+    },
+    {
+        "id":  "314",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "东东想用计算器计算\"125×32\"，但计算器上的\"3\"键坏了，请你帮他设计一种方案，用算式表示操作的过程是()。",
+        "options":  [
+
+                    ],
+        "answer":  "125×8×4",
+        "solution":  "【分析】将32分解为8×4，避免输入数字3，再用乘法结合律改写.\n【详解】125×32=125×(8×4)=125×8×4。",
+        "knowledge_points":  [
+                                 "计算器的初步认识与使用",
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·山东菏泽·期中"
+    },
+    {
+        "id":  "315",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "已知a+b=14,a×b=45，那么b×8×a=(),a×15+15×b=()。",
+        "options":  [
+
+                    ],
+        "answer":  "360;210",
+        "solution":  "【分析】b×8×a=(a×b)×8;a×15+15×b=(a+b)×15.\n【详解】b×8×a=45×8=360;a×15+15×b=14×15=210。",
+        "knowledge_points":  [
+                                 "积的变化规律(整数乘法)",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·河北邢台·期中"
+    },
+    {
+        "id":  "316",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "算式\"98×a+98×b\"可以利用乘法分配律进行简便计算，结果是9800，那么a是(),b是()。",
+        "options":  [
+
+                    ],
+        "answer":  "2;98",
+        "solution":  "【分析】按分配律形式应有相同因数，故b取98，式子看作98×a+98×98=98×(a+98).\n【详解】9800=98×100，所以a+98=100,a=2;b=98。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·山东临沂·期中"
+    },
+    {
+        "id":  "317",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果△+○=198,600-△-○的结果是()。",
+        "options":  [
+
+                    ],
+        "answer":  "402",
+        "solution":  "【分析】连续减两个数等于减这两个数的和:600-△-○=600-(△+○).\n【详解】600-198=402。",
+        "knowledge_points":  [
+                                 "整数减法的性质"
+                             ],
+        "source":  "24-25四年级下·山东菏泽·期中"
+    },
+    {
+        "id":  "318",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用简便方法计算(47×25)×4时，为了计算简便，可以先算(),这样计算是根据()律。",
+        "options":  [
+
+                    ],
+        "answer":  "25×4;乘法结合",
+        "solution":  "【分析】先算25×4可凑整，依据乘法结合律.\n【详解】(47×25)×4=47×(25×4)=47×100=4700。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "24-25四年级下·山东菏泽·期中"
+    },
+    {
+        "id":  "319",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "动物园上午售出231张门票，下午比上午多售出169张门票。全天一共售出()张门票，每张门票30元，门票总金额是()元。",
+        "options":  [
+
+                    ],
+        "answer":  "631;18930",
+        "solution":  "【分析】下午票数=231+169;全天=231+下午。总金额=票数×单价.\n【详解】231+231+169=231+(231+169)=631(张);631×30=18930(元)。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "整数加法结合律"
+                             ],
+        "source":  "24-25四年级下·山东济南·期中"
+    },
+    {
+        "id":  "320",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "45×199=45×(□-□);361-68-132=361-(□+□)。",
+        "options":  [
+
+                    ],
+        "answer":  "200;1;68;132",
+        "solution":  "【分析】199可拆成200-1;连续减两个数可写成减去它们的和.\n【详解】45×199=45×(200-1);361-68-132=361-(68+132)。",
+        "knowledge_points":  [
+                                 "整数减法的性质",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·山东菏泽·期中"
+    },
+    {
+        "id":  "321",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在计算199×101时，可以转化为()或()进行简便计算。",
+        "options":  [
+
+                    ],
+        "answer":  "(200-1)×101;199×(100+1)",
+        "solution":  "【分析】可把199看作200-1，或把101看作100+1，均可用乘法分配律简算.\n【详解】199×101=(200-1)×101=200×101-101;也可=199×(100+1)=199×100+199，结果都为20099。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·广东东莞·期中"
+    },
+    {
+        "id":  "322",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在横线上填上合适的数。(1)a+(30+8)=(□+□)+8;(2)□+82=□+18;(3)125×(4+8)=□*□+□*□;(4)45*□=99*□。",
+        "options":  [
+
+                    ],
+        "answer":  "a;30;18;82;125;4;125;8;99;45",
+        "solution":  "【分析】分别对应加法结合律，加法交换律，乘法分配律，乘法交换律.\n【详解】(1)a+(30+8)=(a+30)+8;(2)18+82=82+18;(3)125×(4+8)=125×4+125×8;(4)45×99=99×45。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数乘法交换律",
+                                 "整数加法结合律",
+                                 "整数加法交换律"
+                             ],
+        "source":  "24-25四年级下·广东东莞·期中"
+    },
+    {
+        "id":  "323",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果□-☆=10，那么47*□-47*☆=()。",
+        "options":  [
+
+                    ],
+        "answer":  "470",
+        "solution":  "【分析】用乘法分配律逆运算:47*□-47*☆=47×(□-☆).\n【详解】47×(□-☆)=47×10=470。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·重庆大渡口·期中"
+    },
+    {
+        "id":  "324",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在○里填上合适的运算符号，在□填上合适的数。87+268+132=87+(□○□);102×64-64×2=□*(□○□);207-35-65=207-(□○□)。",
+        "options":  [
+
+                    ],
+        "answer":  "268;132;+;64;102;2;-;35;65;+",
+        "solution":  "【分析】分别运用加法结合律，乘法分配律，减法性质.\n【详解】87+268+132=87+(268+132);102×64-64×2=64×(102-2);207-35-65=207-(35+65)。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数减法的性质",
+                                 "整数加法结合律"
+                             ],
+        "source":  "23-24四年级下·重庆大渡口·期中"
+    },
+    {
+        "id":  "325",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "25×(40+4)=25×40+25×4，运用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法分配律",
+        "solution":  "【分析】一个数与两个数的和相乘，等于分别相乘再相加，是乘法分配律.\n【详解】25×(40+4)=25×40+25×4，运用了乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·湖南岳阳·期中"
+    },
+    {
+        "id":  "326",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号内填上适当的数。615+16+184=615+(□+□);1200÷25/4=1200÷□;368-187-13=368-(□+□);125×88=125×□+125×□=□。",
+        "options":  [
+
+                    ],
+        "answer":  "16;184;100;187;13;80;8;11000",
+        "solution":  "【分析】依次用加法结合律，除法性质，减法性质，乘法分配律.\n【详解】615+16+184=615+(16+184);1200÷25/4=1200÷(25×4)=1200÷100;368-187-13=368-(187+13);125×88=125×(80+8)=125×80+125×8=11000。",
+        "knowledge_points":  [
+                                 "整数除法的性质",
+                                 "整数乘法分配律",
+                                 "整数乘法结合律",
+                                 "整数减法的性质"
+                             ],
+        "source":  "24-25四年级下·河南南阳·期中"
+    },
+    {
+        "id":  "327",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据运算定律填空。36×25=(□×□)×9=900;125×(8+□)=125×8+125×4。",
+        "options":  [
+
+                    ],
+        "answer":  "4;25;4",
+        "solution":  "【分析】36可拆成4×9并结合25先算;第二问按分配律逆向识别括号内数.\n【详解】36×25=(4×25)×9=900;125×(8+4)=125×8+125×4。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数乘法结合律",
+                                 "整数乘法交换律"
+                             ],
+        "source":  "24-25四年级下·湖南岳阳·期中"
+    },
+    {
+        "id":  "328",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算33×125×8时，为了使计算简便，可以先算(),这样计算是根据()。",
+        "options":  [
+
+                    ],
+        "answer":  "125×8;乘法结合律",
+        "solution":  "【分析】先算125×8可凑整为1000，依据乘法结合律.\n【详解】33×125×8=33×(125×8),可先算125×8，运用乘法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "23-24四年级下·河南许昌·期中"
+    },
+    {
+        "id":  "329",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "35×125×8=35×(125×8),这是运用了乘法()律。",
+        "options":  [
+
+                    ],
+        "answer":  "结合",
+        "solution":  "【分析】三个数相乘，改变先算顺序而积不变，属于乘法结合律.\n【详解】35×125×8与35×(125×8)仅改变结合方式，运用了乘法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "23-24四年级下·湖北荆州·期末"
+    },
+    {
+        "id":  "330",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "1000-424-176=1000-(□○□);69×a-59×a=(□○□)×a。",
+        "options":  [
+
+                    ],
+        "answer":  "424;+;176;69;-;59",
+        "solution":  "【分析】第一空用减法性质a-b-c=a-(b+c);第二空用乘法分配律逆运算.\n【详解】1000-424-176=1000-(424+176);69×a-59×a=(69-59)×a。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数减法的性质"
+                             ],
+        "source":  "23-24四年级下·河南濮阳·期末"
+    },
+    {
+        "id":  "331",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "125+456+175+144=(125+175)+(456+144)运用的是()律和()律。",
+        "options":  [
+
+                    ],
+        "answer":  "加法交换律;加法结合律",
+        "solution":  "【分析】先交换加数位置把可凑整的数放一起，再改变结合顺序，分别是交换律和结合律.\n【详解】125+456+175+144→125+175+456+144→(125+175)+(456+144)。",
+        "knowledge_points":  [
+                                 "整数加法结合律",
+                                 "整数加法交换律"
+                             ],
+        "source":  "23-24四年级下·河北衡水·期末"
+    },
+    {
+        "id":  "332",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在横线填上合适的数，再写出所应用的运算律。(1)125×(8×□)=(125×□)×23()律;(2)65×(100+10)=65×□+65×□()律。",
+        "options":  [
+
+                    ],
+        "answer":  "23;8;乘法结合律;100;10;乘法分配律",
+        "solution":  "【分析】(1)按(a×b)×c=a×(b×c)调整因数结合方式;(2)按(a+b)×c=a×c+b×c展开.\n【详解】(1)125×(8×23)=(125×8)×23，乘法结合律;(2)65×(100+10)=65×100+65×10，乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数乘法结合律"
+                             ],
+        "source":  "23-24四年级下·河北沧州·期末"
+    },
+    {
+        "id":  "333",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "为改善环境，红兴村修建了垃圾清理站。第一季度清理47吨生活垃圾，第二季度清理66吨，第三季度清理53吨，第四季度清理34吨，这一年一共清理()吨生活垃圾。",
+        "options":  [
+
+                    ],
+        "answer":  "200",
+        "solution":  "【分析】全年总量等于四个季度总和，可用交换律，结合律凑整计算.\n【详解】47+66+53+34=(47+53)+(66+34)=100+100=200(吨)。",
+        "knowledge_points":  [
+                                 "整数加法结合律",
+                                 "整数加法交换律",
+                                 "两，三位数的连加运算"
+                             ],
+        "source":  "23-24四年级下·甘肃武威·期末"
+    },
+    {
+        "id":  "334",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用计算器算\"680÷40\"时，发现数字键\"4\"坏了，如果还用这个计算器，你会怎么样计算。请写出算式()。",
+        "options":  [
+
+                    ],
+        "answer":  "680÷5/8",
+        "solution":  "【分析】把40分成5×8，再利用除法的性质：一个数连续除以两个数，等于除以这两个数的乘积.\n【详解】680÷40=680÷(5×8)=680÷5/8，所以可写成680÷5/8。",
+        "knowledge_points":  [
+                                 "整数除法的性质"
+                             ],
+        "source":  "23-24四年级下·河南焦作·期末"
+    },
+    {
+        "id":  "335",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "a-b-c=a-();(a+b)×c=()+()。",
+        "options":  [
+
+                    ],
+        "answer":  "b+c;a×c;b×c",
+        "solution":  "【分析】根据减法性质和乘法分配律直接填空.\n【详解】a-b-c=a-(b+c);(a+b)×c=a×c+b×c。",
+        "knowledge_points":  [
+                                 "整数减法的性质",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·山西长治·期末"
+    },
+    {
+        "id":  "336",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在计算25×88时，小旭是这样想的:25×88=(25×4)×22=100×22=2200，这里运用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法结合律",
+        "solution":  "【分析】把88拆成4×22后，先算25×4，再乘22，改变了连乘的结合顺序.\n【详解】25×88=25×(4×22)=(25×4)×22=2200，运用了乘法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "23-24四年级下·贵州黔西·期末"
+    },
+    {
+        "id":  "337",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据\u0027水果店运进38箱苹果和22箱梨，每箱重量都是15千克\u0027可知:(38+22)×15表示()。(38-22)×15表示()。",
+        "options":  [
+
+                    ],
+        "answer":  "水果店运进苹果和梨一共多少千克;水果店运进的苹果比梨多多少千克",
+        "solution":  "【分析】(38+22)表示总箱数，乘15得总质量;(38-22)表示多出的箱数，乘15得多出的质量.\n【详解】(38+22)×15表示苹果和梨一共多少千克;(38-22)×15表示苹果比梨多多少千克。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·河北沧州·期末"
+    },
+    {
+        "id":  "338",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "根据589+320=909，可知:909-()=320;320+589=()。",
+        "options":  [
+
+                    ],
+        "answer":  "589;909",
+        "solution":  "【分析】由加减法关系：一个加数=和-另一个加数;并用加法交换律.\n【详解】909-589=320;320+589=589+320=909。",
+        "knowledge_points":  [
+                                 "整数加法交换律",
+                                 "加，减法的意义和各部分间的关系"
+                             ],
+        "source":  "23-24四年级下·河北沧州·期末"
+    },
+    {
+        "id":  "339",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "m×16+16×3=(m+□)*□。",
+        "options":  [
+
+                    ],
+        "answer":  "3;16",
+        "solution":  "【分析】提取公因数16，使用乘法分配律逆运算.\n【详解】m×16+16×3=(m+3)×16。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·河北唐山·期末"
+    },
+    {
+        "id":  "340",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明把56×(2+□)错算成56*□+2，他得到的结果与正确的结果相差()。",
+        "options":  [
+
+                    ],
+        "answer":  "110",
+        "solution":  "【分析】正确式展开为56×2+56*□,错误式是56*□+2，相差56×2-2.\n【详解】56×(2+□)-(56*□+2)=112-2=110。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "2023四年级·湖南·竞赛"
+    },
+    {
+        "id":  "341",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在计算62×25×4时，可运用()使计算简便，用字母a,b,c表示为()。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法结合律;a×b×c=a×(b×c)",
+        "solution":  "【分析】先算25×4=100更简便，对应乘法结合律.\n【详解】62×25×4=62×(25×4)=62×100，运用乘法结合律，字母式为a×b×c=a×(b×c)。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "23-24四年级下·云南楚雄·期中"
+    },
+    {
+        "id":  "342",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "99×67+67=67×(99+1)应用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法分配律",
+        "solution":  "【分析】把67看作67×1，得到99×67+67×1，再合并为(99+1)×67.\n【详解】99×67+67=(99+1)×67，应用了乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级上·安徽芜湖·期末"
+    },
+    {
+        "id":  "343",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "周村烧饼是山东省名优特产之一。一箱周村烧饼有12盒，每盒售价4元。买25箱周村烧饼一共需要()元。",
+        "options":  [
+
+                    ],
+        "answer":  "1200",
+        "solution":  "【分析】先算每箱价钱12×4，再乘25箱;也可用结合律先算4×25.\n【详解】12×4×25=12×(4×25)=12×100=1200(元)。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "23-24四年级下·海南省直辖县级单位·期中"
+    },
+    {
+        "id":  "344",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "45+13=13+(),运用了加法()律，用字母表示是()。",
+        "options":  [
+
+                    ],
+        "answer":  "45;交换;a+b=b+a",
+        "solution":  "【分析】交换两个加数位置，和不变，是加法交换律.\n【详解】45+13=13+45，运用了加法交换律，字母表示为a+b=b+a。",
+        "knowledge_points":  [
+                                 "整数加法交换律"
+                             ],
+        "source":  "23-24四年级下·新疆阿克苏·期中"
+    },
+    {
+        "id":  "345",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "游泳池长50米，小力每次都坚持游9个来回，他每次游()米。",
+        "options":  [
+
+                    ],
+        "answer":  "900",
+        "solution":  "【分析】1个来回是2个50米,9个来回是9×2个50米.\n【详解】9×2×50=9×(2×50)=9×100=900(米)。",
+        "knowledge_points":  [
+                                 "整十，整百，整千数与一位数的乘法",
+                                 "整数乘法结合律"
+                             ],
+        "source":  "23-24四年级下·广西百色·期中"
+    },
+    {
+        "id":  "346",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "店庆期间，原价4999元的一部5G手机，降价578元，刘阿姨有贵宾卡还能再优惠222元，最后刘阿姨花()元买到这部手机。",
+        "options":  [
+
+                    ],
+        "answer":  "4199",
+        "solution":  "【分析】总价减去两次优惠，可用减法性质先合并减数.\n【详解】4999-578-222=4999-(578+222)=4999-800=4199(元)。",
+        "knowledge_points":  [
+                                 "整数减法的性质"
+                             ],
+        "source":  "23-24四年级下·广西百色·期中"
+    },
+    {
+        "id":  "347",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "(25×5)×2=25×(5×2)这里运用了(),用字母表示是()。(4+2)×25=4×25+2×25这里运用了(),用字母表示是()。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法结合律;(a×b)×c=a×(b×c);乘法分配律;(a+b)×c=a×c+b×c",
+        "solution":  "【分析】第一式改变了乘法的结合顺序;第二式是和乘一个数拆成两个乘积相加.\n【详解】(25×5)×2=25×(5×2)用乘法结合律;(4+2)×25=4×25+2×25用乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数乘法结合律"
+                             ],
+        "source":  "23-24四年级下·山东菏泽·期中"
+    },
+    {
+        "id":  "348",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "为了让学生们开拓视野，学到更多的知识，图书馆购进了科技书，科幻书和故事书共875本。其中科技书有234本，故事书有366本。科幻书有()本。",
+        "options":  [
+
+                    ],
+        "answer":  "275",
+        "solution":  "【分析】总本数减去科技书和故事书的本数，利用减法性质简算.\n【详解】875-234-366=875-(234+366)=875-600=275(本)。",
+        "knowledge_points":  [
+                                 "整数减法的性质",
+                                 "两，三位数的加减混合运算"
+                             ],
+        "source":  "23-24四年级下·海南省直辖县级单位·期中"
+    },
+    {
+        "id":  "349",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "简便计算时,16×25×4=□*(□*□),这是运用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "16;25;4;乘法结合律",
+        "solution":  "【分析】连乘中把后两个因数先结合，符合乘法结合律.\n【详解】16×25×4=16×(25×4),运用了乘法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "23-24四年级下·陕西西安·期中"
+    },
+    {
+        "id":  "350",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据运算定律，在横线上填上适当的数。(1)125×(8×□)=(125×□)×13;(2)172×3+172×5+172×2=172×□。",
+        "options":  [
+
+                    ],
+        "answer":  "13;8;10",
+        "solution":  "【分析】(1)用乘法结合律;(2)提取公因数172，用分配律逆运算.\n【详解】(1)125×(8×13)=(125×8)×13;(2)172×3+172×5+172×2=172×(3+5+2)=172×10。",
+        "knowledge_points":  [
+                                 "整数乘法结合律",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·河北唐山·期中"
+    },
+    {
+        "id":  "351",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在○里和□里填写相应的运算符号和数。(1)25+58+75=(□+□)○□;(2)14×53=53×□;(3)49×68+68=(□+□)○□。",
+        "options":  [
+
+                    ],
+        "answer":  "25;75;+;58;14;49;1;×;68",
+        "solution":  "【分析】分别对应加法交换结合，乘法交换，乘法分配律逆运算.\n【详解】(1)25+58+75=(25+75)+58;(2)14×53=53×14;(3)49×68+68=(49+1)×68。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数乘法交换律",
+                                 "整数加法交换律"
+                             ],
+        "source":  "23-24四年级下·陕西西安·期中"
+    },
+    {
+        "id":  "352",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在方框内填上合适的数。(1)36+859+64=36+□+859;(2)12×23+23×88=23×(□+88)。",
+        "options":  [
+
+                    ],
+        "answer":  "64;12",
+        "solution":  "【分析】(1)交换加数位置便于先凑整;(2)提取公因数23.\n【详解】(1)36+859+64=36+64+859;(2)12×23+23×88=23×(12+88)。",
+        "knowledge_points":  [
+                                 "整数加法交换律",
+                                 "整数加法结合律",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·贵州六盘水·期中"
+    },
+    {
+        "id":  "353",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "来福超市运来了3种蔬菜，其中土豆有157千克，南瓜有206千克，胡萝卜有143千克，来福超市一共运来了()千克蔬菜。",
+        "options":  [
+
+                    ],
+        "answer":  "506",
+        "solution":  "【分析】把3种蔬菜重量相加，可先将157与143凑整.\n【详解】157+206+143=157+143+206=300+206=506(千克)。",
+        "knowledge_points":  [
+                                 "整数加法交换律",
+                                 "两，三位数的连加运算"
+                             ],
+        "source":  "23-24四年级下·陕西西安·期中"
+    },
+    {
+        "id":  "354",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "简便计算，在方框填数字。26+26×99=26×(□+□);365-98=365-□+□;25×32=25×□×□;8×a+a×5=(□+□)×□。",
+        "options":  [
+
+                    ],
+        "answer":  "1;99;100;2;4;8;8;5;a",
+        "solution":  "【分析】依次用分配律逆运算，凑整减法，分解32，分配律逆运算.\n【详解】26+26×99=26×(1+99);365-98=365-100+2;25×32=25×4×8;8×a+a×5=(8+5)×a。",
+        "knowledge_points":  [
+                                 "整数减法的性质",
+                                 "整数乘法结合律",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·山东菏泽·阶段练习"
+    },
+    {
+        "id":  "355",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填\"\u003e\"\"\u003c\"或\"=\"。15×(60+4)()15×60+4;8×(19-7×2)()8×19-7×2;32×25()25×4×8;300+240÷30×2()(300+240)÷(30×2)。",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;\u003c;=;\u003e",
+        "solution":  "【分析】按运算顺序分别计算或比较，第三个利用32=4×8.\n【详解】15×(60+4)=960,15×60+4=904，所以\u003e;8×(19-7×2)=40,8×19-7×2=138，所以\u003c;32×25=(4×8)×25=25×4×8，所以=;300+240÷30×2=316,(300+240)÷(30×2)=9，所以\u003e。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "有括号的运算顺序",
+                                 "无括号的运算顺序",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "23-24四年级下·河南南阳·阶段练习"
+    },
+    {
+        "id":  "356",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "根据运算律，在下面的方框内填适当的数。(1)28×75=75*□;(2)25×17×8=17×(□*□);(3)5×(20+42)=5*□+□*□。",
+        "options":  [
+
+                    ],
+        "answer":  "28;25;8;20;5;42",
+        "solution":  "【分析】(1)乘法交换律;(2)交换并结合;(3)乘法分配律展开.\n【详解】28×75=75×28;25×17×8=17×(25×8);5×(20+42)=5×20+5×42。",
+        "knowledge_points":  [
+                                 "整数乘法交换律",
+                                 "整数乘法结合律",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·河南南阳·阶段练习"
+    },
+    {
+        "id":  "357",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "服装店出售一款套装，每件上衣275元，每条裤子185元，卖14套这样的套装需要收多少钱。列式为()。",
+        "options":  [
+
+                    ],
+        "answer":  "(275+185)×14",
+        "solution":  "【分析】一套价格是275+185，再乘14套;也可分开乘后相加.\n【详解】275×14+185×14=(275+185)×14=460×14=6440(元)。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·河南南阳·开学考试"
+    },
+    {
+        "id":  "358",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "三个数相加，先把前两个数相加，或者先把后两个数相加，和不变，这叫做()。用字母表示是()。",
+        "options":  [
+
+                    ],
+        "answer":  "加法结合律;(a+b)+c=a+(b+c)",
+        "solution":  "【分析】题干是加法结合律的定义.\n【详解】这叫加法结合律，字母表示为a+b+c=a+(b+c)。",
+        "knowledge_points":  [
+                                 "整数加法结合律"
+                             ],
+        "source":  "23-24四年级下·山东菏泽·阶段练习"
+    },
+    {
+        "id":  "359",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\"\u003e\"\"\u003c\"或\"=\"。540÷5/6()540÷(5×6);18+90÷18()(18+90)÷18。",
+        "options":  [
+
+                    ],
+        "answer":  "=;\u003e",
+        "solution":  "【分析】第一式用除法性质;第二式分别计算比较大小.\n【详解】540÷5/6=540÷(5×6),填=;18+90÷18=23,(18+90)÷18=6，所以填\u003e。",
+        "knowledge_points":  [
+                                 "除加，除减混合运算",
+                                 "带有小括号的混合运算",
+                                 "整数除法的性质"
+                             ],
+        "source":  "23-24四年级下·广东广州·阶段练习"
+    },
+    {
+        "id":  "360",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "\"草船借箭\"的故事为同学们所熟知，足智多谋的诸葛亮利用曹操的多疑，用草船诱敌，\"借\"了十万余支箭。假如诸葛亮一共调了25条船，在每条船上都安排了115个草垛。等到他们满载而归时，平均每个草垛上有40支箭，那么诸葛亮一共\"借\"到了()支箭。",
+        "options":  [
+
+                    ],
+        "answer":  "115000",
+        "solution":  "【分析】总箭数=船数×每船草垛数×每草垛箭数，连乘可交换因数简算.\n【详解】25×115×40=25×40×115=1000×115=115000(支)。",
+        "knowledge_points":  [
+                                 "整数乘法交换律"
+                             ],
+        "source":  "23-24四年级下·河南南阳·阶段练习"
+    },
+    {
+        "id":  "361",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "魾鱼馆第一天卖出门票364张，第二天上午卖出门票173张，下午卖出门票236张。两天一共卖出门票()张。",
+        "options":  [
+
+                    ],
+        "answer":  "773",
+        "solution":  "【分析】总张数是364+173+236，可先把364与236凑整.\n【详解】364+173+236=(364+236)+173=600+173=773(张)。",
+        "knowledge_points":  [
+                                 "整数加法结合律",
+                                 "整数加法交换律"
+                             ],
+        "source":  "23-24四年级下·山东菏泽·期中"
+    },
+    {
+        "id":  "362",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用字母表示下列运算定律。加法交换律()。乘法结合律()。",
+        "options":  [
+
+                    ],
+        "answer":  "a+b=b+a;a×b×c=a×(b×c)",
+        "solution":  "【分析】按运算定律的标准字母形式填写.\n【详解】加法交换律:a+b=b+a;乘法结合律:a×b×c=a×(b×c)。",
+        "knowledge_points":  [
+                                 "整数乘法结合律",
+                                 "整数加法交换律"
+                             ],
+        "source":  "23-24四年级下·山东菏泽·阶段练习"
+    },
+    {
+        "id":  "363",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "(42+103)+17=42+(□+□),这里运用了()。",
+        "options":  [
+
+                    ],
+        "answer":  "103;17;加法结合律",
+        "solution":  "【分析】三个数相加，改变括号位置而和不变，是加法结合律.\n【详解】(42+103)+17=42+(103+17),运用了加法结合律。",
+        "knowledge_points":  [
+                                 "整数加法结合律"
+                             ],
+        "source":  "23-24四年级下·吉林白山·阶段练习"
+    },
+    {
+        "id":  "364",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "☆+□=□+☆,这是利用了加法()律，用字母表示为:a+b=()+()。",
+        "options":  [
+
+                    ],
+        "answer":  "交换;b;a",
+        "solution":  "【分析】交换两个加数的位置，和不变，是加法交换律.\n【详解】☆+□=□+☆,对应a+b=b+a。",
+        "knowledge_points":  [
+                                 "整数加法交换律"
+                             ],
+        "source":  "23-24四年级下·吉林白山·阶段练习"
+    },
+    {
+        "id":  "365",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上合适的数。306×2+306×3+306×5=306×();17×225-4×225-3×225=()×225;84×7+5×84-84×2=()×()。",
+        "options":  [
+
+                    ],
+        "answer":  "10或2+3+5;10或17-4-3;84;10",
+        "solution":  "【分析】三式都可提取公因数，使用乘法分配律逆运算.\n【详解】306×2+306×3+306×5=306×(2+3+5)=306×10;17×225-4×225-3×225=(17-4-3)×225=10×225;84×7+5×84-84×2=84×(7+5-2)=84×10。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·海南海口·单元测试"
+    },
+    {
+        "id":  "366",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "44×25=40×25+4×25这是根据()律简便计算。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法分配律",
+        "solution":  "【分析】44拆成40+4，和乘一个数拆成两个乘积之和.\n【详解】44×25=(40+4)×25=40×25+4×25，依据乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·海南海口·单元测试"
+    },
+    {
+        "id":  "367",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "用简便方法计算397+286+14，要先算(),这样计算是根据()律。",
+        "options":  [
+
+                    ],
+        "answer":  "286+14;加法结合律",
+        "solution":  "【分析】286+14可凑整百，先算更简便，体现加法结合律.\n【详解】397+286+14=397+(286+14)=397+300=697，所以先算286+14，依据加法结合律。",
+        "knowledge_points":  [
+                                 "整数加法结合律"
+                             ],
+        "source":  "24-25四年级下·海南海口·单元测试"
+    },
+    {
+        "id":  "368",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用自己喜欢的方式表示下面运算定律：乘法交换律();加法结合律();乘法分配律()。",
+        "options":  [
+
+                    ],
+        "answer":  "a×b=b×a;a+b+c=a+(b+c);(a+b)×c=a×c+b×c",
+        "solution":  "【分析】按三条运算律的标准字母形式填写.\n【详解】乘法交换律:a×b=b×a;加法结合律:a+b+c=a+(b+c);乘法分配律:(a+b)×c=a×c+b×c。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数乘法交换律",
+                                 "整数加法结合律"
+                             ],
+        "source":  "24-25四年级下·海南海口·单元测试"
+    },
+    {
+        "id":  "369",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\"\u003e\"\"\u003c\"或\"=\"。480÷24/2()480÷(24÷2);102×6+4()102×(6+4);(65-25)×4()65×4-25×4;(18+27)×3()18×3+27;496-120-230()496-(120+230);192-36+64()192-(36+64)。",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003c;=;\u003e;=;\u003e",
+        "solution":  "【分析】分别计算左右两边并比较;其中第三，第五可直接用分配律和减法性质判断.\n【详解】480÷24/2=10,480÷(24/2)=40，所以\u003c;102×6+4=616,102×(6+4)=1020，所以\u003c;(65-25)×4=65×4-25×4，所以=;(18+27)×3=135,18×3+27=81，所以\u003e;496-120-230=496-(120+230),所以=;192-36+64=220,192-(36+64)=92，所以\u003e。",
+        "knowledge_points":  [
+                                 "两，三位数的不退位减法",
+                                 "乘加，乘减混合运算",
+                                 "带有小括号的混合运算",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·海南海口·单元测试"
+    },
+    {
+        "id":  "370",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "明明在计算(27+38)×3时，漏看了小括号，错算成了27+38×3，这样计算与正确结果相差()。",
+        "options":  [
+
+                    ],
+        "answer":  "54",
+        "solution":  "【分析】根据乘法分配律,(27+38)×3=27×3+38×3。错算成27+38×3，相当于少算了2个27.\n【详解】相差=27×(3-1)=27×2=54。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·广东江门·开学考试"
+    },
+    {
+        "id":  "371",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "下面的等式应用了哪个运算定律。填一填。(1)15×38+15×12=15×(38+12)()律;(2)254+183=183+254()律;(3)25×(4×8)=(25×4)×8()律;(4)389+127+73=389+(127+73)()律;(5)65×124=124×65()律。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法分配律;加法交换律;乘法结合律;加法结合律;乘法交换律",
+        "solution":  "【分析】按各式特征判断：和乘法展开/合并为分配律;交换加数或因数位置为交换律;改变三个数加法或乘法的结合顺序为结合律.\n【详解】对应依次为：乘法分配律，加法交换律，乘法结合律，加法结合律，乘法交换律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数乘法结合律",
+                                 "整数乘法交换律",
+                                 "整数加法结合律"
+                             ],
+        "source":  "23-24四年级下·山东菏泽·阶段练习"
+    },
+    {
+        "id":  "372",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在\"()\"里填上\"\u003e\"\"\u003c\"或\"=\"。96-25-45()96-(25+45);679-291-109()679-(291-109);56×207+3()56×(207+3);48÷12+48÷4()48/(12+4)。",
+        "options":  [
+
+                    ],
+        "answer":  "=;\u003c;\u003c;\u003e",
+        "solution":  "【分析】第1题用减法性质可直接判断相等;其余分别计算两边结果比较大小.\n【详解】96-25-45=96-(25+45);679-291-109=279,679-(291-109)=497，所以\u003c;56×207+3=11595,56×(207+3)=11760，所以\u003c;48÷12+48÷4=16,48÷(12+4)=3，所以\u003e。",
+        "knowledge_points":  [
+                                 "整数减法的性质",
+                                 "带有小括号的混合运算",
+                                 "除加，除减混合运算"
+                             ],
+        "source":  "23-24四年级下·山东菏泽·阶段练习"
+    },
+    {
+        "id":  "373",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "131×17+51×123=()。",
+        "options":  [
+
+                    ],
+        "answer":  "8500",
+        "solution":  "【分析】把51看成17×3，提取公因数17，再计算.\n【详解】131×17+51×123=131×17+17×3×123=(131+3×123)×17=(131+369)×17=500×17=8500。",
+        "knowledge_points":  [
+                                 "乘加，乘减混合运算",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25六年级下·北京海淀·开学考试"
+    },
+    {
+        "id":  "374",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "要使\"3□5×5=1500+25\",□里应填()。",
+        "options":  [
+
+                    ],
+        "answer":  "0",
+        "solution":  "【分析】将三位数拆成(300+十位+5)后乘5，与1500+25对照可知十位贡献应为0.\n【详解】3□5×5=(300+□+5)×5=1500+□×5+25。要等于1500+25，则□×5=0，所以□=0。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25三年级上·浙江杭州·期末"
+    },
+    {
+        "id":  "375",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.4,
+        "question":  "(7×1-3×1)+(7×3-3×2)+(7×5-3×3)+。+(7×49-3×25)=()。",
+        "options":  [
+
+                    ],
+        "answer":  "3400",
+        "solution":  "【分析】按项整理为7×(1+3+。+49)-3×(1+2+。+25),再分别求和.\n【详解】原式=7×(1+3+。+49)-3×(1+2+。+25)=7×((1+49)×25÷2)-3×((1+25)×25÷2)=7×625-3×325=4375-975=3400。",
+        "knowledge_points":  [
+                                 "整数减法的性质",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25六年级下·北京海淀·开学考试"
+    },
+    {
+        "id":  "376",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "308×6=()+48;125×8=()×8+25×8;148×6-48×6=()。",
+        "options":  [
+
+                    ],
+        "answer":  "1800;100;600",
+        "solution":  "【分析】分别用乘法分配律拆分与合并.\n【详解】308×6=(300+8)×6=1800+48;125×8=(100+25)×8=100×8+25×8;148×6-48×6=(148-48)×6=100×6=600。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25三年级上·浙江·期末"
+    },
+    {
+        "id":  "377",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "365×283÷29+365×297÷29=()。",
+        "options":  [
+
+                    ],
+        "answer":  "7300",
+        "solution":  "【分析】先把两项都改写成365÷29乘一个数，再提取公因数.\n【详解】365×283÷29+365×297÷29=365÷29×283+365÷29×297=365÷29×(283+297)=365÷29×580=20×365=7300。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "2021五年级·全国·竞赛"
+    },
+    {
+        "id":  "378",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "26×15×20=26×(15×20),运用了乘法()律。25×44=25×()×11=1100。",
+        "options":  [
+
+                    ],
+        "answer":  "结合;4",
+        "solution":  "【分析】第1空是改变连乘结合顺序;第2空把44分解成4×11后便于计算.\n【详解】26×15×20=26×(15×20)用乘法结合律;25×44=25×4×11=100×11=1100。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "22-23四年级下·四川宜宾·期末"
+    },
+    {
+        "id":  "379",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据运算律，在□上填数，在○填运算符号。781-115-85=781-(□○□);(40+4)×25=(□×□)○(□×□);350÷14/5=350÷(□○□)。",
+        "options":  [
+
+                    ],
+        "answer":  "115;+;85;40;×;25;+;4;×;25;14;×;5",
+        "solution":  "【分析】分别应用减法性质，乘法分配律，除法性质.\n【详解】781-115-85=781-(115+85);(40+4)×25=(40×25)+(4×25);350÷14/5=350÷(14×5)。",
+        "knowledge_points":  [
+                                 "整数除法的性质",
+                                 "整数乘法分配律",
+                                 "整数减法的性质"
+                             ],
+        "source":  "23-24四年级下·四川遂宁·期末"
+    },
+    {
+        "id":  "380",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "计算89+75+25时，可运用加法()律，先算()+()。这个运算律用字母表示可以是(a+b)+c=()。",
+        "options":  [
+
+                    ],
+        "answer":  "结合;75;25;a+(b+c)",
+        "solution":  "【分析】75+25先凑成100，体现加法结合律.\n【详解】89+75+25=89+(75+25)=189，所以用加法结合律，先算75+25;字母表示为(a+b)+c=a+(b+c)。",
+        "knowledge_points":  [
+                                 "用字母表示运算定律及计算公式",
+                                 "整数加法结合律"
+                             ],
+        "source":  "23-24四年级下·四川遂宁·期末"
+    },
+    {
+        "id":  "381",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "33×8×125=33×(8×125),运用了(),该运算律用字母表示是()。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法结合律;a×b×c=a×(b×c)",
+        "solution":  "【分析】三个数连乘，改变括号位置积不变，是乘法结合律.\n【详解】33×8×125=33×(8×125),运用了乘法结合律，字母表示a×b×c=a×(b×c)。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "22-23四年级下·四川绵阳·期末"
+    },
+    {
+        "id":  "382",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据运算定律填空。28+236+64=28+(),应用了()律。27×a+a×13=a×(),应用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "(236+64);加法结合律;(27+13)或(13+27);乘法分配律",
+        "solution":  "【分析】第一式将后两个加数先结合;第二式提取公因数a.\n【详解】28+236+64=28+(236+64),应用加法结合律;27×a+a×13=a×(27+13),应用乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数加法结合律"
+                             ],
+        "source":  "23-24四年级下·四川德阳·期末"
+    },
+    {
+        "id":  "383",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "27×()=35×()。这里运用了()律，用字母表示为()。",
+        "options":  [
+
+                    ],
+        "answer":  "35;27;乘法交换;a×b=b×a",
+        "solution":  "【分析】交换两个因数位置，积不变.\n【详解】27×35=35×27，运用了乘法交换律，字母式a×b=b×a。",
+        "knowledge_points":  [
+                                 "整数乘法交换律"
+                             ],
+        "source":  "22-23四年级下·四川德阳·期末"
+    },
+    {
+        "id":  "384",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用计算器计算264×79时，发现键\"7\"坏了。如果用这个计算器算，你会怎样算。请写出算式()。",
+        "options":  [
+
+                    ],
+        "answer":  "264×60+264×19",
+        "solution":  "【分析】把79写成60+19，避开数字7，再用乘法分配律.\n【详解】264×79=264×(60+19)=264×60+264×19。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·四川德阳·期末"
+    },
+    {
+        "id":  "385",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "小红把9×(a+3)错算成9×a+3，她得到的结果与正确的答案相差()。",
+        "options":  [
+
+                    ],
+        "answer":  "24",
+        "solution":  "【分析】正确式是9×a+27，错误式是9×a+3，比较差值即可.\n【详解】(9×a+27)-(9×a+3)=24。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "22-23四年级下·四川南充·期末"
+    },
+    {
+        "id":  "386",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "明明把7×(8+□)错算成7*□+8，他得到的结果与正确结果相差()。",
+        "options":  [
+
+                    ],
+        "answer":  "48",
+        "solution":  "【分析】正确式为7×8+7*□,错误式为7*□+8，差在常数部分.\n【详解】相差=56-8=48。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "22-23四年级下·四川绵阳·期中"
+    },
+    {
+        "id":  "387",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算125×48时，可以用(125*□)*□,这是根据()律。",
+        "options":  [
+
+                    ],
+        "answer":  "8;6;乘法结合或整数乘法结合",
+        "solution":  "【分析】把48分解成8×6，先算125×8更简便，依据乘法结合律.\n【详解】125×48=125×(8×6)=(125×8)×6。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "22-23四年级下·四川绵阳·期中"
+    },
+    {
+        "id":  "388",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "加法结合律用字母可以表示为(),a×b=b×a表示的是()律。",
+        "options":  [
+
+                    ],
+        "answer":  "(a+b)+c=a+(b+c);乘法交换律",
+        "solution":  "【分析】第一空写加法结合律字母式;第二空识别乘法交换律.\n【详解】加法结合律:(a+b)+c=a+(b+c);a×b=b×a表示乘法交换律。",
+        "knowledge_points":  [
+                                 "整数加法结合律",
+                                 "整数乘法交换律"
+                             ],
+        "source":  "23-24四年级上·河南商丘·期末"
+    },
+    {
+        "id":  "389",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "对25×44进行简便运算时，可以写成25×(□+□)或25×□×□。",
+        "options":  [
+
+                    ],
+        "answer":  "40;4;4;11",
+        "solution":  "【分析】44可拆成40+4，也可拆成4×11，分别对应分配律与结合律简算.\n【详解】25×44=25×(40+4)=25×4×11。",
+        "knowledge_points":  [
+                                 "整数乘法结合律",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级上·河南商丘·期末"
+    },
+    {
+        "id":  "390",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "235×4=200×4+30×4+5×4是运用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法分配律",
+        "solution":  "【分析】把235分成200+30+5，再分别与4相乘后相加，是乘法分配律.\n【详解】235×4=(200+30+5)×4=200×4+30×4+5×4。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·湖北黄冈·期中"
+    },
+    {
+        "id":  "391",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据运算定律填空。(1)84×12+16×12=(□+□)*□;(2)(30+□)×10=□*□+84*□;(3)202×40=200×40+□*40。",
+        "options":  [
+
+                    ],
+        "answer":  "84;16;12;84;30;10;10;2",
+        "solution":  "【分析】都依据乘法分配律进行合并或拆分.\n【详解】(1)84×12+16×12=(84+16)×12;(2)(30+84)×10=30×10+84×10;(3)202×40=200×40+2×40。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·重庆南岸·期末"
+    },
+    {
+        "id":  "392",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算40×(□+3),要先算()法，再算()法，小马虎在计算时不小心错算成40*□+3，他得到的结果与正确结果相差()。",
+        "options":  [
+
+                    ],
+        "answer":  "加;乘;117",
+        "solution":  "【分析】有括号先算加法再算乘法;错算少算了3个40中的一部分.\n【详解】正确比错误多(3×40-3)=120-3=117。所以先算加法，再算乘法，相差117。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "23-24四年级下·重庆南岸·期末"
+    },
+    {
+        "id":  "393",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "计算(10+2)×25=10×25+2×25时运用了()律;计算3×4×25=3×(4×25)时运用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法分配律;乘法结合律",
+        "solution":  "【分析】第一式是和乘一个数拆成两个乘积相加;第二式是改变连乘结合顺序.\n【详解】(10+2)×25=10×25+2×25用乘法分配律;3×4×25=3×(4×25)用乘法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数乘法结合律"
+                             ],
+        "source":  "23-24四年级下·重庆巴南·期末"
+    },
+    {
+        "id":  "394",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "(68+25)+32=□+(68+□),算式中运用了( )两条运算律。",
+        "options":  [
+
+                    ],
+        "answer":  "25;32;加法交换律和加法结合律",
+        "solution":  "【分析】先交换68和25的位置，再改变三个数的结合顺序.\n【详解】(68+25)+32=25+(68+32),用到了加法交换律和加法结合律。",
+        "knowledge_points":  [
+                                 "整数加法结合律",
+                                 "整数加法交换律"
+                             ],
+        "source":  "23-24四年级下·广东潮州·期末"
+    },
+    {
+        "id":  "395",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算:65×(100+10)=65×100+65×()=(),运用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "10;7150;乘法分配律",
+        "solution":  "【分析】按乘法分配律展开后计算.\n【详解】65×(100+10)=65×100+65×10=6500+650=7150，运用了乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·陕西渭南·期末"
+    },
+    {
+        "id":  "396",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据○+△=23，填一填。23-○=();4×○+△×4=4×()=()。",
+        "options":  [
+
+                    ],
+        "answer":  "△;23;92",
+        "solution":  "【分析】和减去一个加数等于另一个加数;第二问提取公因数4.\n【详解】由○+△=23得23-○=△。4×○+△×4=4×(○+△)=4×23=92。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "加，减法的意义和各部分间的关系"
+                             ],
+        "source":  "23-24四年级下·陕西渭南·期末"
+    },
+    {
+        "id":  "397",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "103+101+100+97+99=()×()=()。",
+        "options":  [
+
+                    ],
+        "answer":  "100;5;500",
+        "solution":  "【分析】通过拆分与凑整，把各项整理成5个100.\n【详解】103+101+100+97+99=(100+3)+(100+1)+100+97+99=100+100+100+(97+3)+(99+1)=100×5=500。",
+        "knowledge_points":  [
+                                 "整数加法结合律",
+                                 "整数加法交换律",
+                                 "初步认识乘法意义"
+                             ],
+        "source":  "2024六年级·全国·竞赛"
+    },
+    {
+        "id":  "398",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "计算:1346+3164+4631+6413=()。",
+        "options":  [
+
+                    ],
+        "answer":  "15554",
+        "solution":  "【分析】利用加法结合律把前两项，后两项分别相加.\n【详解】1346+3164+4631+6413=(1346+3164)+(4631+6413)=4510+11044=15554。",
+        "knowledge_points":  [
+                                 "整数加法结合律"
+                             ],
+        "source":  "2009四年级·全国·竞赛"
+    },
+    {
+        "id":  "399",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在下面括号里填上合适的数。(125+30)×8=()×8+()×8。",
+        "options":  [
+
+                    ],
+        "answer":  "125;30",
+        "solution":  "【分析】根据乘法分配律，括号内两个加数分别与8相乘.\n【详解】(125+30)×8=125×8+30×8。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·广东肇庆·期末"
+    },
+    {
+        "id":  "400",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "12×101=12×100+12，这个算式运用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法分配律",
+        "solution":  "【分析】把101拆成100+1，按乘法分配律展开.\n【详解】12×101=12×(100+1)=12×100+12，运用了乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·广东江门·期末"
+    },
+    {
+        "id":  "401",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "57×99=57×100-57是运用了( )律，要使25*□+75*□=6000,□里应该填( )(两个□里的数相同)。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法分配律;60",
+        "solution":  "【分析】根据乘法分配律,57×99可写成57×(100-1)=57×100-57;25×□+75×□可提取公因数□,变成(25+75)×□,再求□.\n【详解】57×99=57×(100-1)=57×100-57×1。又因为25×□+75×□=(25+75)×□=100×□=6000，所以□=60。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "402",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "洋洋用计算器计算3215×49时，发现\"4\"这个键坏了。如果还用这个计算器，你猜洋洋会怎么计算。请写出算式()。",
+        "options":  [
+
+                    ],
+        "answer":  "3215×7×7",
+        "solution":  "【分析】把49拆成不含数字4的因数乘积，再利用乘法结合律计算.\n【详解】49=7×7，所以3215×49可写成3215×7×7。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "23-24四年级下·河北衡水·期末"
+    },
+    {
+        "id":  "403",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "34567+66667+99999+65433=()",
+        "options":  [
+
+                    ],
+        "answer":  "266666",
+        "solution":  "【分析】先交换，结合凑整:34567与65433先相加，再处理99999.\n【详解】34567+66667+99999+65433=34567+65433+99999+66667=100000+(100000-1)+66667=266666。",
+        "knowledge_points":  [
+                                 "加减法的速算与巧算"
+                             ],
+        "source":  "2010四年级·全国·竞赛"
+    },
+    {
+        "id":  "404",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小红用计算器计算1234×49时，发现计算器上的按键\"9\"坏了，如果仍用这个计算器，他该怎样操作才能得出正确结果。请用算式表示计算过程()。",
+        "options":  [
+
+                    ],
+        "answer":  "1234×7×7或1234×48+1234或1234×50-1234",
+        "solution":  "【分析】把49改写为7×7,48+1或50-1，避开按键9.\n【详解】可写成1234×7×7;或1234×(48+1)=1234×48+1234;或1234×(50-1)=1234×50-1234。",
+        "knowledge_points":  [
+                                 "整数乘法结合律",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·河北秦皇岛·期末"
+    },
+    {
+        "id":  "405",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "根据运算律填上合适的数或字母.\n(1)576+305+224+95=(576+□)+(95+□)\n(2)99×32=□×32-□\n(3)m×n×y=m×□×n",
+        "options":  [
+
+                    ],
+        "answer":  "224;305;100;32;y",
+        "solution":  "【分析】(1)先交换再结合;(2)把99看成100-1并分配;(3)交换因数位置.\n【详解】(1)576+305+224+95=(576+224)+(95+305);(2)99×32=(100-1)×32=100×32-32;(3)m×n×y=m×y×n。",
+        "knowledge_points":  [
+                                 "整数加法交换律",
+                                 "整数加法结合律",
+                                 "整数乘法交换律",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·河北张家口·期末"
+    },
+    {
+        "id":  "406",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "54×6+54×14可以用()律使计算简便。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法分配律",
+        "solution":  "【分析】提取公因数54.\n【详解】54×6+54×14=54×(6+14)=54×20，所以用乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·河北保定·期末"
+    },
+    {
+        "id":  "407",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "363+999×999+636=()。",
+        "options":  [
+
+                    ],
+        "answer":  "999000",
+        "solution":  "【分析】先把363与636凑成999，再用分配律.\n【详解】363+999×999+636=999+999×999=999×(1+999)=999×1000=999000。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数加法交换律"
+                             ],
+        "source":  "2010四年级·全国·竞赛"
+    },
+    {
+        "id":  "408",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "佳佳在做作业时，由于马虎，把60×(a-8)错算成60×a-8，她得到的结果与正确结果相差()。",
+        "options":  [
+
+                    ],
+        "answer":  "472",
+        "solution":  "【分析】正确应为60×a-60×8，与60×a-8比较差值即可.\n【详解】相差=(60×a-8)-(60×a-480)=480-8=472(取差的绝对值为472)。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·河北沧州·期末"
+    },
+    {
+        "id":  "409",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在计算12×25×4时，可以先算()使计算简便，根据是乘法()律。",
+        "options":  [
+
+                    ],
+        "answer":  "25×4;结合律",
+        "solution":  "【分析】先算25×4可凑整百，利用乘法结合律.\n【详解】12×25×4=12×(25×4)=12×100。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "23-24四年级下·河北沧州·期末"
+    },
+    {
+        "id":  "410",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在计算(125+50)×8时，误算成了125×8+50，这样算的结果与正确相差()。",
+        "options":  [
+
+                    ],
+        "answer":  "350",
+        "solution":  "【分析】正确是125×8+50×8，误算少算了7个50.\n【详解】差值=50×8-50=400-50=350。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·河北保定·期末"
+    },
+    {
+        "id":  "411",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "如果A×B=8，那么125×A×7×B=()",
+        "options":  [
+
+                    ],
+        "answer":  "7000",
+        "solution":  "【分析】交换并结合成125×(A×B)×7.\n【详解】125×A×7×B=125×(A×B)×7=125×8×7=1000×7=7000。",
+        "knowledge_points":  [
+                                 "整数乘法结合律",
+                                 "整数乘法交换律"
+                             ],
+        "source":  "23-24四年级下·河北保定·期末"
+    },
+    {
+        "id":  "412",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "丁丁把10×(□+3)错看成10×□+3，得数与正确结果相差( )。",
+        "options":  [
+
+                    ],
+        "answer":  "27",
+        "solution":  "【分析】正确应是10×□+30，与10×□+3作差.\n【详解】相差=(10×□+30)-(10×□+3)=27。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数减法的性质"
+                             ],
+        "source":  "23-24四年级下·河北保定·期末"
+    },
+    {
+        "id":  "413",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据运算定律在()里填数.\n(1)526+()=327+()(加法交换律)\n(2)86×()+()=(99+1)×86(乘法分配律)",
+        "options":  [
+
+                    ],
+        "answer":  "327;526;99;86",
+        "solution":  "【分析】(1)交换加数位置;(2)把(99+1)×86展开.\n【详解】(1)526+327=327+526;(2)(99+1)×86=86×99+86×1=86×99+86。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数加法交换律"
+                             ],
+        "source":  "23-24四年级下·甘肃天水·期末"
+    },
+    {
+        "id":  "414",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用计算器计算124×36时，发现键\"3\"坏了，如果还用这个计算器，你会怎么计算。请写出算式()。",
+        "options":  [
+
+                    ],
+        "answer":  "124×4×9",
+        "solution":  "【分析】把36拆成不含\"3\"的因数乘积.\n【详解】36=4×9，所以124×36可写成124×4×9(也可18×2,6×6等)。",
+        "knowledge_points":  [
+                                 "计算器的初步认识与使用",
+                                 "整数乘法结合律"
+                             ],
+        "source":  "23-24四年级下·河北保定·期末"
+    },
+    {
+        "id":  "415",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "3×5+□=22,□是( );125×( )=8×( )。",
+        "options":  [
+
+                    ],
+        "answer":  "7;8;125",
+        "solution":  "【分析】先算3×5=15，再求补数;第二空用乘法交换律.\n【详解】22-15=7;125×8=8×125。",
+        "knowledge_points":  [
+                                 "整数乘法交换律",
+                                 "乘加，乘减混合运算"
+                             ],
+        "source":  "23-24四年级下·河北保定·期末"
+    },
+    {
+        "id":  "416",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用计算器在计算201×53时，发现数字键\"1\"不能用了，如果只能使用这个计算器，你准备怎么计算。()",
+        "options":  [
+
+                    ],
+        "answer":  "200×53+53",
+        "solution":  "【分析】把201拆成200+1，用分配律并避开按键1.\n【详解】201×53=(200+1)×53=200×53+53。",
+        "knowledge_points":  [
+                                 "计算器的初步认识与使用",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·山东济南·期末"
+    },
+    {
+        "id":  "417",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "8×53×125=53×(8×125)运用了()律和()律。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法交换律;乘法结合律",
+        "solution":  "【分析】先交换8和53位置，再改变运算顺序.\n【详解】由8×53×125到53×(8×125),依次用了乘法交换律和乘法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法结合律",
+                                 "整数乘法交换律"
+                             ],
+        "source":  "23-24四年级下·山东济南·期末"
+    },
+    {
+        "id":  "418",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "125×25×8×4=(125×8)×(25×4)运用了()律和()律。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法交换律;乘法结合律",
+        "solution":  "【分析】将25和8交换，再分组计算.\n【详解】125×25×8×4→125×8×25×4→(125×8)×(25×4),用了交换律和结合律。",
+        "knowledge_points":  [
+                                 "整数乘法结合律",
+                                 "整数乘法交换律"
+                             ],
+        "source":  "23-24四年级下·甘肃平凉·期末"
+    },
+    {
+        "id":  "419",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "根据运算律填空:263+()+58=()+(42+())。",
+        "options":  [
+
+                    ],
+        "answer":  "42;263;58",
+        "solution":  "【分析】利用加法结合律重组.\n【详解】263+42+58=263+(42+58)。",
+        "knowledge_points":  [
+                                 "整数加法结合律"
+                             ],
+        "source":  "23-24四年级下·甘肃平凉·期末"
+    },
+    {
+        "id":  "420",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "69×99+69=69×(99+1),这是根据()进行简便计算的。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法分配律",
+        "solution":  "【分析】把后一个69看成69×1，提取公因数69.\n【详解】69×99+69=69×99+69×1=69×(99+1)。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·海南省直辖县级单位·期末"
+    },
+    {
+        "id":  "421",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明把4×(a+9)错算成4×a+9，他的计算结果与正确结果相差()。",
+        "options":  [
+
+                    ],
+        "answer":  "27",
+        "solution":  "【分析】正确应为4×a+4×9，与错式比较.\n【详解】相差=(4×a+36)-(4×a+9)=27。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·广西河池·期末"
+    },
+    {
+        "id":  "422",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在算式35×4×25=35×(4×25)中运用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法结合律",
+        "solution":  "【分析】三个数相乘，改变运算顺序，积不变.\n【详解】35×4×25=35×(4×25)应用了乘法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "23-24四年级下·广西南宁·期末"
+    },
+    {
+        "id":  "423",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算25×39时，小丽是这样计算的:25×39=25×(40-1)=25×40-25×1，这样算的依据是()律。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法分配律",
+        "solution":  "【分析】把39看成40-1，使用a×(b-c)=a×b-a×c.\n【详解】25×39=25×(40-1)=25×40-25×1，依据乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·广西南宁·期末"
+    },
+    {
+        "id":  "424",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "5×6+6×15=□*(□+□)。",
+        "options":  [
+
+                    ],
+        "answer":  "6;5;15",
+        "solution":  "【分析】两个乘积有共同因数6，提取公因数.\n【详解】5×6+6×15=6×5+6×15=6×(5+15)。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·四川德阳·期末"
+    },
+    {
+        "id":  "425",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "已知◆+★=10(◆和★都代表非0自然数),那么29×◆+29×★这个算式的结果是( ),用字母表示运用的运算律是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "290;a×b+a×c=a×(b+c)",
+        "solution":  "【分析】提取公因数29.\n【详解】29×◆+29×★=29×(◆+★)=29×10=290，运用乘法分配律a×b+a×c=a×(b+c)。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·四川乐山·期末"
+    },
+    {
+        "id":  "426",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在横线上填上\"\u003e\"\"\u003c\"或\"=\".\n500÷4/25( )500÷(4×25)\n72-4×3÷2( )(72-4)×3÷2",
+        "options":  [
+
+                    ],
+        "answer":  "=;\u003c",
+        "solution":  "【分析】第一题用除法性质，第二题按运算顺序分别计算比较.\n【详解】500÷4/25=500÷(4×25);72-4×3÷2=66,(72-4)×3÷2=102，所以66\u003c102。",
+        "knowledge_points":  [
+                                 "整数除法的性质",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "23-24四年级下·四川南充·期末"
+    },
+    {
+        "id":  "427",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "69+126+74=69+(126+74)运用了()律，请用字母表示这个运算律()。",
+        "options":  [
+
+                    ],
+        "answer":  "加法结合律;(a+b)+c=a+(b+c)",
+        "solution":  "【分析】三个数相加，先加后两个数，和不变.\n【详解】这是加法结合律，字母表示为(a+b)+c=a+(b+c)。",
+        "knowledge_points":  [
+                                 "整数加法结合律"
+                             ],
+        "source":  "23-24四年级下·四川南充·期末"
+    },
+    {
+        "id":  "428",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "天天在计算55×(□+6)时，连犯了两次错误，请帮他解答疑惑.\n(1)天天把算式算成了55×□+6，这样与原来的答案相差了( ).\n(2)天天把算式看成了55×6+□,算出的结果比正确答案少了108，正确答案应该是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "324;440",
+        "solution":  "【分析】(1)比较55×(□+6)与55×□+6;(2)由\"少108\"建立差值方程求□,再回代.\n【详解】(1)55×(□+6)-(55×□+6)=55×6-6=324.\n(2)55×(□+6)-(55×6+□)=108，即54×□=108,□=2。正确值=55×(2+6)=440。",
+        "knowledge_points":  [
+                                 "除数是两位数的笔算除法",
+                                 "带有小括号的混合运算",
+                                 "整数减法的性质",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·贵州遵义·期末"
+    },
+    {
+        "id":  "429",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.4,
+        "question":  "计算:765÷18×213+765×327÷18=()。",
+        "options":  [
+
+                    ],
+        "answer":  "22950",
+        "solution":  "【分析】把第二项改写为765÷18×327，再提取公因数765÷18.\n【详解】765÷18×213+765×327÷18=765÷18×213+765÷18×327=765÷18×(213+327)=765÷18×540=765×(540÷18)=765×30=22950。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "四则混合运算的速算与巧算"
+                             ],
+        "source":  "2007四年级·全国·竞赛"
+    },
+    {
+        "id":  "430",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算:99999+9999+999+99+9=()。",
+        "options":  [
+
+                    ],
+        "answer":  "111105",
+        "solution":  "【分析】各项分别写成10的整次幂减1后再合并.\n【详解】(100000-1)+(10000-1)+(1000-1)+(100-1)+(10-1)=111110-5=111105。",
+        "knowledge_points":  [
+                                 "整数加法交换律",
+                                 "加减法的速算与巧算"
+                             ],
+        "source":  "2007四年级·全国·竞赛"
+    },
+    {
+        "id":  "431",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "填一填.\n(1)190+310=□+190\n(2)96+475+204=475+(□+□)\n(3)278+298=278+300-□\n(4)451-247-53=451-(□+□)",
+        "options":  [
+
+                    ],
+        "answer":  "310;96;204;2;247;53",
+        "solution":  "【分析】分别运用加法交换律，结合律与减法性质.\n【详解】(1)190+310=310+190;(2)96+475+204=475+(96+204);(3)278+298=278+(300-2)=278+300-2;(4)451-247-53=451-(247+53)。",
+        "knowledge_points":  [
+                                 "整数加法交换律",
+                                 "整数加法结合律",
+                                 "整数减法的性质"
+                             ],
+        "source":  "23-24四年级上·河南周口·期中"
+    },
+    {
+        "id":  "432",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算125×43×8时用()律比较简便;计算59×102时用()律比较简便。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法交换律;乘法分配律",
+        "solution":  "【分析】125×43×8先交换43和8凑整千;59×102把102拆成100+2.\n【详解】125×43×8=125×8×43=1000×43;59×102=59×(100+2)=59×100+59×2。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数乘法交换律"
+                             ],
+        "source":  "23-24四年级上·湖南岳阳·期中"
+    },
+    {
+        "id":  "433",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "下面的算式分别运用了什么运算律。用字母表示出来.\n(1)39×37=37×39,(乘法交换律)(a×b=b×a)\n(2)65+52+48=65+(52+48),()()()\n(3)25×44=25×40+25×4,()()",
+        "options":  [
+
+                    ],
+        "answer":  "加法结合律a+b+c=a+(b+c);乘法分配律;a×(b+c)=a×b+a×c",
+        "solution":  "【分析】(2)是三个数相加改变结合顺序;(3)是把44分解后分别相乘再相加.\n【详解】(2)65+52+48=65+(52+48)用加法结合律，字母式a+b+c=a+(b+c);(3)25×44=25×(40+4)=25×40+25×4，用乘法分配律，字母式a×(b+c)=a×b+a×c。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数加法结合律"
+                             ],
+        "source":  "23-24四年级下·全国·课后作业"
+    },
+    {
+        "id":  "434",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "下面的算式分别运用了什么乘法运算律。\n(1)34×25×8=34×(25×8),乘法()律.\n(2)99×32+32=(99+1)×32，乘法()律.\n(3)102×24=100×24+2×24，乘法()律.\n(4)4×25×125×2=(4×25)×(125×2),乘法()律。",
+        "options":  [
+
+                    ],
+        "answer":  "结合律;分配律;分配律;结合律",
+        "solution":  "【分析】(1)(4)是改变乘法运算顺序，属于乘法结合律;(2)(3)是把一个数拆成和后分别相乘，属于乘法分配律.\n【详解】(1)34×25×8=34×(25×8),乘法结合律;(2)99×32+32=99×32+1×32=(99+1)×32，乘法分配律;(3)102×24=(100+2)×24=100×24+2×24，乘法分配律;(4)4×25×125×2=(4×25)×(125×2),乘法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数乘法结合律"
+                             ],
+        "source":  "23-24四年级下·全国·课后作业"
+    },
+    {
+        "id":  "435",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "72×9×25=9×(*)运用了乘法()律。",
+        "options":  [
+
+                    ],
+        "answer":  "72;25;交换律和结合律",
+        "solution":  "【分析】先交换72和9的位置，再把72和25结合先算.\n【详解】72×9×25=9×(72×25),运用了乘法交换律和乘法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法结合律",
+                                 "整数乘法交换律"
+                             ],
+        "source":  "2024四年级下·全国·专题练习"
+    },
+    {
+        "id":  "436",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "55+127+145=127+(55+145)运用了()律和()律。",
+        "options":  [
+
+                    ],
+        "answer":  "加法交换律;加法结合律",
+        "solution":  "【分析】先交换55和127的位置，再把55和145先结合.\n【详解】55+127+145=127+55+145=127+(55+145),运用了加法交换律和加法结合律。",
+        "knowledge_points":  [
+                                 "整数加法结合律",
+                                 "整数加法交换律"
+                             ],
+        "source":  "2024四年级下·全国·专题练习"
+    },
+    {
+        "id":  "437",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "下面的算式分别运用了什么乘法运算律。\n(1)14×25=10×25+4×25，运用了乘法()律\n(2)17×25×8=17×(25×8),运用了乘法()律\n(3)4×25×8×5=(4×25)×(8×5),运用了乘法()律\n(4)105×22=100×22+5×22，运用了乘法()律\n(5)25×(4×9)=(25×4)×9，运用了乘法()律",
+        "options":  [
+
+                    ],
+        "answer":  "分配;结合;结合;分配;结合",
+        "solution":  "【分析】把一个因数拆成和后分别相乘是分配律;改变乘法分组顺序是结合律.\n【详解】(1)14=(10+4),14×25=10×25+4×25，分配律;(2)17×25×8=17×(25×8),结合律;(3)4×25×8×5=(4×25)×(8×5),结合律;(4)105=(100+5),105×22=100×22+5×22，分配律;(5)25×(4×9)=(25×4)×9，结合律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数乘法结合律"
+                             ],
+        "source":  "23-24四年级下·全国·课后作业"
+    },
+    {
+        "id":  "438",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "已知28×a+72×a=500，则a=()。",
+        "options":  [
+
+                    ],
+        "answer":  "5",
+        "solution":  "【分析】提取公因数a,利用乘法分配律逆运算.\n【详解】28×a+72×a=(28+72)×a=100a=500，所以a=5。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·全国·课后作业"
+    },
+    {
+        "id":  "439",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在横线上填上合适的字母.\n(1)(a+b)×c=□*c+□*□\n(2)a×b×c=□*(□*c)\n(3)a×c-b×c=(□-□)*□\n(4)a×(b+1)=a×b+□",
+        "options":  [
+
+                    ],
+        "answer":  "a;b;c;a;b;a;b;c;a",
+        "solution":  "【分析】分别应用乘法分配律与乘法结合律的字母形式.\n【详解】(1)(a+b)×c=a×c+b×c;(2)a×b×c=a×(b×c);(3)a×c-b×c=(a-b)×c;(4)a×(b+1)=a×b+a。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数乘法结合律"
+                             ],
+        "source":  "23-24四年级下·全国·课后作业"
+    },
+    {
+        "id":  "440",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "下面的算式分别运用了什么运算律。\n(1)356+(644+159)=(356+644)+159()\n(2)79×99=79×(100-1)=79×100-79()\n(3)325×125×8=325×(125×8)()\n(4)150+22+78=150+(22+78)()",
+        "options":  [
+
+                    ],
+        "answer":  "加法结合律;乘法分配律;乘法结合律;加法结合律",
+        "solution":  "【分析】根据式子变化判断：加法分组变化是加法结合律;把99拆成100-1后展开是乘法分配律;乘法分组变化是乘法结合律.\n【详解】对应分别为：加法结合律，乘法分配律，乘法结合律，加法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数乘法结合律",
+                                 "整数加法结合律"
+                             ],
+        "source":  "23-24四年级下·全国·单元测试"
+    },
+    {
+        "id":  "441",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "35×8=8×()运用了乘法()律。",
+        "options":  [
+
+                    ],
+        "answer":  "35;交换律",
+        "solution":  "【分析】交换两个因数的位置，积不变.\n【详解】35×8=8×35，运用了乘法交换律。",
+        "knowledge_points":  [
+                                 "整数乘法交换律"
+                             ],
+        "source":  "23-24四年级下·全国·期末"
+    },
+    {
+        "id":  "442",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "实验小学图书室有故事书86本，今天借出58本，还回14本。现在图书室有故事书()本。",
+        "options":  [
+
+                    ],
+        "answer":  "42",
+        "solution":  "【分析】现有本数=原有-借出+还回.\n【详解】86-58+14=86+14-58=100-58=42(本)。",
+        "knowledge_points":  [
+                                 "两，三位数的加减混合运算"
+                             ],
+        "source":  "24-25四年级下·全国·期中"
+    },
+    {
+        "id":  "443",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "检验365+474=839的计算结果是否正确，可以用474+365=839进行检验，这种验算方法是根据加法()律。",
+        "options":  [
+
+                    ],
+        "answer":  "交换律",
+        "solution":  "【分析】加法交换律：交换两个加数的位置，和不变.\n【详解】474+365与365+474和相同，验算依据是加法交换律。",
+        "knowledge_points":  [
+                                 "整数加法交换律"
+                             ],
+        "source":  "2024四年级下·全国·专题练习"
+    },
+    {
+        "id":  "444",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据运算律，填一填.\n(1)74+28=28+□\n(2)45×7×2=□*□*7\n(3)36×5+36×35=(□+□)*□\n(4)8×10×125×4=(□*□)×(□*□)\n(5)(32+25)×2=□*2+□*2",
+        "options":  [
+
+                    ],
+        "answer":  "74;45;2;5;35;36;8;125;10;4;32;25",
+        "solution":  "【分析】分别应用加法交换律，乘法交换律，乘法分配律，乘法结合律，乘法分配律.\n【详解】(1)74+28=28+74;(2)45×7×2=45×2×7;(3)36×5+36×35=(5+35)×36;(4)8×10×125×4=(8×125)×(10×4);(5)(32+25)×2=32×2+25×2。",
+        "knowledge_points":  [
+                                 "整数加法交换律",
+                                 "整数乘法交换律",
+                                 "整数乘法结合律",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·全国·课后作业"
+    },
+    {
+        "id":  "445",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上\"\u003e\"\"\u003c\"或\"=\".\n34×32()32×34;26×12×5()(12×5)×16;36×28×11()11×39×28;45×(23×32)()(45×23)×32",
+        "options":  [
+
+                    ],
+        "answer":  "=;\u003e;\u003c;=",
+        "solution":  "【分析】第1,4题可直接用乘法交换律和结合律;第2,3题先化成可比形式再比较因数大小.\n【详解】34×32=32×34;26×12×5=(12×5)×26，所以\u003e(12×5)×16;36×28×11=11×36×28，而36\u003c39，所以\u003c11×39×28;45×(23×32)=(45×23)×32。",
+        "knowledge_points":  [
+                                 "整数乘法结合律",
+                                 "整数乘法交换律"
+                             ],
+        "source":  "23-24四年级下·全国·课后作业"
+    },
+    {
+        "id":  "446",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "填一填.\n(1)21×24=24*□,用了乘法()律.\n(2)45×4×5=□*(4*□),用了乘法()律.\n(3)(80×25)×4=80×(□*□),用了乘法()律.\n(4)(5×125)×8=5×(□*□),用了乘法()律.\n![](images/55da973d55b249c15b37d2ccfc9b6810dedc12583713761a2b5204148c00ebb5.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "21;交换;45;5;结合;25;4;结合;125;8;结合",
+        "solution":  "【分析】(1)交换因数位置;(2)(3)(4)改变三个数相乘的运算顺序.\n【详解】(1)21×24=24×21，用乘法交换律;(2)45×4×5=45×(4×5),用乘法结合律;(3)(80×25)×4=80×(25×4),用乘法结合律;(4)(5×125)×8=5×(125×8),用乘法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "23-24四年级下·全国·课后作业"
+    },
+    {
+        "id":  "447",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "125×4-25×4=(125-25)×4，运用了()律，用字母表示是()。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法分配律;ab±ac=a×(b±c)",
+        "solution":  "【分析】两项有公因数4，可提取公因数，是分配律逆用.\n【详解】125×4-25×4=(125-25)×4，属于乘法分配律，字母形式可写为ab±ac=a×(b±c)。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·全国·期中"
+    },
+    {
+        "id":  "448",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "25×44=(25×4)×11运用了( )进行简便运算;25×44=25×40+25×4运用了( )进行简便运算。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法结合律;乘法分配律",
+        "solution":  "【分析】第一种把44拆成4×11并重组;第二种把44拆成40+4并展开.\n【详解】25×44=25×(4×11)=(25×4)×11，用乘法结合律;25×44=25×(40+4)=25×40+25×4，用乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法结合律",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·全国·课后作业"
+    },
+    {
+        "id":  "449",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "根据运算律在横线上填上合适的数.\n(1)85+35=35+□\n(2)48+a+52=(□+□)+a\n(3)101+542+99=(101+□)+□\n(4)(180+70)+130=□+(□+□)",
+        "options":  [
+
+                    ],
+        "answer":  "85;48;52;99;542;180;70;130",
+        "solution":  "【分析】主要运用加法交换律和加法结合律.\n【详解】(1)85+35=35+85;(2)48+a+52=(48+52)+a;(3)101+542+99=(101+99)+542;(4)(180+70)+130=180+(70+130)。",
+        "knowledge_points":  [
+                                 "整数加法结合律",
+                                 "整数加法交换律"
+                             ],
+        "source":  "23-24四年级下·全国·课后作业"
+    },
+    {
+        "id":  "450",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用字母表示乘法分配律为(),c×25×4=c×(25×4)是根据乘法()律进行简算的。",
+        "options":  [
+
+                    ],
+        "answer":  "a×(b+c)=a×b+a×c;结合律",
+        "solution":  "【分析】第一空写分配律字母式;第二空是改变乘法分组顺序.\n【详解】乘法分配律:a×(b+c)=a×b+a×c;c×25×4=c×(25×4)依据乘法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数乘法结合律"
+                             ],
+        "source":  "23-24四年级下·陕西西安·期末"
+    },
+    {
+        "id":  "451",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "8×325+12×325=(8+12)×325，这里应用了( )律;如果□+△=100，那么78×□+78×△=( )。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法分配律;7800",
+        "solution":  "【分析】前式是提取公因数325;后式同理提取公因数78.\n【详解】8×325+12×325=(8+12)×325，应用乘法分配律;78×□+78×△=78×(□+△)=78×100=7800。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·山西晋中·期末"
+    },
+    {
+        "id":  "452",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果a×b=30，那么b×a=();240/a/b=()。",
+        "options":  [
+
+                    ],
+        "answer":  "30;8",
+        "solution":  "【分析】前者用乘法交换律，后者用除法性质.\n【详解】b×a=a×b=30;240/a/b=240÷(a×b)=240÷30=8。",
+        "knowledge_points":  [
+                                 "整数除法的性质",
+                                 "整数乘法交换律"
+                             ],
+        "source":  "23-24四年级下·山西长治·期末"
+    },
+    {
+        "id":  "453",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把这三个算式960÷15=64,64-36=28,28×30=840，合并成一个综合算式是( );如果□+△=100，那么45×□+45×△=( )。",
+        "options":  [
+
+                    ],
+        "answer":  "(960÷15-36)×30;4500",
+        "solution":  "【分析】按运算先后合并分步算式;第二问提取公因数45.\n【详解】综合算式为(960÷15-36)×30=840;45×□+45×△=45×(□+△)=45×100=4500。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "有括号的运算顺序",
+                                 "无括号的运算顺序",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "23-24四年级下·云南玉溪·期末"
+    },
+    {
+        "id":  "454",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用计算机计算\"1234×63\"时，发现键\"6\"坏了，如果还是用这个计算器，你会怎么计算。请写出算式:()。",
+        "options":  [
+
+                    ],
+        "answer":  "1234×7×9",
+        "solution":  "【分析】63=7×9，改写成不含数字6的乘法式.\n【详解】1234×63=1234×7×9(答案不唯一)。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "23-24四年级下·山西阳泉·期末"
+    },
+    {
+        "id":  "455",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小童在做作业时，将35×(□+9)错算成35×□+9，这样计算的得数比正确的得数少( )。",
+        "options":  [
+
+                    ],
+        "answer":  "306",
+        "solution":  "【分析】正确应为35×□+35×9，与错式作差.\n【详解】相差=(35×□+315)-(35×□+9)=315-9=306。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·云南楚雄·期末"
+    },
+    {
+        "id":  "456",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在计算84×25时，聪聪的方法是84×25=(21×4)×25=21×(4×25)=2100，聪聪运用了乘法( )律;明明的方法是84×25=(80+4)×25=80×25+4×25=2100，明明运用了乘法( )律。",
+        "options":  [
+
+                    ],
+        "answer":  "结合;分配",
+        "solution":  "【分析】聪聪是重组乘法顺序，明明是把84拆成80+4后展开.\n【详解】聪聪运用乘法结合律;明明运用乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数乘法结合律"
+                             ],
+        "source":  "23-24四年级下·吉林松原·期末"
+    },
+    {
+        "id":  "457",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在简便运算56×99=56×100-56中，应用的运算定律是()。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法分配律",
+        "solution":  "【分析】把99写成100-1，再按分配律展开.\n【详解】56×99=56×(100-1)=56×100-56，应用了乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·贵州铜仁·期末"
+    },
+    {
+        "id":  "458",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "□和△分别代表一个数，如果□+△=25，那么□×8+△×8=( )。",
+        "options":  [
+
+                    ],
+        "answer":  "200",
+        "solution":  "【分析】提取公因数8.\n【详解】□×8+△×8=(□+△)×8=25×8=200。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·贵州黔南·期末"
+    },
+    {
+        "id":  "459",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "(()+126)+b=a+()+b。",
+        "options":  [
+
+                    ],
+        "answer":  "a;126",
+        "solution":  "【分析】根据加法结合律，三个加数不变，只改变结合方式.\n【详解】(a+126)+b=a+(126+b)。",
+        "knowledge_points":  [
+                                 "整数加法结合律"
+                             ],
+        "source":  "23-24四年级下·贵州黔南·期末"
+    },
+    {
+        "id":  "460",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "125×7+125=125×(□+□)。",
+        "options":  [
+
+                    ],
+        "answer":  "7;1",
+        "solution":  "【分析】把125看作125×1，再提取公因数125.\n【详解】125×7+125=125×7+125×1=125×(7+1)。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·贵州黔南·期末"
+    },
+    {
+        "id":  "461",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据所学的运算定律和运算性质填一填，使计算简便.\n(1)3000÷125/8=□○(□○□)\n(2)101×m-m=(□○□)○□",
+        "options":  [
+
+                    ],
+        "answer":  "3000÷(125×8);(101-1)×m",
+        "solution":  "【分析】(1)用除法性质a/b/c=a/(b×c);(2)用乘法分配律逆运算.\n【详解】(1)3000÷125/8=3000÷(125×8)=3000÷1000=3;(2)101×m-m=101×m-1×m=(101-1)×m。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数除法的性质"
+                             ],
+        "source":  "23-24四年级下·云南昭通·期末"
+    },
+    {
+        "id":  "462",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "根据乘法运算律，计算125×7×8时，可以转化成125×()×()。",
+        "options":  [
+
+                    ],
+        "answer":  "8;7",
+        "solution":  "【分析】交换7和8的位置，便于先算125×8.\n【详解】125×7×8=125×8×7。",
+        "knowledge_points":  [
+                                 "整数乘法交换律"
+                             ],
+        "source":  "23-24四年级下·贵州铜仁·期末"
+    },
+    {
+        "id":  "463",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "根据95-60=35,490÷35=14,288+14=302列出综合算式是:()。",
+        "options":  [
+
+                    ],
+        "answer":  "288+490÷(95-60)",
+        "solution":  "【分析】先把35替换为95-60，再把14替换为490÷35.\n【详解】综合算式:288+490÷(95-60)。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "除加，除减混合运算"
+                             ],
+        "source":  "23-24四年级下·贵州六盘水·期末"
+    },
+    {
+        "id":  "464",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "吴平在计算25×(★+4)时，算成25×★+4，他算出的结果与正确的结果相差()。",
+        "options":  [
+
+                    ],
+        "answer":  "96",
+        "solution":  "【分析】正确应为25×★+25×4，与错式作差.\n【详解】相差=(25×★+100)-(25×★+4)=96。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·贵州黔东南·期末"
+    },
+    {
+        "id":  "465",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "明明家到明明妈妈工作的单位步行有800米，妈妈每天上下班都要走2个来回，妈妈每天上下班一共要走()米。",
+        "options":  [
+
+                    ],
+        "answer":  "3200",
+        "solution":  "【分析】一个来回是2个单程,2个来回共4个单程.\n【详解】800×2×2=3200(米)。",
+        "knowledge_points":  [
+                                 "整数乘法结合律",
+                                 "整十，整百，整千数与一位数的乘法"
+                             ],
+        "source":  "23-24四年级下·四川凉山·期末"
+    },
+    {
+        "id":  "466",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小虎把107-(m+21)算成了107-m+21，他这样计算的结果与正确结果相差()。",
+        "options":  [
+
+                    ],
+        "answer":  "42",
+        "solution":  "【分析】正确应为107-m-21，错式为107-m+21，二者相差2个21.\n【详解】相差=21+21=42。",
+        "knowledge_points":  [
+                                 "整数减法的性质"
+                             ],
+        "source":  "23-24四年级下·河南省直辖县级单位·期末"
+    },
+    {
+        "id":  "467",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算35×200+35可以运用的运算律是(),它是求()个35的和。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法分配律;201",
+        "solution":  "【分析】提取公因数35.\n【详解】35×200+35=35×(200+1)=35×201，所以是求201个35的和。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·河南漯河·期末"
+    },
+    {
+        "id":  "468",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用计算器计算\"1235×72\"时，发现键\"7\"坏了。如果还用这个计算器，你会怎样计算。请写出综合算式:()",
+        "options":  [
+
+                    ],
+        "answer":  "1235×8×9",
+        "solution":  "【分析】将72拆成8×9，避开数字7键.\n【详解】1235×72=1235×8×9(答案不唯一)。",
+        "knowledge_points":  [
+                                 "计算器的初步认识与使用",
+                                 "整数乘法结合律"
+                             ],
+        "source":  "23-24四年级下·河南省直辖县级单位·期末"
+    },
+    {
+        "id":  "469",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "(8+4)×25=8×25+4×25运用了乘法()律。",
+        "options":  [
+
+                    ],
+        "answer":  "分配律",
+        "solution":  "【分析】和乘一个数，等于分别相乘再相加.\n【详解】(8+4)×25=8×25+4×25，运用了乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·河南洛阳·期末"
+    },
+    {
+        "id":  "470",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一套演出服装的上衣65元，裤子35元，学校合唱队订购了45套这样的服装，淘气列式为:45×(65+35),算式中的\u002765+35\u0027是要先算()的价格。",
+        "options":  [
+
+                    ],
+        "answer":  "一套服装",
+        "solution":  "【分析】先算一件上衣和一条裤子的总价，再乘45套。也可理解为65×45+35×45，与45×(65+35)等价，体现乘法分配律.\n【详解】65+35表示一套服装(上衣+裤子)的价格。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·河南洛阳·期末"
+    },
+    {
+        "id":  "471",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用简便方法计算25×44时，张冬把算式改写为(25×4)×11是运用了乘法()律;李红把算式改写为25×40+25×4是运用了乘法()律。",
+        "options":  [
+
+                    ],
+        "answer":  "结合律;分配律",
+        "solution":  "【分析】25×44可把44拆成4×11，或拆成40+4，分别对应结合律与分配律.\n【详解】(25×4)×11用乘法结合律;25×40+25×4用乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法结合律",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·河南开封·期末"
+    },
+    {
+        "id":  "472",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "赵雪在口算32×2时，先算30×2，再算2×2，然后把这两个积相加。这个过程实际上是运用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法分配律",
+        "solution":  "【分析】把32拆成30+2，再分别与2相乘后相加.\n【详解】32×2=(30+2)×2=30×2+2×2，运用了乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·河南开封·期末"
+    },
+    {
+        "id":  "473",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "张乐在计算36×48+36×52时是这样想的:36×48+36×52=36×(48+52),他依据的运算律是()。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法分配律",
+        "solution":  "【分析】两个乘法算式有共同因数36，可提取公因数.\n【详解】36×48+36×52=36×(48+52),依据乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·河南信阳·期末"
+    },
+    {
+        "id":  "474",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.4,
+        "question":  "要使34×□+66×□=9900，则□=( );要使54×△+78×☆=7800，且可运用乘法分配律计算，则△=( ),☆=( )。",
+        "options":  [
+
+                    ],
+        "answer":  "99;78;46",
+        "solution":  "【分析】34×□+66×□=(34+66)×□=100×□;第二问要能用分配律，两个乘法算式要有相同因数78.\n【详解】□=9900÷100=99;54×△+78×☆可取△=78，则54×78+78×☆=(54+☆)×78=7800=100×78，所以☆=46。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·河南商丘·期末"
+    },
+    {
+        "id":  "475",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "龙龙根据3570÷35=102，直接写出了算式3570÷102=35，他这样写的依据是( );根据☆×27+☆×73=2800，推算出☆=( ),他推算时应用了( )律。",
+        "options":  [
+
+                    ],
+        "answer":  "被除数÷商=除数;28;乘法分配律",
+        "solution":  "【分析】除法各部分关系：被除数÷商=除数。另一个算式可提取公因数☆.\n【详解】3570÷35=102，所以3570÷102=35;☆×27+☆×73=☆×(27+73)=☆×100=2800，所以☆=28，应用了乘法分配律。",
+        "knowledge_points":  [
+                                 "乘，除法的意义和各部分间的关系",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·河南许昌·期末"
+    },
+    {
+        "id":  "476",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "轩轩在计算45×(△+2)时，把括号丢了，算成了45×△+2，这样得到的结果与正确结果相差( )。",
+        "options":  [
+
+                    ],
+        "answer":  "88",
+        "solution":  "【分析】正确应为45×△+45×2，与45×△+2作差.\n【详解】(45×△+90)-(45×△+2)=88。",
+        "knowledge_points":  [
+                                 "乘加，乘减混合运算",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·河南信阳·期末"
+    },
+    {
+        "id":  "477",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "四(3)共42人，这学期每人新买了一套校服，其中每件上衣55元，每条裤子45元，全班买校服一共花了( )元。",
+        "options":  [
+
+                    ],
+        "answer":  "4200",
+        "solution":  "【分析】先求一套校服单价，再乘人数.\n【详解】(55+45)×42=100×42=4200(元)。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·河南信阳·期末"
+    },
+    {
+        "id":  "478",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "简算793-214-356时，先算(),再算()。",
+        "options":  [
+
+                    ],
+        "answer":  "214+356;793-570",
+        "solution":  "【分析】利用减法性质:a-b-c=a-(b+c).\n【详解】793-214-356=793-(214+356)=793-570，所以先算214+356，再算793-570。",
+        "knowledge_points":  [
+                                 "整数减法的性质"
+                             ],
+        "source":  "23-24四年级下·河南信阳·期末"
+    },
+    {
+        "id":  "479",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "125×88=125×80+125×8运用了(),125×88=125×8×11运用了()。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法分配律;乘法结合律",
+        "solution":  "【分析】把88拆成80+8时用分配律;拆成8×11并重组时用结合律.\n【详解】125×88=125×(80+8)=125×80+125×8;125×88=125×(8×11)=125×8×11。",
+        "knowledge_points":  [
+                                 "整数乘法结合律",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·河南新乡·期末"
+    },
+    {
+        "id":  "480",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算808×125时，贝贝这样做:808×125=800×125+8×125，她运用了()律，妮妮这样做:808×125=101×(8×125),她运用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法分配律;乘法结合律",
+        "solution":  "【分析】前者把808拆成800+8并展开;后者把808写成101×8并改变结合顺序.\n【详解】808×125=800×125+8×125用乘法分配律;808×125=101×(8×125)用乘法结合律。",
+        "knowledge_points":  [
+                                 "整数乘法结合律",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "481",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "天天在计算45×(□+6)时，把算式看成了45×6+□,算出的结果比正确答案少了88，正确答案应该是()。",
+        "options":  [
+
+                    ],
+        "answer":  "360",
+        "solution":  "【分析】由\"正确-错误=88\"先求□,再代回原式.\n【详解】45×(□+6)-(45×6+□)=88，即45×□-□=88,44×□=88,□=2;正确值=45×(2+6)=360。",
+        "knowledge_points":  [
+                                 "乘加，乘减混合运算",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "482",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在学习了乘法分配律之后，小辉同学做了这样的计算5×(a+3)=5a+3，我认为()(填\u0027错\u0027或\u0027对\u0027)。如果错了，是()(\u0027多\u0027或\u0027少\u0027)了，多(或少)了()。",
+        "options":  [
+
+                    ],
+        "answer":  "错;少了;12",
+        "solution":  "【分析】正确应为5×(a+3)=5a+15，与5a+3比较差值.\n【详解】5a+15-(5a+3)=12，所以该做法错误，少了12。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·河南新乡·期末"
+    },
+    {
+        "id":  "483",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算56×102，小明的算法是56×100+56×2，他依据的运算律是(),用字母表示为()。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法分配律;a×(b+c)=a×b+a×c",
+        "solution":  "【分析】把102拆成100+2后分别与56相乘.\n【详解】56×102=56×(100+2)=56×100+56×2，依据乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "484",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "用计算器计算\u0027472×56\u0027时，发现按键\u00275\u0027坏了。如果还继续使用这个计算器你会怎样计算。请列出算式()。",
+        "options":  [
+
+                    ],
+        "answer":  "472×7×8",
+        "solution":  "【分析】把56拆成7×8，避开数字5.\n【详解】472×56=472×7×8(答案不唯一)。",
+        "knowledge_points":  [
+                                 "整数乘法结合律"
+                             ],
+        "source":  "23-24四年级下·辽宁盘锦·期末"
+    },
+    {
+        "id":  "485",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用计算器计算\u00271245×39\u0027时，发现键\u00273\u0027坏了。如果还用这个计算器，你会怎样计算。请写出算式:()。",
+        "options":  [
+
+                    ],
+        "answer":  "1245×40-1245",
+        "solution":  "【分析】39=40-1，用乘法分配律展开且不需要按键3.\n【详解】1245×39=1245×(40-1)=1245×40-1245。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "计算器的初步认识与使用"
+                             ],
+        "source":  "23-24四年级下·江西宜春·期末"
+    },
+    {
+        "id":  "486",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果○+△=24，那么20×○+20×△的结果是( );如果○×△=5，那么○×(△×11)的结果是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "480;55",
+        "solution":  "【分析】前者用分配律，后者用结合律.\n【详解】20×○+20×△=20×(○+△)=20×24=480;○×(△×11)=(○×△)×11=5×11=55。",
+        "knowledge_points":  [
+                                 "整数乘法结合律",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·广东广州·期末"
+    },
+    {
+        "id":  "487",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算12×25时，聪聪是这样计算的:12×25=3×(4×25)=3×100=300，他运用了()律;明明是这样计算的:12×25=(10+2)×25=10×25+2×25=250+50=300，他运用了()律。",
+        "options":  [
+
+                    ],
+        "answer":  "乘法结合律;乘法分配律",
+        "solution":  "【分析】聪聪把12写成3×4并重组;明明把12写成10+2并展开.\n【详解】前者运用乘法结合律，后者运用乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法结合律",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·广东广州·期末"
+    },
+    {
+        "id":  "488",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "10×(37+19)=10×()+10×(),这里运用的是乘法()律。",
+        "options":  [
+
+                    ],
+        "answer":  "37;19;分配",
+        "solution":  "【分析】一个数乘和，可分别相乘再相加.\n【详解】10×(37+19)=10×37+10×19，运用乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·新疆巴音郭楞·期末"
+    },
+    {
+        "id":  "489",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "167×28-67×28，比较简便的算式是(()-())×28，这样的计算是根据()。",
+        "options":  [
+
+                    ],
+        "answer":  "167;67;乘法分配律",
+        "solution":  "【分析】两项有公因数28，可提取公因数.\n【详解】167×28-67×28=(167-67)×28，依据乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "22-23四年级下·广东汕头·期末"
+    },
+    {
+        "id":  "490",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "张华在计算235-99时，他先算235-100，为了得到正确的结果，还需将算出来的结果再()。",
+        "options":  [
+
+                    ],
+        "answer":  "加1",
+        "solution":  "【分析】99=100-1，减99等于先减100再加1.\n【详解】235-99=235-(100-1)=235-100+1，所以还需再加1。",
+        "knowledge_points":  [
+                                 "整数减法的性质"
+                             ],
+        "source":  "23-24四年级下·湖北襄阳·期末"
+    },
+    {
+        "id":  "491",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明把9×(3+□)错算成9×□+3，他得到的结果与正确结果相差()。",
+        "options":  [
+
+                    ],
+        "answer":  "24",
+        "solution":  "【分析】正确应为9×3+9×□,与9×□+3比较.\n【详解】相差=9×3-3=27-3=24。",
+        "knowledge_points":  [
+                                 "乘加，乘减混合运算",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·江西吉安·期末"
+    },
+    {
+        "id":  "492",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小马虎在计算35×(□+6)时，不小心算成35*□+6，他的结果比正确结果少( )。",
+        "options":  [
+
+                    ],
+        "answer":  "204",
+        "solution":  "【分析】正确应为35*□+35×6，与错式作差.\n【详解】(35*□+210)-(35*□+6)=204。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·福建福州·期末"
+    },
+    {
+        "id":  "493",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "已知a+b=13,a×b=36，那么a×7×b=(),a×7+7×b=()。",
+        "options":  [
+
+                    ],
+        "answer":  "252;91",
+        "solution":  "【分析】前者用乘法交换/结合把a×b合并;后者提取公因数7.\n【详解】a×7×b=(a×b)×7=36×7=252;a×7+7×b=(a+b)×7=13×7=91。",
+        "knowledge_points":  [
+                                 "整数乘法交换律",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·浙江嘉兴·期末"
+    },
+    {
+        "id":  "494",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果☆×△=16，那么320÷☆/△=( );如果※-⊙=10，那么12×※-12×⊙=( )。",
+        "options":  [
+
+                    ],
+        "answer":  "20;120",
+        "solution":  "【分析】第一问用除法性质，第二问用乘法分配律逆运算.\n【详解】320÷☆/△=320÷(☆×△)=320÷16=20;12×※-12×⊙=12×(※-⊙)=12×10=120。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数除法的性质",
+                                 "等量代换"
+                             ],
+        "source":  "23-24四年级下·浙江湖州·期末"
+    },
+    {
+        "id":  "495",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "\u0027加法交换律\u0027用字母表示是a+b=b+a,下面式子中的a和b分别表示几。\n(1)2.7+7.63=7.63+2.7,a是();b是().\n(2)12+3.5+5.22=12+5.22+3.5,a是();b是()。",
+        "options":  [
+
+                    ],
+        "answer":  "2.7;7.63;3.5;5.22",
+        "solution":  "【分析】按交换律a+b=b+a对应交换位置的两个加数.\n【详解】(1)可对应a=2.7,b=7.63;(2)交换的是3.5与5.22，所以a=3.5,b=5.22(互换写法也可)。",
+        "knowledge_points":  [
+                                 "整数加法交换律"
+                             ],
+        "source":  "23-24四年级下·浙江宁波·期末"
+    },
+    {
+        "id":  "496",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小马虎由于粗心大意，把70×(★+5)错写成70×★+5。请你帮忙算一算，他得到的结果与正确结果相差( )。",
+        "options":  [
+
+                    ],
+        "answer":  "345",
+        "solution":  "【分析】正确应为70×★+70×5，与错式相减.\n【详解】(70×★+350)-(70×★+5)=345。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·浙江台州·期末"
+    },
+    {
+        "id":  "497",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小丽在计算6×(□+3)时漏看了小括号，则小丽算得的结果与正确的相差( )。",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】正确式比漏括号后的式子多出6×3-3.\n【详解】差值=6×3-3=18-3=15。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "整数减法的性质",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·浙江杭州·期末"
+    },
+    {
+        "id":  "498",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "A=29×31,B=28×32，聪聪用下面方法比较A和B的大小。你能看懂吗。请把下边算式补充完整，并比较A和B的大小.\n因为:\nA=29×31\n=(28+1)×31\n=28×31+1×31\nB=28×32\n=()\n=()\n所以:A( )B。",
+        "options":  [
+
+                    ],
+        "answer":  "28×(31+1);28×31+28×1;\u003e",
+        "solution":  "【分析】把B中的32写成31+1再分配展开，与A对应项比较.\n【详解】B=28×32=28×(31+1)=28×31+28×1;A=28×31+1×31，且1×31\u003e28×1，所以A\u003eB。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·浙江杭州·期末"
+    },
+    {
+        "id":  "499",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "如果A×B=61,C×A=145，那么A×(B+C)=()",
+        "options":  [
+
+                    ],
+        "answer":  "206",
+        "solution":  "【分析】按题目给出的解法，先算A×(B+C)=A×B+A×C.\n【详解】A×(B+C)=61+145=206。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·浙江温州·期末"
+    },
+    {
+        "id":  "500",
+        "unit":  "第三单元运算律",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "计算325×42时，可以把它想成325×40+325×2，这是运用了()。(填运算律)",
+        "options":  [
+
+                    ],
+        "answer":  "乘法分配律",
+        "solution":  "【分析】把42拆成40+2后分别与325相乘再相加.\n【详解】325×42=325×(40+2)=325×40+325×2，运用了乘法分配律。",
+        "knowledge_points":  [
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·浙江温州·期末"
+    }
+]
diff --git "a/\345\233\233\345\271\264\347\272\247\351\242\230\351\233\206/\345\233\233\345\271\264\347\272\247\344\270\213/\345\233\233\345\271\264\347\272\247\344\270\213\347\254\254\344\272\214\345\215\225\345\205\203\350\247\202\345\257\237\347\211\251\344\275\223.json" "b/\345\233\233\345\271\264\347\272\247\351\242\230\351\233\206/\345\233\233\345\271\264\347\272\247\344\270\213/\345\233\233\345\271\264\347\272\247\344\270\213\347\254\254\344\272\214\345\215\225\345\205\203\350\247\202\345\257\237\347\211\251\344\275\223.json"
new file mode 100644
index 0000000000000000000000000000000000000000..3b6ed8c4348e3787e7d67d0a83570fc0fb980d2a
--- /dev/null
+++ "b/\345\233\233\345\271\264\347\272\247\351\242\230\351\233\206/\345\233\233\345\271\264\347\272\247\344\270\213/\345\233\233\345\271\264\347\272\247\344\270\213\347\254\254\344\272\214\345\215\225\345\205\203\350\247\202\345\257\237\347\211\251\344\275\223.json"
@@ -0,0 +1,3243 @@
+[
+    {
+        "id":  "1",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "观察由3个小正方体拼成的立体图形，下列说法错误的是()。",
+        "options":  [
+                        "A. 从前面看可能是2个正方形",
+                        "B. 从上面看可能是3个正方形",
+                        "C. 从任何方向看都是3个正方形"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】![](images/55bfae0d52952dde75bddbe2bb5d56f45c2a6558ab04205684522232d25faee5.jpg)从前面看是2个正方形;![](images/fc637084e531f1c2fcdb7e7fe9f01fbac4bb01d52f977bc46fa8279a5f556fcf.jpg)从上面看是3个正方形;无论3个小正方体如何排列，从某些方向观察时，至少有一个正方体被遮挡，所以不存在从任何方向看都是3个正方形。据此解答.\n【详解】A。从前面看可能是2个正方形，举例:![](images/2026-03-06%20110135.jpg),正确;B。从上面看可能是3个正方形，举例:![](images/2026-03-06%20110147.jpg),正确;C。从任何方向看都是3个正方形，错误。故答案为:C",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "25-26三年级上·宁夏吴忠·期中"
+    },
+    {
+        "id":  "2",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "分别用5个小正方体搭成如图的三个立体图形，从()看这三个立体图形，看到的形状是完全相同的。![](images/c3f69f799dce87472c31eb410057c4ff.png)",
+        "options":  [
+                        "A. 正面",
+                        "B. 左面",
+                        "C. 上面",
+                        "D. 右面"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】分别分析从正面、左面、上面、右面观察三个立体图形的形状，找出形状完全相同的观察方向。\n【详解】A。从正面看，三个立体图形均为下层3个正方形，上层1个正方形靠左边，即:![](images/2026-03-06%20110300.jpg)，形状完全相同，符合题意。B。从左面看，第一个和第二个立体图形是下层2个正方形，上层1个正方形靠左边，即![](images/2026-03-06%20110318.jpg)。第三个立体图形是下层2个正方形，上层1个正方形靠右边，形状不同，不符合题意。C。从上面看，第一个立体图形是上面一层3个正方形，下面一层1个正方形靠右边，即![](images/2026-03-06%20110352.jpg)。第二个立体图形是上面一层3个正方形，下面一层1个正方形靠中间，即![](images/0fab049d0799aaf917c57f0478292661f9634dd49e257c3cca2e8d83b905404b.jpg)。第三个立体图形是上面一层3个正方形，下面一层1个正方形靠左边，即![](images/2026-03-06%20110415.jpg)。形状不同，不符合题意。D。从右面看，第一个和第二个立体图形是下层2个正方形，上层1个正方形靠右边，即![](images/2026-03-06%20110435.jpg)。第三个立体图形是下层2个正方形，上层1个正方形靠左边，即![](images/2026-03-06%20110449.jpg)。形状不同，不符合题意。故答案为:A",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "2025·河北石家庄·小升初真题"
+    },
+    {
+        "id":  "3",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "选择题",
+        "difficulty":  0.4,
+        "question":  "小红先向东偏北45°方向走了50米，又向南偏东45°方向走了50米，她现在所在的位置是在起点的()方向。",
+        "options":  [
+                        "A. 正东",
+                        "B. 正北",
+                        "C. 正西"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】首先，明确方向的基本概念(东，南，西，北及方位角)。小红先向东偏北45°走50米，再向南偏东45°走50米。我们可以通过画图来直观理解：东偏北45°和南偏东45°的路径形成了一个等腰直角三角形，两段路程长度相等(均为50米),最终的位置在起点的正东方向。本题主要考查根据方向和距离确定物体的位置，以及路径的综合分析，涉及方位角(如东偏北45°,南偏东45°)的理解与应用。解题关键是准确理解方位角的定义，并能通过画图或几何分析，明确两段路径的空间关系，从而判断最终位置相对于起点的方向.\n【详解】先画方位图：以起点为原点，画出东偏北45°方向的线段(长50米),再从该端点画出南偏东45°方向的线段(长50米)。![](images/014e3096f6e2e3cf4c3d8e7281f2300fcb3232f75866913ad7563e888e0017d5.jpg)再分析方向关系：东偏北45°与南偏东45°的夹角为45°,且两段路程长度相同，因此最终位置在起点的正东方向。故答案为:A.",
+        "knowledge_points":  [
+                                 "根据方向，角度和距离确定物体的位置，东南，西南，东北，西北方向"
+                             ],
+        "source":  "25-26六年级上·河北邯郸·阶段练习"
+    },
+    {
+        "id":  "4",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "从不同位置观察同一立体图形，我们看到的形状()。",
+        "options":  [
+                        "A. 可能相同",
+                        "B. 一定相同",
+                        "C. 一定不同",
+                        "D. 不可能相同"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】根据题意，对一般的物体来说，从不同的位置观察，所看到的形状是不一样的;但也有特殊情况，例如：这个物体是球体时，不管从什么方向看都是圆形，也就是看到的形状都一样，由此解答.\n【详解】根据分析可知:A。可能相同，球体：无论从哪个位置观察(如正面，侧面或上面),看到的形状始终是圆形。正确。B。一定相同，长方体从不同位置观察不一定相同，错误。C。一定不同，错误。D。不可能相同，错误。从不同位置观察同一立体图形，我们看到的形状可能相同。故答案为:A",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体，物体三视图的认识"
+                             ],
+        "source":  "25-26四年级上·云南曲靖·月考"
+    },
+    {
+        "id":  "5",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "下边的三个图形分别是从什么方向看到的。填一填。![](images/8dbc0fa04f0b880c44d417bf7bf124445da73e71dba2b1b8134cd0ea65159b74.jpg)![](images/359e0f41a5fe8662d5be9b6048eb0d20.png)从()面看;从()面看;从()面看",
+        "options":  [
+
+                    ],
+        "answer":  "上;右;前",
+        "solution":  "【分析】根据所给的立体图形，从不同方位观察物体形状：从前面看，能看到两排小正方形，下面一排有3个小正方形，上面一排左右两边各有1个小正方形，且两边对齐;从右面看，能看到两排小正方形，下面一排有4个小正方形，上面一排中间有1个小正方形，靠右边还有1个小正方形，且靠右对齐;从上面看，看到四排小正方形，第二排有3个小正方形，下面一排左右两边各有1个小正方形，且两边对齐，上面两排靠右边各有1个小正方形，且靠右对齐。据此解答。\n【详解】依次是从上面看、从右面看、从前面看。![](images/2026-03-06%20110830.jpg)",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "25-26四年级下·全国·课前预习"
+    },
+    {
+        "id":  "6",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个由相同的小正方体组成的立体图形，从任何一个方向看到的图形都是田字形，这个立体图形最少由()个小正方体组成。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】要满足从任何方向看到的图形都是田字形，即每个方向的正视图均为2×2的四个小正方形。通过合理摆放小正方体，使得每个方向的行和列均有两层，且小正方体位置被多个方向共享，从而减少总数。如图所示:![](images/d7b1013b7e0e370e9aa1937dc07a52bf9ba03f12aeed23e8fcfd840a5d138ae9.jpg).\n【详解】一个由相同的小正方体组成的立体图形，从任何一个方向看到的图形都是田字形，这个立体图形最少由6个小正方体组成。",
+        "knowledge_points":  [
+                                 "通过三视图会摆放立体图"
+                             ],
+        "source":  "24-25四年级下·浙江宁波·期中"
+    },
+    {
+        "id":  "7",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "小正方体的个数越多，从不同方向看到的形状就越复杂。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】小正方体的个数增加时，若摆放方式不同，从不同方向看到的形状可能复杂也可能简单，因此不一定更复杂.\n【详解】小正方体的个数增多并不必然导致从不同方向看到的形状更复杂。例如：将多个小正方体竖直叠成一列，无论数量多少，从正面，侧面和上面观察都只能看到一个正方形，形状简单;若将小正方体分散摆放，数量增加可能使形状复杂。但题干未限定摆放方式，因此结论不成立。故答案为: ×",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "25-26四年级上·云南曲靖·月考"
+    },
+    {
+        "id":  "8",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "从前面，上面和左面观察同一立体图形，看到的形状一定不同。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】从前面，上面和左面观察同一立体图形，看到的形状一般是不同的，但也有特殊的立体图形，看到的形状相同。例如，正方体从这三个方向观察到的形状均为正方形，形状相同;球从这三个方向观察到的形状均为圆形，形状相同.\n【详解】当立体图形为正方体时，从前面，上面和左面观察到的形状均为相同的正方形，因此这三个方向的形状相同，而非一定不同，原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体，物体三视图的认识"
+                             ],
+        "source":  "25-26四年级上·云南曲靖·月考"
+    },
+    {
+        "id":  "9",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "从不同的位置观察同一个物体，看到的图形不可能相同。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】从不同的位置观察同一个物体，看到的图形有时相同，有时不同，这与物体的形状有关。\n【详解】例如，长方体从不同位置观察可能看到长方形或正方形，而球体无论从哪个位置观察都只能看到圆形。因此，原题中\"不可能相同\"的说法不正确。故答案为: ×",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "24-25四年级下·广东汕头·期末"
+    },
+    {
+        "id":  "10",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "同一个物体，从不同的位置观察，所看到的图形不可能相同。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】同一个物体，从不同的位置观察，所看到的图形可能相同，也可能不相同，可举例说明.\n【详解】同一个物体，从不同的位置观察，所看到的图形可能相同，也可能不相同，例如一个正方体，从上面，下面，前面，后面和侧面看到的都是相同的正方形。故答案为: ×",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "11",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "正方体从前面和左面看到的形状相同。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】正方体的六个面都是完全相同的正方形。从前面和左面观察时，若正对这两个面，看到的都是正方形，形状相同.\n【详解】正方体每个面都是正方形，且所有面大小相等。从前面观察时，看到的是前面的正方形;从左面观察时，看到的是左面的正方形。由于正方形的形状和大小完全相同，因此从前面和左面看到的形状相同。故答案为:√",
+        "knowledge_points":  [
+                                 "正方体的特征，从不同位置观察单个物体"
+                             ],
+        "source":  "24-25四年级下·新疆克孜勒苏·期末"
+    },
+    {
+        "id":  "12",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "从不同方向观察同一个物体，看到的图形可能相同，也可能不同。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】根据观察物体的知识，不同方向观察同一物体时，视图可能相同或不同，具体取决于物体的形状和观察方向。例如，球体各方向视图相同，而长方体不同方向视图可能不同.\n【详解】当观察球体时，无论从哪个方向观察，看到的图形都是相同的圆形;而观察长方体时，从前面和侧面看到的图形不同，但左右面可能相同。因此，从不同方向观察同一物体，看到的图形可能相同，也可能不同。故答案为:√",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "24-25四年级下·贵州黔南·期末"
+    },
+    {
+        "id":  "13",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "观察同一物体，在不同的位置观察，看到的图形一定不相同。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】观察同一物体，在不同的位置观察，看到的图形可能相同，也可能不同。例如，观察一个正方体，从正方体的前面，后面，左面，右面，上面，下面观察，看到的图形都是正方形，即看到的图形相同;再如，观察一个圆柱，从圆柱的上面和下面观察，看到的图形是圆，从圆柱的侧面观察，看到的图形是长方形或正方形，即看到的图形不同.\n【详解】题目中说\"观察同一物体，在不同的位置观察，看到的图形一定不相同\"，这与实际情况不符，因为存在看到的图形相同的情况。所以该说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "24-25四年级上·贵州毕节·期末"
+    },
+    {
+        "id":  "14",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "如果两个立体图形从前面和右面看到的图形相同，那么这两个立体图形一定相同。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】为了验证原说法是否正确，我们来看具体的例子。这里给出两个立体图形，展示如下:![](images/29e355c1591389e0c4660f467c5c9bb7.png)从前面和右面观察这两个立体图形，可以看到它们从正面，右面看到的形状相同，都是:![](images/125304073df5f8de31619cc44da4490e884afacc7b39440267dcdb1d9b2945ea.jpg)虽然这两个立体图形从前面和右面看到的图形一样，但是很明显，左，右两个立体图形的形状是不同的.\n【详解】根据分析可知：\"如果两个立体图形从前面和右面看到的图形相同，那么这两个立体图形一定相同.\"这个说法是错误的。故答案为: ×",
+        "knowledge_points":  [
+                                 "通过三视图还原立体图"
+                             ],
+        "source":  "24-25四年级下·湖南张家界·期末"
+    },
+    {
+        "id":  "15",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "从不同的角度去观察同一个物体，所观察到的图形一定不一样。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】同一物体从不同角度观察到的图形可能相同，也可能不同。例如，球体无论从哪个角度观察，图形都是圆形，据此判断即可。\n【详解】当观察球体时，无论从任何方向观察，看到的图形都是相同的圆形。因此，从不同角度观察同一物体，所看到的图形可能一样，也可能不一样，原题中的\"一定不一样\"说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体，物体三视图的认识"
+                             ],
+        "source":  "24-25四年级下·湖北省直辖县级单位·期中"
+    },
+    {
+        "id":  "16",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "不同的物体分别从不同的角度观察，看到的形状有可能相同，也可能不同。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】根据观察物体的知识，不同物体从不同角度观察时，由于物体形状和观察角度的差异，所看到的形状可能相同(如圆柱和球体从上方观察均为圆形),也可能不同(如长方体和球体从正面观察结果不同).\n【详解】不同的物体分别从不同角度观察，看到的形状有可能相同，也可能不同。题干描述正确。故答案为:√",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "22-23四年级下·新疆乌鲁木齐·期末"
+    },
+    {
+        "id":  "17",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "从前面，左面，上面观察一个长方体，看到的形状可能完全相同。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】根据题意，长方体可能有三个面是正方形的情况，即当长，宽，高相等时，此时长方体为正方体。正方体属于长方体的特殊形式，从前面，左面，上面观察时，看到的均为大小相同的正方形，形状完全相同.\n【详解】根据分析可知：当长方体的长，宽，高相等时，它实际上是一个正方体。正方体的每个面都是完全相同的正方形。从前面，左面，上面观察时，视线分别垂直于对应面，看到的形状均为正方形，且大小和形状完全相同。因此，这种情况是可能的。因此原说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "24-25四年级下·湖南长沙·期中"
+    },
+    {
+        "id":  "18",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "观察同一个物体，从不同方向看到的形状可能相同。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】对一般的物体来说，从不同的位置观察物体，所看到的形状是不同的;但有特殊情况，例如正方体，那么从正面，上面，左面等不同位置，看到的都是完全一样的正方形，即看到的形状一样，据此解答即可.\n【详解】![](images/2026-03-06%20111119.jpg)如上图，观察一个物体，从不同位置观察到的形状可能相同;原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "24-25四年级下·内蒙古通辽·期中"
+    },
+    {
+        "id":  "19",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "从前面，上面，左面观察一个正方体，看到的形状都相同。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】正方体的每个面都是完全相同的正方形。从前面，上面，左面观察时，视线均垂直于对应面，看到的形状均为正方形，且大小和形状完全相同。原说法正确。故答案为:√\n【详解】正方体的6个面都是大小相等的正方形。从前面观察时，看到的是前面的正方形;从上面观察时，看到的是上面的正方形;从左面观察时，看到的是左面的正方形。由于所有面的大小和形状相同，因此从这三个方向观察到的形状均为正方形，且完全相同。据此判断。",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "24-25四年级下·湖南岳阳·期中"
+    },
+    {
+        "id":  "20",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "从正面，左面，上面观察正方体，看到的图形一定是正方形。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】根据正方体的六个面都是正方形，结合题意可知从正面，左面，上面观察正方体，看到的图形一定是正方形，据此解答即可.\n【详解】根据分析可知：从正面，左面，上面观察正方体，看到的图形一定是正方形。原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "24-25四年级下·湖北武汉·期中"
+    },
+    {
+        "id":  "21",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "从不同位置观察用小正方体搭成的一个几何体，看到的图形可能相同，也可能不同。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】可以举例子，例如这样一个几何体![](images/2026-03-06%20111203.jpg),画出从前面，右面，上面观察到的图形，看它们的形状是否相同.\n【详解】![](images/1c4d06f13d5c4beb2b465fb01b3dca889f4aebdafd3b29fd0cf9bd5b100bcede.jpg)这个几何体从前面看到的形状是![](images/2026-03-06%20111217.jpg),从右面看到的形状是![](images/2026-03-06%20111229.jpg),从上面看到的形状是![](images/2026-03-06%20111246.jpg)。从前面和右面看到的形状相同，从上面和从右面看到的形状不同。所以，从不同位置观察用小正方体搭成的一个几何体，看到的图形可能相同，也可能不同。题干说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "三视图的画法，物体三视图的认识"
+                             ],
+        "source":  "24-25四年级下·山西忻州·期中"
+    },
+    {
+        "id":  "22",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "从不同方向观察一个物体，看到的图形可能相同。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】当这个物体是一个球体，那么从不同方向看到的形状都一样;但是对于一般的物体，从不同的位置观察，所看到的形状是不同的;所以从不同方向观察同一个物体，看到的形状有可能相同。以此答题即可。\n【详解】根据分析可知：从不同方向观察一个物体，看到的图形可能相同。例如：一个球体，从不同方向看到的形状都是圆。原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "24-25四年级下·江西赣州·期中"
+    },
+    {
+        "id":  "23",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "同一位置，最多能看到物体的三个面。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】我们观察一个物体，如观察一个长方体，从它一个面看，只能看到它的一个面，从它的一条棱看，可看到它的两个面，从它的一个顶点看，能看到它的三个面，因此，站一个位置上观察物体，最少可看到一个面，最多可看到三个面.\n【详解】根据分析可得：同一位置，最多能看到物体的三个面，是正确的;故答案为:√",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体，物体三视图的认识"
+                             ],
+        "source":  "20-21五年级下·全国·期中"
+    },
+    {
+        "id":  "24",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "如果两个物体从前面和左面看到的图形相同，那么这两个物体一定相同。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】为了验证原说法是否正确，我们来看具体的例子。这里给出两个立体图形，展示如下:![](images/29e355c1591389e0c4660f467c5c9bb7.png)从前面和左面观察这两个立体图形，可以看到它们从正面、左面看到的形状相同，都是:![](images/a4fc7bdae49020fe0d13c198bf53bb2b63fa078672249fa6702e2900456af0e9.jpg)虽然这两个立体图形从前面和左面看到的图形一样，但是它们的整体形状不同。\n【详解】根据分析可知：\"如果两个立体图形从前面和左面看到的图形相同，那么这两个立体图形一定相同。\"这个说法是错误的。故答案为: ×",
+        "knowledge_points":  [
+                                 "三视图的画法"
+                             ],
+        "source":  "24-25四年级下·陕西西安·期中"
+    },
+    {
+        "id":  "25",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "从上面和左面看起来图形相同的物体它的整体形状一定相同。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】从上面和左面看起来图形相同的物体它的整体形状不一定相同，举例解答即可.\n【详解】例如![](images/2026-03-06%20111341.jpg)这两个物体，从上面看都是![](images/2026-03-06%20111355.jpg),从左面看都是![](images/2026-03-06%20111406.jpg),这两个物体，从上面和左面看起来图形相同，但是两个物体它的整体形状不相同，所以题干说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "24-25四年级下·广东汕头·期中"
+    },
+    {
+        "id":  "26",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "观察一个物体，最多可以看到3个面，最少可以看到1个面。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】根据题意，例如：我们可以把一个长方体放在桌子上，从不同的角度去观察，最多能看到3个面，最少能看到1个面，据此判断.\n【详解】根据分析可得：如我们把一个长方体放在桌子上，从不同的角度去观察，最多能看到3个面，最少能看到1个面，所以原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "24-25四年级下·河南南阳·期中"
+    },
+    {
+        "id":  "27",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "从前面，上面和左面观察一个正方体，看到的图形都是正方形。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】正方体的每个面都是完全相同的正方形。从前面观察：正对前面，看到的是正方形。从上面观察：正对顶面，看到的仍是正方形。从左面观察：正对左侧面，同样看到正方形。无论从哪个正对方向观察正方体，看到的图形都是正方形.\n【详解】从前面，上面和左面观察一个正方体，看到的图形都是正方形。说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "24-25四年级下·河北唐山·期中"
+    },
+    {
+        "id":  "28",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "从同一位置观察不同的物体，图形可能相同，也可能不同。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】观察下列两个图形，分别对它们从左面和前面看到的形状进行判断即可。![](images/f7d7a13036bfe3fe88ca5183eb550c3618a3795d0bef3300c30997b994b55ad2.jpg)和![](images/2026-03-06%20111440.jpg),从左面看到的图形分别是:![](images/2026-03-06%20111451.jpg),![](images/2026-03-06%20111502.jpg)看到的图形相同;从正面看到的图形分别是:![](images/2026-03-06%20111524.jpg),![](images/2026-03-06%20111531.jpg)看到的图形不相同.\n【详解】从同一位置观察不同的物体，图形可能相同，也可能不同。原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "23-24四年级下·河北张家口·期中"
+    },
+    {
+        "id":  "29",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "从一个方向观察长方体，最多能看到它的五个面。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】从长方体的一面观察，只能看到1个面;从长方体的一条棱(面与面相交形成的线)观察，只能看到长方体的2个面;从长方体的一个顶点观察，只能看到长方体的3个面。即站在任意位置观察同一长方体最多能看到长方体的3个面.\n【详解】根据分析可知，从一个方向观察长方体最多能看到它的三个面，原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "23-24四年级下·山东菏泽·月考"
+    },
+    {
+        "id":  "30",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "我们站在同一位置，最多能看到正方体状物体的3个面。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】从正方体的一面观察，只能看到1个面;从正方体两个面相交的那条线(棱)观察，只能看到正方体的2个面;从正方体三个面相交的那个点(顶点)观察，只能看到正方体的3个面。即站在任意位置观察同一正方体最少能看到正方体的1个面，最多能看到正方体的3个面。据此判断即可.\n【详解】根据分析可知：我们站在同一位置，最多能看到正方体状物体的3个面，原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "22-23四年级下·四川凉山·期末"
+    },
+    {
+        "id":  "31",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "用4个相同的小正方体可以搭建成一个大的正方体。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】由题意得，用4个相同的小正方体搭建的几何体如下:![](images/6f08d36249f833ccb81b22d582ae1e22.png)\n【详解】由分析得，用4个相同的小正方体可以搭建成一个大长方体，不能搭建成一个大的正方体。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "立体图形的切拼(长方体，正方体的体积)"
+                             ],
+        "source":  "2022·山西太原·小升初真题"
+    },
+    {
+        "id":  "32",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "一个立体图形，从前面看到的图形是![](images/2026-03-06%20111602.jpg),这个立体图形一定是由2个小正方体摆成的。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】一个立体图形，从前面看到的图形是![](images/2026-03-06%20111621.jpg),这个立体图形可能由2个小正方体摆成，如图:![](images/2026-03-06%20111636.jpg)。这个立体图形也可能由3个小正方体摆成，如图:![](images/2026-03-06%20111705.jpg)。这个立体图形也可能由4个小正方体摆成，如图:![](images/2026-03-06%20111838.jpg)在![](images/2026-03-06%20111846.jpg)的前面和后面添加小正方体，看到的都是![](images/2026-03-06%20111857.jpg).\n【详解】一个立体图形，从前面看到的图形是，这个立体图形可能是由2个小正方体摆成的，也有可能是由3个小正方体摆成的，也可能是由4个小正方体摆成的，也可能是由5个小正方体摆成的。小正方体个数不确定。故答案为: ×",
+        "knowledge_points":  [
+                                 "通过三视图会摆放立体图"
+                             ],
+        "source":  "19-20四年级下·全国·单元测试"
+    },
+    {
+        "id":  "33",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "用几个正方体搭成一个物体，从前面看到的图形是![](images/2026-03-06%20111948.jpg),这个物体一定是由2个正方体搭成的。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】本题主要考查物体三视图的相关知识。一个物体，从前面看到的图形是![](images/2026-03-06%20112006.jpg),这个物体可能的情况如下:![](images/3feeeb55c5bd76391af706b106501c56.png)等.\n【详解】由分析得，这个物体有可能是2个正方体搭成的，也有可能是3个正方体搭成的。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "通过三视图会摆放立体图"
+                             ],
+        "source":  "23-24四年级下·吉林四平·期末"
+    },
+    {
+        "id":  "34",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "一个正方体最多可以同时观察到四个面。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】由于正方体有六个面，每个面都是正方形，而从一个位置最多只能看到一个物体的三个面，因此从任何角度最多只能看到三个面。据此判断。\n【详解】根据分析得：一个正方体最多能同时观察到3个面。故答案为: ×",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "23-24四年级下·新疆和田·期末"
+    },
+    {
+        "id":  "35",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "一个几何体从不同方向看到的形状一定不相同。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】本题可以举例子来说明原题说法的正确性。一个几何体，如下图:\n![](images/a1b78292fadb5b9dc0c88500c89229989bed297a7ae16b65ed8c08e106c3f13f.jpg)\n这个几何体从前面和左面看到的图形都是![](images/2026-03-06%20112117.jpg).\n【详解】由分析得，一个几何体从不同方向看到的形状可能相同。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "23-24四年级下·湖南益阳·期末"
+    },
+    {
+        "id":  "36",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "从同一位置观察不同的物体，看到的图形一定不一样。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】可以通过举例子来说明原题说法的正确性。两个不同的几何体如图:\n![](images/c39a1391c4dcbb45833572281f9608e0.png)\n两个几何体，都从正面来看，看到的图形都是![](images/93d29eb8beb4a36ec94f1ce3e5706af43fdeb03c63382c551899a0a1897b931a.jpg)\n【详解】由分析得：从同一位置观察不同的物体，看到的图形可能一样。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "18-19四年级下·全国·期中"
+    },
+    {
+        "id":  "37",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "\"横看成岭侧成峰\"表示从两个方向看庐山得到的结果会不同。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】根据\"横看成岭侧成峰\"这句诗可以知道庐山是座丘壑纵横，峰峦起伏的大山，游人所处的位置不同，看到的景物也各不相同，据此解答.\n【详解】根据分析：\"横看成岭侧成峰\"表示从两个方向看庐山得到的结果会不同，原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "23-24四年级下·湖北·期末"
+    },
+    {
+        "id":  "38",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "观察一个物体，从不同位置观察到的形状可能相同。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】对一般的物体来说，从不同的位置观察物体，所看到的形状是不同的;但有特殊情况，例如正方体，那么从正面，上面，左面等不同位置，看到的都是完全一样的正方形，即看到的形状一样，据此解答即可.\n【详解】![](images/2026-03-06%20112231.jpg)所以，观察一个物体，从不同位置观察到的形状可能相同。原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "23-24四年级下·江西南昌·期末"
+    },
+    {
+        "id":  "39",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "站在一个位置上观察一个正方体，最多只能看到它的3个面。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】站在一个位置上观察一个正方体，角度合适可以最多看到前面，上面和侧面;但看到上面看不到下面，看到前面看不到后面，看到左面看不到右面，据此判断.\n【详解】站在一个位置上观察一个正方体，角度合适可以最多看到前面，上面和侧面3个面，原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "23-24四年级下·河南新乡·期中"
+    },
+    {
+        "id":  "40",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.94,
+        "question":  "从不同方向观察同一个物体，看到的形状有可能相同。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】采用举例的方法进行判断。对于一般的物体，例如一只玩具熊，从不同的位置观察，所看到的形状是不同的;再例如一个球，从不同的方向看到的形状是相同的;据此判断即可。\n【详解】当这个物体是一个球体时，从不同方向看到的形状都一样;但是对于一般的物体，从不同的位置观察，所看到的形状是不同的;所以从不同方向观察同一个物体，看到的形状有可能相同。故答案为:√",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "23-24四年级下·河南开封·期中"
+    },
+    {
+        "id":  "41",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "无论从什么角度观察一个标准的乒乓球，看到的图形都是圆。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】球的主要特征：从正面看，从左面看，从上面看或其它方向看，看到的图形都是一个圆。据此解答.\n【详解】一个标准的乒乓球是一个球，它从各个角度来看，看到的图形都是圆。故答案为:√",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "23-24四年级下·河南新乡·期中"
+    },
+    {
+        "id":  "42",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "用3个小正方体摆成的物体，从哪个方向看到的形状都是相同的。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】根据题意，图形![](images/2026-03-06%20112324.jpg),从上面共看2行，上面1行1个小正方体，下面1行2个小正方体，左对齐;从前面看共1行,2个小正方体;从左面看共看2行，上面1行1个小正方体，下面1行1个小正方体;从右面看共看2行，上面1行1个小正方体，下面1行1个小正方体，从上面和前面看图形是不同的，从左面和右面看图形是相同的，据此解答即可.\n【详解】图形![](images/2026-03-06%20112336.jpg)从上面看![](images/2026-03-06%20112348.jpg),从前面看![](images/2026-03-06%20112359.jpg),从左面看![](images/2026-03-06%20112409.jpg),从右面看![](images/2026-03-06%20112422.jpg)。用3个小正方体摆成的物体，从哪个方向看到的形状不都是相同的。所以原题干说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "三视图的画法",
+                                 "物体三视图的认识"
+                             ],
+        "source":  "23-24四年级下·广东东莞·期中"
+    },
+    {
+        "id":  "43",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "从不同的角度观察一个长方体的茶叶盒，一次最多可以看到4个面。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】站在不同位置观察同一个长方体，可能会看到1个面，也可能看到2个面，最多看到3个面;据此解答.\n【详解】根据分析：从不同的角度观察一个长方体的茶叶盒，一次最多可以看到3个面，原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "23-24四年级下·山西长治·期中"
+    },
+    {
+        "id":  "44",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "观察一个物体，从不同位置观察到的形状一定不相同。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】![](images/8335d23d9a9c46eb1add3e4d9984283b73cbad7d6216150dd10ddfb9fc7b6a32.jpg)从前面看，能看到4个面共3列，第1列1个面，第2列1个面与第1列的面对齐，第3列2个面，最下面的面与第2列的面对齐;从左边看，能看到2个面共1列;从右面看，能看到2个面共1列.\n【详解】从前面看:![](images/2026-03-06%20112500.jpg)从左面看:![](images/2026-03-06%20112512.jpg) 从右面看:![](images/2026-03-06%20112523.jpg)\n观察一个物体，从不同位置观察到的形状可能相同，也可能不相同，所以原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "18-19六年级上·辽宁·单元测试"
+    },
+    {
+        "id":  "45",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "一个物体，如果从前面，上面和左面看到的形状都相同，那么这个物体一定是一个正方体。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】对于一个正方体来说，从前面，上面和左面看到的形状都相同且都是正方形。除了正方体，球从前面，上面和左面看到的形状都相同且都是圆.\n【详解】由分析可知：一个物体，如果从前面，上面和左面看到的形状都相同，那么这个物体可能是一个正方体，而不是一定是一个正方体。故答案为: ×",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "23-24四年级下·湖北武汉·期中"
+    },
+    {
+        "id":  "46",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "从不同方向观察立体图形，看到的形状一定不相同。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】从不同方向观察一个立体图形，所看到的形状可能不相同，也可能相同;如\n![](images/5dedebd69a976df9529d34a14a7cab116ae41319a571a08a1a8d9cb174e7589c.jpg)\n该立体图形从前面看到的图形是：第一层有2个正方形，第二层有1个，右对齐;从上面看到的图形是：第一层有2个正方形，第二层有1个，右对齐;从右面看到的图形是：第一层有2个正方形，第二层有1个，右对齐;据此判断即可.\n【详解】例如:![](images/828a9047d3bd6a8a24f87552f07ae31c428ff3604971d7ee08173816850371cb.jpg)从正面看到的图形是![](images/e2e7bda10483ccc9521bbf6ee39a5adc15501024ecc6b0bbbcdb59cce91a5e00.jpg),从上面看到的图形是![](images/2026-03-06%20112605.jpg),从右面看到的图形是![](images/2026-03-06%20112618.jpg)。所以，从不同方向观察立体图形，看到的形状一定不相同，此说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "23-24四年级下·湖南张家界·期中"
+    },
+    {
+        "id":  "47",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "观察一个物体，从不同位置观察到的图形可能相同。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】对一般的物体来说，从不同的位置观察物体，所看到的形状是不同的;但有特殊情况，这个物体是正方体，那么从正面，上面，左面看到的都是完全一样的正方形，即看到的形状一样。据此判断.\n【详解】如图:![](images/2026-03-06%20112711.jpg)所以观察一个物体，从不同位置观察到的图形可能相同。故答案为:√",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "18-19二年级下·浙江·课后作业"
+    },
+    {
+        "id":  "48",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "几个不同的物体，从不同角度观察，看到的形状可能相同。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】从上面观察圆柱体时，看到的形状是圆形。从任何方向观察球体，看到的形状都是圆形，因此几个不同的物体，从某些角度观察，看到的形状可能相同。据此判断。\n【详解】由分析得：几个不同的物体，从不同角度观察，看到的形状可能相同。如观察圆柱体和球体时，都可能看到圆形，原说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "20-21四年级下·内蒙古巴彦淖尔·期末"
+    },
+    {
+        "id":  "49",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "从不同的位置观察同一个物体，看到的图形可能相同。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】对于一般的物体，从不同的位置观察物体，所看到的形状是不同的;但有特殊情况，如果这个物体是球体，那么从不同的方向看到的形状一样;据此判断即可.\n【详解】从不同的位置观察同一个的物体，看到的图形可能相同。原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "19-20三年级上·辽宁·月考"
+    },
+    {
+        "id":  "50",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "观察一个物体，从任何面看到的形状都相同。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】根据对三视图的认识进行判断，因此可观察下列一个物体从前面，上面和左面看到的形状，然后再判断即可.\n【详解】![](images/6dc4de18d8c96d9b74c154ad59e1794aef7fdb75c1bfa6f674f36015dbfe0717.jpg)此图从前面看到的形状是:![](images/2026-03-06%20112836.jpg);从上面看到的形状是:![](images/2026-03-06%20112852.jpg);从左面看到的形状是:![](images/2026-03-06%20112904.jpg),所以观察一个物体，从任何面看到的形状可能相同，也可能不相同，故原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "22-23四年级下·宁夏固原·月考"
+    },
+    {
+        "id":  "51",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "从同一个位置观察不同的物体，看到的图形可能不一样。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】物体各方向的形状不一样则看到的不一样，若物体各方位完全一致，则看到的图形也是一致的.\n【详解】从同一位置观察不同的物体，观察到的图形有可能一样也有可能不一样。故答案为:√",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "22-23四年级下·河南三门峡·期中"
+    },
+    {
+        "id":  "52",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "几个不同的立体图形，从同一方向看到的图形可能相同。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】如下图，三个不同的几何体，从左面观察，看到的都是![](images/2026-03-06%20112924.jpg),所以几个不同的立体图形，从同一方向看到的图形可能相同，据此即可解答.\n![](images/3283ad6b1d1ffb5d527a3dfa764a6df6.png)\n【详解】根据分析可知，几个不同的立体图形，从同一方向看到的图形可能相同，原说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "22-23四年级下·山西晋中·期中"
+    },
+    {
+        "id":  "53",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "从前面看和上面看一个物体都是4个正方形，这个物体一定是由4个正方体摆成的。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】从前面看和上面看一个物体都是4个正方形，说明这个物体可能有两层，下一层4个正方体摆成1个大正方体，上一层2个或4个正方体，那么一共至少有6个正方体;也可能只有一层，一行4个小正方体;据此解答.\n【详解】根据分析：从前面看和上面看一个物体都是4个正方形，这个物体不一定一定是由4个正方体摆成的，原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "通过三视图会摆放立体图"
+                             ],
+        "source":  "22-23四年级下·四川绵阳·期中"
+    },
+    {
+        "id":  "54",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "从同一个位置观察不同的物体，看到的形状不可能相同。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】从同一个方向观察不同的物体，如观察圆柱体和球体时，看到的形状可能都是圆形，正方体从哪面看都是正方形，所以从同一位置观察不同的物体，看到的形状可能相同也可能不同;据此解答即可.\n【详解】从同一个位置观察不同的物体，看到的形状可能相同。故答案为: ×",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体",
+                                 "从不同位置观察两个物体的相互关系",
+                                 "物体三视图的认识"
+                             ],
+        "source":  "22-23四年级下·江西赣州·期中"
+    },
+    {
+        "id":  "55",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "从同一位置观察不同的物体，看到的图形不可能相同。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】从同一位置观察不同的物体，比如都从左面观察不同的图形，看到的图形可能相同，也可能不相同，举例解答.\n【详解】![](images/2026-03-06%20113020.jpg)如图所示，从左面观察ABCF这4个图形，看到的都是![](images/46153fee62656b62980b30b45c2c6c9dd239dc1614d2e708401d222342d51295.jpg)。但从左面观察DE这2个图形，看到的都是![](images/12f16d70448882abb4ba16856ab04834c346dd76af7823fdbfccd208629162e3.jpg)也就是从同一位置观察不同的物体，看到的图形可能相同，也可能不相同。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "22-23四年级下·广东汕尾·期中"
+    },
+    {
+        "id":  "56",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "从不同方向观察立体图形看到的形状可能是相同的。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】一般情况下，从不同方向观察立体图形看到的形状是不同的，但是对于规则对称的图形如球体和正方体等，从不同方向看到的形状却是相同的，据此解答即可.\n【详解】从分析可知，从不同方向观察立体图形看到的形状可能是相同的。原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "22-23四年级下·新疆喀什·期中"
+    },
+    {
+        "id":  "57",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "几个不同的物体不管从什么角度观察，看到的图形一定不同。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】几个不同的物体不管从什么角度观察，看到的图形可能相同，也可能不同。可举例子说明.\n【详解】如球和圆柱体，虽然是不同的物体，但是球不管从什么角度看到的都是圆。而圆柱体，如果立在那里，从正面看，就是一个长方形。从上面看，就是一个圆。所以几个不同的物体不管从什么角度观察，看到的图形可能相同，也可能不同。故此说法错误。",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "22-23四年级下·云南楚雄·期中"
+    },
+    {
+        "id":  "58",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "观察一个物体，从不同方向看到的形状可能一样也可能不一样。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【详解】对于一般的物体，从不同方向看，看到的图形一般不同;如图:![](images/2026-03-06%20142208.jpg)如果这个物体是正方体或球，不论从哪一个方向看，所看到的图形都是一样的;所以观察一个物体，从不同方向看到的形状可能一样也可能不一样。故答案为:√",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "22-23四年级下·河南南阳·期中"
+    },
+    {
+        "id":  "59",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "不同的立体图形，从同一位置看到的形状可能是相同的。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】观察物体的形状时，从不同位置观察立体图形的形状，主要包括前面，上面和左面观察。辨认，从同一位置观察不同的立体图形，看到的形状可能相同，也可能不相同，如：长方体的侧面和圆柱体侧面，观察到的图形却是相同的，是长方形，因此得解.\n【详解】不同的立体图形，从同一位置看到的形状可能是相同的。原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体",
+                                 "物体三视图的认识"
+                             ],
+        "source":  "22-23四年级下·山东济宁·期中"
+    },
+    {
+        "id":  "60",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "观察一个物体，同时看到的最多三面。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】在观察物体时，由于视线会受到物体的遮挡，从一个地方看过去，并不能完全看到这个物体的各个面。观察一个长方体时，当眼睛在长方体的一个顶点时，能看到的面最多，最多只有3个面，所以我们观察物体时，要从多个角度去观察。据此解答。本题主要考查对物体的观察，属于基础知识，要熟练掌握.\n【详解】![](images/3338d173f6e06c6730d9f9ebe7fa8ddefd463ba9640eba935a0c21c7a8242ad4.jpg)如上图所示，观察一个物体，同时看到的最多三面，题目说法是正确的。故答案为:√",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "22-23四年级下·全国·单元测试"
+    },
+    {
+        "id":  "61",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "把3个完全一样的小正方体拼成一个长方体，从左面和从正面看到的物体形状一定相同。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】如图![](images/304467a466d8f9c2531e166b0583e74c0f8cef94fbd2b96d8a283aef60b89e6b.jpg),两种摆放方式，从左面和从正面看到的物体形状都不相同，据此分析。关键是具有一定的空间想象能力，能确定从不同方向观察物体看到的形状.\n【详解】把3个完全一样的小正方体拼成一个长方体，从左面看到一个小正方形，从从正面看到一行3个小正方形;也可能从左面看到一行3个小正方形，从正面看到一个小正方形，所以看到的物体形状不相同，所以原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "20-21五年级下·海南省直辖县级单位·期中"
+    },
+    {
+        "id":  "62",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "把一个正方体放在桌面上，最多只能看到它的一个面。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】一个正方体，从它的1个面看，只能看到1个面，从它的1条棱看，能看到2个面，从它的一个顶点看，能看到3个面，也就是一个正方体无论在哪个位置观察，最多只能看到3个面，据此解题。此题主要考查正方体的特征，感受观察范围随观察点，观察角度的变化而变化，并能利用所学知识解释生活中的一些现象.\n【详解】把正方体放在桌面上，最多能看到3个面，原说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "正方体的特征",
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "21-22四年级下·广东阳江·期末"
+    },
+    {
+        "id":  "63",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "从同一位置观察不同的几何体，看到的形状可能相同，也可能不同。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】观察下列两个图形，分别对它们从左面和前面看到的形状进行判断即可。图①,②从左面看，都可看到2层，第1层都可看到2个小正方形，第2层都可看到1个小正方形，都是左齐;图①从前面看，可看到2层，第1层都可看到3个小正方形，第2层都可看到1个小正方形，居中对齐;图②从前面看，可看到2层，第1层都可看到1个小正方形，第2层都可看到1个小正方形，左齐，依此判断。熟练掌握对三视图的认识是解答此题的关键.\n![](images/bcefd941a9ff3157bdffd7ad4196838db7988487d08a19122f438caf997662aa.jpg)\n【详解】根据分析可知，从同一位置观察不同的几何体，看到的形状可能相同，也可能不同。故答案为:√",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "21-22四年级下·河北廊坊·期末"
+    },
+    {
+        "id":  "64",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "观察一个物体，从不同位置观察到的形状不可能相同。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】根据对三视图的认识进行判断，因此可观察下列一个物体从左面，前面，上面看到的形状，然后再判断即可。熟练掌握对三视图的认识是解答此题的关键.\n![](images/ce78ed0cdb7703438b8b917c0e7f5c3b6241ee3ce0150c191067606dd7352ad2.jpg)\n【详解】![](images/647ea6a71a8f123aece620f4164fdd495055f3097d9315492b499e62446ad878.jpg)此图从左面看到的形状是:![](images/2026-03-06%20142636.jpg);从前面![](images/82fe8be1f0726f2c154623fdf57665979452c1659598a2a34512d6156c046d1a.jpg)看到的形状是:;从上面看到的形状是:![](images/916d5672b877b3d8296fe2107fe1d2618966f634032fdd6247ca116d77bd716e.jpg),此图从左面和前面看到的形状相同，和上面看到的形状不相同。因此观察一个物体，从不同位置观察到的形状可能相同，也可能不相同。故答案为: ×",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "21-22四年级下·河南三门峡·期末"
+    },
+    {
+        "id":  "65",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "几个不同的物体，从相同的位置观察，看到的形状可能相同。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】从同一个方向观察不同的物体，如观察圆柱体和球体时，看到的形状可能都是圆形;据此解答即可。本题考查观察物体，旨在考查学生的空间想象和推理能力.\n【详解】从上面观察圆柱体时，看到的形状是圆形。从任何方向观察球体，看到的形状都是圆形，则几个不同的物体，从相同的位置观察，看到的形状可能相同。说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "21-22四年级下·河北秦皇岛·期末"
+    },
+    {
+        "id":  "66",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "一个搭建成的几何体图形，随意添加一个小正方体不一定会影响观察的平面图形的形状。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】只要在观察方向上与原来的小正方体重合，则在此面看到的形状就不变。例![](images/5fe50160231065f20e7f3329dac679c7e1ef4d4ffd5a303394a321ecfd8cc824.jpg):从上面看到的是![](images/2026-03-06%20142854.jpg)。在这个立体图形的上面添加一个小正方体，如图:![](images/9489955b9bfb8c5920e994f6304dac03fe030a6be56b82b908a80f7a279f4a28.jpg),![](images/987ca00f5b0ae4ecfb11fcb5c039258a70acf9cebc9434f40dc213554cac7c05.jpg)从上面看到的还是![](images/745f1805ca30663ea8ecce3b8c64920381affc7f304e445189b7a701f8e03f3b.jpg)。此题考查了从不同方向观察物体和几何体，锻炼了学生的空间想象力和抽象思维能力.\n【详解】由分析可得：在几何体图形，随意添加一个小正方体不一定会影响观察的平面图形的形状，所以原题说法正确;故答案为:√",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "21-22五年级下·贵州铜仁·期末"
+    },
+    {
+        "id":  "67",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "从不同的位置观察，才能更全面的认识一个物体。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】因为从不同的角度观察到的图形可能是不一样的，据此判断即可。本题考查从不同的方向观察物体知识点，从不同的角度观察物体才能更全面认识物体.\n【详解】例如![](images/038ea67034d8be6b304def6c66c421f86a94ea0601433a334e4cd5c462583535.jpg)从上面看到的是![](images/2026-03-06%20143001.jpg);![](images/8e8cdd60a4f714d119259d5cbc5711bd10bde7137792959ed9f6610e87866d3e.jpg)从左面看到的是![](images/2026-03-06%20143049.jpg),从正面看到的是![](images/2026-03-06%20143112.jpg);从不同的角度观察到的图形可能相同，也可能不同;所以，观察一个物体时，要从不同的角度去观察才能全面认识物体。故答案为:√",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "21-22四年级下·湖南衡阳·期末"
+    },
+    {
+        "id":  "68",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "观察物体时只从上面和正面观察。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】一个物体可以从正面，侧面和上面三个不同的角度去观察，据此判断即可。此题考查了从不同方向观察物体和几何体的方法，锻炼了学生的空间想象力和抽象思维能力.\n【详解】观察物体时不是只从上面和正面观察，还可以从侧面进行观察。所以原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "21-22四年级下·全国·假期作业"
+    },
+    {
+        "id":  "69",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "从不同方向观察同一个物体，看到的形状是不完全相同的。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】对于一般的物体，从不同的位置观察物体，所看到的形状是不同的;但有特殊情况，如果这个物体是球体，那么从不同的方向看到的形状一样;据此判断即可。解答此题的关键：根据题意，找出反例，进行分析，进而得出结论.\n【详解】由分析可知：当这个物体是一个球体，那么从不同方向看到的形状都一样;所以原题的说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "21-22四年级下·全国·假期作业"
+    },
+    {
+        "id":  "70",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "由相同的小正方体摆成的图形，从前面和左面看到的形状一定相同。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】由相同的小正方体摆成的图形，摆放方法不同，则从前面和左面看到的形状不一定相同，举例解答即可。本题考查观察物体，需要学生有较强的空间想象和推理能力，可通过举反例的方法解决此类问题.\n【详解】例如均由4个小正方体摆成的图形![](images/fff9e5c487f8ddbaee079ab054899bd22fe41ca856bdaee56c0ef63d9ee425f0.jpg)从前面看是![](images/2026-03-06%20143224.jpg),从左面看到的是![](images/2026-03-06%20143237.jpg);![](images/2026-03-06%20143304.jpg)从前面看到的是![](images/2026-03-06%20143315.jpg),从左面看到的是![](images/2026-03-06%20143332.jpg)。即从前面和左面看到的形状不一定相同。故答案为: ×",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "20-21四年级下·湖北荆州·期中"
+    },
+    {
+        "id":  "71",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "从上面看到是正方形的物体一定是正方体。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】如果长方体上面的面是一个正方形，则从上面观察，看到的形状是正方形，据此即可解答。通过看到的形状推断物体的形状，培养学生的方位感和空间想象力.\n【详解】根据分析可知，从上面看到的是正方形的物体不一定是正方体，也可能是长方体，原说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "20-21四年级下·甘肃平凉·期中"
+    },
+    {
+        "id":  "72",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "从同一个方向观察不同的物体，看到的形状可能是相同的。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】从同一个方向观察不同的物体，看到的形状可能相同，也可能不同，举例说明即可。\n【详解】从同一个方向观察不同的物体，看到的形状可能是相同的。例如：从左面观察下面两个物体\n![](images/12088003a7f579794ca616b0d4552474.png)\n图一从左面看到的形状是：\n![](images/586aa9d58fba5f4ae68f6b26269ec8f2eb53d12fa036a4d180dbc623eca1a56d.jpg)\n图二从左面看到的形状也是这个图形，因此从同一个方向观察不同的物体，看到的形状可能相同。故答案为:√",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "17-18四年级下·全国·单元测试"
+    },
+    {
+        "id":  "73",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "一个长方体无论从哪个面观察，看到的形状都是长方形。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】根据题意可知，长方体从不同面观察，看到的形状可能是长方形，也可能是正方形;据此判断。本题主要考查从不同方向观察物体和几何图形，关键是培养学生的观察能力.\n【详解】一个长方体无论从哪个面观察，看到的形状不一定都是长方形，也可能看到正方形，所以原题的说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "20-21四年级下·广西玉林·期中"
+    },
+    {
+        "id":  "74",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "如果两个立体图形从前面和左面看到的图形相同，那么这两个立体图形一定相同。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】若两个立体图形从前面，左面和上面看到的图形相同，则这两个立体图形形状可能相同，单从前面，左面看到的形状相同，这两个立体图形形状不一定相同。要确定一个物体的形状，要通过这个物体的三视图才能确定，而二视图，一视图都不能确定其形状.\n【详解】如图所示:![](images/604632d4946faf039407ffad88ac188a0cd702be35840ce9fb1d068963ccf180.jpg)\n左，右两个立体图形从前面，左面看到的形状相同都是个\"田\"字型，如图所示:![](images/42a08f67840afa9415f3c30c369b413f5ebee38544c3bff2ee75ca797be4d1d3.jpg)\n而左，右两个立体图形的形状却不同，所以原题的说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "18-19五年级下·山东菏泽·期中"
+    },
+    {
+        "id":  "75",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "一个立体图形，从前面，后面，侧面看到的形状不可能相同。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】可以举出反例来说明即可。本题是考查从不同方向观察物体和几何图形，主要培养学生的观察能力和想象能力.\n【详解】正方体，从前面，后面，侧面看到的形状都是正方形，所以判断错误。",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "21-22四年级下·全国·单元测试"
+    },
+    {
+        "id":  "76",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "只要两个物体不同，在每个面看到的图形一定不相同。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】利用列举法进行说明：圆柱和球都可以看到圆的面，所以说两个物体不同，在每个面看到的图形可能相同，也可能不同。本题是考查从不同方向观察物体和几何图形，关键是掌握几何图形的特征做题.\n【详解】圆柱和球是两个不同的物体，都可以看到圆的面，所以判断错误。",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "21-22四年级下·全国·单元测试"
+    },
+    {
+        "id":  "77",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "从一个立体图形的各个方向看，看到的图形一定不相同。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】一般情况下，从不同方向观察立体图形看到的形状是不同的，但是个别图形如正方体，球体，在不同的方向观察的图形却是相同的，因此得解。此题考查了从不同方向观察物体和几何体，锻炼了学生的空间想象力和创新思维能力.\n【详解】据分析可知：从不同方向观察球体和正方体等，从不同方向看到的形状却是相同的，原题说法是错误的。故答案为: ×。",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "20-21四年级下·全国·单元测试"
+    },
+    {
+        "id":  "78",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "从上面，前面看到的是大小相同的正方形，这个物体不一定是正方体。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】可能是正方体的一半，也就是直三棱柱。理由：做一个橡皮泥的正方体，再将它沿左右两个侧面的对角线切下来，该正方体平分为两半，即两个一模一样的直三棱柱，据此判断。本题考查了从不同方向观察物体和几何体，意在培养学生的空间想象能力.\n【详解】据分析可得：从上面，前面看到的是大小相同的正方形，这个物体不一定是正方体，此说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "20-21四年级下·全国·单元测试"
+    },
+    {
+        "id":  "79",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "\"横看成岭侧成峰\"这句诗是从两个方向看庐山的。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】横看成岭侧成峰的意思是说：从正面，侧面看庐山山岭连绵起伏，山峰耸立，据此即可解答。首先读懂诗句的意思是解答本题的关键.\n【详解】根据分析可知，\"横看成岭侧成峰\"这句诗是从正面和侧面来观察庐山的，所以判断正确。",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "2020四年级下·湖南怀化·专题练习"
+    },
+    {
+        "id":  "80",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "看同一个物体，从不同的方向看到的形状是相同的。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】一般地，看同一个物体，从不同的方向观察所看到的形状不完全相同，但对于规则的图形如球体和正方体等，从不同的方向看到的形状却是相同的，据此解答。本题考查了从不同方向观察物体，需要学生们全面考虑，认真判断.\n【详解】根据分析可知：看同一个物体，从不同的方向看到的形状不一定相同的。故原题干说法是错误的。故答案为: ×",
+        "knowledge_points":  [
+                                 "三视图的画法",
+                                 "从不同位置观察单个物体",
+                                 "物体三视图的认识",
+                                 "根据三视图确认几何体"
+                             ],
+        "source":  "17-18四年级下·全国·课后作业"
+    },
+    {
+        "id":  "81",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "无论从上面，右面还是前面看一个正方体，看到的都是正方形。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】正方体有6个面,6个面都是正方形，所以从哪个角度看都是正方形，据此解答。此题考查了正方体的三视图的辨别，要注意生活的经验积累，熟练掌握并灵活应用.\n【详解】无论从上面，右面还是前面看一个正方体，看到的都是正方形。故判断正确。",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体",
+                                 "物体三视图的认识"
+                             ],
+        "source":  "21-22四年级上·河南洛阳·期中"
+    },
+    {
+        "id":  "82",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "我们从同一个方向观察几个物体，如果看到的图形相同，则这几个物体的形状也不一定完全一样。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】我们从同一个方向观察几个物体，如果看到的图形相同，则这几个物体的形状也不一定完全一样。可以通过举例说明。本题主要考查了从同一个方向观察不同的物体或图形，解题时可以举例说明.\n【详解】如图:\n![](images/222b9001b08ece11b41e4670c790b9f9.png)\n这三个不同的图形，从上面看到的图形都是:\n![](images/4b16ca1ccf0ab81b69cdf9f7e959ae5d5dd1c2e1203b1adbcee2f7d134ab9f2c.jpg)\n所以，我们从同一个方向观察几个物体，如果看到的图形相同，则这几个物体的形状也不一定完全一样。故答案为:√",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体",
+                                 "从不同位置观察两个物体的相互关系",
+                                 "物体三视图的认识"
+                             ],
+        "source":  "20-21四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "83",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "观察物体，在不同的角度看到的形状一定不同。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】对于一般的物体，从不同的位置观察物体，所看到的形状是不同的;但有特殊情况，如果这个物体是正方体，那么从正面，侧面和上面看到的都是一个正方形，即看到的形状一样;如果这个物体是球，那么从正面，侧面和上面看到的都是一个圆形，即看到的形状一样;据此判断即可。本题考查了从不同的方向观察物体。解答此题的关键是根据题意找出反例，进行分析，进而得出结论.\n【详解】由分析知，当这个物体是一个正方体或球体，那么从正面，侧面和上面看到的形状都一样;故答案为: ×。",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "19-20四年级下·重庆万州·期中"
+    },
+    {
+        "id":  "84",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "观察正方体时一次最多能看到3个面。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】从正方体的一个面看(视线垂直于这个面)只能看到1个面，从正方体的条棱看(视线垂直于这条棱)能看到2个面，从正方体的一个顶点看，能看到3个面，由此可知，从不同的角度观察正方体，最多能看到3个面。此题主要从不同方向考查观察物体，可以用实物进行演示加深印象。培养学生的空间想象能力和观察能力.\n【详解】![](images/6c8caea944d30f482967adbcb6be93881c83da7d3ed0f37c73f3b98fd4e0a652.jpg)\n由分析可知，如图所示，观察正方体时一次最多能看到3个面。故答案为:√。",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "19-20四年级上·江苏·单元测试"
+    },
+    {
+        "id":  "85",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "两个立体图形，从正面和侧面看形状相同，这两个图形相同。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】如下图所示，这两个立体图形，从正面和侧面看形状相同，但这两个图形却不相同。本题主要考查学生的方法感和空间想象力.\n![](images/fbf2730960114aabd122b9ecff619609cd924b03180bc0bb5ca811e1e292a700.jpg)\n【详解】两个立体图形，从正面和侧面看形状相同，但这两个图形不一定相同，所以判断错误。",
+        "knowledge_points":  [
+                                 "根据三视图确认几何体"
+                             ],
+        "source":  "20-21四年级下·全国·单元测试"
+    },
+    {
+        "id":  "86",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "站在不同的位置观察同一个物体，一定能看到它的三个面。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】当正对一个物体的一个面的时候，只能看到一个面，而最多只能看到三个面。观察物体时，关键是位置的确定，观察同一物体，站在不同的位置，所看的形状也会有所不同.\n【详解】站在不同的位置观察同一个物体，最多只能看到它的三个面，如果正对着它的正面或其它任意一面则只能看到一个面，所以判断错误。",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "20-21四年级下·全国·单元测试"
+    },
+    {
+        "id":  "87",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "同一个立体图形，从不同的方向看到的形状不一定相同。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】一般地，一个立体图形，从不同的方向观察所看到的形状不完全相同，但对于规则的图形如球体和正方体等，从不同的方向看到的形状却是相同的，据此解答。本题考查了从不同方向观察立体图形，需要学生们全面考虑，认真判断.\n【详解】根据分析可知：同一个立体图形，从不同的方向看到的形状不一定相同。说法是正确的。故答案为:√。",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "19-20四年级下·吉林长春·期末"
+    },
+    {
+        "id":  "88",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "观察同一个物体，如果两次看到的形状相同，则观察者位置未发生变化。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】根据对物体在各个面观察的认识，从不同的位置观察一个物体，有可能看到的形状是相同的，据此判断即可。本题考查的是从不同角度观察物体，两次看到形状相同，并不就是因为位置没发生变化，有些物体从几个面观察都是一样的形状.\n【详解】假设这个物体是一个球体，那么从不同方向看到的形状都一样，所以原题说法错误。故判断错误。",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "19-20四年级下·四川成都·期末"
+    },
+    {
+        "id":  "89",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "站在一个位置上观察物体，可能只看到2个面。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】根据题意，站在一个位置上观察物体，可能看到1个面、2个面或3个面，最多可能看到3个面，进而完成判断。本题考查从不同方向观察物体和几何图形，培养学生的观察能力.\n【详解】站在一个位置上观察物体，可能看到1个面、2个面或3个面，所以原题判断正确。故答案为:√",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "19-20四年级下·全国·期末"
+    },
+    {
+        "id":  "90",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "一个物体，从前面观察，一定可以看到它的所有的面。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】一个物体无论从哪个面观察都不可能看到它的所有面，最多能同时看到它的三个面，据此判断。本题考查的是对物体的观察需要从多面，一面无法全面了解物体.\n【详解】因为一个物体无论从哪一面观察都不可能看到它的所有面，所以原题干说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "19-20四年级下·全国·期末"
+    },
+    {
+        "id":  "91",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "一个正方体无论从什么方向看，都是正方形。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】根据正方体的特征,6个面是完全相同的正方形，所以无论从哪个方向看，看到的都是正方形。本题考查的是对物体的观察需要从多面，一面无法全面了解物体.\n【详解】一个正方体，无论从哪个方向看，看到的都是正方形。故答案为:√",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "17-18四年级下·全国·课后作业"
+    },
+    {
+        "id":  "92",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "从不同的角度观察同一物体，所看到的物体的形状是不一样的。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】一般情况下，从不同的角度观察同一物体，看到的形状是不同的，但是对于规则对称的物体如球体和正方体等，从不同的角度看到的形状却是相同的。本题考查观察立体图形推理能力和空间想象能力。需要学生具体问题具体分析.\n【详解】观察不规则对称的物体时，从不同方向观察立体图形看到的形状是不同的。观察规则对称图形时，从不同的角度观察同一物体看到的形状是相同的。故答案为: ×",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "18-19四年级下·全国·期中"
+    },
+    {
+        "id":  "93",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "从上面，正面，左面看到的图形都相同。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】\n![](images/9df4884ea46495fc26d0f5e8ee654762.png)\n例如，这一几何体从上面、正面、左面看到的图形都不相同。此题考查了从不同方向观察物体和几何体，锻炼了学生的空间想象力和抽象思维能力.\n【详解】从上面、正面、左面看到的图形不一定都相同，此说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "19-20五年级下·全国·开学考试"
+    },
+    {
+        "id":  "94",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "从不同方向观察立体图形看到的形状是相同的。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】一般情况下，从不同方向观察立体图形看到的形状是不同的，但是对于规则对称的图形如球体和正方体等，从不同方向看到的形状却是相同的。本题考查观察立体图形推理能力和空间想象能力。需要学生具体问题具体分析.\n【详解】观察规则对称图形时，从不同方向观察立体图形看到的形状是相同的；观察不规则图形时，从不同方向观察立体图形看到的形状可能不同。所以原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "19-20四年级下·全国·期中"
+    },
+    {
+        "id":  "95",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "从同一角度观察不同形状的立体图形的平面图形也一定不相同。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】本题考查从同一角度观察不同形状的立体图形。从同一角度观察不同形状的立体图形，得到的平面图形可能相同，也可能不同。例如，圆柱和圆锥从上面观察，看到的都是圆形;正方体和长方体从某个方向观察，可能都看到正方形。因此，原题说法错误.\n【详解】从同一角度观察不同形状的立体图形，得到的平面图形不一定不相同，有可能相同。例如，从上面观察圆柱和圆锥，看到的都是圆形。所以原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "观察物体"
+                             ],
+        "source":  "19-20四年级下·全国·单元测试"
+    },
+    {
+        "id":  "96",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "从一个立体图形的各个方向看，看到的形状均不相同。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】本题考查从不同方向观察立体图形。从一个立体图形的各个方向看，看到的形状可能相同，也可能不同，取决于立体图形的形状。例如，球体从任何方向看都是圆形，形状相同;而长方体从不同方向看可能看到不同的面。所以原题说法错误.\n【详解】从一个立体图形的各个方向看，看到的形状不一定均不相同，有可能相同(如球体)。所以原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "19-20四年级下·全国·单元测试"
+    },
+    {
+        "id":  "97",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "两个相同的小正方体无论怎样摆放，从不同的位置看到的形状都是相同的。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】本题考查从不同位置观察两个相同小正方体的组合图形。两个相同的小正方体可以有不同的摆放方式(如并排，上下叠放，前后错开等),从不同的位置观察，看到的形状可能相同，也可能不同。例如，两个小正方体并排放置，从前面看是两个正方形并排，从侧面看是一个正方形;如果上下叠放，从前面看是两个正方形上下排列，从侧面看也是两个正方形上下排列(如果叠放方向与侧面平行)。所以摆放方式和观察角度都会影响看到的形状，原题说法过于绝对.\n【详解】两个相同的小正方体摆放方式不同，从不同的位置观察，看到的形状不一定相同。例如，并排放置和上下叠放，从前面看形状就不同。所以原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "从不同位置观察两个物体的相互关系"
+                             ],
+        "source":  "19-20四年级下·全国·单元测试"
+    },
+    {
+        "id":  "98",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "在任意位置都能同时看到物体的所有面。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】我们可以把一个长方体放在桌子上进行观察，从而得到最多能看到几个面，最少能看到几个面，据此即可判断解答。感受观察的范围随观察点，观察角度的变化而改变，并能利用所学的知识解释生活中的一些现象.\n【详解】由题意知，把一个长方体放在桌子上进行观察，从不同的角度去观察最多能看到3个面，最少能看到1个面，所以原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "19-20四年级下·全国·单元测试"
+    },
+    {
+        "id":  "99",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "从正面，上面，左面三个方向观察两个不同的立体图形，所看到的图形一定不相同。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】本题考查从不同方向观察不同形状的立体图形。从正面，上面，左面三个方向观察两个不同的立体图形，所看到的图形有可能相同，有可能不同。例如，一个由4个小正方体组成的L形图形和一个由4个小正方体组成的T形图形，从某个方向看可能看到相同的形状。所以原题说法错误.\n【详解】从正面，上面，左面三个方向观察两个不同的立体图形，所看到的图形不一定不相同，有可能相同。所以原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "19-20五年级下·全国·单元测试"
+    },
+    {
+        "id":  "100",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.82,
+        "question":  "由 5 个小正方体搭成的立体图形，从正面看到的形状是 hline \u0026 \u0026 \n hline ,这个立体图形的形状是唯一的。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】本题考查根据单一方向观察到的形状推测立体图形。仅根据从一个方向(如正面)看到的形状，无法确定小正方体的具体排列方式，尤其是前后位置的深度信息。例如，当从正面看到 3 个横向排列的小正方体时，剩下的 2 个小正方体可以叠在这些方块的上方，下方(若基准面不同),或者藏在它们的后面。由于存在多种搭法，所以形状不唯一.\n【详解】从正面看到一个形状，只能确定该立体图形在左右和上下方向的投影，无法确定其前后的摆放位置。由 5 个小正方体搭成且正面看是 3 个横向方块的情况有多种可能性，原题中\"形状是唯一\"的说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "根据三视图还原几何体",
+                                 "物体的相对位置"
+                             ],
+        "source":  "23-24五年级下·全国·单元测试"
+    },
+    {
+        "id":  "101",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一个物体，从前面看到的图形是BoxBox ,从左边看到的图形是BoxBoxBoxBox ,摆这样一个物体最少需要( )个小正方体。",
+        "options":  [
+                        "A. 4",
+                        "B. 5",
+                        "D. 8"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】要使从前面看到的图形是 BoxBox,可以2个小正方体都放在从前数的第 1行，也可以1个小正方体放在第1行的第1个的位置，另1个小正方体放在第2行从左边数的第2个的位置;同时要使从左边看到的图形是BoxBoxBoxBox ,求摆这样一个物体最少需要多少个小正方体，按从左往右，从前往后的顺序，那么就是第 1行的第1个的位置放1个小正方体，第2行第2个的位置放1个小正方体，第3行第1个或第2个的位置放1个小正方体，第4行第1个或第2个的位置放1个小正方体;据此解答.\n【详解】根据分析: 1 + 1 + 1 + 1 = 4 (个),所以摆这样一个物体最少需要4个小正方体。故答案为:A",
+        "knowledge_points":  [
+                                 "通过三视图会摆放立体图"
+                             ],
+        "source":  "23-24 四年级下·河南信阳·期末"
+    },
+    {
+        "id":  "102",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "观察同一物体( )不同，看到的( )也不相同。",
+        "options":  [
+
+                    ],
+        "answer":  "观察角度;形状",
+        "solution":  "【分析】本题考查观察物体的基本常识。从不同的方向(角度)观察同一个物体，由于视角的变化，所看到的平面图形的形状通常是不同的.\n【详解】观察同一个物体时，观察的角度不同，那么看到的形状也不相同。",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "18-19 四年级下·全国·单元测试"
+    },
+    {
+        "id":  "103",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "从一个角度观察一个立体图形，一次最多只能看到三个面。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】考察立体图形的视觉特性。无论从哪个角度观察长方体或正方体，视线最多能同时触及三个相邻的面.\n【详解】站在一个位置观察立体图形(如长方体),最多只能看到它的三个面(例如：正面，上面和侧面),不可能同时看到相对的两个面。",
+        "knowledge_points":  [
+                                 "观察物体"
+                             ],
+        "source":  "2019·全国·小升初模拟"
+    },
+    {
+        "id":  "104",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小华在班上的位置用数对表示是 (3, 3),小兰在班上的位置用数对表示是 (3, 4),小芳在班上的位置用数对表示是 (4, 4),( ) 坐在 ( ) 的前面,( ) 坐在 ( ) 的左面。",
+        "options":  [
+
+                    ],
+        "answer":  "小华;小兰;小兰;小芳",
+        "solution":  "【分析】数对第一个数表示列，第二个数表示行。同列不同行时，行数小的在前面;同行不同列时，列数小的在左面(通常视教师讲台为观察基准).\n【详解】小华 (3, 3) 和小兰 (3, 4) 都在第 3 列，小华在第 3 行，小兰在第 4 行，所以小华在小兰前面。小兰 (3, 4) 和小芳 (4, 4) 都在第 4 行，小兰在第 3 列，小芳在第 4 列，所以小兰在小芳左面。",
+        "knowledge_points":  [
+                                 "数对与位置"
+                             ],
+        "source":  "18-19 四年级下·全国·期中"
+    },
+    {
+        "id":  "105",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果要把算式 120÷30-2×5 的运算顺序改成先算减法，再算除法，最后算乘法，那么算式应改为 ( )。",
+        "options":  [
+
+                    ],
+        "answer":  "120÷(30-2)×5",
+        "solution":  "【分析】考查括号的使用。括号可以改变运算顺序，小括号内的运算最先进行。\n【详解】要先算减法，需给减法部分加上小括号，然后再按从左往右的顺序进行除法和乘法，所以算式应改为 120÷(30-2)×5。",
+        "knowledge_points":  [
+                                 "整数的四则运算"
+                             ],
+        "source":  "18-19 四年级下·全国·单元测试"
+    },
+    {
+        "id":  "106",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "( ) - 56 = 130;89×( ) = 356;980 ÷ ( ) = 28",
+        "options":  [
+
+                    ],
+        "answer":  "186;4;35",
+        "solution":  "【分析】利用加减乘除各部分间的关系求解：被减数 = 差 + 减数；因数 = 积 ÷ 另一个因数；除数 = 被除数 ÷ 商。\n【详解】130 + 56 = 186；356÷89 = 4；980÷28 = 35。",
+        "knowledge_points":  [
+                                 "整数的四则运算(逆运算)"
+                             ],
+        "source":  "18-19 四年级下·全国·单元测试"
+    },
+    {
+        "id":  "107",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "根据乘，除法各部分间的关系，写出另外两个算式(以 18×20 = 360 为例):( ),( )。",
+        "options":  [
+
+                    ],
+        "answer":  "360÷18 = 20;360÷20 = 18",
+        "solution":  "【分析】积 / 一个因数 = 另一个因数.\n【详解】已知 18×20 = 360，根据逆运算关系可得 360÷18 = 20 和 360÷20 = 18。",
+        "knowledge_points":  [
+                                 "乘，除法各部分间的关系"
+                             ],
+        "source":  "18-19 四年级下·全国·单元测试"
+    },
+    {
+        "id":  "108",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "367 比 ( ) 多 89,247 比 ( ) 少 156。",
+        "options":  [
+
+                    ],
+        "answer":  "278;403",
+        "solution":  "【分析】第一个空：已知一个数比另一个数多多少，求另一个数用减法；第二个空：已知一个数比另一个数少多少，求另一个数用加法。\n【详解】367 - 89 = 278；247 + 156 = 403。",
+        "knowledge_points":  [
+                                 "整数的四则运算"
+                             ],
+        "source":  "18-19 四年级下·全国·单元测试"
+    },
+    {
+        "id":  "109",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.4,
+        "question":  "8 分 = ( ) 元;5.2 米 = ( ) 厘米;5400 米 = ( ) 千米;8 米 6 厘米 = ( ) 厘米;2060 千克 = ( ) 吨 ( ) 千克;3.02 千克 = ( ) 克;4.06 平方千米 = ( ) 平方千米 ( ) 公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "0.08;520;5.4;806;2;60;3020;4;6",
+        "solution":  "【分析】大单位变小单位乘进率，小单位变大单位除以进率。\n【详解】8÷100 = 0.08(元)；5.2×100 = 520(厘米)；5400÷1000 = 5.4(千米)；8×100 + 6 = 806(厘米)；2060千克 = 2吨60千克；3.02×1000 = 3020(克)；4.06平方千米 = 4平方千米6公顷。",
+        "knowledge_points":  [
+                                 "常见的量，长度，面积，质量单位的换算"
+                             ],
+        "source":  "17-18 四年级下·四川·期中"
+    },
+    {
+        "id":  "110",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个正方体，从正面看到的形状是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "正方形",
+        "solution":  "【分析】正方体的六个面都是完全相同的正方形，因此从任何一个垂直于面的方向看去，形状都是正方形.\n【详解】一个正方体，从正面，侧面或上面看到的形状都是正方形。",
+        "knowledge_points":  [
+                                 "观察物体"
+                             ],
+        "source":  "17-18 四年级下·全国·课后作业"
+    },
+    {
+        "id":  "111",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个正方体，最多可以看见 ( ) 个面。",
+        "options":  [
+
+                    ],
+        "answer":  "3",
+        "solution":  "【分析】考察观察点的视野范围.\n【详解】从不同的角度观察正方体，看到的平面数量不同。当视线正对一个面时看到 1 个面;在棱上看到 2 个面;在顶点处最多可以看到 3 个面。",
+        "knowledge_points":  [
+                                 "观察物体"
+                             ],
+        "source":  "17-18 四年级下·全国·课后作业"
+    },
+    {
+        "id":  "112",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "从一个方向最多可以看到长方体的 ( ) 个面。",
+        "options":  [
+
+                    ],
+        "answer":  "3",
+        "solution":  "【分析】根据观察的范围随观察点，观察角度的变化而改变.\n【详解】从一个方向观察长方体，最多可以看到它的 3 个面(通常是相交于同一个顶点的三个面)。",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "15-16 四年级·全国·期中"
+    },
+    {
+        "id":  "113",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.64,
+        "question":  "在横线中填上适当的单位名称：义务献血者每次献血量一般为 200 ( );小明身高 145 ( );一个热水瓶能盛水 2 ( )。",
+        "options":  [
+
+                    ],
+        "answer":  "毫升;厘米;升",
+        "solution":  "【分析】联系生活实际，根据数值大小匹配合适的计量单位.\n【详解】献血量通常用较小的容量单位，填\"毫升\"。四年级学生身高通常在 1.4 米左右，即 145 \"厘米\"。热水瓶容量较大，填\"升\"。",
+        "knowledge_points":  [
+                                 "长度单位，容积单位的选择"
+                             ],
+        "source":  "14-15 四年级·全国·期中"
+    },
+    {
+        "id":  "114",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.64,
+        "question":  "北偏西30°，还可以说成 ( )。小强看小林在北偏东50°，小林看小强在 ( )。",
+        "options":  [
+
+                    ],
+        "answer":  "西偏北60°;南偏西50°",
+        "solution":  "【分析】北偏西30°和西偏北60°表示的是同一个方向；观测点互换时，方向相反，角度不变。\n【详解】因为北和西之间是90°，所以北偏西30°也可以描述为西偏北60°。小林看小强的方向与小强看小林的方向完全相反，因此北偏东50°变成南偏西50°。",
+        "knowledge_points":  [
+                                 "位置与方向，位置的相对性"
+                             ],
+        "source":  "14-15 四年级·全国·期中"
+    },
+    {
+        "id":  "115",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.64,
+        "question":  "一桶水有 8 ( ),一瓶饮料有 250 ( )。",
+        "options":  [
+
+                    ],
+        "answer":  "升;毫升",
+        "solution":  "【分析】根据生活经验选择单位。大容器通常用升，小容器通常用毫升.\n【详解】一桶水的容积较大，数值为 8，对应单位是\"升\";一瓶饮料容积较小，数值为 250，对应单位是\"毫升\"。",
+        "knowledge_points":  [
+                                 "容积单位的选择"
+                             ],
+        "source":  "15-16 四年级·全国·阶段练习"
+    },
+    {
+        "id":  "116",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.64,
+        "question":  "观察正方体时，一般情况下最多能看到正方体的() 面。",
+        "options":  [
+
+                    ],
+        "answer":  "3",
+        "solution":  "【分析】观察一个正方体，最多能看到3个面。如下图所示，由此即可解答.\n【详解】如图所示:![](images/a226e07ad9448bb04ed640d9ffa2bba1a38396b25cc008609136bd588e850522.jpg)观察一个正方体，一般情况下最多能看到正方体的3个面。故答案为3",
+        "knowledge_points":  [
+                                 "图形与变换"
+                             ],
+        "source":  "15-16 四年级·全国·期末"
+    },
+    {
+        "id":  "117",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "( ) 是乘法的逆运算。",
+        "options":  [
+
+                    ],
+        "answer":  "除法",
+        "solution":  "【分析】考查四则运算之间的逻辑关系。已知两个因数的积与其中一个因数，求另一个因数的运算叫做除法，因此除法是乘法的逆运算。\n【详解】因为 a×b = c（b不等于0），那么 c÷b = a。故答案为：除法。",
+        "knowledge_points":  [
+                                 "乘，除法的意义和各部分间的关系"
+                             ],
+        "source":  "14-15 四年级下·全国·课后作业"
+    },
+    {
+        "id":  "118",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "将一个长方体摆在桌面上，最多能同时看到它的 ( ) 个面。",
+        "options":  [
+
+                    ],
+        "answer":  "3",
+        "solution":  "【分析】站在长方体的一个顶点附近观察，视线可以同时覆盖相交于这个顶点的三个面.\n【详解】从不同的位置观察长方体，看到的面的个数是不同的，但无论从哪个角度看，最多只能同时看到 3 个面(如前面，上面，右面)。故答案为:3。",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "18-19 四年级下·甘肃陇南·期中"
+    },
+    {
+        "id":  "119",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "将一个正方体摆在桌子上，最多能同时看到它的 ( ) 个面。",
+        "options":  [
+
+                    ],
+        "answer":  "3",
+        "solution":  "【分析】正方体是特殊的长方体，其观察规律与长方体一致.\n【详解】站在正方体的斜上方观察，一次最多能看到它的 3 个面。故答案为:3。",
+        "knowledge_points":  [
+                                 "观察物体"
+                             ],
+        "source":  "18-19 四年级下·内蒙古巴彦淖尔·阶段练习"
+    },
+    {
+        "id":  "120",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个立体图形，从上面看到的形状是 ![](images/73261bd66f6e5dfe40440ca10c795588.png)，从左边看到的形状是![](images/192657ae1bb59cbaa861c5d047165e75.png)，摆这个立体图形，至少需要 ( ) 个小正方体，最多可以摆 ( ) 个小正方体。",
+        "options":  [
+
+                    ],
+        "answer":  "5;7",
+        "solution":  "【分析】从上面看到的形状决定了底层小正方体的个数和排列方式；从左边看到的形状决定了每一列的最大高度。最少情况是在底层4个的基础上，再堆1个；最多情况是在不改变左视图和俯视图的前提下，让尽可能多的位置达到最大高度。\n【详解】至少需要：4 + 1 = 5（个）；最多可以摆：4 + 3 = 7（个）。故答案为：5;7。",
+        "knowledge_points":  [
+                                 "根据三视图确定几何体的小正方体数量"
+                             ],
+        "source":  "19-20 四年级上·全国·期中"
+    },
+    {
+        "id":  "121",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "![](images/4528d34889ba0c54d199ca9c00f88d40afca706e02320440f51c9d88a3fe2e14.jpg),从正面看到的形状是( ),从左边面看是( ),从上面看是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "图见详解",
+        "solution":  "【分析】本题主要考察从不同方向观察几何体的形状。正面视图：观察物体的长和高。图中底层有 2 个正方形，上层右边有 1 个正方形。左面视图：观察物体的宽(深)和高。从左边看，只能看到一列，由于物体总高度为 2 层，所以看到的是 2 个竖直排列的正方形。上面视图：观察物体的长和宽。从上面往下看，只能看到底层的分布，即一排水平排列的 2 个正方形.\n【详解】根据对立体图形的观察：从正面看：可以看到 3 个正方形的面，底层 2 个，上层 1 个(靠右),形状呈\"L\"形。从左面看：由于左右位置重合，视线穿过左侧的小正方体可以看到右侧上方的正方体，因此呈现出一列 2 个正方形的形状。从上面看：只能看到物体在平面上的投影，即底层并排放置的 2 个正方形的面，呈水平排列。故答案依次填入对应的平面图形即可。",
+        "knowledge_points":  [
+                                 "三视图的画法"
+                             ],
+        "source":  "19-20 二年级上·全国·期末"
+    },
+    {
+        "id":  "122",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在一道有余数的除法算式 ○÷□ = 6……18 中，□ 最小是 ( )，当 □ 最小时，○ 等于 ( )。",
+        "options":  [
+
+                    ],
+        "answer":  "19;132",
+        "solution":  "【分析】在有余数的除法中，余数必须比除数小。已知余数是18，那么除数□必须大于18。求被除数○用公式：被除数 = 商×除数 + 余数。\n【详解】因为余数是18，所以除数□最小取18 + 1 = 19。当除数是19时，被除数○ = 6×19 + 18 = 132。故答案为:19;132。",
+        "knowledge_points":  [
+                                 "有余数的除法;除法各部分间的关系"
+                             ],
+        "source":  "19-20 四年级下·全国·单元测试"
+    },
+    {
+        "id":  "123",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个数加上 0 得 ( ),一个数减去 0 得 ( )。",
+        "options":  [
+
+                    ],
+        "answer":  "原数;原数",
+        "solution":  "【分析】考察 0 的加减法性质：任何数与 0 相加或相减，其数值保持不变.\n【详解】 设原数为 a,则 a + 0 = a,a - 0 = a。故答案为：原数;原数。",
+        "knowledge_points":  [
+                                 "0 在四则运算中的特性"
+                             ],
+        "source":  "19-20 四年级下·全国·单元测试"
+    },
+    {
+        "id":  "124",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "给算式(如 20×3 + 5)添上括号，使计算时先算加法，添上括号后的算式为 ( )。",
+        "options":  [
+
+                    ],
+        "answer":  "20×(3+5)",
+        "solution":  "【分析】在没有括号的混合运算中，通常先乘除后加减。要改变顺序先算加法，必须给加法部分加上小括号。\n【详解】给加法部分加上小括号后，算式变为 20×(3+5)，这样就会先算加法。",
+        "knowledge_points":  [
+                                 "括号在四则混合运算中的作用"
+                             ],
+        "source":  "19-20 四年级下·全国·单元测试"
+    },
+    {
+        "id":  "125",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "减数 = ( ) - 差",
+        "options":  [
+
+                    ],
+        "answer":  "被减数",
+        "solution":  "【分析】根据减法的定义：被减数 - 减数 = 差.\n【详解】通过公式变形可得：减数 = 被减数 - 差。故答案为：被减数。",
+        "knowledge_points":  [
+                                 "减法各部分间的关系"
+                             ],
+        "source":  "19-20 四年级下·全国·单元测试"
+    },
+    {
+        "id":  "126",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "观察物体时，可以从物体的 ( ) 面,( ) 面和 ( ) 面进行观察。",
+        "options":  [
+
+                    ],
+        "answer":  "前;上;左",
+        "solution":  "【分析】考察三视图的基础概念，通常从三个相互垂直的方向观察物体.\n【详解】习惯上，我们从前面(正面),上面，左面(或右面)来全面观察一个物体的形状。故答案为：前;上;左。",
+        "knowledge_points":  [
+                                 "观察物体"
+                             ],
+        "source":  "19-20 四年级下·全国·单元测试"
+    },
+    {
+        "id":  "127",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把两个数合并成 ( ) 的运算，叫做 ( )。",
+        "options":  [
+
+                    ],
+        "answer":  "一个数;加法",
+        "solution":  "【分析】考察加法的定义.\n【详解】加法就是把两个或两个以上的数合并成一个数的运算。故答案为：一个数;加法。",
+        "knowledge_points":  [
+                                 "加法的意义"
+                             ],
+        "source":  "19-20 四年级下·全国·单元测试"
+    },
+    {
+        "id":  "128",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "将一个球放在桌面上，无论从哪个方向上观察，都只能看到 ( ) 形。",
+        "options":  [
+
+                    ],
+        "answer":  "圆",
+        "solution":  "【分析】球体具有高度的对称性，其在任何平面上的投影都是相等的圆.\n【详解】无论从前面，上面还是侧面观察球体，看到的平面图形都是直径相等的圆。故答案为：圆。",
+        "knowledge_points":  [
+                                 "观察物体;球体的特性"
+                             ],
+        "source":  "19-20 四年级下·全国·单元测试"
+    },
+    {
+        "id":  "129",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "圆柱从前面看到的图形是 ( ) 形，从上面看到的图形是 ( ) 形。",
+        "options":  [
+
+                    ],
+        "answer":  "长方;圆",
+        "solution":  "【分析】考察圆柱的三视图特性。圆柱由一个侧面和两个底面组成.\n【详解】从前面(正面)观察，圆柱的曲面会投影成一个长方形(或正方形)。从上面观察，看到的是圆柱的底面，形状是圆形。故答案为：长方;圆。",
+        "knowledge_points":  [
+                                 "观察圆柱体"
+                             ],
+        "source":  "19-20 四年级下·全国·单元测试"
+    },
+    {
+        "id":  "130",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "同样高度的物体，在同一光源下，离光源越近，这个物体的影子就越( )。",
+        "options":  [
+
+                    ],
+        "answer":  "短",
+        "solution":  "【分析】根据\"同样高的物体离光源越近，影子就越短;离光源越远，影子就越长\"进行解答即可.\n【详解】观察下图可得：同样高的物体在同一光源下离光源越近，它的影子就越短。![](images/a109089458724daf9090e9c2df7a1518aec04a4ce7d1e0aa3c59659f3778fc17.jpg)",
+        "knowledge_points":  [
+                                 "观察的范围(视野与盲区)"
+                             ],
+        "source":  "19-20四年级下·全国·期中"
+    },
+    {
+        "id":  "131",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "有一个立体图形，从上面和侧面能看到形状(倒\"L\"型)。搭这样一个的立体图形，最多需要( )个小正方体。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】根据题干，从上面看到的图形是倒\"L\"形，说明这个物体最底层有4个小正方体；从侧面看到的图形也是倒\"L\"形，说明这个物体一共有2层。要使小正方体最多，就让符合条件的位置尽量叠成2层，据此解答。本题是从不同方向观察物体和几何体，意在训练学生观察能力和分析判断能力。\n【详解】根据题干分析画图如下:![](images/ef2738d171c96642609208e4cbbdba7dcdfda560c9b4f7acabed38a762c69a35.jpg)\n4 + 2 = 6（个）。",
+        "knowledge_points":  [
+                                 "根据三视图确认几何体"
+                             ],
+        "source":  "19-20四年级下·全国·期中"
+    },
+    {
+        "id":  "132",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "站在任一位置观察某正方体，最多能看到它的( )个面，最少能看到它的( )个面。",
+        "options":  [
+
+                    ],
+        "answer":  "三;一",
+        "solution":  "【分析】从正方体顶点方向观察正方体，可以看到三个面，从正方体正面观察，只能看到一个面，据此填空。本题考查了观察物体，是课堂学习内容，要理解并掌握.\n【详解】站在任一位置观察某正方体，最多能看到它的三个面，最少能看到它的一个面。",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "19-20 五年级下·全国·单元测试"
+    },
+    {
+        "id":  "133",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "从一个方向观察物体，最少能看到物体的( )个面，最多能看到物体的( )个面。",
+        "options":  [
+
+                    ],
+        "answer":  "1;3",
+        "solution":  "【分析】从物体的一面观察，只能看到1个面；从物体的一条棱附近观察，能看到物体的2个面；从物体的一个顶点附近观察，能看到物体的3个面。即站在任意位置观察同一物体，最少能看到物体的1个面，最多能看到物体的3个面。此题考查从不同方向观察物体。\n【详解】从一个方向观察物体，最少能看到物体的1个面，最多能看到物体的3个面。",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "18-19 四年级下·云南临沧·期末"
+    },
+    {
+        "id":  "134",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "观察长方体时，最多可以看到( )个面，最少可以看到( )个面。",
+        "options":  [
+
+                    ],
+        "answer":  "3;1",
+        "solution":  "【分析】从长方体的一个面看(视线垂直于这个面)只能看到1个面，从长方体的一条棱看(视线垂直于这条棱)能看到2个面，从长方体的一个顶点看，能看到3个面，由此可知，从不同的角度观察长方体，最多能看到 3 个面，最少能看到 1 个面。此题主要考查从不同方向观察物体，可以用实物进行演示加深印象，培养学生的空间想象能力和观察能力.\n【详解】观察长方体时，最多可以看到 3 个面，最少可以看到 1 个面。",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "15-16五年级·全国·期末"
+    },
+    {
+        "id":  "135",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "站在不同位置观察一个长方体形状的盒子，每次最多能看到( )个面。",
+        "options":  [
+
+                    ],
+        "answer":  "3",
+        "solution":  "【分析】以长方体一个面为视线中心，可看到1个面;以长方体的一条棱为视线中心，可看到2个面;以一个顶点为视线中心，可看到3个面，以此作答。本题考查不同方向观察立体图形，培养学生的空间思维能力.\n【详解】根据分析，以一个顶点为视线中心时，可看到的面最多，为3个面。",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "20-21三年级上·河南郑州·期末"
+    },
+    {
+        "id":  "136",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.4,
+        "question":  "甲，乙，丙三位同学用 5 个同样大小的小正方体摆图形，甲说：\"我摆的最高.\"乙说：\"我摆的最长.\"丙说：\"我摆的是字母 L.\"从上面看到是□的是( )摆的。",
+        "options":  [
+
+                    ],
+        "answer":  "甲",
+        "solution":  "【分析】![](images/21d7cd4f9ed90acd000f63a943d8e0c4.png)解答此题的关键是，根据他们三个人说的话，摆出各个图形.\n【详解】![](images/31e0377ac5f3b06c9d1b3d24c862c1ae.png)",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "20-21 四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "137",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "足球不管是从哪个方位看，都看到是( ),圆柱从侧面看是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "圆形;长方形",
+        "solution":  "【分析】观察一个足球，从前面，上面还是侧面看都是圆形。观察一个圆柱，从上面看到的是圆形，从侧面看到的都是长方形。据此解答.\n【详解】足球不管是从哪个方位看，都看到是圆形，圆柱从侧面看是长方形。",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "2020 四年级下·湖南怀化·专题练习"
+    },
+    {
+        "id":  "138",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "把一个文具盒放在桌子上，最多能看到( )个面，最少能看到( )个面。",
+        "options":  [
+
+                    ],
+        "answer":  "3;1",
+        "solution":  "【分析】观察物体，无论从哪个角度看，最多只能看到物体的三个面，最少可看到1个面，据此即可解答。此题考查从不同方向观察物体，意在培养学生观察物体和空间思维的能力.\n【详解】据分析可知：把一个文具盒放在桌子上，最多可看到3个面，最少可看到1个面。",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "20-21 四年级下·全国·单元测试"
+    },
+    {
+        "id":  "139",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "圆柱从侧面看看到是( );从上面或底面看到的图形是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "长方形;圆形",
+        "solution":  "【分析】从侧面观察圆柱看到的是矩形。观察圆柱时从上面和下面看是一个圆形，从正面看时看到的是一个长方形，上下两条边应该是直的而不是弯曲的，从侧面看也是一样的，也是长方形。此题考查的特征：圆柱是由两个大小相等，相互平行的圆形以及连接两个底面的一个曲面围成的几何体.\n【详解】据分析可得：圆柱从侧面看看到是长方形;从上面或底面看到的图形是圆形。",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "2020四年级下·湖南怀化·专题练习"
+    },
+    {
+        "id":  "140",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "我们在观察物体时，观察点距离被观察物体越近，观察到的物体越，观察到的范围越 。",
+        "options":  [
+
+                    ],
+        "answer":  "大;小",
+        "solution":  "【分析】根据生活经验，通过观察物体可知：观察物体时，离得越远看到景物的范围越大，看到的物体越小，越模糊;离得越近看到景物的范围越小，看到的物体越大，越清晰。此题考查了观察物体的方法，要结合生活经验进行解答.\n【详解】我们在观察物体时，观察点距离被观察物体越近，观察到的物体越大，观察到的范围越小。",
+        "knowledge_points":  [
+                                 "观察的范围(视野与盲区)"
+                             ],
+        "source":  "12-13四年级下·浙江衢州·期中"
+    },
+    {
+        "id":  "141",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个正方体，无论从哪个位置看都是( )形。",
+        "options":  [
+
+                    ],
+        "answer":  "正方",
+        "solution":  "【分析】根据正方体的特征,6 个面是完全相同的正方形，所以无论从哪个方向看，看到的都是正方形，由此即可解答。图形观察题，主要考查学生的方位感和空间想象力.\n【详解】根据分析可知，一个正方体，无论从哪个位置看都是正方形。",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体",
+                                 "物体三视图的认识"
+                             ],
+        "source":  "20-21 四年级下·甘肃平凉·期中"
+    },
+    {
+        "id":  "142",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "从同一个方向观察不同的物体，看到的图形可能( )也可能不同。",
+        "options":  [
+
+                    ],
+        "answer":  "相同",
+        "solution":  "【分析】分别举例说明从同一个方向观察不同的物体，看到的图形可能相同，也可能不同。本题考查观察物体，旨在考查学生的空间想象和推理能力。\n【详解】从同一个方向观察不同的物体，看到的图形可能相同，也可能不同。故答案为：相同。",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "20-21四年级下·黑龙江七台河·期末"
+    },
+    {
+        "id":  "143",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "站在不同的位置观察一个长方体，最多可以看到()个面。",
+        "options":  [
+
+                    ],
+        "answer":  "3",
+        "solution":  "【分析】可以举例说明：把一个长方体放在桌子上进行观察，当眼睛在长方体的一个顶点时，能看到的面最多，此时看到3个面。此题考查了从不同方向观察物体和几何体，锻炼了学生的空间想象力和抽象思维能力.\n【详解】根据分析可知：站在不同的位置观察一个长方体，最多能看到长方体的3个面。",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "15-16三年级·广东湛江·期末"
+    },
+    {
+        "id":  "144",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个圆锥从正面看到的是( ),从上面看到的是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "三角形;圆",
+        "solution":  "【分析】本题考查了从不同位置观察圆锥，所看到的形状是有所不同的。一般从正面和侧面看，看到的是三角形；从上面看，看到的是圆。本题主要考查了从不同位置观察物体。\n【详解】一个圆锥从正面看到的是三角形，从上面看到的是圆。",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "21-22四年级下·全国·假期作业"
+    },
+    {
+        "id":  "145",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "学校的围墙外有一座12层的楼房，小明站在12楼比站在 4 楼看到的校园内的范围( )。(填\"大\"或\"小\")",
+        "options":  [
+
+                    ],
+        "answer":  "大",
+        "solution":  "【分析】以小明的眼睛为端点，过学校院墙顶端向学校内的地面画射线，射线以上的范围就是小明可以看到的范围，依此画图并填空即可。本题主要考查观察的范围(视野与盲区),应理解\"站得高方能看得远\".\n【详解】如图![](images/e1e0e91716ebf78570c7567d6e7627c59e16e3ae1dd2da3c3c55d143b3c5167d.jpg)根据画图可知，小明站在12楼比站在4楼看到的校园内的范围大。",
+        "knowledge_points":  [
+                                 "观察的范围(视野与盲区)"
+                             ],
+        "source":  "21-22四年级下·全国·假期作业"
+    },
+    {
+        "id":  "146",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "填一填。![](images/82ceb5907409a048db558e2376df4a400e71e81a0222e327119eadb151944a59.jpg)\n(1)从前面有( )层，能看到( )个正方形。 (2)从上面有( )排，前排能看到( )个正方形，后排能看到( )个正方形。 (3)从左面有( )列，从左往右第一列( )个正方形，第二列( )个正方形。 (4)我发现：从不同方向观察立体图形，得到的平面图形可能是( )的，也可能是( )的。",
+        "options":  [
+
+                    ],
+        "answer":  "一或1;4;两或2;2;2;两或2;1;1;相同;不同",
+        "solution":  "【分析】观察从不同方向看到的排数，层数，列数，以及数全指定位置看到的正方形个数;不同方向观察立体图形，得到的平面图形可能相同也可能不同。本题主要考查的是不同方向观察物体的层数，列数，排数和组成的数量.\n【详解】(1)从前面有一层，能看到 4 个正方形。(2)从上面有两排，前排能看到 2 个正方形，后排能看到 2 个正方形。 (3)从左面有两列，从左往右第一列 1 个正方形，第二列 1 个正方形。 (4)从不同方向观察立体图形，得到的平面图形可能是相同的，也可能是不同的。",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "22-23 四年级下·全国·课前预习"
+    },
+    {
+        "id":  "147",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "我知道从一个位置观察长方体时，最多能看到( )个面。",
+        "options":  [
+
+                    ],
+        "answer":  "3",
+        "solution":  "【分析】我们可以把一个长方体放在桌子上进行观察，从而得到最多能看到几个面，最少能看到几个面。此题考查了从不同方向观察物体的方法，可以实际操作一下.\n【详解】如图所示:![](images/cb5ebca350c284fb2bef4f8b9e3c8b8d53b1c6430bbfeba59867e96e2a939f44.jpg)从一个位置观察一个长方体最多能看到它的 3 个面，最少能看到它的 1 个面。",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "22-23四年级下·全国·单元测试"
+    },
+    {
+        "id":  "148",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明，小天和小亮各用四个同样大的正方体摆图形，小明说：\"我摆的最高.\"小天说：\"我摆的最长.\"小亮说：\"我摆的是英文字母L.\"从上面看到的图形是正方形的是( )摆的。",
+        "options":  [
+
+                    ],
+        "answer":  "小明",
+        "solution":  "【分析】根据题干描述可知![](images/10867c644e4b6565f46b960f0316f936.png)本题是考查从不同方向观察物体和几何图形，关键是培养学生的观察能力.\n【详解】根据分析，小明摆的几何体从上面看到的图形是正方形;小天和小亮摆的几何体从上面看到都是长方形;所以，从上面看到的图形是正方形的是小明摆的。",
+        "knowledge_points":  [
+                                 "物体三视图的认识",
+                                 "根据三视图确认几何体"
+                             ],
+        "source":  "20-21 四年级下·山东菏泽·期末"
+    },
+    {
+        "id":  "149",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "从前面，左面和上面观察由小正方体组成的不同的图形，结果可能( )也可能( )。判断是从哪个方向看到的图形时，需要转换到要观察的方向去思考。",
+        "options":  [
+
+                    ],
+        "answer":  "相同;不同",
+        "solution":  "【分析】从前面、左面和上面观察由小正方体组成的不同图形，结果可能相同，也可能不同。判断是从哪个方向看到的图形时，需要转换到要观察的方向去思考。\n【详解】例如:![](images/153dd838888450f92a6d2555fc343fb6cf9a16a8f7b3b292507c090761164001.jpg) ①![](images/12bb1e869d9f1c6b413b11a4f97af8c6e00898f3fceb1cbf8aff0e5d2bf902a0.jpg) ②![](images/49001477624109becd2497751fbdddf6c2df0952bf38e52bc7ee65ce6c50953b.jpg) ③。①和②从前面、左面和上面观察到的图形相同，①、②、③从前面、左面和上面观察到的图形也可能不同。",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "22-23 四年级下·全国·课前预习"
+    },
+    {
+        "id":  "150",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "观察正方体时，最多能看到( )个面，每个面都是完全( )的。",
+        "options":  [
+
+                    ],
+        "answer":  "3;相同",
+        "solution":  "【分析】观察一个正方体，从一个面看，可以看到1个面；从一条棱附近看，可以看到2个面；从一个顶点附近看可以看到3个面。每个面都是完全相同的正方形。据此解答。本题考查观察物体，需要学生有较强的空间想象和推理能力。\n【详解】观察正方体时，最多能看到3个面，每个面都是完全相同的。",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "22-23四年级上·安徽黄山·期中"
+    },
+    {
+        "id":  "151",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "从不同方向观察一个立体图形，所看到的形状( )。",
+        "options":  [
+
+                    ],
+        "answer":  "可能;不同",
+        "solution":  "【分析】一般情况下，从不同方向观察立体图形看到的形状是不同的，但是对于规则对称的图形如球体和正方体等，从不同方向看到的形状却是相同的，据此解答即可.\n【详解】由分析可知，从不同方向观察一个立体图形，所看到的形状可能不同。",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体",
+                                 "物体三视图的认识"
+                             ],
+        "source":  "22-23 四年级下·安徽铜陵·期中"
+    },
+    {
+        "id":  "152",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "芳芳用 5 个同样的正方体摆成一个组合体，从上面只看到了 4 个正方形，那么她从右面观察会看到两个正方形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】1。 已知条件：总共 5 个正方体，俯视图有 4 个正方形。推论：这说明底层摆了 4 个，剩下的 1 个摆在了其中一个的上面(即有一列是 2 层高)。右视图不确定性：从右面看能看到几个正方形，取决于那\"多出来的一个\"摆放的具体位置。如果它被前面的正方体挡住了，或者摆放的位置在右视图中只体现为一列，结果可能不同.\n【详解】从上面看是 4 个，说明底层占了 4 个位置，由于总数是 5 个，所以有一处叠放了 2 个。但从右面观察时，如果叠放的那一列被挡住，或者叠放位置不同，看到的正方形个数不一定是 2 个(可能是 3 个,4 个或更多，取决于摆放形状).\"一定\"或\"会看到\"这种绝对表述是错误的。故答案为: ×。",
+        "knowledge_points":  [
+                                 "观察物体",
+                                 "三视图的逻辑推导"
+                             ],
+        "source":  "19-20 四年级下·全国·单元测试"
+    },
+    {
+        "id":  "153",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "从正面，上面，左面三个方向观察两个不同的立体图形，所看到的图形一定不相同。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】考察\"不同形状\"与\"相同视图\"之间的关系.\n【详解】不同的立体图形，其三视图是有可能完全相同的。例如：一个外部尺寸相同的空心长方体和实心长方体(假设壁厚不影响外形投影),或者某些特定组合的积木块，虽然内部构造或局部细节不同，但在正，上，左三个投影面上的形状可以完全一致。故答案为: ×。",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "19-20 五年级下·全国·单元测试"
+    },
+    {
+        "id":  "154",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "足球从任何角度观察，看到的都是圆形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】足球可以近似看作一个球体.\n【详解】球体的特性是各向同性。无论从正面，上面还是侧面，甚至斜着观察，其在平面上的正投影始终是一个半径相等的圆。故答案为:√。",
+        "knowledge_points":  [
+                                 "观察物体",
+                                 "球体的几何特性"
+                             ],
+        "source":  "19-20 五年级上·全国·期末"
+    },
+    {
+        "id":  "155",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "几个不同的物体，不管从什么角度观察，看到的形状一定不同。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】考察不同几何体在特定视角下的投影重合.\n【详解】不同的物体在某些角度下看到的形状可能相同。例如：从侧面观察一个圆柱体(平放)和观察一个长方体，看到的可能都是长方形;从正面观察圆锥和等腰三角形框架，看到的可能都是三角形。故答案为: ×。",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "18-19 三年级上·辽宁·单元测试"
+    },
+    {
+        "id":  "156",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "一个几何体，从上面看到的和从正面看到的图形是一样的。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】考察命题的普遍性。只要能举出一个反例，命题就不成立.\n【详解】虽然有些几何体(如正方体，球体)的俯视图和正视图是一样的，但并不是所有几何体都如此。例如圆柱体，正视图是长方形，俯视图是圆形;再如长方体，其长，宽，高如果不等，两个视图的尺寸也会不同。故答案为: ×。",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "18-19 五年级下·吉林白城·期末"
+    },
+    {
+        "id":  "157",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "从不同的位置观察同一个物体，观察到的结果是不一样的。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】考察观察结果的对称性。\n【详解】虽然大多数情况下位置不同结果不同，但对于具有对称性的物体，结果可能一样。最典型的例子是正方体：从正面、上面、左面、右面看，看到的都是完全一样的正方形；再如球体，任何角度观察结果都相同。因此，“结果是不一样的”这一绝对说法错误。故答案为: ×。",
+        "knowledge_points":  [
+                                 "观察物体"
+                             ],
+        "source":  "18-19 四年级下·云南临沧·阶段练习"
+    },
+    {
+        "id":  "158",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "一个立体图形从前面看到的是由2个小正方形组成的图形，说明这个立体图形是由两个小正方体组成的。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】三视图是从上面、左面、正面等不同角度观察同一个几何体得到的图形。题中只是从前面看到的图形，观察并不全面。仅凭从一个方向看到的平面图形，不能确定组成该图形的小正方体个数。\n【详解】从前面看到2个小正方形，只能说明这个立体图形从前面看有2列，至于前后摆了几层无法确定，所以不能说这个立体图形一定是由两个小正方体组成的。故答案为: ×。",
+        "knowledge_points":  [
+                                 "物体三视图的认识",
+                                 "根据三视图确认几何体"
+                             ],
+        "source":  "18-19四年级下·四川南充·期末"
+    },
+    {
+        "id":  "159",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "![](images/32504ef4ba883f5c3144aa8980b8ae12fb30f2d8bcd41ee44f318793a8965a87.jpg)\n这三个物体，从上面看图形相同，从左面看图形也相同。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】根据从不同的方向观察到正方体的个数和正方体的位置作答即可.\n【详解】这三个物体从上面看都是由三个正方形排成的一行;从左面看都是由两个正方形排成的一列，所以这三个物体，从上面看，图形相同：从左面看，图形也相同。故答案为:√",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "18-19 四年级下·山东菏泽·期中"
+    },
+    {
+        "id":  "160",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "在观察物体时，从前面观察与左面观察看到的图形有可能是一样的。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】在观察物体时，从前面观察与左面观察看到的图形有可能是一样的。可举例证，如一个站立的圆柱，从前面，左边看的形状是相同的，都是一个长方形(或正方形).\n【详解】在观察物体时，从前面观察与左面观察看到的图形有可能是一样的。此说法正确。故答案为:√。",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "18-19四年级下·广东东莞·期中"
+    },
+    {
+        "id":  "161",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "从远到近观察景物，看到的范围越来越小。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】考察观察位置与观察范围的关系。视觉原理：当观察点距离物体较远时，视角较小，能够包容更多的背景，即\"视野开阔\"。位置变化：随着观察点逐渐向景物靠近，原本远处的景物在视野中变大，占据了更多的空间，从而遮挡了更多的背景，导致看到的范围(视野)变小.\n【详解】这是一个典型的\"近大远小\"和\"视野变化\"规律。就像拍照时变焦放大，虽然细节更清晰了，但画面容纳的内容变少了。因此，从远到近观察，看到的范围确实越来越小。 \n故答案为:√。",
+        "knowledge_points":  [
+                                 "观察物体",
+                                 "观察范围的变化"
+                             ],
+        "source":  "2019 六年级下·全国·专题练习"
+    },
+    {
+        "id":  "162",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "几个不同的物体，从同一个方向看到的图形可能一样。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】考察三视图中\"深度信息丢失\"的现象。投影特性：三视图是将三维物体压缩成二维平面的过程，在这个过程中，物体的厚度或内部构造可能会被遮挡或重叠。寻找实例：只要能找到两个形状不同但在某个方向投影相同的例子即可证明.\n【详解】这种现象在几何体观察中非常普遍。例如：从侧面观察一个圆柱体和观察一个长方体，看到的图形都可能是长方形。从正面观察一个圆锥和观察一个等腰三角形薄片，看到的图形都是三角形。因此，不同的物体在同一个方向看到的图形完全可能是一样的。故答案为:√",
+        "knowledge_points":  [
+                                 "观察物体",
+                                 "三视图的特性"
+                             ],
+        "source":  "17-18 四年级下·湖南株洲·期中"
+    },
+    {
+        "id":  "163",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "从不同的方向观察同一个立体图形，看到的图形一定是不同的。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】球体，不管从什么方向看，看到的图形一定相同.\n【详解】解：从不同方向观察同一物体，看到的图形可能相同。原题说法错误。故答案为: ×。",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体",
+                                 "从不同方向观察简单物体"
+                             ],
+        "source":  "18-19四年级下·全国·单元测试"
+    },
+    {
+        "id":  "164",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "从任何位置看小正方体最多只能看到三个面。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】考察观察立体图形的视觉极值。单一视角限制：当我们正对一个面时，看到 1 个面;站在棱上时，看到 2 个面;站在顶点处时，最多看到相交于该顶点的 3 个面。遮挡原理：由于正方体是不透明的，相对的面(如前与后，上与下，左与右)永远不可能同时出现在视线内.\n【详解】无论如何变换角度，视线最多只能同时触及三个相邻的面。故答案为:√。",
+        "knowledge_points":  [
+                                 "观察物体",
+                                 "正方体的视觉特性"
+                             ],
+        "source":  "17-18 四年级上·江苏南京·阶段练习"
+    },
+    {
+        "id":  "165",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "从不同的位置观察一个物体，可以更全面地认识它。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】考察\"多维观察\"在认识事物中的作用。片面性：单一方向的观察只能得到物体的一个平面投影(二维信息),会丢失深度或背面的信息。全面性：通过从正面，侧面，上面等多个维度观察，我们可以获取物体的长，宽，高以及各个面的细节，从而在脑海中构建出完整的三维模型.\n【详解】俗话说\"横看成岭侧成峰\"，只有通过多角度观察，才能消除视觉盲点，真正了解物体的全貌。故答案为:√。",
+        "knowledge_points":  [
+                                 "观察物体",
+                                 "空间观念"
+                             ],
+        "source":  "18-19 四年级下·全国·单元测试"
+    },
+    {
+        "id":  "166",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "不同物体，从某个方向看到的形状有可能相同。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】考察三维到二维投影过程中的信息重叠.\n【详解】不同形状的物体在特定角度下的投影可能完全一致。例如：从正面观察一个圆柱体(横放)和一个长方体，看到的可能都是长方形。从上面观察一个球体和一个圆柱体(竖放),看到的都是圆形。故答案为:√。",
+        "knowledge_points":  [
+                                 "观察物体",
+                                 "三视图的重合性"
+                             ],
+        "source":  "18-19 四年级下·全国·单元测试"
+    },
+    {
+        "id":  "167",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "一个物体从上面，左面，前面看到的都是 BoxBox (由 4 个小正方形组成的 2×2 方阵),那么这个物体一定由 4 个小正方体拼成。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】考察\"三视图不能唯一确定物体内部结构\"的特点。陷阱点：题目中的\"一定\"过于绝对。反例寻找：如果一个物体是 2×2×2 的大正方体(由 8 个小正方体组成),它从正面，左面，上面看到的依然是 2×2 的方阵.\n【详解】当三视图都是 2×2 的方阵时，最少的情况确实可能是 4 个(呈空心的架子状摆放),但最多的情况可以是 8 个(摆成一个实心的大正方体)。因为我们无法通过三视图看到被外层遮挡住的内部空间。故答案为: ×。",
+        "knowledge_points":  [
+                                 "观察物体",
+                                 "根据三视图确定物体组成"
+                             ],
+        "source":  "18-19 四年级下·全国·单元测试"
+    },
+    {
+        "id":  "168",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "同一个物体，不论从哪个方向看形状不变。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】一般地，看同一个物体，从不同的方向观察所看到的形状不完全相同，但对于规则的图形如球体和正方体等，从不同的方向看到的形状却是相同的，据此解答.\n【详解】根据分析可知：看同一个物体，从不同的方向看到的形状不一定相同的。原题干说法是错误的。故答案为: ×",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "18-19 四年级下·全国·单元测试"
+    },
+    {
+        "id":  "169",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "任何物体从不同方向去看形状都不同。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】对于一般物体，从不同方向观察到的形状可能不同，也可能相同；如球体从不同方向看到的形状都相同。\n【详解】任何物体从不同方向去看，形状不一定都不同。例如球体从不同方向看都是圆形，所以原题说法错误。故答案为: ×。",
+        "knowledge_points":  [
+                                 "观察物体"
+                             ],
+        "source":  "18-19 四年级下·全国·期中"
+    },
+    {
+        "id":  "170",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "观察一个物体，同时最多能看到三个面。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【详解】如图所示:![](images/17851cebdbe52fa3c3ea710485d7351bb02c5a8b4afde60426677737fd5ac393.jpg)当物体是一个长方体时，同时看到的最多是三个面。![](images/4ed30bddd61081a4134a914aa42a9fae84f8068e0816ed834db58d9164688104.jpg)当物体是凸形物体并且大于6个面时，可能看到的面大于3个。故答案为: ×",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "17-18 三年级上·福建泉州·阶段练习"
+    },
+    {
+        "id":  "171",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "相同个数的几个小立方体可以拼成多个不同形状的大立体图形。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【详解】相同个数的几个小立方体可以通过不同的摆放拼成多个不同形状的大立体图形",
+        "knowledge_points":  [
+                                 "观察物体"
+                             ],
+        "source":  "18-19四年级下·全国·阶段练习"
+    },
+    {
+        "id":  "172",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "一个立体图形，如果观察点合适，可以看到3个以上的面。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【详解】正方体最多只能看到3个面",
+        "knowledge_points":  [
+                                 "观察的范围(视野与盲区)"
+                             ],
+        "source":  "18-19四年级下·全国·阶段练习"
+    },
+    {
+        "id":  "173",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "把一个文具盒放在桌子上，站在一个位置观察，最多能看到三个面。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】观察长方体或正方体，无论从哪个角度看，最多只能看到物体的三个面：前面，上面和一个侧面，据此即可解答.\n【详解】把一个文具盒放在桌子上，站在一个位置观察，最多能看到三个面，此题说法正确。故答案为:√。",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "18-19 四年级下·全国·单元测试"
+    },
+    {
+        "id":  "174",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "长颈鹿长得高，它站在同一个位置就可以看到物体所有的面。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】从不同的方向观察到的物体的形状是不同的，从一个方向观察物体，不可能观察到物体的每一个面.\n【详解】长颈鹿虽然高，但是它不可能看到物体所有的面，原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "观察物体"
+                             ],
+        "source":  "18-19 四年级下·全国·单元测试"
+    },
+    {
+        "id":  "175",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "观察物体时，只能从前面和后面观察。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】从不同的方向观察到的物体形状的不同的，观察物体可以从前后左右或者上面来观察.\n【详解】观察物体时，不是只能从前面和后面观察，原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "18-19 四年级下·全国·单元测试"
+    },
+    {
+        "id":  "176",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "在同一方位，看一个物体最多可以看到多少个面。看到一个面。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】1。 理解题意：题目中后半句\"看到一个面\"实际上是该命题给出的结论，即认为\"最多只能看到一个面\"。2。 逻辑推导：以长方体或正方体为例，当我们站在物体的斜上方(顶点位置)观察时，可以同时看到物体的正面，侧面和上面.\n【详解】从一个位置观察长方体或正方体，由于视角的改变，最多可以同时看到 3 个面，而不是 1 个面。因此原命题关于\"最多只能看到一个面\"的说法是错误的。故答案为: ×。",
+        "knowledge_points":  [
+                                 "观察物体",
+                                 "视角的局限性"
+                             ],
+        "source":  "18-19 四年级下·全国·单元测试"
+    },
+    {
+        "id":  "177",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "观察一个物体，从一个角度一次最多能看到3个面。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】1。 几何特性：对于常见的立体图形(如长方体，正方体),它们有 6 个面，且相对的面是平行的。遮挡原理：因为物体是不透明的，当我们看到一个面时，它相对的那个面一定会被遮挡。所以，我们不可能同时看到\"前和后\"，\"上和下\"或\"左和右\".\n【详解】在同一个角度观察时，我们的视线最多只能交汇于一个顶点，而每一个顶点连接着 3 个相邻的面。因此，一次最多只能看到 3 个面。故答案为:√。",
+        "knowledge_points":  [
+                                 "观察物体"
+                             ],
+        "source":  "18-19 五年级上·河北秦皇岛·期中"
+    },
+    {
+        "id":  "178",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "观察一个物体时，最多能看到两个面。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】观察一个物体时，可能看到物体的一个面，也可能看到两个面，还可能看到三个面，据此解答。本题主要考查了学生对观察立体图形知识的掌握情况，从不同方向观察立体图形时，因观察的方向不同，看到物体的形状也不相同.\n【详解】由分析可得：观察一个物体时，最多能看到三个面，所以原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "18-19 五年级上·全国·期中"
+    },
+    {
+        "id":  "179",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "从同一方向观察不同的物体看到的形状可能相同，也可能不同。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】本题考察的是三维物体在二维平面投影时的特征。形状相同的情况：不同形状的物体在某个特定视角下的投影可能会重合。例如，从正面看一个圆柱体(横放)和一个长方体，看到的都是长方形。形状不同的情况：在大多数情况下，由于物体几何结构不同，其投影也是不同的。例如，从上面看一个球体(圆形)和一个正方体(正方形)。由于命题使用了\"可能。也可能。.\"这一严谨的表述，涵盖了所有可能性，因此是正确的.\n【详解】根据观察物体的经验：相同示例：观察一个圆柱(底面朝前)和一个球，从上面看都是圆形;不同示例：观察一个正方体和一个圆锥，从正面看一个是正方形，一个是三角形。综上所述，命题表述符合事实。故答案为:√",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体",
+                                 "观察物体的多重性"
+                             ],
+        "source":  "17-18 四年级下·全国·期末"
+    },
+    {
+        "id":  "180",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "一个正方体，无论在哪个位置观察，最多能看到6个面。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】一个正方体，从它的1个面看，只能看到1个面，从它的1条棱看，只能看到2个面，从它的 1 个顶点看，只能看到 3 个面，也就是说一个正方体无论在哪个位置观察，最多只能看到3个面.\n【详解】一个正方体，无论在哪个位置观察，最多能看到 3 个面。故答案为: ×。",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "2014五年级·全国·课后作业"
+    },
+    {
+        "id":  "181",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "从同一个位置观察不同的物体，看到的图形可能一样。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】对于一般物体，从不同的位置观察同一个物体，所看到的形状往往不同；但题目说的是从同一个位置观察不同的物体，这时看到的图形可能一样，也可能不一样，据此判断即可。\n【详解】从同一个位置观察不同的物体，看到的图形可能一样。例如，以下三个立体图形从同一方向观察，看到的图形就可能相同。![](images/38ae53bc3888e613de32c419c067ba918f5f043d618a527cc758856256b7acd1.jpg)故答案为:√。",
+        "knowledge_points":  [
+                                 "物体三视图的认识"
+                             ],
+        "source":  "15-16 四年级下·全国·月考"
+    },
+    {
+        "id":  "182",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "球体从任何角度看都是圆形。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】球体是一个特殊的几何体，无论从哪个角度观察得到的图形都是全等的圆形，据此即可解答。考查三视图的有关知识，注意球体是一个特殊的几何体，无论从哪个角度观察得到的图形都是全等的圆形.\n【详解】球体无论从哪个角度观察，看到的平面图形都是一个全等的圆形，所以原题说法正确。",
+        "knowledge_points":  [
+                                 "立体图形的认识，从不同位置观察单个物体"
+                             ],
+        "source":  "13-14 五年级上·内蒙古乌海·期中"
+    },
+    {
+        "id":  "183",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "从正面，侧面，上面观察一个物体，不可能看到一样的图形。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】此题考查图形的三视图，要注意特殊图形：如正方体和球体，从正面，侧面，上面观察看到的图形都一样，据此即可判断。此题考查了从不同方向观察物体和几何体，锻炼了学生的空间想象力和抽象思维能力.\n【详解】正方体和球体从正面，侧面，上面观察看到的图形都一样，原题说法错误。故答案为: ×。",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "14-15四年级·全国·期末"
+    },
+    {
+        "id":  "184",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.7,
+        "question":  "观察一个长方体，从不同的位置观察，最多能看到3个面。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】观察一个长方体时，视线与物体成一定角度，由于物体不透明，一次最多只能看到它的三个面(即正面，侧面和上面)。此题考察空间观察的基本常识，帮助学生建立初步的空间方位感.\n【详解】在观察长方体时，无论从哪个角度看，被遮挡的面至少有三个。因此最多只能看到三个面。原题说法正确。故答案为:√。",
+        "knowledge_points":  [
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "22-23四年级下·全国·期末"
+    },
+    {
+        "id":  "185",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.75,
+        "question":  "由 4 个小正方体摆成的几何体，从正面看到的是 hline \u0026 \u0026 \n hline ,这个几何体只有一种摆法。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】仅从一个方向观察到的平面图形，不能唯一确定几何体的空间形状，因为无法确定小正方体在深度方向(前后)的位置。此题旨在说明单视图的局限性，理解\"三视图\"才能确定物体的空间形态.\n【详解】正面看到横向三个方块，第四个小正方体可以摆在其中任意一个的前面或后面。摆法有多种，原题说法错误。故答案为: ×。",
+        "knowledge_points":  [
+                                 "根据单一方向视图还原几何体"
+                             ],
+        "source":  "21-22五年级下·山东·单元测试"
+    },
+    {
+        "id":  "186",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.8,
+        "question":  "从左面观察一个圆柱体，看到的图形一定是长方形。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】圆柱体的观察结果取决于摆放方式。如果是立着的圆柱，侧面看是长方形;如果是横着的，侧面看可能是圆。考查观察物体时的动态思维，注意几何体摆放姿态对视图的影响.\n【详解】当圆柱底面朝向左侧时，从左面看到的图形是圆。原题没有说明摆放位置，说法太绝对。故答案为: ×。",
+        "knowledge_points":  [
+                                 "圆柱体的三视图"
+                             ],
+        "source":  "19-20五年级下·全国·期末"
+    },
+    {
+        "id":  "187",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.6,
+        "question":  "两个不同的物体，从同一个方向观察，看到的图形可能相同。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】不同形状的立体图形在某一特定方向上的投影(视图)是有可能重合的.\n【详解】例如：一个正方体和一个圆柱(底面直径等于高),从正面看都是正方形。所以原题说法正确。故答案为:√。",
+        "knowledge_points":  [
+                                 "不同物体在同一方向的投影"
+                             ],
+        "source":  "20-21四年级下·江苏·单元测试"
+    },
+    {
+        "id":  "188",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.82,
+        "question":  "用 5 个同样的正方体搭成一个几何体，从上面看到的是 hline \u0026 \n hline \u0026 hline ,这个几何体至少有两层。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】考察结合正方体数量和视图进行逻辑推理的能力，上面看到 4 个方块组成的正方形，说明底层已经占用了 4 个正方体。总共有 5 个，多出的 1 个必须放在第二层.\n【详解】底层用了 4 个，第 5 个必须放在这 4 个中的任意一个上面，因此该几何体一定是两层。故答案为:√。",
+        "knowledge_points":  [
+                                 "根据视图判断几何体层数"
+                             ],
+        "source":  "23-24五年级下·广东·期末"
+    },
+    {
+        "id":  "189",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "一个物体从正面看到的是 hline \u0026 \n hline ,从左面看到的是 hline \u0026 \n hline ,这个物体至少由 2 个小正方体组成。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】此类题属于基础推理，需考虑最少和最多的极端情况。正面两个方块说明左右至少有两列;左面两个方块说明前后至少有两排.\n【详解】当两个正方体呈对角线摆放时，正面看是 2 个，左面看也是 2 个。因此至少需要 2 个。故答案为:√。",
+        "knowledge_points":  [
+                                 "根据两视图判断最小组成数"
+                             ],
+        "source":  "22-23五年级下·北京·单元测试"
+    },
+    {
+        "id":  "190",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.55,
+        "question":  "从任何方向观察球体，看到的图形都是圆。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】球体具有完美的中心对称性，其在任何平面上的投影均为该球的大圆。球体是三视图最特殊的几何体，需牢记其特性.\n【详解】无论从正面，上面还是侧面，球的轮廓始终是一个半径相同的圆。故答案为:√。",
+        "knowledge_points":  [
+                                 "球体的三视图"
+                             ],
+        "source":  "20-21四年级下·全国·月考"
+    },
+    {
+        "id":  "191",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.78,
+        "question":  "如果从正面看到的是 hline \u0026 \n hline ,那么这个几何体一定是由 2 个小正方体组成的。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】区分\"看到的数量\"与\"实际组成的数量\"，正面图只能确定物体的宽度和高度，无法确定其深度(厚度).\n【详解】在正面看到的两个方块后面，可以隐藏任意数量的小正方体而不改变正面视图。故答案为: ×。",
+        "knowledge_points":  [
+                                 "视图与物体组成数量的关系"
+                             ],
+        "source":  "18-19五年级下·河南·期末"
+    },
+    {
+        "id":  "192",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.7,
+        "question":  "从上面观察一个物体，看到的图形是正方形，这个物体一定是正方体。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】多种几何体从上面看都是正方形，不能仅凭一个视图断定物体形状。考查逻辑推理的严密性，注意\"一定\"这种绝对化的措辞.\n【详解】例如长方体(底面是正方形)或圆柱体(横放且长等于底面直径)从上面看也可能是正方形。故答案为: ×。",
+        "knowledge_points":  [
+                                 "几何体的识别与视图分析"
+                             ],
+        "source":  "21-22五年级下·四川·月考"
+    },
+    {
+        "id":  "193",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.6,
+        "question":  "观察由 6 个小正方体搭成的几何体，看到的形状不可能是 7 个小正方形组成的图形。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】每一个方向看到的图形，其组成的小正方形个数不可能超过原物体的小正方体总数。理解视图面积与实体体积(数量)的限制关系.\n【详解】因为只有 6 个正方体，无论如何重叠或排列，在平面视图中最多只能看到 6 个面。故答案为:√。",
+        "knowledge_points":  [
+                                 "视图中图形面积的关系"
+                             ],
+        "source":  "22-23五年级下·浙江·期末"
+    },
+    {
+        "id":  "194",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "在任一位置观察一个长方体，都不可能看到它的底面。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】打破思维定式，明确观察物体时\"上面，下面\"是相对观察者的位置而言的。观察者的位置是灵活的。如果我们将长方体举高，或者人蹲下来看，就能看到底面.\n【详解】题目中\"任一位置\"说法太绝对，只要观察点低于长方体底面所在平面，就可以看到。故答案为: ×。",
+        "knowledge_points":  [
+                                 "观察位置与视角"
+                             ],
+        "source":  "23-24四年级下·山东·期末"
+    },
+    {
+        "id":  "195",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.68,
+        "question":  "从左面和正面观察同一个物体，看到的图形一定不相同。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】考查特殊几何体的三视图特征，对于某些具有对称性的物体(如正方体，球体，长等于宽的长方体),不同方向的视图完全可以一致.\n【详解】参考正方体，其六个面的视图均为完全相同的正方形。故答案为: ×。",
+        "knowledge_points":  [
+                                 "不同方向视图的对比"
+                             ],
+        "source":  "17-18五年级下·全国·单元测试"
+    },
+    {
+        "id":  "196",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.74,
+        "question":  "由若干个相同的小正方体搭成的几何体，从正面看和左面看形状相同，则这个几何体最少由 2 个小正方体组成。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】如果只有1个小正方体，那么从正面看和从左面看形状也相同，都是1个正方形。因此“最少由2个小正方体组成”说法不对。\n【详解】由若干个相同的小正方体搭成的几何体，从正面看和左面看形状相同，最少只需要1个小正方体。所以原题说法错误。故答案为: ×。",
+        "knowledge_points":  [
+                                 "构建几何体的最少需求量"
+                             ],
+        "source":  "20-21五年级下·湖南·期末"
+    },
+    {
+        "id":  "197",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.5,
+        "question":  "观察一个正方体，从任何一个面看到的形状都一样。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】正方体是六个面完全相同的正方形组成的立体图形.\n【详解】无论从正面，上面还是侧面看，得到的投影都是大小相等的正方形。故答案为:√。",
+        "knowledge_points":  [
+                                 "正方体的特征"
+                             ],
+        "source":  "19-20四年级下·江苏·期末"
+    },
+    {
+        "id":  "198",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.62,
+        "question":  "从正面观察这几个物体，看到的形状是相同的。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】如果几个物体在正前方的投影形状和大小完全一致，那么从正面看到的图形就相同。\n【详解】从正面观察这几个物体，看到的图形相同，所以原题说法正确。故答案为:√。",
+        "knowledge_points":  [
+                                 "不同几何体的正面视图"
+                             ],
+        "source":  "21-22五年级下·全国·同步练习"
+    },
+    {
+        "id":  "199",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.5,
+        "question":  "观察一个正方体，从任何一个面看到的形状都一样。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】本题考察正方体的几何特性。正方体是由 6 个完全相同的正方形面组成的立体图形，具有高度的对称性.\n【详解】由于正方体的每一个面都是大小完全相等的正方形，因此无论从正面，上面，下面还是左，右侧面进行正视观察，得到的平面投影图形都是完全一样的正方形。故答案为:√。",
+        "knowledge_points":  [
+                                 "正方体的特征",
+                                 "观察物体"
+                             ],
+        "source":  "19-20 四年级下·江苏·期末"
+    },
+    {
+        "id":  "200",
+        "unit":  "第二单元 观察物体",
+        "question_type":  "判断题",
+        "difficulty":  0.62,
+        "question":  "从正面观察三个物体(如：棱长，底面直径与高均相等的正方体，圆柱和球),看到的形状是相同的。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】本题考查不同立体图形在同一方向上投影形状的比较。正方体从正面看是正方形；圆柱在底面直径等于高时，从正面看也可以是正方形；但球体从正面看是圆形。\n【详解】因为球体从正面看到的是圆形，而正方体和满足条件的圆柱从正面看到的是正方形，所以这三个物体从正面看到的形状并不相同。故答案为: ×。",
+        "knowledge_points":  [
+                                 "不同几何体的正面视图",
+                                 "三视图的重合性"
+                             ],
+        "source":  "21-22 五年级下·全国·同步练习"
+    }
+]
diff --git "a/\345\233\233\345\271\264\347\272\247\351\242\230\351\233\206/\345\233\233\345\271\264\347\272\247\344\270\213/\345\233\233\345\271\264\347\272\247\344\270\213\347\254\254\344\272\224\345\215\225\345\205\203\344\270\211\350\247\222\345\275\242.json" "b/\345\233\233\345\271\264\347\272\247\351\242\230\351\233\206/\345\233\233\345\271\264\347\272\247\344\270\213/\345\233\233\345\271\264\347\272\247\344\270\213\347\254\254\344\272\224\345\215\225\345\205\203\344\270\211\350\247\222\345\275\242.json"
new file mode 100644
index 0000000000000000000000000000000000000000..f9eb84f4ec56b90b8a60570f1e6144720b7a1544
--- /dev/null
+++ "b/\345\233\233\345\271\264\347\272\247\351\242\230\351\233\206/\345\233\233\345\271\264\347\272\247\344\270\213/\345\233\233\345\271\264\347\272\247\344\270\213\347\254\254\344\272\224\345\215\225\345\205\203\344\270\211\350\247\222\345\275\242.json"
@@ -0,0 +1,8827 @@
+[
+    {
+        "id":  "1",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "梯形的4个角的度数和是( )度。",
+        "options":  [
+
+                    ],
+        "answer":  "360",
+        "solution":  "【分析】根据题意，梯形是四边形的一种，所有四边形的内角和都是360度。这是因为任何四边形都可以被分割成两个三角形，而每个三角形的内角和是180度，所以两个三角形的内角和加起来就是360度。以此答题即可.\n【详解】根据分析可知：梯形的4个角的度数和是360度。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "梯形的概念及特点"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "2",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个等腰三角形的顶角是92度，它的一个底角是( )度，这个三角形按角分类属于( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "44;钝角",
+        "solution":  "【分析】三角形内角和180度，等腰三角形两底角相等，一个底角的度数=(三角形内角和-顶角度数)÷2，根据这个三角形最大内角的度数，确定这个三角形按角分的类型即可.\n【详解】(180-92)÷2=88÷2=44(度);92度的角是钝角。所以一个等腰三角形的顶角是92度，它的一个底角是44度，这个三角形按角分类属于钝角三角形。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "3",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个等腰三角形的底角是55°,它的顶角是( )°,按角分这个三角形是一个( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "70;锐角",
+        "solution":  "【分析】任意三角形的内角和都是180°,等腰三角形的两个底角相等。已知一个等腰三角形的底角是55°,则另一个底角也是55°,用三角形的内角和减去两个底角的度数，求出它的顶角的度数;再根据按角分类的方法，得出这个三角形的类型。锐角三角形：三个角都是锐角的三角形;直角三角形：有一个角是直角的三角形;钝角三角形：有一个角是钝角的三角形.\n【详解】180°-55°-55°=70°;55°\u003c90°,70°\u003c90°,这个等腰三角形的三个内角都是锐角，所以它是锐角三角形。所以它的顶角是70°,按角分这个三角形是一个锐角三角形。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "4",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个直角三角形的三条边分别是3厘米,4厘米和5厘米。这个三角形底是3厘米时，它的高是( )厘米;这个三角形的底是4厘米时，它的高是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "4;3",
+        "solution":  "【分析】从三角形任一顶点向它的对边或者对边的延长线作垂线，从顶点到垂足间的线段叫做三角形的高。这个顶点所对的边叫做三角形的底。直角三角形中，两条直角边都小于斜边。直角三角形中，两条直角边互为底和高.\n【详解】根据分析：因为3\u003c4\u003c5，所以这个三角形的两条直角边分别是3厘米和4厘米，那么3厘米和4厘米的直角边互为底和高。所以这个三角形底是3厘米时，它的高是4厘米;这个三角形的底是4厘米时，它的高是3厘米。",
+        "knowledge_points":  [
+                                 "三角形的高及画法"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "5",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "图中有( )个锐角三角形,( )个直角三角形,( )个钝角三角形.\n![](images/b46e69505a7031a6f782349dc60327caeb2e4b9f2470554bc59d7e3b4d2a8935.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "1;4;5",
+        "solution":  "【分析】三个角都是锐角的三角形是锐角三角形;有一个角是钝角的三角形是钝角三角形;有一个角是直角的三角形是直角三角形.\n【详解】如图:![](images/0118be064f49bf4847dbe12710fc523a1f882ddd7a7a8f8268fae55360d42966.jpg)\n锐角三角形有:③+④,共1个;直角三角形有:③,④,②+③,①+②+③,共4个;钝角三角形有:①,②,①+②,②+③+④,①+②+③+④,共5个。因此，图中有1个锐角三角形,4个直角三角形,5个钝角三角形。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "数图形"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "6",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在三角形中,∠1=45°,∠2=38°,∠3=( )°,它是一个( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "97;钝角",
+        "solution":  "【分析】需要用到三角形内角和的知识，即三角形三个内角的度数之和是180度。已知∠1和∠2的度数，用180度减去∠1与∠2的度数和，即可得到∠3的度数;大于90度小于180度的角是钝角，有一个角是钝角的三角形是钝角三角形。因为∠3的度数是97度,97度大于90度，所以可以判断这个三角形是钝角三角形.\n【详解】∠3:180°-45°-38°=97°;所以，在三角形中,∠1=45°,∠2=38°,∠3=97°,它是一个钝角三角形。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "三角形的内角和"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "7",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形，它的一个角是30°,那么它的另两个角可能是( )°和( )°,也可能是( )°和( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "30;120;75;75",
+        "solution":  "【分析】等腰三角形的两腰相等，两个底角相等，三角形的内角和是180°,当30°为底角时，则另一个角也是30°,而用180°减去另外两个角的度数之和，就是顶角的度数;当30°为顶角时，用180°减去30°后，再除以2，就是两个底角的度数，依此解答.\n【详解】当30°为底角时:180°-30°-30°=120°;当30°为顶角时:(180°-30°)÷2=150°/2=75°。所以一个等腰三角形，它的一个角是30°,那么它的另两个角可能是30°和120°,也可能是75°和75°。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "8",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "梯形的一条对角线把它分成( )个三角形，每个三角形的内角和是( )°,所以梯形的内角和是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "2;180;360",
+        "solution":  "【分析】梯形是四边形，连接其一条对角线，可将其分割为2个三角形;三角形的内角和固定为180°,梯形的内角和等于这2个三角形的内角和之和.\n【详解】对角线分割的三角形数量:2个;每个三角形的内角和:180°;梯形的内角和:180°*2=360°。所以梯形的一条对角线把它分成2个三角形，每个三角形的内角和是180°,所以梯形的内角和是360°。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "梯形的概念及特点",
+                                 "多边形的内角和"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "9",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "乐乐买了一个等腰三角形的风铃，它的一个底角是40°,顶角是( )°。按角分，这个风铃是一个( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "100;钝角",
+        "solution":  "【分析】等腰三角形的两个底角相等，且三角形内角和为180°。先利用\"内角和-两个底角的度数和\"求出顶角;再根据顶角的大小判断三角形类型(顶角大于90°则为钝角三角形).\n【详解】求顶角的度数:180°-40°*2=180°-80°=100°;判断三角形类型：顶角为100°(大于90°),因此这个三角形是钝角三角形。所以乐乐买了一个等腰三角形的风铃，它的一个底角是40°,顶角是100°。按角分，这个风铃是一个钝角三角形。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "10",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在三角形ABC中,∠A=48°,∠C=63°,∠B=( )°。按角分，这是一个( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "69;锐角",
+        "solution":  "【分析】三角形内角和为180°,用内角和减去已知的∠A和∠C的度数，可求出∠B;再根据三个角的大小判断三角形类型(三个角均为锐角则是锐角三角形).\n【详解】求∠B的度数:180°-48°-63°=69°;判断三角形类型:∠A=48°,∠B=69°,∠C=63°,三个角均小于90°,因此这是一个锐角三角形。所以在三角形ABC中,∠A=48°,∠C=63°,∠B=69°。按角分，这是一个锐角三角形。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "三角形的内角和"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "11",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "有4cm和9cm长的2根小棒，如果要围成一个三角形，第3根小棒最长是( )cm,最短是( )cm。(小棒长度为整厘米数)",
+        "options":  [
+
+                    ],
+        "answer":  "12;6",
+        "solution":  "【分析】结合题意得知，两根小棒的长度分别为4厘米,9厘米，那么根据三角形的特征：任意两边之和大于第三边,4+9=13，那么可以知道第三边最长是多少厘米;根据三角形两边之差小于第三边,9-4=5，那么由此就可以得出第三边最短是多少厘米.\n【详解】9-4\u003c第三边\u003c9+4，所以:5\u003c第三边\u003c13，即第三边的取值在5厘米~13厘米之间(不包括5厘米和13厘米),因为三根小棒都是整厘米数，所以第三根小棒最长为:13-1=12(厘米),最短为:5+1=6(厘米)。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "12",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "∠1,∠2,∠3是一个三角形的三个内角，已知∠1=25°,∠2=55°,则∠3=( )°,这是一个( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "100;钝角",
+        "solution":  "【分析】三角形的内角和为180°,∠3=180°-∠1-∠2。根据∠3是100°可知∠3是钝角。根据\"有一个钝角的三角形叫做钝角三角形\"，可知这是一个钝角三角形。据此解答.\n【详解】180°-25°-55°=100°;因为100°\u003e90°,∠3是钝角，所以这个三角形是钝角三角形。所以∠1,∠2,∠3是一个三角形的三个内角，已知∠1=25°,∠2=55°,∠3=100°,这是一个钝角三角形。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "三角形的内角和"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "13",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三角形的两个角分别是36°和45°,这个三角形是( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "钝角",
+        "solution":  "【分析】根据三角形内角和为180°可算出第三个角的度数，再根据三角形按角的分类确定这个三角形的类型。三角形按角分类：三个角都是锐角的三角形是锐角三角形，有一个角是直角的三角形是直角三角形，有一个角是钝角的三角形是钝角三角形.\n【详解】180°-36°-45°=99°;99°为钝角，所以这个三角形是钝角三角形。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "三角形的内角和"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "14",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三角形三条边的长度都是整厘米数，其中有两条边长分别是9厘米,4厘米，那么这个三角形的周长最短是( )厘米，最长是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "19;25",
+        "solution":  "【分析】由题意可知：三角形的周长最短，即第三条边最短，三角形的周长最长，即第三条边最长，根据\"任意两边之差\u003c第三边\u003c任意两边之和\"得出第三边最长，最短是多少厘米，再把三角形的三边加起来即可求解.\n【详解】9-4=5(厘米),9+4=13(厘米),5厘米\u003c第三边\u003c13厘米，则第三条边最长是12厘米，最短是6厘米。9+4+6=19(厘米),9+4+12=25(厘米)。那么这个三角形的周长最短是19厘米，最长是25厘米。",
+        "knowledge_points":  [
+                                 "三角形的周长",
+                                 "三角形三边关系"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "15",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三角形沿一条直线剪去一个角后，剩余图形的内角和可能是( )°或( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "180;360",
+        "solution":  "【分析】根据题意，一个三角形沿一条直线剪去一个角后，剩余图形的形状可能变为三角形或四边形。当剪的直线经过一个顶点和其对边上的某一点时，剩余图形仍为三角形。当剪的直线连接两条边上的点(不经过顶点)时，剩余图形变为四边形。明确三角形的内角和为180°;若变为四边形，可以分成两个三角形，则内角和为180°*2=360°。以此答题即可.\n【详解】根据分析可知:180°*2=360°,所以一个三角形沿一条直线剪去一个角后，剩余图形的内角和可能是180°或360°。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "多边形的内角和"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "16",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形的一个角是70°,那么另外两个角可能是( )°,( )°或( )°,( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "55;55;70;40",
+        "solution":  "【分析】三角形内角和等于180°。等腰三角形两底角相等。已知给出了一个内角是70°,没有明确是顶角还是底角，所以要进行分类讨论。当70°的角是顶角时，用180°减去70°,再除以2即可求出另外两个底角的度数;当70°的角是底角时，另一个底角也是70°,用180°减去70°,再减去70°即可求出顶角，据此解答.\n【详解】①当70°的角是顶角时，底角为:(180°-70°)÷2=55°;②70°的角是底角，另一个底角是70°,顶角为:180°-70°-70°=40°。所以一个等腰三角形的一个角是70°,那么另外两个角可能是55°,55°或70°,40°。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "17",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "有2厘米和5厘米的小棒各1根，如果再选一根小棒围成一个三角形，这根小棒最长是( )厘米(取整厘米数);如果要围成一个等腰三角形，这根小棒的长度是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "6;5",
+        "solution":  "【分析】在一个三角形中，任意两边之和大于第三边，任意两边之差小于第三边。那么比2厘米和5厘米的和小1厘米即为第三边最长的长度;如果要围成一个等腰三角形，假设第三边为2厘米,2+2=4(厘米)\u003c5厘米，不能围成三角形，只能选择第三边是5厘米.\n【详解】5+2-1=6(厘米),如果再选一根小棒围成一个三角形，这根小棒最长是6厘米;由分析可知：如果要围成一个等腰三角形，这根小棒的长度是5厘米，则三边长是5厘米,5厘米,2厘米符合等腰三角形特征。所以有2厘米和5厘米的小棒各1根，如果再选一根小棒围成一个三角形，这根小棒最长是6厘米;如果要围成一个等腰三角形，这根小棒的长度是5厘米。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形三边关系"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "18",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三角形的三条边长相等，这个三角形按边分是一个( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "等边",
+        "solution":  "【分析】根据三角形按边分类的标准，三条边都相等的三角形称为等边三角形。题目中明确三条边长相等，因此按边分属于等边三角形.\n【详解】数学定义中，将\"三条边长度都相等的三角形\"称为等边三角形，已知\"三条边长相等\"，符合等边三角形的唯一定义，因此该三角形按边分属于等边三角形。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "19",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个三角形中最大的角是直角，这个三角形按角分是( )三角形，若另一个角是45°,第三个角的度数是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "直角;45",
+        "solution":  "【分析】有一个角是直角的三角形是直角三角形。三角形的内角和是180°,用180°减去已知两个角的度数，就是第三个角的度数。据此解答.\n【详解】180°-90°-45°=45°。所以，一个三角形中最大的角是直角，这个三角形按角分是直角三角形，若另一个角是45°,第三个角的度数是45°。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "三角形的内角和"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "20",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "已知一个三角形的两条边的长度分别是7cm和13cm,则它的第三条边最长是( )cm,最短是( )cm。(长度取整厘米数)",
+        "options":  [
+
+                    ],
+        "answer":  "19;7",
+        "solution":  "【分析】根据三角形三边关系\"任意两边之和大于第三边，任意两边之差小于第三边\"，结合已知边7厘米和13厘米，确定第三边的范围，再取整厘米数即可.\n【详解】根据分析可知:7+13=20(厘米),13-7=6(厘米),20-1=19(厘米),6+1=7(厘米)。所以第三条边最长是19cm,最短是7cm。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "21",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三角形的两边长分别为5cm和8cm,第三边最长是( )cm,最短是( )cm。(取整数)",
+        "options":  [
+
+                    ],
+        "answer":  "12;4",
+        "solution":  "【分析】三角形两边之和大于第三边，三角形两边之差小于第三边，所以三角形的第三边长要小于已知的两条边之和，且大于已知的两条边之差.\n【详解】8-5=3(cm),5+8=13(cm),所以3cm\u003c第三边长\u003c13cm。因此，第三边最长是12cm,最短是4cm。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "22",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形的周长是60cm,底是28cm,它的一条腰长( )cm,在这个三角形中一个底角是70°,顶角是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "16;40",
+        "solution":  "【分析】已知等腰三角形的周长和底边长度，可用周长减去底边得到两腰总长，再除以2，得每条腰长。根据等腰三角形两底角相等的性质，用三角形内角和180°减去两个底角的度数之和，即可求出顶角。列式计算即可.\n【详解】根据分析可知:60-28=32(厘米),32÷2=16(厘米);180°-70°*2=180°-140°=40°。所以一个等腰三角形的周长是60cm,底是28cm,它的一条腰长16cm,在这个三角形中一个底角是70°,顶角是40°。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和",
+                                 "三角形的周长"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "23",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "房屋的屋顶采用三角形设计，是利用了三角形的( )性。",
+        "options":  [
+
+                    ],
+        "answer":  "稳定",
+        "solution":  "【分析】本题考查三角形的特性。三角形具有不易变形的特点，这种特性被称为稳定性，即三角形的三条边长度确定后，三角形的形状和大小就固定不变，不易变形。房屋的屋顶需要稳定的结构，三角形的这种特性能够满足屋顶结构稳定的需求.\n【详解】房屋的屋顶采用三角形设计，是利用了三角形的稳定性。故答案为：稳定",
+        "knowledge_points":  [
+                                 "三角形的稳定性及应用"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "24",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形的底角是45°,它的顶角是( )°;有两根木条，分别是5厘米,10厘米，再用一根长( )厘米的木条，就可以围成一个等腰三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "90;10",
+        "solution":  "【分析】根据等腰三角形的两个底角相等，用三角形的内角和180°减去两个底角的度数即可求出三角形顶角的度数。等腰三角形有两条边相等，任意三角形的两边之和必须大于第三边，求出两边之和与第三边比较，满足三边关系的即可.\n【详解】180°-45°*2=90°;当腰长是5厘米时,5+5=10(厘米),10=10，所以5厘米,5厘米,10厘米不能组成等腰三角形。当腰长是10厘米时,10+5=15(厘米),15\u003e10，所以5厘米,10厘米,10厘米能组成等腰三角形。所以一个等腰三角形的底角是45°,它的顶角是90°;有两根木条，分别是5厘米,10厘米，再用一根长10厘米的木条，就可以围成一个等腰三角形。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和",
+                                 "三角形三边关系"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "25",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形的两个内角分别是50°,80°,则第三个内角是( )°,按角分这个三角形是( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "50;锐角",
+        "solution":  "【分析】三角形内角和是180°,用180°减去两个已知角的度数，即可求出第三个内角是多少度;根据三角形按角的分类，有一个角是钝角的三角形是钝角三角形，有一个角是直角的三角形是直角三角形，三个角都是锐角的三角形是锐角三角形，据此判断是什么三角形即可.\n【详解】180°-50°-80°=50°;三个角都是锐角，所以这个三角形是锐角三角形。所以一个等腰三角形的两个内角分别是50°,80°,则第三个内角是50°,按角分这个三角形是锐角三角形。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "三角形的内角和"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "26",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形的两条边分别是5厘米和10厘米，这个等腰三角形的第三边的长是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "10",
+        "solution":  "【分析】在等腰三角形中，相等的两条边叫做腰，另一条边叫做底。三角形中两边之和大于第三边。根据题意，分情况讨论。第一种情况：把5厘米的边作为腰,5+5=10厘米,10=10，两边之和等于第三边，不能组成三角形。第二种情况：把10厘米的边作为腰,10+5=15厘米,15\u003e10，能组成三角形.\n【详解】因为5厘米+5厘米=10厘米，不能组成三角形，所以这个等腰三角形的第三边的长是10厘米。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形三边关系"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "27",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "明明用一根24分米的铁丝正好围成了一个等边三角形，这个三角形的每个边长是( )分米，每个角是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "8;60",
+        "solution":  "【分析】等边三角形的三条边长度相等，三个角也相等。用铁丝总长度除以3可得边长，根据三角形内角和定理，每个角的度数为180°/3，据此计算作答.\n【详解】24÷3=8(分米),180°/3=60°。因此，这个三角形的每个边长是8分米，每个角是60°。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和",
+                                 "三角形的周长"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "28",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形的两条边是2cm和5cm,则它的周长是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "12cm",
+        "solution":  "【分析】等腰三角形的特点是两腰相等。三角形的三边关系：任意两边之和大于第三边，当2cm为腰长时,2加2得4，而4小于5，所以2cm不能为腰,5cm为腰长，底边长为2cm,再把三边的长度相加，即可求出周长.\n【详解】5+5+2=12(cm)。所以一个等腰三角形的两条边是2cm和5cm,则它的周长是12cm。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的周长",
+                                 "三角形三边关系"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "29",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "已知三角形ABC中.\n(1)若∠A=36°,∠B=53°,则按角分，三角形ABC是( )三角形.\n(2)三角形ABC是一个等腰三角形，它有两条边的长度分别是1厘米和3厘米，则它的周长是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "钝角;7",
+        "solution":  "【分析】(1)三个角都是锐角的三角形是锐角三角形，有1个角是直角的三角形是直角三角形，有1个钝角的三角形是钝角三角形，据此判断即可;(2)根据三角形三条边的关系：两边之和大于第三条边，得出等腰三角形的三条边长度，最后把它们加起来就可以求出周长.\n【详解】(1)180°-36°-53°=91°,91°\u003e90°,则按角分，三角形ABC是钝角三角形。(2)假如这个等腰三角形的腰长是1厘米,1+1\u003c3，两边之和小于第三边了，拼不成一个三角形，所以，腰长只能是3厘米。1+3+3=7(厘米),所以，它的周长是7厘米。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和",
+                                 "三角形三边关系"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "30",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三角形，一个内角是28°,另一个内角是52°,按角分，这是一个( )三角形;一个等腰三角形的顶角度数是一个底角的4倍，这个等腰三角形的一个底角是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "钝角;30",
+        "solution":  "【分析】有一个角大于90度的三角形是钝角三角形，三角形的内角和为180度;等腰三角形中两个底角相等，因此假设底角的度数为1份，则顶角的度数为4份，一共是(1+1+4)份，那么可用180°除以总份数，即可计算出底角的度数.\n【详解】180°-28°-52°=100°,100°\u003e90°,所以这个三角形是钝角三角形。180°/(1+1+4)=30°。所以一个等腰三角形的顶角度数是一个底角的4倍，这个等腰三角形的一个底角是30°。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "31",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "三角形的两条边分别是4cm和9cm,另一条边可能是( )cm;如果这个三角形是等腰三角形，它的另一条边是( )cm。(取整厘米数)",
+        "options":  [
+
+                    ],
+        "answer":  "6、7、8、9、10、11、12;9",
+        "solution":  "【分析】三角形的三边关系为三角形的两边之和大于第三边，三角形的两边之差一定小于第三边;等腰三角形两腰相等，分析两种情况，据此解答即可.\n【详解】4+9=13(cm),9-4=5(cm),5cm\u003c第三边长\u003c13cm,第三边可以取6,7,8,9,10,11,12。当4cm是腰长时,4+4=8(cm),8\u003c9，因此4cm不能是腰长。当9cm是腰长时,9+9=18(cm),18\u003e4,9-9=0(厘米),0\u003c4，因此9cm是腰长。所以三角形的两条边分别是4cm和9cm,另一条边可能是6,7,8,9,10,11,12cm;如果这个三角形是等腰三角形，它的另一条边是9cm。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形三边关系"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "32",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "等边三角形也叫正三角形，它也是( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "锐角",
+        "solution":  "【分析】因为等边三角形的三条边都相等，三个内角都相等，都是60°。60°是锐角，三个角都是锐角的三角形是锐角三角形.\n【详解】等边三角形三个角都是锐角，所以，等边三角形也叫正三角形，它也是锐角三角形。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "33",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个直角三角形，它的一个角是37°,另一个角是( )°。一个三角形，它既是等腰三角形又是钝角三角形，它的一个角是40°,另外两个角分别是( )°和( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "53;40;100",
+        "solution":  "【分析】直角三角形的一个角为90°,已知另一个角是37°,根据三角形内角和为180°,第三个角为180°-90°-37°=53°。钝角三角形有一个角大于90°,等腰三角形有两个相等的角，若40°为底角，则顶角为180°-40°*2=100°(钝角),符合条件;若40°为顶角，则底角为(180°-40°)÷2=70°,此时无钝角，不符合条件。因此另外两个角为40°和100°.\n【详解】180°-90°-37°=53°;180°-40°*2=100°。所以一个直角三角形，它的一个角是37°,另一个角是53°。一个三角形，它既是等腰三角形又是钝角三角形，它的一个角是40°,另外两个角分别是40°和100°。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "34",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三角形，两个内角的和等于第三个内角，按角分，这是( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "直角",
+        "solution":  "【分析】三角形的内角和是180°,两个内角的度数和等于第三个内角的度数，那么180°里面包括2个第三个内角的度数，用180°/2即可求出第三个内角的度数;三角形的分类：三个角都是锐角的三角形是锐角三角形;有一个角是直角的三角形是直角三角形;有一个角是钝角的三角形是钝角三角形，据此解答.\n【详解】根据分析可知:180°/2=90°,所以这个三角形是直角三角形。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "三角形的内角和"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "35",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形，其中一个底角是42°,它的顶角是( )°,按角分它是一个( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "96;钝角",
+        "solution":  "【分析】因为等腰三角形的两个底角相等，已知一个底角是42°,所以另一个底角也是42°。根据三角形内角和是180°,可计算出顶角的度数，据此判断三角形类型即可.\n【详解】180°-42°-42°=96°,因为96°\u003e90°,所以这个角是钝角。因为有一个角是钝角的三角形是钝角三角形，所以这个等腰三角形按角分是钝角三角形。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "三角形的内角和"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "36",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "用长度分别为11cm,6cm和( )cm的小棒一定能围成一个三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "10",
+        "solution":  "【分析】根据三角形的三边关系(两边之和大于第三边，两边之差小于第三边),先计算两边之差，再计算两边之和，则第三边的长度范围是：两边之差\u003c第三边\u003c两边之和，只要是在这个范围内的数都满足.\n【详解】11-6=5(cm),11+6=17(cm),所以第三边的长度只要满足大于5cm且小于17cm即可。所以用长度分别为11cm,6cm和10cm(答案不唯一)的小棒一定能围成一个三角形。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "37",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个三角形，其中两个内角分别是35°和55°,第三个内角的度数是( )°;按角分，这个三角形是( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "90;直角",
+        "solution":  "【分析】根据三角形的内角和是180°可知，第三个内角的度数等于180°减去第一个角，再减去第二个角;根据计算出的第三个角的大小进行分类，等于90°是直角，小于90°是锐角，大于90°是钝角.\n【详解】由分析可得：第三个角:180°-35°-55°=90°,第三个角等于90°是直角，这个三角形是直角三角形。所以一个三角形，其中两个内角分别是35°和55°,第三个内角的度数是90°;按角分，这个三角形是直角三角形。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "三角形的内角和"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "38",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "建筑工地的塔吊上有许多三角形，这是利用了三角形具有( )性。",
+        "options":  [
+
+                    ],
+        "answer":  "稳定",
+        "solution":  "【分析】三角形稳定性是指三角形具有稳定性，有着稳固，坚定，耐压的特点。起重机，三角形吊臂，屋顶，三角形钢架，钢架桥和埃菲尔铁塔都以三角形形状建造.\n【详解】由分析知：建筑工地的塔吊上有许多三角形，这是利用了三角形具有稳定性。",
+        "knowledge_points":  [
+                                 "三角形的稳定性及应用"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "39",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明做了一个等腰三角形的风筝，已知两条边长分别是55厘米,27厘米，那么这个等腰三角形风筝的周长是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "137",
+        "solution":  "【分析】先根据等腰三角形的特征确定第三条边长可能的情况，再根据\"任意两边之和必须大于第三边，否则无法构成三角形\"的三角形三边关系排除无效情况，确定最终的边长长度，最后再将三条边的边长相加计算出周长，据此解答.\n【详解】等腰三角形有两条边长度相等，因此第三条边有两种可能：第一种可能：与55厘米相等，此时三边长为55厘米,55厘米,27厘米,55厘米+55厘米=110厘米\u003e27厘米,55厘米+27厘米=82厘米\u003e55厘米，满足三边关系，有效;第二种可能：与27厘米相等，此时三边长为55厘米,27厘米,27厘米,27厘米+27厘米=54厘米\u003c55厘米，不满足三边关系，无效。因此等腰三角形三条边的边长分别是55厘米,55厘米,27厘米，其周长为:55+55+27=137(厘米)。所以小明做了一个等腰三角形的风筝，已知两条边长分别是55厘米,27厘米，那么这个等腰三角形风筝的周长是137厘米。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的周长",
+                                 "三角形三边关系"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "40",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形的顶角是50°,它的一个底角是( )°,这个三角形按角分是一个( )三角形。一个三角形三条边的长度分别是6cm,9cm,6cm,这个三角形是( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "65;锐角;等腰",
+        "solution":  "【分析】等腰三角形的两个底角相等。三角形的内角和是180°,用180°减去等腰三角形的顶角，求得两个底角的度数和，再用两个底角的度数和除以2，即可求得它的底角度数。按角分，锐角三角形：三个角都是锐角的三角形，直角三角形：有一个角是直角的三角形，钝角三角形：有一个角是钝角的三角形。一个三角形三条边的长度分别是6cm,9cm,6cm,等腰三角形：有两条边相等的三角形，据此解答.\n【详解】(180°-50°)÷2=65°,因此，一个等腰三角形的顶角是50°,它的一个底角是65°,这个三角形按角分是一个锐角三角形。因为这个三角形三条边的长度分别是6cm,9cm,6cm,其中有两条边相等，因此这个三角形是等腰三角形。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和",
+                                 "三角形三边关系"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "41",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小亮用一根48厘米长的铁丝围了一个等边三角形，等边三角形的每条边长是( )厘米，还可以用这根铁丝围成一个底边是18厘米，腰是( )厘米的等腰三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "16;15",
+        "solution":  "【分析】三角形的三边之和是三角形的周长;等边三角形三条边长度相等，用48除以3，即可算出这个等边三角形的每条边长度是多少;等腰三角形的两条腰长度相等;用这根铁丝的长度减去18，算出三角形另外两条边的长度之和，再除以2，即可算出这个等腰三角形的腰是多少。据此解答.\n【详解】48÷3=16(厘米);(48-18)÷2=15(厘米)。所以等边三角形的每条边长是16厘米，还可以用这根铁丝围成一个底边是18厘米，腰是15厘米的等腰三角形。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的周长"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "42",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "明明用三根小棒围成了一个等腰三角形，第一根长5厘米，第二根长11厘米，这个三角形的周长是( )厘米;按角分，这是一个( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "27;锐角",
+        "solution":  "【分析】等腰三角形的两条腰长度相等。由题意得，等腰三角形的两条边分别是5厘米和11厘米，那么可以假设5厘米或11厘米长的边为腰，然后利用三角形三边的关系(任意两边之和大于第三边)来验证假设是否成立。最后用满足题意的三条边算出等腰三角形的周长即可;三角形按角来分，分为锐角三角形，直角三角形和钝角三角形。三个角都是锐角的三角形叫作锐角三角形，有一个角是直角的三角形叫作直角三角形，有一个角是钝角的三角形叫作钝角三角形。可以根据三角形三边的长度画出这个等腰三角形，然后看这个三角形属于哪类三角形即可.\n【详解】假设5厘米长的边为腰，那么另一条腰的长度也是5厘米。5+5=10(厘米),10厘米\u003c11厘米，即这三边无法围成三角形，该假设不成立。假设11厘米长的边为腰，那么另一条腰的长度也是11厘米。5+11=16(厘米),16厘米\u003e11厘米，即这三边可以围成三角形，该假设成立。5+11+11=27(厘米)。根据三角形的三边作图可知![](images/2b004f80cc31398b86af2dab43a3f5228b2d05baebf5ccda71fe1a97a19fb836.jpg),这个三角形的三个角都是锐角，这个三角形是锐角三角形。所以明明用三根小棒围成了一个等腰三角形，第一根长5厘米，第二根长11厘米，这个三角形的周长是27厘米;按角分，这是一个锐角三角形。",
+        "knowledge_points":  [
+                                 "三角形三边关系",
+                                 "三角形的周长",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的分类"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "43",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "若三角形的一个角是80°,另外两个角是( )°和( )°时，它是等腰三角形;如果一个等腰三角形的周长是32cm,其中一条边长8cm,那么另外两条边分别长( )cm和( )cm。",
+        "options":  [
+
+                    ],
+        "answer":  "50;50;80;20;12;12",
+        "solution":  "【分析】根据题意，明确等腰三角形的两底角相等，三角形的内角和是180°,如果80°的角是顶角，用180°减去80°,再除以2，就是底角的度数;如果80°的角是底角，用180°减去2个80°,就是顶角的度数。根据等腰三角形的两腰相等和三角形任意两边之和大于第三边，如果8厘米的边是底边，用32-8=24(厘米),24÷2=12(厘米),8+12=20\u003e12,12+12=24\u003e8，可知另外两条边是12厘米和12厘米;如果8厘米的边是腰,32-8×2=16(厘米),8+8=16，不能构成三角形;解答此题即可.\n【详解】根据分析可知:(180°-80°)÷2=50°,180°-80°*2=20°。所以若三角形的一个角是80°,另外两个角可能是50°和50°,也可能是80°和20°。32-8=24(厘米),24÷2=12(厘米),8+12=20\u003e12,12+12=24\u003e8;32-8×2=16(厘米),8+8=16，不能构成三角形。所以另外两条边分别长12cm和12cm。",
+        "knowledge_points":  [
+                                 "三角形三边关系",
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "44",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个正三角形的周长是3.6厘米，它的边长是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "1.2",
+        "solution":  "【分析】正三角形就是等边三角形，它的三条边都是相等的，可以发现,3个1.2相加等于3.6，所以1.2是边长，据此解答.\n【详解】根据分析:1.2+1.2+1.2=3.6(厘米),所以它的边长是1.2厘米。",
+        "knowledge_points":  [
+                                 "三角形的周长",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "45",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三角形的一个内角是80°,是另一个内角度数的2倍，那么第三个内角的度数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "60°或60度",
+        "solution":  "【分析】用一个内角的度数除以2，就是另一个内角的度数。三角形的内角和是180°,用180°减去一个内角度数，再减去另一个内角度数，就是第三个内角的度数.\n【详解】80°/2=40°,180°-80°-40°=60°。那么第三个内角的度数是60°。",
+        "knowledge_points":  [
+                                 "三角形的内角和"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "46",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一根铁丝恰好围成一个边长为12cm的等边三角形，如果把它改围成有一条边长为8cm的等腰三角形，那么另外两条边长是( )cm和( )cm。",
+        "options":  [
+
+                    ],
+        "answer":  "14;14",
+        "solution":  "【分析】等边三角形的三条边都相等，依此计算出这根铁丝的长度，再根据这根铁丝的长度,8cm,以及三角形三边的关系确定出另外两条边的长度即可。三角形任意两边的长度之和大于第三边，任意两边的长度之差小于第三边，依此解答.\n【详解】12×3=36(cm);假设8cm为腰长:36-8-8=20(cm),8+8=16(cm),16cm\u003c20cm,因此8cm不能为腰长。所以8cm为底长:36-8=28(cm),28÷2=14(cm),该等腰三角形另两条边是14cm和14cm。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的周长",
+                                 "三角形三边关系"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "47",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形两条边的长度分别为5厘米和11厘米，那么第三条边的长度是( )厘米，如果这个等腰三角形的一个底角是70°,它的顶角是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "11;40",
+        "solution":  "【分析】根据题意，等腰三角形的两腰相等，所以第三条边的长度是5厘米或11厘米。再根据三角形任意两边之和都大于第三条边，确定第三条边的最终长度。三角形的内角和是180°,等腰三角形的两个底角相等。用180°减去两个70°,就是顶角的度数.\n【详解】5+5\u003c11，所以5厘米,5厘米,11厘米不能围成三角形。5+11\u003e11,11+11\u003e5，所以5厘米,11厘米,11厘米能围成三角形。那么第三条边的长度是11厘米。180°-70°-70°=40°,它的顶角是40°。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和",
+                                 "三角形三边关系"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "48",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三角形三条边的长度都是整厘米数，其中两条边的长度分别是3厘米和8厘米，第三条边最长是( )厘米，最短是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "10;6",
+        "solution":  "【分析】根据三角形的三边关系：任意两边之和大于第三边，任意两边之差小于第三边，即可解答.\n【详解】根据分析可得:8+3=11(厘米),8-3=5(厘米),5厘米\u003c第三边\u003c11厘米，因为三条边的长度都是整厘米数，所以第三边最短是6厘米，最长是10厘米。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "49",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "根据中国地震局及历史统计数据，中国大陆平均每年发生20余次5级以上地震。发生地震时，我们应尽量寻找\"救命三角\"也就是选择室内结实，能掩护身体的物体旁，易于形成三角空间的地方，这主要是利用了三角形的( )。",
+        "options":  [
+
+                    ],
+        "answer":  "稳定性",
+        "solution":  "【分析】三角形具有稳定性，三角形稳定，因为它的三条边首尾相接，形成了稳定结构。三角形的三个角固定了三条边的位置，使得整个结构在受力时，力的分布更加均匀，不易发生变形。其次，三角形的三条边相互支撑，一旦其中一条边受到拉力或压力，其他两条边会产生反作用力，相互抵消，从而保持整个结构的稳定.\n【详解】由分析知：发生地震时，我们应尽量寻找\"救命三角\"也就是选择室内结实，能掩护身体的物体旁，易于形成三角空间的地方，这主要是利用了三角形的稳定性。",
+        "knowledge_points":  [
+                                 "三角形的稳定性及应用"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "50",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在工地上经常能看见三角形支架，这是运用三角形的( )性以增强施工的安全性。",
+        "options":  [
+
+                    ],
+        "answer":  "稳定",
+        "solution":  "【分析】三角形稳定性是指三角形具有稳定性，有着稳固，坚固，耐压的特点，如埃及金字塔，钢轨，三角形框架，起重机，三角形吊臂，屋顶，三角形钢架，钢架桥和埃菲尔铁塔都以三角形形状建造.\n【详解】在工地上经常能看见三角形支架，这是运用三角形的稳定性以增强施工的安全性。",
+        "knowledge_points":  [
+                                 "三角形的稳定性及应用"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "51",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果一个等腰三角形的顶角是50°,那么它的一个底角是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "65",
+        "solution":  "【分析】等腰三角形的两个底角相等，三角形的内角和为180°。由题意得，等腰三角形的顶角是50°,直接用180°减去50°先算出两个底角的度数之和，再除以2即可算出一个底角的度数。据此解答.\n【详解】(180°-50°)÷2=65°。所以如果一个等腰三角形的顶角是50°,那么它的一个底角是65°。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "52",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形的两条边长是4cm和8cm,这个等腰三角形的周长是( )cm。",
+        "options":  [
+
+                    ],
+        "answer":  "20",
+        "solution":  "【分析】根据题意，明确等腰三角形的性质：两腰相等，两底角相等;三角形三边关系：三角形中，任意两边之和大于第三边，任意两边之差小于第三边;当腰长为4厘米时，则底边长为8厘米，根据等腰三角形两腰相等可知另外一条腰也为4厘米,4+4=8，此时不满足三角形三边关系;当腰长为8厘米时，则底边长为4厘米，根据等腰三角形两腰相等可知另外一条腰也为8厘米,4+8=12\u003e8,8+8=16\u003e4，满足三角形三边关系，则利用周长等于三边之和求解即可.\n【详解】根据分析可知:4+4=8，不满足三角形三边关系;4+8=12\u003e8,8+8=16\u003e4，满足三角形三边关系。8+8+4=20(厘米)。所以这个等腰三角形的周长是20cm。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形三边关系"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "53",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "有5根小棒，长度分别是2cm,3cm,5cm,7cm和8cm。从中选择3根首尾相接围成三角形。要想围成的三角形周长最短，应选择的小棒长度分别是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "3cm,5cm,7cm",
+        "solution":  "【分析】要解决这个问题，需要根据三角形的三边关系进行判断。三角形的任意两边之和大于第三边，任意两边之差小于第三边，先从中选取能围成三角形的小棒，再找出周长最短的三角形。据此解答.\n【详解】从中选取三根小棒，有如下情况:①选2cm,3cm,5cm,因为2+3=5，不满足任意两边之和大于第三边，不能组成三角形。②选2cm,3cm,7cm,因为2+3=5\u003c7，不满足任意两边之和大于第三边，不能组成三角形。③选2cm,3cm,8cm,因为2+3=5\u003c8，不满足任意两边之和大于第三边，不能组成三角形。④选2cm,5cm,7cm,因为2+5=7，不满足任意两边之和大于第三边，不能组成三角形。⑤选2cm,5cm,8cm,因为2+5=7\u003c8，不满足任意两边之和大于第三边，不能组成三角形。⑥选2cm,7cm,8cm,因为2+7=9\u003e8，满足任意两边之和大于第三边，能组成三角形。⑦选3cm,5cm,7cm,因为3+5=8\u003e7，满足任意两边之和大于第三边，能组成三角形。⑧选3cm,5cm,8cm,因为3+5=8，不满足任意两边之和大于第三边，不能组成三角形。⑨选3cm,7cm,8cm,因为3+7=10\u003e8，满足任意两边之和大于第三边，能组成三角形。⑩选5cm,7cm,8cm,因为5+7=12\u003e8，满足任意两边之和大于第三边，能组成三角形。所以,2cm,7cm,8cm;3cm,5cm,7cm;3cm,7cm,8cm;5cm,7cm,8cm四种情况可以围成三角形，周长分别为17cm,15cm,18cm,20cm,15cm\u003c17cm\u003c18cm\u003c20cm,周长最短为15cm,此时三根小棒长度为3cm,5cm,7cm。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "54",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一根铁丝可以围成一个边长为6厘米的正方形。若改围成一个等边三角形，则边长为( )cm;若改围成其中一条边长为6厘米的等腰三角形，则另两条边长为( )cm和( )cm。",
+        "options":  [
+
+                    ],
+        "answer":  "8;9;9",
+        "solution":  "【分析】已知正方形边长为6厘米，根据正方形周长=边长*4可得铁丝长度，因为等边三角形三条边长度相等，铁丝长度除以3得出等边三角形的边长。当6厘米为腰长时：等腰三角形两腰相等，所以另一条腰长也为6厘米，求出三角形底边的长度;当6厘米为底边时，算出三角形的腰长。最后根据三角形的三边关系：三角形任意两边的和大于第三边，任意两边的差小于第三边。判断三条线段能否围成三角形，把较短的两条线段相加的和与最长的线段相比较，若大于最长的线段，则能围成三角形，反之则不能。据此解答.\n【详解】6×4÷3=8(厘米);当腰长为6厘米时，底边长为:24-6×2=12(厘米),但是6+6=12，不满足三角形任意两边之和大于第三边的性质，所以这种情况不成立。当底边的长为6厘米时，腰长为:(24-6)÷2=9(厘米),此时9+9\u003e6,9+6\u003e9，满足三角形三边关系。所以若改围成一个等边三角形，则边长为8厘米;若改围成其中一条边长为6厘米的等腰三角形，则另两条边长为9厘米和9厘米。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "正方形的周长",
+                                 "三角形的周长",
+                                 "三角形三边关系"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "55",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在探究六边形的内角和时，可以把六边形分成( )个三角形，得出六边形的内角和是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "4;720°",
+        "solution":  "【分析】从六边形的一个顶点向相对的顶点连线，可以将六边形分成4个三角形，如图![](images/2026-03-10%20105709.jpg)每个三角形的内角和是180°,所以计算六边形的内角和就是计算4个三角形的内角和。据此解答.\n【详解】6-2=4(个),4×180°=720°。所以在探究六边形的内角和时，可以把六边形分成4个三角形，得出六边形的内角和是720°。",
+        "knowledge_points":  [
+                                 "多边形的内角和"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "56",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在一个直角三角形中，其中一个锐角的度数是另一个锐角度数的2倍。在这个直角三角形中，较大的那个锐角的度数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "60°",
+        "solution":  "【分析】有一个角是直角的三角形是直角三角形，三角形的内角和是180°,在直角三角形中，两个锐角的度数和是90°,\"其中一个锐角的度数是另一个锐角的2倍\"，根据和倍问题，其中一个锐角的度数是另一个锐角的(2+1)倍，用90°除以(2+1)求出较小的那个锐角的度数，再用较小的那个锐角的度数乘2，即可求出较大的那个锐角的度数.\n【详解】90°/(2+1)=30°,30°*2=60°。所以较大的那个锐角的度数是60°。",
+        "knowledge_points":  [
+                                 "和差倍问题",
+                                 "三角形的分类",
+                                 "三角形的内角和",
+                                 "直角",
+                                 "钝角",
+                                 "锐角的认识及特征"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "57",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把一根长24厘米的铁丝围成等边三角形框架，框架的一条底边长( )厘米;如果把它围成一条腰长7厘米的等腰三角形，它的底边长( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "8;10",
+        "solution":  "【分析】等边三角形的三条边都相等，用铁丝的长度除以3求出一条边的长度即为框架的一条底边长度;等腰三角形两条腰相等，用一条腰的长度乘2求出两条腰的长度，再用铁丝的长度减去两条腰的长度即可求出底边的长度.\n【详解】24÷3=8(厘米);24-7×2=10(厘米)。所以等边三角形框架的一条底边长8厘米;围成等腰三角形，它的底边长10厘米。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的周长"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "58",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "有两根小棒分别长4厘米和8厘米，如果再增加一根小棒围成一个三角形(取整厘米数),增加的这根小棒可以有( )种不同的长度，其中最长可以是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "7;11",
+        "solution":  "【分析】根据三角形的三边关系，两边之和大于第三边，两边之差小于第三边，可以确定第三边的取值范围，又因为第三边取整厘米数，可以列举出能取的值，据此作答.\n【详解】8-4=4(厘米),8+4=12(厘米),因此第三边大于4厘米，小于12厘米，取整厘米数可以是5厘米,6厘米,7厘米,8厘米,9厘米,10厘米和11厘米。因此增加的这根小棒可以有7种不同的长度，其中最长可以是11厘米。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "59",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个直角三角形，其中一个锐角比另一个锐角小20°,这两个锐角分别是( )°和( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "35;55",
+        "solution":  "【分析】直角三角形中有一个角是直角，三角形的内角和等于180°,先用180°减去90°,求出另外两个锐角的和，再减去20°即可得到最小锐角度数的2倍，然后除以2即可求出这个最小锐角的度数，再用这个锐角度数加上20°即可求出另一个锐角的度数.\n【详解】180°-90°=90°,(90°-20°)÷2=35°,35°+20°=55°。则这两个锐角的度数分别是35°,55°。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "三角形的内角和",
+                                 "直角",
+                                 "钝角",
+                                 "锐角的认识及特征"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "60",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把一根12cm长的吸管剪成3段(每段都是整厘米数),围成一个三角形，这个三角形三条边的长度可能分别是( )cm,( )cm和( )cm,还可能分别是( )cm,( )cm和( )cm。",
+        "options":  [
+
+                    ],
+        "answer":  "3;4;5;4;4;4",
+        "solution":  "【分析】根据三角形三边关系，任意两边之和大于第三边，且三边总长为12cm。最长边必须小于12÷2=6cm,即不超过5cm。列举满足条件的整厘米数组合，验证每组是否满足三角形的三边关系.\n【详解】总长12cm,最长边必须小于12÷2=6cm,因此最长边小于6cm。组合一(3cm,4cm,5cm):3+4+5=12(cm),且3+4\u003e5，满足三角形三边关系;组合二(4cm,4cm,4cm):4+4+4=12(cm),且4+4\u003e4，满足三角形三边关系;组合三(2cm,5cm,5cm):2+5+5=12(cm),且2+5\u003e5，满足三角形三边关系。有三种可能，题目要求两组可能的组合，因此可任选两组。即：这个三角形三条边的长度可能分别是3cm,4cm和5cm,还可能分别是4cm,4cm和4cm。(答案不唯一)",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "61",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明准备用图钉固定硬纸条做一个三角形。他最少应该准备( )根硬纸条，准备( )个图钉。如果有2根硬纸条分别长4厘米和6厘米，那么另一根硬纸条最长为( )厘米。(硬纸条的长度是整厘米数)",
+        "options":  [
+
+                    ],
+        "answer":  "3;3;9",
+        "solution":  "【分析】三角形有三条边，所以最少应该准备3根硬纸条;每两根纸条要用一个图钉固定住，所以至少准备3个图钉;根据三角形两边之和大于第三边，先将已知2条边之和算出=4+6=10(厘米),另一条应该小于10厘米，最大为9厘米.\n【详解】根据分析，他最少应该准备3根硬纸条;准备3个图钉;如果有2根硬纸条分别长4厘米和6厘米，那么另一根硬纸条最长为9厘米。",
+        "knowledge_points":  [
+                                 "三角形的概念及表示方法",
+                                 "三角形三边关系"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "62",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小东有两根长都是4厘米的小棒，他想再用一根小棒围三角形，这根小棒的长度一定要小于( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "8",
+        "solution":  "【分析】根据三角形三边关系，任意两边之和大于第三边。已知两根小棒均为4厘米,4+4=8(厘米),因此第三根小棒的长度必须小于8厘米.\n【详解】4+4=8(厘米),所以这根小棒的长度一定要小于8厘米。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "63",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "下图中,∠1=( )°.\n![](images/bda728cc8f235a12f5ab1cbabd3411a0f07a902cf9f1a9db41d7be12e3eee297.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "92",
+        "solution":  "【分析】观察图中可知,∠2和130°的角组成一个平角，平角=180°,用180°减去130°,即为∠2的度数，进而结合四边形的内角和为360°,即可求得∠1的度数.\n【详解】∠2=180°-130°=50°,360°-123°-95°-50°=92°。因此上图中,∠1=92°。",
+        "knowledge_points":  [
+                                 "平角",
+                                 "周角的认识及特征",
+                                 "多边形的内角和"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "64",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形的顶角是50°,那么其中的一个底角是( )°,它还是一个( )三角形;如果一个底角是40°,顶角是( )°,它就是一个( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "65;锐角;100;钝角",
+        "solution":  "【分析】(1)三角形内角和为180°,等腰三角形两底角相等，一个等腰三角形的顶角是50°,底角=(180°-50°)÷2，三个角都是锐角的三角形是锐角三角形;如果一个底角是40°,顶角=180°-40°*2，一个角是钝角的三角形是钝角三角形，据此解答.\n【详解】底角:(180°-50°)÷2=65°;顶角:180°-40°*2=100°。所以一个等腰三角形的顶角是50°,那么其中的一个底角是65°,它还是一个锐角三角形;如果一个底角是40°,顶角是100°,它就是一个钝角三角形。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "65",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形，其中两条边的长度分别是4cm和10cm,第三条边的长度是( )cm,这个三角形的周长是( )cm。",
+        "options":  [
+
+                    ],
+        "answer":  "10;24",
+        "solution":  "【分析】三角形任意两边之和大于第三边，据此确定第三条边的长度，等腰三角形的周长=腰长*2+底长，据此分析.\n【详解】4+4\u003c10，不符合三角形三边关系;10+4\u003e10，符合三角形三边关系。所以第三条边的长度是10cm。10×2+4=24(cm),这个三角形的周长是24cm。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的周长",
+                                 "三角形三边关系"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "66",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形，有两条边长分别是3厘米和6厘米，它的周长一定是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】一个等腰三角形，有两条边长分别是3厘米和6厘米，如果这个等腰三角形的腰是3厘米，则3+3=6(厘米),根据三角形的两边之和大于第三边，所以这个等腰三角形的腰长不可能是3厘米，而是6厘米;把这个等腰三角形三边相加，所得和即为它的周长，据此解答.\n【详解】由分析可知，这个等腰三角形的三边分别是3厘米,6厘米,6厘米。3+6+6=15(厘米)。所以它的周长一定是15厘米。",
+        "knowledge_points":  [
+                                 "三角形三边关系",
+                                 "三角形的周长",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "67",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个三角形的一个内角是120°,这是一个( )三角形，一个等腰三角形的一个底角50°,它的顶角是( )°。我知道了，一个三角形最多有( )个钝角，但最多是有( )个锐角。",
+        "options":  [
+
+                    ],
+        "answer":  "钝角;80;1;3",
+        "solution":  "【分析】三角形的内角和是180°,有一个角是钝角的三角形是钝角三角形，有一个角是直角的三角形是直角三角形，三个角都是锐角的三角形是锐角三角形。120°是钝角，所以一个三角形的一个内角是120°,这是一个钝角三角形;等腰三角形底角相等，一个等腰三角形的一个底角50°,它的顶角就用180°-50°-50°;一个三角形最多有1个钝角，但最多是有3个锐角.\n【详解】180°-50°-50°=80°。所以一个三角形的一个内角是120°,这是一个钝角三角形，一个等腰三角形的一个底角50°,它的顶角是80°。我知道了，一个三角形最多有1个钝角，但最多是有3个锐角。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和",
+                                 "直角",
+                                 "钝角",
+                                 "锐角的认识及特征"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "68",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果一个三角形的两条边分别是3厘米和8厘米，第三条边不是最长的，也是整厘米数，最短是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】根据三角形三边关系：三角形的两边之和大于第三边，三角形的两边的差小于第三边即可解答.\n【详解】3+8=11(厘米),8-3=5(厘米),由于第三条边不是最长的，且8厘米是最长边，因此第三边必须小于8厘米。5厘米\u003c第三条边的长\u003c8厘米，第三条边最短是6厘米。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "69",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个三角形中，已知∠1=45°,∠2=60°,那么∠3=( )°,如果按角分类，这是一个( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "75;锐角",
+        "solution":  "【分析】三角形内角和是180°,用180°减去∠1的度数，再减去∠2的度数，即可算出∠3的度数。三个角都是锐角的三角形是锐角三角形，有一个角是直角的三角形是直角三角形，有一个角是钝角的三角形是钝角三角形。据此解答.\n【详解】180°-45°-60°=75°。在一个三角形中，如果∠1=45°,∠2=60°,那么∠3=75°,按角分类这是锐角三角形。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "三角形的内角和"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "70",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "三角形的两条边分别是3厘米和4厘米，另一条边最长是( )厘米，最短是( )厘米。(取整厘米数)",
+        "options":  [
+
+                    ],
+        "answer":  "6;2",
+        "solution":  "【分析】根据三角形的三边关系，三角形的两边之和大于第三边，确定第三边的最大长度;任意两边之差小于第三边，确定第三边的最小长度;据此解答.\n【详解】4-3=1(厘米),4+3=7(厘米),1厘米\u003c第三边的长度\u003c7厘米。所以三角形的两条边分别是3厘米和4厘米，另一条边最长是6厘米，最短是2厘米。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "71",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果有一个三角形的两条边分别长4厘米和7厘米，另一条边可能是( )厘米(取整厘米数)。",
+        "options":  [
+
+                    ],
+        "answer":  "4,5,6,7,8,9,10",
+        "solution":  "【分析】根据三角形的三边关系，任意两边之和大于第三边，任意两边之差小于第三边。据此解答即可.\n【详解】7厘米-4厘米\u003c第三边\u003c7厘米+4厘米，所以3厘米\u003c第三边\u003c11厘米，即第三边在3厘米至11厘米之间，不包括3厘米和11厘米，即另一条边可能是4,5,6,7,8,9,10厘米。所以另一条边可能是4,5,6,7,8,9,10厘米。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "72",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "当地震来临时，倒塌物与坚固的大型物体周边会形成三角空间(被称为\"生命三角区\")。这主要是利用了三角形( )的特性，任意三角形的内角和是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "稳定;180°或180度",
+        "solution":  "【分析】三角形具有稳定性，三角形稳定，因为它的三条边首尾相接，形成了稳定结构。三角形的三个角固定了三条边的位置，使得整个结构在受力时，力的分布更加均匀，不易发生变形。其次，三角形的三条边相互支撑，一旦其中一条边受到拉力或压力，其他两条边会产生反作用力，相互抵消，从而保持整个结构的稳定。这种力的平衡状态使得三角形结构在受力时能够均匀分散压力，提高了结构的抗压能力和稳定性。三角形的内角和是180°,据此解答即可.\n【详解】当地震来临时，倒塌物与坚固大型物体周边会形成三角空间(被称为\"生命三角区\")。这主要是利用了三角形稳定的特性，任意三角形的内角和是180°。",
+        "knowledge_points":  [
+                                 "三角形的稳定性及应用",
+                                 "三角形的内角和"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "73",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "风筝起源于中国，其制作技艺被列为国家非物质文化遗产。美美想做一个等腰三角形形状的风筝，它的一个底角比直角小40°,请你帮她算一算这个风筝的顶角是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "80°或80度",
+        "solution":  "【分析】等腰三角形的两个底角相等，三角形内角和是180°,直角是90°。根据一个底角比直角小40°,用90°减去40°即可求出底角的度数;用180°减去两个底角的和，即可求出其顶角的度数.\n【详解】底角的度数:90°-40°=50°;顶角的度数:180°-(50°+50°)=80°。所以这个风筝的顶角是80°。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "74",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "三角形ABC中,∠A=65°,∠B=45°,∠C=( )°,这是一个( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "70;锐角",
+        "solution":  "【分析】三角形按角来分，分为锐角三角形，直角三角形和钝角三角形。三个角都是锐角的三角形叫作锐角三角形，有一个角是直角的三角形叫作直角三角形，有一个角是钝角的三角形叫作钝角三角形。三角形的内角和为180°。由题意得，三角形ABC中,∠A=65°,∠B=45°,直接用180°减去∠A和∠B的度数即可算出∠C的度数。然后根据三个角的大小来判断三角形的类型即可.\n【详解】180°-65°-45°=70°,三个角都是锐角，所以这个三角形是锐角三角形。所以三角形ABC中,∠C=70°,这是一个锐角三角形。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "三角形的内角和"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "75",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形其中的两条边的长度分别是4厘米和9厘米，那么这个三角形的周长是( )厘米，合( )米。",
+        "options":  [
+
+                    ],
+        "answer":  "22;0.22",
+        "solution":  "【分析】如果4厘米为三角形的腰，那么4+4=8(厘米),8厘米\u003c9厘米，不符合三角形的三边关系，因此只有9厘米的边为等腰三角形的腰，三角形的周长等于三边之和;计算出周长之后再根据1米=100厘米，把厘米为单位的数转换成以米为单位的数.\n【详解】由分析可得:9+9+4=22(厘米);22厘米=22÷100=0.22米。所以这个三角形的周长是22厘米，合0.22米。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的周长",
+                                 "三角形三边关系",
+                                 "运用小数点移动解决小数的单位换算问题"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "76",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "红领巾作为少先队员的标志，常为红色的三角形;它的两个底角都是30°,顶角是120°,按角分这是一个( )三角形;腰长60厘米，底边长100厘米，周长为( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "钝角;220",
+        "solution":  "【分析】三角形按角来分，分为锐角三角形，直角三角形和钝角三角形。三个角都是锐角的三角形叫作锐角三角形，有一个角是直角的三角形叫作直角三角形，有一个角是钝角的三角形叫作钝角三角形;有两条边相等的三角形叫作等腰三角形，等腰三角形的两个底角的度数相等。把三角形的三条边相加，即可求出三角形的周长。据此解答.\n【详解】120°的角是一个钝角，所以这个三角形是钝角三角形。30°=30°,所以红领巾是一个等腰三角形，它的两条腰的长度相等，都是60厘米。60+60+100=220(厘米)。所以红领巾按角分这是一个钝角三角形，周长为220厘米。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的周长"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "77",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "五边形的内角和是( )°,一个多边形内角和是720°,这是( )边形。一个等腰三角形，底角是52°,它的顶角是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "540;六;76",
+        "solution":  "【分析】五边形可以分成3个三角形，三角形的内角和为180°,用3乘180°可算出五边形内角和。一个多边形内角和是720°,因为180°*4=720°,说明这个多边形可以分成4个三角形，所以是六边形。两腰相等的三角形为等腰三角形，等腰三角形的底角相等，根据三角形内角和为180°,用180°-52°-52°可算出顶角.\n【详解】180°*3=540°,所以五边形的内角和是540°。720°/180°=4,4+2=6，所以一个多边形内角和是720°,这是六边形。180°-52°-52°=76°,所以它的顶角是76°。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和",
+                                 "多边形的内角和"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "78",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "三角形ABC中,∠A=65°,∠B=45°,∠C=( )°,这是一个( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "70;锐角",
+        "solution":  "【分析】根据题意，明确三角形内角和为180°,用180°减去已知的两个角的度数即可求出∠C的度数。按角分类：三角形按角分为锐角三角形(三个角均小于90°),直角三角形(一个角为90°)和钝角三角形(一个角大于90°)。再根据三个角的度数判断三角形的类型：若三个角都是锐角，则为锐角三角形.\n【详解】180°-65°-45°=70°,三个角都是锐角，所以三角形ABC中,∠C=70°,这是一个锐角三角形。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "三角形的内角和"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "79",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上适当的数.\n①1.67km=( )m ②钝角三角形里有( )个锐角 ③2500g=( )kg ④3.5的1÷( )是0.035",
+        "options":  [
+
+                    ],
+        "answer":  "1670;2或两;2.5;100",
+        "solution":  "【分析】①1km=1000m,高级单位化成低级单位乘进率;②钝角三角形，有一个角是钝角的三角形，另外两个角是锐角;③1kg=1000g,低级单位化成高级单位除以进率;④根据小数点的移动规律，小数点向左移动两位，就缩小为原来的1÷100.\n【详解】①1.67×1000=1670(m),所以1.67km=1670m;②钝角三角形里有2个锐角;③2500÷1000=2.5(kg),所以2500g=2.5kg;④3.5的1÷100是0.035。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "小数点向左移动引起小数大小变化的规律",
+                                 "千米和米之间的进率与换算",
+                                 "三角形的分类"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "80",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "等腰三角形的周长是5米，一条腰长20分米，它的底边长是( )分米。",
+        "options":  [
+
+                    ],
+        "answer":  "10",
+        "solution":  "【分析】根据题意,1米=10分米，把5米转换成50分米。三角形的周长就是它的三条边的长度之和。等腰三角形的两腰相等，用周长减去两条腰的长度就是它的底边长.\n【详解】5米=50分米,50-20-20=10(分米)。所以，它的底边长是10分米。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的周长",
+                                 "毫米，厘米，分米，米之间的进率与换算"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "81",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三角形三条边的长度均为整厘米数，其中两条边的长度分别为5厘米和9厘米，则第三条边的长度最短为( )厘米，最长为( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "5;13",
+        "solution":  "【分析】三角形的三边关系为三角形的两边之和大于第三边，三角形的两边之差一定小于第三边;据此解答即可.\n【详解】5+9=14(厘米),9-5=4(厘米),14-1=13(厘米),4+1=5(厘米)。所以第三条边的长度最短为5厘米，最长为13厘米。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "82",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明在探究六边形的内角和时，他从六边形的一个顶点依次连接与这个顶点不相连的其它顶点，这样就把六边形分成( )个三角形，因为三角形的内角和是180°,所以小明推算出六边形的内角和是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "4;720",
+        "solution":  "【分析】根据题意，从一个顶点依次连接与这个顶点不相连的其它顶点，可以将六边形分成几个三角形，三角形内角和是180°,用180°乘分成的三角形个数，即可求出六边形的内角和是多少度.\n【详解】如图:![](images/0ce8ad6ba5627d837fcb3a035bcd3bd3717753890f0e3c9db71f7a1c3c858a65.jpg) 180°*4=720°。所以小明把六边形分成4个三角形，推算出六边形的内角和是720°。",
+        "knowledge_points":  [
+                                 "多边形的内角和",
+                                 "三角形的内角和"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "83",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "平行四边形的对角( ),等腰梯形的两个底角( )。",
+        "options":  [
+
+                    ],
+        "answer":  "相等;相等",
+        "solution":  "【分析】平行四边形的特点是两组对边分别平行，且两组对边分别相等，其对角是相等的。等腰梯形的两腰相等，所以其两个底角也是相等的.\n【详解】平行四边形的对角相等，等腰梯形的两个底角相等。",
+        "knowledge_points":  [
+                                 "直角梯形和等腰梯形的概念及特点",
+                                 "平行四边形的概念及特点"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "84",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "从两根3厘米，两根6厘米和两根8厘米的小棒中选出三根围成一个等腰三角形，围成等腰三角形的周长最大是( )厘米，最小是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "22;15",
+        "solution":  "【分析】构成三角形需要两边之和大于第三边，两边之差小于第三边。要想周长最大，三边选最长的8厘米,8厘米,6厘米,8+6=14(厘米),14\u003e8，符合构成三角形的条件。要想周长最小，三边选最短的3厘米,3厘米,6厘米,3+3=6(厘米),6=6，不符合构成三角形的条件;选择6厘米,6厘米,3厘米,6+3=9(厘米),9\u003e6，符合构成三角形的条件，计算出两个三角形的周长即可解答.\n【详解】根据分析可得,8+8+6=22(厘米),6+6+3=15(厘米)。所以围成等腰三角形的周长最大是22厘米，最小是15厘米。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的周长"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "85",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个多边形的边数增加1条，则它的内角和增加( )。",
+        "options":  [
+
+                    ],
+        "answer":  "180°",
+        "solution":  "【分析】多边形的内角和与它的边数有关。边数越多，内角和越大。每增加一条边，就相当于在多边形中增加了一个三角形(因为多边形可以分割成三角形),而每个三角形的内角和是180°。所以，当边数增加一条时，内角和就增加180°.\n【详解】设原多边形的边数为n,则其内角和为(n-2)×180°,边数增加一条后，新边数为n+1，新内角和为(n+1-2)×180°=(n-1)×180°,内角和的增加量为:(n-1)×180°-(n-2)×180°=180°。因此，一个多边形的边数增加1条，则它的内角和增加180°。",
+        "knowledge_points":  [
+                                 "多边形的内角和"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "86",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "王老师准备了5cm和8cm的木棒各一根，现在他想拼成一个三角形，第三根木棒最长是( )cm,最短是( )cm。(填整厘米数)",
+        "options":  [
+
+                    ],
+        "answer":  "12;4",
+        "solution":  "【分析】根据三角形的三边关系，两边之和大于第三边，两边之差小于第三边，第三根木棒最长不能超出已知两根木棒之和，最短不能小于已知两根木棒之差.\n【详解】5+8=13(cm),8-5=3(cm),3cm\u003c第三根木棒\u003c13cm,且取整厘米数。所以第三根木棒最长是12cm,最短是4cm。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "87",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形的底角是30°,那么它的顶角是( )°,它还是一个( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "120;钝角",
+        "solution":  "【分析】已知等腰三角形的底角是30°,根据等腰三角形两底角相等的性质，另一个底角也是30°。利用三角形内角和为180°,可求出顶角的度数。再根据三个角都是锐角的三角形是锐角三角形，有一个角是直角的三角形是直角三角形，有一个角是钝角的三角形是钝角三角形;据此解答.\n【详解】计算顶角:180°-30°*2=120°;判断三角形类型：顶角为120°,是一个钝角，因此这个三角形是钝角三角形。所以一个等腰三角形的底角是30°,那么它的顶角是120°,它还是一个钝角三角形。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "88",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三角形的两条边分别是5cm和7cm,第三条边最短是( )cm,最长是( )cm。(取整厘米数)",
+        "options":  [
+
+                    ],
+        "answer":  "3;11",
+        "solution":  "【分析】根据题意，三角形三条边的关系是：任意两边的长度之和大于第三边，任意两边的长度之差小于第三边，依此计算并填空.\n【详解】5+7=12(cm),7-5=2(cm),2cm\u003c第三条边\u003c12cm,2+1=3(cm),12-1=11(cm)。则第三条边(取整厘米数)最短是3cm,最长是11cm。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "89",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三角形中两个内角是56°和74°,另一个内角是( )°,这是一个( )三角形，一个等腰三角形中，一个底角是75°,它的顶角是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "50;锐角;30",
+        "solution":  "【分析】(1)根据三角形的内角和是180°,用180°连续减去两个内角的和，求出剩下的角的度数后，再进行判断;(2)三角形根据角的分类可以分为锐角三角形，直角三角形和钝角三角形。锐角三角形是三个内角都是锐角的三角形，直角三角形是其中一个角为直角的三角形，钝角三角形是其中一个角是钝角的三角形。(3)结合等腰三角形的特征可知，等腰三角形的两个底角相等，根据三角形的内角和是180°,用180°连续减去两个底角的度数就是顶角的度数.\n【详解】(1)180°-56°-74°=50°,三个内角都是锐角，故是锐角三角形。(2)180°-75°-75°=30°。所以另一个内角是50°,这是一个锐角三角形，顶角是30°。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "90",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三角形的两条边分别是9m和5m,它的第三条边最长是( )m。(取整米数)",
+        "options":  [
+
+                    ],
+        "answer":  "13",
+        "solution":  "【分析】三角形两边之和大于第三条边，先求出这个三角形已知的两边之和，第三条边必须是小于此数，取最大整数即可.\n【详解】9+5=14(m),9-5=4(m),所以4m\u003c第三边长度\u003c14m,14-1=13(m)。所以第三条边最长是13m。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "91",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形的顶角是120°,那么这个三角形的底角是( )°,按角分它是一个( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "30;钝角",
+        "solution":  "【分析】三角形按角来分，分为锐角三角形，直角三角形和钝角三角形。三个角都是锐角的三角形叫作锐角三角形，有一个角是直角的三角形叫作直角三角形，有一个角是钝角的三角形叫作钝角三角形;等腰三角形的两个底角相等，三角形的内角和为180°。由题意得，等腰三角形的顶角的度数是120°,那么直接用180°减去这个顶角的度数即可算出两个底角的度数之和，再除以2即可算出一个底角的度数。然后根据三个角的大小来判断三角形的类型即可.\n【详解】(180°-120°)÷2=30°,120°的角是一个钝角，所以这个三角形是钝角三角形。所以底角是30°,按角分它是一个钝角三角形。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "92",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在数学实践活动课上，小英将一根18厘米长的彩带围成等边三角形，等边三角形的边长是( )厘米。如果把这根彩带围成一个其他三角形(边长都是整厘米数),则新围成的三角形的边最长是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "6;8",
+        "solution":  "【分析】三条边都相等的三角形叫作等边三角形。由题意得，小英将一根18厘米长的彩带围成等边三角形，那么等边三角形的周长就为18厘米，直接用18厘米除以3即可算出等边三角形的边长;如果把这根彩带围成一个其他三角形(边长都是整厘米数),由三角形三边的关系(任意两边之和大于第三边)可知，三角形最长的边的长度不能超过它周长的一半，用18除以2可以算出三角形最长的边的长度不能超过多少厘米。据此解答.\n【详解】18÷3=6(厘米),即等边三角形的边长是6厘米。18÷2=9(厘米),即最长的边不能超过9厘米，所以新围成的三角形的边最长是8厘米。所以等边三角形的边长是6厘米，新围成的三角形的边最长是8厘米。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的周长",
+                                 "三角形三边关系"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "93",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形的顶角是80°,它的一个底角是( )°;如果一个等腰三角形其中两条边分别是5cm和10cm,第三条边一定是( )cm。",
+        "options":  [
+
+                    ],
+        "answer":  "50;10",
+        "solution":  "【分析】根据题意，明确三角形的内角和是180°,等腰三角形的两条腰相等，两个底角相等，已知一个等腰三角形的顶角是80°,用180°减去80°,再除以2，就是它的一个底角的度数;根据三角形的三边关系：任意两边之和必须大于第三边，如果一个等腰三角形其中两条边分别是5厘米和10厘米，有两种情况，当5厘米是腰时,5+5=10，不能构成三角形，所以不成立;当10厘米是腰时,10+10=20\u003e5，可以构成三角形，所以判断第三条边一定是10厘米。以此答题即可.\n【详解】根据分析可知:(180°-80°)÷2=50°;5+5=10，不能构成三角形;10+10=20\u003e5，可以构成三角形。所以一个等腰三角形的顶角是80°,它的一个底角是50°;如果一个等腰三角形其中两条边分别是5cm和10cm,第三条边一定是10cm。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和",
+                                 "三角形三边关系"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "94",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形其中一个角是80°,这个三角形按角分类是( )三角形，底角是( )°或( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "锐角;50;80",
+        "solution":  "【分析】等腰三角形两个底角相等，根据三角形的内角和等于180°。如果80°是底角，那么另一个底角也是80°,顶角的度数可以用180°减去两个底角的度数。如果80°是顶角，用180°减去80°的结果再除以2，得到底角的度数。再根据三个角都是锐角的三角形是锐角三角形，有一个角是直角的三角形是直角三角形，有一个角是钝角的三角形是钝角三角形，判断它的类型.\n【详解】如果80°是底角，那么另一个底角也是80°,顶角:180°-80°-80°=20°,三个角是80°,80°,20°,是锐角三角形。如果80°是顶角，底角:(180°-80°)÷2=50°,三个角是80°,50°,50°,也是锐角三角形。所以这个三角形按角分类是锐角三角形，底角是50°或80°。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "95",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "用同样长的小棒围图形,3根可以围成( )个三角形,4根可以围成( )个形状不同的四边形，拉动围成的图形，发现三角形具有( )性，生活中的( )就是利用三角形的特性。",
+        "options":  [
+
+                    ],
+        "answer":  "1;无数;稳定;自行车架",
+        "solution":  "【分析】由三条线段首位相接围成的图形是三角形。三角形具有稳定性，不管怎么拉都不会变形。所以三根小棒只能围成一种三角形。生活中一些物体就是利用三角形的稳定性，例如相机的三脚架，自行车架等。同一平面内的四条线段首位相接围成的图形是四边形。四边形容易变形，像伸缩门就利用四边形的容易变形的特点。所以4根小棒可以围成不同形状的四边形.\n【详解】根据分析，用同样长的小棒围图形,3根可以围成1个三角形,4根可以围成无数个形状不同的四边形，拉动围成的图形，发现三角形具有稳定性，生活中的自行车架就是利用三角形的特性。(答案不唯一)",
+        "knowledge_points":  [
+                                 "平行四边形的不稳定性及应用",
+                                 "三角形的概念及表示方法",
+                                 "三角形的稳定性及应用"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "96",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "等腰三角形的一个底角是30°,它的另外两个角分别是( )°和( )°,这个三角形又是( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "30;120;钝角",
+        "solution":  "【分析】等腰三角形的底角相等，说明等腰三角形中有两个角都为30°,再根据三角形的内角和是180°,用180°减去两个30°角，从而可以求出顶角的度数，再根据三个角的度数，三个角都是锐角的三角形叫做锐角三角形，有一个角为直角的三角形叫做直角三角形，有一个角是钝角的三角形叫做钝角三角形，据此判定这个三角形的类别.\n【详解】180°-30°-30°=120°,所以另外两个角分别是30°和120°。由于有一个角大于90°为钝角，这个三角形为钝角三角形。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "97",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "自行车的车架通常做成三角形的形状，是因为( );学校大门口的伸缩门做成平行四边形的形状，是因为( )。",
+        "options":  [
+
+                    ],
+        "answer":  "三角形具有稳定性;平行四边形具有不稳定性",
+        "solution":  "【分析】三角形具有稳定性，有着稳固，坚定，耐压的特点，自行车的车架通常做成三角形的形状就是运用了三角形的这种特性。而平行四边形具有不稳定性，容易变形，学校大门口的伸缩门工作时，伸缩门中的平行四边形会变形，所以伸缩门的设计就是运用了它的这种特性.\n【详解】自行车的车架通常做成三角形的形状，是因为三角形具有稳定性;学校大门口的伸缩门做成平行四边形的形状，是因为平行四边形具有不稳定性。",
+        "knowledge_points":  [
+                                 "平行四边形的不稳定性及应用",
+                                 "三角形的稳定性及应用"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "98",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一根铁丝恰好可以围成一个底边长为5厘米，腰长为8厘米的等腰三角形，那么用这根铁丝围成一个最大的等边三角形，这个等边三角形的边长是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "7",
+        "solution":  "【分析】根据等腰三角形的周长=底边长+腰长*2，求出铁丝长度;三边相等的三角形是等边三角形，用铁丝长度除以3求出等边三角形的边长，据此解答.\n【详解】铁丝长度:5+8×2=21(厘米),等边三角形的边长:21÷3=7(厘米)。所以这个等边三角形的边长是7厘米。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的周长"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "99",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "三角形任意两边长度的和一定大于( )边;生活中的很多三角形框架，都是利用了三角形不易变形的( )性。",
+        "options":  [
+
+                    ],
+        "answer":  "第三;稳定",
+        "solution":  "【分析】三角形三边的关系：任意两边之和大于第三边，任意两边之差小于第三边;三角形具有稳定性，它有着稳固，坚定，耐压的特点，如：埃及金字塔，钢轨，三角形框架，起重机，三角形吊臂，屋顶，三角形钢架，钢架桥和埃菲尔铁塔都以三角形形状建造.\n【详解】三角形任意两边长度的和一定大于第三边;生活中的很多三角形框架，都是利用了三角形不易变形的稳定性。",
+        "knowledge_points":  [
+                                 "三角形的稳定性及应用",
+                                 "三角形三边关系"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "100",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "等腰三角形的一个底角是40°,顶角是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "100",
+        "solution":  "【分析】根据一个三角形的内角和为180°,然后又根据等腰三角形的两个底角相等。求等腰三角形的顶角，首先计算出两个底角的度数，就用2乘40°;然后再用180°减去两个底角的度数即可.\n【详解】180°-2×40°=100°,所以顶角是100°。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和"
+                             ],
+        "source":  ""
+    },
+    {
+        "id":  "101",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "有一种三角形，任意两个内角的和大于第三个角。按角来分，这种三角形是( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "锐角",
+        "solution":  "【分析】根据三角形的内角和为180°可知，若三角形中两个内角和等于第三个内角，则第三个内角应为180°/2=90°。而这个三角形中的任意两个内角之和大于第三个内角，第三个内角应小于90°,即三个内角均小于90°,这个三角形是锐角三角形。据此解答即可.\n【详解】由分析知，一个三角形任意两个内角的和大于第三个内角，三个内角都是锐角，它一定是锐角三角形。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·浙江嘉兴·期末"
+    },
+    {
+        "id":  "102",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形三条边都是整厘米数，已知其中两条边的长分别是5厘米,11厘米，那么第三条边的长度是( )厘米，周长是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "11;27",
+        "solution":  "【分析】根据题意，等腰三角形的两腰相等。由此可知第三条边长可能是5厘米，也可能是11厘米。再根据三角形任意两边之和大于第三边，确定第三边的长度。最后把三条边的长度相加，就是三角形的周长.\n【详解】当第三条边长是5厘米时:5+5\u003c11，所以，不能围成三角形。当第三条边长11厘米时:5+11\u003e11,11+11\u003e5，所以，等腰三角形的第三条边的长度是11厘米。11+11+5=27(厘米),那么，周长是27厘米。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的周长",
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·浙江嘉兴·期末"
+    },
+    {
+        "id":  "103",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "神舟二十号的太阳能板的形状是三角形，其中两条边的长度分别是4米和6米。第三条边的长度最长是( )米(取整米数)。",
+        "options":  [
+
+                    ],
+        "answer":  "9",
+        "solution":  "【分析】根据三角形的三边关系，第三条边必须满足：两边之和大于第三边，且两边之差小于第三边。已知两条边分别为4米和6米，则第三条边要小于已知的两条边之和，且大于已知的两条边之差，据此解答.\n【详解】4+6=10(米),6-4=2(米),因此,2米\u003c第三条边\u003c10米;当取整米数，第三条边最长是9米。因此，神舟二十号的太阳能板的形状是三角形，其中两条边的长度分别是4米和6米。第三条边的长度最长是9米。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "104",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如图，将一张长方形的纸折起一角，已知∠1=70°,则∠2=( )°,∠4=( )°.\n![](images/66e00d94382e5e565525de33af18299aab23039e4af02e041bdb6644015743e0.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "55;35",
+        "solution":  "【分析】观察图形可知，平角=180°,∠2+∠2+∠1=180°,三角形的内角和是180°,则∠4+∠2+90°=180°,据此解答.\n【详解】根据分析:∠2=(180°-70°)÷2=110°/2=55°;∠4=180°-90°-55°=90°-55°=35°。所以∠2=55°,∠4=35°。",
+        "knowledge_points":  [
+                                 "平角",
+                                 "周角的认识及特征",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·浙江嘉兴·期末"
+    },
+    {
+        "id":  "105",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如图，三角形ABC是一个等腰三角形,∠2=∠3;∠1=60°,∠2=( )°,如果按角分类，三角形ABC是( )三角形.\n![](images/62ff75cef50a3767ec10aaea95bc85a4b6c714e2a4467b1684f35a340e63ca09.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "30;钝角",
+        "solution":  "【分析】根据题意可知,∠ABC与∠1组成一个平角，所以180°减去∠1等于∠ABC,三角形内角和等于180°,所以180°减去∠ABC等于∠2与∠3的度数和，又因为∠2=∠3，所以∠2与∠3的度数和除以2等于∠2的度数，再根据三角形的分类知识判断三角形ABC是什么三角形;据此即可解答.\n【详解】∠ABC=180°-∠1=180°-60°=120°;180°-∠ABC=180°-120°=60°;∠2=∠3=60°/2=30°;∠ABC等于120°,是钝角，所以三角形ABC是钝角三角形。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·辽宁抚顺·期末"
+    },
+    {
+        "id":  "106",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用三根小棒搭三角形，其中的两根分别是5cm和7cm,那么第三根小棒最长可以是( )cm,最短是( )cm。(长度为整厘米数)",
+        "options":  [
+
+                    ],
+        "answer":  "11;3",
+        "solution":  "【分析】根据三角形三边关系，任意两边之和大于第三边，任意两边之差小于第三边。已知两根小棒分别为5cm和7cm,第三根小棒长度需满足:5cm+7cm\u003e第三根小棒的长度,7cm-5cm\u003c第三根小棒的长度，据此解答.\n【详解】确定第三边范围:5+7=12(cm),7-5=2(cm),2cm\u003c第三根小棒的长度\u003c12cm,所以第三根小棒最长可以是11cm,最短是3cm。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·浙江杭州·期末"
+    },
+    {
+        "id":  "107",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "有一张等腰三角形的大桌子，桌子的顶角是52°,它的一个底角是( )°,这还是一个( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "64;锐角",
+        "solution":  "【分析】根据题意，三角形的内角和是180°,等腰三角形的两个底角相等，用180°减去52°就是两个底角的度数，然后再除以2就是一个底角的度数。三个角都是锐角的三角形是锐角三角形。据此解答.\n【详解】(180°-52°)÷2=128°/2=64°,它的一个底角是64°。因为64°,64°,52°都是锐角，所以这还是一个锐角三角形。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·贵州黔南·期末"
+    },
+    {
+        "id":  "108",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小温有两根小棒，长度分别是6厘米和13厘米。如果再添加一根小棒(取整厘米数)搭成一个三角形，小棒最短是( )厘米，最长是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "8;18",
+        "solution":  "【分析】根据三角形的特征：任意两边之和大于第三边，任意两边之差小于第三边;把已知的两个边长相加再减去1即可求出最长是多少，两个边长相减再加1即可求出最短是多少.\n【详解】13-6=7(厘米),7+1=8(厘米);13+6-1=19-1=18(厘米)。如果再添加一根小棒(取整厘米数)搭成一个三角形，小棒最短是8厘米，最长是18厘米。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·河南漯河·期末"
+    },
+    {
+        "id":  "109",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个等腰三角形，顶角是96°,底角是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "42",
+        "solution":  "【分析】等腰三角形的两个底角相等，根据三角形内角和为180°,用内角和减去顶角度数，得到两个底角的和，再除以2即可求出每个底角的度数.\n【详解】(180°-96°)÷2=84°/2=42°。一个等腰三角形，顶角是96°,底角是42°。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·黑龙江绥化·期末"
+    },
+    {
+        "id":  "110",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形的底角是65°,它的顶角是( )°,按角分它是个( )三角形;如果这个等腰三角形的两条边分别是5cm,11cm,它的周长是( )cm,它的对称轴有( )条。",
+        "options":  [
+
+                    ],
+        "answer":  "50;锐角;27;1",
+        "solution":  "【分析】根据等腰三角形的两个底角相等，三角形的内角和是180°,可以求出顶角的度数;再根据三角形三个角的度数，判断按角分它是什么三角形;根据三角形三边关系\"任意两边之和大于第三边\"来判断等腰三角形的腰长，从而求出周长;![](images/a1b5c4ccdd610f2c999026908c7603f7f62bae71957e446959b571ba3567f637.jpg)从图中可以看出，等腰三角形对称轴的条数.\n【详解】①已知一个底角是65°,则另一个底角也是65°,所以顶角为180°-65°-65°=50°;②因为三角形的三个角分别是65°,65°,50°,三个角都小于90°,所以按角分它是个锐角三角形;③已知这个等腰三角形的两条边分别是5厘米,11厘米，如果腰长是5厘米，那么5+5=10(厘米),10\u003c11，不满足三边关系;如果腰长是11厘米，那么11+5=16(厘米),16\u003e11，所以这个三角形的腰长是11厘米，底边长是5厘米，周长为11+11+5=27(厘米);④等腰三角形有1条对称轴，这条对称轴是底边上的高所在的直线。所以一个等腰三角形的底角是65°,它的顶角是50°,按角分它是个锐角三角形;如果这个等腰三角形的两条边分别是5cm,11cm,它的周长是27厘米，它的对称轴有1条。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和",
+                                 "三角形的周长"
+                             ],
+        "source":  "24-25四年级下·内蒙古鄂尔多斯·期末"
+    },
+    {
+        "id":  "111",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "任意一个四边形四个角的度数和是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "360",
+        "solution":  "【分析】根据多边形内角和公式:n边形的内角和=(n-2)×180°,计算四边形的内角和.\n【详解】(4-2)×180°=2×180°=360°。所以，任意一个四边形的度数和是360°。",
+        "knowledge_points":  [
+                                 "多边形的内角和"
+                             ],
+        "source":  "25-26四年级上·海南海口·单元测试"
+    },
+    {
+        "id":  "112",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "有一个三角形，已知其中两个角分别是90°和45°,那第三个角的度数是( )°,这个三角形按边分是( )三角形，按角分是( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "45;等腰;直角",
+        "solution":  "【分析】三角形的内角和是180°,用180°减去已知两个角的度数，就是第三个角的度数。按边分，有两个角相等的三角形是等腰三角形。按角分，有一个角是直角的三角形是直角三角形。据此解答.\n【详解】180°-90°-45°=45°,所以第三个角的度数是45°。这个三角形按边分是等腰三角形，按角分是直角三角形。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·新疆克孜勒苏·期末"
+    },
+    {
+        "id":  "113",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三角形有两条边的长分别是8厘米和11厘米，第三条边的长(整厘米)最长是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "18",
+        "solution":  "【分析】三角形三边的关系：任意两边之和大于第三边，任意两边之差小于第三边。由题意得，一个三角形有两条边的长分别是8厘米和11厘米，那么第三条边的长度大于两边之差，小于两边之和。据此解答.\n【详解】两边之差\u003c第三条边的长度\u003c两边之和,11-8\u003c第三条边的长度\u003c11+8,3\u003c第三条边的长度\u003c19，第三条边的长是整厘米，所以第三条边最长是18厘米。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·甘肃兰州·期中"
+    },
+    {
+        "id":  "114",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在一个三角形中，最大的角是75°,这个三角形是( )角三角形;三条边长度相等的三角形叫做( )三角形，它的每个内角都是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "锐;等边;60",
+        "solution":  "【分析】三个角都是锐角的三角形为锐角三角形，三条边相等的三角形为等边三角形，等边三角形的每个内角都是60°,据此解答.\n【详解】在一个三角形中，最大的角是75°,这个三角形是锐角三角形;三条边长度相等的三角形叫做等边三角形，它的每个内角都是60°。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·甘肃兰州·期中"
+    },
+    {
+        "id":  "115",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "如果一个三角形的两条边分别长4cm和7cm,那么它的第三条边最长是( )cm。(取整厘米数)",
+        "options":  [
+
+                    ],
+        "answer":  "10",
+        "solution":  "【分析】三角形任意两边之和大于第三边，任意两边之差小于第三边，求出已知两边的长度和，根据任意两边之和大于第三边判断出第三条边最长是多少厘米，据此即可解答.\n【详解】4+7=11(cm),11-1=10(cm),第三条边(取整厘米数)小于11cm,所以它的第三条边最长是10cm。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·辽宁抚顺·期末"
+    },
+    {
+        "id":  "116",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "明明家，公园，学校三个地点连起来正好是一个三角形。已知明明家到公园的距离是5千米，学校到公园的距离是3千米，那么明明家到学校的距离最远是( )千米，最近是( )千米。(三地之间的距离都是整千米数)",
+        "options":  [
+
+                    ],
+        "answer":  "7;3",
+        "solution":  "【分析】三角形任意两边之和大于第三边，任意两边之差小于第三边，明明家，公园，学校三个地点连起来正好是一个三角形，所以明明家到学校的距离大于5-3=2(千米),小于5+3=8(千米),据此即可解答.\n【详解】5-3=2(千米),5+3=8(千米),所以，明明家到学校的距离大于2千米，小于8千米，最远是7千米，最近是3千米。(三地之间的距离都是整千米数)",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·辽宁葫芦岛·期末"
+    },
+    {
+        "id":  "117",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形的周长是30厘米，底边长是12厘米，它的一条腰长是( )厘米。在这个三角形中，其中一个底角是40°,它的顶角是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "9;100",
+        "solution":  "【分析】等腰三角形的两腰相等，等腰三角形的周长减去底边长度等于两腰的长度和，再除以2，即等于一条腰长;等腰三角形的两个底角相等，三角形的内角和等于180°,180°减去两底角的度数和，即等于顶角的度数.\n【详解】(30-12)÷2=18÷2=9(厘米);180°-40°*2=180°-80°=100°。所以一条腰长是9厘米，它的顶角是100°。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和",
+                                 "三角形的周长"
+                             ],
+        "source":  "24-25四年级下·内蒙古呼伦贝尔·期末"
+    },
+    {
+        "id":  "118",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小云用4根相同长度的小棒围四边形，能围成多个不同形状的四边形;小明用3根相同长度的小棒围三角形，他能围成( )种三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "1",
+        "solution":  "【分析】根据三角形具有稳定性，三角形3条边的长度确定，三角形的形状和大小也就确定，所以他能围成1种三角形，即可解题.\n【详解】由分析可知：小云用4根相同长度的小棒围四边形，能围成多个不同形状的四边形;小明用3根相同长度的小棒围三角形，他能围成1种三角形。",
+        "knowledge_points":  [
+                                 "三角形的稳定性及应用"
+                             ],
+        "source":  "24-25四年级下·云南昆明·期末"
+    },
+    {
+        "id":  "119",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "李老师做了一个等腰三角形的教具，已知其中的两条边分别是20厘米,9厘米，这个等腰三角形教具的周长是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "49",
+        "solution":  "【分析】根据等腰三角形的两腰相等，三角形3条边的关系是：任意两边的长度之和大于第三边，任意两边的长度之差小于第三边，依此确定出这个等腰三角形教具的腰长，再计算出这个等腰三角形教具三条边的总长度，即可解题.\n【详解】当腰长为9厘米时,9+9=18(厘米),18厘米\u003c20厘米，因此腰长不能为9厘米;当腰长为20厘米时,20+9=29(厘米),20-9=11(厘米),29厘米\u003e20厘米,11厘米\u003c20厘米，因此腰长为20厘米。20+20+9=49(厘米),所以这个等腰三角形教具的周长是49厘米。",
+        "knowledge_points":  [
+                                 "三角形三边关系",
+                                 "三角形的周长",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·云南昆明·期末"
+    },
+    {
+        "id":  "120",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个三角形中的两个角分别是50°和80°,那么第三个角是( )°;按角分，这是( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "50;锐角",
+        "solution":  "【分析】三角形的内角和是180°,三角形的内角和-一个内角的度数-另一个内角的度数=第三个内角的度数;三角形按角分，三个角都是锐角的三角形是锐角三角形，有个角是直角的三角形是直角三角形，有一个角是钝角的三角形是钝角三角形;据此作答.\n【详解】180°-80°-50°=50°,第三个角是50°。三个角都小于90°是锐角三角形，这个三角形是锐角三角形。因此第三个角是50°,这个三角形是锐角三角形。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·浙江温州·期末"
+    },
+    {
+        "id":  "121",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "房屋的屋架设计成三角形是利用了三角形的( )性。停车场的推拉门利用了平行四边形的( )性。",
+        "options":  [
+
+                    ],
+        "answer":  "稳定;不稳定",
+        "solution":  "【分析】三角形具有稳定性，因此屋架设计成三角形;平行四边形具有不稳定性，推拉门利用这一特性实现灵活开合.\n【详解】根据分析，房屋的屋架设计成三角形是利用了三角形的稳定性;停车场的推拉门利用了平行四边形的不稳定性。",
+        "knowledge_points":  [
+                                 "平行四边形的不稳定性及应用",
+                                 "三角形的稳定性及应用"
+                             ],
+        "source":  "24-25四年级下·云南玉溪·期末"
+    },
+    {
+        "id":  "122",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个直角三角形的一个锐角是35°,另一个锐角是( )°。一个等腰三角形的顶角是100°,它的一个底角是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "55;40",
+        "solution":  "【分析】根据三角形的内角和是180°可得，一个直角三角形的一个锐角是35°,另一个锐角应该用\"180°-90°-35°\";等腰三角形的特点可知：等腰三角形的两个底角相等，以及三角形的内角和是180°,用180°减去顶角的100°,即可求出两个底角的和，那么再除以2，即为一个底角的度数，据此填空.\n【详解】根据分析可得:180°-90°-35°=55°;(180°-100°)÷2=80°/2=40°。所以一个直角三角形的一个锐角是35°,另一个锐角是55°。一个等腰三角形的顶角是100°,它的一个底角是40°。",
+        "knowledge_points":  [
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·云南玉溪·期末"
+    },
+    {
+        "id":  "123",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形的两条边长分别是6cm和3cm,它的周长是( )cm。(请填写所有可能的结果)",
+        "options":  [
+
+                    ],
+        "answer":  "15",
+        "solution":  "【分析】等腰三角形两条腰长度相等。已知两条边的长度，可假设其中一条为腰，然后利用三角形三边的关系(任意两边之和大于第三边)来验证假设是否成立，如果假设成立，把这三条边的长度相加就是周长。如果假设不成立，则假设另一条边为腰，然后利用三角形三边的关系(任意两边之和大于第三边)来验证假设是否成立，如果假设成立，把这三条边的长度相加就是周长.\n【详解】当6cm长的边为腰,3cm长的边为底时,6+3\u003e6满足三角形三边关系，能组成三角形，周长为:6+6+3=15(cm);当3cm长的边为腰,6cm长的边为底时,3+3=6不满足三角形三边关系，不能组成三角形。综上所述：一个等腰三角形的两条边长分别是6cm和3cm,它的周长是15cm。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的周长",
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·云南玉溪·期末"
+    },
+    {
+        "id":  "124",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "等腰三角形的底角是65°,顶角是( )°。等边三角形每个内角是( )°。直角三角形的一个锐角是35°,另一个锐角是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "50;60;55",
+        "solution":  "【分析】根据三角形内角和是180°,等腰三角形两个底角相等，顶角=180°-2*底角;等边三角形三个内角都相等，每个内角=180°/3;直角三角形的另外两个锐角和是90°,另一个锐角=90°-一个锐角的度数.\n【详解】180°-65°*2=180°-130°=50°;180°/3=60°;90°-35°=55°。所以等腰三角形的底角是65°,顶角是50°。等边三角形每个内角是60°。直角三角形的一个锐角是35°,另一个锐角是55°。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·吉林白城·期末"
+    },
+    {
+        "id":  "125",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用一根18厘米长的铁丝围一个等边三角形，每条边的长是( )厘米;如果围成其它三角形，那么最长的一条边要小于( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "6;9",
+        "solution":  "【分析】根据题意，等边三角形的边长为(18÷3)厘米;如果围成其他三角形，根据三角形的任意两边之和大于第三边，所以最长的一条边要小于(18÷2)厘米，据此解答.\n【详解】根据分析可得:18÷3=6(厘米),18÷2=9(厘米)。用一根18厘米长的铁丝围一个等边三角形，每条边的长是6厘米;如果围成其它三角形，那么最长的一条边要小于9厘米。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·河南新乡·期末"
+    },
+    {
+        "id":  "126",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三角形中最少有( )个锐角，最多有( )个锐角。",
+        "options":  [
+
+                    ],
+        "answer":  "2;3",
+        "solution":  "【分析】三角形按角来分，分为锐角三角形，直角三角形和钝角三角形。三个角都是锐角的三角形叫作锐角三角形，有一个角是直角的三角形叫作直角三角形，有一个角是钝角的三角形叫作钝角三角形。直角为90度的角，锐角为大于0度小于90度的角，钝角为大于90度，小于180度的角;三角形内角和为180度;据此分析解答.\n【详解】当这个三角形为锐角三角形时，三个角都是锐角;当这个三角形为直角三角形时，其中一个角为90度，另外两个角之和为:180°-90°=90°,另外两个角必然都小于90度，都是锐角;当这个三角形为钝角三角形时，其中一个角大于90度，另外两个角之和必定小于90度，这两个角必然都小于90度，都是锐角。综上可知，一个三角形中最少有2个锐角，最多有3个锐角。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "三角形的内角和",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·新疆阿勒泰·期末"
+    },
+    {
+        "id":  "127",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形的两条边长是4.5cm和9cm,这个等腰三角形的周长是( )cm。",
+        "options":  [
+
+                    ],
+        "answer":  "22.5",
+        "solution":  "【分析】等腰三角形任意两边之和大于第三边，先判断出腰长和底长各是多少，然后把三边的长度相加即等于等腰三角形的周长，据此即可解答.\n【详解】4.5cm+4.5cm=9cm,不符合三角形三边之间的关系，腰长不能是4.5cm,所以腰长为9cm,底长为4.5cm。9+9+4.5=22.5(cm)。这个等腰三角形的周长是22.5cm。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的周长",
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·浙江·期末"
+    },
+    {
+        "id":  "128",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个直角三角形中一个锐角是67°,它的另一个锐角是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "23",
+        "solution":  "【分析】三角形的内角和是180°,直角三角形里有一个角是直角，是90°,用180°减去90°再减去67°,就是另一个锐角的度数.\n【详解】180°-90°-67°=23°,所以，它的另一个锐角是23°。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·广东肇庆·期末"
+    },
+    {
+        "id":  "129",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把一根21cm长的铁丝折成一个等边三角形铁框，铁框的一边长( )cm,若折成一条腰是6cm的等腰三角形铁框，铁框的底边长( )cm。",
+        "options":  [
+
+                    ],
+        "answer":  "7;9",
+        "solution":  "【分析】由题意可知，铁丝的长度就是等边三角形的周长，用铁丝的长度除以3即可求出铁框的边长;等腰三角形的两条腰的长度相等，用铁丝的长度减去两条腰的长度即可求出铁框底边长.\n【详解】21÷3=7(cm);21-6×2=21-12=9(cm)。则把一根21cm长的铁丝折成一个等边三角形铁框，铁框的边长是7cm,若折成一个腰长是6cm的等腰三角形铁框，铁框的底边长9cm。",
+        "knowledge_points":  [
+                                 "三角形的周长",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "130",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个直角三角形的一个锐角是38°,它的另一个锐角是( )°;一个等腰三角形的顶角是40°,它的一个底角是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "52;70",
+        "solution":  "【分析】根据直角三角形的特点：有一个直角(90°)和两个锐角，用180°减90°减38°即可求出另一个锐角的度数;再根据等腰三角形的特点：两个底角相等，先用180°减40°求出两个底角之和，再除以2，即可求出一个底角的度数.\n【详解】180°-90°-38°=52°;(180°-40°)÷2=140°/2=70°。所以，一个直角三角形的一个锐角是38°,它的另一个锐角是52°,一个等腰三角形的顶角是40°,它的一个底角是70°。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "131",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形的底角是50°,它的顶角是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "80",
+        "solution":  "【分析】三角形内角和为180°,等腰三角形两个底角相等，用180°减去两个底角的度数即可求出它的顶角是多少度.\n【详解】180°-50°-50°=80°,所以，一个等腰三角形的底角是50°,它的顶角是80°。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·广东阳江·期末"
+    },
+    {
+        "id":  "132",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三角形三条边的长度都是整厘米数，其中两条边分别是8cm和13cm,则第三条边最短是( )cm,最长是( )cm。",
+        "options":  [
+
+                    ],
+        "answer":  "6;20",
+        "solution":  "【分析】三角形的三边关系：两边之和大于第三边，两边之差小于第三边;据此求出这个三角形的第三条边的取值范围，再根据三角形三条边的长度都是整厘米数，找到第三边最短和最长长度即可.\n【详解】13+8=21(cm),13-8=5(cm),所以三角形的第三边大于5cm小于21cm。因为三角形三条边的长度都是整厘米数，所以三角形的第三边最短是6cm,最长是20cm。即一个三角形三条边的长度都是整厘米数，其中两条边分别是8cm和13cm,则第三条边最短是6cm,最长是20cm。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·广东肇庆·期末"
+    },
+    {
+        "id":  "133",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一根铁丝可以围成一个边长为3厘米的正方形，这个正方形的周长是( )厘米，如果改围一个等边三角形，那么等边三角形的边长是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "12;4",
+        "solution":  "【分析】正方形周长=边长*4，代入计算即可;等边三角形的周长=边长*3，所以用铁丝的长度除以3可以算出等边三角形边长.\n【详解】3×4=12(厘米),12÷3=4(厘米)。所以，正方形的周长为12厘米，等边三角形的边长是4厘米。",
+        "knowledge_points":  [
+                                 "三角形的周长",
+                                 "正方形的周长",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·新疆昌吉·期末"
+    },
+    {
+        "id":  "134",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形的底角是50°,它的顶角是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "80",
+        "solution":  "【分析】三角形内角和为180°,等腰三角形两个底角相等，用180°减去两个底角的度数即可求出它的顶角是多少度.\n【详解】180°-50°-50°=80°,所以，一个等腰三角形的底角是50°,它的顶角是80°。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·广东阳江·期末"
+    },
+    {
+        "id":  "135",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三角形三条边的长度都是整厘米数，其中两条边分别是8cm和13cm,则第三条边最短是( )cm,最长是( )cm。",
+        "options":  [
+
+                    ],
+        "answer":  "6;20",
+        "solution":  "【分析】三角形的三边关系：两边之和大于第三边，两边之差小于第三边;据此求出这个三角形的第三条边的取值范围，再根据三角形三条边的长度都是整厘米数，找到第三边最短和最长长度即可.\n【详解】13+8=21(cm),13-8=5(cm),所以三角形的第三边大于5cm小于21cm。因为三角形三条边的长度都是整厘米数，所以三角形的第三边最短是6cm,最长是20cm。即一个三角形三条边的长度都是整厘米数，其中两条边分别是8cm和13cm,则第三条边最短是6cm,最长是20cm。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·广东肇庆·期末"
+    },
+    {
+        "id":  "136",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一根铁丝可以围成一个边长为3厘米的正方形，这个正方形的周长是( )厘米，如果改围一个等边三角形，那么等边三角形的边长是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "12;4",
+        "solution":  "【分析】正方形周长=边长*4，代入计算即可;等边三角形的周长=边长*3，所以用铁丝的长度除以3可以算出等边三角形边长.\n【详解】3×4=12(厘米),12÷3=4(厘米)。所以，正方形的周长为12厘米，等边三角形的边长是4厘米。",
+        "knowledge_points":  [
+                                 "三角形的周长",
+                                 "正方形的周长",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·新疆昌吉·期末"
+    },
+    {
+        "id":  "137",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个直角三角形中的一个锐角是31°,另一个锐角是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "59",
+        "solution":  "【分析】直角=90°,三角形内角和为180°,所以用180°-90°-31°可以算出另一个锐角.\n【详解】180°-90°-31°=90°-31°=59°。所以另一个锐角是59°。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·辽宁鞍山·期末"
+    },
+    {
+        "id":  "138",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三角板，有一个锐角是60度，另一个锐角是( )度。",
+        "options":  [
+
+                    ],
+        "answer":  "30",
+        "solution":  "【分析】根据三角板的特征，有一个角是直角，再根据三角形内角和等于180°,用180°减去90°,再减去60°,列式计算即可.\n【详解】根据分析可知:180°-90°-60°=30°。一个三角板，有一个锐角是60度，另一个锐角是30度。",
+        "knowledge_points":  [
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "139",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "两根木条分别长5厘米和10厘米，再用一根长( )厘米的木条，就可以拼成一个等腰三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "10",
+        "solution":  "【分析】根据三角形的特性：两边之和大于第三边，三角形的两边的差一定小于第三边;等腰三角形有两条边相等，据此解题.\n【详解】10-5=5(厘米),10+5=15(厘米),另一边在5厘米和15厘米之间,10厘米符合题意，是一个等腰三角形。所以再用一根10厘米长的木条，就可以拼成一个等腰三角形。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·湖北恩施·期末"
+    },
+    {
+        "id":  "140",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个等腰三角形的底角是43°,顶角是( )°,按角分它是( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "94;钝角",
+        "solution":  "【分析】已知等腰三角形的一个底角是43°,根据等腰三角形两底角相等的性质，可知另一个底角也是43°。因为三角形内角和是180°,所以用180°减去两个底角的度数，就可以得到顶角的度数。根据三角形按角分类的方法判断三角形的分类。三个角都是锐角的三角形是锐角三角形;有一个角是直角的三角形是直角三角形;有一个角是钝角的三角形是钝角三角形.\n【详解】180°-43°-43°=94°,顶角是94°,94°是钝角，所以它是钝角三角形。所以，一个等腰三角形的底角是43°,顶角是94°,按角分它是钝角三角形。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的分类"
+                             ],
+        "source":  "23-24四年级下·河南安阳·期末"
+    },
+    {
+        "id":  "141",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个直角三角形中的一个锐角是42°,它的另一个锐角是( )°;一个等腰三角形的一个顶角是80°,它的底角是( )°,这个三角形按角分类是( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "48;50;锐角",
+        "solution":  "【分析】有一个角是直角的三角形是直角三角形，直角是90°。根据三角形内角和为180°,用180°减去90°再减去42°可以算出另一个锐角的度数;等腰三角形的两个底角相等，根据三角形内角和为180°,用180°减去80°之后再除以2可以得到底角的度数;三角形按角分类：三个角都是锐角的三角形是锐角三角形，有一个角是直角的三角形是直角三角形，有一个角是钝角的三角形是钝角三角形.\n【详解】180°-90°-42°=48°,所以它的另一个锐角是48°;(180°-80°)÷2=100°/2=50°,所以它的底角是50°;这个三角形三个角都是锐角，所以按角分类是锐角三角形。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的分类"
+                             ],
+        "source":  "23-24四年级下·吉林松原·期末"
+    },
+    {
+        "id":  "142",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个三角形的两条边分别是7厘米和10厘米，这个三角形的第三条边最长是( )厘米，最短是( )厘米。(三条边长都是整厘米数)",
+        "options":  [
+
+                    ],
+        "answer":  "16;4",
+        "solution":  "【分析】结合题意，然后根据三角形三边的关系，任意两条边之和大于第三边，任意两边之差小于第三边。据此解答.\n【详解】7+10=17(厘米),第三边必须小于17厘米，因此，最长是16厘米。10-7=3(厘米),第三边必须大于3厘米，因此，最短是4厘米。即，这个三角形的第三条边最长是16厘米，最短是4厘米。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·广东阳江·期末"
+    },
+    {
+        "id":  "143",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "将一根长9厘米的小棒剪两次，剪成三段，围成一个三角形。第一次一定不能剪在( )厘米处。",
+        "options":  [
+
+                    ],
+        "answer":  "4.5",
+        "solution":  "【分析】根据三角形的三边关系：三角形任意两边之和大于第三边。若第一次剪在4.5厘米处，会出现一段小棒长度等于另外两段小棒长度和，不满足任意两边之和大于第三边，无法围成三角形，故根据此关系判断不能剪在4.5厘米处.\n【详解】9-4.5=4.5(厘米),所以第一次一定不能剪在4.5厘米处。",
+        "knowledge_points":  [
+                                 "一位小数的进位加法，退位减法",
+                                 "三角形三边关系"
+                             ],
+        "source":  "23-24四年级下·浙江嘉兴·期末"
+    },
+    {
+        "id":  "144",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "桌面上有一些三角形卡片，这些三角形中一共有36个锐角,5个钝角,4个直角，那么共有( )个锐角三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】钝角三角形里面有2个锐角，直角三角形里面有2个锐角，锐角三角形里面有3个锐角;锐角三角形的个数=[锐角的总个数-(钝角三角形的个数+直角三角形的个数)×2]/每个锐角三角形里面锐角的个数.\n【详解】36-(5+4)×2=36-18=18(个),18÷3=6(个)。桌面上有一些三角形卡片，这些三角形中一共有36个锐角,5个钝角,4个直角，那么共有6个锐角三角形。",
+        "knowledge_points":  [
+                                 "三角形的分类"
+                             ],
+        "source":  "23-24四年级下·浙江嘉兴·期末"
+    },
+    {
+        "id":  "145",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个用绳子围成的等腰三角形，两条邻边分别长8厘米,6厘米，围成这个三角形至少需要( )厘米长的绳子。",
+        "options":  [
+
+                    ],
+        "answer":  "20",
+        "solution":  "【分析】等腰三角形有两条边相等。已知两条邻边分别为8厘米和6厘米，需确定哪条是腰，哪条是底边。等腰三角形的腰长为8厘米，底边为6厘米。三边分别为8厘米,8厘米,6厘米。验证三边关系:8+8=16\u003e6,8+6=14\u003e8，符合三角形任意两边之和大于第三边。把三边相加求出周长。等腰三角形的腰长为6厘米，底边为8厘米。三边分别为8厘米,6厘米,6厘米。验证三边关系:6+6=12\u003e8,8+6=14\u003e8，符合三角形任意两边之和大于第三边。把三边相加求出周长。最后判断周长少的一个答题即可.\n【详解】根据分析可知:8+8=16\u003e6,8+6=14\u003e8，符合三角形任意两边之和大于第三边,8+8+6=22(厘米);6+6=12\u003e8,8+6=14\u003e8，符合三角形任意两边之和大于第三边,6+6+8=20(厘米);20\u003c22。所以围成这个三角形至少需要20厘米长的绳子。",
+        "knowledge_points":  [
+                                 "三角形三边关系",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·广东江门·期末"
+    },
+    {
+        "id":  "146",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "等腰三角形的顶角是80°,它的底角是( )°;按角分，这个三角形是( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "50;锐角",
+        "solution":  "【分析】根据等腰三角形底角相等的性质，利用三角形内角和为180°,用180°减去顶角再除以2可以得到每个底角的度数。根据三个角的度数判断三角形的类型，三角形按角分类：三个角都是锐角的三角形是锐角三角形，有一个角是直角的三角形是直角三角形，有一个角是钝角的三角形是钝角三角形.\n【详解】(180°-80°)÷2=100°/2=50°,所以它的底角是50°。它的三个角分别为50°,50°,80°,小于90°的角为锐角，这三个角都是锐角，所以这个三角形是锐角三角形。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·安徽安庆·期末"
+    },
+    {
+        "id":  "147",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个边长均为整厘米数的三角形的两条边长分别是4cm和7cm,那么第三条边最长是( )cm,最短是( )cm。",
+        "options":  [
+
+                    ],
+        "answer":  "10;4",
+        "solution":  "【分析】根据三角形三边关系定理，第三条边必须满足两边之和大于第三边且两边之差小于第三边。已知一个边长均为整厘米数的三角形的两条边长分别是4cm和7cm,用4+7=11(cm),7-4=3(cm),以此求出第三条边最长，最短各是多少即可.\n【详解】根据分析可知:4+7=11(cm),11-1=10(cm),7-4=3(cm),3+1=4(cm)。所以第三条边最长是10cm,最短是4cm。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·河南驻马店·期末"
+    },
+    {
+        "id":  "148",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "等腰三角形的一个底角是28°,那么这个等腰三角形的顶角是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "124",
+        "solution":  "【分析】由已知等腰三角形的一个角是28°,利用等腰三角形的性质可知另一个底角也是28°,结合三角形内角和定理，用180°减去两个底角的和，即可求出顶角的度数.\n【详解】180°-28°*2=180°-56°=124°。所以这个等腰三角形的顶角是124°。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·河南驻马店·期末"
+    },
+    {
+        "id":  "149",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "三角形的内角和是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "180",
+        "solution":  "【分析】任意一个三角形的三个内角的度数之和都是180°,据此解答.\n【详解】根据分析可知，三角形的内角和是180°。",
+        "knowledge_points":  [
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·河南驻马店·期末"
+    },
+    {
+        "id":  "150",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "由3条( )围成的图形(每相邻两条线段的端点相连)叫做三角形。三角形有( )个角，有( )个顶点。",
+        "options":  [
+
+                    ],
+        "answer":  "线段;3;3",
+        "solution":  "【分析】如图:![](images/2026-03-10%20215115.jpg)三角形的特点：三条线段首尾相连，三角形有3条边,3个角和3个顶点。据此解答.\n【详解】由分析可知：由3条线段围成的图形(每相邻两条线段的端点相连)叫做三角形。三角形有3个角，有3个顶点。",
+        "knowledge_points":  [
+                                 "三角形的概念及表示方法"
+                             ],
+        "source":  "24-25四年级下·海南省直辖县级单位·期末"
+    },
+    {
+        "id":  "151",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三角形的两个内角分别是45°和62°,第三个内角是( )°,它是一个( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "73;锐角",
+        "solution":  "【分析】三角形的内角和为180°,用180°减去已知两个角可以算出第三个内角的度数;三角形按角分类：三个角都是锐角的三角形是锐角三角形，有一个角是直角的三角形是直角三角形，有一个角是钝角的三角形是钝角三角形，据此判断.\n【详解】180°-45°-62°=73°,一个三角形的两个内角分别是45°和62°,第三个内角是73°,这三个角都小于90°,都是锐角，所以它是一个锐角三角形。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·湖南岳阳·期末"
+    },
+    {
+        "id":  "152",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形的底角是80°,它的顶角是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "20",
+        "solution":  "【分析】等腰三角形两底角相等，已知底角是80°,则两底角和为160°。用三角形内角和减去两底角和，即可求出顶角度数.\n【详解】180°-(80°+80°)=180°-160°=20°,所以该等腰三角形的顶角是20°。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·广东佛山·期末"
+    },
+    {
+        "id":  "153",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "高压线杆的支架焊成三角形是因为三角形具有( );请你再举出一个生活中应用三角形这种特性的例子( )。",
+        "options":  [
+
+                    ],
+        "answer":  "稳定性;塔吊的横梁",
+        "solution":  "【分析】三角形具有稳定性，即在受到外力作用时，其形状不易改变。高压线杆的支架焊成三角形正是利用了这种特性，使得支架更加稳固。生活中应用三角形稳定性的例子有很多，如塔吊的横梁，屋顶的三角结构.\n【详解】根据三角形的特性，三角形具有稳定性，因此高压线杆的支架焊成三角形是为了利用这一特性，使结构更稳固。生活中应用三角形稳定性的例子可以是塔吊的横梁。塔吊在作业时需要承受较大的重量，其利用了三角形结构，能够有效增强稳定性，防止变形。所以高压线杆的支架焊成三角形是因为三角形具有稳定性;生活中塔吊的横梁现象也是运用了这一项原理。(答案不唯一)",
+        "knowledge_points":  [
+                                 "三角形的稳定性及应用"
+                             ],
+        "source":  "23-24四年级下·湖南长沙·期末"
+    },
+    {
+        "id":  "154",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小红参加了一个风筝制作挑战活动，她决定制作一个三角形的风筝，并已经测量了两条边的长度，分别是1.5米和2米。为了确保风筝的稳定性，她需要计算出第三条边的长度范围(整数或一位小数),即必须小于( )米且大于( )米。",
+        "options":  [
+
+                    ],
+        "answer":  "3.5;0.5",
+        "solution":  "【分析】三角形的三边关系：两边之和大于第三边，两边之差小于第三边;据此解答.\n【详解】1.5+2=3.5(米),2-1.5=0.5(米)。所以第三条边必须小于3.5米且大于0.5米。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·吉林·期末"
+    },
+    {
+        "id":  "155",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "有一个等腰三角形的两条边分别是6cm和13cm,那么这个等腰三角形的另一条边的长度是( )cm,这个等腰三角形的周长是( )cm。",
+        "options":  [
+
+                    ],
+        "answer":  "13;32",
+        "solution":  "【分析】根据三角形三边关系，三角形任意两边之和大于第三边，等腰三角形有两条腰的长度相等，据此求出第三条边的长度，将三条边的长度相加即可求出周长.\n【详解】当腰为6cm时，三角形三条边为:6cm,6cm,13cm,6+6\u003c13，不能构成三角形;所以腰为13cm,三角形三条边为:13cm,13cm,6cm,13+6\u003e13，能构成三角形。6+13+13=32(cm)。所以这个等腰三角形的另一条边的长度是13cm,周长是32cm。",
+        "knowledge_points":  [
+                                 "三角形三边关系",
+                                 "三角形的周长",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·广东汕头·期末"
+    },
+    {
+        "id":  "156",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "有一个直角三角形，它的一个锐角是32°,则另一个锐角是( )°;一个六边形的内角和是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "58;720",
+        "solution":  "【分析】(1)直角三角形中，直角为90°,两个锐角之和为90°,已知一个锐角是32°,可求出另一个锐角。(2)多边形内角和公式为(n-2)×180°,六边形的边数n=6，代入公式即可.\n【详解】(1)90°-32°=58°;(2)(6-2)×180°=4×180°=720°。",
+        "knowledge_points":  [
+                                 "多边形的内角和",
+                                 "三角形的内角和",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·广东汕头·期末"
+    },
+    {
+        "id":  "157",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "等腰三角形的一个底角是48°,它的顶角是( )°;如果这个三角形的两条边分别是9cm和4cm,那么第三条边是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "84;9cm或9厘米",
+        "solution":  "【分析】根据题意，明确三角形的内角和为180°,等腰三角形两个底角相等，已知底角为48°,顶角=180°-48°-48°=84°。根据三角形三边关系，两边之和大于第三边，在等腰三角形中，有两条边相等。已知两条边分别为9cm和4cm(长度不等),因此这两条边中一条是腰(相等的边),一条是底边(第三边)。如果两腰为4cm,4cm,4+4=8\u003c9，不成立,9+9=18\u003e4,9+4=13\u003e9，所以第三条边是9cm。以此答题即可.\n【详解】根据分析可知：等腰三角形的一个底角是48°,它的顶角是84°;如果这个三角形的两条边分别是9cm和4cm,那么第三条边是9cm。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和",
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·山东临沂·期末"
+    },
+    {
+        "id":  "158",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "把一个大三角形分成完全相同的两个小三角形，这两个小三角形的内角和都是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "180°",
+        "solution":  "【分析】三角形的内角和等于180°,所以只要是三角形，它的内角和都等于180°,据此即可解答.\n【详解】根据分析可知，把一个大三角形分成完全相同的两个小三角形，这两个小三角形的内角和都是180°。",
+        "knowledge_points":  [
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·广东佛山·期末"
+    },
+    {
+        "id":  "159",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一名长跑运动员腿长1米，在一次比赛中他大约跑了520步，他参加的可能是( )(填\u0027800米\u0027或\u00271500米\u0027)的比赛。",
+        "options":  [
+
+                    ],
+        "answer":  "800米",
+        "solution":  "【分析】三角形三边的关系：任意两边之和大于第三边(较短两边之和大于第三边)。由题意得，一名长跑运动员腿长1米，那么他的两条腿和地面围成了一个三角形。如图:![](images/16449b631ff3d33bbf5c57fa2371873189e8c75c81e60663af6f43f3df5a6791.jpg)根据三角形三边的关系可知，他的步长(两脚之间的距离)应该小于两条腿的长度之和，即这名长跑运动员的步长小于2米。然后据此推算出他跑520步对应的距离，从而来判断他参加的是多少米的比赛.\n【详解】1+1=2(米),即这名长跑运动员的步长小于2米。2×520=1040(米),即这名长跑运动员跑520步对应的距离小于1040米。所以他参加的是800米的比赛。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·内蒙古包头·期末"
+    },
+    {
+        "id":  "160",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三角形的其中两条边都是5cm,有一个角是40°,那另外两个角分别是( )°和( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "40或70;100或70",
+        "solution":  "【分析】一个三角形的其中两条边都是5cm,说明这个三角形是等腰三角形，如果40°的角是底角，则另一个底角也是40°,三角形内角和等于180°,则顶角为180°-40°-40°=100°;如果40°的角是顶角，则两底角的度数和为180°-40°=140°,底角为140°/2=70°;据此即可解答.\n【详解】一个三角形的其中两条边都是5cm,说明这个三角形是等腰三角形。(1)如果40°的角是底角，则另一个底角也是40°,180°-40°-40°=100°,另外两个角分别是40°和100°。(2)如果40°的角是顶角,180°-40°=140°,140°/2=70°,另外两个角分别是70°和70°。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·内蒙古赤峰·期末"
+    },
+    {
+        "id":  "161",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "等腰三角形在建筑设计中常用于增强结构稳定性，有一座屋顶在建造时就设计成了等腰三角形，它的两条边分别是18米,36米，那么这个等腰三角形的周长是( )米。",
+        "options":  [
+
+                    ],
+        "answer":  "90",
+        "solution":  "【分析】等腰三角形两条腰长相等，根据三角形三边关系，任意两边之和大于第三边，任意两边之差小于第三边，两条边分别是18米,36米，据此先找出第三条边的长度，相加即可求出这个等腰三角形的周长是多少米.\n【详解】18米,18米,36米:18+18=36(米),两边之和等于第三边，不能围成三角形;18米,36米,36米:18+36=54(米),54\u003e36,36-18=18(米),18\u003c36，能围成三角形。18+36+36=90(米)。所以这个等腰三角形的周长是90米。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的周长",
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "162",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在一个等腰三角形中，其中一个角是100°,另外两个角都是( )°,这个三角形也叫做( )三角形(填\u0027锐角\u0027\u0027直角\u0027或\u0027钝角\u0027)。",
+        "options":  [
+
+                    ],
+        "answer":  "40;钝角",
+        "solution":  "【分析】等腰三角形的两个底角相等，三角形内角和等于180°,所以100°的角不是底角，是顶角,180°减去100°等于两个底角的度数和，再除以2即等于底角的度数;由于100°的角是钝角，所以这个三角形也是钝角三角形;据此即可解答.\n【详解】(180°-100°)÷2=80°/2=40°,所以，在一个等腰三角形中，其中一个角是100°,另外两个角都是40°,这个三角形也叫做钝角三角形。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·贵州黔东南·期末"
+    },
+    {
+        "id":  "163",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "三角形的内角和是( )°,等边三角形的每个角都是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "180;60",
+        "solution":  "【分析】等边三角形的三条边相等，三个内角也相等，三角形的内角和是180°,用三角形的内角和除以3，就是等边三角形每个内角的度数.\n【详解】180°/3=60°,三角形的内角和是180°,等边三角形的每个角都是60°。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·河北张家口·期末"
+    },
+    {
+        "id":  "164",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在一个三角形中，若∠1=72°,∠2=48°,那么∠3=( )°,是一个( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "60;锐角",
+        "solution":  "【分析】根据三角形的三个内角之和为180°。已知两个角的度数，用180°减去这两个角的度数即可求出第三个角的度数。再根据三个角的大小判断三角形的类型：三个角都是锐角的三角形是锐角三角形;有一个角是直角的三角形是直角三角形;有一个角是钝角的三角形是钝角三角形。据此解答.\n【详解】180°-72°-48°=60°,在一个三角形中，若∠1=72°,∠2=48°,那么∠3=60°,是一个锐角三角形。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·广东东莞·期末"
+    },
+    {
+        "id":  "165",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个三角形的两个内角分别是35°和55°,这个三角形是( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "直角",
+        "solution":  "【分析】三角形的内角和为180°,已知两个内角分别是35°和55°,内角和180°减去这两个角得到另一个角。根据三个角的度数判断是什么三角形.\n【详解】180°-35°-55°=90°,根据\u0027有一个角为90°的三角形是直角三角形\u0027得这是一个直角三角形。所以这个三角形是直角三角形。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·黑龙江佳木斯·期末"
+    },
+    {
+        "id":  "166",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "等腰三角形两边分别长18dm和20dm,这个三角形的周长是( )dm或( )dm。",
+        "options":  [
+
+                    ],
+        "answer":  "56;58",
+        "solution":  "【分析】根据题意，明确等腰三角形的两腰相等，如果腰长18分米，三条边为18分米,18分米,20分米。根据三角形的三边关系(任意两边之和大于第三边)进行验证，再计算周长;如果腰长20分米，三条边为18分米,20分米,20分米。根据三角形的三边关系(任意两边之和大于第三边)进行验证，再计算周长;以此答题即可.\n【详解】根据分析可知:18+18=36\u003e20,18+20=38\u003e18,18分米,18分米,20分米三边能组成三角形。18+18+20=56(分米);20+20=40\u003e18,18+20=38\u003e20,18分米,20分米,20分米三边能组成三角形。18+20+20=58(分米)。所以这个三角形的周长是56dm或58dm。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·辽宁鞍山·期末"
+    },
+    {
+        "id":  "167",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明剪出了一个直角三角形纸片，这个三角形中一个锐角是29°,另一个锐角是( )°。他再把这个直角三角形纸片的直角剪去，得到一个四边形，这个四边形内角和是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "61;360",
+        "solution":  "【分析】根据题意，这个三角形是直角三角形，则有一个角是90°,三角形内角和是180°,用180°减90°再减去其中一个锐角的度数，即可求出另一个锐角是多少度;所有的四边形都可以分成两个三角形，用180°*2即可求出四边形的内角和是多少度.\n【详解】180°-90°-29°=61°,180°*2=360°。所以另一个锐角是61°,这个四边形内角和是360°。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "三角形的内角和",
+                                 "多边形的内角和"
+                             ],
+        "source":  "24-25四年级下·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "168",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "上饶美食果木炭烤鸭的包装盒是一个等腰三角形，它的顶角是50°,每个底角是( )°。按角分，它是( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "65;锐角",
+        "solution":  "【分析】因为三角形的内角和是180°,等腰三角形的顶角是50°,等腰三角形的两个底角相等，用180°减去顶角的度数求出两个底角的度数，再除以2得出每个底角的度数;如果一个三角形有一个角是钝角，则是钝角三角形，有一个角是直角，则是直角三角形，如果三个角都是锐角，则是锐角三角形，再根据三角形按照角的大小分类情况进行判断即可.\n【详解】(180°-50°)÷2=130°/2=65°,这个等腰三角形的三个角分别是50°,65°,65°,三个角都小于90°,都是锐角，所以这个等腰三角形是锐角三角形。所以，每个底角是65°,按角分，它是锐角三角形。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·江西上饶·期末"
+    },
+    {
+        "id":  "169",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果一个三角形的两条边分别是30cm和40cm,这个三角形的第三条边最长是( )厘米，最短是( )厘米。(取整厘米数)",
+        "options":  [
+
+                    ],
+        "answer":  "69;11",
+        "solution":  "【分析】因为第三边长度是整厘米数，根据两边之和大于第三边以及两边之差小于第三边，由此可求出第三条边的长度范围，进而得出第三条边最短和最长是多少厘米.\n【详解】30+40=70(厘米),40-30=10(厘米),10厘米\u003c第三条边的长度\u003c70厘米，最短为10+1=11(厘米),最长为70-1=69(厘米)。所以第三条边最长是69厘米，最短是11厘米。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·海南三亚·期末"
+    },
+    {
+        "id":  "170",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "\"儿童散学归来早，忙趁东风放纸鸢.\"风筝是中国古代劳动人民发明的。在风筝节上，有一个等腰三角形的风筝，其中两条边分别长6米和4米，这个等腰三角形风筝的周长是( )米或( )米。",
+        "options":  [
+
+                    ],
+        "answer":  "16;14",
+        "solution":  "【分析】等腰三角形有两条边长相等，根据三角形三边关系，两边之和大于第三边，两边之差小于第三边，先判断三条边的长度，相加即为周长，据此解答.\n【详解】如果腰为6米，底边为4米:4+6\u003e6，可以构成三角形，此时三角形的周长是:6+6+4=16(米);如果腰为4米，底边为6米:4+4\u003e6，可以构成三角形，此时三角形的周长是:4+4+6=14(米)。所以这个等腰三角形风筝的周长是16米或14米。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的周长",
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·海南三亚·期末"
+    },
+    {
+        "id":  "171",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等边三角形的边长是5厘米，它的周长是( )厘米，它的内角和是( )度。",
+        "options":  [
+
+                    ],
+        "answer":  "15;180度",
+        "solution":  "【分析】根据题意，明确等边三角形的3条边长相等,3个角都是60度，已知一个等边三角形的边长是5厘米，用5乘3，就是等边三角形的周长;三角形的内角和就是60°乘3，列式计算即可.\n【详解】根据分析可知:5×3=15(厘米),60°*3=180°。所以周长是15厘米，内角和是180度。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和",
+                                 "三角形的周长"
+                             ],
+        "source":  "24-25四年级下·内蒙古赤峰·期末"
+    },
+    {
+        "id":  "172",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在三角形ABC中，如果∠A=∠B=∠C,按边分类，三角形ABC是一个( )三角形;如果∠A=∠B+∠C,按角分类，三角形ABC是一个( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "等边;直角",
+        "solution":  "【分析】三角形的内角和是180°,三个角都相等，说明每个角都是60°,三个角都是60°的三角形是等边三角形。180°/2=90°,直角三角形的直角等于剩余两个内角度数的和。据此解答.\n【详解】三个角都相等的三角形是等边三角形，按边分类，三角形ABC是一个等边三角形。如果∠A=∠B+∠C,按角分类，三角形ABC是一个直角三角形。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·浙江杭州·期末"
+    },
+    {
+        "id":  "173",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三角形的三条边长度都是整厘米数，其中的两条边长度分别是7cm和21cm,第三条边最长是( )cm,最短是( )cm。",
+        "options":  [
+
+                    ],
+        "answer":  "27;15",
+        "solution":  "【分析】根据三角形三边之间的关系可知，第三条边小于另外两条边之和，大于另外两条边之差，求出已知两条边的和与差即可知第三条边最长是多少cm,最短是多少cm;据此即可解答.\n【详解】7+21=28(cm),28-1=27(cm),21-7=14(cm),14+1=15(cm),第三条边大于14cm,小于28cm,所以第三条边最长是27cm,最短是15cm。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·河南濮阳·期末"
+    },
+    {
+        "id":  "174",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "直角三角形的一个锐角是53°,另一个锐角是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "37",
+        "solution":  "【分析】三角形的内角和为180°,直角三角形的一个角为90°,已知一个锐角是53°,用180°-90°-53°即可求得另一个锐角的度数.\n【详解】180°-90°-53°=37°,所以另一个锐角是37°。",
+        "knowledge_points":  [
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·陕西西安·期末"
+    },
+    {
+        "id":  "175",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "室外空调主机的支架做成三角形，是因为三角形具有( )性;正三角形的每一个内角是( )度。",
+        "options":  [
+
+                    ],
+        "answer":  "稳定;60",
+        "solution":  "【分析】三角形不易变形，具有稳定性，例如自行车的三脚架，篮球架，相机三脚架，屋顶，斜拉桥等都是利用了三角形稳定性，根据正三角形概念：等边三角形(又称正三角形),为三边相等的三角形，其三个内角相等，均为60°,它是锐角三角形的一种。据此解答即可.\n【详解】室外空调主机的支架做成三角形，是因为三角形具有稳定性;正三角形的每一个内角是60度。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和",
+                                 "三角形的稳定性及应用"
+                             ],
+        "source":  "24-25四年级下·陕西西安·期末"
+    },
+    {
+        "id":  "176",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "三角形的内角度数和是( )°。一个多边形的内角度数和是720°,该多边形是一个( )边形。",
+        "options":  [
+
+                    ],
+        "answer":  "180°;六",
+        "solution":  "【分析】三角形的内角度数和是180°;一个多边形的内角和是720°,720÷180=4，说明这个多边形可以分成4个三角形，所以这个多边形是六边形。如图:![](images/a10a9c7e936d5caa018338fbf1c5373f6922f60fd923388ae77e12ef2f9ce189.jpg)\n【详解】三角形的内角度数和是180°;720°/180°=4，说明这个多边形可以分成4个三角形,4+2=6，所以该多边形是个六边形。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "多边形的内角和"
+                             ],
+        "source":  "24-25四年级下·云南楚雄·期末"
+    },
+    {
+        "id":  "177",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形的三条边长都是整厘米数，其中两条边长都是9cm,这个三角形的第三条边长最短是( )cm,最长是( )cm。",
+        "options":  [
+
+                    ],
+        "answer":  "1;17",
+        "solution":  "【分析】根据题意，明确三角形两边之和大于第三边，两边之差小于第三边，据此即可解答.\n【详解】根据分析可知:9-9=0(cm),9+9=18(cm),所以第三边大于0，小于18，那么这个三角形的第三条边长最短是1cm,最长是17cm。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·云南楚雄·期末"
+    },
+    {
+        "id":  "178",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在一个三角形中，如果最小的角的度数是45°,那么这个三角形可能是( )三角形或者( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "等腰直角;锐角",
+        "solution":  "【分析】根据题意可推导出其他两角必须均大于或等于45°,且三个角之和为180°,另外两个角和为135°。由此可以分情况先判断出三角形的三个角是什么角，再根据三角形的分类知识判断是什么三角形，据此作答.\n【详解】当其中一个角为45°时，另一个角为90°,此时三角形为等腰直角三角形;当其中一个角大于45°时，另一个角小于90°,此时三角形为锐角三角形;因此符合条件的三角形可能是等腰直角三角形或者锐角三角形。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·云南楚雄·期末"
+    },
+    {
+        "id":  "179",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形的顶角是40°,它的一个底角是( )°;直角三角形中，一个锐角是45°,另一个锐角是( )°;△OBM是等腰三角形,∠O=∠M=35°,那么∠B=( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "70;45;110",
+        "solution":  "【分析】三角形内角和是180°,等腰三角形两底角相等，内角和为180°,顶角已知为40°,则底角为(180°-40°)÷2;直角三角形两锐角之和为90°,已知一锐角为45°,另一锐角为90°-45°;△OBM中,∠O和∠M均为35°,说明这两个角是底角，顶角∠B为180°-35°*2，据此解题.\n【详解】(180°-40°)÷2=70°;90°-45°=45°;180°-35°*2=180°-70°=110°。所以一个等腰三角形的顶角是40°,它的一个底角是70°;直角三角形中，一个锐角是45°,另一个锐角是45°;∠B=110°。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·甘肃庆阳·期末"
+    },
+    {
+        "id":  "180",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "等腰三角形的周长是20厘米，一条边的长是6厘米，则其它两边的长可能是( )和( )或( )和( )。",
+        "options":  [
+
+                    ],
+        "answer":  "6厘米;8厘米;7厘米;7厘米",
+        "solution":  "【分析】根据等腰三角形的性质(两腰相等)和三角形三边关系(任意两边之和大于第三边),分别讨论当6厘米的边是腰时和当6厘米的边是底边时两种情况，进而求出其他两边的长度.\n【详解】情况一：当6厘米的边是腰时，则另一条腰的长度也是6厘米。20-6-6=8(厘米),6+6=12\u003e8,6+8=14\u003e6，满足三角形三边关系，此情况成立。情况二：当6厘米的边是底边时，腰长为:(20-6)÷2=7(厘米),则两条腰的长度均为7厘米。7+7=14\u003e6,7+6=13\u003e7，满足三角形三边关系，此情况成立。所以其它两边的长可能是6厘米和8厘米或7厘米和7厘米。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·甘肃庆阳·期末"
+    },
+    {
+        "id":  "181",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明观察一个三角形，下面是一些他的研究记录.\n(1)它的三个角很特别,∠1+∠2=∠3，由此可知，这是一个( )三角形.\n(2)以(1)中这个三角形的最长边作为底画高，沿着高剪开得到两个完全相同的三角形。由此可知，这个三角形是( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "直角;等腰直角",
+        "solution":  "【分析】(1)三角形内角和是180°,∠1+∠2=∠3，则用180°/2即可求出∠3的度数，根据有一个角是直角的三角形是直角三角形，有一个角是钝角的三角形是钝角三角形，三个角都是锐角的三角形是锐角三角形，据此判断这个三角形是什么三角形即可。(2)根据题意，这个三角形的最长边作为底画高，则在直角的对边上画高，沿着高剪开得到两个完全相同的三角形，首先高垂直于底边，则分成的两个三角形一定是直角三角形，因为两个三角形完全相同，则大三角形的两条直角边一样长，这是一个等腰直角三角形，据此填空即可.\n【详解】(1)∠3:180°/2=90°,它的三个角很特别,∠1+∠2=∠3，由此可知，这是一个直角三角形。(2)如图:![](images/e5be395da831b6bef9124b2abffb141c423aed9fb30b28c2ffcaeee766a44979.jpg)以(1)中这个三角形的最长边作为底画高，沿着高剪开得到两个完全相同的三角形。由此可知，这个三角形是等腰直角三角形。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和",
+                                 "三角形的高及画法"
+                             ],
+        "source":  "24-25四年级下·广东广州·期末"
+    },
+    {
+        "id":  "182",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.4,
+        "question":  "把一根长24米木头锯成3段(每段都是整米数),围成一个三角形，有( )种不同的围法。",
+        "options":  [
+
+                    ],
+        "answer":  "12",
+        "solution":  "【分析】根据三角形的三边关系可知，最长边应小于这根木头长度的一半，即小于12米，当最长边为11米时，其余两条边的长度和是13米，可以是11米和2米,10米和3米,9米和4米,8米和5米,7米和6米，共5种围法;当最长边是10米时，其余两条边的长度和是14米，可以是10米和4米,9米和5米,8米和6米,7米和7米，共4种围法;当最长边为9米时，其余两条边的长度和是15米，可以是9米和6米,8米和7米，共2种围法;当最长边为8米时，其余两条边的长度和是16米，可以是8米和8米，共1种围法。据此解答.\n【详解】据分析可知:5+4+2+1=12(种),即将一根长24米的木头锯成3段(每段都是整米数),围成一个三角形，有12种不同的围法。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·甘肃平凉·期末"
+    },
+    {
+        "id":  "183",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在三角形ABC中,∠A=45°,∠B=90°,∠C=( )°,它是( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "45;等腰直角",
+        "solution":  "【分析】根据三角形的内角和为180°,可以用三角形内角和减去已知的两个角的度数，得到∠C的度数。再根据有个角是直角(即90°)的三角形是直角三角形，有两个角相等的三角形是等腰三角形对三角形ABC的分类进行判断.\n【详解】∠C=180°-∠A-∠B=180°-45°-90°=45°,因为∠B=90°,∠A=∠C,所以三角形ABC是等腰直角三角形。所以，在三角形ABC中,∠C=45°,它是等腰直角三角形。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·甘肃平凉·期末"
+    },
+    {
+        "id":  "184",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一条红领巾的一个底角是25°,它的顶角是( )°,按角分它是一个( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "130;钝角",
+        "solution":  "【分析】因为红领巾是等腰三角形，等腰三角形的两个底角相等，已知一个底角是25°,那么另一个底角也是25°。根据三角形内角和是180°,所以顶角的度数为180°减去两个底角的度数;有一个角是钝角的三角形是钝角三角形，有一个角是直角的三角形是直角三角形，三个角都是锐角的三角形是锐角三角形。据此解答.\n【详解】180°-25°-25°=130°,因为这个三角形的顶角是130°,130°\u003e90°,根据钝角三角形的定义，有一个角是钝角的三角形是钝角三角形，所以按角分它是一个钝角三角形。所以，一条红领巾的一个底角是25°,它的顶角是130°,按角分它是一个钝角三角形。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·甘肃陇南·期末"
+    },
+    {
+        "id":  "185",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在一个直角三角形中，有一个锐角是36°,另一个锐角是( )°;在一个等腰三角形中，其中两条边的长分别是5cm,2cm,则这个等腰三角形的周长是( )cm。",
+        "options":  [
+
+                    ],
+        "answer":  "54;12",
+        "solution":  "【分析】直角三角形锐角计算：在直角三角形中，直角为90°,两个锐角之和为90°,已知一个锐角是36°,据此计算另一个锐角。等腰三角形周长计算：等腰三角形有两条边相等，需判断哪两条边相等。若两条5cm边相等，第三边为2cm,满足三角形三边关系(5+5\u003e2,5+2\u003e5);若两条2cm边相等，第三边为5cm,不满足三边关系(2+2\u003c5)。因此三边分别为5cm,5cm和2cm,据此计算周长.\n【详解】(1)另一个锐角为90°-36°=54°。(2)若两条相等的边为5cm,第三边为2cm,验证三边关系:5+5\u003e2(成立),周长为:5+5+2=12(cm);若两条相等的边为2cm,第三边为5cm,验证三边关系:2+2\u003c5(不成立),无法构成三角形。因此最终周长为12cm。即另一个锐角是54°,这个等腰三角形的周长是12cm。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和",
+                                 "三角形的周长",
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·甘肃定西·期末"
+    },
+    {
+        "id":  "186",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如下图，用6个相同的直角三角形拼成一个大长方形，这个长方形的长是( )厘米，每个直角三角形的面积是( )平方厘米.\n![](images/e51c8b78712ebfaf8c0ad51c9635f02c9a3c9a0d779b9ae6608df4a9d4677b1b.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "24;64",
+        "solution":  "【分析】这个长方形的长=直角三角形较长的直角边长度+较短的直角边长度;每个直角三角形的面积=大长方形的长×宽/直角三角形的个数.\n【详解】长方形的长:16+8=24(厘米);24×16÷6=384÷6=64(平方厘米)。这个长方形的长是24厘米，每个直角三角形的面积是64平方厘米。",
+        "knowledge_points":  [
+                                 "长方形的面积",
+                                 "用同样的图形拼接"
+                             ],
+        "source":  "24-25四年级下·浙江宁波·期末"
+    },
+    {
+        "id":  "187",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "用一根铁丝围成一个等腰三角形，相邻的两条边分别长4厘米和9厘米，围成这个三角形需要( )厘米长的铁丝。",
+        "options":  [
+
+                    ],
+        "answer":  "22",
+        "solution":  "【分析】等腰三角形的两条腰相等，则第三条边可能是4厘米或者9厘米。根据三角形的三边关系：三角形的两边之和大于第三边，三角形的两边之差小于第三边，明确第三条边的长度，再将三条边的长度相加，即可求出这根铁丝的长度，据此作答.\n【详解】第三边:4厘米,4+4=8(厘米),8厘米\u003c9厘米，不能围成三角形。第三边:9厘米,4+9=13(厘米),9-4=5(厘米),13厘米\u003e9厘米\u003e5厘米，能围成等腰三角形。周长:4+9+9=22(厘米)。所以围成这个三角形需要22厘米长的铁丝。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的周长",
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·贵州黔西·期末"
+    },
+    {
+        "id":  "188",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "轩轩从3cm,4cm,7cm,10cm的小棒中选出3根，拼成一个三角形，这个三角形的周长是( )cm。",
+        "options":  [
+
+                    ],
+        "answer":  "21",
+        "solution":  "【分析】根据三角形三边之间的关系：三角形任意两边之和大于第三边，任意两边之差小于第三边，只能选择4cm,7cm,10cm这三根小棒，根据三角形的周长就是三条边的总和，即把4cm,7cm,10cm加起来，即可算出这个三角形的周长。据此解答.\n【详解】选择4cm,7cm,10cm,4+7\u003e10，能拼成三角形，那么这个三角形周长为:4+7+10=21(cm)。所以这个三角形的周长是21cm。",
+        "knowledge_points":  [
+                                 "三角形的周长",
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·辽宁盘锦·期末"
+    },
+    {
+        "id":  "189",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "科学家们通过激光测距法测量出月球到地球的平均距离约384400千米，改写成用\u0027万\u0027作单位是( )万千米，精确到十分位是( )万千米。嫦娥六号探测器测得某陨石坑深度四舍五入后是5.7米，这个两位小数最大是( )米，最小是( )米。一个装有等腰三角形的太阳能板支架探测器，其三角形底角为38°,顶角是( )°,按角分它属于( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "38.44;38.4;5.74;5.65;104;钝角",
+        "solution":  "【分析】把一个非整万数改写成用\u0027万\u0027作单位的数，只需要找到万位，在万位数的右下角点上小数点，然后把小数末尾的0去掉，最后在数的后面写上\u0027万\u0027字;一个小数，精确到十分位或保留一位小数时，需要看百分位上数的大小，用\u0027四舍五入\u0027法求它的近似数;三角形的内角和为180°。等腰三角形的两个底角的度数相等。由题意得，等腰三角形的底角为38°,直接用180°减去两个底角的度数即可算出顶角的度数。然后根据三个角的大小来判断三角形的类型即可.\n【详解】384400=38.44万,38.44的百分位上是4，需要舍去，所以38.44≈38.4，即38.44万≈38.4万;一个两位小数四舍五入后是5.7，要使这个小数最大，说明原数\u0027四舍\u0027后得到5.7，那么百分位上的数必须小于5，百分位上最大是4，即这个小数最大是5.74;要使这个小数最小，说明原数\u0027五入\u0027后得到5.7，那么百分位上的数必须大于或等于5，百分位上最小是5，十分位上是6，个位上是5，即这个小数最小是5.65.180°-38°-38°=104°,104°的角是钝角，所以按角分它属于钝角三角形。所以月球到地球的平均距离约384400千米，改写成用\u0027万\u0027作单位是38.44万千米，精确到十分位是38.4万千米。陨石坑深度这个两位小数最大是5.74米，最小是5.65米。顶角是104°,按角分它属于钝角三角形。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和",
+                                 "小数的改写",
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25四年级下·山东日照·期末"
+    },
+    {
+        "id":  "190",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "乐乐测量一个等腰三角形中的一个底角是45°,那么按角分这是一个( )三角形;乐乐用一根铁丝刚好围成一个等腰三角形，测量其中的两条边，分别长8cm,20cm,算一算这根铁丝长( )cm。",
+        "options":  [
+
+                    ],
+        "answer":  "直角;48",
+        "solution":  "【分析】等腰三角形的底角相等，所以另一个底角也是45°,根据三角形内角和为180°,用180°减去两个底角的度数可以得出顶角的度数，等于90°的是直角，小于90°的是锐角，大于90°小于180°的为钝角，三角形按角分类：三个角都是锐角的三角形，有一个角是直角的三角形，有一个角是钝角的三角形，据此判断;等腰三角形两条腰相等，如果腰为8cm,则三边长度分别为:8cm,8cm,20cm,如果腰为20cm,则三边长度分别为:8cm,20cm,20cm,再根据三角形三边关系：两边之和大于第三边判断这两种情况是否能围成三角形.\n【详解】180°-45°-45°=90°,90°是直角，所以这个三角形按角分是直角三角形;如果腰为8cm,则三边长度分别为:8cm,8cm,20cm,8+8\u003c20，所以不能围成三角形;如果腰为20cm,则三边长度分别为:8cm,20cm,20cm,8+20\u003e20，可以围成三角形,8+20+20=48(cm),所以这根铁丝长48cm。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和",
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·河南洛阳·期末"
+    },
+    {
+        "id":  "191",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "三角形ABC中,∠A=45°,∠B=55°,∠C=( )°,按角分这是一个( )角三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "80;锐角",
+        "solution":  "【分析】三角形内角和是180°,用180°减去∠A和∠B的度数，即可求出∠C的度数;有一个角是钝角的三角形是钝角三角形，有一个角是直角的三角形是直角三角形，三个角都是锐角的三角形是锐角三角形，据此判断这个三角形是什么三角形即可.\n【详解】∠C=180°-∠A-∠B=180°-45°-55°=80°,三角形ABC中,∠A=45°,∠B=55°,∠C=80°,按角分这是一个锐角三角形。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·甘肃兰州·期末"
+    },
+    {
+        "id":  "192",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形其中两条边分别长5厘米,11厘米，第三条边长( )厘米;这个三角形的一个底角是77°,按角分，它属于( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "11;锐角",
+        "solution":  "【分析】等腰三角形的两条腰长度相等。由题意得，等腰三角形的两条边分别长5厘米和11厘米，可以假设5厘米或11厘米长的边为腰，然后利用三角形三边的关系(任意两边之和大于第三边)来验证假设是否成立;三角形的内角和为180°。等腰三角形的两个底角的度数相等。由题意得，等腰三角形的一个底角是77°,那么另一个底角也是77°,直接用180°减去两个底角的度数即可算出顶角的度数。然后根据三角形三个角的大小来判断三角形的类型即可.\n【详解】假设5厘米长的边为腰，那么另一条腰的长度也是5厘米。5+5=10(厘米),10厘米\u003c11厘米，即这三边无法围成三角形。这种假设不成立。假设11厘米长的边为腰，那么另一条腰的长度也是11厘米。5+11=16(厘米),16厘米\u003e11厘米，即这三边可以围成三角形，这种假设成立。所以等腰三角形的第三条边长11厘米。180°-77°-77°=26°,这个三角形的三个角都是锐角，所以这是一个锐角三角形。所以第三条边长11厘米，按角分它属于锐角三角形。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和",
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·河南三门峡·期末"
+    },
+    {
+        "id":  "193",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小海做了一个等腰三角形的风筝，其中的两条边长度分别是25厘米和50厘米，这个三角形的周长是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "125",
+        "solution":  "【分析】需根据等腰三角形的性质和三角形三边关系来确定三角形的边长，进而计算周长。等腰三角形性质：两条边长度相等，这两条相等的边称为腰，另一条边称为底边。已知两条边分别为25厘米和50厘米，需分两种情况讨论(25厘米为腰或50厘米为腰),再结合三边关系判断哪种情况成立，最后计算周长.\n【详解】已知等腰三角形的两条边分别为25厘米和50厘米，分两种情况分析：情况1：假设25厘米为腰,50厘米为底边，此时三条边的长度分别为25厘米,25厘米,50厘米。25+25=50(厘米),两边之和等于第三边，不符合\u0027任意两边之和大于第三边\u0027的条件，无法构成三角形，此情况不成立。情况2：假设50厘米为腰,25厘米为底边，此时三条边的长度分别为50厘米,50厘米,25厘米。50+50=100(厘米),100\u003e25;50+25=75(厘米),75\u003e50，满足任意两边之和大于第三边，可以构成三角形，此情况成立。该等腰三角形的三条边分别为50厘米,50厘米,25厘米，周长为三条边长度之和:50+50+25=125(厘米)。所以这个三角形的周长是125厘米。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的周长",
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·广东广州·期末"
+    },
+    {
+        "id":  "194",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "直角三角形的一个锐角是28°,则另一个锐角是( )°;把这样两个完全一样的三角形拼成一个大三角形，大三角形的内角和是( )°;如果拼成一个平行四边形，平行四边形的内角和是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "62;180;360",
+        "solution":  "【分析】在直角三角形中，两个锐角的度数和是90°,所以用90°减去28°就是另一个锐角的度数，三角形内角和为180°,不论几个三角形拼成一个大三角形，这个大三角形的内角和都是180°,不论几个三角形拼成一个平行四边形，因为平行四边形是四边形，所有四边形的内角和都是360°,据此解答即可.\n【详解】90°-28°=62°,所以另一个锐角是62°;把这样两个完全一样的三角形拼成一个大三角形，大三角形的内角和是180°;如果拼成一个平行四边形，平行四边形的内角和是360°。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "三角形的内角和",
+                                 "多边形的内角和"
+                             ],
+        "source":  "24-25四年级下·河南郑州·期末"
+    },
+    {
+        "id":  "195",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "直角三角形中，一个锐角是47°,另一个锐角是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "43",
+        "solution":  "【分析】有一个角是直角的三角形是直角三角形,1个直角=90°,三角形内角和是180°,用180°减去直角和一个锐角的度数，即可求出另一个锐角的度数。据此解答.\n【详解】180°-90°-47°=43°。所以另一个锐角是43°。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·甘肃天水·期末"
+    },
+    {
+        "id":  "196",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "林林用三根小棒围成一个三角形，其中两条小棒的长度分别是15厘米和10厘米，围成的三角形的周长最少是( )厘米，最长是( )厘米。(小棒长度均为整厘米数)",
+        "options":  [
+
+                    ],
+        "answer":  "31;49",
+        "solution":  "【分析】三角形三边的关系：任意两边之和大于第三边，任意两边之差小于第三边。由题意得，两条小棒的长度分别是15厘米和10厘米，即三角形的两边长度分别为15厘米和10厘米，可以根据三角形三边的关系来推算出第三条边(小棒)的长度。然后把三条边的长度全部加起来即可算出三角形的周长.\n【详解】两边之差\u003c第三条边的长度\u003c两边之和,15-10\u003c第三条边的长度\u003c15+10,5\u003c第三条边的长度\u003c25，小棒的长度为整厘米数，所以第三根小棒最短是6厘米，最长是24厘米。周长最短为:15+10+6=31(厘米),周长最长为:15+10+24=49(厘米)。故围成的三角形的周长最少是31厘米，最长是49厘米。",
+        "knowledge_points":  [
+                                 "三角形的周长",
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "197",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "等腰三角形的一个底角是40°,它另外的两个角分别是( )°和( )°,这个三角形又叫( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "40;100;钝角",
+        "solution":  "【分析】根据题意，等腰三角形的底角相等，说明等腰三角形中有两个角都为40°,再根据三角形的内角和是180°,用180°减去两个40°角，从而可以求出顶角的度数，再根据三个角的度数，三个角都是锐角的三角形叫做锐角三角形，有一个角为直角的三角形叫做直角三角形，有一个角是钝角的三角形叫做钝角三角形，即可判定这个三角形的类别.\n【详解】180°-40°-40°=100°,所以它另外的两个角分别是40°和100°,这个三角形又叫钝角三角形。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "198",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果一个三角形中有两个锐角度数加起来的和是89°,那么这个三角形是( )三角形。如果把完全相同的两个这样的三角形拼成一个四边形，这个四边形的内角和是( )度。",
+        "options":  [
+
+                    ],
+        "answer":  "钝角;360",
+        "solution":  "【分析】根据题意，三角形内角和为180°,已知两个锐角的和为89°,可用180°减去89°,求出第三个角的度数;按角分类：三角形按角分为锐角三角形(三个角均小于90°),直角三角形(一个角为90°)和钝角三角形(一个角大于90°)。判断三角形即可。无论用两个完全相同的三角形如何拼接，所得四边形的内角和均为两个180°相加。列式计算即可.\n【详解】根据分析可知:180°-89°=91°,91°\u003e90°,所以这个三角形是钝角三角形。180°*2=360°,所以这个四边形的内角和是360度。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·重庆·期末"
+    },
+    {
+        "id":  "199",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "五边形和六边形的内角和分别为( )°和( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "540;720",
+        "solution":  "【分析】计算多边形的内角和时，可以把这个多边形分成多个三角形，然后利用三角形的内角和进行计算;一个n边形可以分成(n-2)个三角形，所以n边形的内角和=(n-2)×180°,据此解答即可.\n【详解】(5-2)×180°=3×180°=540°;(6-2)×180°=4×180°=720°。所以五边形和六边形的内角和分别为540°和720°。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "多边形的内角和"
+                             ],
+        "source":  "24-25四年级下·新疆喀什·期末"
+    },
+    {
+        "id":  "200",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "三角形的三条边都是整厘米数，其中两条边分别是5厘米和8厘米，另一条边最短可以是( )厘米，最长可以是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "4;12",
+        "solution":  "【分析】根据题意，三角形的三边关系为三角形的两边之和大于第三边，三角形的两边之差一定小于第三边;据此解答即可.\n【详解】5+8=13(厘米),13-1=12(厘米),8-5=3(厘米),3+1=4(厘米)。所以另一条边最短可以是4厘米，最长可以是12厘米。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "201",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明要做一个等腰三角形的风筝骨架，他已经准备了一根6dm的竹签，还要从6dm,15dm,12dm,9dm中选( )dm和( )dm的竹签做骨架，这样选的理由是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "6;9;这两根与已有的6dm组成等腰三角形，且满足三角形三边关系",
+        "solution":  "【分析】根据等腰三角形的定义，需有两条边相等。小明已有1根6dm的竹签，若再选一根6dm,则两条边均为6dm,第三条边需满足三角形三边关系(任意两边之和大于第三边)。验证各选项后发现，只有选6dm和9dm时，三边为6dm,6dm,9dm,满足6+6\u003e9，且符合等腰三角形条件.\n【详解】等腰三角形条件：必须有两条边相等。小明已有1根6dm的竹签，因此需再选一根6dm,使两条边均为6dm,第三条边为另一根长度。验证三边关系：若选6dm和15dm,三边为6dm,6dm,15dm,6+6=12\u003c15，不满足三角形三边关系。若选6dm和12dm,三边为6dm,6dm,12dm,6+6=12，等于第三边，不满足三角形三边关系。若选6dm和9dm,三边为6dm,6dm,9dm,6+6=12\u003e9，且6+9\u003e6,9+6\u003e6，满足三角形三边关系。所以小明要做一个等腰三角形的风筝骨架，他已经准备了一根6dm的竹签，还要从6dm,15dm,12dm,9dm中选6dm和9dm的竹签做骨架，这样选的理由是这两根与已有的6dm组成等腰三角形，且满足三角形三边关系。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "202",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形的底角是42°,它的顶角是( )°,按角分类这个三角形是( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "96;钝角",
+        "solution":  "【分析】三角形的内角和是180°;在等腰三角形中,2个底角是相等的，用180°减去2个42°就是等腰三角形的顶角的度数;根据三角形的分类可知，一个内角为钝角的三角形是钝角三角形，三个内角都为锐角的三角形是锐角三角形，一个内角为直角的三角形是直角三角形.\n【详解】顶角:180°-42°*2=180°-84°=96°,96°\u003e90°,即最大的角是钝角。所以一个等腰三角形的底角是42°,它的顶角是96°,按角分类这个三角形是钝角三角形。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·安徽芜湖·期末"
+    },
+    {
+        "id":  "203",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把一个等边三角形对折后可以得到两个同样大小的( )三角形，对折后得到的三角形的3个内角分别是( )°,( )°,( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "直角;30;60;90",
+        "solution":  "【分析】把一个等边三角形对折后图形为:![](images/2026-03-10%20223318.jpg),即得到两个同样大小的直角三角形。对折后的一个底角仍是60°,另一个底角是90°,顶角是原来顶角的一半，也就是60°的一半是30°.\n【详解】把一个等边三角形对折后可以得到两个同样大小的直角三角形，对折后得到的三角形的3个内角分别是30°,60°,90°。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·浙江绍兴·期末"
+    },
+    {
+        "id":  "204",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形的周长是30厘米，它的一条腰长是底边长的2倍，这个三角形的底边长是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】等腰三角形的两腰相等，已知它的一条腰长是底边长的2倍，那么(2+2+1)倍的底边长就是等腰三角形的周长，用等腰三角形的周长除以5，就是这个三角形底边的长度，据此解答即可.\n【详解】30÷(2+2+1)=30÷5=6(厘米)。这个三角形的底边长是6厘米。",
+        "knowledge_points":  [
+                                 "已知一个数的几倍是多少，求这个数",
+                                 "三角形的周长",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "205",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用小棒以首尾相接的方式围一个三角形，现有8厘米和6厘米两根小棒，还需要一根小棒(取整厘米数),这根小棒最长可以是( )厘米，最短可以是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "13;3",
+        "solution":  "【分析】三角形三边的关系：任意两边之和大于第三边，任意两边之差小于第三边。由题意得，三角形的两边为8厘米和6厘米，可以根据三角形三边的关系来推算第三条边(第三根小棒)的长度.\n【详解】8+6=14(厘米),8-6=2(厘米),两边之差\u003c第三根小棒\u003c两边之和，则2\u003c第三根小棒\u003c14，小棒的长度是整厘米数，所以这根小棒最长可以是13厘米，最短可以是3厘米。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "206",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "乐乐参加了一个风筝制作挑战活动，她决定制作一个三角形的风筝，并已经测量了两条边的长度，分别是5分米和7分米。为了确保风筝的稳定性和安全性，她计算出第三条边的长度至少是( )分米。(取整分米数)",
+        "options":  [
+
+                    ],
+        "answer":  "3",
+        "solution":  "【分析】根据三角形的特征：任意两边之和大于第三边，任意两边之差小于第三边;由此解答即可.\n【详解】5+7=12(分米),7-5=2(分米),2分米\u003c第三条边\u003c12分米，所以第三条边的长度至少是3分米。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·陕西宝鸡·期末"
+    },
+    {
+        "id":  "207",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形的顶角是96°,那么它的底角是( )°,这个三角形按角分是( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "42;钝角",
+        "solution":  "【分析】根据等腰三角形角的特征可知，等腰三角形的两个底角相等;用(180°-顶角度数)÷2=底角度数。三个角中如果有一个角大于90°就是钝角三角形，如果有一个角等于90°就是直角三角形，三个角都小于90°就是锐角三角形.\n【详解】(180°-96°)÷2=84°/2=42°,96°\u003e90°,所以，它的底角是42°,这个三角形按角分是钝角三角形。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "208",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小杰想用木棒为他的手工黏土制作一个三角形底座，现有5根木棒，长度分别是6cm,8cm,9cm,13cm,22cm,他将从中选择三根，有( )种选法，其中,( )cm,( )cm,( )cm长的小棒围成的三角形周长最大。",
+        "options":  [
+
+                    ],
+        "answer":  "4或四;8;9;13",
+        "solution":  "【分析】先列举出5根木棒中选取3根木棒的长度进行组合有哪些可能:①6cm,8cm,9cm,②6cm,8cm,13cm,③6cm,8cm,22cm,④8cm,9cm,13cm,⑤8cm,9cm,22cm,⑥9cm,13cm,22cm,⑦6cm,9cm,13cm,⑧6cm,9cm,22cm,⑨6cm,13cm,22cm,⑩8cm,13cm,22cm。再根据三角形三边关系\u0027任意两边之和大于第三边，任意两边之差小于第三边\u0027,对木棒长度进行组合判断。最后将符合条件的组合内的木棒长度相加即为该组合三角形周长，再进行比较即可判断哪种组合小棒围成的三角形周长最大.\n【详解】由分析得，因为6+8\u003c22，所以③不可以构成三角形;因为8+9\u003c22，所以⑤不可以构成三角形;因为9+13=22，所以⑥不可以构成三角形;因为6+9\u003c22，所以⑧不可以构成三角形;因为6+13\u003c22，所以⑨不可以构成三角形;因为8+13\u003c22，所以⑩不可以构成三角形。所以组合③⑤⑥⑧⑨⑩不可以构成三角形，组合①②④⑦可以构成三角形。所以有4种选法。①6+8+9=23(cm),②6+8+13=27(cm),④8+9+13=30(cm),⑦6+9+13=28(cm),23\u003c27\u003c28\u003c30。所以，小杰想用木棒为他的手工黏土制作一个三角形底座，现有5根木棒，长度分别是6cm,8cm,9cm,13cm,22cm,他将从中选择三根，有4种选法，其中,8cm,9cm,13cm长的小棒围成的三角形周长最大。",
+        "knowledge_points":  [
+                                 "三角形三边关系",
+                                 "三角形的周长"
+                             ],
+        "source":  "24-25四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "209",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个等边三角形的周长是18厘米，它的边长是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】等边三角形的三条边都相等，所以边长=周长/3;据此解答.\n【详解】18÷3=6(厘米)。所以它的边长是6厘米。",
+        "knowledge_points":  [
+                                 "三角形的周长",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·广东广州·期末"
+    },
+    {
+        "id":  "210",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "三角形按角分类关注的是三角形中最大角的度数，比如一个三角形中最大的角是钝角，那这个三角形就是( )三角形;其实我们也可以换个角度来判断，比如一个三角形中两个锐角和大于90°,那么这个三角形就一定是( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "钝角;锐角",
+        "solution":  "【分析】如果三角形的三个角都是锐角，说明是锐角三角形，如果有一个角是钝角，说明是钝角三角形，如果有一个角是直角，说明是直角三角形，据此解题.\n【详解】三角形按角分类关注的是三角形中最大角的度数，比如一个三角形中最大的角是钝角，那这个三角形就是钝角三角形;其实我们也可以换个角度来判断，比如一个三角形中两个锐角和大于90°,那么这个三角形就一定是锐角三角形。",
+        "knowledge_points":  [
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "211",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形的一个底角是45°,它的顶角是( )°。这个三角形也是( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "90;直角",
+        "solution":  "【分析】等腰三角形有两个底角和一个顶角，且两个底角的度数相等，已知一个底角的度数，再根据三角形的内角和是180°可知，用180°减去两个底角的度数，即可求出顶角的度数;根据三角形按照角的大小分类：如果一个三角形有一个角是钝角，则这个三角形是钝角三角形，有一个角是直角，则这个三角形是直角三角形，如果三个角都是锐角，则这个三角形是锐角三角形，据此判断这个等腰三角形是哪种三角形.\n【详解】180°-45°-45°=90°,所以它的顶角是90°;90°是直角，所以这个三角形是直角三角形。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·河南洛阳·期末"
+    },
+    {
+        "id":  "212",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "明明想用三条整分米数的木条围一个三角形相框(接头处不计),他选择了7分米和4分米的木条，那么可选择的第三根木条最短是( )分米，最长是( )分米。",
+        "options":  [
+
+                    ],
+        "answer":  "4;10",
+        "solution":  "【分析】根据三角形的三边关系，任意两边之和大于第三边，任意两边之差小于第三边;由此求出第三边的范围，再根据长度是整分米数，据此即可得出最长和最短的长度.\n【详解】7分米-4分米\u003c第三边\u003c7分米+4分米，所以3分米\u003c第三边\u003c11分米，即第三边在3分米至11分米之间，不包括3分米和11分米，第三条边最短是3+1=4(分米),最长是11-1=10(分米)。所以可选择的第三根木条最短是4分米，最长是10分米。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·山东济南·期末"
+    },
+    {
+        "id":  "213",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "婷婷折纸时，一个三角形卡纸中最小的角是46°,这是个( )三角形;一个四边形，它有互相平行的对边，有2个锐角,2个钝角，对折后左右两边可以完全重合，它是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "锐角;等腰梯形",
+        "solution":  "【分析】需要用到三角形内角和以及不同类型三角形的特点，还有四边形的边，角和平移重合的性质知识。已知三角形中最小的角是46°,那么另外两个角一定大于或等于46°。假设第二小的角是46°,求出最大的角，再判断三角形的形状;因为四边形有互相平行的对边，所以它可能是平行四边形，梯形等。又因为它有2个锐角,2个钝角，且对折后左右两边可以完全重合，符合等腰梯形的特征.\n【详解】180°-46°-46°=88°,因为这个三角形的三个角都小于90°,所以这个三角形是锐角三角形。有一组对边平行，有2个锐角和2个钝角，且对折后左右两边可以完全重合，符合等腰梯形的特征。所以，婷婷折纸时，一个三角形卡纸中最小的角是46°,这是个锐角三角形;一个四边形，它有互相平行的对边，有2个锐角,2个钝角，对折后左右两边可以完全重合，它是等腰梯形。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "三角形的内角和",
+                                 "梯形的概念及特点",
+                                 "直角梯形和等腰梯形的概念及特点"
+                             ],
+        "source":  "24-25四年级下·山东济南·期末"
+    },
+    {
+        "id":  "214",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "西西想用m米长的竹条做一个等腰三角形的风筝框架，这个风筝的一条腰长n米，那么这个风筝底边长( )米。如果这个风筝的顶角是112°,那它的一个底角是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "m-2n;34",
+        "solution":  "【分析】等腰三角形的两腰相等，因此用竹条长(等腰三角形的周长)减2个腰长，就是底边的长度;等腰三角形的两个底角相等，内角和是180°,因此用180°减112°,就是两个底角的度数，再除以2，就是一个底角的度数.\n【详解】(180°-112°)÷2=68°/2=34°。所以这个风筝底边长(m-2n)米，它的一个底角是34°。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和",
+                                 "三角形的周长",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·山东济南·期末"
+    },
+    {
+        "id":  "215",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "爸爸给小红买了一个等腰三角形的风筝。风筝的一个底角是70°,风筝的顶角是( )°,按角分这是一个( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "40;锐角",
+        "solution":  "【分析】等腰三角形两个底角相等，三角形内角和为180°,用180°减去两个底角的度数，即可求出顶角是多少度;根据三角形的分类，有一个角是直角的三角形是直角三角形，有一个角是钝角的三角形是钝角三角形，三个角都是锐角的三角形是锐角三角形，据此判断是什么三角形即可.\n【详解】180°-70°-70°=40°,40°,70°,70°三个角都是锐角。所以风筝的顶角是40°,按角分这是一个锐角三角形。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·辽宁鞍山·期末"
+    },
+    {
+        "id":  "216",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "每一个三角形都有( )条高，三角形具有( )性。",
+        "options":  [
+
+                    ],
+        "answer":  "3或三;稳定",
+        "solution":  "【分析】根据题意，三角形有三个顶点和三条边。从每个顶点向它的对边作一条垂线，这条垂线就是三角形的一条高。因此，每个三角形都有3条高。三角形在结构上具有稳定性。这意味着当三角形受到外力时，它的形状不易变形。以此答题即可.\n【详解】根据分析可知：每一个三角形都有3条高，三角形具有稳定性。",
+        "knowledge_points":  [
+                                 "三角形的高及画法",
+                                 "三角形的稳定性及应用"
+                             ],
+        "source":  "24-25四年级下·陕西延安·期末"
+    },
+    {
+        "id":  "217",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在一个等腰三角形中，有两条边的长度分别是5cm和12cm,那么第三条边的长度是( )cm。",
+        "options":  [
+
+                    ],
+        "answer":  "12",
+        "solution":  "【分析】根据题意，三角形的三条边长度关系，任意两边之和大于第三边，任意两边之差小于第三边;等腰三角形的两腰相等，如果5厘米是等腰三角形的腰，这个等腰三角形三条边的长度是5厘米,5厘米,12厘米，因为5+5=10(厘米)\u003c12(厘米),所以不满足任意两边之和大于第三边，不能围成三角形。如果12厘米是等腰三角形的腰，这个等腰三角形三条边的长度是5厘米,12厘米,12厘米，因为5+12=17(厘米)\u003e12(厘米),第三条边的长度是12厘米。以此答题即可.\n【详解】根据分析可知:5+5=10(厘米)\u003c12(厘米)(排除);5+12=17(厘米)\u003e12(厘米),所以第三条边的长度是12cm。",
+        "knowledge_points":  [
+                                 "三角形三边关系",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·浙江温州·期末"
+    },
+    {
+        "id":  "218",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "少先队员佩戴的红领巾是等腰三角形，其中的一个底角是30°,它的顶角是( )°,按角分，这个三角形是( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "120;钝角",
+        "solution":  "【分析】根据题意，两腰相等的三角形叫等腰三角形，等腰三角形的两个底角度数相等;三角形的内角和为180°,用180°减去两个30°,就是顶角的度数;有一个角是钝角的三角形叫钝角三角形。据此解答.\n【详解】根据分析可知:180°-30°*2=180°-60°=120°,所以它的顶角是120°,按角分，这个三角形是一个钝角三角形。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·山东德州·期末"
+    },
+    {
+        "id":  "219",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形的两条边长分别是5厘米和12厘米，这个等腰三角形的周长是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "29",
+        "solution":  "【分析】等腰三角形有两边相等，根据三角形的特性：两边之和大于第三边，三角形的两边的差一定小于第三边;判断出该三角形的腰为12厘米，进而根据三角形的周长等于三边长度之和解答即可.\n【详解】5+5=10(厘米),10\u003c12，腰长5厘米不符合;12+12=24(厘米),24\u003e5，腰长12厘米符合;12+12+5=29(厘米)。所以这个等腰三角形的周长是29厘米。",
+        "knowledge_points":  [
+                                 "三角形三边关系",
+                                 "三角形的周长",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "220",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "聪聪用木条做了一个等腰三角形，其中两条边分别长3厘米和7厘米，这个三角形的周长是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "17",
+        "solution":  "【分析】等腰三角形是指至少有两边相等的三角形，相等的两个边称为这个三角形的腰。根据三角形的三边关系，任意两边之和大于第三边，任意两边之差小于第三边，据此求出等腰三角形第三条边长，然后把三条边长相加，即可求出这个三角形的周长是多少厘米.\n【详解】假如第三条边是3厘米,3厘米+3厘米\u003c7厘米，不符合三边关系。假如第三条边是7厘米,3厘米+7厘米\u003e7厘米，符合三边关系。3+7+7=17(厘米)。所以这个三角形的周长是17厘米。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的周长",
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·河南信阳·期末"
+    },
+    {
+        "id":  "221",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "任意一个四边形的四个角剪下来都可以拼成一个( )角，四边形的内角和是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "周;360°或360度",
+        "solution":  "【分析】已知任意三角形的内角和为180°,利用过多边形一个顶点向与它不相邻的顶点依次画线段，把多边形分割成三角形的办法，将分成的几个三角形的内角和相加，就是多边形的内角和。多边形内角和=180°*(边数-2),据此即可解答.\n【详解】连接四边形的一条对角线，将其分为两个三角形，每个三角形的内角和为180°,所以四边形内角和为2×180°=360°。四个角拼合后总度数为360°,正好形成一个周角。所以任意一个四边形的四个角剪下来都可以拼成一个周角，四边形的内角和是360°。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "多边形的内角和",
+                                 "平面图形的分割"
+                             ],
+        "source":  "24-25四年级下·河南信阳·期末"
+    },
+    {
+        "id":  "222",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在一个直角三角形中，一个锐角36°,另一个锐角是( )°;一个等腰三角形的一个底角是45°,它的顶角是( )°,这个三角形还是( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "54;90;直角",
+        "solution":  "【分析】直角为90°,三角形内角和为180°,在直角三角形中，一个锐角36°,用180°-90°-36°即可;等腰三角形两底角相等，用180°-45°-45°即可求出顶角;求出顶角为90°,有90°的三角形为直角三角形，据此解答.\n【详解】180°-90°-36°=54°;180°-45°-45°=90°。所以另一个锐角是54°,它的顶角是90°,这个三角形还是直角三角形。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·内蒙古呼伦贝尔·期末"
+    },
+    {
+        "id":  "223",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三角形的任意两个内角之和都大于90°,则这个三角形一定是( )三角形。(填\"直角\"\"钝角\"或\"锐角\")",
+        "options":  [
+
+                    ],
+        "answer":  "锐角",
+        "solution":  "【分析】三角形按角来分，分为锐角三角形，直角三角形和钝角三角形。三个角都是锐角的三角形叫作锐角三角形，有一个角是直角的三角形叫作直角三角形，有一个角是钝角的三角形叫作钝角三角形。三角形的内角和为180°。据此解答.\n【详解】由题意得，一个三角形的任意两个内角之和都大于90°,180°-90°=90°,那么这个三角形的三个内角都小于90°。所以这个三角形是锐角三角形。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·广东广州·期末"
+    },
+    {
+        "id":  "224",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小红买了一个等腰三角形的风筝，风筝的一个底角是70°,风筝顶角的度数是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "40",
+        "solution":  "【分析】等腰三角形的特点是底角相等;已知小红买了一个等腰三角形的风筝的一个底角是70°,三角形的内角和为180°,则用180°减去2个70°即可;据此解答.\n【详解】70°+70°=140°,180°-140°=40°。所以风筝顶角的度数是40°。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·河南焦作·期末"
+    },
+    {
+        "id":  "225",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形的两条边分别是7厘米和10厘米，那么它的周长可能是( )厘米;也可能是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "24;27",
+        "solution":  "【分析】等腰三角形的两条腰相等，第三条边可能是7厘米或10厘米。三角形的三边关系为三角形的两边之和大于第三边，三角形的两边之差一定小于第三边。据此判断第三条边的长度，再将三条边的长度相加求出周长.\n【详解】7+7\u003e10，第三条边可能是7厘米。7+10\u003e10，第三条边可能是10厘米。7+7+10=24(厘米),7+10+10=27(厘米)。所以它的周长可能是24厘米，也可能是27厘米。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的周长",
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·重庆黔江·期末"
+    },
+    {
+        "id":  "226",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三角形至少有( )个锐角;在一个直角三角形中，一个锐角是35°,另一个锐角是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "2;55°或55度",
+        "solution":  "【分析】根据题意，三角形的内角和等于180°,三个角中最多有一个直角或钝角，所以至少有两个锐角;用180°减去90°再减去35°,求出另一个锐角的度数，据此解答即可.\n【详解】根据分析可知:180°-90°-35°=55°。所以一个三角形至少有两个锐角;在一个直角三角形中，一个锐角是35°,另一个锐角是55°。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·湖北十堰·期末"
+    },
+    {
+        "id":  "227",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "∠1,∠2,∠3是三角形的三个内角,∠1=63°,∠2=40°,∠3=( )°。一个等腰三角形中顶角是80°,它的一个底角是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "77;50",
+        "solution":  "【分析】(1)根据三角形的内角和180°,用180°分别减去∠1和∠2的度数，就是∠3的度数。(2)等腰三角形的特征：两腰相等，两底角也相等。再根据三角形内角和是180°,用180°减去顶角的度数，就能求出两个底角的度数和，再除以2即可求出它的一个底角的度数.\n【详解】(1)180°-(63°+40°)=180°-103°=77°,∠3的度数是77°。(2)180°-80°=100°,100°/2=50°,它的一个底角是50°。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·江苏南京·期末"
+    },
+    {
+        "id":  "228",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "有7根小棒，分别为4厘米,5厘米,5厘米,6厘米,7厘米,10厘米,11厘米，用其中3根作为等腰三角形的边，可以搭出( )种不同的等腰三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "3",
+        "solution":  "【分析】等腰三角形：有两条边相等的三角形;三角形三边之间的关系：三角形两边之和大于第三边，三角形两边之差小于第三边;第一种:4厘米,5厘米,5厘米，第二种:5厘米,5厘米,6厘米，第三种:5厘米,5厘米,7厘米，第四种:5厘米,5厘米,10厘米，第五种:5厘米,5厘米,11厘米，找出其中符合三角形三边关系的即可，据此解答.\n【详解】根据分析：第一种:4厘米,5厘米,5厘米,4+5=9(厘米),9\u003e5，符合三角形的三边关系;第二种:5厘米,5厘米,6厘米,5+5=10(厘米),10\u003e6，符合三角形的三边关系;第三种:5厘米,5厘米,7厘米,5+5=10(厘米),10\u003e7，符合三角形的三边关系;第四种:5厘米,5厘米,10厘米,5+5=10(厘米),10=10，不符合三角形的三边关系;第五种:5厘米,5厘米,11厘米,5+5=10(厘米),10\u003c11，不符合三角形的三边关系。所以用其中3根作为等腰三角形的边，可以搭出3种不同的等腰三角形。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·浙江宁波·期末"
+    },
+    {
+        "id":  "229",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三角形的一条边是8厘米，另一条边是6厘米，围成这个三角形还需要一根至少( )厘米长的小棒。(长度都是整厘米)",
+        "options":  [
+
+                    ],
+        "answer":  "3",
+        "solution":  "【分析】根据三角形三边关系：任意两边之和大于第三边，任意两边之差小于第三边，因此还需要一根小棒的长度应比(8-6)厘米大，且比(8+6)厘米小，符合这样的数有3厘米,4厘米,5厘米,6厘米,7厘米,8厘米,9厘米,10厘米,11厘米,12厘米,13厘米，其中最小的是3厘米，据此填空.\n【详解】根据分析可得:8-6=2(厘米),8+6=14(厘米),符合比2大且比14小的数有3,4,5,6,7,8,9,10,11,12,13，其中最小的是3。所以围成这个三角形还需要一根至少3厘米长的小棒。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·湖南岳阳·期末"
+    },
+    {
+        "id":  "230",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "三角形的内角和是180°,四边形的内角和是360°,十边形的内角和是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "1440°",
+        "solution":  "【分析】三角形的内角和是180°;从四边形的一个顶点可以引出1条对角线，把四边形分成2个三角形，那么它的内角和是180°*2=360°;从五边形的一个顶点可以引出2条对角线，把五边形分成3个三角形，那么它的内角和是180°*3=540°。;从十边形的一个顶点可以引出8条对角线，把十边形分成(10-2)个三角形，那么它的内角和是180°*(10-2)。据此计算.\n【详解】180°*(10-2)=180°*8=1440°。所以十边形的内角和是1440°。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "多边形的内角和"
+                             ],
+        "source":  "24-25四年级下·湖南岳阳·期末"
+    },
+    {
+        "id":  "231",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "三角形中有一个角是30°,另一个角是它的4倍，第三个角是( )°,按角分它是一个( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "30;钝角",
+        "solution":  "【分析】三角形中有一个角是30°,另一个角=30°*4=120°,再根据三角形内角和是180°,依次减去前两个角的度数，即为第三个角度数。在这三个角里有一个角比90°大是钝角，所以有一个角是钝角的三角形是钝角三角形.\n【详解】30°*4=120°,180°-30°-120°=30°,120°\u003e90°,所以第三个角是30°,按角分它是一个钝角三角形。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·吉林四平·期末"
+    },
+    {
+        "id":  "232",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个直角三角形，其中一个锐角是50°,那么另一个锐角是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "40",
+        "solution":  "【分析】本题涉及直角三角形的内角和定理，即三角形的内角和是180°,直角三角形有一个角是90°。我们已知一个锐角的度数，通过三角形内角和减去直角和已知锐角的度数，就能求出另一个锐角的度数.\n【详解】180°-90°-50°=40°,所以另一个锐角是40°。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·河北承德·期末"
+    },
+    {
+        "id":  "233",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等边三角形的一条边长是15cm,它的周长是( )cm。一个三边长均为整数厘米的三角形，有两条边的长度分别是4cm和8cm,它的周长最长是( )cm。",
+        "options":  [
+
+                    ],
+        "answer":  "45;23",
+        "solution":  "【分析】等边三角形的三条边都相等，用15乘3，求出这个等边三角形的周长是多少厘米;三角形的三边关系：两边之和大于第三条边，两边之差小于第三条边;据此求出第三条边的最大的长度，再把三条边都相加，据此即可解答.\n【详解】15×3=45(cm);8-4\u003c第三条边的长度\u003c8+4,4\u003c第三条边的长度\u003c12，则第三条边最长是11cm,该三角形的周长为4+8+11=23(cm)。所以等边三角形的周长是45cm,另一个三角形的周长最长是23cm。",
+        "knowledge_points":  [
+                                 "三角形三边关系",
+                                 "三角形的周长",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·湖北孝感·期末"
+    },
+    {
+        "id":  "234",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一根铁丝可以围成一个边长6厘米的正方形，如果改围成一个等边三角形，那么这个等边三角形的边长是( )厘米;如果改围成一个腰长7厘米的等腰三角形，那么这个等腰三角形的底是( )米。",
+        "options":  [
+
+                    ],
+        "answer":  "8;0.1",
+        "solution":  "【分析】由题意得，一根铁丝可以围成一个边长6厘米的正方形，那么这根铁丝的长度就等于正方形的周长。正方形的周长=边长*4，直接把数据代入即可算出这根铁丝的长度。如果改围成一个等边三角形，等边三角形的三条边长度相等，那么直接用这根铁丝的长度除以3即可算出等边三角形的边长;如果改围成一个腰长7厘米的等腰三角形，那么另一条腰的长度是7厘米，直接用这根铁丝的总长度减去两条腰的长度即可算出等腰三角形底边的长度。最后将底边的长度转化为多少米即可.\n【详解】6×4=24(厘米),24÷3=8(厘米),24-7-7=10(厘米),10÷100=0.1，所以10厘米=0.1米。故这个等边三角形的边长是8厘米;如果改围成一个腰长7厘米的等腰三角形，那么这个等腰三角形的底是0.1米。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "正方形的周长",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·湖南株洲·期末"
+    },
+    {
+        "id":  "235",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "明明剪了两个三角形，一个是等腰三角形，它的底角70°,顶角是( )°。一个是等腰直角三角形，它的锐角是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "40;45",
+        "solution":  "【分析】等腰三角形的两个底角相等，三角形的内角和为180°。由题意得，等腰三角形的一个底角的度数是70°,那么另一个底角的度数也是70°,直接用180°减去两个底角的度数之和即可算出顶角的度数;有一个角是直角的三角形叫作直角三角形。由题意得，一个三角形是等腰直角三角形，那么直接用180°减去90°算出两个底角的度数，然后再除以2即可算出一个底角的度数.\n【详解】180°-70°*2=180°-140°=40°;(180°-90°)÷2=90°/2=45°。所以等腰三角形的顶角是40°,等腰直角三角形的锐角是45°。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·湖南株洲·期末"
+    },
+    {
+        "id":  "236",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用三根小棒围三角形，已知其中两根小棒分别是10厘米和5厘米，第三根小棒最长是( )厘米，最短是( )厘米。(小棒长度取整厘米数)",
+        "options":  [
+
+                    ],
+        "answer":  "14;6",
+        "solution":  "【分析】根据题意，任意两边之和必须大于第三边，任意两边之差必须小于第三边。已知两根小棒分别为10厘米和5厘米，可计算出(10-5)厘米\u003c第3根小棒长度\u003c(10+5)厘米，即5厘米\u003c第3根小棒长度\u003c15厘米。由此推断出第三根小棒最长和最短各是多少厘米;以此答题即可.\n【详解】根据分析可知:10-5=5(厘米),10+5=15(厘米),5+1=6(厘米),15-1=14(厘米)。所以第三根小棒最长是14厘米，最短是6厘米。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·湖南株洲·期末"
+    },
+    {
+        "id":  "237",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形的顶角是80°,它的一个底角是( )°。按角分，这是一个( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "50;锐角",
+        "solution":  "【分析】根据题意，等腰三角形的两腰相等，两个底角都相等，三角形的内角和为180°,因此用180°减去80°,即可得到两个底角的度数，再除以2，就是1个底角的度数;根据三角形的分类标准填空，有一个角是钝角的三角形是钝角三角形，有一个角是直角的三角形是直角三角形，三个角都是锐角的三角形是锐角三角形。以此答题即可.\n【详解】根据分析可知:(180°-80°)÷2=100°/2=50°,三角形的三个角是80°,50°,50°,三个角都是锐角。所以它的一个底角是50°,按角分，这是一个锐角三角形。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·湖南株洲·期末"
+    },
+    {
+        "id":  "238",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三角形的边长都是8cm,若把它分成两个直角三角形后，直角三角形的两个锐角分别是( )°和( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "30;60",
+        "solution":  "【分析】三条边长度相等的三角形是等边三角形，等边三角形的三个角都是60°。把它分成两个直角三角形。也就是沿着它的高平均分成两个直角三角形。那么其中一个锐角还是60°,另一个锐角可以用60°除以2.\n【详解】60°/2=30°。所以直角三角形的两个锐角分别是30°和60°。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·辽宁抚顺·期末"
+    },
+    {
+        "id":  "239",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在一个等腰三角形中，顶角的度数是一个底角度数的4倍，那么这个等腰三角形顶角的度数是( )°,一个底角的度数是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "120;30",
+        "solution":  "【分析】等腰三角形两个底角相等，顶角的度数是一个底角度数的4倍，则这个三角形的内角和是底角度数的(4+1+1)倍，三角形内角和是180°,用180°除以倍数，即可求出底角是多少度;用底角的度数乘4即可求出顶角的度数.\n【详解】180°/(4+1+1)=180°/6=30°,4×30°=120°。所以顶角的度数是120°,一个底角的度数是30°。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·山东济南·期末"
+    },
+    {
+        "id":  "240",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个等边三角形的周长是24厘米，每条边的长度是( )厘米，它的内角和是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "8;180",
+        "solution":  "【分析】因为等边三角形的三条边长度相等，已知该等边三角形的周长是24厘米，所以用周长除以3就能得到每条边的长度;根据三角形的内角和是180°来完成第二空.\n【详解】24÷3=8(厘米)。所以每条边的长度是8厘米，它的内角和是180°。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·湖南株洲·期末"
+    },
+    {
+        "id":  "241",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "三角形按角分类可分为( )三角形,( )三角形和( )三角形。一个三角形有( )条高，至少有( )个锐角。",
+        "options":  [
+
+                    ],
+        "answer":  "锐角;直角;钝角;3;2",
+        "solution":  "【分析】根据三角形中最大的角可以把三角形分为三类，最大角是直角就是直角三角形;最大角是钝角就是钝角三角形;最大角是锐角就是锐角三角形。从三角形的顶点作对边的垂线段即为三角形的高，三角形有3个顶点，所以有3条高。根据三角形的分类可知，锐角三角形有3个锐角;直角三角形有1个直角和2个锐角;钝角三角形有1个钝角和2个锐角，所以一个三角形中至少有2个锐角。据此解答即可.\n【详解】三角形按角分类可分为锐角三角形，直角三角形和钝角三角形。一个三角形有3条高，至少有2个锐角。",
+        "knowledge_points":  [
+                                 "三角形的高及画法",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·湖南株洲·期末"
+    },
+    {
+        "id":  "242",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在一个三角形中，最大的内角的度数是最小的内角的3倍，另一个内角的度数是最小的内角的2倍，这个三角形三个内角的度数分别是( )°,( )°,( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "30;60;90",
+        "solution":  "【分析】三角形的内角和是180°,以最小角为标准，看作1份，则最大角是3份，另一个角是2份，三个角共有1+3+2=6(份),与180°相对应，用180°除以6可以先求出1份，也就是最小的角，进而再求出其他两角.\n【详解】最小的角:180°/(3+2+1)=180°/6=30°;最大的角:30°*3=90°;另一个角:30°*2=60°。所以这个三角形三个内角的度数分别是30°,60°,90°。",
+        "knowledge_points":  [
+                                 "和差倍问题",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·湖北省直辖县级单位·期末"
+    },
+    {
+        "id":  "243",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "有两个小棒分别长10厘米,14厘米，现在剪断其中一根，用它们拼三角形，可以拼出( )个不同的三角形。(三角形的三条边均为整厘米数)",
+        "options":  [
+
+                    ],
+        "answer":  "5",
+        "solution":  "【分析】三角形三边的关系：任意两边之和大于第三边(较短两边之和大于第三边)。由题意得，如果剪断10厘米的小棒，那么最长的小棒是14厘米，其余两根小棒的长度之和为10厘米。10厘米\u003c14厘米，即两边之和小于第三边的长度，不满足围成三角形的条件。那么只能剪断14厘米的小棒,14厘米的小棒可以分成1厘米和13厘米,2厘米和12厘米,3厘米和11厘米,4厘米和10厘米,5厘米和9厘米,6厘米和8厘米,7厘米和7厘米。然后根据三角形三边的关系来判断几种剪法能否围成三角形即可.\n【详解】10\u003c14，所以不能剪断10厘米的小棒。把14厘米的小棒分成1厘米和13厘米的小棒:1+10=11(厘米),11厘米\u003c13厘米，无法围成三角形。把14厘米的小棒分成2厘米和12厘米的小棒:2+10=12(厘米),12厘米=12厘米，无法围成三角形。把14厘米的小棒分成3厘米和11厘米的小棒:3+10=13(厘米),13厘米\u003e11厘米，可以围成三角形。把14厘米的小棒分成4厘米和10厘米的小棒:4+10=14(厘米),14厘米\u003e10厘米，可以围成三角形。把14厘米的小棒分成5厘米和9厘米的小棒:5+9=14(厘米),14厘米\u003e10厘米，可以围成三角形。把14厘米的小棒分成6厘米和8厘米的小棒:6+8=14(厘米),14厘米\u003e10厘米，可以围成三角形。把14厘米的小棒分成7厘米和7厘米的小棒:7+7=14(厘米),14厘米\u003e10厘米，可以围成三角形。综上所述，一共有5种不同的剪法。所以可以拼出5个不同的三角形。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·湖北省直辖县级单位·期末"
+    },
+    {
+        "id":  "244",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "已知一个等腰三角形的两条边分别是6厘米和14厘米，这个三角形的周长是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "34",
+        "solution":  "【分析】这个等腰三角形的两条边可能为两条腰或一条腰和一条底边。根据三角形三边关系，若两条腰为6厘米，则底边14厘米无法满足两边之和大于第三边，因此腰必须为14厘米，底边为6厘米，此时三边满足条件，即可求周长.\n【详解】情况一：若6厘米为腰，底边为14厘米，则三边为6厘米,6厘米,14厘米。验证三角形三边关系:6+6=12,12\u003c14，不符合三角形的任意两边之和大于第三边的条件，故不成立。情况二：若14厘米为腰，底边为6厘米，则三边为14厘米,14厘米,6厘米。验证三角形三边关系:14+14=28,28\u003e6,14+6=20,20\u003e14，满足三角形三边关系，故成立。所以这个等腰三角形的边长为14厘米,14厘米,6厘米;周长为:14+14+6=34(厘米)。所以这个三角形的周长是34厘米。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的周长",
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·湖北武汉·期末"
+    },
+    {
+        "id":  "245",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形的一条边是4厘米，另一条边是9厘米，则第三条边是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "9",
+        "solution":  "【分析】题意没有说明等腰三角形的腰长是4厘米还是9厘米，所以要考虑2种情况，再根据能不能围成三角形的判断方法：三角形两条短边之和必须大于第三边，据此解答即可.\n【详解】第一种情况：当腰长为4厘米时,4+4\u003c9，等腰三角形的腰长不可能是4厘米。第二种情况：当腰长为9厘米时,4+9\u003e9，等腰三角形的腰长应该是9厘米。所以第三条边是9厘米。",
+        "knowledge_points":  [
+                                 "三角形三边关系",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·广东广州·期末"
+    },
+    {
+        "id":  "246",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三角形，两个内角分别是65°和39°,另一个内角是( )°;一个直角三角形，其中一个锐角是38°,另一个锐角是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "76;52",
+        "solution":  "【分析】三角形的内角和为180°。有一个角是直角的三角形叫直角三角形;由题意得，一个三角形的两个内角分别是65°和39°,那么直接用180°减去已知的两个角的度数即可算出第三个内角的度数;直角三角形的一个锐角是38°,那么直接用180°减去90°再减去38°即可算出另一个锐角的度数.\n【详解】180°-65°-39°=76°;180°-90°-38°=52°。所以另一个内角是76°,另一个锐角是52°。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·湖南邵阳·期末"
+    },
+    {
+        "id":  "247",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个等腰三角形的周长是20厘米，已知其中一条边长是4厘米，另外两条边长是( )厘米和( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "8;8",
+        "solution":  "【分析】根据题目可得：等腰三角形的周长是20厘米，已知其中一条边长是4厘米，据此可计算出等腰三角形另外两条边长之和为:20-4=16(厘米),根据等腰三角形的两腰相等和三角形三边关系(任意两边之和大于第三边),分两种情况讨论:4厘米为底边或腰，排除不符合条件的情况即可.\n【详解】等腰三角形另外两条边长之和为:20-4=16(厘米)。若4厘米为等腰三角形的底边长：则三角形的腰长为:16÷2=8(厘米),此时，等腰三角形的三边长为4厘米,8厘米,8厘米，满足8+4\u003e8,8+8\u003e4，符合三角形三边关系。若4厘米为等腰三角形的腰长：则三角形的底边长度为:16-4=12(厘米),此时，等腰三角形的三边长为4厘米,4厘米,12厘米，但4+4=8，而8\u003c12，不符合三角形三边关系，因此舍去。因此，另外两条边长是8厘米和8厘米。",
+        "knowledge_points":  [
+                                 "三角形的周长",
+                                 "三角形三边关系",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·湖北黄冈·期末"
+    },
+    {
+        "id":  "248",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "看图填一填.\n![](images/2026-03-10%20223629.jpg)\n∠1=( )°,树高( )米。",
+        "options":  [
+
+                    ],
+        "answer":  "45;8.3",
+        "solution":  "【分析】根据题图可知，这个三角形有一个直角，根据三角形的内角和为180°可知,∠1=180°-90°-45°=45°,则这个三角形的两个底角相等，是等腰直角三角形，两条腰相等。树高即为8.3米。据此解答.\n【详解】180°-90°-45°=45°,所以∠1=45°,树高8.3米。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "角度的计算"
+                             ],
+        "source":  "24-25四年级下·湖南郴州·期末"
+    },
+    {
+        "id":  "249",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如图，已知:∠2=∠3，那么∠4比∠1小( )°.\n![](images/3d256af1b90f16abd4ac52cfeafe4ff2c53b5584cab9f9a374f99560182a7047.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "40",
+        "solution":  "【分析】三角形的内角和是180°,所以∠1+∠2+80°=180°,∠3+∠4+120°=180°;又因为∠2=∠3，所以用120°减去80°即可得到∠4比∠1小多少.\n【详解】因为∠1+∠2+80°=180°,∠3+∠4+120°=180°;所以∠1+∠2=100°,∠3+∠4=60°;所以∠1=100°-∠2,∠4=60°-∠3;所以∠1-∠4=100°-∠2-(60°-∠3)=100°-∠2-60°+∠3=40°。所以∠4比∠1小40°。",
+        "knowledge_points":  [
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·湖北鄂州·期末"
+    },
+    {
+        "id":  "250",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一根铁丝刚好围成一个边长是5cm的等边三角形，若把它围成一个等腰三角形，使它的一条边是3cm,那么另外两条边分别是( )cm和( )cm。",
+        "options":  [
+
+                    ],
+        "answer":  "6;6",
+        "solution":  "【分析】根据题意，等边三角形的3条边长相等，用边长乘3，计算出铁丝的长度，等腰三角形两腰相等，三角形任意两边之和大于第三边，据此计算出等腰三角形的另外两条边;据此解答.\n【详解】根据分析可知:5×3=15(cm)。假设3cm为底边:(15-3)÷2=12÷2=6(cm),另外两条边都是6cm;3+6\u003e6，可以围成三角形。假如3cm为腰,15-3×2=15-6=9(cm),另外两条边分别是3cm和9cm;3+3\u003c9，不能围成一个三角形。所以另外两条边分别是6cm和6cm。",
+        "knowledge_points":  [
+                                 "三角形三边关系",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·河北保定·期末"
+    },
+    {
+        "id":  "251",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形的顶角是80°,它的一个底角是( )°,它也是( )三角形。一个等腰三角形的两条边的长分别是4厘米和9厘米，则它的周长是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "50;锐角;22",
+        "solution":  "【分析】三角形内角和是180°,等腰三角形两个底角相等，用180°减去顶角的度数，再除以2即可求出底角的度数;根据三角形的分类，有一个角是钝角的三角形是钝角三角形，有一个角是直角的三角形是直角三角形，三个角都是锐角的三角形是锐角三角形，据此判断这是一个什么三角形即可;等腰三角形两条腰长相等，根据三角形三边关系，任意两边之和大于第三边，任意两边之差小于第三边，据此先找出腰长是多少厘米，将三条边的长度相加即可求出周长是多少厘米.\n【详解】(180°-80°)÷2=100°/2=50°。当腰长是4厘米时:4+4=8(厘米),8\u003c9，两边之和小于第三边，不能围成三角形;当腰长是9厘米时:4+9=13(厘米),13\u003e9,9-4=5(厘米),5\u003c9，能围成三角形。4+9+9=22(厘米)。所以它的一个底角是50°,它也是锐角三角形，它的周长是22厘米。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和",
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·青海果洛·期末"
+    },
+    {
+        "id":  "252",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三角形中，若∠1+∠2=80°,则∠3=( )°,按角分，这个三角形是( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "100°或100度;钝角",
+        "solution":  "【分析】根据题意，根据三角形内角和定理，三角形的内角和为180°,已知两个角的和，可用180°减去80°,求出第三个角的度数，再根据角的度数判断三角形的类型，三角形按角分为锐角三角形(三个角均小于90°),直角三角形(一个角为90°)和钝角三角形(一个角大于90°).\n【详解】∠3=180°-80°=100°,100°\u003e90°,所以这个三角形是钝角三角形。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·湖南永州·期末"
+    },
+    {
+        "id":  "253",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "屋顶钢架常用三角形，因为三角形具有( )性，一个三角形中最多有( )个直角。",
+        "options":  [
+
+                    ],
+        "answer":  "稳定;1",
+        "solution":  "【分析】三角形的性质：三角形具有稳定性;直角为90°,如果一个三角形中有两个直角，那么就是90°*2=180°,不符合三角形的定义，所以一个三角形中最多有1个直角.\n【详解】屋顶钢架常用三角形，因为三角形具有稳定性，一个三角形中最多有1个直角。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "三角形的稳定性及应用"
+                             ],
+        "source":  "24-25四年级下·湖南永州·期末"
+    },
+    {
+        "id":  "254",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个三角形的三条边长都是整厘米数，其中两条边的长度分别是6cm和10cm,第三条边最短是( )cm,最长是( )cm。",
+        "options":  [
+
+                    ],
+        "answer":  "5;15",
+        "solution":  "【分析】三角形两边之和大于第三边，两边之差小于第三边，分别计算出已知两条边的和，差，最短不小于差，最长不大于和的整数即可.\n【详解】6+10=16(cm),10-6=4(cm),4cm\u003c第三边长\u003c16cm,又因为边长是整厘米数，所以最短是5cm,最长是15cm。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·湖北孝感·期末"
+    },
+    {
+        "id":  "255",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "小美想用彩纸制作一个三角形书签。她已经剪出了两个角，其中一个角是50°,另一个角是80°,第三个角是( )°。这个三角形书签按角分是( )三角形，按边分是( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "50;锐角;等腰",
+        "solution":  "【分析】根据三角形的内角和为180°,用180°减去已知的两个内角的度数，即可求出第三个角的度数。再根据三角形的分类中三个内角都小于90°的三角形为锐角三角形，等腰三角形至少有两个内角相等，即可解答此题.\n【详解】180°-50°-80°=50°,三个角分别是50°,80°,50°,其中50°和50°相等，三个角都是锐角。所以第三个角是50°,这个三角形书签按角分是锐角三角形，按边分是等腰三角形。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·湖南常德·期末"
+    },
+    {
+        "id":  "256",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "最少用( )个等边三角形可以拼成一个等腰梯形。",
+        "options":  [
+
+                    ],
+        "answer":  "3",
+        "solution":  "【分析】根据题意，如图:![](images/1a44c354c278b17d7cd8b2d4eb92c32775d9737190d69c0ced793e2878479214.jpg)要拼成等腰梯形，需满足上下底平行且两腰相等。等边三角形每个内角为60°,通过组合调整角度和边长，可构造出符合条件的等腰梯形。以此答题即可.\n【详解】根据分析可知：将两个等边三角形并排作为下底(总长2个单位),再将一个等边三角形倒置作为上底(长1个单位),与下底平行。最少用3个等边三角形可以拼成一个等腰梯形。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "直角梯形和等腰梯形的概念及特点"
+                             ],
+        "source":  "24-25四年级下·河北保定·期末"
+    },
+    {
+        "id":  "257",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把一根长12厘米的铁丝剪成4厘米,( )厘米,( )厘米的三段就能围成一个三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "4;4",
+        "solution":  "【分析】根据三角形的三边关系，任意两边之和大于第三边。已知铁丝总长12厘米，其中一段为4厘米，剩余两段之和为8厘米。7+1=8(厘米),6+2=8(厘米),5+3=8(厘米),4+4=8(厘米),但4+1=5\u003c7,4+2=6,4+3=7\u003e5,4+4=8\u003e4，另外两边的长度排除7厘米和1厘米,6厘米和2厘米，所以另外两条边可以是4厘米,4厘米,5厘米,3厘米，以此答题即可.\n【详解】根据分析可知:7+1=8(厘米),6+2=8(厘米),5+3=8(厘米),4+4=8(厘米)。4+1=5\u003c7,4+2=6,4+3=7\u003e5,4+4=8\u003e4。所以把一根长12厘米的铁丝剪成4厘米,4厘米,4厘米的三段就能围成一个三角形。(答案不唯一)",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·河北保定·期末"
+    },
+    {
+        "id":  "258",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把一根长15cm的吸管剪成3段(每段的长度都是整厘米),用这3段吸管能围成一个等腰三角形，可以有( )种不同的剪法。",
+        "options":  [
+
+                    ],
+        "answer":  "4",
+        "solution":  "【分析】等腰三角形：有两条边相等的三角形;三角形三边之间的关系：三角形两边之和大于第三边，三角形两边之差小于第三边;等边三角形：三条边都相等的三角形，等边三角形是特殊的等腰三角形;根据等腰三角形的特性,15=1+1+13=2+2+11=3+3+9=4+4+7=5+5+5=6+6+3=7+7+1，分别计算出这些情况中哪些符合三角形三边的关系，哪些就是正确的剪法;据此解答.\n【详解】根据分析：第一种：三边分别为1cm,1cm和13cm,1+1=2(cm),2\u003c13，不符合三角形三边关系，不能围成三角形;第二种：三边分别为2cm,2cm和11cm,2+2=4(cm),4\u003c11，不符合三角形三边关系，不能围成三角形;第三种：三边分别为3cm,3cm和9cm,3+3=6(cm),6\u003c9，不符合三角形三边关系，不能围成三角形;第四种：三边分别为4cm,4cm和7cm,4+4=8(cm),8\u003e7，符合三角形三边关系，能围成等腰三角形;第五种：三边分别为5cm,5cm和5cm,5+5=10(cm),10\u003e5，符合三角形三边关系，能围成等边三角形，一定是等腰三角形;第六种：三边分别为6cm,6cm和3cm,3+6=9(cm),9\u003e6，符合三角形三边关系，能围成等腰三角形;第七种：三边分别为7cm,7cm和1cm,7+1=8(cm),8\u003e7，符合三角形三边关系，能围成等腰三角形。所以用这3段吸管能围成一个等腰三角形，可以有4种不同的剪法。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级上·湖南邵阳·期末"
+    },
+    {
+        "id":  "259",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三角形的其中两个内角分别是80°,50°,这个三角形按角分是( )三角形，按边分是( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "锐角;等腰",
+        "solution":  "【分析】三角形的内角和是180°,其中两个内角分别是80°,50°,另一个内角为180°-80°-50°=50°,三个角都是锐角，按角分，这个三角形是锐角三角形;其中两个角相等，为等腰三角形，所以按边分是等腰三角形.\n【详解】180°-80°-50°=50°,按角分，这个三角形是锐角三角形，按边分，这个三角形是等腰三角形。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级上·湖南邵阳·期末"
+    },
+    {
+        "id":  "260",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "胡夫金字塔是埃及吉萨高原上的一座古代金字塔，也是世界七大奇迹之一。它从四面看都呈等腰三角形，很像汉语中的\"金\"字。它的顶角大约是52°,它的一个底角是( )°,如果按角分，它是一个( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "64;锐角",
+        "solution":  "【分析】等腰三角形的两个底角相等，三角形的三个内角和等于180°,所以180°减去顶角的52°等于两个底角的度数和，再除以2，即等于一个底角的度数;三个角都是锐角的三角形是锐角三角形，有一个角是直角的三角形是直角三角形，有一个角是钝角的三角形是钝角三角形，据此判断这个等腰三角形按角分是什么三角形.\n【详解】(180°-52°)÷2=128°/2=64°,52°,64°,64°都是锐角，所以三角形是锐角三角形。所以它的一个底角是64°,按角分它是一个锐角三角形。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·湖北十堰·期末"
+    },
+    {
+        "id":  "261",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "计算一个六边形内角和时，可以把它分成4个三角形，它的内角和就是180°*4=720°。像这样，计算八边形的内角和可以用( )。",
+        "options":  [
+
+                    ],
+        "answer":  "180°*6=1080°",
+        "solution":  "【分析】把八边形的一个顶点与不相邻的顶点用线段连接起来，这样八边形就被分成了6个三角形，求八边形的内角和就变成了求6个三角形的内角和，即等于180°*6=1080°,据此即可解答.\n【详解】根据分析可知，计算八边形的内角和可以用180°*6=1080°。",
+        "knowledge_points":  [
+                                 "多边形的内角和",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·湖北十堰·期末"
+    },
+    {
+        "id":  "262",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "有两根长度分别是5厘米和9厘米的小棒，再添上一根小棒(长度取整厘米数)可以摆成一个三角形，则这个小棒的长度最短是( )厘米，最长是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "5;13",
+        "solution":  "【分析】三角形三边的关系：任意两边之和大于第三边，任意两边之差小于第三边。由题意得，有两根长度分别是5厘米和9厘米的小棒，再添上一根小棒可以摆成一个三角形，那么这三根小棒的长度需要满足三角形三边的关系。据此解答.\n【详解】两边之差\u003c第三根小棒的长度\u003c两边之和,9-5\u003c第三根小棒的长度\u003c9+5,4\u003c第三根小棒的长度\u003c14，小棒的长度是整厘米数，所以第三根小棒最短是5厘米，最长是13厘米。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·湖北十堰·期末"
+    },
+    {
+        "id":  "263",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三角形，有两个角的大小分别为44°和68°,它的第三个角是( )°,这个三角形按角分是( )三角形，按边分是( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "68;锐角;等腰",
+        "solution":  "【分析】根据三角形内角和为180°,用180°减去44°,再减去68°,可求出第三个角的度数。再根据角的大小判断三角形的类型，按角分类：锐角三角形(三个角均小于90°),直角三角形(一个角为90°)和钝角三角形(一个角大于90°)。根据边的相等情况判断是否为等腰或等边三角形。以此答题即可.\n【详解】根据分析可知:180°-44°-68°=68°,三个角分别为44°,68°,68°,均小于90°,因此这个三角形是锐角三角形。按边分类：两个角均为68°,这两个角对应的边相等，因此这个三角形是等腰三角形。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·湖北孝感·期末"
+    },
+    {
+        "id":  "264",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "等腰三角形中的一个底角是80°,它的顶角是( )°;两根木条分别长5厘米和10厘米，再用一根长( )厘米的木条，就可以钉成一个等腰三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "20;10",
+        "solution":  "【分析】等腰三角形两个底角相等，三角形内角和是180°,用180°减去两个底角的度数，即可求出顶角是多少度;等腰三角形两条腰相等，根据三角形三边关系，两边之和大于第三边，两边之差小于第三边，已知两条边的长度，据此判断第三条边的长度即可.\n【详解】180°-80°-80°=20°。假设腰长是5厘米:5+5=10(厘米),10=10，两边之和等于第三边，不能围成三角形;假设腰长是10厘米:5+10=15(厘米),15\u003e10,10-5=5(厘米),5\u003c10，能围成三角形。所以它的顶角是20°,再用一根长10厘米的木条，就可以钉成一个等腰三角形。",
+        "knowledge_points":  [
+                                 "三角形三边关系",
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·湖北荆州·期末"
+    },
+    {
+        "id":  "265",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "等边三角形是特殊的( )三角形，每个内角的度数都是( )度。",
+        "options":  [
+
+                    ],
+        "answer":  "等腰;60",
+        "solution":  "【分析】根据题意，明确等边三角形具有三条边相等，三个内角相等的性质。根据三角形分类，等边三角形属于特殊的等腰三角形。每个内角度数可通过三角形内角和180°除以3，以此答题即可.\n【详解】根据分析可知:180°/3=60°,所以等边三角形是特殊的等腰三角形，每个内角的度数都是60度。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·湖北襄阳·期末"
+    },
+    {
+        "id":  "266",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在一个直角三角形中，一个锐角是36°,另一个锐角是( )°。用2个完全一样这样的三角形正好能拼成一个更大的三角形，这个大三角形的内角和是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "54;180",
+        "solution":  "【分析】已知三角形的三个内角和是180°,三个内角有两个已经知道，要求未知的那一个内角，用180°连续减去两个内角的度数即可.\n【详解】180°-90°-36°=54°,用2个完全一样这样的三角形正好能拼成一个更大的三角形，这个大三角形的内角和是180°。",
+        "knowledge_points":  [
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·湖北武汉·期末"
+    },
+    {
+        "id":  "267",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "有两根长度分别为5厘米和3厘米的小棒，如果再添一根小棒围成一个三角形，那么这根小棒最短是( )厘米。(长度为整厘米数)",
+        "options":  [
+
+                    ],
+        "answer":  "3",
+        "solution":  "【分析】根据三角形的三边关系\"任意两边之和大于第三边，任意两边之差小于第三边\"来确定第三根小棒的长度范围，进而求出第三条边最短的值。已知两根小棒长度分别为5厘米和3厘米，根据三边关系可得:5-3\u003c第三条边的长度\u003c5+3.\n【详解】5-3=2(厘米),5+3=8(厘米),因为小棒长度为整厘米数，所以第三条边的长度可以为3厘米,4厘米,5厘米,6厘米,7厘米。所以这根小棒最短是3厘米。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·湖北武汉·期末"
+    },
+    {
+        "id":  "268",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个钝角三角形是等腰三角形，有一个内角是40°,这个三角形另外两个内角分别是( )°和( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "40;100",
+        "solution":  "【分析】钝角三角形有一个角大于90°,等腰三角形有两个角相等。已知一个内角为40°,需确定它是顶角还是底角。三角形的内角和为180°,若40°为顶角，则底角为(180°-40°)÷2=70°,此时顶角40°及底角70°都不是钝角，不符合条件。因此40°必为底角，另一个底角也是40°,顶角为180°-40°-40°=100°,符合钝角条件.\n【详解】钝角三角形中有一个角大于90°,等腰三角形两个底角相等。已知一个内角为40°,分两种情况讨论：若40°为顶角：底角为(180°-40°)÷2=70°,此时顶角40°及底角70°都不是钝角，不符合钝角三角形的定义。若40°为底角：另一个底角也为40°,顶角为180°-40°-40°=100°,100°是钝角，符合条件。所以另外两个内角分别是40°和100°。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·湖北武汉·期末"
+    },
+    {
+        "id":  "269",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用三根小棒围三角形，第一根长6厘米，第二根长3厘米，第三根小棒的长度必须小于( )厘米且大于( )厘米才能围成一个三角形。(取整厘米数)",
+        "options":  [
+
+                    ],
+        "answer":  "9;3",
+        "solution":  "【分析】三角形的三边关系为三角形的两边之和大于第三边，三角形的两边之差一定小于第三边;据此解答即可.\n【详解】6+3=9(厘米),6-3=3(厘米),则用三根小棒围三角形，第三根小棒的长度必须小于9厘米且大于3厘米才能围成一个三角形。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·湖北孝感·期末"
+    },
+    {
+        "id":  "270",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用三根小棒围三角形，第一根长6厘米，第二根长3厘米，第三根小棒的长度必须小于( )厘米且大于( )厘米才能围成一个三角形。(取整厘米数)",
+        "options":  [
+
+                    ],
+        "answer":  "9;3",
+        "solution":  "【分析】三角形的三边关系为三角形的两边之和大于第三边，三角形的两边之差一定小于第三边;据此解答即可.\n【详解】6+3=9(厘米),6-3=3(厘米),则用三根小棒围三角形，第三根小棒的长度必须小于9厘米且大于3厘米才能围成一个三角形。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·湖北孝感·期末"
+    },
+    {
+        "id":  "271",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三角形的两边长分别为3分米和8分米，如果三条边的长度均为整数，那么第三边最长是( )分米，最短是( )分米。",
+        "options":  [
+
+                    ],
+        "answer":  "10;6",
+        "solution":  "【分析】本题考查三角形三边关系的应用。三角形三边关系是指：三角形任意两边之和大于第三边，任意两边之差小于第三边。据此解答。已知两边长分别为3分米和8分米，且第三边为整数，直接通过这两个关系确定第三边的范围，再找出范围内的最大和最小整数.\n【详解】根据\u0027三角形任意两边之和大于第三边\u0027:第三边的长度必须小于3分米与8分米的和。即第三边\u003c3+8=11(分米)。根据\u0027三角形任意两边之差小于第三边\u0027:第三边的长度必须大于8分米与3分米的差。即第三边\u003e8-3=5(分米)。综上，第三边的长度范围是:5分米\u003c第三边\u003c11分米。因为第三边的长度是整数，且小于11分米，所以符合条件的最大整数是10分米。又因为第三边的长度大于5分米，所以符合条件的最小整数是6分米。所以第三边最长是10分米，最短是6分米。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·山东济南·期末"
+    },
+    {
+        "id":  "272",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形其中一个底角是35°,则这个三角形的顶角是( )°,按角分，这是一个( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "110;钝角",
+        "solution":  "【分析】(1)等腰三角形的两个底角相等，根据三角形的内角和是180°,用180°减去两个底角的度数，即可求出顶角的度数;(2)三角形按角分类：三个角都是锐角的三角形叫锐角三角形;有一个角是直角的三角形叫直角三角形;有一个角是钝角的三角形叫钝角三角形;据此解答.\n【详解】180°-35°-35°=110°,所以这个三角形的顶角是110°;110°是一个钝角，所以按角分这是一个钝角三角形。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·湖北咸宁·期末"
+    },
+    {
+        "id":  "273",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "风筝又名\"纸鸢\"，丽丽做了一个等腰三角形的风筝，它的两条边分别长4dm和8dm,那么第三条边长( )dm。",
+        "options":  [
+
+                    ],
+        "answer":  "8",
+        "solution":  "【分析】根据等腰三角形的定义，有两条边相等。题目给出两条边分别为4dm和8dm,第三条边可能是4dm或8dm。需验证这两种情况是否满足三角形三边关系定理：任意两边之和大于第三边。若第三条边为4dm,则三边为4dm,4dm,8dm,此时4+4=8，不满足两边之和大于第三边，无法构成三角形。因此第三条边只能是8dm.\n【详解】若第三条边为4dm:三边为4dm,4dm,8dm,验证三角形三边关系:4+4=8，不满足\u0027任意两边之和大于第三边\u0027,因此无法构成三角形。若第三条边为8dm:三边为8dm,8dm,4dm,验证三角形三边关系:8+8\u003e4,8+4\u003e8，均满足条件，可以构成三角形。所以第三条边长为8dm。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·重庆铜梁·期末"
+    },
+    {
+        "id":  "274",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在一个直角三角形中，一个锐角是55°,另一个锐角是( )°;乐乐用一根长42厘米的铁丝围成一个等腰三角形，它的底比一条腰长6厘米，这个等腰三角形的底长( )厘米，一条腰长( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "35;18;12",
+        "solution":  "【分析】有一个角是直角的三角形是直角三角形，在一个直角三角形中，一个锐角是55°,三角形内角和是180°,用180°-90°-55°即可求出另一个锐角的度数;等腰三角形两条腰相等，一个等腰三角形，它的底比一条腰长6厘米，周长是42厘米，先用42-6即可求出腰长的3倍是多少厘米，除以3即可求出腰长多少厘米，用腰长加6即可求出底长多少厘米.\n【详解】180°-90°-55°=35°;腰:(42-6)÷3=36÷3=12(厘米),底:12+6=18(厘米)。所以另一个锐角是35°,这个等腰三角形的底长18厘米，一条腰长12厘米。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和",
+                                 "三角形的周长"
+                             ],
+        "source":  "24-25四年级下·河北保定·期末"
+    },
+    {
+        "id":  "275",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三角形的三条边都是整厘米数，其中有两条边分别是15厘米和18厘米。那么第三条边最长是( )厘米，最短是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "32;4",
+        "solution":  "【分析】根据三角形三边关系，两边之和大于第三边，两边之差小于第三边，据此分析第三条边的长度范围后填空即可.\n【详解】15+18=33(厘米),18-15=3(厘米),3厘米\u003c第三条边\u003c33厘米,3+1=4(厘米),33-1=32(厘米)。所以第三条边最长是32厘米，最短是4厘米。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·河北保定·期末"
+    },
+    {
+        "id":  "276",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把一个三角形平均分成两个小三角形，每个小三角形内角和为( )°,五边形内角和为( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "180;540",
+        "solution":  "【分析】无论形状和大小，任何一个三角形的内角和都是180°;已知任意三角形的内角和为180°,利用过多边形一个顶点向与它不相邻的顶点依次画线段，把多边形分割成三角形的办法，将分成的几个三角形的内角和相加，就是多边形的内角和。多边形内角和=180°*(边数-2),据此即可解答.\n【详解】180°*(5-2)=180°*3=540°。所以每个小三角形内角和为180°,五边形内角和为540°。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "多边形的内角和"
+                             ],
+        "source":  "24-25四年级下·湖南张家界·期末"
+    },
+    {
+        "id":  "277",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "红领巾是一个等腰三角形，它的一个底角为30°,顶角为( )°,按角分，红领巾还是一个( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "120;钝角",
+        "solution":  "【分析】等腰三角形两个底角相等，它的一个底角为30°,三角形的内角和为180°,顶角等于180°减去两个底角的度数，即180°-30°-30°=120°,按角分,120°是钝角，所以红领巾还是一个钝角三角形.\n【详解】180°-30°-30°=120°,按角分，红领巾还是一个钝角三角形。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·湖南张家界·期末"
+    },
+    {
+        "id":  "278",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "两根小棒的长度分别为7cm和10cm,如果再选一根小棒拼成一个三角形，选的这根小棒应比( )cm长，比( )cm短。",
+        "options":  [
+
+                    ],
+        "answer":  "3;17",
+        "solution":  "【分析】根据三角形两边之和大于第三边，两边之差小于第三边，算出已知两条边长的和，差，第三条边长应比和短，比差长.\n【详解】10+7=17(cm),10-7=3(cm),所以选的这根小棒应比3cm长，比17cm短。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·湖南张家界·期末"
+    },
+    {
+        "id":  "279",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三角形最大角的度数是最小角的3倍，中间角的度数是最小角的2倍，最大的一个角是( )°,这是个( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "90;直角",
+        "solution":  "【分析】三角形的内角和是180°,把最小角的度数看作1份，则最大角的度数是3份，中间角的度数是2份，所以180°里面包括6份最小角的度数，用180°除以6，即可求出最小角的度数;再用最小角的度数乘3，即可求出最大角的度数;三角形的分类，一个内角为钝角的三角形是钝角三角形，三个内角都为锐角的三角形是锐角三角形，一个内角为直角的三角形是直角三角形，据此解答.\n【详解】最小角的度数:180°/6=30°,最大角的度数:30°*3=90°,所以最大的一个角是90°,这是个直角三角形。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·湖南益阳·期末"
+    },
+    {
+        "id":  "280",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个多边形的内角和是1800°,这个多边形是( )边形。",
+        "options":  [
+
+                    ],
+        "answer":  "十二",
+        "solution":  "【分析】根据多边形的内角和公式:(n-2)×180°,则用1800°除以180°,再加上2，就能求出这个图形是几边形.\n【详解】1800°/180°=10,10+2=12，这个多边形是十二边形。",
+        "knowledge_points":  [
+                                 "多边形的内角和"
+                             ],
+        "source":  "24-25四年级下·湖南邵阳·期末"
+    },
+    {
+        "id":  "281",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形的周长是22cm,底边长6cm,它的一条腰长( )cm。",
+        "options":  [
+
+                    ],
+        "answer":  "8",
+        "solution":  "【分析】已知等腰三角形两腰相等，其中一底边长是6cm,用周长减去底边的长度，即得到这个三角形的两条腰长。据此解答.\n【详解】(22-6)÷2=16÷2=8(cm)。所以，这个三角形的一条腰长是8cm。",
+        "knowledge_points":  [
+                                 "三角形的周长",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·湖南长沙·期末"
+    },
+    {
+        "id":  "282",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "∠1,∠2,∠3是一个三角形的3个内角。如果∠1=47°,∠2=86°,那么∠3=( )度;这是一个( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "47;锐角",
+        "solution":  "【分析】根据三角形内角和为180°,已知∠1=47°,∠2=86°,先求出第三个角的度数，按角分类：三角形按角分为锐角三角形(三个角均小于90°),直角三角形(一个角为90°)和钝角三角形(一个角大于90°)。再根据最大角的度数判断三角形的类型.\n【详解】根据分析可知:∠3=180°-47°-86°=47°,三个角分别为47°,86°,47°,其中最大角是86°,小于90°,属于锐角。因此，这个三角形是锐角三角形。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·重庆梁平·期末"
+    },
+    {
+        "id":  "283",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在一个等腰三角形中，它的一个底角是一个顶角的2倍，这个三角形的顶角是( )°,一个底角是( )°,按角分它是一个( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "36;72;锐角",
+        "solution":  "【分析】三个内角和一共是180°,假设顶角占1份，一个底角是一个顶角的2倍，根据题意一个底角占2份，又因为是等腰三角形，两底角相等，所以另一个底角也占2份，总共将内角和分成1+2+2=5份，将180°/5=36°,就是顶角的度数，然后便可确定三角形的类型。三角形按角的大小，可分为三种：三个角都是锐角的三角形，叫锐角三角形;有一个角是直角的三角形，叫直角三角形;有一个角是钝角的三角形，叫钝角三角形.\n【详解】180°/(1+2+2)=180°/5=36°,36°*2=72°,36°\u003c72°\u003c90°,所以这个三角形的顶角是36°,一个底角是72°,按角分它是一个锐角三角形。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "284",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形的一边是7厘米，另一边是15厘米，这个等腰三角形的周长是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "37",
+        "solution":  "【分析】根据等腰三角形的两个腰相等，因此分两种情况:①当7厘米是腰时，此时三角形的三边分别是7厘米,7厘米,15厘米，此时7+7=14,14\u003c15;不符合三角形三边关系：任意两边之和大于第三边，所以7厘米不能是腰;②当15厘米是腰时，此时三角形的三边分别是15厘米,15厘米,7厘米，此时7+15=22,22\u003e15;符合三角形三边关系;所以这个等腰三角形的三边分别是15厘米,15厘米,7厘米，求三角形的周长就是把三条边的长加起来;据此解答.\n【详解】根据分析可知，这个等腰三角形的三边分别是15厘米,15厘米,7厘米,15+15+7=37(厘米)。所以这个等腰三角形的周长是37厘米。",
+        "knowledge_points":  [
+                                 "三角形三边关系",
+                                 "三角形的周长",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·重庆铜梁·期末"
+    },
+    {
+        "id":  "285",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用三根小棒围成一个等腰三角形，其中两根小棒的长为4cm和8cm,那么第三根小棒长为( )cm。",
+        "options":  [
+
+                    ],
+        "answer":  "8",
+        "solution":  "【分析】根据题意可知，等腰三角形的两个腰相等，因此分两种情况:①当4cm为腰时，此时三角形的三边分别为4cm,4cm,8cm,此时4+4=8，不符合三角形三边关系：任意两边之和大于第三边，所以4cm不能是腰;②当8cm为腰时，此时三角形的三边分别为8cm,8cm,4cm,8+4=12,12\u003e8，符合三角形三边关系;所以第三边的长是8cm;据此解答.\n【详解】根据分析可知，用三根小棒围成一个等腰三角形，其中两根小棒的长为4cm和8cm,那么第三根小棒长为8cm。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·湖南怀化·期末"
+    },
+    {
+        "id":  "286",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用一根长13厘米的铁丝围成一个等腰三角形，它的顶角是34°,那么底角是( )°;其中一条腰是5厘米，那么底边长是( )cm。",
+        "options":  [
+
+                    ],
+        "answer":  "73;3",
+        "solution":  "【分析】根据三角形的特点和等腰三角形的特征，等腰三角形底角的度数=(三角形的内角和-顶角的度数)÷2;等腰三角形底边的长度=等腰三角形周长-腰长*2，由此解答此题.\n【详解】(180°-34°)÷2=146°/2=73°;13-5×2=13-10=3(厘米)。所以底角是73°,底边长是3cm。",
+        "knowledge_points":  [
+                                 "三角形三边关系",
+                                 "三角形的周长",
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·浙江温州·期末"
+    },
+    {
+        "id":  "287",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个七边形的内角和为( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "900",
+        "solution":  "【分析】多边形的内角和计算公式为：多边形的内角和=(边数-2)×180°,据此可解答此题.\n【详解】(7-2)×180°=5×180°=900°,所以一个七边形的内角和为900°。",
+        "knowledge_points":  [
+                                 "多边形的内角和"
+                             ],
+        "source":  "24-25四年级下·浙江台州·期末"
+    },
+    {
+        "id":  "288",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形的两条边分别是4.8cm和9.6cm,围成这个等腰三角形至少需要( )cm长的绳子。",
+        "options":  [
+
+                    ],
+        "answer":  "24",
+        "solution":  "【分析】已知等腰三角形的两条腰长相等。已知一个等腰三角形的两条边分别是4.8cm和9.6cm,根据三角形任意两边之和必须大于第三边，因此需要考虑两种情况：两条腰长为4.8cm,底边为9.6cm;两条腰长为9.6cm,底边为4.8cm。根据三角形的三边关系判断哪种情况能围成等腰三角形，再把三条边相加，即是围成这个等腰三角形至少需要绳子的全长.\n【详解】情况一：等腰三角形的两条腰长为4.8cm;4.8+4.8=9.6(cm),9.6=9.6，不符合三角形的三边关系，所以4.8cm,4.8cm,9.6cm不能构成三角形。情况二：等腰三角形的两条腰长为9.6cm;9.6+4.8=14.4(cm),14.4\u003e9.6，符合三角形的三边关系，所以4.8cm,9.6cm,9.6cm能构成三角形。等腰三角形的周长:9.6+9.6+4.8=24(cm)。所以，围成这个等腰三角形至少需要24cm长的绳子。",
+        "knowledge_points":  [
+                                 "三角形三边关系",
+                                 "三角形的周长",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·湖北随州·期末"
+    },
+    {
+        "id":  "289",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形的顶角是80°,它的一个底角是( )°,按角分类，它是( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "50;锐角",
+        "solution":  "【分析】根据题意，等腰三角形的两个底角相等，已知顶角为80°,利用三角形内角和180°减去顶角，再除以2，即可得到一个底角的度数。按角分类：三角形按角分为锐角三角形(三个角均小于90°),直角三角形(一个角为90°)和钝角三角形(一个角大于90°)。以此答题即可.\n【详解】根据分析可知:(180°-80°)÷2=100°/2=50°,三个角分别为80°,50°,50°,都是锐角，所以它的一个底角是50°,按角分类它是锐角三角形。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·湖南长沙·期末"
+    },
+    {
+        "id":  "290",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "过多边形一个顶点的所有对角线将这个多边形分成三个三角形，这个多边形是( )边形，内角和是( )度。",
+        "options":  [
+
+                    ],
+        "answer":  "五;540",
+        "solution":  "【分析】在多边形中，有一个重要的规律:n边形过一个顶点的所有对角线和边所组成的三角形的个数是n-2。这里设这个多边形的边数为n,题目中说过多边形一个顶点的所有对角线将这个多边形分成3个三角形，所以根据上述规律可以列出等式n-2=3。因为过多边形一个顶点的所有对角线将这个多边形分成了3个三角形，而我们知道一个三角形的内角和是180°,那么这个五边形的内角和就等于这3个三角形内角和之和.\n【详解】3+2=5,180°*3=540°,所以这个多边形是五边形，内角和是540度。",
+        "knowledge_points":  [
+                                 "多边形的内角和"
+                             ],
+        "source":  "24-25四年级下·辽宁鞍山·期末"
+    },
+    {
+        "id":  "291",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形的一条边是7厘米，另一条边是5厘米，围成这个等腰三角形至少需要( )厘米长的绳子。",
+        "options":  [
+
+                    ],
+        "answer":  "17",
+        "solution":  "【分析】等腰三角形的两条腰相等，则第三条边长5厘米或7厘米，根据三角形任意两边之和大于第三边，任意两边之差小于第三边，可知5+5\u003e7,5+7\u003e7，则等腰三角形三条边的长分别为5厘米,5厘米,7厘米或7厘米,7厘米,5厘米，要使围成这个等腰三角形需要的绳子最少，就要使它的腰比底边短，所以要使它的腰长是5厘米,5厘米，底边长是7厘米，将这三条边长度相加求和即可.\n【详解】5+5+7=17(厘米),围成这个等腰三角形至少需要17厘米长的绳子。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的周长",
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·山东菏泽·期末"
+    },
+    {
+        "id":  "292",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把一个三角形的三个角剪下来，可以拼成一个( )角，三角形的内角和是( )°,\"n\"边形的内角和是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "平;180;(n-2)×180°",
+        "solution":  "【分析】三角形的内角和是180°,平角等于180°,所以把一个三角形的三个角剪下来，可以拼成一个平角;如图,![](images/aa01e76469f94500b05fee700aec111f72e4a2bdc4bbe72baa20add62ab3c64b.jpg)![](images/45ac781a81f4c0ad40721ccfd15b3bdae0650562b64f7285e6f5588576e50540.jpg)四边形可以分割成2个三角形，所以四边形的内角和是2个三角形内角和相加，又因为四边形的边数是4，所以四边形的内角和也就是(边数-2)个180°的和;五边形可以分割成3个三角形，所以五边形的内角和是3个三角形内角和相加，又因为五边形的边数是5，所以五边形的内角和也就是(边数-2)个180°的和;通过观察发现，多边形的内角和=(多边形的边数-2)×180°,据此解答.\n【详解】根据分析可知，把一个三角形的三个角剪下来，可以拼成一个平角，三角形的内角和是180°,\"n\"边形的内角和是(n-2)×180°。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "多边形的内角和"
+                             ],
+        "source":  "24-25四年级下·山东菏泽·期末"
+    },
+    {
+        "id":  "293",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形的一条边长是10厘米，另一条边长是5厘米，这个等腰三角形的周长是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "25",
+        "solution":  "【分析】等腰三角形的两腰相等，那么第三条边是5厘米或10厘米，当是5厘米时，两边之和5加5得10厘米，与另一条边相等，这样的三个线段不能组成三角形，那么第三条边只能是10厘米,10与5的和15厘米大于另一条边，所以能组成三角形，即这个三角形的三条边长是5厘米,10厘米,10厘米，把三条边的长度相加，即可求出周长.\n【详解】根据分析可知：这个等腰三角形的三条边长是5厘米,10厘米,10厘米;10+10+5=25(厘米)。所以这个等腰三角形的周长是25厘米。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的周长"
+                             ],
+        "source":  "24-25四年级下·山东菏泽·期末"
+    },
+    {
+        "id":  "294",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三角形的边长都是整厘米数，其中两条边长分别为2厘米和7厘米，第三条边最长是( )厘米，此时周长是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "8;17",
+        "solution":  "【分析】根据三角形的两边之和大于第三边，以及两边之差小于第三边，据此求出第三边的范围，再根据边长是整数厘米，由此可知第三边最长是多少;然后将三边长度相加，即可得出三角形的周长.\n【详解】2+7=9(厘米),7-2=5(厘米),所以5厘米\u003c第三条边的长度\u003c9厘米，由于是整数厘米，所以第三条边最大为8厘米。2+7+8=17(厘米)。综上可知，第三条边最长是8厘米，此时周长是17厘米。",
+        "knowledge_points":  [
+                                 "三角形三边关系",
+                                 "三角形的周长"
+                             ],
+        "source":  "24-25四年级下·江西赣州·期末"
+    },
+    {
+        "id":  "295",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个直角三角形的一个锐角是46°,则另一个锐角是( )°;一个等腰三角形的两条边分别是60厘米和25厘米，这个等腰三角形的周长是( )厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "44;145",
+        "solution":  "【分析】根据题意，直角三角形里面有一个角是直角，是90°。三角形的内角和是180°,用180°减去90°,再减去46°,就是另一个锐角的度数。等腰三角形的两边相等，三角形的任意两边之和大于第三条边，据此可知第三条边长多少厘米。再把三条边加起来就是三角形的周长.\n【详解】180°-90°-46°=44°;60+25\u003e60,60+60\u003e25,25+25\u003c60,25+60\u003e25，所以，等腰三角形的第三条边长是60厘米。60+60+25=145(厘米)。因此，另一个锐角是44°,这个等腰三角形的周长是145厘米。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和",
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·江西吉安·期末"
+    },
+    {
+        "id":  "296",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一块等腰三角形菜地，周长是32米，腰长10米，底边长( )米。",
+        "options":  [
+
+                    ],
+        "answer":  "12",
+        "solution":  "【分析】根据等腰三角形的定义：两条边相等的三角形叫作等腰三角形，相等的两边叫作腰。三角形的周长就是三条边的总长，因此用周长减去两边腰的长度，就是底边的长度。据此进行解答即可.\n【详解】32-10×2=32-20=12(米)。因此底边长12米。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的周长"
+                             ],
+        "source":  "24-25四年级下·湖北黄石·期末"
+    },
+    {
+        "id":  "297",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在一个三角形中，第一条边长是3厘米，第二条边长是9厘米，第三条边最短是( )厘米。(填整厘米数)",
+        "options":  [
+
+                    ],
+        "answer":  "7",
+        "solution":  "【分析】根据三角形三边关系，两边之和大于第三边，两边之差小于第三边，据此找出第三条边最短的长度即可.\n【详解】9-3=6(厘米),9+3=12(厘米),6厘米\u003c第三条边\u003c12厘米,6+1=7(厘米)。所以第三条边最短是7厘米。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·河南驻马店·期末"
+    },
+    {
+        "id":  "298",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "笑笑在纸上画了一个三角形，量得其中两个内角之和为80°,这个三角形是( )三角形，另一个内角是( )度。",
+        "options":  [
+
+                    ],
+        "answer":  "钝角;100",
+        "solution":  "【分析】三角形的内角和是180°,用180°减去两个内角的和80°,就是剩下的一个角的度数。三角形中有一个角是钝角，那么这个三角形就是钝角三角形.\n【详解】180°-80°=100°,100°是一个钝角。所以这个三角形是钝角三角形，另一个内角是100度。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·河南驻马店·期末"
+    },
+    {
+        "id":  "299",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "东方红广场的一个花坛近似三角形，它的一个内角是38°,另一个内角是72°,那么第三个内角是( )°,这个三角形是( )三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "70;锐角",
+        "solution":  "【分析】根据三角形内角和为180°,已知两个内角分别为38°和72°,可求出第三个内角;再根据三个内角的大小判断三角形类型.\n【详解】三角形内角和为180°,因此第三个内角为:180°-38°-72°=70°。三个内角分别为38°,72°,70°,均小于90°,因此该三角形为锐角三角形。综上，第三个内角是70°,这个三角形是锐角三角形。",
+        "knowledge_points":  [
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·湖南湘潭·期末"
+    },
+    {
+        "id":  "300",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "世界著名的跨江，跨海大桥都有三角形，这是应用了三角形的( ),生活中应用三角形这一特性的还有( )。",
+        "options":  [
+
+                    ],
+        "answer":  "稳定性;自行车架",
+        "solution":  "【分析】三角形具有稳定性;三角形稳定，因为它三条边首尾相接，形成了稳定结构。三角形的三个角固定了三条边的位置，使得整个结构在受力时，力的分布更加均匀，不易发生形变。其次，三角形的三条边相互支撑，一旦其中一条边受到拉力或压力，其他两条边会产生反作用力，相互抵消，从而保持整个结构的稳定。这种力的平衡状态使得三角形结构在受力时能够均匀分散压力，提高了结构的抗压能力和稳定性。据此解答即可.\n【详解】世界著名的跨江，跨海大桥都有三角形，这是应用了三角形的稳定性，生活中应用三角形这一特性的还有自行车架。",
+        "knowledge_points":  [
+                                 "三角形的稳定性及应用"
+                             ],
+        "source":  "24-25四年级下·云南玉溪·期末"
+    },
+    {
+        "id":  "301",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "一个直角三角形两个锐角的和是()。",
+        "options":  [
+                        "A. 90°",
+                        "B. 小于90°",
+                        "C. 大于90°",
+                        "D. 不能确定"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】有一个角是直角的三角形叫作直角三角形，则直角三角形有一个内角是90°,一个直角三角形两个锐角的和=三角形的内角和-90°,据此解答.\n【详解】三角形的内角和是180°。180°-90°=90°,所以，一个直角三角形两个锐角的和是90°。故答案为:A",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "三角形的分类"
+                             ],
+        "source":  "2025·内蒙古通辽·小升初真题"
+    },
+    {
+        "id":  "302",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面不是利用三角形稳定性的是()。",
+        "options":  [
+                        "A. ![](images/65eca79058fa5c782e4624d05b56a2cd9396200541843148c97850dbb814f299.jpg)伸缩门",
+                        "B. ![](images/2026-03-11%20183004.jpg)房顶钢架",
+                        "C. ![](images/2026-03-11%20183031.jpg)固定树木",
+                        "D. ![](images/2026-03-11%20183057.jpg)人字梯"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】平行四边形容易变形，三角形具有稳定的特性，据此找出图中的三角形即可.\n【详解】A。伸缩门利用了平行四边形容易变形;B。房顶钢架利用了三角形稳定性;C。固定树木利用了三角形稳定性;D。人字梯利用了三角形稳定性。不是利用三角形稳定性的是A。故答案为:A",
+        "knowledge_points":  [
+                                 "平行四边形的不稳定性及应用",
+                                 "三角形的稳定性及应用"
+                             ],
+        "source":  "24-25五年级上·北京房山·期末"
+    },
+    {
+        "id":  "303",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一个三角形两条边的长度分别是9厘米和4厘米，第三条边的长度可能是()。",
+        "options":  [
+                        "A. 4厘米",
+                        "B. 5厘米",
+                        "C. 9厘米",
+                        "D. 13厘米"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】三角形任意两边的和大于第三边，两边之差小于第三边。据此解答.\n【详解】4+9=13(厘米),9-4=5(厘米),因此第三条边的长度大于5厘米，小于13厘米，选项中9厘米符合题意。故答案为:C",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25五年级上·北京房山·期末"
+    },
+    {
+        "id":  "304",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面图形能单独密铺的有()个。![](images/d1f4a837a78cd71bdae9e044b03b52d9.png)",
+        "options":  [
+                        "A. 1",
+                        "B. 2",
+                        "C. 3",
+                        "D. 4"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】用一种或几种全等图形进行拼接，图形之间没有空隙，也不重复，叫图形的密铺。密铺时公共顶点处角度和应为360°,据此判断.\n【详解】图一圆不能单独密铺;图二正六边形内角120°,360°/120°=3，能单独密铺;图三正八边形内角135°,360°/135°不能整除，不能;图四正五边形内角108°,360°/108°不能整除，不能。即能单独密铺的有1个。故答案为:A",
+        "knowledge_points":  [
+                                 "多边形的内角和",
+                                 "图形的密铺"
+                             ],
+        "source":  "24-25五年级上·河北邢台·期末"
+    },
+    {
+        "id":  "305",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "如图，有一个梯形ABCD,如果点B沿AB所在的直线慢慢往左移动，与A点重合后停止运动。这个图形的变化过程是()。![](images/5cc0babb09f29fd278dac9637269229069d56b07207976a2f94681ce1c4543b6.jpg)",
+        "options":  [
+                        "A. 梯形→平行四边形→梯形",
+                        "B. 梯形→三角形→平行四边形→梯形",
+                        "C. 梯形→平行四边形→三角形",
+                        "D. 梯形→平行四边形→梯形→三角形"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】只有一组对边平行的是梯形;两组对边平行的是平行四边形;三条线段围成的是三角形。判断点B移动过程中图形变化.\n【详解】初始是梯形ABCD;B向左移动至AB=DC时变为平行四边形;继续左移又变回梯形;当B与A重合时形成三角形ACD。所以过程是：梯形→平行四边形→梯形→三角形。故答案为:D",
+        "knowledge_points":  [
+                                 "三角形的概念及表示方法",
+                                 "梯形的概念及特点",
+                                 "平行四边形的概念及特点"
+                             ],
+        "source":  "24-25五年级上·重庆合川·期末"
+    },
+    {
+        "id":  "306",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "等腰三角形的两边长分别是4cm和2cm,则这个三角形的周长是()。",
+        "options":  [
+                        "A. 10cm",
+                        "B. 8cm",
+                        "C. 6cm",
+                        "D. 10cm或8cm"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】分情况讨论等腰三角形的腰长：若腰为2cm,则2+2=4，不满足三角形三边关系;若腰为4cm,则可构成三角形.\n【详解】可行边长为4cm,4cm,2cm,周长=4+4+2=10(cm)。故答案为:A",
+        "knowledge_points":  [
+                                 "三角形三边关系",
+                                 "三角形的周长",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "2025·四川凉山·小升初真题"
+    },
+    {
+        "id":  "307",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一个三角形最小的内角是50°,关于这个三角形，正确的说法是()。",
+        "options":  [
+                        "A. 可能是钝角三角形",
+                        "B. 可能是直角三角形",
+                        "C. 一定是锐角三角形"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】三角形内角和180°。最小角为50°,其余两角都不小于50°,据此判断最大角范围.\n【详解】其余两角和为180°-50°=130°。若其中一个也取最小50°,另一个为80°,最大角最多80°\u003c90°,所以一定是锐角三角形。故答案为:C",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "三角形的分类"
+                             ],
+        "source":  "2025·江西上饶·小升初真题"
+    },
+    {
+        "id":  "308",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一个三角形的最小的内角是46°,那么这个三角形是()三角形。",
+        "options":  [
+                        "A. 锐角",
+                        "B. 直角",
+                        "C. 钝角",
+                        "D. 无法确定"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】最小角46°,则其余两角都不少于46°,最大角最大值可由180°-46°-46°求得.\n【详解】最大角最大为88°,小于90°,所以三个角都为锐角，是锐角三角形。故答案为:A",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "三角形的内角和",
+                                 "三角形的分类"
+                             ],
+        "source":  "2025·河南开封·小升初真题"
+    },
+    {
+        "id":  "309",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面4组数据，表示的都是一个三角形中的其中两个角的度数。哪组的三角形是等腰三角形？()",
+        "options":  [
+                        "A. 40°,100°",
+                        "B. 30°,60°",
+                        "C. 45°,55°",
+                        "D. 50°,60°"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】三角形内角和180°,若三个角中有两个相等则是等腰三角形.\n【详解】A第三角=40°,出现40°与40°,是等腰;其余选项第三角分别为90°,80°,70°,均无两个角相等。故答案为:A",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "2025·重庆梁平·小升初真题"
+    },
+    {
+        "id":  "310",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面4组数据，表示的都是小棒的长度。哪组的小棒不能拼成三角形。()",
+        "options":  [
+                        "A. 4cm,4cm,6cm",
+                        "B. 3cm,7cm,10cm",
+                        "C. 8cm,8cm,8cm",
+                        "D. 5cm,9cm,12cm"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】判定是否能成三角形，检查较短两边之和是否大于最长边.\n【详解】A:4+4\u003e6可;B:3+7=10，不满足大于;C:8+8\u003e8可;D:5+9\u003e12可。不能拼成的是B。故答案为:B",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "2025·重庆梁平·小升初真题"
+    },
+    {
+        "id":  "311",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "三角形中最小的一个角是50°,这是一个()三角形。",
+        "options":  [
+                        "A. 锐角",
+                        "B. 直角",
+                        "C. 钝角",
+                        "D. 不能确定"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】最小角是50°,其余两角都\u003e=50°,据三角形内角和求最大角最大值.\n【详解】最大角最多为180°-50°-50°=80°,小于90°,所以一定是锐角三角形。故答案为:A",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "三角形的内角和",
+                                 "三角形的分类"
+                             ],
+        "source":  "2025·山东临沂·小升初真题"
+    },
+    {
+        "id":  "312",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "两个同样大小的等腰直角三角形，可以拼成一个()。",
+        "options":  [
+                        "A. 梯形",
+                        "B. 平行四边形或正方形",
+                        "C. 直角三角形或钝角三角形"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】依据等腰直角三角形可拼组图形的常见结果判断.\n【详解】两个同样大小的等腰直角三角形可拼成平行四边形或正方形。故答案为:B",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "2025·四川南充·小升初真题"
+    },
+    {
+        "id":  "313",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "爸爸不小心把一块三角形玻璃打碎了(如图所示),现在要去重新配一块同样的玻璃，他只需要带()号玻璃去即可。![](images/c8f88b4d11f05749819f47843487a71ffe85e4064977d9fed3ab6f1d8b433ba0.jpg)",
+        "options":  [
+                        "A. ①",
+                        "B. ②",
+                        "C. ③"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】重配同样三角形，需要能确定三角形形状的关键要素(至少两个角和一条边等信息).\n【详解】③保留了两个角和一条边及其余边部分信息，能确定原三角形;①②信息不足。故答案为:C",
+        "knowledge_points":  [
+                                 "三角形的概念及表示方法"
+                             ],
+        "source":  "24-25四年级下·河北邢台·期末"
+    },
+    {
+        "id":  "314",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "爸爸不小心把一块三角形玻璃打碎了(如图所示),现在要去重新配一块同样的玻璃，他只需要带()号玻璃去即可。![](images/c66268ade7f924807a615e99e60d8631cf2525b27cc4f0f9763d63755ed7c5ba.jpg)",
+        "options":  [
+                        "A. ①",
+                        "B. ②",
+                        "C. ③"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】确定三角形需足够约束条件，两个角与一条边可确定.\n【详解】③同时保留两个角和一条边，能配同样玻璃;①②不能。故答案为:C",
+        "knowledge_points":  [
+                                 "三角形的概念及表示方法"
+                             ],
+        "source":  "24-25四年级下·河北邢台·期末"
+    },
+    {
+        "id":  "315",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "小明把一副三角尺像如图这样重叠在一起，形成了新的角,∠1是()。![](images/0d272109d94d69881c972cba61941b69f56985e0ca451bc016a67e37d35aeebf.jpg)",
+        "options":  [
+                        "A. 15°",
+                        "B. 75°",
+                        "C. 105°",
+                        "D. 135°"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】利用三角尺常见角度与三角形内角和180°求解.\n【详解】∠1=180°-(45°+30°)=105°。故答案为:C",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "角度的计算"
+                             ],
+        "source":  "24-25六年级下·四川南充·期末"
+    },
+    {
+        "id":  "316",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "一个三角形中有两个内角的和等于90°,这个三角形()是直角三角形。",
+        "options":  [
+                        "A. 一定",
+                        "B. 可能",
+                        "C. 不一定",
+                        "D. 不可能"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】三角形内角和是180°.\n【详解】第三个角=180°-90°=90°,所以该三角形一定是直角三角形。故答案为:A",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25五年级上·广东汕头·期中"
+    },
+    {
+        "id":  "317",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "如果等腰三角形的两条边分别长5厘米和7厘米，那么这个等腰三角形的周长是()。",
+        "options":  [
+                        "A. 17厘米",
+                        "B. 19厘米",
+                        "C. 17厘米或19厘米"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】等腰三角形两边相等，需分\"腰=5\"与\"腰=7\"两种情况，并满足三边关系.\n【详解】5,5,7可成，周长17;7,7,5可成，周长19。所以周长可能是17厘米或19厘米。故答案为:C",
+        "knowledge_points":  [
+                                 "三角形三边关系",
+                                 "三角形的周长",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级上·山东菏泽·期中"
+    },
+    {
+        "id":  "318",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "古人的智慧是无穷的，在商周时期建造窗户时，木工师傅都需要在框架上再钉一根木条，使框架更牢固，下面符合木工师傅做法的是()。![](images/7e4df7a9d09f559bc638a47dfd3ef5b9c9182b34890d5cf74cea620ce6d42002.jpg)",
+        "options":  [
+                        "A. ![](images/82b5df2342602e38e5ed01440941f56fc9f9b8d1a3e22043a83c27ba099fe262.jpg)",
+                        "B. ![](images/aacf283753d3df81dd2e134d0ad262b9de9f12e227c7206b941dd0e3e14644b7.jpg)",
+                        "C. ![](images/0d5212b6f56faed581091c19dc43bad4f5e562b58b924a981e10f81421e0932d.jpg)"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】三角形稳定，四边形易变形，需看是否形成三角形加固.\n【详解】A形成两个三角形，最牢固;B,C形成四边形，易变形。故答案为:A",
+        "knowledge_points":  [
+                                 "三角形的稳定性及应用"
+                             ],
+        "source":  "24-25四年级下·河南郑州·期末"
+    },
+    {
+        "id":  "319",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "在平行四边形框架上钉一根木条进行加固，加固效果最差的是()。",
+        "options":  [
+                        "A. ![](images/bd41ee11e2390305c0b297a2038cc44919164ecb773671b94316cb65adcc93ba.jpg)",
+                        "B. ![](images/e3164fc8e78e6c491d7f103a75739787665bd3101ee6b407202d4ccb111abe4a.jpg)",
+                        "C. ![](images/ded76c122000beaa3fc09907b734de112a9bce2cd5b6f0917089db1abb8e5b83.jpg)",
+                        "D. ![](images/8836d7b43ef67df205b966581ab344a50d10155665abe63568757b8d84ccb8d7.jpg)"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】加固效果好坏关键在于是否形成三角形稳定结构.\n【详解】A,B,C都能形成三角形,D不能形成三角形，所以加固效果最差。故答案为:D",
+        "knowledge_points":  [
+                                 "三角形的稳定性及应用"
+                             ],
+        "source":  "24-25四年级下·河北唐山·期末"
+    },
+    {
+        "id":  "320",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "一扇窗户打开后用窗钩将其固定(如图),运用到的数学知识是()。![](images/4265dcf2b6be5d2d092332b4079b7c76a0b90579bbf019e2d5de304c60a347a0.jpg)",
+        "options":  [
+                        "A. 两点之间线段最短",
+                        "B. 平行四边形的不稳定性",
+                        "C. 三角形具有稳定性",
+                        "D. 图形平移后形状大小不变，位置变化。"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】窗钩，窗户和窗沿形成三角形结构，利用三角形稳定性固定.\n【详解】运用到的是三角形具有稳定性。故答案为:C",
+        "knowledge_points":  [
+                                 "三角形的稳定性及应用"
+                             ],
+        "source":  "24-25四年级下·四川南充·期末"
+    },
+    {
+        "id":  "321",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "如图是由边长分别为3厘米和5厘米的正方形组成的，三角形ABC中与底AB对应的高是()厘米。![](images/6a31d76341dc34c279e4c05f02b919bc3e8e9305e65d51d4f5135d0eac57c148.jpg)",
+        "options":  [
+                        "A. 2",
+                        "B. 3",
+                        "C. 5",
+                        "D. 8"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】从顶点向底AB作垂线即为对应高，结合图中正方形边长判断.\n【详解】底AB对应的高与小正方形边长相同，为3厘米。故答案为:B",
+        "knowledge_points":  [
+                                 "三角形的高及画法"
+                             ],
+        "source":  "24-25四年级下·四川南充·期末"
+    },
+    {
+        "id":  "322",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "有一根20cm长的木棒，现要把它截成3段，并将截成后的3根木棒围成一个三角形，下列的截法中，能围成三角形的是()。",
+        "options":  [
+                        "A. 1cm,9cm,10cm",
+                        "B. 4cm,14cm,2cm",
+                        "C. 5cm,8cm,7cm",
+                        "D. 2cm,12cm,6cm"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】检验三角形三边关系：任意两边和大于第三边.\n【详解】A有1+9=10不行;B有2+4\u003c14不行;C各组两边和均大于第三边可行;D有2+6\u003c12不行。故答案为:C",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25六年级下·广东东莞·期末"
+    },
+    {
+        "id":  "323",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "有一个三角形纸片被撕掉一个角(如图),按角分，这是一个()三角形。![](images/71877903b2bf52471a212865021604ea7072b256bd2b524f0113d1d468fc7745.jpg)",
+        "options":  [
+                        "A. 锐角",
+                        "B. 等边",
+                        "C. 直角",
+                        "D. 钝角"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】利用三角形内角和180°求出被撕掉角，再按角分类.\n【详解】第三角=180°-46°-67°=67°,三个角都小于90°,是锐角三角形。故答案为:A",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·陕西商洛·期末"
+    },
+    {
+        "id":  "324",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面三组小棒中，能摆成三角形的一组是()。",
+        "options":  [
+                        "A. 1厘米,2厘米,4厘米",
+                        "B. 2厘米,2厘米,5厘米",
+                        "C. 4厘米,5厘米,9厘米",
+                        "D. 3厘米,4厘米,5厘米"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】比较较短两边和与最长边，需满足大于关系.\n【详解】A:1+2\u003c4不行;B:2+2\u003c5不行;C:4+5=9不行;D:3+4\u003e5可行。故答案为:D",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·河北廊坊·期末"
+    },
+    {
+        "id":  "325",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面各组线段中，不能围成三角形的是()。",
+        "options":  [
+                        "A. 3厘米,4厘米,7厘米",
+                        "B. 5厘米,5厘米,5厘米",
+                        "C. 3厘米,4厘米,5厘米"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】依据三角形任意两边之和大于第三边判断.\n【详解】A中3+4=7，不满足大于，不能围成;B,C均满足可围成。故答案为:A",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·河南驻马店·期末"
+    },
+    {
+        "id":  "326",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一根磁力棒由14节同样的圆柱形小磁铁连接而成。要把这根磁力棒分成三部分，首尾相接围成一个三角形。如图所示，小芳已经把磁力棒分为两部分，接下来在()处分开，一定能围成三角形。![](images/951de3fed9fd11772a5af93414ad6d5b61261c65cba612cc7fd51cb4464d7f0f.jpg)![](images/41c5c6f23dd2ede94ef67c028d7504ef47e312a41b0ff173bd87828e71fda725.jpg)",
+        "options":  [
+                        "A. A",
+                        "B. B",
+                        "C. C"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】按各分点得到三段长度，逐一验证三角形三边关系.\n【详解】A:3,3,8,3+3\u003c8不行;B:6,3,5，各两边和都大于第三边可行;C:6,7,1,6+1=7不行。故答案为:B",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·河北廊坊·期末"
+    },
+    {
+        "id":  "327",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "有长度分别为6厘米,11厘米的两根小棒，要想围成一个三角形，第三根小棒的长度可以是()。",
+        "options":  [
+                        "A. 5厘米",
+                        "B. 9厘米",
+                        "C. 17厘米"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】第三边应满足两边差与和之间:11-6\u003c第三边\u003c11+6.\n【详解】即5\u003c第三边\u003c17。A等于5不行,C等于17不行,B=9符合。故答案为:B",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·陕西安康·期末"
+    },
+    {
+        "id":  "328",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面说法正确的是()。",
+        "options":  [
+                        "A. 一个三角形只有一条高",
+                        "B. 等边三角形一定是锐角三角形",
+                        "C. 一个三角形有一个角是锐角，那么这个三角形是锐角三角形"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】逐项依据三角形高，等边三角形性质及锐角三角形定义判断.\n【详解】A错，三角形有3条高;B对，等边三角形每角60°,是锐角三角形;C错，仅有一个锐角不能判定。故答案为:B",
+        "knowledge_points":  [
+                                 "三角形的高及画法",
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·河南驻马店·期末"
+    },
+    {
+        "id":  "329",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形，顶角是86°,其中一个底角是()。",
+        "options":  [
+                        "A. 47°",
+                        "B. 86°",
+                        "C. 94°"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】等腰三角形两底角相等，底角和=180°-顶角.\n【详解】底角和=180°-86°=94°,一个底角=94°/2=47°。故答案为:A",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·河北唐山·期末"
+    },
+    {
+        "id":  "330",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下列图形中最具有稳定性的是()。",
+        "options":  [
+                        "A. ![](images/b4ce354eb687faf0a1c46f3986dad2890b1cac20a7daaa40568353137991bd9d.jpg)",
+                        "B. ![](images/8ea7ebe65c03e79fe29dbff2af84ca281990cc54dfeee56192b650c343acd9f3.jpg)",
+                        "C. ![](images/7d8ebf610d1f29e7eedb670683ffcf40d410c73b6facbca3b0b370e89b5700dd.jpg)"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】三角形稳定，平行四边形类结构易变形.\n【详解】A内部含多个三角形，稳定性强;B,C主要是长方形/正方形结构，稳定性较差。故答案为:A",
+        "knowledge_points":  [
+                                 "三角形的稳定性及应用"
+                             ],
+        "source":  "24-25四年级下·山西吕梁·期末"
+    },
+    {
+        "id":  "331",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面三组线段中可以围成三角形的是()(单位：厘米)。",
+        "options":  [
+                        "A. 2,1,1",
+                        "B. 2,3,5",
+                        "C. 7,8,9"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】应用三角形任意两边和大于第三边.\n【详解】A:1+1=2不行;B:2+3=5不行;C:7+8\u003e9且其余也成立，可围成。故答案为:C",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·山西吕梁·期末"
+    },
+    {
+        "id":  "332",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "把一根长24dm的木棍截成3段围成三角形，一定不能截在()处。![](images/3e847c791fca2d4693f0cfef99f86aec2fd739ccfdc50139a8727dce3537340f.jpg)",
+        "options":  [
+                        "A. A",
+                        "B. B",
+                        "C. C",
+                        "D. D"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】若第一刀在中点，会出现一段等于其余两段和，不能成三角形.\n【详解】24÷2=12(dm),在12dm处截会使两段和等于第三段，不满足三角形三边关系，所以一定不能截在C处。故答案为:C",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·湖北十堰·期末"
+    },
+    {
+        "id":  "333",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "哪组小棒不能围成三角形(单位:cm):()。",
+        "options":  [
+                        "A. 3,4,5",
+                        "B. 3,3,3",
+                        "C. 2,2,6",
+                        "D. 4,5,6"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】依据三角形任意两边之和大于第三边判断.\n【详解】A,B,D均满足;C中2+2\u003c6，不满足条件，不能围成三角形。故答案为:C",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·湖南长沙·期末"
+    },
+    {
+        "id":  "334",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "在一个等腰三角形中，已知一个角是20°,这个三角形不可能是()。",
+        "options":  [
+                        "A. 钝角三角形",
+                        "B. 直角三角形",
+                        "C. 锐角三角形"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】等腰三角形内角和是180°。若20°是顶角，则两个底角相等;若20°是底角，则另一个底角也是20°,据此分别判断三角形类型.\n【详解】若20°是顶角，则底角为(180°-20°)÷2=80°,三角形为锐角三角形;若20°是底角，则顶角为180°-20°×2=140°,三角形为钝角三角形。所以不可能是直角三角形。故答案为:B",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·山西晋中·期末"
+    },
+    {
+        "id":  "335",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面各组线段中,()不能围成三角形。",
+        "options":  [
+                        "A. 4cm,4cm,4cm",
+                        "B. 2cm,3cm,7cm",
+                        "C. 5cm,5cm,2cm",
+                        "D. 3cm,4cm,5cm"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据三角形三边关系：任意两边之和大于第三边.\n【详解】A中4+4\u003e4，可围成;B中2+3=5\u003c7，不能围成;C中5+2\u003e5，可围成;D中3+4\u003e5，可围成。故答案为:B",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·陕西安康·期末"
+    },
+    {
+        "id":  "336",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "以下说法正确的是()。",
+        "options":  [
+                        "A. 锐角三角形的内角和小于钝角三角形",
+                        "B. 梯形具有稳定性",
+                        "C. 直角三角形的两个锐角之和一定是90°",
+                        "D. 等边三角形不是等腰三角形"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】逐项根据三角形内角和，图形稳定性，直角三角形性质和等腰三角形定义判断.\n【详解】A错，任意三角形内角和都为180°;B错，梯形(四边形)不具稳定性;C对，直角三角形两个锐角和=180°-90°=90°;D错，等边三角形是特殊的等腰三角形。故答案为:C",
+        "knowledge_points":  [
+                                 "三角形的稳定性及应用",
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·广东深圳·期末"
+    },
+    {
+        "id":  "337",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "《交通法》规定驾驶自行车必须年满12周岁。自行车构造中的三角架之所以做成三角形，其中很重要原因就是利用三角形的()。",
+        "options":  [
+                        "A. 内角和是180°",
+                        "B. 稳定性",
+                        "C. 容易变形"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】三角形在结构上不易变形，常用于需要稳固的框架.\n【详解】自行车三角架用三角形是为了增强结构稳定性，不是利用内角和，也不是\"容易变形\"。故答案为:B",
+        "knowledge_points":  [
+                                 "三角形的稳定性及应用",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·安徽淮北·期末"
+    },
+    {
+        "id":  "338",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一个四边形，沿一条直线剪一刀后得到的新图形的内角和是()。",
+        "options":  [
+                        "A. 180°",
+                        "B. 360°",
+                        "C. 540°",
+                        "D. 以上都有可能"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】切割方式不同，得到的新图形可能是三角形，四边形或五边形.\n【详解】三角形内角和180°,四边形内角和360°,五边形内角和540°,都可能出现。故答案为:D",
+        "knowledge_points":  [
+                                 "多边形的内角和",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·辽宁盘锦·期末"
+    },
+    {
+        "id":  "339",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "为了加固这个巨型魔方墙，需要在它的后面支起一根钢管，具体如下图所示，那么钢管的长度不可能是()。![](images/53be0762d9fcb361ab4959b52a105f270086e5b6bab90f8b8cd0ca66881e19ef.jpg)",
+        "options":  [
+                        "A. 15分米",
+                        "B. 30分米",
+                        "C. 28分米"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】由图可构成三角形，钢管长度为第三边，满足三角形三边关系.\n【详解】已知两边约为22分米和15分米时，第三边应满足22-15\u003c第三边\u003c22+15，即7\u003c第三边\u003c37。结合题意钢管应大于22分米,15分米明显不可能。故答案为:A",
+        "knowledge_points":  [
+                                 "三角形三边关系",
+                                 "三角形的稳定性及应用"
+                             ],
+        "source":  "24-25四年级下·河南郑州·期末"
+    },
+    {
+        "id":  "340",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "用下面6根小棒，你能围出()种三角形(单位:cm)。![](images/5e37c28f308e45cef15b69a58e93a20417d2f1bac7589ab22a3bf91da2e3cccb.jpg)![](images/1404ece27c807f5d2b2637c9cd86e585c6ad1f51961d33a7e0de1b6a7eb7983e.jpg)",
+        "options":  [
+                        "A. 2",
+                        "B. 3",
+                        "C. 4",
+                        "D. 5"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】按三角形三边关系枚举可行边长组合.\n【详解】可组成的不同组合为:2,5,6;2,6,6;5,6,6;6,6,6，共4种。故答案为:C",
+        "knowledge_points":  [
+                                 "三角形三边关系",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·河南许昌·期末"
+    },
+    {
+        "id":  "341",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "如图，若∠1=30°,则∠2=()。![](images/ed97ab994cf567a53687a2a0a946749dad7d8090daba7fe9f71ab05d1f5fb373.jpg)",
+        "options":  [
+                        "A. 60°",
+                        "B. 100°",
+                        "C. 120°",
+                        "D. 140°"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】先在直角三角形中求另一个锐角，再利用平角求∠2.\n【详解】另一个锐角=180°-90°-30°=60°,与∠2互补，所以∠2=180°-60°=120°。故答案为:C",
+        "knowledge_points":  [
+                                 "平角，周角的认识及特征",
+                                 "三角形的内角和",
+                                 "角度的计算"
+                             ],
+        "source":  "24-25六年级下·内蒙古赤峰·期末"
+    },
+    {
+        "id":  "342",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "一个三角形被遮住了两个角(如下图),这个三角形一定是()。![](images/4a983a608c6af58e056a877ce489f1ea5a0fbb820d5028a496e5d38224293eaf.jpg)",
+        "options":  [
+                        "A. 锐角三角形",
+                        "B. 直角三角形",
+                        "C. 钝角三角形"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】按角分类：有一个钝角就是钝角三角形.\n【详解】图中露出的角是钝角，因此该三角形一定是钝角三角形。故答案为:C",
+        "knowledge_points":  [
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·河南郑州·期末"
+    },
+    {
+        "id":  "343",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "计算一个五边形的内角和时，我们可以把它分成3个三角形(如图),它的内角和就是180°×3=540°。像这样，计算八边形内角和也可以把它分成几个三角形，那么八边形的内角和是()。![](images/0c9c093582b446f67e1baba5575cc6d39d64997dd1dedd95d1262deb75edb2fc.jpg)",
+        "options":  [
+                        "A. 900°",
+                        "B. 1080°",
+                        "C. 1260°"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】n边形可分成(n-2)个三角形，内角和=(n-2)×180°.\n【详解】八边形内角和=(8-2)×180°=6×180°=1080°。故答案为:B",
+        "knowledge_points":  [
+                                 "多边形的内角和",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·河南洛阳·期末"
+    },
+    {
+        "id":  "344",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面是3个被盖住的三角形，不能直接判断出三角形的类型的是()。",
+        "options":  [
+                        "A. ![](images/7df7391bd68bff4f50a9570be565c3e15692d6a0ee698230fc712a03a069ba8e.jpg)",
+                        "B. ![](images/39f634cba8d1cd1c14acb84e269b2385a05c178eae2c88e3ccda9cd44f0d1028.jpg)",
+                        "C. ![](images/9f7c1ccf8de74f1e1fbb83b27ecbf96f137cd69af794dbf0a91e932cad0df9a9.jpg)"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】已知一个钝角可判钝角三角形，已知一个直角可判直角三角形;仅知一个锐角无法唯一确定类型.\n【详解】A能判为钝角三角形,C能判为直角三角形;B只露出锐角，可能是锐角/直角/钝角三角形。故答案为:B",
+        "knowledge_points":  [
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·河南洛阳·期末"
+    },
+    {
+        "id":  "345",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "(如图)巩义市街心公园宣传展示栏的支架应用了三角形的()。![](images/7e519b1b9a5e54978d70788f3af5528b6dcc4bb81927e48053f7c3aa7bb3dc13.jpg)",
+        "options":  [
+                        "A. 稳定性",
+                        "B. 内角和是180°",
+                        "C. 任意两边的和大于第三边"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】支架结构追求牢固，利用的是三角形稳定性.\n【详解】宣传栏支架应用了三角形的稳定性。故答案为:A",
+        "knowledge_points":  [
+                                 "三角形的稳定性及应用"
+                             ],
+        "source":  "24-25四年级下·河南郑州·期末"
+    },
+    {
+        "id":  "346",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "做手工模型时，聪聪想用一根长24cm的木棒剪2刀，用得到的三根木棒首尾相接围一个三角形。他第一刀不能从()处剪开。(24cm长的木棒如图，图中每段![](images/2026-03-11%20213634.jpg)一样长)![](images/f23cecde8dcb708eddfae64df1ca05398bd28f6b866ca63e6ad7692d4274a381.jpg)",
+        "options":  [
+                        "A. A",
+                        "B. B",
+                        "C. C",
+                        "D. D"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】第一刀若剪在中点，后续无论怎么剪，都会出现两段和等于第三段，不能成三角形.\n【详解】24÷2=12(cm),B点是中点位置，所以第一刀不能从B处剪开。故答案为:B",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·河南郑州·期末"
+    },
+    {
+        "id":  "347",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "古都西安，文物古迹俯拾皆是。方方想真切感受西安的人文气息，从西安城墙出发，经过大明宫国家遗址公园，来到永兴坊，然后直接从永兴坊回到西安城墙，正好走了三角形的三条边，从大明宫国家遗址公园到永兴坊大约有()米。![](images/547f6d14451b6b37804523fefb0f6a3349aa749844be880a36dea4cdc9207cd9.jpg)",
+        "options":  [
+                        "A. 890",
+                        "B. 3200",
+                        "C. 4980"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】已知两边2900米和2000米，第三边应满足两边差与和之间.\n【详解】第三边范围:2900-2000\u003c第三边\u003c2900+2000，即900\u003c第三边\u003c4900。A小于900不行,C大于4900不行,B符合。故答案为:B",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·河南郑州·期末"
+    },
+    {
+        "id":  "348",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面三角形被遮住了一部分，按角分，不能判断出一定是什么三角形的是()。",
+        "options":  [
+                        "A. ![](images/529076a60b927f9610b99cc233f8fa02c6ef4c163f73e9cfa7ce26994ed158ad.jpg)",
+                        "B. ![](images/4ec53dc92b2cc66df6afce8381cbe1ffd8e18050b239d060d29877a3cc2bc3ab.jpg)",
+                        "C. ![](images/2b92fdba755f4f5e1ad340db1faea7a6edf239d546bcfb2d69dd6768022a9a9d.jpg)",
+                        "D. ![](images/3626277e35500c195d750625d453c22cb1fcec449b16c738c9b2fae61bfbeb47.jpg)"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】依据露出角信息判断是否可唯一确定三角形类型.\n【详解】A露出钝角，可判钝角三角形;B露出直角，可判直角三角形;D可判锐角三角形;C信息不足，不能确定。故答案为:C",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "三角形的内角和",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·山东日照·期末"
+    },
+    {
+        "id":  "349",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下图是一台升降机，工人站在升降机上放心工作是因为()。![](images/ed2fefc8e500c8063306f729b40107a137fc7f5e57b5a24eb6b2a7fadad98f9e.jpg)",
+        "options":  [
+                        "A. 升降机上的三角形具有稳定性",
+                        "B. 三角形有三条边",
+                        "C. 三角形任意两边之和大于第三边"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】升降机结构稳固关键在于三角形不易变形.\n【详解】工人可放心工作，是因为升降机上的三角形结构具有稳定性。故答案为:A",
+        "knowledge_points":  [
+                                 "三角形的稳定性及应用"
+                             ],
+        "source":  "24-25四年级下·河北衡水·期末"
+    },
+    {
+        "id":  "350",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "油纸伞是中国国家级非物质文化遗产，其伞骨结构巧妙运用了数学原理。观察图中油纸伞的伞骨展开图，伞骨设计成许多小三角形的原因是()。![](images/68a6cf77ca616da1bb21c37a1202f4bdd6958292ff65887b0140d5790aee502d.jpg)",
+        "options":  [
+                        "A. 三角形数量多显得更精致",
+                        "B. 竹条只能做成三角形形状",
+                        "C. 三角形图案更符合传统审美",
+                        "D. 三角形具有稳定性，能让伞面牢固不易变形"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】结构设计核心是受力稳定，三角形不易变形.\n【详解】伞骨设计成许多小三角形是因为三角形具有稳定性，能让伞面牢固不易变形。故答案为:D",
+        "knowledge_points":  [
+                                 "三角形的稳定性及应用"
+                             ],
+        "source":  "24-25四年级下·福建龙岩·期末"
+    },
+    {
+        "id":  "351",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "小文家位于点A处，体育场位于点B处，小文从家去体育场锻炼身体，按路线()走最近。![](images/aa37acbf595be10ce7a60ccb103363a050df22c22e446caed74eef6a74d99731.jpg)",
+        "options":  [
+                        "A. A→C→B",
+                        "B. A→E→C→B",
+                        "C. A→D→B",
+                        "D. A→F→D→B"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】两点之间线段最短.\n【详解】从A到B走最短路径对应题图中的A→C→B路线。故答案为:A",
+        "knowledge_points":  [
+                                 "两点间线段最短与两点间的距离"
+                             ],
+        "source":  "24-25四年级下·河北邢台·期末"
+    },
+    {
+        "id":  "352",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "有四根木条，长度分别为4cm,5cm,9cm,13cm,如果取其中的三根做成一个三角形框架，那么这个三角形框架的周长应该是()。",
+        "options":  [
+                        "A. 18cm",
+                        "B. 26cm",
+                        "C. 27cm"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】先枚举任取三根的组合，再用三角形三边关系筛选.\n【详解】可行组合只有5cm,9cm,13cm(5+9\u003e13)。周长=5+9+13=27cm。故答案为:C",
+        "knowledge_points":  [
+                                 "三角形三边关系",
+                                 "三角形的周长"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "353",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面每组中的三个角，不可能是同一个三角形的三个角的是()。",
+        "options":  [
+                        "A. 15°,79°,86°",
+                        "B. 120°,35°,25°",
+                        "C. 60°,60°,60°",
+                        "D. 27°,103°,70°"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】三角形内角和必须是180°.\n【详解】A,B,C三组和均为180°;D组27°+103°+70°=200°,不可能。故答案为:D",
+        "knowledge_points":  [
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·四川广元·期末"
+    },
+    {
+        "id":  "354",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面各组小棒中，能围成三角形的是()。",
+        "options":  [
+                        "A. 4cm,5cm,9cm",
+                        "B. 6dm,6dm,10dm",
+                        "C. 6m,13m,20m",
+                        "D. 2cm,5cm,8cm"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】判断较短两边和是否大于最长边.\n【详解】A中4+5=9不行;B中6+6\u003e10可行;C中6+13\u003c20不行;D中2+5\u003c8不行。故答案为:B",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·四川广元·期末"
+    },
+    {
+        "id":  "355",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面是四组线段(单位：厘米),其中不能围成三角形的是()。",
+        "options":  [
+                        "A. 3,4,5",
+                        "B. 6,7,8",
+                        "C. 4,4,4",
+                        "D. 4,6,10"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】三角形任意两边和大于第三边.\n【详解】D中4+6=10，不满足大于关系，不能围成三角形;其余都可围成。故答案为:D",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·安徽淮北·期末"
+    },
+    {
+        "id":  "356",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "我国桥梁总数屡破世界纪录。其中杨浦大桥和苏通大桥是大跨度斜拉桥的代表。斜拉桥的外观设计中运用了三角形的知识。这是因为()。![](images/ea938e720d90d3d1dee2025fce9ffe89b87b726a1e7b91cbf79e4ed978e04729.jpg)",
+        "options":  [
+                        "A. 三角形有稳定性",
+                        "B. 三角形任意两边的和大于第三边",
+                        "C. 三角形内角和是180°"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】桥梁结构重在稳定抗变形，三角形最稳定.\n【详解】斜拉桥运用三角形知识主要是利用三角形稳定性。故答案为:A",
+        "knowledge_points":  [
+                                 "三角形的稳定性及应用"
+                             ],
+        "source":  "24-25四年级下·山西晋中·期末"
+    },
+    {
+        "id":  "357",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面说法正确的是()。",
+        "options":  [
+                        "A. 等腰三角形一定是锐角三角形",
+                        "B. 等腰三角形是特殊的等边三角形",
+                        "C. 直角三角形可能是等腰三角形",
+                        "D. 钝角三角形不可能是等腰三角形"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】等腰三角形可为锐角，直角或钝角;等边三角形是特殊的等腰三角形.\n【详解】A错，等腰不一定锐角;B错，关系应反过来;C对，如45°-45°-90°;D错，钝角三角形也可能是等腰。故答案为:C",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·浙江杭州·期末"
+    },
+    {
+        "id":  "358",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在三角形ABC中∠1=∠2,∠3=∠4,∠A=70°。那么∠5等于()。![](images/ae6b11f3c0183cb4fc7b01514d87d1c854e68a113e6bcf895d634254834860c9.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "125",
+        "solution":  "【分析】先求∠1+∠2+∠3+∠4=180°-70°=110°。又因∠1=∠2,∠3=∠4，所以∠2+∠4=110°/2=55°。再用平角求∠5.\n【详解】∠5=180°-55°=125°,所以∠5等于125°。",
+        "knowledge_points":  [
+                                 "三角形的内角和"
+                             ],
+        "source":  "2024三年级上·全国·竞赛"
+    },
+    {
+        "id":  "359",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如图,∠1=()°,∠2=()°。![](images/bd5a1b5fbce28a9bbfb573c2d50947055eafeaf42a1f8fc2c6985b32957b6cdf.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "45;45",
+        "solution":  "【分析】∠2与135°组成平角，先求∠2;再用直角三角形内角和求∠1.\n【详解】∠2=180°-135°=45°。∠1=180°-90°-45°=45°。所以∠1=45°,∠2=45°。",
+        "knowledge_points":  [
+                                 "平角，周角的认识及特征",
+                                 "三角形的内角和",
+                                 "角度的计算"
+                             ],
+        "source":  "24-25四年级上·河北石家庄·期中"
+    },
+    {
+        "id":  "360",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在一个三角形中,∠1=48°,∠2=82°,∠3=()°;在一个等腰三角形中，一个底角是36°,顶角是()°。",
+        "options":  [
+
+                    ],
+        "answer":  "50;108",
+        "solution":  "【分析】三角形内角和是180°。先求第一个三角形的第三个角;等腰三角形两个底角相等，再求顶角.\n【详解】∠3=180°-48°-82°=50°。若一个底角是36°,则顶角=180°-36°-36°=108°。答案为50;108。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·广东汕头·期末"
+    },
+    {
+        "id":  "361",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "下面三角形是一个顶角为40°的等腰三角形，沿虚线剪掉一个三角形后得到一个四边形，在剩下的四边形中,∠3+∠4=()°。![](images/ca6a82ff005c7442e5d715f9542eb42767cf6e0be1548ec49d0dc9c6a0584ffd.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "250",
+        "solution":  "【分析】先求等腰三角形底角，再用四边形内角和360°减去已知两个角.\n【详解】底角=(180°-40°)÷2=70°。所以∠3+∠4=360°-(40°+70°)=250°。",
+        "knowledge_points":  [
+                                 "多边形的内角和",
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·山东日照·期末"
+    },
+    {
+        "id":  "362",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "由下图可知，五边形的内角和是()°,六边形的内角和是()°。![](images/863941a627017eb5bc91ba0e265a59754d9b03540ab4a73e149d72a01324240c.jpg)![](images/3462fd9bed42eec1db65d48791eb71ebf79395979709a846a09b30521cae19a1.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "540;720",
+        "solution":  "【分析】五边形可分3个三角形，六边形可分4个三角形，三角形内角和均为180°.\n【详解】五边形内角和=3×180°=540°;六边形内角和=4×180°=720°。",
+        "knowledge_points":  [
+                                 "多边形的内角和"
+                             ],
+        "source":  "24-25四年级下·山西长治·期末"
+    },
+    {
+        "id":  "363",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如图三角形，已知∠1=65°,则∠2=()°;沿图中虚线剪下一个小三角形后，剩余部分的内角和是()°。![](images/22759620d6db268513f4e77155986c4cd235bf9ffe213b5f949cf46b84b8a1c5.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "25;360",
+        "solution":  "【分析】先用三角形内角和求∠2;剪去一个角后剩余图形是四边形，内角和360°.\n【详解】∠2=180°-(90°+65°)=25°。剩余四边形内角和=(4-2)×180°=360°。",
+        "knowledge_points":  [
+                                 "多边形的内角和",
+                                 "三角形的内角和",
+                                 "角度的计算"
+                             ],
+        "source":  "24-25四年级下·河北唐山·期末"
+    },
+    {
+        "id":  "364",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如图:∠1=35°,∠2=105°,∠3=()°。![](images/f382b0e65726488b90a5c9955f6e72c67b71045338acb3a26027a92901805367.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "140",
+        "solution":  "【分析】先在三角形内求第三个内角，再利用平角关系求∠3.\n【详解】三角形第三角=180°-35°-105°=40°,所以∠3=180°-40°=140°。",
+        "knowledge_points":  [
+                                 "平角，周角的认识及特征",
+                                 "三角形的内角和",
+                                 "角度的计算"
+                             ],
+        "source":  "24-25四年级下·河南驻马店·期末"
+    },
+    {
+        "id":  "365",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如图，一块三角形纸片被撕去了一个角，这个角是()度，原来这块纸片的形状是()三角形，也是()三角形。![](images/baceae95c57481eb01c316e93cf2087bd156c09c2bdb4e7ff6c32cff80af5491.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "67;等腰;锐角",
+        "solution":  "【分析】用三角形内角和求被撕去角;再按角和按边分类判断.\n【详解】被撕去角=180°-67°-46°=67°。三个角分别为67°,67°,46°,有两个角相等，对应等腰三角形;且三个角都小于90°,也是锐角三角形。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·陕西宝鸡·期末"
+    },
+    {
+        "id":  "366",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "三角形ABD和等腰三角形ADC组成一个直角三角形(如图)。已知∠1=60°,那么∠2=()°,∠3=()°。![](images/7ac4f3ddee723986f88ee95e493ae985a8d2226733f00c68274d3c0cc7408eac.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "30;60",
+        "solution":  "【分析】先由平角求等腰三角形ADC顶角，再求其底角∠2;最后在直角三角形中求∠3.\n【详解】ADC顶角=180°-60°=120°,所以∠2=(180°-120°)÷2=30°。又∠3=180°-90°-30°=60°。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "角度的计算"
+                             ],
+        "source":  "24-25四年级下·浙江宁波·期末"
+    },
+    {
+        "id":  "367",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在探究五边形的内角和时，小丽把五边形分成了5个三角形(如图),所以她认为：五边形的内角和是900°。小丽说得()(填\u0027对\u0027或\u0027不对\u0027)。因为()。![](images/12b8c701c2cd01b49904553d54173f5925acad4aec5c7cce595d48a008635fbb.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "不对;五边形的内角和是540°,不是900°",
+        "solution":  "【分析】多边形内角和公式为(n-2)×180°.\n【详解】五边形内角和=(5-2)×180°=540°,所以\u0027900°\u0027说法错误，应填\u0027不对\u0027,原因为五边形内角和是540°。",
+        "knowledge_points":  [
+                                 "多边形的内角和",
+                                 "四边形，五边形及多边形的初步认识"
+                             ],
+        "source":  "24-25四年级下·山西晋中·期末"
+    },
+    {
+        "id":  "368",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一根10cm长的小棒，要把它剪成三段(长度都是整cm),围成一个三角形。如果第一刀剪在2cm处，那么第二刀应剪在()cm处。![](images/690e76084af12fed67e22fd535623d73749b06eeefc9308e629792ce0fef0160.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】第一刀后已有一段2cm,剩余8cm再切成两段，逐个位置检验是否满足三角形三边关系.\n【详解】若第二刀在3cm、4cm、5cm处，得到的三边分别会出现1+2\u003c7,2+2\u003c6,2+3=5，都不能成三角形;第二刀在6cm处得到2,4,4，满足两边和大于第三边。故应剪在6cm处。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·浙江杭州·期末"
+    },
+    {
+        "id":  "369",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如下图,∠A=()°,这个三角形按角分是()三角形。![](images/18e2f315c03944bee9add2e3d9cc8d4a7f3fb1df96abbefbbaf55aa3ebb1935a.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "80;锐角",
+        "solution":  "【分析】先用平角求∠ABC,再用三角形内角和求∠A,最后按角分类.\n【详解】∠ABC=180°-135°=45°,∠A=180°-45°-55°=80°。三个角都小于90°,所以是锐角三角形。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·浙江杭州·期末"
+    },
+    {
+        "id":  "370",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "如图所示,∠1=(),∠2=(),这个三角形按角分是()三角形，按边分是()三角形。![](images/ecffdab5a55e8bb0f0e0c04995d9262f84bb1625ef772fc73ba108a85d3b1716.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "30°;60°;钝角;等腰",
+        "solution":  "【分析】三角形内角和是180°,可先用180°减去已知两个角得到∠1;∠2与120°组成平角，可用180°减120°求得。再根据最大角判断按角分类，根据有两角相等判断按边分类.\n【详解】∠1=180°-120°-30°=30°;∠2=180°-120°=60°。该三角形最大角为120°,所以是钝角三角形;又有两个角都是30°,对应两边相等，所以是等腰三角形。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·四川乐山·期末"
+    },
+    {
+        "id":  "371",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如图，一张三角形纸片被撕去了一个角，这个角是()度，原来这张纸片的形状按边分是()三角形。![](images/75a8a22b42a1bbbc2a28affa382685efc0cf4bd8e94bf86f870fc31e616a9354.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "66;等腰",
+        "solution":  "【分析】根据三角形内角和是180°,用180°减去已知两个角得到被撕去角。再根据有两个角相等可判定为等腰三角形.\n【详解】被撕去角=180°-48°-66°=66°。三个角中有两个66°,对应两边相等，所以按边分是等腰三角形。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·河南焦作·期末"
+    },
+    {
+        "id":  "372",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "下图是文化节的起始位置到结束位置的三条路线，最近的是()号路线，用线段的知识解释是因为(),用三角形的三边关系来解释是因为()。![](images/368b9759c58b2d4499c3e5d22ceefbbee31e71b271464d5d0b35d5fec669a027.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "②;两点之间线段最短;两边之和大于第三边",
+        "solution":  "【分析】两点之间线段最短;同时从三角形三边关系看，折线路径对应两边之和，大于直接连线的第三边.\n【详解】最近的是②号路线。线段知识解释为\"两点之间线段最短\";三角形三边关系解释为\"两边之和大于第三边\"。",
+        "knowledge_points":  [
+                                 "三角形三边关系",
+                                 "两点间线段最短与两点间的距离"
+                             ],
+        "source":  "24-25四年级下·河南洛阳·期末"
+    },
+    {
+        "id":  "373",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如图，三角形ABC是等边三角形，三角形BCD是等腰三角形。三角形ABC的周长最大可能是()cm。(所有边的长度都是整厘米数)![](images/d241eefa236931547ce2ea42c986401ed6bb1ce45a1b0cc8bedbcb03550a1fd0.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "57",
+        "solution":  "【分析】在三角形BCD中，已知两边为10cm,10cm,第三边BC满足0\u003cBC\u003c20，且为整数。要使等边三角形ABC周长最大，需BC取最大整数.\n【详解】BC最大可取19cm,则等边三角形ABC周长=19×3=57(cm)。",
+        "knowledge_points":  [
+                                 "三角形三边关系",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·河南安阳·期末"
+    },
+    {
+        "id":  "374",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如图，一个等腰三角形和一个正方形拼成一个梯形，若梯形的上底是7厘米，则下底是()厘米，高是()厘米。![](images/91592ae28819fbc021502cfc5855ce312a96ed7f89dc3bf58c016ae00b2eb269.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "14;7",
+        "solution":  "【分析】由图可知正方形边长即上底7厘米，等腰三角形两腰与正方形边长对应。下底为上底加一条腰，高与正方形边长相同.\n【详解】下底=7+7=14(cm),高=7(cm)。",
+        "knowledge_points":  [
+                                 "梯形的概念及特点",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "正方形的概念及特点"
+                             ],
+        "source":  "24-25四年级下·河南洛阳·期末"
+    },
+    {
+        "id":  "375",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个西瓜味的三角形冰棒被咬了一个角(如图),已知其他两个角都是75°,则被咬掉的角是()°。按角分，原来的冰棒形状是一个()三角形。![](images/d99224390e7c56ffc848ce0d603c5374db5d16160ac562c806940be0023c1938.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "30;锐角",
+        "solution":  "【分析】三角形内角和180°,先求缺角;再按角分类判断三角形类型.\n【详解】缺角=180°-75°-75°=30°。三个角都小于90°,所以是锐角三角形。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·河北衡水·期末"
+    },
+    {
+        "id":  "376",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在数学活动课上，同学们准备把一根长10cm的吸管剪成3段围成三角形。如果第一刀剪在3cm处(如图),第二刀可以剪在()处。(选填字母a,b,c)![](images/457b0311054ecf60b524d0831d12be55398c8ce437d8f021858bd3efb9a3aa98.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "c",
+        "solution":  "【分析】分别检验第二刀在a,b,c处时三段长度是否满足三角形两边之和大于第三边.\n【详解】剪在a处得到1,3,6,1+3\u003c6，不行;剪在b处得到2,3,5,2+3=5，不行;剪在c处得到3,3,4，满足三边关系，可以围成三角形。所以选c。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·福建漳州·期末"
+    },
+    {
+        "id":  "377",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "已知下图三角形ABC是直角三角形,∠1+∠2=()度。![](images/e372bc92a2ece68df145accf8c5f7c03acf447f2f46a8a1e910e95050fb07c41.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "270",
+        "solution":  "【分析】∠1与∠ACB互补,∠2与∠ABC互补;又因直角三角形中∠ACB+∠ABC=90°.\n【详解】∠1+∠2=(180°-∠ACB)+(180°-∠ABC)=360°-(∠ACB+∠ABC)=360°-90°=270°。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "角度的计算"
+                             ],
+        "source":  "24-25四年级下·福建福州·期末"
+    },
+    {
+        "id":  "378",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "下图三角形ABC是一个等腰三角形，三角形ADC也是一个等腰三角形，已知∠1=30°,则∠C=()°,∠2=()°。![](images/59509a2bcae4a3242b31b8daa46f98808ce171ffed4277242d750541c0b16ba0.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "30;120",
+        "solution":  "【分析】等腰三角形底角相等。由ABC等腰得∠C=∠1;在ADC中再用内角和求顶角.\n【详解】∠C=30°。在三角形ADC中两底角各30°,所以∠2=180°-30°-30°=120°。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·福建福州·期末"
+    },
+    {
+        "id":  "379",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形，其中一个底角是42°,它的顶角是(),按角分它是一个()三角形。",
+        "options":  [
+
+                    ],
+        "answer":  "96°;钝角",
+        "solution":  "【分析】等腰三角形两个底角相等，先求顶角，再按角分类.\n【详解】顶角=180°-42°-42°=96°。因96°\u003e90°,所以是钝角三角形。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "380",
+        "unit":  "第五单元三角形",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "用相同的正五边形可以密铺。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】能密铺要求拼接点角度和为360°.\n【详解】正五边形每个内角108°,360°不能被108°整除，所以不能单独密铺。原说法错误。",
+        "knowledge_points":  [
+                                 "多边形的内角和",
+                                 "图形的密铺"
+                             ],
+        "source":  "24-25五年级上·河北保定·期末"
+    },
+    {
+        "id":  "381",
+        "unit":  "第五单元三角形",
+        "question_type":  "判断题",
+        "difficulty":  0.94,
+        "question":  "所有的三角形都有3条高。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】三角形每个顶点都可向对边作高.\n【详解】三角形有3个顶点，因此都有3条高，锐角，直角，钝角三角形都成立。原说法正确。",
+        "knowledge_points":  [
+                                 "三角形的高及画法"
+                             ],
+        "source":  "24-25五年级上·广东河源·期末"
+    },
+    {
+        "id":  "382",
+        "unit":  "第五单元三角形",
+        "question_type":  "判断题",
+        "difficulty":  0.94,
+        "question":  "三角形的底与对应的高互相垂直。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】高是从顶点向对边或其延长线作的垂线段.\n【详解】既然是垂线段，则底与对应的高成直角，互相垂直。原说法正确。",
+        "knowledge_points":  [
+                                 "三角形的高及画法",
+                                 "垂直的特征"
+                             ],
+        "source":  "25-26五年级上·广东揭阳·期中"
+    },
+    {
+        "id":  "383",
+        "unit":  "第五单元三角形",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "用5厘米,8厘米,3厘米长的三根小棒，不能围成一个三角形。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】检验较短两边和是否大于第三边.\n【详解】5+3=8，等于第三边，不满足\"两边和大于第三边\"，不能围成三角形。原说法正确。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "2025·湖南邵阳·小升初模拟"
+    },
+    {
+        "id":  "384",
+        "unit":  "第五单元三角形",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "三角形的内角和是180°,四边形的内角和是240°。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】四边形可分成2个三角形.\n【详解】三角形内角和180°正确;四边形内角和应为2×180°=360°,不是240°。原说法错误。",
+        "knowledge_points":  [
+                                 "多边形的内角和",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·河南驻马店·期末"
+    },
+    {
+        "id":  "385",
+        "unit":  "第五单元三角形",
+        "question_type":  "判断题",
+        "difficulty":  0.94,
+        "question":  "一个三角形的三条边的长度分别是5cm,8cm和13cm。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】三角形任意两边之和大于第三边.\n【详解】5+8=13，不满足大于关系，不能构成三角形。原说法错误。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25六年级下·贵州黔东南·期末"
+    },
+    {
+        "id":  "386",
+        "unit":  "第五单元三角形",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "一个三角形中的两个内角的和是130度，这个三角形一定是钝角三角形。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】先求第三角，再讨论另外两个角的可能情况.\n【详解】第三角=180°-130°=50°。前两个角可能有钝角，也可能都为锐角(如80°和50°),因此不一定是钝角三角形。原说法错误。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·河北廊坊·期末"
+    },
+    {
+        "id":  "387",
+        "unit":  "第五单元三角形",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "等腰三角形不一定是等边三角形。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】等腰三角形是至少两边相等，等边三角形是三边都相等.\n【详解】如边长4,4,5是等腰但不是等边，所以\"等腰三角形不一定是等边三角形\"正确。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级上·山东菏泽·期中"
+    },
+    {
+        "id":  "388",
+        "unit":  "第五单元三角形",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "三角形的两条边长分别是3厘米,4厘米，第三条边长可能是7厘米。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】第三边需满足大于两边差且小于两边和.\n【详解】3+4=7，第三边若为7则等于两边和，不能构成三角形。原说法错误。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级上·山东菏泽·期中"
+    },
+    {
+        "id":  "389",
+        "unit":  "第五单元三角形",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "一个三角形里虽然有两个锐角，但这个三角形不一定是锐角三角形。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】第三个角可能是锐角，直角或钝角.\n【详解】若第三角是直角则是直角三角形，若第三角是钝角则是钝角三角形，所以不一定是锐角三角形。原说法正确。",
+        "knowledge_points":  [
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·河南驻马店·期末"
+    },
+    {
+        "id":  "390",
+        "unit":  "第五单元三角形",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "三根木条分别长4厘米,5厘米,9厘米，用它们可以摆一个三角形。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】较短两边和需大于第三边.\n【详解】4+5=9，等于第三边，不能摆成三角形。原说法错误。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·河南驻马店·期末"
+    },
+    {
+        "id":  "391",
+        "unit":  "第五单元三角形",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "一个三角形中最大的角是91°,这个三角形一定是钝角三角形。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】有一个角大于90°即为钝角三角形.\n【详解】最大角91°是钝角，所以该三角形一定是钝角三角形。原说法正确。",
+        "knowledge_points":  [
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·辽宁盘锦·期末"
+    },
+    {
+        "id":  "392",
+        "unit":  "第五单元三角形",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "两个完全一样的三角形可以拼成一个平行四边形。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】两个全等三角形沿对应边拼接，可得到两组对边分别平行且相等的四边形.\n【详解】将一个三角形翻转后与另一个按对应边拼接，能拼成平行四边形。原说法正确。",
+        "knowledge_points":  [
+                                 "用同样的图形拼接",
+                                 "三角形的概念及表示方法",
+                                 "平行四边形的概念及特点"
+                             ],
+        "source":  "24-25四年级下·安徽芜湖·期末"
+    },
+    {
+        "id":  "393",
+        "unit":  "第五单元三角形",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "顶角是60°的等腰三角形是等边三角形。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】等腰三角形两底角相等，利用内角和180°求底角.\n【详解】底角=(180°-60°)÷2=60°,三个角都为60°,对应三边都相等，所以是等边三角形。原说法正确。",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·安徽淮北·期末"
+    },
+    {
+        "id":  "394",
+        "unit":  "第五单元三角形",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "每个三角形中最少有两个内角是锐角。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】直角三角形和钝角三角形除去一个直角或钝角后，其余两角和小于等于90°,都为锐角;锐角三角形三个角都锐.\n【详解】无论哪类三角形，至少有两个锐角。原说法正确。",
+        "knowledge_points":  [
+                                 "三角形的分类"
+                             ],
+        "source":  "17-18四年级下·全国·课后作业"
+    },
+    {
+        "id":  "395",
+        "unit":  "第五单元三角形",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "在一个直角三角形中，其中的一个锐角是54度，求另一个锐角的度数是多少度。",
+        "options":  [
+
+                    ],
+        "answer":  "36度",
+        "solution":  "【分析】直角三角形有一个角是90°,三角形内角和180°,另一个锐角=180°-90°-54°.\n【详解】180-90-54=36(度)。答：另一个锐角是36度。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "三角形的内角和",
+                                 "三角形的分类"
+                             ],
+        "source":  "25-26四年级上·河南驻马店·期中"
+    },
+    {
+        "id":  "396",
+        "unit":  "第五单元三角形",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一根铁丝可以围成一个腰长15分米，底边长24分米的等腰三角形。如果改围成一个等边三角形，那么等边三角形的每条边长多少分米。",
+        "options":  [
+
+                    ],
+        "answer":  "18分米",
+        "solution":  "【分析】同一根铁丝长度不变。先求等腰三角形周长，再除以3得到等边三角形边长.\n【详解】原周长=15×2+24=54(分米)。等边三角形每边=54÷3=18(分米)。答：每条边长18分米。",
+        "knowledge_points":  [
+                                 "三角形的周长",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·河北邢台·期末"
+    },
+    {
+        "id":  "397",
+        "unit":  "第五单元三角形",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一根铁丝可以围成一个腰长15分米，底边长24分米的等腰三角形。如果改围成一个等边三角形，那么等边三角形的每条边长多少分米。",
+        "options":  [
+
+                    ],
+        "answer":  "18分米",
+        "solution":  "【分析】铁丝总长不变，即新旧图形周长相等。先求等腰三角形周长，再平均分成3份.\n【详解】(15+15+24)÷3=54÷3=18(分米)。答：等边三角形每条边长18分米。",
+        "knowledge_points":  [
+                                 "三角形的周长",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·河北邢台·期末"
+    },
+    {
+        "id":  "398",
+        "unit":  "第五单元三角形",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "学校准备用篱笆围一块等腰三角形花圃。已知一条腰长8米，篱笆总长22米。这个花圃的底边长是多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "6米",
+        "solution":  "【分析】等腰三角形两条腰相等，篱笆总长即周长。底边=周长-2*腰长.\n【详解】底边=22-8×2=22-16=6(米)。答：底边长6米。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·湖南长沙·期末"
+    },
+    {
+        "id":  "399",
+        "unit":  "第五单元三角形",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "等腰三角形的周长是98厘米，底边长26厘米，这个三角形的腰长是多少厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "36厘米",
+        "solution":  "【分析】等腰三角形两条腰相等，先用周长减底边得到两腰和，再除以2.\n【详解】腰长=(98-26)÷2=72÷2=36(厘米)。答：腰长36厘米。",
+        "knowledge_points":  [
+                                 "三角形的周长",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·安徽淮北·期末"
+    },
+    {
+        "id":  "400",
+        "unit":  "第五单元三角形",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "在三角形ABC中，已知∠A=102°,∠A是∠B的3倍，则∠B=(),∠C=()。![](images/a400442a7c25df68039b577c26d8945e0b26160093edde1356deab69dccc32f4.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "34°;44°",
+        "solution":  "【分析】已知∠A是∠B的3倍，可先求∠B;再由三角形内角和180°求∠C.\n【详解】∠B=102°/3=34°。∠C=180°-102°-34°=44°。答:∠B=34°,∠C=44°。",
+        "knowledge_points":  [
+                                 "已知一个数的几倍是多少，求这个数",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·四川南充·期末"
+    },
+    {
+        "id":  "401",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "等腰三角形的底角一定是()。",
+        "options":  [
+                        "A. 锐角",
+                        "B. 直角",
+                        "C. 钝角",
+                        "D. 无法确定"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】等腰三角形两个底角相等，三角形内角和是180°。若底角是直角或钝角，则两个底角和会大于或等于180°,不成立.\n【详解】所以等腰三角形的底角一定是锐角。故答案为:A.",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·河南驻马店·期末"
+    },
+    {
+        "id":  "402",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "在一个平面上，可能有()的三角形。",
+        "options":  [
+                        "A. 只有一个锐角",
+                        "B. 三边长度不变，形状能改变",
+                        "C. 两角之和等于第三角"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】三角形至少有2个锐角;三角形三边确定后形状也确定。等腰直角三角形中45°+45°=90°,可能出现两角和等于第三角.\n【详解】A,B错误,C正确。故答案为:C.",
+        "knowledge_points":  [
+                                 "三角形的稳定性及应用",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·河北廊坊·期末"
+    },
+    {
+        "id":  "403",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面说法正确的是()。",
+        "options":  [
+                        "A. 所有的等腰三角形都是锐角三角形",
+                        "B. 一个三角形中最多有2个直角",
+                        "C. 等边三角形是特殊的等腰三角形"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】等腰三角形可能是锐角，直角或钝角;三角形内角和180°,不可能有2个直角;等边三角形三边都相等，属于特殊等腰三角形.\n【详解】A,B错误,C正确。故答案为:C.",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·河北廊坊·期末"
+    },
+    {
+        "id":  "404",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "两个()的三角形一定可以拼成一个平行四边形。",
+        "options":  [
+                        "A. 周长相等",
+                        "B. 周长和面积分别相等",
+                        "C. 等底等高",
+                        "D. 完全相同"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】能拼成平行四边形要求两三角形形状和大小完全对应。仅周长，面积，等底等高都不能保证可拼.\n【详解】只有两个完全相同的三角形一定可以拼成平行四边形。故答案为:D.",
+        "knowledge_points":  [
+                                 "用同样的图形拼接",
+                                 "三角形的概念及表示方法",
+                                 "平行四边形的概念及特点"
+                             ],
+        "source":  "24-25四年级下·湖南永州·期末"
+    },
+    {
+        "id":  "405",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "在一个等腰三角形中，已知一个内角是40°,另外两个内角分别()。",
+        "options":  [
+                        "A. 都是70°或40°和100°",
+                        "B. 是100°",
+                        "C. 70°和100°",
+                        "D. 140°"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】40°可能是顶角，也可能是底角，分情况讨论.\n【详解】若40°是底角，另两角为40°和100°;若40°是顶角，另两角为70°和70°。故答案为:A.",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·湖南永州·期末"
+    },
+    {
+        "id":  "406",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一个直角三角形的一个锐角是45°,这也是一个()。",
+        "options":  [
+                        "A. 锐角三角形",
+                        "B. 钝角三角形",
+                        "C. 等腰三角形",
+                        "D. 等边三角形"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】直角三角形内角和180°,第三个角=180°-90°-45°=45°,有两个角相等.\n【详解】该三角形是等腰直角三角形，按边分是等腰三角形。故答案为:C.",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·河北邢台·期末"
+    },
+    {
+        "id":  "407",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一张三角形纸片被撕下一个角(如图所示),原来这张三角形纸片的形状是()三角形。![](images/4d615aaf0dec92c9b647b8ef3cbca50db6b556d751870e46eb6b4074757013a5.jpg)",
+        "options":  [
+                        "A. 直角",
+                        "B. 锐角",
+                        "C. 钝角"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】三角形内角和180°,已知两个角35°和43°.\n【详解】第三角=180°-35°-43°=102°,大于90°,是钝角三角形。故答案为:C.",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·河南省直辖县级单位·期末"
+    },
+    {
+        "id":  "408",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "如图，在池塘的一侧选取一点O,测得OA长6m,OB长12m。那么A,B两点之间的距离可能是()。![](images/1be8dda23862b66b3332040a41df88151b05b9d8b474991cb59cbf6c4b8e65ee.jpg)",
+        "options":  [
+                        "A. 6m",
+                        "B. 15m",
+                        "C. 20m"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】由三角形三边关系得:|12-6|\u003cAB\u003c12+6，即6\u003cAB\u003c18.\n【详解】6m不行,20m不行,15m可行。故答案为:B.",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "2025·河北衡水·小升初真题"
+    },
+    {
+        "id":  "409",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "一个三角形的两条边分别是6cm和14cm,第三条边不可能是()。",
+        "options":  [
+                        "A. 8cm",
+                        "B. 9cm",
+                        "C. 19cm"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】第三边x需满足14-6\u003cx\u003c14+6，即8\u003cx\u003c20.\n【详解】8cm等于下界，不可能;9cm和19cm都可能。故答案为:A.",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·河南郑州·期末"
+    },
+    {
+        "id":  "410",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "在锐角三角形中，任意两个锐角的和至少大于()。",
+        "options":  [
+                        "A. 90°",
+                        "B. 100°",
+                        "C. 110°",
+                        "D. 120°"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】任意两个锐角之和=180°-第三个锐角;第三个锐角\u003c90°.\n【详解】因此任意两个锐角之和一定大于90°。故答案为:A.",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "三角形的内角和",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·河南许昌·期末"
+    },
+    {
+        "id":  "411",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "如图，以线段AB为底边，画高是2厘米的三角形，能画出()个。![](images/00e6f6e44cf516073f0c8b3a142fc51f87771a43814b9926cbe45b6cf95af037.jpg)",
+        "options":  [
+                        "A. 2",
+                        "B. 3",
+                        "C. 无数"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】高为2厘米时，顶点可以在线段AB平行的两条高线上的无数位置.\n【详解】满足条件的三角形可画无数个。故答案为:C.",
+        "knowledge_points":  [
+                                 "三角形的高及画法"
+                             ],
+        "source":  "24-25四年级下·河南郑州·期末"
+    },
+    {
+        "id":  "412",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一个三角形的两条边边长分别是12厘米和8厘米，第三条边边长可能是()厘米。",
+        "options":  [
+                        "A. 20",
+                        "B. 6",
+                        "C. 3"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】第三边x满足12-8\u003cx\u003c12+8，即4\u003cx\u003c20.\n【详解】20不行,3不行,6可行。故答案为:B.",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·河南安阳·期末"
+    },
+    {
+        "id":  "413",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "如下图。展览区的展板支架损坏了，需要更换钢条。则更换的钢条的长度可能为()。![](images/bfe8bf3d0fc98abeccfe422eb226385b228637b17f43147e92e051a17ccf7dc0.jpg)",
+        "options":  [
+                        "A. 0.5米",
+                        "B. 1米",
+                        "C. 0.4米",
+                        "D. 0.7米"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】已知两边0.6米,0.8米，第三边应满足0.2\u003cx\u003c1.4;结合图意该边应大于0.8米.\n【详解】选项中只有1米满足。故答案为:B.",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·浙江台州·期末"
+    },
+    {
+        "id":  "414",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "在下图三角形中，底边AB边上的高是()。![](images/80ae1341762900b3a317bd67b6fc3515afb7804b6fab91d9bab55c81fe049453.jpg)",
+        "options":  [
+                        "A. 线段CD",
+                        "B. 线段BF",
+                        "C. 线段EF",
+                        "D. 线段BE"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】底AB上的高应从对顶点作到底AB的垂线段.\n【详解】线段CD满足条件，其他线段不满足。故答案为:A.",
+        "knowledge_points":  [
+                                 "三角形的高及画法"
+                             ],
+        "source":  "24-25四年级下·浙江台州·期末"
+    },
+    {
+        "id":  "415",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "如图，三角形ABC中底边AB上的高是()。![](images/0a3cae2469a709297f2eb316b0f5ed7406d2e656ea41dbfcb1033e4cde0b3f0e.jpg)",
+        "options":  [
+                        "A. AF",
+                        "B. BD",
+                        "C. BE",
+                        "D. CD"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】底边AB上的高应从顶点C向AB作垂线.\n【详解】对应线段是CD。故答案为:D.",
+        "knowledge_points":  [
+                                 "三角形的高及画法"
+                             ],
+        "source":  "24-25四年级下·江西赣州·期末"
+    },
+    {
+        "id":  "416",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "如图，已知虚线l⊥AB,点C可以沿着虚线l上下移动，那么所组成的三角形ABC()。![](images/81376249dae1d174c9514151a987545b2b6858d74608b2612b44c5046211d2d1.jpg)",
+        "options":  [
+                        "A. 一定是钝角三角形",
+                        "B. 一定是锐角三角形",
+                        "C. 一定是直角三角形",
+                        "D. 可能是钝角三角形，也可能是锐角三角形，还可能是直角三角形"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】点C位置变化会改变角度大小.\n【详解】三角形ABC可能为锐角，直角或钝角三角形。故答案为:D.",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·江西上饶·期末"
+    },
+    {
+        "id":  "417",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "科学课上学生们用木条钉成模型，下面各形状中最不容易变形的是()。",
+        "options":  [
+                        "A. ![](images/fd819f680058fdc705b02aff9b50e7bc04528b5c3bd01717be893bcf9519c85e.jpg)",
+                        "B. ![](images/3fb129b3675ddaf4371eff26be1efe1231b84bd7713c38ed241800a8694005a7.jpg)",
+                        "C. ![](images/fc6c7e37b50bb078ad76c1b2eae95a8c6531e26bbecab190616cdf303393341c.jpg)"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】三角形稳定，四边形易变形.\n【详解】B中有三角形结构，最不易变形。故答案为:B.",
+        "knowledge_points":  [
+                                 "三角形的稳定性及应用",
+                                 "平行四边形的不稳定性及应用"
+                             ],
+        "source":  "24-25四年级下·江西上饶·期末"
+    },
+    {
+        "id":  "418",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下图中的三角形，有一个角被撕坏了，这个三角形是()。![](images/2e1331d7ab62fbd7fce4bcb1ac23ac08b0d57a87095c40ba8fb962e1ace44162.jpg)",
+        "options":  [
+                        "A. 锐角三角形",
+                        "B. 直角三角形",
+                        "C. 钝角三角形",
+                        "D. 无法判断"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】已知两角35°,45°,第三角=180°-35°-45°.\n【详解】第三角100°,是钝角，所以为钝角三角形。故答案为:C.",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·贵州遵义·期末"
+    },
+    {
+        "id":  "419",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "已知一个三角形的三条边都是整厘米数，其中的两条边分别长3厘米和5厘米，第三条边最长是()厘米。",
+        "options":  [
+                        "A. 6",
+                        "B. 7",
+                        "C. 8"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】第三边x满足2\u003cx\u003c8，且为整数.\n【详解】最大整数是7。故答案为:B.",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "2025·河北石家庄·小升初真题"
+    },
+    {
+        "id":  "420",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形的周长是26cm,其中一条边长6cm,和它不相等的另一条边的长度是()。",
+        "options":  [
+                        "A. 14cm",
+                        "B. 10cm",
+                        "C. 14cm或10cm",
+                        "D. 无法确定"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】若6是腰，则底=26-12=14，但6+6\u003c14，不成立;故6只能是底.\n【详解】两腰=(26-6)÷2=10cm,所以不相等的另一条边是10cm。故答案为:B.",
+        "knowledge_points":  [
+                                 "三角形三边关系",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25六年级下·山东济南·期末"
+    },
+    {
+        "id":  "421",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "学习了三角形内角和及四边形内角和之后，在探究\u0027六边形的内角和是多少度\u0027的过程中，三位同学给出了如图的方法，你认为正确的有()。![](images/64f258fbb9325f667f4d8501077e75bc.png)",
+        "options":  [
+                        "A. 甲",
+                        "B. 甲和乙",
+                        "C. 甲和丙",
+                        "D. 甲，乙和丙"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】甲，乙，丙都能把六边形转化为若干三角形/四边形计算，结果都为720°.\n【详解】三种方法都正确。故答案为:D.",
+        "knowledge_points":  [
+                                 "多边形的内角和",
+                                 "三角形的内角和",
+                                 "角度的计算"
+                             ],
+        "source":  "24-25四年级下·贵州铜仁·期末"
+    },
+    {
+        "id":  "422",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形有两条边分别是7.85cm和3.6cm,它的周长是()。",
+        "options":  [
+                        "A. 15.05cm",
+                        "B. 19.3cm",
+                        "C. 15.05cm或者19.3cm"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】若腰为3.6，则3.6+3.6\u003c7.85不成立;故腰为7.85.\n【详解】周长=7.85+7.85+3.6=19.3cm。故答案为:B.",
+        "knowledge_points":  [
+                                 "三角形三边关系",
+                                 "三角形的周长",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·江西吉安·期末"
+    },
+    {
+        "id":  "423",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "图中，露出的部分是这个三角形中最小的角，那么这个三角形是()。![](images/8ebbaed350edff85939cd09c511cf1b3ebbba93b3cb95261f6a3b03332e95246.jpg)",
+        "options":  [
+                        "A. 锐角三角形",
+                        "B. 直角三角形",
+                        "C. 钝角三角形",
+                        "D. 等边三角形"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】最小角是50°,其余两角都不小于50°.\n【详解】另外两角和为130°,最大角最多80°,小于90°,故为锐角三角形。故答案为:A.",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·福建漳州·期末"
+    },
+    {
+        "id":  "424",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "如图，点A在∠B的一条边上不动，点C在∠B的另一条边上任意移动，连接AC,则组成的三角形ABC可能是()。![](images/c2199b5a49406478cdaf6f658b214271f50e33804381d180cf941981c422daf8.jpg)",
+        "options":  [
+                        "A. ①②④",
+                        "B. ①②③",
+                        "C. ①③④",
+                        "D. ①②③④"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】∠B固定为45°,可形成锐角，等腰直角，钝角三角形;等边三角形每角60°,不可能.\n【详解】可能是①②③,不可能是④。故答案为:B.",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·福建漳州·期末"
+    },
+    {
+        "id":  "425",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面不可以围成一个三角形的三条线段是()。",
+        "options":  [
+                        "A. 5cm,6cm,7cm",
+                        "B. 5cm,5cm,10cm",
+                        "C. 3cm,4cm,6cm",
+                        "D. 5cm,10cm,10cm"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】较短两边和需大于第三边.\n【详解】B中5+5=10，不满足大于，不能围成三角形;其余可围成。故答案为:B.",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25六年级下·湖南永州·期末"
+    },
+    {
+        "id":  "426",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.4,
+        "question":  "如图,A,C两点均在直线m上，如果点A和点B固定不动，点C沿直线m向右移动，三角形ABC不可能变成()。![](images/e951132212ad6449634e7bb279c382c38b2b2411fa9632e32f41fb3928296fd1.jpg)",
+        "options":  [
+                        "A. 等腰三角形",
+                        "B. 钝角三角形",
+                        "C. 等边三角形"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】C在线m上移动时可出现等腰，钝角等情况;但要成等边需三边同时相等，受位置限制不可能.\n【详解】不可能变成等边三角形。故答案为:C.",
+        "knowledge_points":  [
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·河南新乡·期末"
+    },
+    {
+        "id":  "427",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面是用木条钉的架子，最不易变形的是()。",
+        "options":  [
+                        "A. ![](images/170015687dae82d8c3452cdd9f3d7778f3219eb9487b68c6ebddb927c9668145.jpg)",
+                        "B. ![](images/429583737b64b801a5ca4897431e065716f8b23a94a4b6df3aa99ca102f0429f.jpg)",
+                        "C. ![](images/4771a588fbb24f356a5f68bededb1403bc2c373161711f24872a2e74ae4b7507.jpg)"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】三角形结构最稳定，长方形结构易变形.\n【详解】C由三角形组成，最不易变形。故答案为:C.",
+        "knowledge_points":  [
+                                 "三角形的稳定性及应用"
+                             ],
+        "source":  "24-25四年级下·江西吉安·期末"
+    },
+    {
+        "id":  "428",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "如图，李叔叔要估计池塘两岸A,B之间的距离，因为A,B之间有淤泥，不方便直接测量，所以李叔叔在池塘外选取一点O,测得OA=20米,OB=15米。根据以上信息，李叔叔猜测AB之间的距离可能是()米。![](images/56c45720aa2ff08f9f708f099fbdf9eb3f9868c768c158c790a4c0cc8cc7213d.jpg)",
+        "options":  [
+                        "A. 5",
+                        "B. 35",
+                        "C. 40",
+                        "D. 25"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】三角形三边关系:|20-15|\u003cAB\u003c20+15，即5\u003cAB\u003c35.\n【详解】只有25满足条件。故答案为:D.",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·陕西西安·期末"
+    },
+    {
+        "id":  "429",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "如下图所示，一张三角形的纸片被撕去了一个角，原来这张三角形纸片的形状是()三角形。![](images/50b30bee6ef9fec88afac047fa56dc9581d6f0af172a3bff9a5f06759871d9e3.jpg)",
+        "options":  [
+                        "A. 锐角",
+                        "B. 钝角",
+                        "C. 等腰"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】已知两角30°,55°,第三角=180°-(30°+55°).\n【详解】第三角95°\u003e90°,原三角形为钝角三角形。故答案为:B.",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "三角形的分类",
+                                 "三角形的内角和",
+                                 "直角，钝角，锐角的认识及特征"
+                             ],
+        "source":  "24-25四年级下·江西吉安·期末"
+    },
+    {
+        "id":  "430",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "两个完全相同的等腰直角三角形，拼成一个大三角形，这个大三角形是()三角形。",
+        "options":  [
+                        "A. 锐角",
+                        "B. 等腰直角",
+                        "C. 钝角",
+                        "D. 等边"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】两个完全相同的等腰直角三角形可拼成一个有一个直角且两底角相等的大三角形.\n【详解】该大三角形是等腰直角三角形。故答案为:B.",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "用同样的图形拼接"
+                             ],
+        "source":  "24-25四年级下·福建福州·期末"
+    },
+    {
+        "id":  "431",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "有一个三角形的3个角分别是45°,45°,90°,以下说法不正确的是()。",
+        "options":  [
+                        "A. 它是等腰三角形",
+                        "B. 两个相同的它可以拼成一个正方形",
+                        "C. 它是直角三角形",
+                        "D. 两个相同的它不能拼成一个三角形"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】45°-45°-90°三角形是等腰直角三角形;两个全等的该三角形既可拼正方形，也可拼更大的三角形.\n【详解】A,B,C正确;D错误。故答案为:D.",
+        "knowledge_points":  [
+                                 "用同样的图形拼接",
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·福建福州·期末"
+    },
+    {
+        "id":  "432",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "如果一个等腰三角形的两边分别长4厘米和8厘米，那么这个等腰三角形的周长是()厘米。",
+        "options":  [
+                        "A. 20",
+                        "B. 16",
+                        "C. 20或16",
+                        "D. 23或16"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】若4是腰，则4+4=8，不能成三角形;故8是腰.\n【详解】周长=8+8+4=20厘米。故答案为:A.",
+        "knowledge_points":  [
+                                 "三角形三边关系",
+                                 "三角形的周长",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·福建福州·期末"
+    },
+    {
+        "id":  "433",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面各组线段中，能围成三角形的是()。",
+        "options":  [
+                        "A. 2cm,3cm,5cm",
+                        "B. 3cm,4cm,8cm",
+                        "C. 5cm,6cm,7cm",
+                        "D. 1cm,2cm,3cm"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】判断标准：任意两边之和大于第三边.\n【详解】A中2+3=5不行;B中3+4\u003c8不行;D中1+2=3不行;C中5+6\u003e7且其余也成立，可围成三角形。故答案为:C.",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "25-26五年级上·黑龙江鹤岗·开学考试"
+    },
+    {
+        "id":  "434",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面图形中，虚线表示三角形的高，其中画正确的是()。![](images/3ecb11bf7780e7d16e4f24b6686ff9e7.png)",
+        "options":  [
+                        "A. ①②",
+                        "B. ①③",
+                        "C. ②③",
+                        "D. ①②③"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】三角形的高是从顶点向对边(或对边延长线)作的垂线段，必须经过顶点且与对应底边垂直.\n【详解】图③中虚线不满足\"与对应边垂直\"的条件，不是高;①②正确。故答案为:A.",
+        "knowledge_points":  [
+                                 "三角形的高及画法"
+                             ],
+        "source":  "24-25四年级下·浙江杭州·期末"
+    },
+    {
+        "id":  "435",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "在计算六边形的内角和时，我们可以把它分成4个三角形(如图),它的内角和是180°×4=720°。计算八边形内角和正确的是()。![](images/e1ad15d334f87f756cdc9adbfe851c7726a0957f42d7b6768f58577289887090.jpg)",
+        "options":  [
+                        "A. 180°×6=1080°",
+                        "B. 180°×7=1260°",
+                        "C. 180°×8=1440°"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】八边形从一个顶点可分成6个三角形，内角和=180°×6。\n【详解】八边形内角和为1080°,对应A。故答案为:A.",
+        "knowledge_points":  [
+                                 "多边形的内角和",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·河南洛阳·期末"
+    },
+    {
+        "id":  "436",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "如下图，以线段AB为底边，画高是3厘米的三角形，能画出()个。![](images/f7ffcd636dc8d10c277d311f6006f38208d0aed326c33b0cb77e1192a67286ae.jpg)",
+        "options":  [
+                        "A. 1",
+                        "B. 2",
+                        "C. 3",
+                        "D. 无数"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】以AB为底，顶点只要在与AB平行且距离为3厘米的直线上即可，满足条件的位置有无数个.\n【详解】因此可画无数个这样的三角形。故答案为:D.",
+        "knowledge_points":  [
+                                 "画三角形"
+                             ],
+        "source":  "24-25四年级下·贵州遵义·期末"
+    },
+    {
+        "id":  "437",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面每组线段中，能围成三角形的是()。",
+        "options":  [
+                        "A. 3cm,4cm,10cm",
+                        "B. 6cm,4cm,10cm",
+                        "C. 2cm,1cm,3cm",
+                        "D. 1cm,10cm,10cm"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】判断能否围成三角形看\"较短两边之和是否大于最长边\".\n【详解】A:3+4\u003c10，不行;B:6+4=10，不行;C:2+1=3，不行;D:1+10\u003e10，可行。故答案为:D.",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25六年级下·云南昆明·期末"
+    },
+    {
+        "id":  "438",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "用三根长度为整厘米数的小棒围成一个三角形，如果其中两根小棒分别长5厘米,8厘米，那么第三根小棒最短是()厘米。",
+        "options":  [
+                        "A. 3",
+                        "B. 4",
+                        "C. 7",
+                        "D. 14"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】第三边x满足8-5\u003cx\u003c8+5，即3\u003cx\u003c13.\n【详解】x取整时最小为4。故答案为:B.",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·江西萍乡·期末"
+    },
+    {
+        "id":  "439",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形的一个内角是58°,这个三角形的顶角是()。",
+        "options":  [
+                        "A. 58°",
+                        "B. 64°",
+                        "C. 58°或64°",
+                        "D. 58°或122°"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】58°可能是顶角，也可能是底角.\n【详解】若58°是顶角，则顶角=58°;若58°是底角，则顶角=180°-58°-58°=64°。故答案为:C.",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·福建福州·期末"
+    },
+    {
+        "id":  "440",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "用3厘米,8厘米和第三根小棒(长度取整厘米)首尾相连组成三角形，第三根小棒最长是()。",
+        "options":  [
+                        "A. 8厘米",
+                        "B. 9厘米",
+                        "C. 10厘米",
+                        "D. 11厘米"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】第三边x满足8-3\u003cx\u003c8+3，即5\u003cx\u003c11.\n【详解】取整时最大为10。故答案为:C.",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·福建福州·期末"
+    },
+    {
+        "id":  "441",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "已知三角形ABC为直角三角形,∠B=90°。若沿图中虚线剪去∠B,则∠1+∠2=()。![](images/6a3d1b863e89dd2113e79c4d21928da9534990ab0be38e9b3e17789d1b918a4d.jpg)",
+        "options":  [
+                        "A. 90°",
+                        "B. 135°",
+                        "C. 270°",
+                        "D. 235°"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】直角三角形中∠A+∠C=90°。剪去后形成四边形，内角和是360°.\n【详解】∠1+∠2=360°-(∠A+∠C)=360°-90°=270°。故答案为:C.",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "多边形的内角和"
+                             ],
+        "source":  "24-25四年级下·浙江嘉兴·期末"
+    },
+    {
+        "id":  "442",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面说法错误的是()。",
+        "options":  [
+                        "A. 0.67和0.670的计数单位不同，但大小相同",
+                        "B. 大于2.3且小于2.5的小数只有2.4",
+                        "C. 一个三角形一边长6米，另一边长9米，第三边可能长5米"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】A利用小数性质判断;B未限制小数位数;C用三角形三边关系判断.\n【详解】A正确;B错误，因为2.3与2.5之间有无数个小数;C正确,6+5\u003e9且9+5\u003e6。故答案为:B.",
+        "knowledge_points":  [
+                                 "三角形三边关系",
+                                 "一位小数的大小比较",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·江西上饶·期末"
+    },
+    {
+        "id":  "443",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "用3根小棒围三角形，不能选择()。",
+        "options":  [
+                        "A. 3厘米,4厘米,5厘米",
+                        "B. 5厘米,5厘米,5厘米",
+                        "C. 2厘米,3厘米,5厘米"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】看较短两边和是否大于第三边.\n【详解】A可围成,B可围成,C中2+3=5，不满足大于，不能围成。故答案为:C.",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·江西宜春·期末"
+    },
+    {
+        "id":  "444",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下列现象中，利用了平行四边形不稳定易变形特点的是()。",
+        "options":  [
+                        "A. 塔吊支架",
+                        "B. 自行车架",
+                        "C. 伸缩门"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】塔吊支架和自行车架主要利用三角形稳定性;伸缩门利用平行四边形易变形.\n【详解】故答案为:C.",
+        "knowledge_points":  [
+                                 "平行四边形的不稳定性及应用",
+                                 "三角形的稳定性及应用"
+                             ],
+        "source":  "24-25四年级下·江西宜春·期末"
+    },
+    {
+        "id":  "445",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "刘叔叔有一根20分米长的木条，要做一个三角框架，他可以怎么截取()。",
+        "options":  [
+                        "A. 10分米,5分米,5分米",
+                        "B. 5分米,6分米,9分米",
+                        "C. 7分米,2分米,11分米"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】用三角形三边关系逐项判断.\n【详解】A中5+5=10，不行;B中5+6\u003e9，可行;C中7+2\u003c11，不行。故答案为:B.",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·江西九江·期末"
+    },
+    {
+        "id":  "446",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "用三根长度为整厘米数的小棒围成一个三角形，如果其中两根小棒分别长5cm,8cm,那么第三根小棒最短是()cm。",
+        "options":  [
+                        "A. 3",
+                        "B. 4",
+                        "C. 14"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】第三边x满足3\u003cx\u003c13.\n【详解】最短整数是4。故答案为:B.",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·江西萍乡·期末"
+    },
+    {
+        "id":  "447",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "实验室的工作人员在工作时，不小心把一块三角形玻璃打碎成三块，现在他要到玻璃店去配一块形状完全一样的玻璃，他应该带第()块去。![](images/b865347d9499949f799e3d8c4bee0ff1afdbbdfe045e230ce01b48eea0004537.jpg)",
+        "options":  [
+                        "A. ①",
+                        "B. ②",
+                        "C. ③",
+                        "D. 无法选择"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】要唯一确定三角形，需足够信息(如一边及其两端角).\n【详解】③保留了一条边和该边两端角，能确定原三角形;①②信息不足。故答案为:C.",
+        "knowledge_points":  [
+                                 "三角形的概念及表示方法"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "448",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "如图，点C可以在虚线上左右移动,A,B,C三点所形成的三角形可能是()。![](images/ca79df7e09bce0ac4e4380a839a7c88b1234fc84d551c6fb7831579a2ba75177.jpg)![](images/2026-03-11%20230204.jpg)",
+        "options":  [
+                        "A. 直角三角形",
+                        "B. 锐角三角形",
+                        "C. 钝角三角形",
+                        "D. 以上都有可能"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】C点位置变化可使三角形角度结构发生变化.\n【详解】可能形成锐角，直角，钝角三角形。故答案为:D.",
+        "knowledge_points":  [
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·浙江宁波·期末"
+    },
+    {
+        "id":  "449",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "天文爱好者小明将天文望远镜放在地面上准备观星，望远镜能够稳稳地放在地面，是根据()。![](images/017a6d4f032c5540a24d593963fe94768ae33ce045ce555a3a27ddbe7e711aa2.jpg)",
+        "options":  [
+                        "A. 三角形任意两边之和大于第三边",
+                        "B. 三角形的稳定性",
+                        "C. 三角形的内角和是180°",
+                        "D. 三角形有三个顶点"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】支架稳固依赖三角形结构不易变形.\n【详解】利用的是三角形的稳定性。故答案为:B.",
+        "knowledge_points":  [
+                                 "三角形三边关系",
+                                 "三角形的稳定性及应用",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·浙江台州·期末"
+    },
+    {
+        "id":  "450",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一个三角形的三条边都是整厘米数，其中两条边长分别为7cm和10cm,第三条边最长是()cm。",
+        "options":  [
+                        "A. 4",
+                        "B. 16",
+                        "C. 17",
+                        "D. 18"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】第三边x满足10-7\u003cx\u003c10+7，即3\u003cx\u003c17.\n【详解】取整时最大为16。故答案为:B.",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·江西赣州·期末"
+    },
+    {
+        "id":  "451",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "在探究三角形三边关系时，明明把一根20厘米长的小棒截成三段，围成一个等腰三角形，三段小棒的长度可能是()。",
+        "options":  [
+                        "A. 10厘米,5厘米,5厘米",
+                        "B. 6厘米,6厘米,8厘米",
+                        "C. 4厘米,4厘米,12厘米",
+                        "D. 3厘米,3厘米,14厘米"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】等腰且满足两边和大于第三边.\n【详解】A:5+5=10不行;B:6+6\u003e8可行;C:4+4\u003c12不行;D:3+3\u003c14不行。故答案为:B.",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·江西赣州·期末"
+    },
+    {
+        "id":  "452",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "四条线段的长度分别是2厘米,4厘米,5厘米,7厘米，任意选择其中3条，可以围成()个不同的三角形。",
+        "options":  [
+                        "A. 2",
+                        "B. 3",
+                        "C. 4"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】四种三边组合:2,4,5;2,4,7;2,5,7;4,5,7.\n【详解】可行的是(2,4,5)和(4,5,7)两组，共2个。故答案为:A.",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·广西贵港·期末"
+    },
+    {
+        "id":  "453",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "三角形的特性不包括()。",
+        "options":  [
+                        "A. 稳定性",
+                        "B. 三边关系",
+                        "C. 易变形",
+                        "D. 内角和180°"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】三角形有稳定性，三边关系，内角和180°,不具\"易变形\"特性.\n【详解】故不包括的是C。故答案为:C.",
+        "knowledge_points":  [
+                                 "三角形三边关系",
+                                 "三角形的稳定性及应用",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·贵州遵义·期末"
+    },
+    {
+        "id":  "454",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "三角形的两边长分别为3cm和4cm,第三边不可能是()。",
+        "options":  [
+                        "A. 5cm",
+                        "B. 6cm",
+                        "C. 7cm",
+                        "D. 4cm"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】第三边x满足1\u003cx\u003c7.\n【详解】7cm等于上界，不可能;其余可能。故答案为:C.",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·贵州遵义·期末"
+    },
+    {
+        "id":  "455",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "李师傅是一名很有经验的维修工，有一个三角支架的一条钢条需要更换(如图),需要更换钢条的长度可能为()。![](images/9bd6f46f64a66f547dd313f6c73d9019701944ab0cf8f8fad63573f3ce4941b9.jpg)",
+        "options":  [
+                        "A. 5.4m",
+                        "B. 6.4m",
+                        "C. 0.4m"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】第三边x满足|3.4-3|\u003cx\u003c3.4+3，即0.4\u003cx\u003c6.4.\n【详解】5.4在范围内;6.4和0.4在边界上不行。故答案为:A.",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·贵州黔南·期末"
+    },
+    {
+        "id":  "456",
+        "unit":  "第五单元三角形",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面图形中，是轴对称图形的是()。",
+        "options":  [
+                        "A. 平行四边形",
+                        "B. 梯形",
+                        "C. 等边三角形"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】轴对称图形对折后两部分能完全重合.\n【详解】一般平行四边形和普通梯形不是轴对称图形，等边三角形是轴对称图形。故答案为:C.",
+        "knowledge_points":  [
+                                 "梯形的概念及特点",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "轴对称的认识及辨认",
+                                 "平行四边形的概念及特点"
+                             ],
+        "source":  "25-26三年级上·黑龙江佳木斯·开学考试"
+    },
+    {
+        "id":  "457",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "我们在求四边形的内角和时，把一个四边形如下图(左图)分成了2个三角形，所以四边形的内角和是180°×2=360°。按照这个方法，下图(右图)中这个多边形可以分成( )个三角形，所以这个多边形的内角和是180°×( )。![](images/a101dee5fd91f601e6d9bfd57c42c3a64b822f17f8979e07d0555e0e706656a3.jpg)![](images/dd9179560e79ff720add5afb177e2dfb309316a88f51b01db45327d0f1527733.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "4;4",
+        "solution":  "【分析】右图多边形从一个顶点出发可分成4个三角形.\n【详解】所以可分成4个三角形，内角和表达式是180°×4。",
+        "knowledge_points":  [
+                                 "多边形的内角和"
+                             ],
+        "source":  "24-25四年级下·河北邯郸·期末"
+    },
+    {
+        "id":  "458",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如下图指示牌是三角形的形状(如图),不小心缺失了一个角，缺失的这个角是( )°,这块指示牌的形状是( )三角形，也是( )三角形。![](images/8b994cf21a249da1d679962cfd5a98fe5dfe54f3aece61407ecc861dd5e4b027.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "53;锐角;等腰",
+        "solution":  "【分析】缺失角=180°-74°-53°=53°。三个角都小于90°,且有两个角相等.\n【详解】缺失角是53°;该三角形是锐角三角形，也是等腰三角形。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·江西上饶·期末"
+    },
+    {
+        "id":  "459",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如图，图中共有( )个锐角三角形,( )个钝角三角形;∠1=( )°。![](images/018495f52232c120aa6317ca9b6dec5baca9d978f15530369679b3b11df422bb.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "4;4;37",
+        "solution":  "【分析】按图形计数得锐角三角形4个，钝角三角形4个。先求∠2=180°-65°=115°,再用三角形内角和求∠1.\n【详解】∠1=180°-115°-28°=37°。所以填4,4,37。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·河南许昌·期末"
+    },
+    {
+        "id":  "460",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "客家农具--簸箕(如图),簸箕的底部采用三角形网格设计，避免单点受力导致变形或断裂，从而增强整体承重能力，这是利用了三角形具有( )。![](images/84afaf69a34135b459081f25d12482dd552ce6083e14eba5c60f2a4bbb37a4e2.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "稳定性",
+        "solution":  "【分析】三角形结构不易变形，承重能力更稳定.\n【详解】利用的是三角形的稳定性。",
+        "knowledge_points":  [
+                                 "三角形的稳定性及应用"
+                             ],
+        "source":  "24-25四年级下·江西上饶·期末"
+    },
+    {
+        "id":  "461",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "下图是由边长分别为8分米和5分米的两个正方形组成，在三角形ABC中与底AB边上对应的高是( )分米。![](images/ff4e623c359e0d16de85527184114042649e37a96da0235124a0ce3bf857b361.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "5",
+        "solution":  "【分析】作出底AB对应的高后可见其长度等于小正方形边长.\n【详解】对应高长为5分米。",
+        "knowledge_points":  [
+                                 "三角形的高及画法"
+                             ],
+        "source":  "24-25四年级下·江西九江·期末"
+    },
+    {
+        "id":  "462",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "如图,∠1=( )°,∠2=( )°。![](images/375209885f21aa5eb1ad98bb5bfd315080b786ba219f6cde5dd1cc7522f185ac.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "40;110",
+        "solution":  "【分析】先在直角三角形中用内角和求∠1，再用平角求∠2.\n【详解】∠1=180°-90°-50°=40°;∠2=180°-40°-30°=110°。",
+        "knowledge_points":  [
+                                 "平角，周角的认识及特征",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·河南郑州·期末"
+    },
+    {
+        "id":  "463",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如下图，王叔叔用无人机拍摄池塘全景，小明说:\u0027池塘的宽(即AB的长)是22米。\u0027我认为小明的说法是( )(填\u0027错的\u0027或\u0027对的\u0027)。我的理由是:( )。![](images/8da1149a636cc7e2056f8bfd813d176f797e01dd8b192771b3c10e006e86a02f.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "错的;三角形的三边关系：任意两边的和大于第三条边，池塘的宽22米与13米,9米不能围成三角形。",
+        "solution":  "【分析】若AB=22，则与13和9构成三边有13+9=22，不满足三角形两边和大于第三边.\n【详解】所以小明说法错，理由是违背三角形三边关系。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·浙江台州·期末"
+    },
+    {
+        "id":  "464",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "下图两个椭圆重合的部分是( )三角形。![](images/2026-03-11%20230657.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "等腰锐角",
+        "solution":  "【分析】重合部分同时满足\"等腰\"和\"锐角\"特征.\n【详解】应填等腰锐角三角形。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·江西赣州·期末"
+    },
+    {
+        "id":  "465",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如图,∠1=60°,∠2=22°,那么∠3=( )°,∠4=( )°。![](images/c2fcf7402d4c0b6a7f12307316131c562aee46dcfe64f147ffa28d584f27b7db.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "120;38",
+        "solution":  "【分析】∠1与∠3构成平角，先求∠3;再在三角形内用内角和求∠4.\n【详解】∠3=180°-60°=120°;∠4=180°-22°-120°=38°。",
+        "knowledge_points":  [
+                                 "平角，周角的认识及特征",
+                                 "三角形的内角和",
+                                 "角度的计算"
+                             ],
+        "source":  "24-25四年级下·河南洛阳·期末"
+    },
+    {
+        "id":  "466",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "下图是一块三角形碎玻璃片，打碎角的度数是( )°,按角分它是一个( )三角形。![](images/0f4526cf4ea03142293c97447fa4276943466e431b9fbbac7eed9a731ebd8faf.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "58;锐角",
+        "solution":  "【分析】打碎角=180°-65°-57°=58°.\n【详解】三个角都小于90°,按角分是锐角三角形。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·江西赣州·期末"
+    },
+    {
+        "id":  "467",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如图，一张三角形的纸片被撕去一个角，撕去的这个角是( )度;按角分类，这是一个( )三角形。![](images/e08db7751b4927f68c597eec46351afae9e4375506d7e1dc4b64e92393fe7ea2.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "90;直角",
+        "solution":  "【分析】撕去角=180°-37°-53°=90°.\n【详解】有一个90°角，所以是直角三角形。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·江西上饶·期末"
+    },
+    {
+        "id":  "468",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如图，把长方形纸片的一个角折叠起来。![](images/c182f564f6f20c74b5d7920f5d7eb60efedab1c953f066742c4bf33eef865cfe.jpg)∠1+∠2=( )°,把折起来的这个角剪掉，剩下图形的内角和是( )°。",
+        "options":  [
+
+                    ],
+        "answer":  "90;540",
+        "solution":  "【分析】折起部分形成三角形，且有一个角是90°,所以另外两角和为90°。剪掉后剩余图形是五边形.\n【详解】∠1+∠2=180°-90°=90°;五边形内角和=(5-2)×180°=540°。",
+        "knowledge_points":  [
+                                 "多边形的内角和",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·广西南宁·期末"
+    },
+    {
+        "id":  "469",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "下图中,∠1=( )°。![](images/c98e275d83235f666d1b90f44e373450e4fd6200b84ea28259f9bc7ae61194d4.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "92",
+        "solution":  "【分析】先求与130°互补的∠2，再利用四边形内角和360°求∠1.\n【详解】∠2=180°-130°=50°;∠1=360°-123°-95°-50°=92°。",
+        "knowledge_points":  [
+                                 "多边形的内角和",
+                                 "平角，周角的认识及特征",
+                                 "角度的计算"
+                             ],
+        "source":  "24-25四年级下·河南新乡·期末"
+    },
+    {
+        "id":  "470",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如图△ABC和△ABD都是等腰三角形，已知∠1=40°,则∠2=( )°,∠3=( )°。![](images/133cb1c96b000d41be4882e878b9390837bf686da34f57a52cfd765991a4d361.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "100;60",
+        "solution":  "【分析】△ABC是等腰三角形，可得∠ABC=∠1=40°;△ABD是等腰三角形，可得∠BAD=∠ABC=40°。再利用三角形内角和与平角关系求其余角.\n【详解】∠2=180°-40°-40°=100°;又∠ADC=180°-∠2=80°,所以∠3=180°-40°-80°=60°。故∠2=100°,∠3=60°。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·浙江台州·期末"
+    },
+    {
+        "id":  "471",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如图：一副三角板叠放在一起,∠1=( )°,∠2=( )°。![](images/c6ac2c12153dd624102647a8d366ac76280635e644b2133e8d3a3fb6f8e98e03.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "15;120",
+        "solution":  "【分析】利用一副三角板角度(30°,60°,90°和45°,45°,90°)及四边形内角和进行计算.\n【详解】∠1=45°-30°=15°;∠2=360°-90°-90°-60°=120°。所以∠1=15°,∠2=120°。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "三角形的内角和",
+                                 "多边形的内角和"
+                             ],
+        "source":  "23-24四年级上·湖北鄂州·期末"
+    },
+    {
+        "id":  "472",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "飞船的太阳能板支架平面图是三角形结构，其中有个角被撕掉了(如图),撕去的角是( )°,这是一个( )三角形。![](images/3f83e6524dfc120833e0adc2ef320ebd433f115b5f43681a55901e862e1ea7df.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "98;钝角",
+        "solution":  "【分析】三角形内角和是180°,先求缺角，再按角分类.\n【详解】撕去的角=180°-57°-25°=98°。98°大于90°,所以这是一个钝角三角形。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "三角形的内角和",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "473",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在下图的直角梯形ABCD中一共有( )个三角形;已知∠1=40°,∠2=60°,那么∠3=( )°。![](images/f96bc8a1decebdc88b016086da79d949316388b9753d8054a4dbcdf00f72c628.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "8;110",
+        "solution":  "【分析】先数图中三角形个数，再利用直角，三角形内角和与角之间关系求∠3.\n【详解】图中共有8个三角形。∠ADB=180°-40°-90°=50°;∠ACD=180°-60°-90°=30°;∠BDC=90°-50°=40°;∠3=180°-40°-30°=110°。所以答案是8和110°。",
+        "knowledge_points":  [
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·浙江台州·期末"
+    },
+    {
+        "id":  "474",
+        "unit":  "第五单元三角形",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "下图直角三角形ABC中，已知:∠1=35°,DA=DB。则∠2=( )。![](images/3d24c24e0d1f23ef4283732501110c0170b6edbc3ee1e886a512831fc28da2c6.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "70°",
+        "solution":  "【分析】DA=DB,所以△ABD是等腰三角形，先求∠ADB,再利用平角求∠2.\n【详解】∠DAB=∠1=35°,∠ADB=180°-35°-35°=110°,∠2=180°-110°=70°。",
+        "knowledge_points":  [
+                                 "角度的计算",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·浙江杭州·期末"
+    },
+    {
+        "id":  "475",
+        "unit":  "第五单元三角形",
+        "question_type":  "判断题",
+        "difficulty":  0.94,
+        "question":  "工人叔叔用长3米,3米和6米的三根木料做成了一个三角形木架。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】三角形任意两边之和必须大于第三边.\n【详解】3+3=6，不满足\"大于\"，不能围成三角形。原说法错误。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·河北秦皇岛·期末"
+    },
+    {
+        "id":  "476",
+        "unit":  "第五单元三角形",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "只有两个锐角的三角形一定是直角三角形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】若只有两个锐角，第三个角可能是直角，也可能是钝角.\n【详解】例如30°,40°,110°是钝角三角形，不是直角三角形。原说法错误。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·四川乐山·期末"
+    },
+    {
+        "id":  "477",
+        "unit":  "第五单元三角形",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "在长度为3cm,6cm,7cm,8cm和15cm的这几根小棒中，任选三根小棒都可以围成三角形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】只要存在一种组合不满足三角形三边关系，结论就错误.\n【详解】如3cm,6cm,15cm中,3+6\u003c15，不能围成三角形。原说法错误。",
+        "knowledge_points":  [
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·四川乐山·期末"
+    },
+    {
+        "id":  "478",
+        "unit":  "第五单元三角形",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "一个三角形最小的一个角是47°,这个三角形既可能是锐角三角形，又可能是直角三角形，还可能是钝角三角形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】已知最小角47°,其余角都不小于47°.\n【详解】若是直角三角形，则另一个角为43°,小于47°,矛盾;若是钝角三角形，也会出现小于47°的角。只能是锐角三角形。原说法错误。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·四川乐山·期末"
+    },
+    {
+        "id":  "479",
+        "unit":  "第五单元三角形",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "![](images/92812346fd9b959781224acb2851718477f5211e642a96c4255a6807fff1870e.jpg)左图中∠1的度数是120°。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】三角形内角和是180°.\n【详解】∠1=180°-40°-20°=120°,说法正确。",
+        "knowledge_points":  [
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·陕西渭南·期末"
+    },
+    {
+        "id":  "480",
+        "unit":  "第五单元三角形",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "许多大桥的桥面，索塔和绳索构成了一个三角形，它利用的是三角形的稳定性。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】三角形结构稳固，不易变形.\n【详解】桥梁中采用三角形构造正是利用其稳定性。原说法正确。",
+        "knowledge_points":  [
+                                 "三角形的稳定性及应用"
+                             ],
+        "source":  "24-25四年级下·陕西渭南·期末"
+    },
+    {
+        "id":  "481",
+        "unit":  "第五单元三角形",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "茶几设计成三角形桌腿应用了三角形的稳定性。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】三角形具有稳定性，受力后不易变形.\n【详解】三角形桌腿正是稳定性应用。原说法正确。",
+        "knowledge_points":  [
+                                 "三角形的稳定性及应用"
+                             ],
+        "source":  "24-25四年级下·河南·期末"
+    },
+    {
+        "id":  "482",
+        "unit":  "第五单元三角形",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "![](images/80b4035ee91967c61459a97d1f150d633874838b5df287443ab971d841a4dc4d.jpg)左图中三角形的底对应的高是![](images/4f58d3eff6be0c6326249293cd8c7bd03f64222b994afeaa70e1e30bd60ec871.jpg)。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】高是从顶点向对边作的垂线段.\n【详解】图示线段满足\"过顶点，垂直于底边\"，是对应的高。原说法正确。",
+        "knowledge_points":  [
+                                 "三角形的高及画法"
+                             ],
+        "source":  "24-25四年级下·陕西·期末"
+    },
+    {
+        "id":  "483",
+        "unit":  "第五单元三角形",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "一个三角形中最大的角是100°,这个三角形一定是钝角三角形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】钝角三角形定义：有一个角大于90°.\n【详解】100°\u003e90°,所以一定是钝角三角形。原说法正确。",
+        "knowledge_points":  [
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·甘肃兰州·期中"
+    },
+    {
+        "id":  "484",
+        "unit":  "第五单元三角形",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形的两条边分别是7cm和15cm,那么第三条边可以是7cm,也可以是15cm。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】分两种情况：第三边为7或15，再检验三边关系.\n【详解】若第三边是7，则7+7\u003c15，不成三角形;若第三边是15，则7+15\u003e15，可以。故第三边只能是15cm。原说法错误。",
+        "knowledge_points":  [
+                                 "三角形三边关系",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·河南·期末"
+    },
+    {
+        "id":  "485",
+        "unit":  "第五单元三角形",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "钝角三角形的两个锐角之和一定为锐角。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】两个锐角和=180°-钝角;钝角\u003e90°.\n【详解】所以180°-钝角\u003c90°,两个锐角之和一定是锐角。原说法正确。",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "三角形的内角和",
+                                 "直角，钝角，锐角的认识及特征"
+                             ],
+        "source":  "24-25四年级下·江西宜春·期末"
+    },
+    {
+        "id":  "486",
+        "unit":  "第五单元三角形",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形的两条边分别长8cm和4cm,它的周长是16cm或20cm。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】若腰是8cm,三边8,8,4可成;若腰是4cm,三边4,4,8不成.\n【详解】可行周长只有8+8+4=20cm,不是\"16或20\"。原说法错误。",
+        "knowledge_points":  [
+                                 "三角形三边关系",
+                                 "三角形的周长",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25六年级下·河南省直辖县级单位·期末"
+    },
+    {
+        "id":  "487",
+        "unit":  "第五单元三角形",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "刘晓用三根长分别是6厘米,6厘米,15厘米的小棒，围成了一个等腰三角形。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】检验三角形三边关系.\n【详解】6+6=12\u003c15，不能围成三角形，更不可能是等腰三角形。原说法错误。",
+        "knowledge_points":  [
+                                 "三角形三边关系",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·河南·期末"
+    },
+    {
+        "id":  "488",
+        "unit":  "第五单元三角形",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形的周长是24厘米，已知底边长比腰长多3厘米。这个等腰三角形的腰长和底边长各是多少厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "腰长是7厘米，底边长是10厘米",
+        "solution":  "【分析】设腰长为x厘米，则底边为x+3厘米。周长24厘米:x+x+(x+3)=24.\n【详解】3x+3=24,3x=21,x=7。底边=7+3=10。答：腰长7厘米，底边长10厘米。",
+        "knowledge_points":  [
+                                 "三角形的周长",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·湖南永州·期末"
+    },
+    {
+        "id":  "489",
+        "unit":  "第五单元三角形",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "为了保护正方形造型的花圃，园艺师沿着花圃边围了整一圈隔离带，隔离带长40米。过了一个月，花圃换成了等腰三角形的形状，这条隔离带恰好够围一圈。等腰三角形花圃的一条边长是8米，另两条边的长分别是多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "16米、16米",
+        "solution":  "【分析】等腰三角形周长40米，一条边是8米。若8米是腰，则底边=24米，不满足三边关系;故8米为底边.\n【详解】两腰和=40-8=32米，每条腰=32÷2=16米。答：另外两条边都是16米。",
+        "knowledge_points":  [
+                                 "三角形三边关系",
+                                 "三角形的周长",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·河南郑州·期末"
+    },
+    {
+        "id":  "490",
+        "unit":  "第五单元三角形",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "钱叔叔做了一个等腰三角形的架子，它的顶角是40度，它的底角是多少度。",
+        "options":  [
+
+                    ],
+        "answer":  "70度",
+        "solution":  "【分析】等腰三角形两个底角相等，三角形内角和180°.\n【详解】两个底角和=180°-40°=140°,每个底角=140°/2=70°。答：底角是70度。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "491",
+        "unit":  "第五单元三角形",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "如图，三角形ABC是等腰三角形,∠A,∠B是它的两个底角。求∠B的度数。![](images/ed2f344857fd11a94c768d95ebe22f8a896e3087edb0312e8812d8a9704fc49c.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "67°",
+        "solution":  "【分析】先由平角求∠ACB,再利用等腰三角形底角相等和内角和求∠B.\n【详解】∠ACB=180°-134°=46°。∠A=∠B,所以2∠B+46°=180°,∠B=(180°-46°)÷2=67°。答:∠B=67°。",
+        "knowledge_points":  [
+                                 "平角，周角的认识及特征",
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "角度的计算"
+                             ],
+        "source":  "24-25四年级下·贵州遵义·期末"
+    },
+    {
+        "id":  "492",
+        "unit":  "第五单元三角形",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "王叔叔用一根铁丝围成一个等边三角形框架(如下图),这个三角形框架的边长扩大10倍后周长是27米，那么原来框架的边长是多少分米。![](images/bc0cb558b5ab6ae77bb6c3e475200c2f61324c1ed5ba74cb46c90e716902d787.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "9分米",
+        "solution":  "【分析】边长扩大10倍，周长也扩大10倍。先求扩大前周长，再除以3求边长，并统一单位.\n【详解】27米=270分米;扩大前周长=270÷10=27分米;原边长=27÷3=9分米。答：原来框架边长是9分米。",
+        "knowledge_points":  [
+                                 "三角形的周长",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·陕西渭南·期末"
+    },
+    {
+        "id":  "493",
+        "unit":  "第五单元三角形",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "我们佩戴的红领巾的颜色是国旗红，其形状是等腰三角形，分为小号，大号两种规格。其中小号的尺寸是底边长100厘米，腰长60厘米;大号的尺寸是底边长120厘米，腰长72厘米。(1)算式100+60×2解决的问题是()。(2)如果一条红领巾中最大的角是120°,这条红领巾的另外两个角分别是多少度。",
+        "options":  [
+
+                    ],
+        "answer":  "小号红领巾的周长;30°、30°",
+        "solution":  "【分析】(1)等腰三角形周长=底边+2*腰长.\n(2)等腰三角形两底角相等，且三角形内角和180°.\n【详解】(1)100+60×2表示小号红领巾的周长.\n(2)另外两个角=(180°-120°)÷2=30°,所以另外两个角各是30°。",
+        "knowledge_points":  [
+                                 "三角形的周长",
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·内蒙古呼和浩特·期末"
+    },
+    {
+        "id":  "494",
+        "unit":  "第五单元三角形",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "放风筝是我国民间传统游戏之一。华华想做一个等腰三角形的风筝。(1)如果等腰三角形风筝的顶角是40度，它的两个底角分别是多少度。![](images/e11f77f5f007c0e22fd195c6dbe2934e6c307bb7c69b1dfc1aa229357cafb35b.jpg)(2)如果这个等腰三角形风筝有两条边分别长45厘米和22厘米。那么它的周长是多少厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "70度;70度;112厘米",
+        "solution":  "【分析】(1)等腰三角形两底角相等，三角形内角和180°.\n(2)先判断哪条是腰：若腰为22，则22+22\u003c45，不成三角形;故腰为45.\n【详解】(1)底角=(180°-40°)÷2=70°,两个底角分别是70°,70°.\n(2)周长=45+45+22=112(厘米)。",
+        "knowledge_points":  [
+                                 "三角形三边关系",
+                                 "三角形的周长",
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·四川南充·期末"
+    },
+    {
+        "id":  "495",
+        "unit":  "第五单元三角形",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "妙妙发现窗棂上有一个三角形木雕，量得其中有两条边长都是10厘米，另一条边长是16厘米，这个三角形的周长是多少厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "36厘米",
+        "solution":  "【分析】三角形周长是三边长度之和.\n【详解】10+10+16=36(厘米)。答：周长是36厘米。",
+        "knowledge_points":  [
+                                 "三角形的概念及表示方法",
+                                 "三角形的周长"
+                             ],
+        "source":  "24-25四年级下·福建漳州·期末"
+    },
+    {
+        "id":  "496",
+        "unit":  "第五单元三角形",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "用一根长60厘米的铁丝，围成一个底边是1分米的等腰三角形，这个等腰三角形的一条腰长是多少厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "25厘米",
+        "solution":  "【分析】先统一单位:1分米=10厘米。等腰三角形两腰相等.\n【详解】两腰和=60-10=50(厘米),一条腰=50÷2=25(厘米)。答：一条腰长25厘米。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的周长",
+                                 "毫米，厘米，分米，米之间的进率与换算"
+                             ],
+        "source":  "24-25四年级下·福建福州·期末"
+    },
+    {
+        "id":  "497",
+        "unit":  "第五单元三角形",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "如图，一个直角三角形，一个锐角是54°,另一个锐角是多少度。![](images/753c1418a8adf53541f03e4ae22f6ca8bfc434ac8c88a32a0763c30e0b9e71db.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "36度",
+        "solution":  "【分析】直角三角形中，两个锐角和是90°.\n【详解】另一个锐角=90°-54°=36°。答：另一个锐角是36度。",
+        "knowledge_points":  [
+                                 "直角，钝角，锐角的认识及特征",
+                                 "三角形的内角和",
+                                 "角度的计算"
+                             ],
+        "source":  "24-25四年级下·河南开封·期末"
+    },
+    {
+        "id":  "498",
+        "unit":  "第五单元三角形",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "如图，在等边三角形ABC中,∠1=∠2=∠3=∠4。求∠5的度数。![](images/7a581615d45b0980d14f3c33bd3986f14eed1728d820512aa2af9972b085c937.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "120°",
+        "solution":  "【分析】等边三角形每个角60°。由∠1=∠2=∠3=∠4可得每个是30°,再由平角关系求∠5.\n【详解】∠1=∠2=∠3=∠4=60°/2=30°,∠5=180°-30°-30°=120°。答:∠5=120°。",
+        "knowledge_points":  [
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·福建漳州·期末"
+    },
+    {
+        "id":  "499",
+        "unit":  "第五单元三角形",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "用一条绳子刚好能围成一个边长是9厘米的正方形而没有剩余，用这条绳子围成一个最大的等边三角形，它的边长是多少厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "12厘米",
+        "solution":  "【分析】绳长等于正方形周长:9×4=36厘米。等边三角形三边相等，边长=36÷3.\n【详解】36÷3=12(厘米)。答：边长是12厘米。",
+        "knowledge_points":  [
+                                 "三角形的周长",
+                                 "正方形的周长",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·广西贵港·期末"
+    },
+    {
+        "id":  "500",
+        "unit":  "第五单元三角形",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "丽丽量出红领巾两条边的长度分别为100厘米和60厘米，还量出了红领巾其中两个角都为30°,那么红领巾的三边之和是多少厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "220厘米",
+        "solution":  "【分析】两个角都为30°,说明是等腰三角形，且两腰相等。结合边长100和60可判定腰为60，底为100.\n【详解】三边之和=60+60+100=220(厘米)。答：红领巾的三边之和是220厘米。",
+        "knowledge_points":  [
+                                 "三角形的周长",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·浙江宁波·期末"
+    }
+]
diff --git "a/\345\233\233\345\271\264\347\272\247\351\242\230\351\233\206/\345\233\233\345\271\264\347\272\247\344\270\213/\345\233\233\345\271\264\347\272\247\344\270\213\347\254\254\345\205\253\345\215\225\345\205\203\345\271\263\345\235\207\346\225\260\345\222\214\346\235\241\345\275\242\345\233\276.json" "b/\345\233\233\345\271\264\347\272\247\351\242\230\351\233\206/\345\233\233\345\271\264\347\272\247\344\270\213/\345\233\233\345\271\264\347\272\247\344\270\213\347\254\254\345\205\253\345\215\225\345\205\203\345\271\263\345\235\207\346\225\260\345\222\214\346\235\241\345\275\242\345\233\276.json"
new file mode 100644
index 0000000000000000000000000000000000000000..48fccb1d0f86dba6f5dd7979fc9b8bd735d4169a
--- /dev/null
+++ "b/\345\233\233\345\271\264\347\272\247\351\242\230\351\233\206/\345\233\233\345\271\264\347\272\247\344\270\213/\345\233\233\345\271\264\347\272\247\344\270\213\347\254\254\345\205\253\345\215\225\345\205\203\345\271\263\345\235\207\346\225\260\345\222\214\346\235\241\345\275\242\345\233\276.json"
@@ -0,0 +1,9035 @@
+[
+    {
+        "id":  "1",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "(24-25四年级下·全国·课后作业)甲，乙车间同时加工一批零件，甲车间有14人，平均每人加工25个零件;乙车间有16人，一共加工400个零件。两个车间平均每人加工()个零件。",
+        "options":  [
+                        "A. 25",
+                        "B. 26",
+                        "C. 28",
+                        "D. 30"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】先用甲车间平均每人加工的零件个数乘以人数，求出甲车间加工的零件总数，再加上乙车间加工的零件总数，算出两个车间加工的零件总数，最后除以两个车间的总人数，可得到两个车间平均每人加工的零件个数，据此解答.\n【详解】25×14=350(个)\n(350+400)÷(14+16)\n=750÷30\n=25(个)\n所以两个车间平均每人加工25个零件。故答案为:A",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算，两位数乘两位数的进位乘法，平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "2",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "(24-25四年级下·全国·课后作业)有一组数据的平均数是10，下面()加入这组数据后，整组数据的平均数还是10。",
+        "options":  [
+                        "A. 9;10和11",
+                        "B. 10和11",
+                        "C. 11;12和13"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】要使加入数据后整组数据的平均数还是10，加入的数据的平均数也必须是10。分别计算各选项数据的平均数，与10比较即可.\n【详解】A:(9+10+11)÷3=30÷3=10，平均数是10，符合要求。B:(10+11)÷2=21÷2=10.5，平均数不是10，不符合要求。C:(11+12+13)÷3=36÷3=12，平均数不是10，不符合要求。故答案为A。",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "3",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "(24-25四年级下·全国·课后作业)甲公司第二季度共销售饮料54万箱，第二季度平均每个月销售()万箱。",
+        "options":  [
+                        "A. 54÷2",
+                        "B. 54÷3",
+                        "C. 54÷6"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】第二季度包括4月,5月,6月，共3个月，求平均每个月销售多少万箱，用销售总量除以3即可.\n【详解】第二季度有3个月，所以平均每个月销售:54÷3(万箱)故答案为B。",
+        "knowledge_points":  [
+                                 "平均数的意义及求法，年，月，日的认识及换算"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "4",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(25-26四年级上·河南洛阳·期中)下列说法正确的有()句。①如果A/B=5。3，那么(A×10)÷(B×10)=5。3。②从不同的方向观察同一物体，看到的图形可能相同。③如果被除数的末尾有两个零，那么商的末尾至少有一个零。④小华身高145厘米，河水平均深度是110厘米。河中有的地方比110厘米深，有的地方比110厘米浅，所以小华可能有危险。",
+        "options":  [
+                        "A. 2",
+                        "B. 3",
+                        "C. 4"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】①被除数和除数同时乘或除以相同的数(0除外),商不变，余数也同时乘或除以相同的数。据此进行判断。②![](images/e4aa00f6961f43f6a06fec309e7102c9f83064b5fc5a1587218e896c2c105d50.jpg)先观察出左边的立体图形从正面和侧面看到的图形，然后再进行判断即可。③被除数的末尾有两个零，商的末尾不一定有零。计算100÷25即可判断。④平均数是表示一组数据的平均值，是指在一组数据中所有数据之和再除以这组数据的个数;其特点是比最大数小，比最小数大，由此即可进行判断。解答此题的关键是要熟练掌握商的不变规律，对三视图的认识，以及平均数的意义.\n【详解】①据此如果A/B=5。3，那么(A×10)÷(B×10)=5。30。所以原题说法错误.\n![](images/3af3eca9b5b9044ac973e27d34459a1711673ac4a058eb63ee119e65883385b9.jpg)此图从正面和侧面看到的图形相同，都可看到1层，为2个小正方形;因此同一个物体从不同角度观察，看到的形状可能相同，所以原题说法正确。③100÷25=4，则被除数的末尾有两个零，商的末尾不一定有零。所以原题说法错误。④小华身高145厘米，河水平均深度是110厘米。河中有的地方比110厘米深，有的地方比110厘米浅，所以小华可能有危险，所以原题说法正确。由此可知，说法正确的有2句。故答案为:A",
+        "knowledge_points":  [
+                                 "商不变的规律及应用，物体三视图的认识，商末尾有0的除法，平均数的意义及求法"
+                             ],
+        "source":  "25-26四年级上·河南洛阳·期中"
+    },
+    {
+        "id":  "5",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "(2025·湖北武汉·小升初真题)有5个数，它们的平均数是8，加入数()后，这组数的平均数是9。",
+        "options":  [
+                        "A. 5",
+                        "B. 13",
+                        "C. 14",
+                        "D. 15"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】先计算原来5个数的总和，再计算加入一个数后,6个数的总和。用6个数的总和减去5个数的总和，即可得到加入数的数值.\n【详解】5×8=40;6×9=54,54-40=14，即加入的数为14。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "2025·湖北武汉·小升初真题"
+    },
+    {
+        "id":  "6",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "(25-26六年级·全国·随堂练习)六年级学生的平均身高为153cm,关于六年级学生身高的说法，正确的是()。",
+        "options":  [
+                        "A. 可能每个学生都低于153cm",
+                        "B. 可能每个学生都高于153cm",
+                        "C. 肯定有学生不低于153cm"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】平均数的核心特征是\"代表一组数据的整体水平，数据中既有可能存在高于平均数的数，也有可能存在低于平均数的数\".\n【详解】A。若每个学生都低于153cm,平均数会小于153cm,不符合条件;B。若每个学生都高于153cm,平均数会大于153cm,不符合条件;C。平均数是153cm,说明数据中必然存在不低于153cm的学生(否则平均数会更低),该说法正确。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "25-26六年级·全国·随堂练习"
+    },
+    {
+        "id":  "7",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "(24-25六年级下·广东东莞·期末)某小组6名同学数学测试成绩:85,90,92,88,95,m。若平均分是90，则m是()。",
+        "options":  [
+                        "A. 88",
+                        "B. 90",
+                        "C. 92",
+                        "D. 95"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据平均数=所有数据之和/总数，可先求出所有数据之和=平均数×总数，再减去几个已知数据，即可求得m的值.\n【详解】所有数据之和为90×6=540(分),则m=540-85-90-92-88-95=90。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25六年级下·广东东莞·期末"
+    },
+    {
+        "id":  "8",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25六年级下·广东东莞·期末)5名同学身高(cm):152,155,158,160,a。若中位数是158，则a可能是()。",
+        "options":  [
+                        "A. 150",
+                        "B. 157",
+                        "C. 156",
+                        "D. 162"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】将一组数据按照从小到大(或从大到小)的顺序排列，如果数据的个数是奇数，则处于中间位置的数就是这组数据的中位数;如果数据的个数是偶数，则中间两个数据的平均数是这组数据的中位数。已知数据152,155,158,160，因为这组数据有5个(个数为奇数)，中位数是158，所以将5个数据从小到大排列后，第3个数是158。那么a需要满足a不小于158，这样才能保证158处于第3位。\n【详解】A。150\u003c158，不符合。B。157\u003c158，不符合。C。156\u003c158，不符合。D。162\u003e158，符合。所以a可能是162。故答案为:D.",
+        "knowledge_points":  [
+                                 "中位数的意义及求法"
+                             ],
+        "source":  "24-25六年级下·广东东莞·期末"
+    },
+    {
+        "id":  "9",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "(24-25六年级下·广东中山·期末)某商场去年四个季度销售额分别为：第一季度120万元，第二季度150万元，第三季度180万元，第四季度210万元。全年平均每月销售额是()。",
+        "options":  [
+                        "A. 55万元",
+                        "B. 60万元",
+                        "C. 65万元",
+                        "D. 70万元"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】把四个季度销售额相加得到全年的销售额再除以12个月就是平均每月销售额，据此解答.\n【详解】(120+150+180+210)÷12=660÷12=55(万元)全年平均每月销售额是55万元。故答案为:A",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25六年级下·广东中山·期末"
+    },
+    {
+        "id":  "10",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·山东日照·期末)下面说法正确的是()。",
+        "options":  [
+                        "A. 某班级同学体育测试的平均成绩是93分，可能有一个人的成绩是80分。",
+                        "B. 明明所在班级同学的平均体重是36kg,乐乐所在班级同学的平均体重是41kg,乐乐一定比明明重。",
+                        "C. 统计小东家下半年的用电情况适合用复式统计图。"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】平均数是表示一组数据的平均值，是指在一组数据中所有数据之和再除以这组数据的个数;其特点是比最大数小，比最小数大，不能表示这组数据中各个数据的大小，由此即可进行选择。复式条形统计图能让人清楚地看出数量的多少;但只用于比较多个物体的数量;依此判断.\n【详解】A。某班级同学体育测试的平均成绩是93分，可能有一个人的成绩是80分，即原说法正确。B。明明所在班级同学的平均体重是36kg,乐乐所在班级同学的平均体重是41kg,乐乐和明明的体重无法比较，即原说法错误。C。统计小东家下半年的用电情况不适合用复式统计图，适合用单式统计图，即原说法错误。故答案为:A",
+        "knowledge_points":  [
+                                 "复式条形统计图，平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·山东日照·期末"
+    },
+    {
+        "id":  "11",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "(24-25六年级下·广东东莞·期末)小明记录了一周的体温(单位:°C):36.2,36.5,36.3,36.4,36.6,36.1,36.5，这组数据的中位数是()。",
+        "options":  [
+                        "A. 36.3",
+                        "B. 36.4",
+                        "C. 36.5",
+                        "D. 36.2"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】中位数是将一组数据按照大小顺序排列后，位于中间位置的数值。它能反映数据的集中趋势，且不受极端值的影响。对于一组数据，若数据个数为奇数，则中位数是排序后位于正中间的那个数;若数据个数为偶数，则中位数是排序后中间两个数的平均值。本题中小明记录的体温数据共7个(7为奇数),因此需先将数据排序，再找正中间的第4个数据作为中位数.\n【详解】对数据排序得:36.1,36.2,36.3,36.4,36.5,36.5,36.6\n数据总个数为7，数据个数代入计算:(7+1)÷2=4，即中位数是排序后第4个数据，第4个数据为36.4，因此这组数据的中位数是36.4。故答案为:B",
+        "knowledge_points":  [
+                                 "中位数的意义及求法"
+                             ],
+        "source":  "24-25六年级下·广东东莞·期末"
+    },
+    {
+        "id":  "12",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·河北唐山·期末)五人中身高最高的是1.64m,最矮的是1.42m,他们的平均身高可能是()。",
+        "options":  [
+                        "A. 1.56m",
+                        "B. 1.66m",
+                        "C. 1.40m",
+                        "D. 数据不全，无法判断"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】平均数是表示一组数据的平均值，平均数比最小数大，比最大数小，据此解题.\n【详解】1.42m\u003c平均数\u003c1.64m,1.40m\u003c1.42m\u003c1.56m\u003c1.64m\u003c1.66m,五人中身高最高的是1.64m,最矮的是1.42m,他们的平均身高可能是1.56m。故答案为:A",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·河北唐山·期末"
+    },
+    {
+        "id":  "13",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "(24-25四年级下·安徽淮北·期末)2,8,14,16,20的平均数是()。",
+        "options":  [
+                        "A. 8",
+                        "B. 12",
+                        "C. 9"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据平均数的求法，将五个数相加求得的和除以5即可得到五个数的平均数.\n【详解】(2+8+14+16+20)÷5=60÷5=12,2,8,14,16,20的平均数是12。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·安徽淮北·期末"
+    },
+    {
+        "id":  "14",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·湖南长沙·期末)某环保小组记录了5个社区某月的垃圾分类量(单位：吨):28吨,35吨,42吨,x吨,31吨。已知这5个社区某月平均垃圾分类量为35吨，下列说法错误的是()。",
+        "options":  [
+                        "A. 未知分类量X的值为39",
+                        "B. 若新增一个分类量为35吨的社区，则新的平均分类量仍为35吨",
+                        "C. 若去掉一个最低分类量(28吨),剩余4个社区的平均分类量比原来高",
+                        "D. 若将最高分类量(42吨)提高5吨，新的平均数将超过37吨"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】由题意可知：用平均数×份数=总数量，再用总数量减去已知的4个社区的垃圾分类总量，即可求出未知分类量X的值;根据\"平均数=总数量/份数\"求出新的平均分类量，进行比较即可;据此分析选项即可.\n【详解】A。35×5=175(吨),175-(28+35+42+31)=39(吨),未知分类量X的值为39，说法正确;B。35×5=175(吨),175+35=210(吨),5+1=6(个),210÷6=35(吨),新的平均分类量仍为35吨，说法正确;C。35×5=175(吨),175-28=147(吨),5-1=4(个),147÷4=36.75(吨),36.75\u003e35，剩余4个社区的平均分类量比原来高，说法正确;D。35×5=175(吨),175+5=180(吨),180÷5=36(吨),36\u003c37，新的平均数没有超过37吨，说法错误。故答案为:D",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·湖南长沙·期末"
+    },
+    {
+        "id":  "15",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·河南驻马店·期末)关于\"平均数\"表述错误的是()。",
+        "options":  [
+                        "A. 平均数只能是一个整数",
+                        "B. 平均数一般位于最大数和最小数之间",
+                        "C. 要改变一组数的平均数，可以改变这一组数的任何一个数",
+                        "D. 歌唱比赛，在计算歌手平均得分时，一般要去掉最高分和最低分"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】根据题意，平均数是一组数据的总和除以数据的个数，反映数据的集中趋势。根据平均数的性质，逐一分析选项即可.\n【详解】A。平均数只能是整数。例如，数据3和4的平均数为3.5，是小数，因此平均数可以是小数。选项错误。B。平均数位于最大数与最小数之间。因为平均数不会超过最大值，也不会低于最小值。选项正确。C。改变任何一个数，总和必然变化，平均数也会变化，例如原数2,2,2，其中一个2改为3，平均数就不是2。选项正确。D。在比赛中，计算平均得分时要去掉一个最高分和一个最低分的原因是避免平均数受极端数据的影响，使比赛更公平。选项正确。故答案为:A",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·河南驻马店·期末"
+    },
+    {
+        "id":  "16",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·河北廊坊·期末)下面三位同学对平均数理解正确的是()。",
+        "options":  [
+                        "A. 五年级同学的平均体重是38千克，表示我到了五年级，体重将会长到38千克",
+                        "B. 我们班和四(2)班比跳绳成绩，用平均数比较公平",
+                        "C. 如果五年级有一位同学的体重改变了，他们的平均体重并不会改变"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】平均数是反映一组数据整体水平的统计量，具有虚拟性，每个数据的变化都会影响平均数。据此解答.\n【详解】A。平均数表示整体水平，不代表某个个体一定达到该数值，此选项错误。B。比较两个班级跳绳成绩时，平均数能公平反映整体水平，此选项正确。C。任一数据变化均会影响平均数，此选项错误。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·河北廊坊·期末"
+    },
+    {
+        "id":  "17",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·湖北十堰·期末)下面的说法中，错误的是()。",
+        "options":  [
+                        "A. 等腰三角形不一定是锐角三角形。",
+                        "B. 小数部分相邻两个计数单位之间的进率与整数一样，都是10。",
+                        "C. 给![](images/2026-03-14%20160753.jpg)添上一个小正方体，使其从上面看到的图形不变，有3种不同的添法。",
+                        "D. 典典所在班级的同学平均每人捐款15元，说明典典一定捐了15元。"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】有两条边相等的三角形叫作等腰三角形，三条边都相等的三角形叫作等边三角形。三角形按角来分，可以分为锐角三角形，直角三角形和钝角三角形。小数的计数单位有十分之一，百分之一，千分之一。观察给出的立体图形，要使其从上面看到的图形不变，只能在已有的正方体的上方添加。平均数是一组数据的总和除以这组数据的个数得到的值，它反映的是这组数据的一般水平，并不能代表每个个体的数据。据此解答即可.\n【详解】A。等腰三角形有可能是锐角三角形，也可能是直角三角形，还可能是钝角三角形。所以，该说法正确。B。小数部分相邻两个计数单位，如十分位和百分位之间的进率是10;整数部分相邻两个计数单位，如个位和十位之间的进率也是10。所以，该说法正确。C。从上面观察给出的立体图形，有3个小正方形。要使其添上一个小正方体后，从上面看到的图形不变，就只需在这3个小正方形的正上方添，有3种不同的添法。所以，该说法正确。D。典典所在班级的同学平均每人捐款15元，那么典典捐的钱可能比15元少，也有可能比15元多，还有可能正好等于15元。所以，该说法错误。故答案为:D",
+        "knowledge_points":  [
+                                 "物体三视图的认识，小数的数位和计数单位的认识，平均数的意义及求法，等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·湖北十堰·期末"
+    },
+    {
+        "id":  "18",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·湖南永州·期末)三个连续自然数的平均数是78，其中最大一个数是()。",
+        "options":  [
+                        "A. 80",
+                        "B. 79",
+                        "C. 77",
+                        "D. 26"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】三个连续自然数的平均数是78，说明这个三个数分别为:78-1,78,78+1，即这三个连续的自然数就是77,78,79，其中最大的是79.\n【详解】78-1=77,78+1=79 平均数:(77+78+79)÷3=234÷3=78，所以三个连续自然数的平均数是78，这三个连续的自然数就是77,78,79;77\u003c78\u003c79，所以三个连续自然数的平均数是78，其中最大一个数是79。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·湖南永州·期末"
+    },
+    {
+        "id":  "19",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·河南许昌·期末)下列说法正确的有()个。①在一个轴对称图形上,A的对称点是B,A,B到对称轴的距离相等。②钝角三角形中两个锐角的和一定小于直角三角形中两个锐角的和。③3.56精确到十分位是4。④三角形衣架，大桥的斜拉钢索等，都用到了三角形的稳定性。⑤一组数据的平均数一定大于这组数据中的最小数。",
+        "options":  [
+                        "A. 1",
+                        "B. 2",
+                        "C. 3",
+                        "D. 4"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】一个图形沿一条直线对折，直线两旁的图形完全重合，这样的图形叫做轴对称图形，折痕所在的直线就是对称轴。在轴对称图形中，两个对称点到对称轴的距离相等;三角形按角来分，分为锐角三角形，直角三角形和钝角三角形。三个角都是锐角的三角形叫作锐角三角形，有一个角是直角的三角形叫作直角三角形，有一个角是钝角的三角形叫作钝角三角形。三角形的内角和为180°;一个小数，精确到十分位时，需要看百分位上数的大小，用\"四舍五入\"法求它的近似数;三角形具有稳定性，有着稳固，坚定，耐压的特点;平均数代表的是一组数据的一般水平。它一般比这组数据的最小值大，一般比这组数据的最大值小。据此解答.\n【详解】①根据题意作图如下:![](images/fba1a1a4ac7ec0925f036fb142b5e8fea07501aa9375ea0d3116dfa5103176d4.jpg)由图可知，该图形是一个轴对称图形,A点和B点为对称点，它们到对称轴的距离相等。该选项说法正确。②在钝角三角形中，两个锐角的和=180°-钝角的度数。钝角的度数大于90°,180°-90°=90°,所以两个锐角的和\u003c90°;在直角三角形中，两个锐角的和=180°-直角的度数。直角的度数为90°,180°-90°=90°,所以两个锐角的和是90°。所以钝角三角形中两个锐角的和一定小于直角三角形中两个锐角的和。该选项说法正确。③3.56的百分位上是6，需要进一，所以3.56≈3.6。该选项说法错误。④三角形具有稳定性，有着稳固，坚定，耐压的特点，三角形衣架，大桥的斜拉钢索等都运用了三角形的这种特性。该选项说法正确。⑤如果一组数据是:2,2,2,2,2，那么这组数据的最小值是2，平均值也是2，此时平均值等于这组数据的最小值。该选项说法错误。综上所述，正确的说法有3个。故答案为:C",
+        "knowledge_points":  [
+                                 "三角形的稳定性及应用，平均数的意义及求法，三角形的分类，轴对称的认识及辨认"
+                             ],
+        "source":  "24-25四年级下·河南许昌·期末"
+    },
+    {
+        "id":  "20",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·河南郑州·期末)下列说法正确的是()。",
+        "options":  [
+                        "A. 用长5cm,4cm和1cm的三根小棒能围成一个三角形",
+                        "B. 算式62000÷125/62转化成62000÷62/125进行计算更简便",
+                        "C. 四年级同学的平均身高是1.42m,四一班的聪聪身高一定是1.42m"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】三角形两边之和大于第三边，三角形的两边的差小于第三边;62000÷125/62可以交换125和62的位置,62000÷62/125，此时62000÷62结果是整千数，计算更简便;平均数是表示一组数据集中趋势的量数，是指在一组数据中所有数据之和再除以这组数据的个数。它是反映数据集中趋势的一项指标。平均数是在最小数与最大数之间.\n【详解】A。4+1=5(cm);用长5cm,4cm和1cm的三根小棒不能围成一个三角形;B。算式62000÷125/62转化成62000÷62/125进行计算更简便，说法正确。C。四年级同学的平均身高是1.42m,四一班的聪聪身高一定是1.42m,说法错误，可能低于1.42m,也有可能高于1.42m,或者是等于1.42m。故答案为:B",
+        "knowledge_points":  [
+                                 "三角形三边关系，除数是两位数的连除，平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·河南郑州·期末"
+    },
+    {
+        "id":  "21",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "(24-25四年级下·河南郑州·期末)为了能够更快地拼好这面魔方墙，将同学们平均分成了4组,10分钟内平均每组完成魔方的个数为12个。下列说法中正确的是()。",
+        "options":  [
+                        "A. 10分钟内每组都完成了12个魔方",
+                        "B. 这4组可能有的组完成的魔方比12个多，也可能比12个少",
+                        "C. 如果第一组去休息了，那么剩下的3个组平均10分钟每组完成的魔方数要少于12个"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据题意，平均数是一组数据的总和除以数据个数，它反映的是整体平均水平，不代表每组数据都等于平均数，个别数据可能高于，低于或等于平均数.\n【详解】A。平均每组完成12个是4组完成数量的整体平均水平，不代表每组实际完成数都等于12个。例如，可能存在\"10个,12个,13个,13个\"的情况;B。根据平均数的特性，为了使4组的平均值为12个，必然存在\"部分组高于12个，部分组低于12个\"的情况;C。若第一组完成数高于12个(如15个),则剩下3个组总和为48-15=33(个),平均为33÷3=11(个)(少于12个);若第一组完成数低于12个(如9个),则剩下3个组总和为48-9=39(个),平均为39÷3=13(个)(多于12个)。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·河南郑州·期末"
+    },
+    {
+        "id":  "22",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·河南漯河·期末)实验小学四(1)班全班同学的平均身高原来是150厘米，新学期又转来了两名同学，他们的身高分别是151厘米和154厘米，与原来相比，现在全班的平均身高()。",
+        "options":  [
+                        "A. 增高了",
+                        "B. 降低了",
+                        "C. 不变",
+                        "D. 不能确定"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】一组数据的总和除以这组数据个数所得到的商叫这组数据的平均数，将新来的两名同学的身高和全班同学原来的平均身高比较后选择即可.\n【详解】151厘米\u003e150厘米,154厘米\u003e150厘米，两个人的身高都高于原来的平均身高，现在全班的平均身高增高了。故答案为:A",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·河南漯河·期末"
+    },
+    {
+        "id":  "23",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "(24-25四年级下·河南焦作·期末)在学校组织的数学竞赛中，四(1)班和四(2)班的数学竞赛平均分分别是85分和83分。小贝在四(1)班，小塔在四(2)班，他们的竞赛成绩相比,()。",
+        "options":  [
+                        "A. 小贝的分数高",
+                        "B. 小塔的分数高",
+                        "C. 分数一样高",
+                        "D. 无法比较"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】平均数是表示一组数据集中趋势的量数，它是反映数据集中趋势的一项指标。已知小贝所在班的平均成绩是85分，并不表示小贝的成绩是85分，他的成绩可能比85分高，也可能比85分低，也可能正好是85分;同理，小塔所在班的平均成绩是83分，并不表示小塔的成绩是83分，他的成绩可能比83分高，也可能比83分低，也可能正好是83分;据此解答.\n【详解】由分析可知，小贝所在班的平均成绩是85分，小塔所在班的平均成绩是83分，并不能确定小贝，小塔具体的成绩，所以小贝与小塔的成绩是无法比较的。故答案为:D",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·河南焦作·期末"
+    },
+    {
+        "id":  "24",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·河南郑州·期末)张丽，王芳和李红比赛做一分钟仰卧起坐，她们三人平均做了45个，其中张丽做了42个，王芳做了48个，李红做的个数()。",
+        "options":  [
+                        "A. 小于42个",
+                        "B. 大于48个",
+                        "C. 在42个和48个之间"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】总数=平均个数×人数，用平均个数乘3，求出三人一共做了多少个，再用总数依次减去张丽和王芳做的个数，即可求出李红做的个数，再与每个选项进行比较即可.\n【详解】45×3=135(个),135-42-48=93-48=45(个),A。45\u003e42，不符合题意。B。45\u003c48，不符合题意。C。42\u003c45\u003c48，符合题意。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·河南郑州·期末"
+    },
+    {
+        "id":  "25",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·河南许昌·期末)下面说法正确的是()。",
+        "options":  [
+                        "A. 4.1和4.9之间有7个一位小数",
+                        "B. 河水平均深度是100cm,身高145cm的小明下河不会有危险",
+                        "C. 小数点移动一位，小数就扩大到原来的10倍",
+                        "D. 1.5+0.9-1.5+0.9=2.4-2.4=0"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】A。由题意可知题干中限制了小数的位数，在4.1和4.9之间的一位小数有4.2,4.3,4.4,4.5,4.6,4.7,4.8，据此判断即可;B。平均数是表示一组数据的集中趋势，不是其中每一个数据的特征，它是这组数据所有数字的和除以这组数据的个数得出的。这组数据的数字分布在平均数的上下，里面可能有一些极端数字，如极端大，极端小，所以河水有的地方比100cm深，有的地方比100cm浅，有的地方深度正好是100cm,所以身高145cm的小明下河会有危险;C。小数点位置移动引起数的大小变化规律：一个数的小数点向右移动一位，两位，三位。,相当于把原数乘10,100,1000，小数就扩大到原数的10倍,100倍,1000倍。;小数点向左移动一位，两位，三位。,相当于把原数除以10,100,1000，小数就缩小到原数的1÷10,1÷100,1÷1000;D。根据小数加减法运算的运算顺序进行计算，然后再判断即可.\n【详解】A。4.1和4.9之间的一位小数有:4.2,4.3,4.4,4.5,4.6,4.7,4.8，共7个，所以本选项说法正确;B。根据分析可知，河水的平均深度是100cm,身高145cm的小明下河会有危险，而原题中说不会有危险，所以说法错误;C。把一个小数的小数点移动一位，并没有说明是向右移还是向左移，就不能确定这个小数是扩大到原来的10倍还是缩小到原来的1÷10，所以本选项说法错误;D。1.5+0.9-1.5+0.9=2.4-1.5+0.9=0.9+0.9=1.8,1.8!=0，所以本选项计算错误。故答案为:A",
+        "knowledge_points":  [
+                                 "小数点向右移动引起小数大小变化的规律，小数的加，减法混合运算，一位小数的大小比较，平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·河南许昌·期末"
+    },
+    {
+        "id":  "26",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "(24-25四年级下·河南郑州·期末)四年级同学参加科普知识竞赛。男生组的平均成绩是88分，女生组的平均成绩是86分。男生组第一名与女生组第一名相比,()。",
+        "options":  [
+                        "A. 男生成绩高",
+                        "B. 女生成绩高",
+                        "C. 成绩相等",
+                        "D. 无法确定"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】平均数=总数量/总份数，男生组的平均成绩是88分，男生组的第一名的成绩大于或等于88分。女生组的平均成绩是86分，女生组的第一名的成绩大于或等于86分，可能大于88分，也可能小于88分，还可能等于88分。也就是男生组第一名与女生组第一名相比，可能男生成绩高，也可能女生成绩高，还可能成绩相等，据此分析.\n【详解】平均数只是反映一组数据的整体情况，男生组的平均成绩是88分，女生组的平均成绩是86分。男生组第一名与女生组第一名相比，无法确定谁成绩高。故答案为:D",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·河南郑州·期末"
+    },
+    {
+        "id":  "27",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "(24-25四年级下·河南郑州·期末)学校举行校园安全知识竞赛。聪聪前三轮的答题平均分是90分，第四轮答题公布成绩后，聪聪的平均分下降了2分，聪聪第四轮答题得了()分。",
+        "options":  [
+                        "A. 92",
+                        "B. 90",
+                        "C. 87",
+                        "D. 82"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】先计算出前三轮的答题总分=平均分*3，第四轮答题公布成绩后，聪聪的平均分下降了2分，平均分下降到90-2=88(分),则四轮的答题的平均分*4即为四轮的答题总分，四轮的答题总分-前三轮的答题总分=第四轮的答题的得分。据此解答.\n【详解】90×3=270(分)(90-2)×4=88×4=352(分) 352-270=82(分)所以聪聪第四轮答题得了82分。故答案为:D",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·河南郑州·期末"
+    },
+    {
+        "id":  "28",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "(24-25四年级下·河南郑州·期末)体育用品商店四天售出的足球个数分别是：第一天14个，第二天12个，第三天11个，第四天15个。可以算出平均每天售出13个，方法是:14+12+11+15=52(个),52÷4=13(个)。如果第二天多卖出4个，这时平均每天卖出多少个。下面是三位同学的方法：小兰:14+(12+4)+11+15=56(个),56÷4=14(个)小亮:52+4=56(个),56÷4=14(个)小明:4÷4=1(个),13+1=14(个)下面选项正确的是()。",
+        "options":  [
+                        "A. 只有小兰的方法正确",
+                        "B. 小兰和小亮的方法正确",
+                        "C. 三位同学的方法都正确"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】平均数的概念，即总数除以份数。三位同学分别从不同角度来计算第二天多卖4个后的平均每天售出个数。小兰是先求出第二天多卖4个后四天总共卖出的足球个数。原来四天卖出的个数分别是14个,12个,11个,15个，第二天多卖4个，那么现在四天卖出的个数就是第一天的14个加上第二天变为12+4个，再加上第三天的11个和第四天的15个，即14+(12+4)+11+15=56(个),然后用总数56除以天数4，得到平均每天卖出56÷4=14(个)。小亮是先算出原来四天卖出的总数为52个(前面已算出14+12+11+15=52个),第二天多卖4个，那么总数就增加4个变为52+4=56个，再除以天数4，得到平均每天卖出56÷4=14(个)。小明是先算出多卖的4个平均到四天里，每天多卖的个数为4÷4=1(个),原来平均每天卖出13个，所以现在平均每天卖出13+1=14(个).\n【详解】根据分析可知，三位同学的方法都正确。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·河南郑州·期末"
+    },
+    {
+        "id":  "29",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·福建龙岩·期末)下面选项正确的是()。①有一个角是60°的等腰三角形是等边三角形。②0.7和0.70的大小相等，表示的意义也相同。③求小数的近似数时，保留两位小数就是精确到百分位。④明明班的平均身高是135厘米，明明的身高不可能超过135厘米。⑤先把6.52的小数点向右移动一位，再缩小到1÷100后得到的数是0.652。",
+        "options":  [
+                        "A. ①②③",
+                        "B. ①②⑤",
+                        "C. ①③⑤",
+                        "D. ③④⑤"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】①等腰三角形的两个底角度数相等，三角形的内角和是180°,如果等腰三角形中有一个角是60°,则所有的角都是60°,三个内角相等的等腰三角形是等边三角形。②在小数的末尾添上\"0\"或者去掉\"0\"，不改变小数的大小，只改变小数的性质。③小数点后第一位是十分位，第二位是百分位，求小数的近似数时，保留一位小数也就是精确到十分位，保留两位小数也就是精确到百分位。④平均数反映一组数据的平均情况，在一组数据中，有的数据可能会大于平均数，有的数据可能会小于平均数，有的数据可能会等于平均数。⑤把一个小数的小数点向右移动一位，小数扩大到原数的10倍，把小数点向左移动两位，小数缩小到原数的1÷100;据此解答.\n【详解】①由分析可知，有一个角是60°的等腰三角形是等边三角形。①说法正确。②由分析可知,0.7和0.70的大小相等,0.7的计数单位是0.1,0.70的计数单位是0.01，计数单位不同，所以②说法错误。③由分析可知，求小数的近似数时，保留两位小数就是精确到百分位。③说法正确。④由分析可知，明明班的平均身高是135厘米，明明身高可能是135厘米，也可能比135厘米高，还可能比135厘米矮。④说法错误。⑤6.52小数点向右移动一位后，是65.2,65.2小数点向左移动两位后，是0.652。⑤说法正确。综上可知，说法正确的有①③⑤。故答案为:C",
+        "knowledge_points":  [
+                                 "小数的性质，等腰三角形和等边三角形的认识及特征，平均数的意义及求法，小数点向左移动引起小数大小变化的规律"
+                             ],
+        "source":  "24-25四年级下·福建龙岩·期末"
+    },
+    {
+        "id":  "30",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "(24-25四年级下·福建福州·期末)小丰，小军和小清三人的英语口语测试平均成绩是92分，小丰的成绩比平均分少2分，小军的成绩比平均分多3分，那么小清的成绩是()。",
+        "options":  [
+                        "A. 小于92分",
+                        "B. 等于92分",
+                        "C. 大于92分",
+                        "D. 无法确定"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】平均成绩为三人成绩之和除以3，所以三人成绩之和为92×3=276(分),小丰成绩比平均分少2分，为92-2=90(分),小军的成绩比平均分多3分，为92+3=95(分),所以小清的成绩就是用总成绩减另外两人分数.\n【详解】92×3=276(分)92-2=90(分) 92+3=95(分) 276-90-95=186-95=91(分)91\u003c92那么小清的成绩是小于92分。故答案为:A",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·福建福州·期末"
+    },
+    {
+        "id":  "31",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·福建福州·期末)期末语文，数学，英语三科测试，鹏鹏语文得了88分，数学得了94分，他三门学科的平均分是93分。那么鹏鹏的英语得了()分。",
+        "options":  [
+                        "A. 91",
+                        "B. 93",
+                        "C. 95",
+                        "D. 97"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】由题意得，鹏鹏语文，数学，英语三门学科的平均分是93分，直接用93乘3先算出鹏鹏三门学科的总分。鹏鹏语文得了88分，数学得了94分，直接用前面的得数减去88再减去94即可算出鹏鹏英语得了多少分.\n【详解】93×3=279(分)279-88-94=191-94=97(分) 即鹏鹏的英语得了97分。故答案为:D",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·福建福州·期末"
+    },
+    {
+        "id":  "32",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·江西宜春·期末)小亮语文，数学，英语三科的平均成绩是91分，核对试卷发现数学附加题6分忘记加分，那么小亮加上附加题之后，平均分应是()分。",
+        "options":  [
+                        "A. 93",
+                        "B. 95",
+                        "C. 97"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】根据平均数=总数/总份数，用三门学科平均分乘3，计算出语文，数学，英语三科总成绩是多少，再加上忘记加分的6分，计算出加上附加题分数后的总分，然后再除以3，即可算出加上附加题后的平均分是多少。据此解答.\n【详解】91×3=273(分)273+6=279(分)279÷3=93(分)小亮语文，数学，英语三科的平均成绩是91分，核对试卷发现数学附加题6分忘记加分，那么小亮加上附加题之后，平均分应是93分。故答案为:A",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·江西宜春·期末"
+    },
+    {
+        "id":  "33",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·江西上饶·期末)乔乔读一本连环画，第一天上午读了21页，下午读了16页，第二天读了17页，正好读完，乔乔平均每天读()页。",
+        "options":  [
+                        "A. 17",
+                        "B. 21",
+                        "C. 27",
+                        "D. 无法确定"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】由题意得，乔乔读一本连环画，第一天上午读了21页，下午读了16页，可以先用21加上16算出第一天共读了多少页。第二天读了17页书，然后再用第一天读的总页数加上17算出乔乔两天一共读了多少页。最后再除以2即可算出乔乔平均每天读多少页.\n【详解】(21+16+17)÷2=(37+17)÷2=54÷2=27(页)即乔乔平均每天读27页。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·江西上饶·期末"
+    },
+    {
+        "id":  "34",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·江西赣州·期末)四年级某班男生平均身高142cm,女生平均身高140cm,则该班平均身高()。",
+        "options":  [
+                        "A. 大于142cm",
+                        "B. 小于140cm",
+                        "C. 在140cm-142cm之间"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】根据题意，班级平均身高表示整个班级所有学生身高的整体水平。若男生人数多于女生，班级平均更接近142cm;反之更接近140cm。但无论人数如何，班级平均必介于两者之间。以此答题即可.\n【详解】根据分析可知：四年级某班男生平均身高142cm,女生平均身高140cm,则该班平均身高在140cm-142cm之间。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·江西赣州·期末"
+    },
+    {
+        "id":  "35",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·江西鹰潭·期末)下列说法中，正确的有()个。①小军的身高是136cm,他到一个平均水深110cm的池塘里游泳，没有危险。②24×a+a×35=(24+35)×a,是运用了乘法分配律。③\"坐地日行八万里\"中的\"八万里\"是近似数。④乘数里一共有3个0，那么积的末尾也一定只有3个0。",
+        "options":  [
+                        "A. 1",
+                        "B. 2",
+                        "C. 3",
+                        "D. 4"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】①\"平均水深110cm\"表示池塘各处水深的平均值为110cm,但部分区域水深可能远超过110cm(比如超过小军的身高136cm),因此小军游泳仍有危险。②乘法分配律的公式为:a×c+b×c=(a+b)×c(其中c是相同因数)。题干中,24×a+a×35的相同因数是a,将其提取后得到(24+35)×a,完全符合乘法分配律的定义。③\"坐地日行八万里\"是对地球自转时，人随地球表面运动距离的大致估算，并非精确测量的结果。④乘数末尾0的数量的和，不等于积末尾0的数量，因为非0部分相乘可能额外产生0。例如:200×50=10000(乘数中共3个0，但积末尾有4个0，因2×5=10额外产生1个0).\n【详解】①小军的身高是136cm,他到一个平均水深110cm的池塘里游泳，可能有危险，原说法错误;②24×a+a×35=(24+35)×a,是运用了乘法分配律，说法正确。③\"坐地日行八万里\"中的\"八万里\"是近似数，说法正确。④乘数里一共有3个0，那么积的末尾不一定只有3个0，原说法错误。正确的说法是②和③,共2个。故答案为:B",
+        "knowledge_points":  [
+                                 "亿以内数的近似数，整数乘法分配律，三位数乘两位数，乘数末尾有0，平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·江西鹰潭·期末"
+    },
+    {
+        "id":  "36",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·江西鹰潭·期末)小红期末考试语文，数学，英语三科平均分是96分，现在知道小红的语文科是98分，英语科是94分，那么她的数学科是()分。",
+        "options":  [
+                        "A. 94",
+                        "B. 95",
+                        "C. 96",
+                        "D. 97"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】用三科平均分乘3，求出三科总分。再减去语文和英语的成绩和，求出数学成绩.\n【详解】96×3-98-94=288-192=96(分)那么她的数学科是96分。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·江西鹰潭·期末"
+    },
+    {
+        "id":  "37",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "(24-25四年级下·江西赣州·期末)安安数学，英语的平均分是95分，其中英语是91分，数学是()。",
+        "options":  [
+                        "A. 99分",
+                        "B. 95分",
+                        "C. 92分",
+                        "D. 93分"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】根据平均数的定义，用数学，英语两科的平均成绩乘2就是数学，英语两科的总分数，再用总分数减去英语的分数就是数学的分数.\n【详解】95×2=190(分)190-91=99(分)因此数学是99分。故答案为:A",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·江西赣州·期末"
+    },
+    {
+        "id":  "38",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "(24-25四年级下·江西赣州·期末)红星小学举行体育测评比赛，小兰，小明，小亮体育测评的平均成绩为95分，小云的成绩为99分，他们四个人的平均成绩为()分。",
+        "options":  [
+                        "A. 95",
+                        "B. 96",
+                        "C. 97",
+                        "D. 98"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】将小兰，小明，小亮体育测评的平均成绩乘3，求出这三人体育测评总成绩，再加上小云的体育测评成绩，求出四人体育测评总成绩，再用四人体育测评总成绩除以4，求出四个人的平均成绩.\n【详解】(95×3+99)÷4=(285+99)÷4=384÷4=96(分) 他们四个人的平均成绩为96分。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·江西赣州·期末"
+    },
+    {
+        "id":  "39",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "(24-25四年级下·浙江杭州·期末)男，女生跳绳比赛，各有6位队员。男生平均每人跳128个，女生平均每人跳132个。下列说法正确的是()。",
+        "options":  [
+                        "A. 每位男生一定都跳了128个",
+                        "B. 男生小亮跳的个数一定比女生小丽少",
+                        "C. 一定有某位女生跳了132个",
+                        "D. 男生跳绳总数一定少于女生总数"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】根据平均数的定义，平均数反映整体水平，不能确定每个个体的具体数值。可以通过平均数乘个体数求出一个整体的总数。据此逐个分析选项并作答.\n【详解】A。平均每人跳128个，不代表每位男生都跳128个，可能存在差异，例如其中一个男生可能跳125个，另一个可能跳130个。此选项错误。B。男生平均低于女生，但个体可能跳得更多。例如，男生小亮跳130个，女生小丽跳125个。此选项错误。C。平均数132个不要求某位女生必须跳132个，可通过不同数值组合达到平均。此选项错误。D。男生总数:6×128=768(个),女生总数:6×132=792(个)。768\u003c792，故男生跳绳总数一定少于女生跳绳。此选项正确。故答案为:D",
+        "knowledge_points":  [
+                                 "两，三位数与一位数连续进位的乘法，平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·浙江杭州·期末"
+    },
+    {
+        "id":  "40",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·江西吉安·期末)下面说法中错误的有()个。①大于0.3而小于0.5的一位小数只有一个。②0.5和0.50的大小相等，计数单位也相同。③计算小数加减法时，要像计算整数加减法一样，把末尾对齐。④小明身高132cm,他去平均水深120cm的河里游泳不会有危险。",
+        "options":  [
+                        "A. 1",
+                        "B. 2",
+                        "C. 3"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】①一位小数是指小数点后只有一位的小数。大于0.3而小于0.5的一位小数，只有0.4这一个，所以①正确。②根据小数的性质,0.5=0.50，它们的大小相等。但0.5的计数单位是0.1,0.50的计数单位是0.01，计数单位不同，所以②错误。③计算小数加减法时，要把小数点对齐，也就是相同数位对齐，而不是把末尾对齐，所以③错误。④平均水深120cm的河，并不代表每一处的水深都是120cm,可能存在某些地方水深超过132cm,所以小明去游泳可能会有危险,④错误。综上，数出错误个数即可解答.\n【详解】根据分析可知,②③④说法错误，所以一共有3个说法错误。故答案为:C",
+        "knowledge_points":  [
+                                 "一位小数的进位加法，退位减法，一位小数的大小比较，小数的数位和计数单位的认识，平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·江西吉安·期末"
+    },
+    {
+        "id":  "41",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·江西吉安·期末)四(1)班学生进行1分钟跳绳测试，有两名学生因病没有参加，这时参加测试学生的平均成绩是112次。后来这两名学生进行了补测，成绩分别为109次和113次，重新计算后，学生的平均成绩和112比,()。",
+        "options":  [
+                        "A. 比112次少",
+                        "B. 比112次多",
+                        "C. 正好112次"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】先求出两名补测学生的平均成绩，将补测学生平均成绩与原来参加测试学生的平均成绩比较大小，根据比较结果判断加入补测成绩后整体平均成绩的变化情况.\n【详解】两名补测学生的平均成绩:(109+113)÷2=222÷2=111次，因为111\u003c112，也就是补测学生的平均成绩低于原来参加测试学生的平均成绩，当把补测学生的成绩加入整体中计算新的平均成绩时，新的平均成绩会被拉低，所以重新计算后学生的平均成绩比112少。故答案为:A",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·江西吉安·期末"
+    },
+    {
+        "id":  "42",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25六年级下·河南南阳·期末)希望小学编程社团成员平均年龄为12.5岁，又加入一名新成员后，平均年龄变成了12.25岁，新加入成员的年龄不可能是()岁。",
+        "options":  [
+                        "A. 10",
+                        "B. 11",
+                        "C. 12",
+                        "D. 13"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】由题意可知，又加入一名新成员后，平均年龄变小了，说明加入新成员的年龄小于原来的平均年龄，即新加入成员的年龄不可能大于12.5岁，据此解答.\n【详解】因为12.25岁\u003c12.5岁，则现在的平均年龄小于原来的平均年龄，那么新加入成员的年龄应该小于12.5岁，选项中13岁\u003e12.5岁，所以新加入成员的年龄不可能是13岁。故答案为:D",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25六年级下·河南南阳·期末"
+    },
+    {
+        "id":  "43",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·河南新乡·期末)下面说法正确的是()。",
+        "options":  [
+                        "A. 小明想比较5.7△和5.62□的大小，必须知道□里是几。",
+                        "B. 四一班学生平均身高是140厘米，转走一名同学后，平均身高一定会变化。",
+                        "C. 一般的平行四边形不是轴对称图形。"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】小数大小的比较：先看它们的整数部分，整数部分大的那个数就大;如果整数部分相同，十分位上的数大的那个数就大;如果十分位上相同，百分位上的数大的那个数就大。.\n平均数代表的是一组数据的一般水平。它一般比这组数据的最小值大，一般比这组数据的最大值小;一个图形沿一条直线对折，直线两旁的图形完全重合，这样的图形叫做轴对称图形，折痕所在的直线就是对称轴。据此解答.\n【详解】A。5.7△和5.62□比较大小，整数部分相同，如果△的值比6小，十分位上△\u003c6，那么5.7△\u003c5.62。即无需知道□里的值就能比较两个小数的大小。该选项说法错误.\nB。四一班学生平均身高是140厘米，那么有的学生的身高比140厘米高，有的学生的身高比140厘米矮，有的同学的身高等于140厘米。转走一名同学后，如果转走的同学的身高为140厘米，那么全班的平均身高不会变化。该选项说法错误.\nC。![](images/edcb58042266f8a2b091cf23432b75700ff97b3442ebf87880d5f1c1913fc207.jpg)由图可知，无法找到直线使得平行四边形关于直线对称，它不是轴对称图形。该选项说法正确。故答案为:C",
+        "knowledge_points":  [
+                                 "轴对称的认识及辨认，平均数的意义及求法，多位小数的大小比较"
+                             ],
+        "source":  "24-25四年级下·河南新乡·期末"
+    },
+    {
+        "id":  "44",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(2024五年级·全国·竞赛)2022年北京冬奥会短道速滑500米比赛成绩公布，前十名选手的平均成绩是45秒，前六名选手的平均成绩是42秒，后六名选手的平均成绩是48秒，那么第五名和第六名选手的平均成绩是()秒。",
+        "options":  [
+                        "A. 43",
+                        "B. 45",
+                        "C. 46",
+                        "D. 47"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】前6名与后6名的总成绩是(42+48)×6=540(秒),这里边第5,6名选手成绩都加了2次，所以第5,6名选手的成绩和是540-45×10=90(秒),则第5,6名选手的平均是90÷2=45(秒).\n【详解】解：前6名与后6名的总成绩:(42+48)×6=540(秒),第5,6名选手的成绩和:540-45×10=90(秒),第5,6名选手的平均:90÷2=45(秒)。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "2024五年级·全国·竞赛"
+    },
+    {
+        "id":  "45",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·河南周口·期末)下列说法中，正确的有()个。(1)在一个三角形中，两个内角的和与第三个内角相等，这个三角形一定是钝角三角形。(2)小数5.28和5.280的大小相同，计数单位不同。(3)王聪所在小组同学的平均身高138厘米，张航所在小组同学的平均身高140厘米，从这些数据中不能确定王聪和张航两人谁高。(4)在轴对称图形中，对应点的连线与对称轴互相垂直。",
+        "options":  [
+                        "A. 1",
+                        "B. 2",
+                        "C. 3"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】(1)三角形的内角和是180°,三角形根据角分类，可以分为钝角三角形，直角三角形和锐角三角形，当两个锐角都为45°时，第三个角为90°,这时候正好是一个直角三角形;(2)小数点后第一位是十分位，计数单位是0.1;小数点后第二位是百分位，计数单位是0.01;小数点后第三位是千分位，计数单位是0.001;小数末尾添上0或去掉0，不改变小数大小;(3)平均数是一组数的总和除以这组数的数量。平均数反映的是一组数据的平均水平，平均数会受偏大或偏小数据的影响;(4)因为轴对称图形沿着对称轴折叠，两边的图形能够完全重合，所以对应点的连线与对称轴是垂直的。据此解答.\n【详解】(1)由分析可知，在一个三角形中，两个内角的和与第三个内角相等，这个三角形一定是直角三角形。原题表述错误.\n(2)由分析可知,5.28和5.280的大小一样,5.28的计数单位是0.01,5.280的计数单位是0.001，计数单位不一样。表述正确.\n(3)由分析可知，王聪所在小组同学的平均身高138厘米，并不表示王聪身高138厘米，可能王聪身高比138厘米高，也可能比138厘米矮;同样张航所在小组同学平均身高140厘米，并不表示张航身高140厘米，可能张航身高比140厘米高，也可能比140厘米矮。所以从这些数据中不能确定王聪和张航两人谁高。表述正确.\n(4)由分析可知，在轴对称图形中，对应点的连线与对称轴互相垂直。表述正确.\n综上可知，正确的说法有3个。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法，三角形的分类，小数的性质，轴对称的认识及辨认"
+                             ],
+        "source":  "24-25四年级下·河南周口·期末"
+    },
+    {
+        "id":  "46",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·浙江台州·期末)下面说法中，正确的有()个。①如果☆-351=△,那么△-☆=351。②小红参加\"六一\"儿童节歌唱比赛，前四位评委给出的平均分是83分，第五位评委给出85分，现在她的平均分为84分。③当三角形中两个角的和等于第三个角时，这一定是直角三角形。④7.5和7.8之间有两个小数。",
+        "options":  [
+                        "A. 1",
+                        "B. 2",
+                        "C. 3",
+                        "D. 4"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】①根据被减数-减数=差，减数=被减数-差判断;②平均数的计算方法是：一组数据中所有数据之和再除以这组数据的个数;③根据三角形内角和定理，三角形的内角和是180°,三角形中两个角的和等于第三个角，用180°/2求出第三个角的度数，再判断;④小数大小的比较：先看小数的整数部分，整数部分大的这个数就大;整数部分相同的就看十分位，十分位大的这个数就大，十分位相同的，再看百分位，百分位大的这个数就大。以此类推.\n【详解】①如果☆-351=△,那么☆-△=351，故原说法错误;②(83×4+85)÷5=(332+85)÷5=417÷5=83.4，故原说法错误;③180°/2=90°,第三个角是90°,这一定是直角三角形，故原说法正确;④题目中没有给出是几位小数，大于7.5而小于7.8的一位小数有7.6,7.7;大于7.5而小于7.8的两位小数有7.51,7.52。;大于7.5而小于7.8的三位小数有7.511,7.512。;有无数个，故原说法错误;因此上面说法中，正确的有1个。故答案为:A",
+        "knowledge_points":  [
+                                 "加，减法的意义和各部分间的关系，多位小数的大小比较，平均数的意义及求法，三角形的内角和"
+                             ],
+        "source":  "24-25四年级下·浙江台州·期末"
+    },
+    {
+        "id":  "47",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·浙江台州·期末)聪聪，明明，星星三名同学本学期的阅读情况如下：聪聪阅读的本数比三人的平均数多2，明明阅读的本数比三人的平均数少3，星星阅读的本数比三人的平均数()。",
+        "options":  [
+                        "A. 多1",
+                        "B. 少1",
+                        "C. 多5",
+                        "D. 多2"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】平均的过程其实是\"移多补少\"的过程，大数需要拿出多的部分来补给小数，使它们变得一样多;据此分析解答.\n【详解】聪聪阅读的本数比三人的平均数多2，明明阅读的本数比三人的平均数少3，把聪聪比平均数多的2补给明明比平均数少的3，这时明明阅读的本数比三人的平均数就少3-2=1本，此时明明比平均数还少1，就要让星星多的来补，星星阅读的本数要比三人的平均数多1，即可。故答案为:A",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·浙江台州·期末"
+    },
+    {
+        "id":  "48",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·福建漳州·期末)小丽，小明和小红所在的四年级(1)班，学生的平均身高是1.39米，以下说法合理的是()。",
+        "options":  [
+                        "A. 小丽的身高不可能在1.39米以上",
+                        "B. 小明的身高一定是1.39米",
+                        "C. 小明一定比小红高",
+                        "D. 小红的身高可能是1.42米"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】平均数是反映一组数据的平均水平，并不能反映这组数据中各个数据的大小;班级的平均身高并不能反映出每一位学生的身高，所以他们的身高可能高于1.39米，可能低于1.39米，也可能等于1.39米;据此选择.\n【详解】A。小丽的身高可能在1.39米以上。原说法不合理;B。小明的身高不一定是1.39米，可能高于1.39米，也可能低于1.39米。原说法不合理;C。小明的身高可能是1.4米，小红的身高可能是1.35米，还可能是1.45米，小明不一定比小红高。原说法不合理;D。小红的身高可能是1.42米。说法合理。故答案为:D",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·福建漳州·期末"
+    },
+    {
+        "id":  "49",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·浙江嘉兴·期末)明明练习跳绳，第一次跳了69下，第二次跳了75下。她要想三次的平均成绩达到80下，第三次至少要跳()下。",
+        "options":  [
+                        "A. 96",
+                        "B. 90",
+                        "C. 80",
+                        "D. 75"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】80乘3等于三次跳绳下数的和，减去第一，二次跳的下数即等于第三次至少要跳的下数.\n【详解】80×3=240(下)240-(69+75)=240-144=96(下) 明明练习跳绳，第一次跳了69下，第二次跳了75下。她要想三次的平均成绩达到80下，第三次至少要跳96下。故答案为:A.",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·浙江嘉兴·期末"
+    },
+    {
+        "id":  "50",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·贵州遵义·期末)在学校运动会开幕式中，四(1)班全体同学排成5列纵队入场，个子矮的同学站前面，个子高的同学站在后面。下面更接近全班同学的平均身高的是()。",
+        "options":  [
+                        "A. 第一行同学的平均身高",
+                        "B. 第一列同学的平均身高",
+                        "C. 随机抽取5名同学的平均身高",
+                        "D. 以上均可"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据平均数是反映一组数据的集中趋势的一个量，不能反映具体数据，据此解答即可.\n【详解】A。第一行同学都在最前面，身高普遍较矮，平均身高会低于整体平均身高，不符合题意.\nB。第一列同学包含不同身高层次的同学，平均身高更接近全班同学的平均身高，符合题意.\nC。随机抽取5名同学的平均身高，数据不确定，可能身高普遍较矮，也可能身高普遍较高，平均身高不确定，不符合题意.\nD。选取的同学需要包含不同身高层次的同学，第一行同学的平均身高和随机抽取5名同学的平均身高都不合适，不符合题意。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·贵州遵义·期末"
+    },
+    {
+        "id":  "51",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "(24-25四年级下·吉林·期末)四年三班同学的平均身高是135厘米，新学期又来了一名同学，他的身高是138厘米。这时四年三班的平均身高与原来相比()。",
+        "options":  [
+                        "A. 降低了",
+                        "B. 提高了",
+                        "C. 不变"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据新加入的数据大于原平均数时，整体平均数会提高;反之则会降低的原则来进行解答.\n【详解】因为135\u003c138，所以四年三班的平均身高与原来相比提高了。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·吉林·期末"
+    },
+    {
+        "id":  "52",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·黑龙江绥化·期末)淘淘前3次共拍球87下，后2次共拍球78下，他这5次拍球平均数是()下。",
+        "options":  [
+                        "A. 29",
+                        "B. 33",
+                        "C. 35"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】淘淘前3次共拍球87下，后2次共拍球78下，据此用加法先算出淘淘5次排球的总个数，再除以排球的次数，即可求出他这5次拍球平均数是多少下.\n【详解】(87+78)÷5=165÷5=33(下)即淘淘前3次共拍球87下，后2次共拍球78下，他这5次拍球平均数是33下。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·黑龙江绥化·期末"
+    },
+    {
+        "id":  "53",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(2024五年级·全国·竞赛)国庆节前夕，欧欧和乐乐准备做一些小国旗送给同学们，第一天欧欧5小时，乐乐3小时共做190面，第二天欧欧3小时，乐乐5小时共做210面。那么欧欧和乐乐平均每小时可以合做()面小国旗。",
+        "options":  [
+                        "A. 40",
+                        "B. 50",
+                        "C. 60",
+                        "D. 70"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】两天欧欧和乐乐都共做了5+3=8(小时),做的小国旗的总数为210+190=400(面),可见他们平均每小时可以合做400÷8=50(面).\n【详解】解：两天欧欧和乐乐都做的时间:5+3=8(小时) 两人两天做的小国旗的总数:210+190=400(面)\n他们平均每小时可以合做:400÷8=50(面) 故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法，两人合作的工程问题"
+                             ],
+        "source":  "2024五年级·全国·竞赛"
+    },
+    {
+        "id":  "54",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·湖北省直辖县级单位·期末)雨欣，青青，俊俊三人参加演讲比赛，分别获得前三名。雨欣以99分获得第一名，青青以90分获得第三名。这三位选手的平均成绩应该在()。",
+        "options":  [
+                        "A. 90分以下",
+                        "B. 90分至93分之间",
+                        "C. 93至96分之间",
+                        "D. 96至99分之间"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】根据题意，俊俊的分数最少为91，最大为98，用总分数除以3即可求出平均数，且根据平均数的意义，几个数的平均数一定比其中的最大数小，比最小数大，据此解答.\n【详解】根据分析,90\u003c平均成绩\u003c99A。90分以下不符合90\u003c平均成绩\u003c99，选项错误;B。假设俊俊成绩为98分,(99+98+90)÷3=287÷3≈95.67(分),不符合90分至93分之间，选项错误;C。假设俊俊成绩为98分,(99+98+90)÷3=287÷3≈95.67(分),符合93分至96分之间，选项正确;D。假设俊俊成绩为91分,(99+91+90)÷3=280÷3≈93.33(分),不符合96分至99分之间，选项错误。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·湖北省直辖县级单位·期末"
+    },
+    {
+        "id":  "55",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·吉林白城·期末)一组数据:85,90,88,93，它们的平均数是()。",
+        "options":  [
+                        "A. 88",
+                        "B. 89",
+                        "C. 90"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】先求出以上所有数据的总和，再除以数据的个数，即可求得平均数.\n【详解】(85+90+88+93)÷4=356÷4=89它们的平均数是89。故答案为:B",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算，平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·吉林白城·期末"
+    },
+    {
+        "id":  "56",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "(24-25四年级下·山东济宁·期末)2,8,14,16,20的平均数为()。",
+        "options":  [
+                        "A. 8",
+                        "B. 12",
+                        "C. 9",
+                        "D. 18"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】平均数是一组数的总和除以这组数的数量，将这五个数加在一起再除以5即可.\n【详解】(2+8+14+16+20)÷5=60÷5=12所以2,8,14,16,20的平均数为12。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "57",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·广东云浮·期末)小红的语文，数学，英语三门功课的平均成绩是92分，其中语文90分，数学95分，则小红的英语成绩应该是()分。",
+        "options":  [
+                        "A. 88",
+                        "B. 91",
+                        "C. 93"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据平均成绩×科目的数量=总成绩，用92×3算出语文，数学，英语三门功课的总成绩，再用三门功课的总成绩分别减去语文和数学两门功课的成绩即可求出英语的成绩.\n【详解】三门功课总分为:92×3=276(分)语文和数学的总分:90+95=185(分)英语成绩:276-185=91(分)所以小红的英语成绩应该是91分。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·广东云浮·期末"
+    },
+    {
+        "id":  "58",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·广东肇庆·期末)一组数据最大值是30，最小值是18，这组数据的平均数()。",
+        "options":  [
+                        "A. 一定是24",
+                        "B. 在30到18之间",
+                        "C. 比30大",
+                        "D. 比18小"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】平均数代表的是一组数据的一般水平。它一般比这组数据的最小值大，一般比这组数据的最大值小。据此解答.\n【详解】由题意得，一组数据最大值是30，最小值是18，那么这组数据的平均数应该在30到18之间。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·广东肇庆·期末"
+    },
+    {
+        "id":  "59",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "(23-24四年级下·河南安阳·期末)一组数据中最大数是46，最小数是39，这组数据的平均数可能是()。",
+        "options":  [
+                        "A. 39",
+                        "B. 42",
+                        "C. 46",
+                        "D. 47"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】平均数是反映一组数据的平均情况，在一组数据中，有的数据可能大于平均数，有的数据可能小于平均数，有的可能等于平均数，平均数大于一组数据中的最小值，且小于这组数据的最大值，所以在这组数据中，最大的数是46，最小的数是39，平均数一定是大于39且小于46的数，据此解答.\n【详解】一组数据中最大数是46，最小数是39，这组数据的平均数可能是大于39且小于46的数，结合选项，可能是42。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·河南安阳·期末"
+    },
+    {
+        "id":  "60",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "(24-25四年级下·广东阳江·期末)小明所在班级同学的平均身高是140厘米，小刚所在班级同学的平均身高是143厘米，小明()比小刚矮。",
+        "options":  [
+                        "A. 一定",
+                        "B. 不一定",
+                        "C. 一定不"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】平均数代表的是班级同学身高的总体平均水平，不能据此确定个体的具体身高情况。小明所在班级同学的平均身高是140厘米，这意味着小明班级里同学的身高有高有低，小明的身高可能高于140厘米，比如145厘米;也可能等于140厘米;还可能低于140厘米，比如135厘米。小刚所在班级同学的平均身高是143厘米，同理，小刚班级里同学身高有差异，小刚的身高可能高于143厘米，比如148厘米;可能等于143厘米;也可能低于143厘米，比如140厘米.\n【详解】从上面的分析可以看出，小明和小刚的身高都有多种可能性，所以仅根据班级平均身高不能确定小明一定比小刚矮，也不能确定小明一定比小刚高。所以，小明不一定比小刚矮。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·广东阳江·期末"
+    },
+    {
+        "id":  "61",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·广东江门·期末)四(1)班某次数学测试的最高分是98分，最低分是80分，那么四(1)班此次测试的平均分可能是()分。",
+        "options":  [
+                        "A. 98",
+                        "B. 88",
+                        "C. 80",
+                        "D. 75"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据题意，平均数的计算公式为：平均数=总分/人数。由于所有分数介于80和98之间，因此平均数必然大于80且小于98，以此逐项判断选择即可.\n【详解】A。98=98，不符合题意。B。80\u003c88\u003c98，符合题意。C。80=80，不符合题意。D。75\u003c80，不符合题意。那么四(1)班此次测试的平均分可能是88分。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·广东江门·期末"
+    },
+    {
+        "id":  "62",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·内蒙古包头·期末)一分钟跳绳比赛中，小美组平均每人跳了150下，小乐组平均每人跳了164下。小美和小乐跳绳个数相比()。",
+        "options":  [
+                        "A. 小美多",
+                        "B. 小乐多",
+                        "C. 无法确定"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】平均数是指在一组数据中所有数据之和再除以这组数据的个数，平均数是反映一组数据的平均水平，并不能反映这组数据中各个数据的大小，据此即可解答.\n【详解】根据分析可知，由每组平均跳的下数不确定小美和小乐的跳绳个数，所以无法比较两人跳绳个数的多少。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·内蒙古包头·期末"
+    },
+    {
+        "id":  "63",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·云南昆明·期末)四一班学生平均身高为136厘米，下面说法正确的是()。",
+        "options":  [
+                        "A. 四一班一定有学生身高为136厘米",
+                        "B. 四一班如果转走了两名同学，平均身高一定还是136厘米",
+                        "C. 四一班可能有比136厘米高的同学，也可能有比136厘米矮的同学"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】平均数是表示一组数据平均水平的量，是指在一组数据中所有数据之和再除以这组数据的个数;它不能表示这组数据中各个数据的具体大小，由此即可进行判断.\n【详解】A。四一班可能有学生身高为136厘米，即原说法错误.\nB。四一班如果转走了两名同学，平均身高不一定还是136厘米，即原说法错误.\nC。四一班可能有比136厘米高的同学，也可能有比136厘米矮的同学，即原说法正确.\n故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·云南昆明·期末"
+    },
+    {
+        "id":  "64",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·黑龙江佳木斯·期末)学校篮球队8名队员的平均身高是170厘米，赵强身高168厘米，李明身高174厘米，他们加入球队后，现在篮球队队员的平均身高与原来比,()。",
+        "options":  [
+                        "A. 比原来高",
+                        "B. 比原来低",
+                        "C. 与原来一样",
+                        "D. 无法确定"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】根据题意，平均数是指在一组数据中所有数据之和再除以这组数据的个数，平均数介于最大值和最小值之间，用168加上174，再除以2，求出两人的平均数，再与170进行比较，如果大于170，就比原来高，如果小于170，就比原来低，如果等于170，就与原来一样，据此即可解答.\n【详解】根据分析可知:(168+174)÷2=342÷2=171(厘米) 171\u003e170 学校篮球队8名队员的平均身高是170厘米，赵强身高168厘米，李明身高174厘米，他们加入球队后，现在篮球队队员的平均身高与原来比，比原来高。故答案为:A",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·黑龙江佳木斯·期末"
+    },
+    {
+        "id":  "65",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·辽宁鞍山·期末)下面说法不正确的是()。",
+        "options":  [
+                        "A. 两个完全一样的等腰直角三角形，可以拼成一个正方形。",
+                        "B. 乐乐身高1.48米，他去平均水深为1.4米的水池游泳不会有危险。",
+                        "C. 三角形三边关系：任意两边之和大于第三边;任意两边之差小于第三边。"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】等腰直角三角形的两腰相等，且顶角是90°的角;正方形的四条边都相等，四个角都是直角;将两个完全一样的等腰直角三角形底边重合即可拼成一个正方形;平均数是反映一组数据的集中趋势，它比最大数小，比最小的数大;三角形中，三条边关系为：任意两边之和大于第三边，任意两边之差小于第三边;据此对每个选项进行分析即可解答.\n【详解】A。如图:![](images/2026-03-14%20161027.jpg)两个完全一样的等腰直角三角形可以拼成一个正方形，所以题干中说法正确，不符合题意;B。平均水深为1.4米，不能说明水池所有地方都是1.4米，可能有些地方水很深，超过乐乐的身高，所以题干中说法错误，符合题意;C。三角形三边关系：任意两边之和大于第三边，任意两边之差小于第三边;所以题干中说法正确，不符合题意。故答案为:B",
+        "knowledge_points":  [
+                                 "用同样的图形拼接，三角形三边关系，平均数的意义及求法，等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·辽宁鞍山·期末"
+    },
+    {
+        "id":  "66",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "(24-25四年级下·江西上饶·期末)春节期间，金龙岗灯会表演连续四天分别接待游客120人,150人,130人,100人，在这四天里，平均每天接待游客()人。",
+        "options":  [
+                        "A. 120",
+                        "B. 125",
+                        "C. 130"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据题意，把4天接待的人数相加，算出四天的总人数，再除以4，就是这四天平均每天接待的游客人数.\n【详解】(120+150+130+100)÷4=(270+130+100)÷4=(400+100)÷4=500÷4=125(人)所以，在这四天里，平均每天接待游客125人。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·江西上饶·期末"
+    },
+    {
+        "id":  "67",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·黑龙江佳木斯·期末)小明比小红多18颗糖，小明给小红()颗糖后两人一样多。",
+        "options":  [
+                        "A. 18",
+                        "B. 9",
+                        "C. 36",
+                        "D. 27"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据题意，小明比小红多18颗糖，那么去掉小明这多出来的18颗糖，也就是小明和小红一样多，再把这18颗糖平均分给小明和小红，两人就一样多了，据此解答.\n【详解】把多出来的18颗平均分成2份，也就是18÷2=9(颗),即小明给小红9颗糖后，两人就一样多了。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·黑龙江佳木斯·期末"
+    },
+    {
+        "id":  "68",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·甘肃庆阳·期末)四年级5个班去植树，第一天植了145棵，第二天植了178棵，第三天植了157棵，平均每班植树多少棵。正确列式是()。",
+        "options":  [
+                        "A. (145+178+157)÷5",
+                        "B. (145+178+157)÷4",
+                        "C. (145+178+157)÷3",
+                        "D. (145+178+157)"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】求平均每班植树棵数，首先求出四年级5个班三天共植树的棵数，即145+178+157，然后除以班级个数5即可.\n【详解】(145+178+157)÷5=480÷5=96(棵)平均每班植树96棵，正确列式是(145+178+157)÷5\n故答案为:A",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·甘肃庆阳·期末"
+    },
+    {
+        "id":  "69",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·浙江杭州·期末)乐乐所在的校排球队队员平均身高160厘米，那么乐乐身高()。",
+        "options":  [
+                        "A. 超过160厘米",
+                        "B. 不到160厘米",
+                        "C. 无法判断"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】平均数是指在一组数据中所有数据之和再除以数据的个数;平均数是表示一组数据集中趋势的数，它是反映数据集中趋势的一项指标，不能反映每个数据情况，据此解答即可.\n【详解】根据分析可知，乐乐所在的校排球队队员平均身高160厘米，表示校排球队队员的整体情况，并非每个队员的身高都是160厘米。有的队员的身高比160厘米高，也有的队员的身高比160厘米低，所以，乐乐身高无法判断。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·浙江杭州·期末"
+    },
+    {
+        "id":  "70",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·山东日照·期末)甲，乙两数的平均数是90，丙数是60，这三个数的平均数是()。",
+        "options":  [
+                        "A. 90",
+                        "B. 80",
+                        "C. 60",
+                        "D. 50"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】已知甲，乙两个数的平均数是90，根据平均数×份数=总数，用90乘2求得甲，乙两数的总数;最后再根据总数/份数=平均数，用三个数的和除以3，即得到它们的平均数。据此解答.\n【详解】90×2=180\n180+60=240\n240÷3=80\n所以，这三个数的平均数是80。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·山东日照·期末"
+    },
+    {
+        "id":  "71",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(23-24四年级下·云南昭通·期末)在\"六一\"游园投壶活动中，王冰前后5局投中的成绩分别是12个,16个,16个,13个,18个。用()个来表示王冰投壶的水平比较合适。",
+        "options":  [
+                        "A. 15",
+                        "B. 16",
+                        "C. 18"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】一组数据的总和除以这组数据个数所得到的商叫这组数据的平均数，平均数可以表示一组数据的平均水平，据此求出平均数即可.\n【详解】(12+16+16+13+18)÷5=(28+16+13+18)÷5=(44+13+18)÷5=(57+18)÷5=75÷5=15(个)用15个来表示比较合适。故答案为:A",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·云南昭通·期末"
+    },
+    {
+        "id":  "72",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·云南昭通·期末)四(1)班同学体检，体重最重的是51千克，最轻的是34千克，这个班同学的平均体重可能是()千克。",
+        "options":  [
+                        "A. 34",
+                        "B. 45",
+                        "C. 51",
+                        "D. 52"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】一组数据的总和除以这组数据个数所得到的商叫这组数据的平均数，当这组数据不全相等时，平均数大于最小数，小于最大数，据此判断.\n【详解】A。34等于最轻体重，所以不可能是平均数;B。45大于最轻体重，小于最重体重，所以可能是平均数;C。51是最重体重，所以不可能是平均数;D。52大于最重体重，所以不可能是平均数。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·云南昭通·期末"
+    },
+    {
+        "id":  "73",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "(24-25四年级下·黑龙江佳木斯·期末)小方3分钟跳绳453下，小明2分钟跳了286下,()的速度快。",
+        "options":  [
+                        "A. 小方",
+                        "B. 小明",
+                        "C. 无法确定"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】用跳绳的总次数除以时间分别求出小方和小明每分钟跳绳的个数，也就是他们各自的跳绳速度，然后比较两人速度的大小，速度大的人跳绳就更快.\n【详解】453÷3=151(下)286÷2=143(下) 151\u003e143 所以，小方的速度快。故答案为:A",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·黑龙江佳木斯·期末"
+    },
+    {
+        "id":  "74",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·甘肃平凉·期末)东仁堡小学举行了一次主题为\"红星照耀中国\"手抄报征集展示活动。二年级有三个班，一班征集23幅符合标准的手抄报，二班征集26幅符合标准的手抄报，学校二年级展区需要三个班平均手抄报数量达到30幅，那么三班至少需要征集()幅符合标准的手抄报。",
+        "options":  [
+                        "A. 31",
+                        "B. 41",
+                        "C. 40",
+                        "D. 37"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据题意可知，四年级展区需要三个班平均手抄报幅数要达到30幅，则一共需要手抄报30×3=90幅，再用共需要手抄报的幅数依次减去一班和二班手抄报数量，即可求出三班手抄报数量.\n【详解】30×3-23-26=90-23-26=67-26=41(幅)那么三班至少需要征集41幅符合标准的手抄报。故答案为:B",
+        "knowledge_points":  [
+                                 "乘加，乘减混合运算，平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·甘肃平凉·期末"
+    },
+    {
+        "id":  "75",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·河南郑州·期末)下面说法正确的是()。",
+        "options":  [
+                        "A. 3个百分之一写成小数是0.300",
+                        "B. 小名所在的班级同学平均身高是150厘米，小红所在的班级同学的平均身高为148厘米，所以小名的身高一定比小红高。",
+                        "C. 0.3和0.30计数单位相同，大小也相等。",
+                        "D. 照相机的支架常常设计成三脚架，是利用了三角形具有稳定性这一特性。"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】小数的计数单位是十分之一，百分之一，千分之一。分别写作0.1,0.01,0.001。每相邻两个计数单位之间的进率是10，则3个百分之一写成小数是0.03.\n平均数表示一组数据的趋势，实际数可以大于平均数，可以等于平均数，也可以小于平均数.\n根据小数的性质，小数的末尾填上\"0\"或去掉\"0\"，小数的大小不变,0.3=0.30.\n三角形具有稳定性，生活中常利用这个原理来建构物体.\n【详解】A。3个百分之一表示3在百分位上，是0.03，所以3个百分之一写成小数是0.03，原题说法错误.\nB。平均数是反映一组数据的平均水平，并不能反映这组数据中各个数据的大小。小名所在班级同学的平均身高是150厘米，小名的身高可能比150厘米高，也可能低于150厘米，也可能正好是150厘米，小红所在班级同学的平均身高是148厘米，小红的身高可能比148厘米高，也可能低于148厘米，也可能正好148厘米，所以小名和小红的身高没法比较;原题说法错误.\nC。根据小数的基本性质可知,0.3=0.30，根据小数的意义可知:0.3的计数单位是0.1,0.30的计数单位是0.01，即0.3和0.30的大小相等，但是计数单位不同，所以原题的说法错误.\nD。不容易变形，是三角形的特性，照相机的支架常常做成三角形，是利用了三角形的稳定性，此说法正确.\n故答案为:D",
+        "knowledge_points":  [
+                                 "三角形的稳定性及应用，小数的数位和计数单位的认识，平均数的意义及求法，小数的性质"
+                             ],
+        "source":  "24-25四年级下·河南郑州·期末"
+    },
+    {
+        "id":  "76",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "(24-25四年级下·甘肃兰州·期末)乐乐期末考试的语文，数学，英语三科平均分是93分，语文92分，数学94分，英语()分。",
+        "options":  [
+                        "A. 95",
+                        "B. 93",
+                        "C. 94"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】将三科平均分乘3算出总分，将总分减去语文分数和数学分数得到英语分数，据此解答.\n【详解】总分:93×3=279(分)英语:279-92-94=187-94=93(分)所以英语93分。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·甘肃兰州·期末"
+    },
+    {
+        "id":  "77",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·湖南永州·期末)有一条山路，一辆汽车上山时每小时行30千米，下山原路返回每小时行50千米，则汽车上山往返一次的平均速度是()。",
+        "options":  [
+                        "A. 40千米/时",
+                        "B. 37.5千米/时",
+                        "C. 35千米/时"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】可用赋值法，假设这条山路上山路程为150千米(因为150可以被30和50整除，方便运算),然后根据路程/速度=时间，分别算出上山和下山的时间，再用路程的两倍除以两段时间和即可算出平均速度.\n【详解】假设这条山路上山路程为150千米。150÷30=5(小时)150÷50=3(小时)5+3=8(小时)150+150=300(千米)300÷8=37.5(千米/小时)故答案为:B",
+        "knowledge_points":  [
+                                 "基础行程问题，除数是整数，需要补0的小数除法，平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·湖南永州·期末"
+    },
+    {
+        "id":  "78",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·河南三门峡·期末)下面说法正确的是()。",
+        "options":  [
+                        "A. 几个数中再加入一个数，平均数一定会变化",
+                        "B. 在一个锐角三角形中，任意两个锐角的和都大于90°",
+                        "C. 小数点后面添上\"0\"或者去掉\"0\"，小数的大小不变",
+                        "D. 5.89精确到十分位是6.0"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】一组数据的总和除以这组数据个数所得到的商叫这组数据的平均数，如果加入的数大于或小于平均数，则平均数会变化，如果正好等于平均数，则还是原来的平均数;三角形内角和是180°,锐角三角形三个角都是锐角;小数的性质：小数末尾的0增加或去掉，小数的大小不变;精确到十分位则看百分位上的数，如果小于5则舍去，大于或等于5则向十分位进1，据此分析每个选项选择即可.\n【详解】A。如果加入的数正好等于平均数，则平均数不变，选项说法错误;B。任意两个锐角的和都大于90°,这样第三个角的度数小于90°也是锐角，才符合锐角三角形，选项说法正确;C。3.02去掉小数点后面的0是3.2，大小变了，选项说法错误;D。5.89≈5.9，精确到十分位是5.9，选项说法错误。故答案为:B",
+        "knowledge_points":  [
+                                 "小数的性质，三角形的内角和，平均数的意义及求法，小数的近似数"
+                             ],
+        "source":  "24-25四年级下·河南三门峡·期末"
+    },
+    {
+        "id":  "79",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·广东广州·期末)四(1)班学生的平均体重是33千克，四(2)班学生的平均体重是35千克。关于四(1)班的小明与四(2)班的小亮的体重，说法不正确的是()。",
+        "options":  [
+                        "A. 小明的体重一定比小亮的轻",
+                        "B. 小明的体重可能比小亮的轻",
+                        "C. 小明的体重可能比小亮的重",
+                        "D. 小明的体重可能跟小亮的相同"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】平均数：一组数据的总和除以这组数据个数所得到的商叫这组数据的平均数，平均数比一组数中最大的数小，比最小的数大;逐项分析后进行选择，据此解答.\n【详解】根据分析:A。如果小明的体重是他们班最重的，小亮的体重是他们班最轻的，那么小明的体重可能会比小亮的重，故此说法不正确;B。如果小明的体重是他们班较轻的，小亮的体重是他们班较重的，那么小明的体重可能比小亮的轻，所以此说法正确;C。如果小明的体重是他们班最重的，小亮的体重是他们班最轻的，那么小明的体重可能会比小亮的重，故此说法正确;D。如果小明的体重是他们班较重的，小亮的体重是他们班较轻的，小明的体重可能跟小亮的相同，故此说法正确。故答案为:A",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·广东广州·期末"
+    },
+    {
+        "id":  "80",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "(25-26五年级上·广东潮州·期末)一辆货车载满货物从甲地开出,3小时行驶了120千米到达乙地，卸完货物后原路返回,2小时后到达甲地。这辆货车往返甲乙两地平均每小时行驶( )千米。",
+        "options":  [
+
+                    ],
+        "answer":  "48",
+        "solution":  "【分析】先求出往返的总路程，即甲地到乙地路程的2倍，甲地到乙地的行驶时间是3小时，乙地到甲地的行驶时间是2小时，相加求出往返的总时间，最后根据\"平均速度=总路程/总时间\"求出这辆货车往返甲乙两地平均每小时行驶的路程，据此解答.\n【详解】120×2÷(3+2)=240÷5=48(千米)所以，这辆货车往返甲乙两地平均每小时行驶48千米。",
+        "knowledge_points":  [
+                                 "基础行程问题，带有小括号的混合运算，平均数的意义及求法"
+                             ],
+        "source":  "25-26五年级上·广东潮州·期末"
+    },
+    {
+        "id":  "81",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·全国·课后作业)小军，小林和小华比赛拍皮球，小军2分钟拍了166下，小林4分钟拍了384下，小华3分钟拍了258下。( )拍球的速度最快。",
+        "options":  [
+
+                    ],
+        "answer":  "小林",
+        "solution":  "【分析】用拍球的总次数除以时间，分别算出小军，小林和小华平均每分钟拍球的次数，再比较大小，据此找出谁拍球的速度最快.\n【详解】小军:166÷2=83(下)小林:384÷4=96(下)小华:258÷3=86(下)因为83\u003c86\u003c96\n所以小林拍球的速度最快。",
+        "knowledge_points":  [
+                                 "三位数除以一位数，首位不够除,100以内数的比大小，平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "82",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "(24-25四年级下·全国·课后作业)乐乐的书房有红，黄，蓝，绿4种小书架，平均每种书架上有35本书，其中红书架上有32本，黄书架上有38本，蓝书架上有34本，则绿书架上有( )本。",
+        "options":  [
+
+                    ],
+        "answer":  "36",
+        "solution":  "【分析】根据平均每种书架上有35本书，先求出4种书架上书的总本数，再用总本数依次减去红，黄，蓝书架上的本数，即可得到绿书架上的本数.\n【详解】35×4=140(本)140-32-38-34=108-38-34=70-34=36(本)答：绿书架上有36本。",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "83",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "(24-25四年级下·全国·课后作业)海海记录了自己6分钟的脉搏每分钟跳动次数，分别是76,68,72,75,72,75。这6分钟平均每分钟跳动( )次。",
+        "options":  [
+
+                    ],
+        "answer":  "73",
+        "solution":  "【分析】根据平均数=总数量/总份数，先求出6分钟脉搏跳动的总次数，再除以6即可求出平均每分钟跳动的次数.\n【详解】76+68+72+75+72+75=(76+68)+(72+72)+(75+75)=144+144+150=288+150=438(次)438÷6=73(次)\n答：这6分钟平均每分钟跳动73次。",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "84",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "(24-25四年级下·全国·课后作业)下面是甲，乙两个停车场车辆停放情况统计图，请看图填空.\n![](images/d45bf9b34623615c139d696fc3c3d1ae6caf8da9db61047a699d968f739bb34d.jpg)\n(1)上图是( )统计图。(2)两个停车场都停放了( )种车辆。(3)两个停车场都是停放( )车最多,( )车最少。(4)两个停车场一共停放了( )辆车。",
+        "options":  [
+
+                    ],
+        "answer":  "复式条形统计图;4;轿车;货车;108",
+        "solution":  "【分析】(1)观察统计图的标题及形式，可直接判断统计图类型.\n(2)通过统计图纵轴或图例可知车辆种类数量.\n(3)比较各车辆在两个停车场的停放数量，得出最多和最少的车型.\n(4)将两个停车场各类车辆数量分别相加，再求和得到总车辆数.\n【详解】(1)由图可知，上图是复式条形统计图。(2)货车，大客车，面包车，轿车，两个停车场都停放了4种车辆。(3)甲停车场:4\u003c6\u003c15\u003c33;乙停车场:3\u003c7\u003c12\u003c28;两个停车场都是停放轿车最多，货车最少。(4)(3+4)+(7+6)+(12+15)+(28+33)=7+13+27+61=20+27+61=47+61=108(辆)答：两个停车场一共停放了108辆车。",
+        "knowledge_points":  [
+                                 "统计图表的综合应用，复式条形统计图"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "85",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "(25-26六年级·全国·随堂练习)在某次期末测试中，第一小组有4人，平均成绩是92.5分;第二小组有5人，平均成绩是93.4分。这两个小组的平均成绩是( )分。",
+        "options":  [
+
+                    ],
+        "answer":  "93",
+        "solution":  "【分析】第一小组有4人，平均成绩是92.5分，根据\"总成绩=平均成绩×人数\"，第一小组总成绩为370分;第二小组有5人，平均成绩是93.4分，同理可得第二小组总成绩为467分;两个小组的总成绩之和为837分;两个小组的总人数之和为9人;根据\"平均成绩=总成绩之和/总人数之和\"，两个小组的平均成绩为93分.\n【详解】92.5×4=370(分)93.4×5=467(分) 370+467=837(分) 4+5=9(人) 837÷9=93(分)所以这两个小组的平均成绩是93分。",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "25-26六年级·全国·随堂练习"
+    },
+    {
+        "id":  "86",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "(2025·湖南长沙·小升初真题)有6个数排成一列，这6个数的平均数是11，前4个数的平均数是9，后3个数的平均数是14，则第四个数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "12",
+        "solution":  "【分析】用前4个数的平均数乘4，求出前4个数的和。用后3个数的平均数乘3，求出后3个数的和。用6个数的平均数乘6，求出6个数的和。用前4个数的和加上后3个数的和，再减去6个数的和，即可求出第4个数.\n【详解】(9×4+14×3)-11×6\n=(36+42)-66\n=78-66\n=12\n所以第四个数是12。",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "2025·湖南长沙·小升初真题"
+    },
+    {
+        "id":  "87",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "(24-25五年级上·湖南郴州·期中)小王4分钟做了100道口算题，平均每分钟做( )题。",
+        "options":  [
+
+                    ],
+        "answer":  "25",
+        "solution":  "【分析】用口算题的道数100除以做的分钟数4分钟，即可求出平均每分钟做几道题.\n【详解】100÷4=25(道),即平均每分钟做25道题。",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25五年级上·湖南郴州·期中"
+    },
+    {
+        "id":  "88",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级上·湖南永州·期中)三个连续自然数的平均数是3450，其中最大的一个数是( ),最小的一个数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "3451;3449",
+        "solution":  "【分析】表示物体个数的1,2,3,4,5,6,7,8,9,10,11。都是自然数。一个物体也没有，用0表示,0也是自然数。由题意得，三个连续自然数的平均数是3450，因为连续的自然数相差1，那么最中间的自然数是3450，较小的自然数是(3450-1),较大的自然数是(3450+1)。据此解答.\n【详解】由分析得，最中间的自然数是3450.3450+1=3451,3450-1=3449三个连续自然数的平均数是3450，其中最大一个数是3451，最小一个数是3449。",
+        "knowledge_points":  [
+                                 "自然数的认识，平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级上·湖南永州·期中"
+    },
+    {
+        "id":  "89",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "(2007五年级·全国·竞赛)9个小朋友在2张乒乓球台上进行乒乓球单打比赛练习，他们从上午8:00练习到上午11:00，平均每人练习了( )分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "80",
+        "solution":  "【分析】先计算练习的总时间，再根据球台数量和单打规则，得出2张球台同时练习时的总人次练习时间，最后用总人次练习时间除以人数得到平均每人练习的时长。这类问题要明确活动的时间计算方法以及活动的参与人数规则，通过总人次练习时间与总人数的关系来计算平均每人的练习时长.\n【详解】先计算练习总时长：从上午8:00到上午11:00，时长为11-8=3(小时),因为1小时=60分钟，所以3小时换算为分钟是3×60=180(分钟);然后，计算2张球台单打时的总人次练习时间：乒乓球单打是2人一组,2张球台同时进行，那么同时练习的人数是2×2=4(人)。所以2张球台180分钟的总人次练习时间4×180=720(分钟);最后，计算平均每人练习的时长：总共有9个小朋友，那么平均每人练习的时长为720÷9=80(分钟)。",
+        "knowledge_points":  [
+                                 "时，分，秒有关的计算，平均数的意义及求法"
+                             ],
+        "source":  "2007五年级·全国·竞赛"
+    },
+    {
+        "id":  "90",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·山东济宁·期末)小芳参加小歌手比赛,5位评委的评分分别是97分,97分,95分,96分,93分，比赛计分规则是\"先去掉1个最高分和1个最低分，再算平均分\"，那么，小芳最后的得分是( )分。",
+        "options":  [
+
+                    ],
+        "answer":  "96",
+        "solution":  "【分析】根据题意，五位评委的评分分别是97分,97分,95分,96分,93分。最高分是97分(有两个97分，但只去掉一个)。最低分是93分(只有一个)。去掉一个最高分(97分)和一个最低分(93分)后，剩下的分数是:97分,95分,96分。先求剩余三个分数的和，再除以3，列式计算即可.\n【详解】根据分析可知:97\u003e96\u003e95\u003e93 (97+96+95)÷3=288÷3=96(分)那么，小芳最后的得分是96分。",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "91",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "(24-25四年级下·全国·课后作业)海海4次跳高的平均成绩是1.4m,他每次跳高的成绩肯定都是1.4m。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数是总数量除以总份数。平均成绩为1.4m,表示4次跳高的总高度是1.4×4=5.6m。但这并不要求每次跳高成绩都等于1.4m;实际成绩可能高于或低于1.4m,只要总和为5.6m即可。因此，\"每次跳高的成绩肯定都是1.4m\"的说法不一定成立.\n【详解】因为平均成绩是1.4m,只说明4次跳高的总高度是5.6m,但每次的成绩可能不同，例如可能是1.3m,1.5m等组合，不一定都是1.4m。所以该说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "92",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·陕西安康·期末)西安市上周平均最低气温23℃,所以西安市上周最低气温不可能低于23℃。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均气温是各天最低气温的平均值，可能存在某天的气温低于或高于平均值。因此，平均最低气温为23℃时，不能排除某天的最低气温低于23℃的可能性.\n【详解】平均数是所有数据的总和除以数据的个数。若上周某几天的最低气温低于23℃,而其他几天的最低气温高于23℃,则平均最低气温仍可能为23。例如：假设三天最低气温为21℃,四天为24.5℃,总和为21×3+24.5×4=63+98=161℃,平均为161÷7=23(℃)。此时存在低于23℃的情况。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·陕西安康·期末"
+    },
+    {
+        "id":  "93",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·河南驻马店·期末)1.5米高的小华，过一条平均水深1.2米的河，可能有危险。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】根据题意，平均数是各数据的总和除以数量，不能代表所有具体数值。平均水深1.2米表示整条河水深的平均值，但实际某些区域的水深可能高于或低于1.2米。若存在水深超过1.5米的区域，小华过河就会有危险。以此判断即可.\n【详解】根据分析可知：若河中有部分区域水深1.6米，则小华经过时会被淹没。因此，虽然平均水深1.2米，但小华仍\"可能有危险\"。原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·河南驻马店·期末"
+    },
+    {
+        "id":  "94",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·湖北省直辖县级单位·期中)计算下面各题，怎样简便就怎样计算.\n382×101-382\n58×59+58×42-58\n5400÷12/5\n2+4+6+⋯+50",
+        "options":  [
+
+                    ],
+        "answer":  "38200;5800;90;650",
+        "solution":  "【分析】根据乘法分配律(a+b)×c=a×c+b×c,把原式变为382×(101-1),再按照运算顺序计算即可.\n根据乘法分配律a×c+b×c=(a+b)×c,把原式变为58×(59+42-1),再按照运算顺序计算即可.\n根据除法的运算性质,a/b/c=a/(b×c),把原式变为5400÷(12×5),再按照运算顺序计算即可.\n从1加到50是50个数,2+4+6+⋯+50是其中的双数相加，就是有25个数相加，相邻的两个加数之间相差2，数据个数是25,25是单数，那么最中间的加数26为25个加数的平均数，因此选取最中间的那个数26乘个数25即可简算得解.\n【详解】382×101-382\n=382×(101-1)\n=382×100\n=38200\n58×59+58×42-58\n=58×(59+42-1)\n=58×100\n=5800\n5400÷12/5\n=5400÷(12×5)\n=5400÷60\n=90\n2+4+6+⋯+50\n=26×25\n=650",
+        "knowledge_points":  [
+                                 "整数除法的性质，整数乘法分配律，平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·湖北省直辖县级单位·期中"
+    },
+    {
+        "id":  "95",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "(22-23五年级上·黑龙江齐齐哈尔·期末)列式计算。88,90,89,94,30,92,91，它们的平均数是多少。",
+        "options":  [
+
+                    ],
+        "answer":  "82",
+        "solution":  "【分析】根据平均数=7个数的和/数的总个数，用(88+90+89+94+30+92+91)÷7即可求出平均数.\n【详解】(88+90+89+94+30+92+91)÷7\n=574÷7\n=82\n它们的平均数是82。",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "22-23五年级上·黑龙江齐齐哈尔·期末"
+    },
+    {
+        "id":  "96",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "(2022五年级下·全国·专题练习)五个连续自然数的和是135，这五个连续自然数分别是多少。",
+        "options":  [
+
+                    ],
+        "answer":  "25、26、27、28、29",
+        "solution":  "【分析】中间的一个自然数是这五个连续自然数的平均数，先求出中间的自然数，再根据相邻两个自然数相差1求出其它的自然数;据此解答.\n【详解】中间的自然数:135÷5=27\n27-1=26,26-1=25,27+1=28,28+1=29\n所以，这五个连续自然数分别是25、26、27、28、29。",
+        "knowledge_points":  [
+                                 "平均数的意义及求法，自然数的认识"
+                             ],
+        "source":  "2022五年级下·全国·专题练习"
+    },
+    {
+        "id":  "97",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·全国·课后作业)甲，乙，丙三人的平均年龄是24岁，如果甲，乙的平均年龄是22岁，乙，丙的平均年龄是30岁，那么乙的年龄是多少岁。",
+        "options":  [
+
+                    ],
+        "answer":  "32岁",
+        "solution":  "【分析】如果甲，乙的平均年龄是22岁，那么甲，乙的年龄和是22×2=44(岁);如果乙，丙的平均年龄是30岁，那么乙，丙的年龄和是30×2=60(岁).\n甲，乙，丙三人的平均年龄是24岁，则三人的年龄和是24×3=72(岁),由于44+60多算了一个乙的年龄，因此乙的年龄是44+60-24×3=32(岁).\n【详解】22×2+30×2-24×3\n=44+60-72\n=104-72\n=32(岁)\n答：乙的年龄是32岁。",
+        "knowledge_points":  [
+                                 "年龄问题，平均数问题"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "98",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "(24-25五年级上·上海浦东新·期末)学校举行书画作品展，四年级3个班分别展出24件,27件,20件，五年级4个班共展出188件，四，五年级平均每个班展出多少件。",
+        "options":  [
+
+                    ],
+        "answer":  "37件",
+        "solution":  "【分析】首先求出四年级3个班一共展出多少件，用加法，把三个班展出数量加起来;然后再求出四五年级共展出的总数，最后用四五年级的总数/四五年级的班级总数就求出了四五年级平均每个班展出多少件.\n【详解】根据分析得出:24+27+20=71(件) 71+188=259(件)259÷(3+4)=37(件)答：四，五年级平均每个班展出37件。",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25五年级上·上海浦东新·期末"
+    },
+    {
+        "id":  "99",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "(24-25五年级上·浙江宁波·期中)杭州文三数字街区\"机器人未来世界\"8天接待游客近8万人次，平均每天接待游客约多少万人次。",
+        "options":  [
+
+                    ],
+        "answer":  "1万人次",
+        "solution":  "【分析】用这组数据的和除以数据的个数，就是平均数，则总数量/份数=平均数，平均每天接待游客的人数=接待游客的总人数/接待游客的天数，即8÷8，据此解答.\n【详解】8÷8=1(万人次)\n答：平均每天接待游客约1万人次。",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25五年级上·浙江宁波·期中"
+    },
+    {
+        "id":  "100",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "(25-26五年级上·福建莆田·月考)淘淘和壮壮对自己家2024年各季度电费情况进行了统计，如下图。请根据下面的复式条形统计图回答问题。淘淘家，壮壮家2024年各季度电费情况统计图\n![](images/4cdc7001f80231c75bfc00ed5fdbccd5da19061d6353e9b9d6626058ade3b33b.jpg)\n(1)请根据\"壮壮家2024年全年的电费比淘淘家少\"这条信息把图例补充完整.\n(2)壮壮家平均每季度的电费是多少元。\n(3)你还能从统计图中发现哪些信息。请从不同角度写出两条。",
+        "options":  [
+
+                    ],
+        "answer":  "见详解;155元;见详解",
+        "solution":  "【分析】(1)通过比较发现，四个季度中，白色直条均长于灰色直条，即白色直条代表的电费多于灰色直条代表的电费，所以白色直条图例为\"淘淘家\"，灰色直条图例为\"壮壮家\"。(2)先求出壮壮家的全年电费，再除以4即可得到平均每季度电费。(3)可从季度电费高低角度，两家电费对比角度等不同角度观察统计图找信息。(答案不唯一，合理即可)\n【详解】(1)如图:![](images/11d1457fd853e6a3cc6cceca8b51b0d50f6440d8a73f69f1a4e133ba1ce8830d.jpg)\n(2)(135+210+115+160)÷4\n=(345+115+160)÷4\n=(460+160)÷4\n=620÷4\n=155(元)\n答：壮壮家平均每季度的电费是155元。(3)示例：从季度电费高低角度：淘淘家和壮壮家都是第三季度电费最高。从两家电费对比角度：第一季度淘淘家电费205元，壮壮家电费160元，淘淘家第一季度电费比壮壮家多。(答案不唯一，合理即可)",
+        "knowledge_points":  [
+                                 "复式条形统计图，平均数的意义及求法"
+                             ],
+        "source":  "25-26五年级上·福建莆田·月考"
+    },
+    {
+        "id":  "101",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·河南信阳·期末)小海的语文和数学的平均成绩是93分，英语成绩是99分，他三科平均成绩是()分。",
+        "options":  [
+                        "A. 94",
+                        "B. 95",
+                        "C. 96"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据科目的总分/科目数=平均分，用语文和数学的平均成绩乘2，就可以求出语文和数学的两科总分;再用语文和数学的两科总分加上英语成绩，即可求出三科的总分;最后用三科的总分除以3，即可求出三科平均成绩.\n【详解】语文和数学的总分:93×2=186(分)\n三科总分:186+99=285(分)\n三科平均分:285÷3=95(分)\n所以三科平均成绩是95分。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·河南信阳·期末"
+    },
+    {
+        "id":  "102",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·陕西宝鸡·期末)一个游泳池里，立指示牌最能帮助人们根据身高判断下水是否安全。()",
+        "options":  [
+                        "A. 最浅处水深70cm",
+                        "B. 平均水深140cm",
+                        "C. 最深处水深140cm"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】平均数：一组数据的总和除以这组数据个数所得到的商叫这组数据的平均数，平均数比一组数中最大的数小，比最小的数大;逐项分析后进行选择，据此解答.\n【详解】A.\"最浅处水深70cm\"，只知道最浅处水深70cm,不知道别处水深，也就无法判断人们到别处是否安全;B.\"平均水深140cm\"，知道平均水深140cm,有的地方可能会特别深，无法判断人们进入游泳池是否安全;C.\"最深处水深140cm\"，知道最深处水深140cm,那么人们只需要比较自己的身高在140cm水深处是否是安全的，就可以判断在整个游泳池是否是安全的。所以立指示牌\"最深处水深140cm\"最能帮助人们根据身高判断下水是否安全。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·陕西宝鸡·期末"
+    },
+    {
+        "id":  "103",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·甘肃天水·期末)某班男生平均身高145cm,女生平均身高142cm,全班平均身高()。",
+        "options":  [
+                        "A. 小于142cm",
+                        "B. 大于145cm",
+                        "C. 在142cm~145cm之间"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】全班平均身高由男生和女生的人数及各自平均身高共同决定。由于男生平均身高高于女生，若男生人数多，则全班平均更接近145cm,若女生人数多，则更接近142cm,但无论人数如何，全班平均身高都在两者之间.\n【详解】根据分析，故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·甘肃天水·期末"
+    },
+    {
+        "id":  "104",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·广东广州·期末)6个同学为一组进行一分钟跳绳比赛，跳的成绩最多是184下，最少是138下，猜一猜他们6人的平均成绩可能是()。",
+        "options":  [
+                        "A. 184下",
+                        "B. 138下",
+                        "C. 145下"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】平均数的取值范围应在最大值和最小值之间。根据题目，最大值为184，最小值为138，因此平均数应介于138和184之间.\n【详解】最大值为184，最小值为138，因此平均数应大于138小于184。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·广东广州·期末"
+    },
+    {
+        "id":  "105",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·安徽芜湖·期末)四(1)班体育课中测试1分钟仰卧起坐，男生平均每人做25个，女生平均每人做27个。下面说法正确的是()。",
+        "options":  [
+                        "A. 兰兰是一名女生，她一定做了27个。",
+                        "B. 每个女生仰卧起坐的个数一定比每个男生多。",
+                        "C. 女生仰卧起坐整体水平比男生高。",
+                        "D. 四(1)班全体同学1分钟仰卧起坐的平均成绩一定是26个。"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】平均数：一组数据的总和除以这组数据个数所得到的商叫这组数据的平均数。平均数是反映一组数据的平均水平，并不能反映这组数据中各个数据的大小，据此解答.\n【详解】A。平均数是一个整体水平的统计指标，无法保证每个个体都等于平均数。因此，兰兰作为女生，她的实际成绩可能高于，等于或低于27个，说法错误;B。平均数可以反映整体差异，但无法推断个体间的绝对关系，可能存在男生成绩高于女生或女生成绩低于男生的情况，说法错误;C。女生的平均数(27个)高于男生的平均数(25个),说明女生仰卧起坐整体水平比男生高，说法正确;D。四(1)班全体同学1分钟仰卧起坐的平均成绩不一定是26个，即不一定是(25+27)÷2=26(个),说法错误。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·安徽芜湖·期末"
+    },
+    {
+        "id":  "106",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·安徽芜湖·期末)甲，乙，丙三人中，甲最矮，身高是163cm,丙最高，身高是175cm。甲乙丙三人的平均身高可能是()cm。",
+        "options":  [
+                        "A. 175",
+                        "B. 170",
+                        "C. 160",
+                        "D. 155"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】平均数能代表一组数据的整体水平，一组数据中有一些数比平均数小，有一些数比平均数大，平均数介于这组数据的最大值和最小值之间。甲乙丙三人的平均身高介于163cm和175cm之间.\n【详解】A。175=175，甲乙丙三人的平均身高不可能是175cm.B。163\u003c170\u003c175，甲乙丙三人的平均身高可能是170cm.C。160\u003c163\u003c175，甲乙丙三人的平均身高不可能是160cm.D。155\u003c163，甲乙丙三人的平均身高不可能是155cm。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·安徽芜湖·期末"
+    },
+    {
+        "id":  "107",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·河南南阳·期末)下面说法正确的有()个。①计算99×101+101时，可以直接用100×101来算。②任何三角形中最少有2个锐角，最多有1个直角。③在10.06中，左边的\"0\"不能去掉，右边的\"0\"可以去掉。④慧慧所在的班级某次数学平均分是93分，慧慧的成绩不可能不及格。",
+        "options":  [
+                        "A. 1",
+                        "B. 2",
+                        "C. 3",
+                        "D. 4"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】①利用乘法分配律将99×101+101转化为101×(99+1)=101×100;②三角形内角和是180°,直角是90°,所以最多有1个直角，剩余的两个角都是锐角，钝角是大于90°且小于180°的角，所以一个三角形只能有一个钝角，剩余的两个角都是锐角，所以任何三角形中最少有2个锐角。③根据小数的性质，小数末尾的0增加或者去掉，小数大小不变;④平均分高不代表所有个体都高于及格线，可能存在不及格情况.\n【详解】①计算99×101+101时，可以直接用100×101来算。原题说法正确。②任何三角形中最少有2个锐角，最多有1个直角。原题说法正确。③在10.06中，左边的\"0\"不能去掉，右边的\"0\"也不可以去掉。因为0不在小数的末尾不能去掉，原题说法错误。④慧慧所在的班级某次数学平均分是93分，慧慧的成绩可能不及格。原题说法错误。所以正确的有①②;说法正确的有2个。故答案为:B",
+        "knowledge_points":  [
+                                 "整数乘法分配律，平均数的意义及求法，三角形的内角和，小数的性质"
+                             ],
+        "source":  "24-25四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "108",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "(24-25四年级下·广东广州·期末)四3班的全体同学参加体检，其中身高最高的是148cm,身高最矮的是131cm。这个班同学的平均身高可能是()。",
+        "options":  [
+                        "A. 148cm",
+                        "B. 140cm",
+                        "C. 131cm"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】平均数能代表一组数据的整体水平，一组数据中有一些数比平均数小，有一些数比平均数大，平均数介于这组数据的最大值和最小值之间.\n【详解】平均数介于一组数据的最大值和最小值之间，这个班级的平均身高不会高于148cm,不会低于131cm.\nA。148=148，因此不可能是这个班的平均身高。B。131\u003c140\u003c148，因此可能是这个班的平均身高。C。131=131，因此不可能是这个班的平均身高。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·广东广州·期末"
+    },
+    {
+        "id":  "109",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·天津河北·期末)\"某篮球队队员平均身高是175cm\"根据这条信息判断，下列说法正确的是()。",
+        "options":  [
+                        "A. 每个队员的身高都一定是175cm",
+                        "B. 每个队员的身高都不可能低于175cm",
+                        "C. 有的队员身高可能超过175cm",
+                        "D. 一定有队员的身高是175cm"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】根据平均数的概念：平均数反映的是一组数据的特征，不是其中每一个数据的特征，所以篮球队员的身高可能比175cm高，也可能比175cm矮，不一定都是175cm,据此即可解答.\n【详解】A。每个队员的身高不一定都是175cm,假如篮球队有3名队员，第一名队员身高170cm,第二名队员身高175cm,第三名队员身高180cm,但他们的平均身高还是175cm,故A选项说法错误;B。根据平均数的说法，队伍中可能有队员的身高比175cm低，也有可能刚好等于175cm,也有可能高于175cm,不一定都低于175cm,故B选项说法错误;C。根据平均身高的概念，有的队员身高可能超过175cm,也有的队员身高可能低于175cm,故C选项说法正确;D。假如篮球队有2名队员，第一名队员身高170cm,第二名队员身高180cm,但他们的平均身高还是175cm,不一定有队员的身高是175cm,故D选项说法错误。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·天津河北·期末"
+    },
+    {
+        "id":  "110",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·山东济南·期末)下面说法错误的有()。①2a表示两个a相乘。②大于0.6小于0.8的小数只有1个。③四(1)班同学的身高，最高是165cm,最矮是146cm,四(1)班同学平均身高一定比165cm低，比146cm高。④78×101=78×100+78运用了乘法结合律。",
+        "options":  [
+                        "A. 1个",
+                        "B. 2个",
+                        "C. 3个",
+                        "D. 4个"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】需要逐一分析每个说法所涉及的数学概念，判断其正确性。平均数是指在一组数据中所有数据之和再除以这组数据的个数。在一组数据中，平均数一定介于这组数据的最大值和最小值之间。乘法分配律是a×(b+c)=a×b+a×c,乘法结合律是(a×b)×c=a×(b×c).\n【详解】①2a表示的是2乘以a,即两个a相加，所以说法①错误。②大于0.6小于0.8的小数有无数个，例如0.61,0.611,0.7,0.75等等，只要小数的整数部分是0，十分位是6，百分位及以后有非零数字，都满足条件，所以说②错误。③四(1)班同学身高最高是165cm,最矮是146cm,那么平均身高一定比165cm低，比146cm高，说法③正确。④根据乘法分配律78×101=78×(100+1)=78×100+78×1=78×100+78，这里运用的是乘法分配律，而不是乘法结合律，所以说法④错误.\n所以,①,②,④都是错误的，错误的有3个。故答案为:C",
+        "knowledge_points":  [
+                                 "整数乘法分配律，整数乘法结合律，多位小数的大小比较，平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·山东济南·期末"
+    },
+    {
+        "id":  "111",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·浙江温州·期末)下列说法正确的是()。",
+        "options":  [
+                        "A. 王明3次跳远的总成绩是6.6米，他每次跳远的成绩肯定是2.2米。",
+                        "B. 游泳池的平均深度是140厘米，小明身高154厘米，他下水游泳不会有危险。",
+                        "C. 某小学排球队员的平均身高是160厘米，有的队员身高可能超过160厘米，有的队员身高可能不到160厘米。",
+                        "D. 小明前4单元数学测试平均成绩为90分，第5单元考了92分，他5次测试的平均分为90分。"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】根据平均数的意义，一组数据的平均数一定大于这组数据中的最小数，一定小于这组数据中的最大数。平均数是指在一组数据中所有数据之和再除以这组数据的个数。平均数反映一组数据的集中趋势的量数，它是反映数据集中趋势的一项指标，不能反映某一个数据的情况，据此分析各选项，即可解答.\n【详解】A。王明3次跳远的总成绩是6.6米，他每次跳远的成绩不一定是2.2米。原题说法错误;B。游泳池的平均深度是140厘米，小明身高154厘米，他下水游泳可能会有危险。原题说法错误;C。某小学排球队员的平均身高是160厘米，有的队员身高可能超过160厘米，有的队员身高可能不到160厘米。原题说法正确;D。(90×4+92)÷5=(360+92)÷5=452÷5=90.4(分),原题说法错误。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·浙江温州·期末"
+    },
+    {
+        "id":  "112",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·河南信阳·期末)下面说法正确的是()。",
+        "options":  [
+                        "A. 4.5和4.50大小相同，计数单位也相同。",
+                        "B. 三角形具有稳定性，且只有一条高。",
+                        "C. 68000÷125/68可以变成68000÷68/125进行简便计算。",
+                        "D. \"献爱心\"活动中，四年级同学平均每人捐款255元，那么四年级的小红一定捐了255元。"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】根据计数单位的认识分析判断;根据三角形具有稳定性和高的定义：从一个顶点向对边作垂线，顶点和垂足之间的距离，所以三角形有三条高;算出两个算式的结果再判断;平均数表示一组数据的平均值，不是某一个具体的数据;据此解答.\n【详解】A。4.5和4.50大小相同，但是4.5的计数单位是0.1,4.50的计数单位是0.01，计数单位不相同，所以题干中说法错误;不符合题意;B。三角形具有稳定性，有3条高，所以题干中说法错误，不符合题意;C。68000÷125/68=8,68000÷68/125=8，两个算式的结果相同，且68000÷68/125计算更简便，所以题干中说法正确，符合题意;D。四年级同学平均每人捐款255元，不能说明小红也一定捐款255元，所以题干中说法错误，不符合题意。故答案为:C",
+        "knowledge_points":  [
+                                 "三角形的稳定性及应用，小数的数位和计数单位的认识，平均数的意义及求法，三角形的高及画法"
+                             ],
+        "source":  "24-25四年级下·河南信阳·期末"
+    },
+    {
+        "id":  "113",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·河南信阳·期末)三个数的平均数是15，其中两个数的平均数是17，第三个数是()。",
+        "options":  [
+                        "A. 11",
+                        "B. 13",
+                        "C. 15"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】由题意得，三个数的平均数是15，其中两个数的平均数是17。可以用15乘3算出三个数的和，再用17乘2算出两个数的和。最后再用三个数的和减去两个数的和即可算出第三个数.\n【详解】15×3-17×2=45-34=11，即第三个数是11。故答案为:A",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·河南信阳·期末"
+    },
+    {
+        "id":  "114",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·河南南阳·期末)有甲乙丙三个数，其中甲\u003e乙\u003e丙，这三个数的平均数应该()。",
+        "options":  [
+                        "A. 大于甲",
+                        "B. 大于丙而小于甲",
+                        "C. 小于丙",
+                        "D. 无法确定"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】平均数代表的是一组数据的一般水平。它一般比这组数据的最小值大，一般比这组数据的最大值小。据此解答.\n【详解】由题意得，有甲乙丙三个数，其中甲\u003e乙\u003e丙，即甲是最大的数，丙是最小的数，那么平均数应该大于丙而小于甲。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "115",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·重庆黔江·期末)三位选手参加讲故事比赛，小维以9.9分获得第一名，小刚以9.3分获得第三名，小华获得第二名。这三位选手的平均成绩在()。",
+        "options":  [
+                        "A. 9.8分以上",
+                        "B. 9.3分以下",
+                        "C. 9.4分到9.7分之间",
+                        "D. 无法确定"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】第三名\u003c小华的成绩\u003c第一名，所以小华的成绩为9.4~9.8之间，根据对平均数的认识可知，一组数据的平均数应大于最小值，小于最大值。由题可知，最高分9.9，最低分9.3，所以三人的平均成绩应在9.3-9.9之间。据此作答\n【详解】A。9.3\u003c平均成绩\u003c9.9，小华的成绩为9.4~9.8之间，假设小华成绩最高为9.8，此时三人成绩分别为9.3,9.8,9.9，平均成绩不可能为9.8分以上，排除;B。9.3\u003c平均成绩\u003c9.9，所以平均成绩不可能在9.3分以下，排除;C。9.4分到9.7分之间符合9.3\u003c平均成绩\u003c9.9，符合题意;D。9.3\u003c平均成绩\u003c9.9能确定在一个范围，排除.\n故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·重庆黔江·期末"
+    },
+    {
+        "id":  "116",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·湖南岳阳·期末)下列说法正确的是()。",
+        "options":  [
+                        "A. 钝角三角形只能画一条高",
+                        "B. 96÷(8+4)=96÷8+96÷4",
+                        "C. 小马身高1.45米，他在平均水深1.2米的小河中游泳可能有危险。"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】(1)任何三角形都有三个角，三条边，三条高;(2)乘法分配律，一个数乘两个数的和等于分别乘这两个数再相加，除法没有分配律;(3)平均水深1.2米，说明河水中有比1.2米深的地方，可能有危险.\n【详解】由分析可得:A。钝角三角形也有三条高，其中有两条高画在三角形的外部，此选项说法错误;B。96÷(8+4)=96÷12=8,96÷8+96÷4=12+24=36，结果不相等，除法没有分配律，此选项说法错误;C。平均水深1.2米，有些地方肯定超过1.2米深，有可能超过1.45米，可能有危险，此选项说法正确;故答案为:C",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算，平均数的意义及求法，三角形的高及画法"
+                             ],
+        "source":  "24-25四年级下·湖南岳阳·期末"
+    },
+    {
+        "id":  "117",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·湖南株洲·期末)池塘的平均水深是1.2米，小明的身高是1.4米。小明在这个池塘里游泳()。",
+        "options":  [
+                        "A. 一定没有危险",
+                        "B. 可能有危险",
+                        "C. 一定有危险"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】平均数只能反映一组数据的平均水平，并不能反映这组数据中所有数据的大小，一个池塘的平均水深是1.2米，可能有的地方的水深不超过1.2米，可能有的地方水深超过1.2米，甚至超过1.4米，所以小明在池塘游泳可能有危险，据此解答即可.\n【详解】池塘平均水深1.2米，可能有的地方的水深不超过1.2米，可能有的地方水深超过1.2米，甚至超过1.4米，所以小明在池塘游泳可能有危险。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·湖南株洲·期末"
+    },
+    {
+        "id":  "118",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "(24-25四年级下·河北保定·期末)一分钟跳绳比赛中，健健组平均每人跳了120下，康康组平均每人跳了134下，健健和康康跳绳的个数相比,()跳得多。",
+        "options":  [
+                        "A. 健健",
+                        "B. 康康",
+                        "C. 无法确定"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】平均数只能反映一组数据的集中趋势，不能反映具体数据，据此解答.\n【详解】健健组平均每人跳120下，不能说明健健跳120下;康康组平均每人跳134下，不能说明康康跳134下，因此健健和康康跳的个数不确定，所以无法比较。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·河北保定·期末"
+    },
+    {
+        "id":  "119",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·湖北省直辖县级单位·期末)四(1)班同学的平均身高是1.4米，下列说法正确的有()个。①至少有一位同学的身高是1.4米 ②高于1.4米的人数和低于1.4米的人数一样多 ③去掉最高和最低的同学，平均身高还是1.4米 ④如果有1位同学身高是1.3米，那么肯定有1位同学身高是1.5米",
+        "options":  [
+                        "A. 0",
+                        "B. 1",
+                        "C. 2",
+                        "D. 3"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】平均数代表的是一组数据的一般水平。它一般比这组数据的最小值大，一般比这组数据的最大值小。由题意得，四(1)班同学的平均身高是1.4米，那么有的同学的身高比1.4米高，有的同学的身高比1.4米矮，有的同学可能恰好是1.4米。据此解答.\n【详解】①四(1)班同学的平均身高是1.4米，有的同学的身高比1.4米高，有的同学的身高比1.4米矮，有的同学可能恰好是1.4米，也有可能所有同学的身高都不是1.4米。原说法错误。②四(1)班同学的平均身高是1.4米，有的同学的身高比1.4米高，有的同学的身高比1.4米矮。但高于1.4米的人数和低于1.4米的人数不一定一样多。原说法错误。③去掉身高最高和身高最低的同学，四(1)班同学的平均身高可能会变化。原说法错误。④如果有1位同学身高是1.3米，他的身高比平均身高低，那么肯定有同学的身高比平均身高高，但不一定是1.5米。原说法错误。综上所述，四个说法都不正确。故答案为:A",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·湖北省直辖县级单位·期末"
+    },
+    {
+        "id":  "120",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·湖南郴州·期末)一次投篮比赛四人一组，小明和小军在同一组。小明投中10个，在他们组是投中最少的;小军投中18个，在他们组是投中最多的。这个组的平均成绩可能是()个。",
+        "options":  [
+                        "A. 18",
+                        "B. 16",
+                        "C. 10"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】平均数：反映了一组数据的平均大小，常用来代表一组数据的总体\"平均水平\"，也就是能较好地反映一组数据的\"整体水平\"，并且平均数大于本组数的最小数，小于最大数，据此解答.\n【详解】小明投中10个，在他们组是投中最少的;小军投中18个，在他们组是投中最多的，且10\u003c16\u003c18，所以这个组的平均成绩可能是16个。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·湖南郴州·期末"
+    },
+    {
+        "id":  "121",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·湖北黄冈·期末)甲，乙，丙，丁四个数的平均数是46，减去丁之后，甲乙丙三个数的平均数是35，丁是()。",
+        "options":  [
+                        "A. 81",
+                        "B. 75",
+                        "C. 105",
+                        "D. 79"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】根据平均数=总数/份数可知，平均数×份数=总数;用甲乙丙丁四个数的平均数46乘4可得到四个数的总数，用甲乙丙三个数的平均数35乘3可得到三个数的总数，再把两个总数相减，即得到丁。据此解答.\n【详解】46×4-35×3=184-105=79，所以，甲，乙，丙，丁四个数的平均数是46，减去丁之后，甲乙丙三个数的平均数是35，丁是79。故答案为:D",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·湖北黄冈·期末"
+    },
+    {
+        "id":  "122",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·青海果洛·期末)王叔叔是一名骑手，第一天上午送出外卖21单，下午送出16单，第二天读了25页，正好读完，平均每天读()页。",
+        "options":  [
+                        "A. 17单",
+                        "B. 21单",
+                        "C. 27单"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】先将两天送的单数相加求出一共送了多少单，再除以2即可求出平均每天送的单数.\n【详解】(21+16+17)÷2=54÷2=27(单)王叔叔这两天平均每天送出27单。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·青海果洛·期末"
+    },
+    {
+        "id":  "123",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25五年级下·江西九江·期末)妙想练习1分钟跳绳，前3次分别跳了151下,166下,162下，第4次跳了169下。她的平均成绩会因为第4次成绩而()。",
+        "options":  [
+                        "A. 提高",
+                        "B. 不变",
+                        "C. 降低"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】用这组数据的和除以数据的个数，就是平均数。分别求出前3次跳绳的平均成绩和4次跳绳的平均成绩，再比较大小，即可求得.\n【详解】(151+166+162)÷3=479÷3≈159.67(下) (151+166+162+169)÷4=648÷4=162(下) 因为159.67下\u003c162下，所以她的平均成绩会因为第4次成绩而提高。故答案为:A",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25五年级下·江西九江·期末"
+    },
+    {
+        "id":  "124",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·湖南常德·期末)校运会中，四(1)班5名同学1分钟跳绳成绩分别是148下,155下,160下,145下,152下。下面说法正确的是()。",
+        "options":  [
+                        "A. 这组同学平均成绩是160下。",
+                        "B. 这些同学的平均成绩可能低于145下。",
+                        "C. 如果跳145下的同学超常发挥跳了165下，平均成绩会提高。",
+                        "D. 每个同学的成绩都减去10下，新数据的平均成绩与原数据平均成绩相同。"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】一组数据的总和除以这组数据个数所得到的商叫这组数据的平均数;所以这组数据数量一定时，总和越大，那么平均数越大;平均数大于这组数据中的最小数据，小于最大数据;据此求出这组数据的平均数;然后进行分析.\n【详解】(148+155+160+145+152)÷5=760÷5=152(下)\nA。这组同学平均成绩是152下，原题干说法不正确;B。这些同学的平均成绩高于145下，原题干说法不正确;C。如果跳145下的同学超常发挥跳了165下，那么5个人的总成绩增加，人数不变，平均成绩则会提高;原题干说法正确;D。每个同学的成绩都减去10下，那么平均成绩也会在原有的基础上减去10;原题干说法不正确。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·湖南常德·期末"
+    },
+    {
+        "id":  "125",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·湖北黄冈·期末)在学校安全知识竞赛中，四(2)班派出数名学生组成小队参加比赛，元元也是其中的一名，小队的最终平均分是86分，元元得了96分，该小队不算元元平均分是84分，这个小队一共有()人。",
+        "options":  [
+                        "A. 5",
+                        "B. 6",
+                        "C. 7"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】算上元元，平均分从84分提高到86分，即提高了2分，元元的分数是96分，比平均分84分高12分，计算平均分时，将这12分平均分给每个队员，用12除以2，即可求出人数.\n【详解】(96-84)÷(86-84)=12÷2=6(人)这个小队一共有6人。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·湖北黄冈·期末"
+    },
+    {
+        "id":  "126",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "(24-25四年级下·湖北随州·期末)在一组数据中，最大的数是28，最小的数是12，这组数的平均数不可能是()。",
+        "options":  [
+                        "A. 14",
+                        "B. 25",
+                        "C. 40"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】平均数定义：一组数据的总和除以这组数据个数所得到的商叫这组数据的平均数。平均数反映了一组数据的平均大小，常用来代表一组数据的总体\"平均水平\"，也就是能较好地反映一组数据的\"整体水平\"，它一定大于最小的数，小于最大的数，即12\u003c平均数\u003c28，据此解答.\n【详解】A。12\u003c14\u003c28，有可能是平均数;B。12\u003c25\u003c28，有可能是平均数;C。40\u003e28，不符合平均数小于最大数的性质，不可能是平均数。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·湖北随州·期末"
+    },
+    {
+        "id":  "127",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·湖北武汉·期末)小华参加一分钟跳绳比赛，他四次成绩分别是223个,157个,215个和197个。关于小华跳绳的平均成绩，下面说法不正确的是()。",
+        "options":  [
+                        "A. 平均成绩不会高于223",
+                        "B. 平均成绩不会低于157",
+                        "C. 平均成绩在157~223之间",
+                        "D. 可以采用比最高成绩低，比最低成绩高的197或215作为平均成绩"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】将四人的跳绳成绩相加，再除以人数，求出四人跳绳的平均成绩。再进行解答.\n【详解】总数量:223+157+215+197=792(个)平均数:792÷4=198(个)A。平均成绩不会高于223，说法正确;B。平均成绩不会低于157，说法正确;C。平均成绩在157~223之间，说法正确;D。可以采用比最高成绩低，比最低成绩高的197或215作为平均成绩，说法不正确。故答案为:D",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·湖北武汉·期末"
+    },
+    {
+        "id":  "128",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·湖北武汉·期末)学校运动会上立定跳远比赛前三名平均成绩是176厘米，其中两人的成绩分别是178厘米和170厘米，另外一人的名次是()。",
+        "options":  [
+                        "A. 第一名",
+                        "B. 第二名",
+                        "C. 第三名",
+                        "D. 无法确定"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】根据题意，用平均成绩乘3算出三人的总成绩。再用总成绩减去两人的成绩就是另外一个人的成绩;最后比较三个人的成绩即可.\n【详解】176×3=528(厘米)528-178-170=350-170=180(厘米)180\u003e178\u003e170 所以另外一人的名次是第一名。故答案为:A",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·湖北武汉·期末"
+    },
+    {
+        "id":  "129",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·湖北武汉·期末)下面的说法中，错误的是()。",
+        "options":  [
+                        "A. 等腰三角形不一定是锐角三角形。",
+                        "B. 小数部分相邻两个计数单位之间的进率与整数一样，都是10。",
+                        "C. 一个六边形的内角和是720°。",
+                        "D. 典典所在班级的同学平均每人捐款15元，说明典典一定捐了15元。"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】三角形按角来分，分为锐角三角形，直角三角形和钝角三角形。三个角都是锐角的三角形叫作锐角三角形，有一个角是直角的三角形叫作直角三角形，有一个角是钝角的三角形叫作钝角三角形;有两条边相等的三角形叫作等腰三角形，三条边都相等的三角形叫作等边三角形;小数的数位顺序表如下:\n| | 整数部分 | 小数点 | 小数部分 | | | | | | | | | |\n| -------- | -------- | ------ | -------- | ---- | ---- | ---- | -------- | -------- | -------- | -------- | ------ | ---- |\n| 数位 | 。 | 万位 | 千位 | 百位 | 十位 | 个位 | 。 | 十分位 | 百分位 | 千分位 | 万分位 | 。 |\n| 计数单位 | 。 | 万 | 千 | 百 | 十 | 个 | 十分之一 | 百分之一 | 千分之一 | 万分之一 | 。 | |\n求多边形的内角和，可以把它分成几个三角形。把多边形分成了几个三角形，多边形的内角和就等于几个三角形的内角之和;平均数表示的是这组数据的一般水平，它一般比这组数据的最小值大，一般比这组数据的最大值小。据此解答.\n【详解】A。根据题意作图如下:![](images/455f43a0d20e00d7e8bcad7601ef59d1.png)\n由图可知，等腰三角形可能是锐角三角形，也有可能是直角三角形，还有可能是钝角三角形。该选项说法正确.\nB。根据题意作图如下:![](images/9c3c3924830565c2b0fc856d60af28c28cf32c1173c7427c408496adc960108d.jpg)\n由图可知，小数部分相邻两个计数单位之间的进率与整数一样，都是10。该选项说法正确。C。六边形可以分成4个三角形。180°*4=720°,即六边形的内角和是720°。该选项说法正确。D。典典所在班级的同学平均每人捐款15元，那么典典捐的钱可能比15元少，也有可能比15元多，还有可能等于15元。该选项说法错误。故答案为:D",
+        "knowledge_points":  [
+                                 "多边形的内角和，小数的数位和计数单位的认识，平均数的意义及求法，等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·湖北武汉·期末"
+    },
+    {
+        "id":  "130",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·湖北荆州·期末)下面说法正确的是()。",
+        "options":  [
+                        "A. 0.6和0.60大小相等，计数单位也相同",
+                        "B. 游泳池的平均水深1.3米，明明身高138厘米，他在游泳池里游泳不会有危险",
+                        "C. 图形平移后可以改变图形位置，但没有改变图形的大小"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】根据题意逐项分析,A选项0.6和0.60大小相等，但计数单位不同;B选项平均水深不能保证所有区域均低于1.38米;C选项平移仅改变位置，不改变大小，据此选择.\n【详解】根据分析可得:A。0.6和0.60的数值相等，但0.6的计数单位是0.1(十分位),0.60的计数单位是0.01(百分位),因此计数单位不同，故原说法错误;B。平均数表示的是一组数据的平均水平，并不能表示这组数据中各个数据的大小，游泳池的平均水深为1.3米，而明明身高138厘米(1.38米),若某处水深超过1.38米则存在危险，故原说法错误;C。平移是图形的位置变化，形状和大小均保持不变，故原说法正确。故答案为:C",
+        "knowledge_points":  [
+                                 "小数的数位和计数单位的认识，平均数的意义及求法，平移与平移现象"
+                             ],
+        "source":  "24-25四年级下·湖北荆州·期末"
+    },
+    {
+        "id":  "131",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "(24-25四年级下·湖北咸宁·期末)四(1)班同学的平均身高是135厘米，四(2)班同学的平均身高是137厘米，亮亮在四(1)班，文文在四(2)班，亮亮比文文()。",
+        "options":  [
+                        "A. 矮",
+                        "B. 高",
+                        "C. 无法比较"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】平均数表示的是一组数据的平均水平，并不能表示这组数据中各个数据的大小，由此即可进行选择.\n【详解】135厘米是四(1)班同学的平均身高,137厘米是四(2)班同学的平均身高,135厘米并不一定是亮亮的实际身高,137厘米也不一定是文文的实际身高，因此亮亮与文文谁高无法比较.\n故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·湖北咸宁·期末"
+    },
+    {
+        "id":  "132",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·湖北咸宁·期末)羽毛球队原来有8名队员，平均体重34kg,现增加1名体重是38kg的队员。现在羽毛球队队员的平均体重()。",
+        "options":  [
+                        "A. 是34kg",
+                        "B. 比34kg重",
+                        "C. 比34kg轻"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】平均数是表示一组数据的平均值，是指在一组数据中所有数据之和再除以这组数据的个数;其特点是比最大数小，比最小数大;依此即可选择.\n【详解】原来有8名队员，平均体重34kg,现增加1名体重是38kg的队员，这名队员的体重比原来8名队员的平均体重要重(38kg\u003e34kg),因此现在羽毛球队队员的平均身高比34kg重。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·湖北咸宁·期末"
+    },
+    {
+        "id":  "133",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "(24-25六年级下·湖南长沙·期末)在跳绳比赛中，小明所在小组平均每人跳75个，小红所在小组平均每人跳85个，小明和小红跳的个数相比较,()。",
+        "options":  [
+                        "A. 小明跳得多",
+                        "B. 小红跳得多",
+                        "C. 一样多",
+                        "D. 无法确定"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】平均数只是反映小组的整体水平，不能确定小明和小红个人具体跳的个数。小明所在小组平均每人跳75个，这里的75个是小明所在小组所有人跳绳个数的平均值，小明个人跳的个数可能高于75个，可能等于75个，也可能低于75个;小红所在小组平均每人跳85个，这里的85个是小红所在小组所有人跳绳个数的平均值，小红个人跳的个数可能高于85个，可能等于85个，也可能低于85个。据此解答.\n【详解】已知小明所在小组的平均数是75个，小红所在小组的平均数是85个;平均数反映的是小组整体的平均水平，并不能确定小明或小红个人的跳绳个数;小明可能跳得比75个多，比75个少或者正好是75个，同理，小红可能跳得比85个多，比85个少或者正好是85个。因此，仅根据小组平均数无法确定小明和小红跳的个数谁多谁少。故答案为:D",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25六年级下·湖南长沙·期末"
+    },
+    {
+        "id":  "134",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "(24-25四年级下·河北保定·期末)一次测试，文文语文得了95分，数学得了98分，英语得了95分，他这三科的平均分是()分。",
+        "options":  [
+                        "A. 96",
+                        "B. 95",
+                        "C. 94"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】一共是语文，数学，英语三科，用三科成绩之和除以3即可计算出平均分.\n【详解】(95+98+95)÷3\n=(193+95)÷3\n=288÷3\n=96(分)\n他这三科的平均分是96分。故答案为:A",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·河北保定·期末"
+    },
+    {
+        "id":  "135",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·湖南张家界·期末)四年级3个班，平均每个班捐款36元，五年级4个班一共捐款102元，四五年级平均每个班捐款多少元。列式正确地是()。",
+        "options":  [
+                        "A. (36+102)÷2",
+                        "B. (36+102)÷(3+4)",
+                        "C. (36×3+102)÷(3+4)",
+                        "D. (36×3+102×4)÷(3+4)"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】根据题意，已知四年级3个班，平均每个班捐款36元，五年级4个班一共捐款102元，先用36乘3，求出四年级的捐款数，再加上五年级的102元，求出四，五年级的总捐款数，再除以总班级数:3+4=7(班),就是四五年级平均每个班捐款数，以此答题即可.\n【详解】(36×3+102)÷(3+4)\n=(108+102)÷7\n=210÷7\n=30(元)\n故答案为:C",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·湖南张家界·期末"
+    },
+    {
+        "id":  "136",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "(24-25四年级下·湖南益阳·期末)小玲语文，数学总分182分，英语85分，她三科的平均成绩是()分。",
+        "options":  [
+                        "A. 89",
+                        "B. 90",
+                        "C. 91"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】平均成绩=总成绩/科目数量。语文和数学总分182分，加上英语85分，得到三科总成绩，再除以3即可求出平均分。据此解答.\n【详解】(182+85)÷3\n=267÷3\n=89(分)\n她三科的平均成绩是89分。故答案为:A",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·湖南益阳·期末"
+    },
+    {
+        "id":  "137",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·湖南长沙·期末)5个评委给一个节目的打分分别是93分,96分,95分,85分和91分，规定去掉一个最高分和一个最低分再算平均分作为最后得分。这个节目的最后得分是()分。",
+        "options":  [
+                        "A. 92",
+                        "B. 93",
+                        "C. 96",
+                        "D. 85"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】平均数=这组数据的总和/数据的个数。由题意得,5个评委给这个节目的打分分别是93分,96分,95分,85分和91分。其中，最高分是96分，最低分是85分，去掉这两个分数后再算剩下三个分数的平均分即可得到这个节目的最后得分.\n【详解】去掉96分和85分，剩下93分,95分,91分.\n(93+95+91)÷3\n=(188+91)÷3\n=279÷3\n=93(分)\n即这个节目的最后得分是93分。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·湖南长沙·期末"
+    },
+    {
+        "id":  "138",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·湖北恩施·期末)下面说法不正确的是()。",
+        "options":  [
+                        "A. 三条长度分别为5,5,8的木条，可以围成一个三角形",
+                        "B. 要统计四年级各班的学生人数，可以选择条形统计图",
+                        "C. 聪聪组平均身高比洛洛组多2厘米，但聪聪不一定比洛洛高",
+                        "D. 0.999是最接近1的小数"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】(1)根据三角形的三边关系，两边之长大于第三边，即可判断。(2)条形统计图的特点是能够清晰地反映出数量的多少。统计学校各年级的学生人数，重点是要直观地看出每个年级人数的具体数值，用条形统计图可以很清楚地对比不同年级的人数。(3)平均数是指在一组数据中所有数据之和再除以这组数据的个数。聪聪班平均身高比洛洛班多2厘米，只能说明聪聪所在班级学生身高的总体水平比洛洛所在班级高，但不能就此得出聪聪一定比洛洛高的结论。因为平均数反映的是整体情况，个体之间存在差异，有可能洛洛比聪聪高.\n(4)0.999与1的差是0.001。但存在更接近1的小数，例如:0.9999与1的差是0.0001(小于0.001),0.9995与1的差是0.0005(小于0.001).\n【详解】根据分析可知:A。5+5=10\u003e8,5+8=13\u003e5，三条长度分别为5,5,8的木条，可以围成一个三角形，正确。B。要统计四年级各班的学生人数，可以选择条形统计图，正确。C。聪聪组平均身高比洛洛组多2厘米，但聪聪不一定比洛洛高，正确。D。0.999不是最接近1的小数，原选项错误。故答案为:D",
+        "knowledge_points":  [
+                                 "三角形三边关系",
+                                 "1格表示多个单位的单式条形统计图",
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·湖北恩施·期末"
+    },
+    {
+        "id":  "139",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·重庆梁平·期末)上期期末，体育老师对王洪同学\"1分钟跳绳\"项目进行了6次测试，成绩是(单位：下):108,118,120,126,122,120。哪个数据能代表王洪同学这项测试的成绩。()。",
+        "options":  [
+                        "A. 126",
+                        "B. 120",
+                        "C. 119",
+                        "D. 118"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】平均数=这组数据的总和/数据的个数。由题意得，王洪同学\"1分钟跳绳\"项目6次测试成绩分别是:108,118,120,126,122,120，可以先用加法算出6次跳绳的总成绩，然后再除以6即可算出他的平均成绩，得到的数据可以代表王洪同学这项测试的成绩.\n【详解】(108+118+120+126+122+120)÷6\n=(226+120+126+122+120)÷6\n=(346+126+122+120)÷6\n=(472+122+120)÷6\n=(594+120)÷6\n=714÷6\n=119(下)\n即王洪同学这项测试的成绩是119下。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·重庆梁平·期末"
+    },
+    {
+        "id":  "140",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·重庆铜梁·期末)四(2)班24个男生与26个女生进行一分钟跳绳，男生平均每人跳100个，女生平均每人跳110个，四(2)班平均每人跳多少个。下面算式正确的是()。",
+        "options":  [
+                        "A. (110+100)÷2",
+                        "B. (100×24+110×26)÷50",
+                        "C. (110+100)÷50"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】由题意得，四(2)班24个男生与26个女生进行一分钟跳绳，男生平均每人跳100个，女生平均每人跳110个，可以用男生或女生平均每人跳绳的个数分别乘上男女生的人数算出他们各自跳绳的个数，然后再把得数加起来算出总个数，列式为:100×24+110×26。最后再用总个数除以总人数即可算出四(2)班平均每人跳多少个.\n【详解】24+26=50(人)求四(2)班平均每人跳多少个，列式为:(100×24+110×26)÷50。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·重庆铜梁·期末"
+    },
+    {
+        "id":  "141",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·湖南湘西·期末)下列说法错误的是()。①8.93的小数点先向左移动两位，再向右移动一位，就缩小到原数的1÷10。②长方形和正方形都有4条对称轴。③小亮身高148厘米，他到一个平均水深110厘米的泳池里游泳，所以不会有危险的。④131-63+37=131-(63+37)",
+        "options":  [
+                        "A. ①③④",
+                        "B. ②③",
+                        "C. ②③④",
+                        "D. ①②④"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】根据小数点的移动规律，小数点向右移动一位，小数就扩大到原数的10倍，向右移动两位，小数就扩大到原数的100倍，向右移动三位，小数就扩大到原数的1000倍;小数点向左移动一位，小数就缩小到原数的1÷10，向左移动两位，小数就缩小到原数的1÷100，向左移动三位，小数就缩小到原数的1÷1000.\n轴对称图形的特点，沿一条直线对折图形的两部分能够完全重合的图形是轴对称图形，这条直线是对称轴,\n平均数的概念，是表示一组数据集中趋势的量数，是指在一组数据中所有数据之和再除以这组数据的个数.\n根据四则运算规则:131-63+37按照从左向右的顺序计算，先算减法，再算加法,131-(63+37)要先算括号里面的加法，最后算减法;据此分别计算出两个算式的得数，再做选择即可.\n【详解】①小数点向左移动两位，再向右移动一位，相当于是向左移动了一位，向左移动一位就缩小为原来的1÷10，故该项正确。②能使长方形对折完全重合的是：长方形上下两边的中点连线，或左右两边的中点连线，只有2条对称轴;正方形除了中点连线，还有对角连线，正方形有4条对称轴。故该选项说法错误。③游泳池平均水深110厘米，平均水深不是所有地方都是110厘米，是深水区跟浅水区折合之后的结果，所以深水区肯定不只110厘米，有可能超过小亮的身高。故该选项说法错误。④131-63+37=68+37=105,131-(63+37)=131-100=31,105!=31。故该选项说法错误。错误的说法有②③④。故答案为:C",
+        "knowledge_points":  [
+                                 "小数点向左移动引起小数大小变化的规律",
+                                 "小数点向右移动引起小数大小变化的规律",
+                                 "平均数的意义及求法",
+                                 "对称轴的画法及数量"
+                             ],
+        "source":  "24-25四年级下·湖南湘西·期末"
+    },
+    {
+        "id":  "142",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·辽宁鞍山·期末)三位同学描述\"平均数\"的相关知识，描述正确的是()。",
+        "options":  [
+                        "A. 丽丽3天共看了12页课外书，她每天看的页数一定都是4页。",
+                        "B. 学校四年级平均每个班有学生45人，有的班的学生人数可能比45人多，有的班的学生人数可能比45人少。",
+                        "C. ABCD四个数的平均数是20，其中D一定大于20。"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】平均数是一组数的总和除以这组数的数量。平均数反映的是一组数据的平均水平，平均数会受偏大或偏小数据的影响，据此解答.\n【详解】A。丽丽3天共看了12页书，平均每天看12÷3=4(页),但每天看的页数不一定都是4页;她可能第1天看3页，第2天看5页，第3天看4页，总和仍是12页，平均每天看4页，所以\"一定都是4页\"的说法不正确;B。根据解析可知，四年级平均每个班有学生45人，则有的班的学生人数可能比45人多，有的班的学生人数可能比45人少，也有可能有的班的学生人数正好是45人，所以表达正确;C.ABCD四个数的平均数是20，总和是4×20=80，但D不一定大于20;如果A,B,C,D分别为20,20,20,20时,D等于20，所以\"D一定大于20\"的说法不正确.\n故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·辽宁鞍山·期末"
+    },
+    {
+        "id":  "143",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·重庆渝北·期末)欢欢的数学和英语的平均分为96分，其中数学是94分，英语是()分。",
+        "options":  [
+                        "A. 96",
+                        "B. 98",
+                        "C. 92"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】平均数是表示一组数据的平均值，是指在一组数据中所有数据之和再除以这组数据的个数;因此先用平均分乘2，从而计算出数学和英语的总分，然后再减去数学的分数即可.\n【详解】96×2-94=192-94=98(分)英语是98分。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·重庆渝北·期末"
+    },
+    {
+        "id":  "144",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "(24-25四年级下·山东菏泽·期末)在\"一日捐\"活动中，我校老师平均捐款200元，解释合理的是()。",
+        "options":  [
+                        "A. 每人一定都捐了200元",
+                        "B. 大多数老师捐款数少于150元",
+                        "C. 不可能有人捐了300元",
+                        "D. 200元只是老师捐款数的整体水平"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】平均数只能反映一组数据的平均水平，并不能反映这组数据中所有数据的大小。我校老师平均捐款200元，有人捐款可能小于200元，可能大于200元，也可能等于200元;据此解答即可.\n【详解】A。每人不一定都捐了200元，原题说法错误;B。无法确定\"大多数\"是否少于150元，原题说法错误;C。可能有人捐了300元，原题说法错误;D。200元只是老师捐款数的整体水平，原题说法正确.\n故答案为:D",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·山东菏泽·期末"
+    },
+    {
+        "id":  "145",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·广西河池·期末)小阳，小明和小美的体重分别是36kg,31kg和35kg,他们三人的平均体重是()kg。",
+        "options":  [
+                        "A. 31",
+                        "B. 34",
+                        "C. 36"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】先将三人的体重相加求出三人的体重之和，然后再根据\"平均数=总数/份数\"，用三人的体重之和除以3，即可求出他们三人的平均体重.\n【详解】(36+31+35)÷3\n=(67+35)÷3\n=102÷3\n=34(kg)\n因此他们三人的平均体重是34kg。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·广西河池·期末"
+    },
+    {
+        "id":  "146",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·重庆南岸·期末)小米在期末考试中，语文，数学，英语三科的平均成绩是96分，数学是94分，英语是97分，那么语文是()分。",
+        "options":  [
+                        "A. 94",
+                        "B. 95",
+                        "C. 97",
+                        "D. 96"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】根据总成绩=平均成绩×科目数量，用96×3求出小米期末考试语文，数学，英语三门功课的总成绩，再减去英语，数学的成绩就是语文的成绩.\n【详解】96×3-94-97\n=288-94-97\n=97(分)\n语文是97分。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·重庆南岸·期末"
+    },
+    {
+        "id":  "147",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·黑龙江齐齐哈尔·期末)有7个数，它们的平均数是35，去掉一个数后，余下数的平均数是33，去掉的数是()。",
+        "options":  [
+                        "A. 45",
+                        "B. 46",
+                        "C. 47"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】平均数是指在一组数据中所有数据之和再除以数据的个数。平均数是表示一组数据集中趋势的量数，它是反映数据集中趋势的一项指标。本题可先根据\"平均数×个数=总数\"分别求出7个数的总数和去掉一个数后余下6个数的总数，再用7个数的总数减去6个数的总数，得到去掉的数.\n【详解】已知7个数的平均数是35，根据\"平均数×个数=总数\"，可得这7个数的总数为:7×35=245。去掉一个数后，余下数的平均数是33，此时余下7-1=6(个)数，同样根据\"平均数×个数=总数\"，可得余下6个数的总数为:6×33=198。用7个数的总数减去余下6个数的总数，即去掉的数为245-198=47。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·黑龙江齐齐哈尔·期末"
+    },
+    {
+        "id":  "148",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "(24-25四年级下·山东菏泽·期末)你知道吗。跳远或跳高比赛时，每个运动员的成绩是按哪种方法计分的。正确的一种是()。",
+        "options":  [
+                        "A. 按第一次跳的成绩",
+                        "B. 按最后一次跳的成绩",
+                        "C. 按几次跳的平均成绩",
+                        "D. 按跳最好的一次成绩"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】了解跳远或跳高比赛的计分规则，跳高或跳远比赛是以最好的成绩为最终的成绩。对每个选项进行判断.\n【详解】A。在跳远或跳高比赛中，第一次跳的成绩往往不能代表运动员的最佳水平，比赛不会仅按第一次跳的成绩计分，所以A选项错误。B。最后一次跳的成绩也不一定是运动员的最好成绩，不能单纯以最后一次跳的成绩来计分，所以B选项错误。C。按几次跳的平均成绩不能突出运动员的最佳表现，而比赛更关注运动员的最高水平，所以C选项错误。D。跳远或跳高比赛通常是按运动员跳最好的一次成绩来计分，这样能体现运动员在比赛中的最佳水平，所以D选项正确.\n故答案为:D",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·山东菏泽·期末"
+    },
+    {
+        "id":  "149",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "(24-25四年级下·山东菏泽·期末)歌咏比赛时，计算选手的平均分一般要去掉一个最低分和一个最高分，原因在于()。",
+        "options":  [
+                        "A. 计算方便",
+                        "B. 中间的数接近，有说服力",
+                        "C. 平均数具有敏感性，易受极大值和极小值的影响，这样更能体现选手的真实水平"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】了解平均数的概念，平均数是一组数据的总和除以数据的个数得到的值，它反映了一组数据的总体平均水平。逐项分析.\n【详解】A。去掉一个最高分和一个最低分，实际上是增加了计算的步骤，并没有让计算变得方便。所以选项A不符合实际情况.\nB。中间的数接近有说服力，这种说法比较模糊。平均数的意义是综合反映整体水平，而不是单纯因为中间的数接近就有说服力。而且去掉最高分和最低分的目的不是因为中间数接近，所以选项B也不正确.\nC。平均数具有敏感性，极大值和极小值会对平均数产生较大的影响。例如，在一场歌咏比赛中，有一位选手的其他评委打分都在80-90分之间，但有一个评委打了极低的20分，另一个评委打了极高的99分。如果不去掉这两个分数，那么这个极低分和极高分就会使平均分不能很好地反映该选手的真实水平。去掉一个最高分和一个最低分后，能减少这种极端值对平均分的影响，从而更能体现选手的真实水平，所以选项C是正确的.\n故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·山东菏泽·期末"
+    },
+    {
+        "id":  "150",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "(24-25四年级下·安徽芜湖·期末)小明所在班级同学的平均身高是140厘米，小刚所在班级同学的平均身高是142厘米，小明和小刚的身高相比,()。",
+        "options":  [
+                        "A. 小明高",
+                        "B. 小刚高",
+                        "C. 一样高",
+                        "D. 无法确定谁高"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】本题可根据平均数的概念，分析平均身高与个体身高之间的关系，进而判断小明和小刚身高的情况。平均数是一组数据的总和除以数据个数得到的值，它反映的是一组数据的总体情况，不能代表其中某个个体的具体数值。小明所在班级同学的平均身高是140厘米，这只能说明小明班级同学身高的总体水平，小明的实际身高可能高于140厘米，可能等于140厘米，也可能低于140厘米。同理，小刚所在班级同学的平均身高是142厘米，小刚的实际身高可能高于142厘米，可能等于142厘米，也可能低于142厘米.\n【详解】由于不知道小明和小刚的具体身高，所以无法确定小明和小刚谁高。故答案为:D",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·安徽芜湖·期末"
+    },
+    {
+        "id":  "151",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·江西赣州·期末)下面的说法中，正确的是()。",
+        "options":  [
+                        "A. 一个三角形中最小的角是49°,那么这个三角形一定是锐角三角形。",
+                        "B. 0.8和0.80的大小相同，计数单位也相同。",
+                        "C. 学校举行跳绳比赛，欢欢所在的班级平均每人1分钟跳125下，乐乐所在的班级平均每人1分钟跳132下，所以乐乐跳的下数比欢欢多。"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】A。三个角均小于90°的三角形是锐角三角形。已知最小角为49°(\u003c90°),而其他两个角一定大于等于49°,至少是49°,再算第三个角最大是180°-49°-49°=82°,所以三个角中最大是82°,都小于90°,因此三个角均为锐角。B。小数的末尾添上\"0\"或去掉\"0\"，小数的大小不变;一位小数的计数单位是0.1，两位小数的计数单位是0.01，依此判断;C。平均数是反映一组数据的平均水平，并不能反映这组数据中各个数据的大小，依此判断.\n【详解】A。一个三角形中最小的角是49°,最大可能是180°-49°-49°=82°,这个三角形一定是锐角三角形，符合题意;B。0.8和0.80的大小相等,0.8的计数单位是0.1,0.80的计数单位是0.01，计数单位不同，不符合题意;C。学校举行跳绳比赛，欢欢所在的班级平均每人1分钟跳125次，乐乐所在的班级平均每人1分钟跳132次，欢欢可能高于班级平均分，乐乐可能低于班级平均分，因此无法确定两人具体次数的大小关系，不符合题意;故答案为:A",
+        "knowledge_points":  [
+                                 "小数的数位和计数单位的认识",
+                                 "平均数的意义及求法",
+                                 "三角形的分类",
+                                 "小数的性质"
+                             ],
+        "source":  "24-25四年级下·江西赣州·期末"
+    },
+    {
+        "id":  "152",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·江西赣州·期末)哪吒读《混元仙法》,第一天上午读了19页，下午读了16页，第二天读了25页，正好读完，平均每天读()页。",
+        "options":  [
+                        "A. 17",
+                        "B. 22",
+                        "C. 30"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】先算出两天一共读了几页=第一天上午读了19页+下午读了16页+第二天读了25页，再除以2天求出平均每天读几页.\n【详解】(19+16+25)÷2\n=60÷2\n=30(页)\n平均每天读30页.\n故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·江西赣州·期末"
+    },
+    {
+        "id":  "153",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25六年级下·内蒙古通辽·期末)聪聪所在的六(1)班同学的平均身高是1.52m,明明所在的四(1)班同学的平均身高是1.41m,下列说法正确的是()。",
+        "options":  [
+                        "A. 聪聪比明明高",
+                        "B. 明明比聪聪高",
+                        "C. 聪聪和明明一样高",
+                        "D. 无法判断他们两人谁更高"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】平均数是指几个数值中平均的数值，不能体现出个体的数值大小，只能体现出一组数据的大致水平。据此可得出答案.\n【详解】聪聪所在的六(1)班同学的平均身高是1.52m,明明所在的四(1)班同学的平均身高是1.41m,则无法判断聪聪和明明谁更高。因为只知道身高平均数，不能体现出个体的身高，无法判断.\n故答案为:D",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25六年级下·内蒙古通辽·期末"
+    },
+    {
+        "id":  "154",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25三年级下·河北邢台·期末)甲，乙，丙三个数都是大于0的整数，而且甲比乙大，丙比乙小，这三个数的平均数一定()。",
+        "options":  [
+                        "A. 比甲大",
+                        "B. 比丙小",
+                        "C. 比甲小比乙大",
+                        "D. 比甲小比丙大"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】甲，乙，丙三个数都是大于0的整数，根据平均数的概念，可知平均数应该大于最小的数而小于最大的数，由此可解.\n【详解】由\"甲比乙大，丙比乙小\"可知：甲\u003e乙\u003e丙\n三个数的平均数肯定比最大的数小，比最小的数大，即这三个数的平均数一定比甲小比丙大。故答案为:D",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25三年级下·河北邢台·期末"
+    },
+    {
+        "id":  "155",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·北京丰台·期末)下面有关平均数的说法，正确的有()个。①中国选手以10枪103环的成绩夺冠，他每枪的成绩肯定都是10.3环。②多名裁判打分的比赛中，计算平均成绩时一般要去掉一个最高分和一个最低分，这样做可以提高打分的公正性。③上个星期，妈妈平均每天走8000步，上周她的总步数约56000步。④篮球队上场球员的平均身高是180cm,换下1名身高175cm的球员，换上的球员身高185cm,现在场上球员的平均身高还是180cm。",
+        "options":  [
+                        "A. 1",
+                        "B. 2",
+                        "C. 3",
+                        "D. 4"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】①平均数代表一组数据的整体平均情况，不代表其中每个数值;②去掉极端分数可减少误差，提高公正性;③用平均每天走的步数乘天数，求出的是总数，据此判断是否为56000步;④平均身高=所有人总身高/人数，换人后总和增加，但人数不变，所以平均身高会变化.\n【详解】①总成绩103环，平均10.3环，但各枪成绩不一定都是10.3环，此选项错误;②去掉最高分和最低分可避免极端值影响，提高公正性，此选项正确;③7天总步数:8000×7=56000(步),与题干一致，此选项正确;④原平均身高180cm,换人后总和增加10cm,人数不变，所以平均身高必然大于180cm,此选项错误.\n综上，正确的有②和③,共2个。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·北京丰台·期末"
+    },
+    {
+        "id":  "156",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·江西九江·期末)下面对平均数理解正确的是()。",
+        "options":  [
+                        "A. 王悦5次跳远的总成绩是10m,她每次跳远的成绩肯定是2m。",
+                        "B. 某小学排球队队员的平均身高是162cm,有的队员身高可能超过162cm,有的队员身高可能不到162cm。",
+                        "C. 小东所在班级的平均体重是34kg,小刚所在班级的平均体重是30kg,小东一定比小刚重。"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】平均数是反映一组数据的平均水平，是一组数据中所有数据之和再除以这组数据的个数，平均数介于一组数据中的最大值与最小值之间，据此即可解答.\n【详解】A。王悦5次跳远的总成绩是10m,她每次跳远的成绩不一定是2m,可能大于2m,也可能小于2m。原说法错误;B。某小学排球队队员的平均身高是162cm,有的队员身高可能超过162cm,有的队员身高可能不到162cm。说法正确;C。小东所在班级的平均体重是34kg,小刚所在班级的平均体重是30kg,小东的体重可能是31kg,小刚的体重可能是32kg,此时小东比小刚轻，所以小东不一定比小刚重。原说法错误。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·江西九江·期末"
+    },
+    {
+        "id":  "157",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·天津滨海新·期末)小琪玩拍球游戏，三次的平均成绩是98下。第一次拍了96下，第二次拍了100下。她第三次拍了()下。",
+        "options":  [
+                        "A. 96",
+                        "B. 97",
+                        "C. 98",
+                        "D. 99"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】根据总数=平均数×次数，先算出三次的总成绩，再减去第一次和第二次的成绩即可解答.\n【详解】98×3=294(下)294-96-100=98(下) 所以第三次拍了98下。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·天津滨海新·期末"
+    },
+    {
+        "id":  "158",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·北京朝阳·期末)某校篮球队5名队员的平均身高是157厘米，小奇加入篮球队后，现在队员们的平均身高比原来高了2厘米。下面关于小奇身高说法中正确的是()。",
+        "options":  [
+                        "A. 他的身高不到157厘米",
+                        "B. 他的身高在157厘米-159厘米之间",
+                        "C. 他的身高是159厘米",
+                        "D. 他的身高超过了159厘米"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】先根据原来5名队员的平均身高求出原来5名队员的总身高，再根据现在6名队员的平均身高求出现在6名队员的总身高，最后用现在的总身高减去原来的总身高，即可求出小奇的身高。\n【详解】原来5名队员的总身高:157×5=785(厘米)\n现在6名队员的平均身高:157+2=159(厘米)\n现在6名队员的总身高:159×6=954(厘米)\n小奇的身高:954-785=169(厘米)\n169厘米\u003e159厘米，所以小奇的身高超过了159厘米。故答案为:D",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·北京朝阳·期末"
+    },
+    {
+        "id":  "159",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·天津南开·期末)表格中是四名同学参加跳绳比赛的成绩，王鹏和李杰的成绩被墨水弄脏了，请推测一下李杰跳了()个.\n![](images/2026-03-14%20163322.jpg)",
+        "options":  [
+                        "A. 82",
+                        "B. 89",
+                        "C. 95",
+                        "D. 85"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】已知四名同学的平均成绩是91个，根据\"平均数×总份数=总数\"，可得四名同学跳绳的总成绩为:91×4=364(个),用总成绩减去张军，刘亮两人的成绩就是王鹏和李杰的成绩和，是364-98-92=174(个),李杰成绩的个位是5，则王鹏成绩的个位应是9，即王鹏成绩是89个，用王鹏和李杰的成绩之和减去王鹏的成绩，可得李杰的成绩.\n【详解】91×4=364(个)364-98-92=266-92=174(个) 174-89=85(个)因此，王鹏跳了89个，李杰跳了85个.\n故答案为:D",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·天津南开·期末"
+    },
+    {
+        "id":  "160",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·湖南长沙·期中)四(1)班学生平均身高1.45米，四(2)班平均身高1.42米，小明在四(1)班，小刚在四(2)班。小明和小刚比,()。",
+        "options":  [
+                        "A. 小明一定更高",
+                        "B. 小刚一定更矮",
+                        "C. 无法确定"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】根据平均数的意义，一组数据的平均数一定大于这组数据中的最小数，一定小于这组数据中的最大数。平均数是指在一组数据中所有数据之和再除以这组数据的个数。平均数反映一组数据的集中趋势的量数，它是反映数据集中趋势的一项指标，不能反映某一个数据的情况。小明的身高可能比他所在班同学的平均身高高，也可能比他所在班同学的平均身高低，小刚的身高可能比他所在班同学的平均身高高，也可能比他所在班同学的平均身高低，所以无法确定小明和小刚谁高谁矮。据此解答即可.\n【详解】四(1)班学生平均身高1.45米，四(2)班平均身高1.42米，小明在四(1)班，小刚在四(2)班。小明和小刚比，无法确定。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·湖南长沙·期中"
+    },
+    {
+        "id":  "161",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·天津和平·期末)在下面的叙述中，正确的有()个。①用3根小棒围一个等腰三角形，如果它的一条边是长8cm的小棒，另一条边是长4cm的小棒，围成这个等腰三角形至少还需要一根长4cm的小棒。②当三角形中两个内角的和等于第三个角时，这是一个钝角三角形。③某小学排球队队员的平均身高是160cm,有的队员身高可能超过160cm,有的队员身高可能不到160cm。",
+        "options":  [
+                        "A. 1",
+                        "B. 2",
+                        "C. 3"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】根据题意，等腰三角形两边之和需大于第三边。若两腰为8cm,底为4cm,满足条件;若两腰为4cm,底为8cm,则4+4=8，无法构成三角形。因此第三根小棒需8cm。若两内角和等于第三个角，则第三个角为90°,三角形为直角三角形，而非钝角三角形。平均数代表整体水平，队员身高可能高于或低于平均值。以此选择即可.\n【详解】根据分析可知:①用3根小棒围一个等腰三角形，如果它的一条边是长8cm的小棒，另一条边是长4cm的小棒，围成这个等腰三角形至少还需要一根长8cm的小棒。原说法错误。②当三角形中两个内角的和等于第三个角时，这是一个直角三角形。原说法错误。③某小学排球队队员的平均身高是160cm,有的队员身高可能超过160cm,有的队员身高可能不到160cm。原说法正确。正确的有1个。故答案为:A",
+        "knowledge_points":  [
+                                 "三角形三边关系",
+                                 "平均数的意义及求法",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·天津和平·期末"
+    },
+    {
+        "id":  "162",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(23-24四年级下·浙江杭州·期末)四位同学在复习相关数学知识，说对的有()人。小优：\"运算律帮助我，计算37×99+37可以转化成37×100计算.\"小伊：\"三角形的知识我学得好，我明白一个三角形至少有2个锐角.\"小奇：\"平均数知识告诉我，身高1.4米的我到平均水深1.2米的水池中游泳不一定安全.\"小轩：\"认识了小数，我知道了4.5和4.50大小相等，计数单位相同.\"",
+        "options":  [
+                        "A. 1",
+                        "B. 2",
+                        "C. 3",
+                        "D. 4"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】小优：根据乘法分配律，两个数的和与一个数相乘，可以用这两个数分别和这个数相乘，再把它们的积相加，积不变;把37化成37×1，再利用乘法分配律计算即可;小伊：直角三角形有1个直角和2个锐角，锐角三角形有3个锐角，钝角三角形有1个钝角和2个锐角，所以三角形最少有2个锐角;小奇：平均数能代表一组数据的整体水平，一组数据中有些数比平均数小，有一些数比平均数大。平均水深1.2米的水池，并不是水池所有地方的水深都是1.2米，有的地方有可能比1.2米深得多，也有可能比1.2米浅得多;小轩：小数的性质：在小数的末尾添上0或去掉0，小数的大小不变。一位小数的计数单位是十分之一，两位小数的计数单位是百分之一，三位小数的计数单位是千分之一。,分别写作0.1,0.01,0.001。,据此解答即可.\n【详解】小优:37×99+37=37×99+37×1=37×(99+1)=37×100=3700，说法正确。小伊：一个三角形至少有2个锐角。说法正确。小奇：身高1.4米的我到平均水深1.2米的水池中游泳不一定安全。说法正确。小轩:4.5=4.50,4.5的计数单位是0.1,4.50的计数单位是0.01。原题说法错误。四位同学在复习相关数学知识，说对的有3人。故答案为:C",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "小数的数位和计数单位的认识",
+                                 "平均数的意义及求法",
+                                 "三角形的分类"
+                             ],
+        "source":  "23-24四年级下·浙江杭州·期末"
+    },
+    {
+        "id":  "163",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(22-23四年级下·新疆昌吉·期末)工人叔叔修机器第一天修了3台，第二天修了4台，第三天上午修了3台，下午修了2台。平均每天修()台。",
+        "options":  [
+                        "A. 2",
+                        "B. 3",
+                        "C. 4",
+                        "D. 12"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】根据题意，把第一天，第二天和第三天修的台数相加，求出3天修的总台数，再用3天修的总台数除以3，即可求出平均每天修多少台.\n【详解】(3+4+3+2)÷3\n=12÷3\n=4(台)\n平均每天修4台。故答案为:C",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "22-23四年级下·新疆昌吉·期末"
+    },
+    {
+        "id":  "164",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(2024·新疆吐鲁番·小升初真题)如果1,20和a这三个数的平均数是11，那么a是()。",
+        "options":  [
+                        "A. 10",
+                        "B. 11",
+                        "C. 12"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】已知三个数的平均数是11，用11乘3得出三个数的和，再减去1和20，即可求出a.\n【详解】11×3=33\n33-1-20=32-20\n=12\n所以a是12。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "2024·新疆吐鲁番·小升初真题"
+    },
+    {
+        "id":  "165",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "(24-25四年级下·湖南岳阳·期中)四年级学生平均体重36千克，下列说法正确的是()。",
+        "options":  [
+                        "A. 每个学生体重都是36千克",
+                        "B. 可能有学生体重超过36千克",
+                        "C. 所有学生体重都低于36千克"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】平均数表示一组数据的一般水平，有的数据可能比平均值大，也有可能比平均值小，还有可能等于平均值。据此解答.\n【详解】A。每个学生体重都是36千克，不符合实际情况，不可能所有人体重是相同的，不符合题意;B。因为平均体重的36千克，所以有的人体重低于36千克，有的人体重高于36千克，符合题意;C。所有学生体重都低于36千克，则平均体重也一定会低于36千克，不符合题意。因此，说法正确的是可能有学生体重超过36千克。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·湖南岳阳·期中"
+    },
+    {
+        "id":  "166",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(23-24四年级下·广东湛江·期末)男生小东和女生小丽都是四(1)班的学生，该班男生的平均体重是33千克，女生的平均体重是31千克，关于小东，小丽的体重，下面的说法正确的是()。",
+        "options":  [
+                        "A. 小东比小丽重",
+                        "B. 小东比小丽轻",
+                        "C. 小东和小丽的体重相等",
+                        "D. 不能确定谁重谁轻"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】平均数代表的是一组数据的一般水平。由题意得，四(1)班男生的平均体重是33千克。小东是男生，小东有可能比33千克重，也有可能比33千克轻，还有可能等于33千克。四(1)班女生的平均体重是31千克，小丽是女生，小丽有可能比31千克重，也有可能比31千克轻，还有可能等于31千克。所以无法确定小东和小丽谁重谁轻.\n【详解】由分析得，无法确定小东和小丽谁重谁轻。故答案为:D",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·广东湛江·期末"
+    },
+    {
+        "id":  "167",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(2024四年级·广东深圳·竞赛)有两组数，第一组9个数的和是63，第二组的平均数是11，两个组所有数的平均数是8，那么，第二组有()个数。",
+        "options":  [
+                        "A. 6",
+                        "B. 5",
+                        "C. 4",
+                        "D. 2",
+                        "E. 3"
+                    ],
+        "answer":  "E",
+        "solution":  "【分析】可设第二组有x个数，根据平均数的概念，建立关于x的方程，解方程即可.\n【详解】解：设第二组有x个数.\n63+11x=8×(9+x)\n63+11x=72+8x\n11x-8x=72-63\n3x=9\nx=3\n所以第二组有3个数。故答案选:E.",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "2024四年级·广东深圳·竞赛"
+    },
+    {
+        "id":  "168",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(2025四年级下·海南海口·专题练习)下面正确的是()。",
+        "options":  [
+                        "A. 0.8和0.80的大小相等，计数单位也相同。",
+                        "B. 被减数=减数-差。",
+                        "C. 一个三角形中最大的角是79°,这个三角形一定是锐角三角形。",
+                        "D. 学校举行跳绳比赛，笑笑所在班级平均每人1分钟跳135次，乐乐所在班级平均每人1分钟跳129次，所以笑笑跳的次数比乐乐多。"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】A。小数的末尾添上\"0\"或去掉\"0\"，小数的大小不变;一位小数的计数单位是0.1，两位小数的计数单位是0.01，依此判断;B。根据减法关系：被减数-减数=差，可推导出：被减数=减数+差;C。三个角均小于90°的三角形是锐角三角形。已知最大角为79°(\u003c90°),则其他两个角必然也小于79°,因此三个角均为锐角.\nD。平均数是反映一组数据的平均水平，并不能反映这组数据中各个数据的大小，依此判断.\n【详解】A。0.8和0.80的大小相等，计数单位不同，不符合题意;B。被减数=减数+差，不符合题意;C。一个三角形中最大的角是79°,这个三角形一定是锐角三角形，符合题意;D。学校举行跳绳比赛，笑笑所在班级平均每人1分钟跳135次，乐乐所在班级平均每人1分钟跳129次，笑笑可能低于班级平均分，乐乐可能高于班级平均分，因此无法确定两人具体次数的大小关系。故答案为:C",
+        "knowledge_points":  [
+                                 "加，减法的意义和各部分间的关系",
+                                 "小数的数位和计数单位的认识",
+                                 "平均数的意义及求法",
+                                 "三角形的分类"
+                             ],
+        "source":  "2025四年级下·海南海口·专题练习"
+    },
+    {
+        "id":  "169",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(2024·北京房山·小升初真题)四(1)班男生的平均身高是142厘米，下面说法正确的是()。①亮亮的身高是151厘米，在四(1)班男生中个子是偏高的。②四(1)班一定有某位男生的身高正好是142厘米。③四(1)班新转来一位男同学身高是137厘米，现在全班男生的平均身高会比142厘米低。④四(1)班女生的平均身高是144厘米，红红是女生，她一定比本班所有男同学都高。",
+        "options":  [
+                        "A. 只有①",
+                        "B. 只有①③",
+                        "C. 只有①②③",
+                        "D. 只有①③④"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】①平均身高为142厘米，亮亮身高151厘米明显高于平均水平，属于偏高.\n②平均数的计算并不要求有人的身高恰好等于平均值，可能存在所有人身高都高于或低于平均值的情况.\n③加入一个低于平均值的数(137厘米)会使新的平均值下降.\n④女生平均身高高于男生，并不能保证每个女生都高于所有男生，存在个别差异的可能.\n【详解】①亮亮的身高是151厘米，明显高于平均水平,151厘米属于偏高的，原题说法正确.\n②四(1)班一定有某位男生的身高正好是142厘米。这种情况不一定，所以此项错误.\n③四(1)班新转来一位男同学身高是137厘米,137厘米低于原来的平均身高(142厘米),会使新的平均值下降，原说法正确.\n④四(1)班女生的平均身高是144厘米，并不能保证每个女生都高于所有男生。原题说法不准确.\n所以只有①③正确。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "2024·北京房山·小升初真题"
+    },
+    {
+        "id":  "170",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "(23-24四年级下·海南省直辖县级单位·期末)小乐语文测试94分，数学96分，英语98分，三科的平均分是()分。",
+        "options":  [
+                        "A. 90",
+                        "B. 92",
+                        "C. 94",
+                        "D. 96"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】先用加法求出三科的总分数，进而根据\"总成绩/科目数=平均分\"解答即可.\n【详解】(94+96+98)÷3\n=288÷3\n=96(分)\n三科的平均分是96分。故答案为:D",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·海南省直辖县级单位·期末"
+    },
+    {
+        "id":  "171",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·河北唐山·期末)程诚所在小组同学的平均身高是162厘米，程诚的身高一定是162厘米。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数是所有数据的和除以数据的个数，它反映的是整体水平，并不代表每个个体的数值都等于平均数.\n【详解】根据分析可知，程诚所在小组同学的平均身高是162厘米，并不代表程诚的身高是162厘米，程诚的身高可能高于162厘米，也可能低于162厘米，所以程诚的身高不一定是162厘米，原题表达错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·河北唐山·期末"
+    },
+    {
+        "id":  "172",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·广东汕头·期末)运动员的平均身高是175cm,那么运动员中身高不可能有190cm。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数是所有数据的总和除以数据的个数，并不能限制单个数据的大小。若存在身高较高的运动员，只要其他运动员的身高较低，总和仍可满足平均数。因此，运动员中可能有身高190cm的人.\n【详解】运动员的平均身高是175cm,那么运动员中身高可能有190cm,原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·广东汕头·期末"
+    },
+    {
+        "id":  "173",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·黑龙江·期末)王悦5次跳远的总成绩是10m,她每次跳远的成绩肯定都是2m。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】根据平均数的定义，平均每次跳远的成绩等于总成绩除以跳远次数，题干中，总成绩是10m,共跳了5次，所以用总成绩除以跳远次数可得平均成绩为10÷5=2(m)。平均成绩是所有成绩的平均值，不代表每次的具体成绩，每次成绩可能高于或低于平均值，可能存在某次跳3m,另一次跳1m,其他三次跳2m的情况，总成绩仍为10m,据此解答即可.\n【详解】王悦5次跳远的总成绩是10m,她每次跳远的成绩不一定是2m。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·黑龙江·期末"
+    },
+    {
+        "id":  "174",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "(24-25四年级下·山东日照·期末)小新4次跳高的平均成绩是1.1米，他一定有某一次跳的低于1.1米。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数是表示一组数据的平均值，是指在一组数据中所有数据之和再除以这组数据的个数。这组数据中部分数据可能等于平均数，可能小于平均数，可能大于平均数;据此判断.\n【详解】假设小新4次跳高成绩均为1.1米，则总成绩为4.4米，平均成绩为1.1米。小新4次跳高的平均成绩是1.1米，他可能有某一次跳的低于1.1米，也可能所有成绩都等于1.1米。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·山东日照·期末"
+    },
+    {
+        "id":  "175",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "(24-25四年级下·山西长治·期末)中国男性平均寿命大约是75岁，小吉的爷爷一定会活到75岁。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】根据平均数的特点，它反映一组数据的整体情况，处在这组数据的中间，比最大的数据小，比最小的数据大，也就是说这组数据中有比平均数大的数，也有比平均数小的数。所以，根据题意，中国男性平均寿命大约是75岁，则小吉的爷爷可能活得比75岁长，也可能活得比75岁短。据此判断.\n【详解】根据分析可知：中国男性平均寿命大约是75岁，小吉的爷爷不一定会活到75岁。原题说法错误.\n故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·山西长治·期末"
+    },
+    {
+        "id":  "176",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·河南驻马店·期末)学校绿化带中树木的平均高度是182cm,绿化带中不可能有157cm高的树。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数：一组数据的总和除以这组数据个数所得到的商叫这组数据的平均数。平均数易受极端值的影响，这里的极端值是指偏大或偏小数，当出现偏大数时，平均数将会被抬高，当出现偏小数时，平均数会降低.\n【详解】学校绿化带中树木的平均高度是182cm,绿化带中树木的高度有可能高于182cm,也可能低于182cm,所以绿化带中可能有157cm高的树。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·河南驻马店·期末"
+    },
+    {
+        "id":  "177",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "(24-25四年级下·陕西商洛·期末)乐乐家今年前半年平均每月用水量是8吨，那么乐乐家5月份的用水量不可能比8吨多。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数代表的是一组数据的一般水平。它一般比这组数据的最小值大，一般比这组数据的最大值小。由题意得，乐乐家今年前半年平均每月用水量是8吨，那么乐乐家5月份的用水量可能比8吨多，可能比8吨少，还有可能等于8吨.\n【详解】由分析得，乐乐家今年前半年平均每月用水量是8吨，那么乐乐家5月份的用水量可能比8吨多。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·陕西商洛·期末"
+    },
+    {
+        "id":  "178",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·贵州遵义·期末)平均数与平均分表示的意思是一样的。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数和平均分虽然计算方式相同，但表示的意义不同。平均数是统计量，反映数据整体水平，结果可为小数;平均分是分配结果，通常要求每份为整数，除非题目允许拆分.\n【详解】平均数的定义是一组数据的总和除以数据的个数，用于表示这组数据的整体水平，例如:3人的年龄分别为10岁,12岁,14岁，平均年龄为(10+12+14)÷3=12(岁)。平均分则是将物品均分到若干份，每份数量需为整数(如15个苹果分给3人，每人分15÷3=5(个))。虽然计算过程类似，但平均数可以是虚拟值，而平均分强调实际分配，因此两者意义不同。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·贵州遵义·期末"
+    },
+    {
+        "id":  "179",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "(24-25四年级下·江西吉安·期末)某小组跳绳成绩(单位：下):120,135,140,140,130，若下一个人跳了135下，这组的平均数变小了。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】先用这五人跳绳成绩相加，再除以5，求出原来数据的平均数。若下一个人跳绳成绩大于原来数据的平均数，则这组的平均数变大。若下一个人跳绳成绩等于原来数据的平均数，则这组的平均数不变。若下一个人跳绳成绩小于原来数据的平均数，则这组的平均数变小.\n【详解】(120+135+140+140+130)÷5\n=(255+140+140+130)÷5\n=(395+140+130)÷5\n=(535+130)÷5\n=665÷5\n=133(下)\n135\u003e133，所以，平均数变大了。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·江西吉安·期末"
+    },
+    {
+        "id":  "180",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "(24-25四年级下·河南新乡·期末)四(1)班同学的平均体重是30kg,李明是四(1)班的同学，他的体重不可能超过30kg。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均体重是全体同学体重的总和除以人数，代表整体水平，并不限制个别数据的大小。可能存在体重超过平均数的同学.\n【详解】四(1)班同学的平均体重为30kg,表示全体同学的体重总和除以人数等于30kg。若部分同学体重低于30kg,则必然有同学体重高于30kg才能使平均数保持30kg。例如：若甲，乙，丙三人体重分别为28kg,30kg,32kg,平均体重为(28+30+32)÷3=90÷3=30kg,此时丙的体重超过30kg。因此，李明的体重可能超过30kg,原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·河南新乡·期末"
+    },
+    {
+        "id":  "181",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·河南新乡·期末)跳绳比赛，跳最多的同学可以获得个人冠军，第一组平均成绩是165下，第二组平均成绩是179下，个人冠军一定是第二组中的同学。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数是反映一组数据的整体水平，不能确定个体成绩。第一组平均成绩是165下，第一组有的同学的成绩比165下多，有的比165下少，有的等于165下;第二组平均成绩是179下，第二组有的同学的成绩比179下多，有的比179下少，有的等于179下。所以无法确定个人冠军在哪个组.\n【详解】由分析得，第一组平均成绩是165下，第二组平均成绩是179下，无法确定个人冠军在哪个组。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·河南新乡·期末"
+    },
+    {
+        "id":  "182",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·河北沧州·期末)小亮身高1.5米，要蹚过平均水深为1.3米的河，不会有危险。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数代表的是一组数据的一般水平。它一般比这组数据的最小值大，一般比这组数据的最大值小。由题意得，一条河的平均水深为1.3米，那么有的地方比1.3米浅，有的地方比1.3米深。若某处水深超过1.5米，那么小亮就会有危险.\n【详解】由分析得，一条河的平均水深为1.3米，部分区域的实际水深可能大于1.3米。若某处水深超过小亮的身高1.5米，他将无法安全趟过。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·河北沧州·期末"
+    },
+    {
+        "id":  "183",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·陕西渭南·期末)四年级一班第一小组5个人的平均身高是1.51米，组员王丽的身高一定是1.51米。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数代表的是一组数据的一般水平。它一般比这组数据的最小值大，一般比这组数据的最大值小。由题意得，四年级一班第一小组5个人的平均身高是1.51米，那么组员王丽的身高可能比1.51米高，可能比1.51米矮，还有可能等于1.51米.\n【详解】由分析得，四年级一班第一小组5个人的平均身高是1.51米，那么组员王丽的身高可能比1.51米高，可能比1.51米矮，还有可能等于1.51米。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·陕西渭南·期末"
+    },
+    {
+        "id":  "184",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "(24-25四年级下·江西·期末)小明所在班级的平均身高是140厘米，小强所在班级的平均身高是145厘米，小明一定比小强矮。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均身高反映的是班级整体的平均水平，并不能确定每个个体的具体身高。小明所在班级的平均身高较低，但小明个人的身高可能高于班级平均;同理，小强所在班级的平均身高较高，但小强个人的身高可能低于班级平均。因此，无法通过班级平均身高直接推断两人身高的具体关系。据此进行判断即可.\n【详解】班级平均身高是全体成员身高的平均值，不代表每个人的具体身高。例如：小明班级的平均身高为140厘米，但小明可能身高150厘米;小强班级的平均身高为145厘米，但小强可能身高142厘米。此时小明比小强高。因此，\"小明一定比小强矮\"的结论不成立。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·江西·期末"
+    },
+    {
+        "id":  "185",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·河南新乡·期末)学校为足球队买球鞋，可以按队员平均脚长来买。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数反映一组数据的整体水平，它处在这组数据的中间，比最大的数据小，比最小的数据大;根据题意，如果学校为足球队买球鞋，按队员平均脚长来买，则脚长比平均脚长更长或更短的学生不适合穿这双球鞋，所以不能按队员平均脚长来买。据此判断.\n【详解】根据分析可知：学校为足球队买球鞋，不可以按队员平均脚长来买。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·河南新乡·期末"
+    },
+    {
+        "id":  "186",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·河南周口·期末)四年级某班体育测试的平均分是90分，可能有人得60分。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】平均数是一组数的总和除以这组数的数量。平均数反映的是一组数据的平均水平，平均数会受偏大或偏小数据的影响.\n【详解】根据解析可知，平均成绩是90分，则有人成绩可能比90分高，也有可能比90分低，也有可能正好是90分;所以原表达正确。故答案为:√",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·河南周口·期末"
+    },
+    {
+        "id":  "187",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "(24-25四年级下·河南·期末)在一组数据中，平均数可能比最大的数要大。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数是一组数据的总和除以数据的数量的结果，所以这组数据中的数要么比平均数大，要么比平均数小，或者和平均数相等.\n【详解】例如:3,4,5这三个数的平均数是(3+4+5)÷3=12÷3=4,4\u003c5。所以在一组数据中，平均数不可能比最大的数要大。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·河南·期末"
+    },
+    {
+        "id":  "188",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "(24-25四年级下·江西南昌·期末)\"五一\"假期泰山景区日均游客量8万人次，说明5月1日当天是8万人入园。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数是所有数据的总和除以数据的个数，它反映的是整体水平，不能直接推断某一天的具体数值。题目中给出\"日均游客量8万人次\"，只能说明整个假期的总游客量除以天数等于8万，但无法确定某一天(如5月1日)的游客量是否正好等于平均值.\n【详解】日均游客量8万并不代表5月1日当天的游客量就是8万,5月1日当天的游客量可能高于8万，也可能低于8万。所以，题目说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·江西南昌·期末"
+    },
+    {
+        "id":  "189",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "(24-25六年级下·云南昭通·期中)一个水池平均水深1.1米，这个水池有的地方水深可能超过1.1米，有的地方水深可能不足1.1米。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】已知水池平均水深1.1米，平均水深是通过水池各个地方水深相加后除以测量点数得到的。若存在部分区域深度大于1.1米，则必然存在其他区域深度小于1.1米，才能保证平均值为1.1米.\n【详解】平均数是一组数据的总和除以数据个数所得，平均数反映的是一组数据的总体情况，不代表每个数据都相同。在这个水池中：有的地方水深可能大于1.1米，比如某些较深的区域;有的地方水深可能小于1.1米，比如一些较浅的角落。因此，原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25六年级下·云南昭通·期中"
+    },
+    {
+        "id":  "190",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·黑龙江哈尔滨·期末)平均数一定大于最小数，小于最大数。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数通常在这组数据的最大数和最小数之间，但当一组数据都相等时，平均数就等于最大数，也等于最小数，据此判断.\n【详解】例如:4,4,4这组数据的平均数是(4+4+4)÷3=4，平均数等于最大数，也等于最小数。所以“平均数一定大于最小数，小于最大数”这种说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "191",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "(24-25四年级下·山东济宁·期末)一个水坑平均水深是1.3米，明明身高150厘米，他在水坑里游泳可能会有危险。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】平均水深1.3米表示水坑各处的深度可能高于或低于1.3米。1米=100厘米，据此将明明身高换算成1.5米，若水坑某处深度超过1.5米，则他可能无法触底，存在危险.\n【详解】150厘米=1.5米\n一个水坑平均水深是1.3米，明明身高150厘米，他在水坑里游泳可能会有危险。这句话正确。故答案为:√",
+        "knowledge_points":  [
+                                 "毫米，厘米，分米，米之间的进率与换算，平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "192",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·广东汕头·期末)一条小河平均水深1.2m,李泳身高1.48m,他趟水过河没有危险。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】根据平均数的意义，一组数据的平均数一定大于这组数据中的最小数，一定小于这组数据中的最大数。平均数反映一组数据的集中趋势的量数，它是反映数据集中趋势的一项指标，不能反映某一个数据的情况，小河平均水深1.2m,并不能反映出整个小河中每一处水的深浅，有的地方可能会比1.2m深，有的地方可能会比1.2m浅，有的地方可能等于1.2m,甚至有的地方最深处可能大于1.48m,所以身高1.48m的李泳趟水过河，可能会有危险.\n【详解】一条小河平均水深1.2m,李泳身高1.48m,他趟水过河可能有危险。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·广东汕头·期末"
+    },
+    {
+        "id":  "193",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·甘肃陇南·期末)小强身高1.4米，他在平均水深为1.1米的河里游泳，不会有危险。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均水深1.1米表示整条河水深的平均值，但实际各区域水深可能高于或低于平均值。小强身高1.4米，若某处水深超过1.4米，则存在危险.\n【详解】平均水深1.1米不能保证所有区域水深均小于或等于1.1米，可能存在部分区域水深超过1.4米，因此小强游泳可能有危险。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·甘肃陇南·期末"
+    },
+    {
+        "id":  "194",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·黑龙江哈尔滨·期末)学校篮球队补选新队员。原来老队员的平均身高是170厘米，补选的两名队员身高分别为169厘米和173厘米。补选后篮球队队员的平均身高比原来增高了。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】用这组数据的和除以数据的个数，就是平均数，假设原来老队员有8人，原来老队员的总身高=平均身高×人数，再求出补选后篮球队队员的总身高，再除以现在的总人数，求出补选后篮球队队员的平均身高，最后比较大小，据此解答.\n【详解】假设原来老队员有8人\n170×8=1360(厘米)\n(1360+169+173)÷(8+2)\n=1702÷10\n=170.2(厘米)\n因为170.2厘米\u003e170厘米，所以现在篮球队队员的平均身高增高了。故答案为:√",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "195",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·内蒙古赤峰·期末)某一小组同学的平均体重是36kg,其中飞飞的体重一定是36kg。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数是指在一组数据中所有数据之和再除以这组数据的个数，它反映的是一组数据的平均水平，但并不代表这组数据中的每一个数据都等于平均数。平均体重是36kg表示全组同学的体重总和除以总人数等于36kg,但每个同学的体重可能高于，低于或等于平均数，不能确定飞飞的体重一定是36kg.\n【详解】例如：若小组有3人，体重分别为34kg,36kg,38kg,则平均体重为(34+36+38)÷3=108÷3=36(kg),但飞飞的体重可能是34kg,36kg或38kg中的任意一个。因此，飞飞的体重不一定是36kg。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·内蒙古赤峰·期末"
+    },
+    {
+        "id":  "196",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·浙江宁波·期末)小红在期末考试中语文，数学，英语的平均分是96分，如果学科成绩达到90分以上可以得星，小红一定可以得3颗星。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】小红的总分=三科的平均分*3=288分，假设其中两门都是100分，则第三科的分数=288-100-100=88分\u003c90分，所以小红不一定得3颗星.\n【详解】96×3=288(分)假设其中两门都是100分，则288-100-100=88(分),88分\u003c90分，小红不一定得3颗星。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·浙江宁波·期末"
+    },
+    {
+        "id":  "197",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·广东江门·期末)明明4次跳绳的总成绩是600下，她每次跳的都是150下。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】总成绩除以跳的次数等于平均每次跳绳的成绩，平均数介于一组数据的最大值和最小值之间，并不代表每次的实际成绩都等于这个数值，可能存在某些次数高于或低于平均数的情况，据此解答即可.\n【详解】600÷4=150(下),明明4次跳绳的总成绩是600下，平均每次跳150下，但每次不一定都跳150下。原题说法错误。故答案为: ×。",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·广东江门·期末"
+    },
+    {
+        "id":  "198",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·内蒙古呼伦贝尔·期末)要衡量两个人数不同的小组的数学成绩，可以用平均分来进行比较。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】平均数代表的是一组数据的一般水平。由题意得，两个小组的人数不同，所以他们的总成绩不具备可比性，但平均分将总成绩均摊到每个成员，能有效消除人数差异带来的影响。所以可以通过平均分来比较两个小组的数学成绩.\n【详解】由分析得，要衡量两个人数不同的小组的数学成绩，可以用平均分来进行比较。原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·内蒙古呼伦贝尔·期末"
+    },
+    {
+        "id":  "199",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "(18-19四年级下·江苏·单元测试)用计算器算出下面这组数的平均数\n98,84,120,244,360,188,429,101",
+        "options":  [
+
+                    ],
+        "answer":  "203",
+        "solution":  "【分析】本题考查平均数的计算。平均数等于一组数据的总和除以数据的个数。先将所有数相加，再除以8即可.\n【详解】(98+84+120+244+360+188+429+101)÷8=1624÷8=203",
+        "knowledge_points":  [
+                                 "平均数的意义及求法，计算器与复杂的运算"
+                             ],
+        "source":  "18-19四年级下·江苏·单元测试"
+    },
+    {
+        "id":  "200",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "(2014三年级·全国·课后作业)33,42,20,53,32这5个数的平均数是多少。33,42,20,53,32,36这6个数的平均数是多少。两个平均数相差多少。",
+        "options":  [
+
+                    ],
+        "answer":  "36;36;0",
+        "solution":  "【分析】本题考查平均数的计算。平均数等于一组数据的总和除以数据的个数。先分别计算两组数的总和，再除以对应的个数，最后求两个平均数的差.\n【详解】试题分析:(1)先求出5个数的和，进而根据\"总数/个数=平均数\"进行解答即可;(2)先求出6个数的和，进而根据\"总数/个数=平均数\"进行解答即可;(3)求两个平均数相差多少，用减法解答即可.\n解:(1)(33+42+20+53+32)÷5\n=180÷5\n=36\n答：这5个数的平均数是36。(2)(33+42+20+53+32+36)÷6\n=216÷6\n=36\n答：这6个数的平均数是36。(3)36-36=0\n答：两个平均数相差0。",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "2014三年级·全国·课后作业"
+    },
+    {
+        "id":  "201",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(23-24四年级下·河北保定·期末)第一小组的学生称体重，最重的45千克，最轻的23千克。下面哪个数量有可能是这组学生的平均体重。()。",
+        "options":  [
+                        "A. 46千克",
+                        "B. 45千克",
+                        "C. 32千克",
+                        "D. 23千克"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】根据平均数的意义及求解方法可知，平均数比这组数据的最大数据小，比最小数据大，据此即可选择.\n【详解】A。46千克\u003e45千克，即46千克大于最重体重，因此46千克不可能是平均体重;B。45千克=45千克，即45千克等于最重体重，因此45千克不可能是平均体重;C。32千克\u003c45千克,32千克\u003e23千克，即32千克介于最轻体重和最重体重之间，因此32千克可能是平均体重;D。23千克=23千克，即23千克等于最轻体重，因此23千克不可能是平均体重.\n故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·河北保定·期末"
+    },
+    {
+        "id":  "202",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(23-24四年级下·浙江温州·期末)四(1)班全班同学的平均身高是145厘米。下面说法正确的是()。",
+        "options":  [
+                        "A. 小明同学的身高是145厘米，那么小明是四(1)班的同学",
+                        "B. 四(1)班至少有一个同学的身高是145厘米",
+                        "C. 四(1)班不可能有身高是160厘米的同学",
+                        "D. 如果这时四(1)班调来一位身高150厘米的同学，那么他们的平均身高就增加了"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】平均数表示一组数据的一般水平。由题意得，四(1)班全班同学的平均身高是145厘米，那么四(1)班同学的身高可能比145厘米高，也有可能比145厘米矮，还有可能就等于145厘米。据此解答.\n【详解】A。小明同学的身高是145厘米，他有可能是四(1)班的同学，还有可能是其它班的同学。该选项说法错误.\nB。四(1)班同学的身高可能比145厘米高，也有可能比145厘米矮，有可能全班没有一个人的身高等于145厘米。该选项说法错误.\nC。四(1)班同学的身高可能比145厘米高，即四(1)班可能有身高是160厘米的同学。该选项说法错误.\nD。如果这时四(1)班调来一位身高150厘米的同学，这个同学的身高高于原本的平均身高，所以此时四(1)班的平均身高就增加了。该选项说法正确.\n故答案为:D",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·浙江温州·期末"
+    },
+    {
+        "id":  "203",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·海南省直辖县级单位·期中)小明看一本故事书，前3天看的72页，后5天看了80页，小明平均每天看多少页。正确算式是()。",
+        "options":  [
+                        "A. 72÷3",
+                        "B. 80÷5",
+                        "C. 72÷3+80÷5",
+                        "D. (72+80)÷(5+3)"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】根据题意，用看的总页数除以总天数就是平均每天看多少页。而这本故事书看的总页数是(72+80)页，总天数是(3+5)天。据此解答.\n【详解】根据分析，求小明平均每天看多少页，列算式是(72+80)÷(5+3)。故答案为:D",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·海南省直辖县级单位·期中"
+    },
+    {
+        "id":  "204",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "(24-25四年级下·全国·单元测试)李乐，韦晨，刘微的平均年龄是15岁，其中一人19岁，另外两个人中不可能有()岁的。",
+        "options":  [
+                        "A. 12",
+                        "B. 26",
+                        "C. 24"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据题意，平均数反应的是一组数据的总体情况。求平均数可以用总数/份数=平均数。那么用平均数×份数=总数，据此解答.\n【详解】15×3=45(岁)45-19=26(岁)三人的总岁数是45岁，其中一人19岁，另外两人的岁数之和是26岁。所以，另外两个人中不可能有人是26岁.\n故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·全国·单元测试"
+    },
+    {
+        "id":  "205",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(23-24六年级下·江西南昌·期中)已知一组数据28,a,32,24的平均数是24，那么a的值是()。",
+        "options":  [
+                        "A. 18",
+                        "B. 16",
+                        "C. 22",
+                        "D. 12"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】根据总数/份数=平均数可知：平均数×份数=总数，用平均数24乘数字的个数4，即得到这组数据的总数，再用总数减去已知的3个数，所得的差就是a的值。据此解答.\n【详解】24×4-28-32-24=96-28-32-24=12所以,a的值是12。故答案为:D",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24六年级下·江西南昌·期中"
+    },
+    {
+        "id":  "206",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "(23-24六年级下·江西南昌·期中)小强所在班级学生的平均身高是1.4米，小刚所在班级学生的平均身高是1.5米，小强()比小刚矮。",
+        "options":  [
+                        "A. 一定",
+                        "B. 一定不",
+                        "C. 可能",
+                        "D. 不可能"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】平均数是表示一组数据集中趋势的量数，它是反映数据集中趋势的一项指标，通过平均数并不能知道这组数据中具体的某个数值;据此解答即可.\n【详解】小强所在班级的平均身高为1.4米，但小强个人的身高可能高于或低于1.4米;同理，小刚所在班级的平均身高为1.5米，小刚的身高也可能高于或低于1.5米。因此，存在以下三种情况:1，小强可能比小刚矮：例如小强身高1.4米，小刚身高1.5米;2，小强可能不比小刚矮：例如小强身高1.6米，小刚身高1.45米;3，两人的身高可能相同。所以小强可能比小刚矮。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24六年级下·江西南昌·期中"
+    },
+    {
+        "id":  "207",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·湖南岳阳·期中)小明4次数学测验平均分89分，第5次得99分，他这5次的平均分是()。",
+        "options":  [
+                        "A. 90分",
+                        "B. 91分",
+                        "C. 92分"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据\"小明前4次数学测验的平均成绩为89分\"，可以求出4次的总成绩为89×4，将前面4次总成绩加上第5次总成绩即可得到5次的总成绩，根据\"平均数=总数量/总份数\"求解即可.\n【详解】(89×4+99)÷5=(356+99)÷5=455÷5=91(分)小明4次数学测验平均分89分，第5次得99分，他这5次的平均分是91分。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·湖南岳阳·期中"
+    },
+    {
+        "id":  "208",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(23-24四年级下·福建厦门·期末)小明5次数学考试的成绩各不相同，去掉最高的一次成绩后，剩下的平均成绩跟原来5次的平均成绩相比,()。",
+        "options":  [
+                        "A. 提高了",
+                        "B. 降低了",
+                        "C. 不变",
+                        "D. 都有可能"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】平均的过程其实是\"移多补少\"的过程，大数需要拿出多的部分来补给小数，使它们变的一样多。去掉最高的一次成绩后，就会从第二高的成绩里面拿出多的部分补给低的成绩，使平均数变小.\n【详解】根据分析可知：小明5次数学考试的成绩各不相同，去掉最高的一次成绩后，剩下的平均成绩跟原来5次的平均成绩相比，降低了。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·福建厦门·期末"
+    },
+    {
+        "id":  "209",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(23-24四年级下·宁夏固原·期末)朵朵家今年前半年平均每月用水量是32吨，下列说法中不合理的是()。",
+        "options":  [
+                        "A. 3月份的用水量可能是32吨",
+                        "B. 4月份的用水量可能比32吨少",
+                        "C. 5月份的用水量不可能比32吨多"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】平均数代表的是一组数据的一般水平。它一般比这组数据的最小值大，一般比这组数据的最大值小。据此解答.\n【详解】A。朵朵家今年前半年平均每月用水量是32吨，朵朵家3月份的用水量可能是32吨。该选项说法正确B。朵朵家今年前半年平均每月用水量是32吨，朵朵家4月份的用水量可能比32吨少。该选项说法正确。C。朵朵家今年前半年平均每月用水量是32吨，朵朵家5月份的用水量可能比32吨多，也有可能比32吨少，还有可能等于32吨。该选项说法错误。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·宁夏固原·期末"
+    },
+    {
+        "id":  "210",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(23-24四年级下·河北廊坊·期末)下面信息中适合用复式条形统计图的是()。",
+        "options":  [
+                        "A. 5名同学的身高",
+                        "B. 某超市一周的销售额",
+                        "C. 四年级各班男，女生人数"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】复式条形统计图的特点是可以同时表示多个类别或组的数据，并且能够清晰地对比不同类别或组之间的数据差异.\n【详解】A。只需用单式条形统计图展示这5名同学各自的身高数据即可，不需要对比不同类别，不适合用复式条形统计图。B。通常用单式条形统计图来展示每天的销售额，无需使用复式条形统计图。C。这里有两个类别，即男生和女生，需要同时展示每个班级男女生人数的对比情况，使用复式条形统计图可以清晰地看出各班男女生人数的多少以及班级之间男女生人数的差异，所以适合用复式条形统计图。故答案为:C",
+        "knowledge_points":  [
+                                 "复式条形统计图"
+                             ],
+        "source":  "23-24四年级下·河北廊坊·期末"
+    },
+    {
+        "id":  "211",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·全国·单元测试)李昊对自制的机器人进行攀爬能力测试，发现6分钟到达测试点，每分钟爬150米。原路返回需要4分钟。机器人上山，下山平均每分钟爬()米。",
+        "options":  [
+                        "A. 187",
+                        "B. 160",
+                        "C. 180"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】根据题意，用上山每分钟爬行的米数乘上山的时间就是上山的路程。下山的路程和上山的路程一样。那么总路程就是上山和下山的路程之和，时间就是上山和下山的时间和。用总路程除以时间和就是上山，下山平均每分钟爬多少米.\n【详解】150×6=900(米)900×2=1800(米)6+4=10(分)1800÷10=180(米)所以，机器人上山，下山平均每分钟爬180米。故答案为:C",
+        "knowledge_points":  [
+                                 "被除数和除数末尾都有0的除法",
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·全国·单元测试"
+    },
+    {
+        "id":  "212",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "(24-25四年级下·全国·单元测试)影影所在班级同学的平均身高是150厘米，青青所在班级同学的平均身高是156厘米，则影影和青青的身高相比,()。",
+        "options":  [
+                        "A. 影影比青青高",
+                        "B. 青青比影影高",
+                        "C. 影影和青青一样高",
+                        "D. 无法比较"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】一组数据的总和除以这组数据个数所得到的商叫这组数据的平均数。平均身高是指所有同学的身高相加除以人数，反应的是这个班级的整体情况，并不能知道影影和青青各自的准确身高，无法比较.\n【详解】影影所在班级同学的平均身高是150厘米，青青所在班级同学的平均身高是156厘米，影影的身高和青青的身高无法比较。故答案为:D",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·全国·单元测试"
+    },
+    {
+        "id":  "213",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(2024·江西赣州·小升初真题)小东所在小组同学的平均体重是34千克，下面说法合理的是()。",
+        "options":  [
+                        "A. 不可能有体重50千克的同学。",
+                        "B. 大部分同学的体重在30~90千克之间。",
+                        "C. 小组一半同学的体重低于34千克，一半同学的体重高于34千克。"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】平均数反映一组数据的平均情况，在一组数据中，有的数据可能会大于平均数，有的数据可能会小于平均数，有的数据可能会等于平均数，据此分析解答.\n【详解】A。可能有体重50千克的同学，原题说法错误，故不符合题意;B。大部分同学的体重在30~90千克之间，原题说法正确，故符合题意;C。并不能确定小组一半同学的体重低于34千克，一半同学的体重高于34千克，原题说法错误，故不符合题意。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "2024·江西赣州·小升初真题"
+    },
+    {
+        "id":  "214",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(2024·广西南宁·小升初真题)在30秒快速跳绳测试中，小青前几轮的平均成绩是80次，再测一轮后，所有轮次的平均成绩提高到82次。最后一轮测试的成绩可能是()次。",
+        "options":  [
+                        "A. 78",
+                        "B. 80",
+                        "C. 82",
+                        "D. 90"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】小青前几轮的平均成绩是80次，再测一轮后，所有轮次的平均成绩提高到82次，最后轮的成绩一定高于前几次的平均成绩，并且高出的分数大于2次，据此解答.\n【详解】A。78\u003c80，不符合;B。80=80，不符合;C。82\u003e80，刚好高出2次，不符合;D。90\u003e80，高出的分数大于2次，符合。最后一轮测试的成绩可能是90次。故答案为:D",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "2024·广西南宁·小升初真题"
+    },
+    {
+        "id":  "215",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(23-24六年级下·河北保定·期末)下面是某班同学的身高统计表。估计该班同学的平均身高是().\n| 身高/cm | 141-145 | 146-150 | 151-155 | 156-160 | 161-165 |\n| -------- | -------- | -------- | -------- | -------- | -------- |\n| 男生人数 | 1 | 5 | 7 | 6 | 3 |\n| 女生人数 | 2 | 4 | 8 | 7 | 2 |",
+        "options":  [
+                        "A. 141",
+                        "B. 153",
+                        "C. 165",
+                        "D. 166"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】看表可知，身高在151-155cm,人数最多，据此解答即可.\n【详解】A。141cm在身高141-145cm之间，人数偏少,141cm不可能是该班同学的平均身高;B。153cm在身高151-155cm之间，人数偏多,153cm可能是该班同学的平均身高;C。165cm在身高161-165cm之间，人数偏少,165cm不可能是该班同学的平均身高;D。统计表中没有166cm身高的学生，所以166cm不可能是该班同学的平均身高.\n故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24六年级下·河北保定·期末"
+    },
+    {
+        "id":  "216",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25六年级下·海南海口·期末)小刚是一名篮球队员，他所在的篮球队队员的平均身高是178cm,小刚的身高是()。",
+        "options":  [
+                        "A. 一定比178cm高",
+                        "B. 一定比178cm矮",
+                        "C. 一定是178cm",
+                        "D. 可能比178cm高，也可能比178cm矮，也可能等于178cm。"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】平均数=总数量/总份数，平均数反应了数据的平均水平，单一数据可能比平均数高，也可能比平均数低，也可能恰好和平均数相等.\n【详解】A。小刚的身高可能比178cm高，原说法错误;B。小刚的身高可能比178cm矮，原说法错误;C。小刚的身高可能是178cm,原说法错误;D。小刚的身高可能比178cm高，也可能比178cm矮，也可能等于178cm。原说法正确。故答案为:D",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25六年级下·海南海口·期末"
+    },
+    {
+        "id":  "217",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·海南海口·单元测试)四(3)班数学单元测试最高分是96分，最低分是84分，那么四(3)班单元测试的平均分可能是()分。",
+        "options":  [
+                        "A. 83",
+                        "B. 92",
+                        "C. 97"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】一组数的平均数，比这组数中最大的数小，比这组数中最小的数大，这个班的最高分是96分，最低分是84分，所以这个班的平均分大于84分而小于96分，据此解答.\n【详解】A。83\u003c84，所以83分不可能是平均分。B。96\u003e92\u003e84，所以92可能是平均分。C。97\u003e96，所以97分不可能是平均分。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·海南海口·单元测试"
+    },
+    {
+        "id":  "218",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级下·海南海口·单元测试)两个小组比赛跳绳，记录1分钟成绩如下：第一小组有3人，一共跳了270下，第二小组4人，一共跳了352下,( )小组的成绩好一些。",
+        "options":  [
+                        "A. 第一小组",
+                        "B. 第二小组",
+                        "C. 无法比较"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】平均数=总数量/总份数，第一小组有3人，一共跳了270下，用270除以3即可求出第一组平均每人跳了多少下，同理用352除以4即可求出第二组平均每人跳了多少下，最后把两个商比较即可.\n【详解】270÷3=90(下)352÷4=88(下)90\u003e88第一小组的成绩好一些。故答案为:A",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·海南海口·单元测试"
+    },
+    {
+        "id":  "219",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(22-23四年级下·四川绵阳·期末)丽丽语文，数学，英语三科的平均成绩是94分，其中语文90分，数学96分，英语成绩是()。",
+        "options":  [
+                        "A. 94分",
+                        "B. 95分",
+                        "C. 96分"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】根据题意可知，语文，数学，英语三科的平均成绩*3=语文，数学，英语三科的总成绩，语文，数学，英语三科的总成绩-语文和数学的成绩之和=英语的成绩，依此计算.\n【详解】94×3=282(分)90+96=186(分)282-186=96(分)英语成绩是96分。 故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "22-23四年级下·四川绵阳·期末"
+    },
+    {
+        "id":  "220",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(22-23四年级下·四川南充·期末)连续5个数的平均数是24，那么中间的数是()。",
+        "options":  [
+                        "A. 22",
+                        "B. 24",
+                        "C. 无法确定"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】由题意可知，用平均数乘5，得出5个数的和。因为这5个数是连续的，相邻两个数相差1，所以用5个数的和除以5即可得中间的数，据此解答即可.\n【详解】24×5÷5=120÷5=24所以连续5个数的平均数是24，那么中间的数是24。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "22-23四年级下·四川南充·期末"
+    },
+    {
+        "id":  "221",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "(22-23四年级下·四川广元·期末)甲，乙，丙三个数分别是10,10,40，这三个数的平均数是()。",
+        "options":  [
+                        "A. 15",
+                        "B. 20",
+                        "C. 25",
+                        "D. 30"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】先把三个数相加求出三个数的总和，再用三个数的总和除以3，即可求出这三个数的平均数。据此解答.\n【详解】(10+10+40)÷3=(20+40)÷3=60÷3=20则甲，乙，丙三个数分别是10,10,40，这三个数的平均数是20。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "22-23四年级下·四川广元·期末"
+    },
+    {
+        "id":  "222",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(22-23四年级下·四川德阳·期末)4.1班学生的平均身高是1.42米,4.2班学生的平均身高是1.45米。张亮在4.1班，李强在4.2班，张亮比李强()。",
+        "options":  [
+                        "A. 高",
+                        "B. 矮",
+                        "C. 有可能高，有可能矮，也可能一样高。"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】平均数代表的是一组数据的一般水平。由题意得,4.1班学生的平均身高是1.42米，张亮在4.1班，那么张亮的身高有可能比1.42米高，也有可能比1.42米矮，还有可能等于1.42米。4.2班学生的平均身高是1.45米，李强在4.2班，那么李强的身高有可能比1.45米高，也有可能比1.45米矮，还有可能等于1.45米。所以两人无法确定谁高谁矮.\n【详解】由分析得，无法确定张亮和李强谁高谁矮。张亮与李强相比，有可能高，有可能矮，也可能一样高.\n故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "22-23四年级下·四川德阳·期末"
+    },
+    {
+        "id":  "223",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(23-24四年级下·四川德阳·期末)五个人踢毽子，丽丽踢了39个，明明踢了28个，华华踢了10个，另外两个人踢的个数比明明少，比华华多。这五个人踢毽子的平均数应是()。",
+        "options":  [
+                        "A. 大于28小于39",
+                        "B. 28",
+                        "C. 大于10小于28"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】由题意得，五个人踢毽子，丽丽踢了39个，明明踢了28个，华华踢了10个，另外两个人踢的个数比明明少，比华华多，那么另外两个人最多每个人踢了27个，最少每个人踢了11个。可以把五个人踢的个数加起来，再除以5分别算出他们踢毽子个数平均数的最大和最小情况，再结合选项进行判断.\n【详解】如果另外两个人每个人都踢了27个,(39+28+10+27+27)÷5=131÷5=26.2(个)如果另外两个人每个人都踢了11个,(39+28+10+11+11)÷5=99÷5=19.8(个)所以这五个人踢毽子的平均数应该在19.8个到26.2个之间。A。28\u003e26.2，这五个人踢毽子的平均数大于28小于39，该选项说法错误。B。28\u003e26.2，这五个人踢毽子的平均数等于28，该选项说法错误。C。28\u003e26.2\u003e19.8\u003e10，这五个人踢毽子的平均数大于10小于28，该选项说法正确。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·四川德阳·期末"
+    },
+    {
+        "id":  "224",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(23-24四年级下·四川德阳·期末)要统计四年级五个班男女生最喜欢的课程，最好制作成()。",
+        "options":  [
+                        "A. 单式条形统计图",
+                        "B. 复式条形统计图",
+                        "C. 以上两种都可以"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】要统计四年级五个班男女生最喜欢的课程，最好制作成复式条形统计图。单式条形统计图只能表示一种量的多少，而复式条形统计图可以同时表示多种量的多少，因此更适合用于统计四年级五个班男女生最喜欢的课程。以此答题即可.\n【详解】根据分析可知：要统计四年级五个班男女生最喜欢的课程，最好制作成复式条形统计图.\n故答案为:B",
+        "knowledge_points":  [
+                                 "复式条形统计图"
+                             ],
+        "source":  "23-24四年级下·四川德阳·期末"
+    },
+    {
+        "id":  "225",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(23-24四年级下·四川德阳·期末)工人叔叔做零件，第一小组9人做了420个，第二小组11人做了530个，平均每个小组做了多少个零件。正确列式是()。",
+        "options":  [
+                        "A. (420+530)÷9",
+                        "B. (420+530)÷(9+11)",
+                        "C. (420+530)÷2"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】用加工零件总个数除以小组数，求出平均每个小组加工零件个数。据此解答.\n【详解】两个小组的零件总数是(420+530)个，一共有两个小组。所以求平均每个小组做的零件个数列式为:(420+530)÷2。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·四川德阳·期末"
+    },
+    {
+        "id":  "226",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "(24-25四年级上·湖南邵阳·期中)四年级5个班去植树，第一天植了145棵，第二天植了178棵，第三天植了157棵，平均每班植树()棵，正确列式是()。",
+        "options":  [
+                        "A. (145+178+157)÷4",
+                        "B. (145+178+157)÷3",
+                        "C. (145+178+157)÷5",
+                        "D. (145+178+157)÷(5+3)"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】求平均每班植树棵数，首先求出四年级5个班三天共植树的棵树，即145+178+157，然后除以班级个数5即可.\n【详解】(145+178+157)÷5=480÷5=96(棵)平均每班植树96棵，正确列式是(145+178+157)÷5。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级上·湖南邵阳·期中"
+    },
+    {
+        "id":  "227",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级上·湖南邵阳·期中)第一小组的学生称体重，最重的是48千克，最轻的是25千克，这组学生的平均体重可能是()千克。",
+        "options":  [
+                        "A. 24",
+                        "B. 36",
+                        "C. 48",
+                        "D. 50"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】平均数体现的是这一小组的平均趋势，平均数位于这组数据最低值和最高值之间，第一小组的学生称体重，最重的是48千克，最轻的是25千克，这组学生的平均体重位于25千克和48千克之间.\n【详解】25千克\u003c平均体重\u003c48千克。第一小组的学生称体重，最重的是48千克，最轻的是25千克，这组学生的平均体重可能是36千克。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级上·湖南邵阳·期中"
+    },
+    {
+        "id":  "228",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(24-25四年级上·河南洛阳·期中)下面说法正确的有()个.\n①池塘平均水深1.1米，小强身高1.4米，下水游泳，没有危险.\n②试商时，如果余数比除数大，说明商小了，要调大.\n③从不同方向观察同一个物体，看到的形状一定不同.\n④三位数除以两位数，商有可能是一位数，也有可能是两位数。",
+        "options":  [
+                        "A. 2",
+                        "B. 3",
+                        "C. 4"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】①平均数只能反映一组数据的平均水平，并不能反映这组数据中所有数据的大小。池塘平均水深1.1米，可能有的地方水深超过1.1米，甚至超过1.4米很多，所以小强下水游泳可能有危险.\n②试商时，如果余数大于除数，初商小，应调大;如果商与除数的积大于被除数，初商大，应调小.\n③在不同的位置观察同一个物体，看到的形状有可能不同，也有可能相同。例如:![](images/76664a6fad43e5ad0ff40fc8dcb2906169c2e68033c833164b7e16b9e3039643.jpg)此图从正面和侧面看到的图形相同，都可看到1层，为2个小正方形;因此同一个物体从不同角度观察，看到的形状可能相同.\n④三位数除以两位数，要知道商是几位数，先将被除数百位，十位上的数合起来，如果这个数等于或大于除数时，那么商就是两位数;如果这个数小于除数，那么商就是一位数。据此进行判断.\n【详解】①池塘有的地方水深可能超过小强身高，则他下水游泳可能有危险。原说法错误;②计算除法试商时，如果余数比除数大，也就是余数中还有至少一个除数，也就是商小了，应把商调大。原说法正确;③从不同方向观察同一个物体，看到的形状可能相同，也可能不同。原说法错误;④三位数除以两位数，商有可能是一位数，也有可能是两位数。原说法正确。则说法正确的有2个。故答案为:A",
+        "knowledge_points":  [
+                                 "判断商是几位数(除数是两位数)",
+                                 "物体三视图的认识",
+                                 "多位数除以两位数的试商",
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级上·河南洛阳·期中"
+    },
+    {
+        "id":  "229",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(22-23四年级下·重庆丰都·期末)要想清楚的看出相关联的两组数据不同的地方，应该选用()。",
+        "options":  [
+                        "A. 单式条形统计图",
+                        "B. 复式条形统计图",
+                        "C. 平均数"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】条形统计图上的每个小格表示就是一个长度单位，每个长度单位表示的数量是相同的，由于数据不同，所以画出直条的长短也不同，单式条形统计图中的数据只用一种直条来表示，而复式条形统计图可以同时表示两种或两种以上不同的量。平均数是一组数据的平均值，它反映了这组数据的集中趋势.\n【详解】由分析可知，要想清楚的看出相关联的两组数据不同的地方，应该选用复式条形统计图。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法",
+                                 "复式条形统计图",
+                                 "1格表示一个单位的单式条形统计图"
+                             ],
+        "source":  "22-23四年级下·重庆丰都·期末"
+    },
+    {
+        "id":  "230",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(22-23四年级下·重庆丰都·期末)我们在计算比赛选手的得分时，去掉一个最高分和最低分是()。",
+        "options":  [
+                        "A. 不合理的",
+                        "B. 合理的",
+                        "C. 不确定是否合理"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】计算比赛选手的得分，我们通常采用用平均分。平均数能代表一组数据的整体水平，一组数据中有一些数比平均数小，有一些数比平均数大，但平均数容易受一组数据中过大或过小这些极端数据的影响。评委的评分常带有主观性，去掉一个最高分和最低分，能使评分更具公平性.\n【详解】我们在计算比赛选手的得分时，去掉一个最高分和最低分是合理的。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "22-23四年级下·重庆丰都·期末"
+    },
+    {
+        "id":  "231",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(23-24四年级下·重庆南岸·期末)小丁所在小组的平均身高是153cm,小兰所在小组的平均身高是150cm,小丁和小兰的身高相比()。",
+        "options":  [
+                        "A. 小丁高",
+                        "B. 小兰高",
+                        "C. 两人同样高",
+                        "D. 无法确定谁高"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】平均数是表示一组数据集中趋势的量数，是指在一组数据中所有数据之和再除以这组数据的个数。它是反映数据集中趋势的一项指标。解答平均数应用题的关键在于确定\"总数量\"以及和总数量对应的总份数。而平均数是在最小数和最大数之间.\n【详解】小丁所在小组的平均身高是153cm,小丁可能身高大于153cm,可能等于153cm,也可能小于153cm;小兰所在小组的平均身高是150cm,小兰可能身高大于150cm,可能等于150cm,也可能小于150cm,两人身高无法比较。故答案为:D",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·重庆南岸·期末"
+    },
+    {
+        "id":  "232",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(23-24四年级下·广东肇庆·期末)在投球比赛中，乐乐4次投中球的个数分别是5个,4个,6个和5个，乐乐平均每次投中()个球。",
+        "options":  [
+                        "A. 6",
+                        "B. 5.5",
+                        "C. 5",
+                        "D. 4"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】根据题意，把乐乐4次投中球的个数相加，求出乐乐4次投中球的总个数，再用乐乐4次投中球的总个数除以4，即可求出乐乐平均每次投中多少个球.\n【详解】(5+4+6+5)÷4=(9+6+5)÷4=(15+5)÷4=20÷4=5(个)乐乐平均每次投中5个球。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·广东肇庆·期末"
+    },
+    {
+        "id":  "233",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "(23-24四年级下·陕西渭南·期末)如果1,20和a这三个数的平均数是11，那么a是()。",
+        "options":  [
+                        "A. 12",
+                        "B. 11",
+                        "C. 10"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】已知这三个数的平均数是11，根据平均数=总数量/总份数，可以求出总数量=平均数×总份数;据此求出总和，再减去已知数字，最后求出a为多少.\n【详解】11×3-1-20=33-1-20=32-20=12所以a是12。故答案为:A",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·陕西渭南·期末"
+    },
+    {
+        "id":  "234",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "三名同学练习踢毽子，平均每人踢了35个。已知其中两名同学踢的都超过了35个。估一估：第三名同学踢的()。",
+        "options":  [
+                        "A. 比35个多",
+                        "B. 比35个少",
+                        "C. 正好35个"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据题意，因为平均每人踢了35个，那么3个人共踢了35×3=105(个),已知其中两名同学踢的都超过了35个，假设都是36个，那么用105减去2个36，即可求出第三名同学踢的数量，据此计算解答.\n【详解】35×3=105(个)假设其中两名同学踢的都是36个;第三名同学踢了105-36×2\n=105-72\n=33(个)\n33\u003c35\n因此第三名同学踢的比35少。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·广东云浮·期末"
+    },
+    {
+        "id":  "235",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "要统计同学们每天睡眠时间，你认为要收集的信息是()。",
+        "options":  [
+                        "A. 调查同学们每天睡眠的姿势和环境",
+                        "B. 调查同学们每天入睡的时刻",
+                        "C. 调查同学们每天睡眠的总时间"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】由题意得，需要逐个分析题目中的信息能否统计出同学们每天的睡眠时间.\n【详解】A。调查同学们每天睡眠的姿势和环境无法统计出同学们每天的睡眠时间，不满足题意;B。单独调查同学们每天入睡的时刻无法统计出同学们每天的睡眠时间，不满足题意;C。调查同学们每天睡眠的总时间可以统计出同学们每天的睡眠时间，满足题意.\n故答案为:C",
+        "knowledge_points":  [
+                                 "数据的搜集与整理"
+                             ],
+        "source":  "23-24四年级下·广东江门·期末"
+    },
+    {
+        "id":  "236",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "体育课上，第三小组的5位同学依次测试1分钟仰卧起坐。前4位同学仰卧起坐的平均成绩是44个。如果要使小组的平均成绩达到45个，那么最后一位同学的成绩应达到()个。",
+        "options":  [
+                        "A. 45",
+                        "B. 48",
+                        "C. 49"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】先用45乘5求出5人1分钟仰卧起坐的总数，再用44乘4求出前4人1分钟仰卧起坐的总数，最后相减，即可求出最后一位同学的成绩应达到几个.\n【详解】45×5=225(个) 44×4=176(个) 225-176=49(个) 即最后一位同学的成绩应达到49个。 故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·河北廊坊·期末"
+    },
+    {
+        "id":  "237",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面的数学问题能用算式\"850÷17\"的是()。①买17棵玉兰树苗需要850元，每棵树苗价值多少元。②一艘轮船17小时行驶850千米，这艘轮船行驶的速度是多少千米/时。③17名同学参加1分钟跳绳测试，一共跳了850个，他们的平均成绩是多少个。",
+        "options":  [
+                        "A. ①②",
+                        "B. ②③",
+                        "C. ①②③"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】①单价=总价/数量，买17棵玉兰树苗需要850元，求每棵树苗价值多少元，用除法计算。②速度=路程/时间。一艘轮船17小时行驶850千米，求这艘轮船行驶的速度是多少千米/时，用除法计算。③17名同学参加1分钟跳绳测试，一共跳了850个，求他们的平均成绩是多少个，直接用总个数除以人数即可解答.\n【详解】①求每棵树苗价值多少元，列式为:850÷17。②求这艘轮船行驶的速度是多少千米/时，列式为:850÷17。③求这些同学的平均成绩是多少个，列式为:850÷17。故能用算式\"850÷17\"解决的问题有:①②③.\n故答案为:C",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "平均数的意义及求法",
+                                 "基础行程问题"
+                             ],
+        "source":  "23-24四年级下·河北廊坊·期末"
+    },
+    {
+        "id":  "238",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "小明家今年上半年的平均每月用水量是20吨，下面说法正确的是()。",
+        "options":  [
+                        "A. 小明家二月份的用水量一定是20吨",
+                        "B. 小明家上半年1至6月每个月的用水量都是20吨",
+                        "C. 小明家三月份的用水量可能比20吨少",
+                        "D. 小明家五月份的用水量不可能比20吨少"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】根据平均数的特点可知：平均数反映一组数据的整体情况，它处于这组数据的中间，比最大的数据小，比最小的数据大;小明家今年上半年的平均每月用水量是20吨，说明上半年可能有些月份比20吨多，可能有些月份比20吨少，还有些月份可能等于20吨;据此判断解答.\n【详解】A。小明家二月份的用水量不一定是20吨，原题说法错误;B。小明家上半年1至6月每个月的用水量不一定都是20吨，原题说法错误;C。小明家三月份的用水量可能比20吨少，原题说法正确;D。小明家五月份的用水量可能比20吨少，原题说法错误;所以，说法正确的是：小明家三月份的用水量可能比20吨少。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·河北廊坊·期末"
+    },
+    {
+        "id":  "239",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "某小学五(1)班学生平均身高是150厘米，下面说法正确的是()。",
+        "options":  [
+                        "A. 有一半学生身高低于150厘米，一半学生身高高于150厘米。",
+                        "B. 所有学生身高在140-160厘米之间。",
+                        "C. 不可能有身高135厘米的学生。",
+                        "D. 可能有身高165厘米的学生。"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】平均数反映一组数据的平均水平，并不能反映这组数据里各个数据的大小，平均数会受极端数据的影响，因此某小学五(1)班学生平均身高是150厘米，并不能反映出小学五(1)班每一个学生的身高，有的学生会高一些，有的学生会矮一些，据此解答.\n【详解】A。不一定是一半低于150厘米，一半高于150厘米，太过绝对，表达错误;B。不一定所有学生身高在140-160厘米之间，太过绝对，表达错误;C。可能有身高135厘米的学生，表达错误;D。可能有身高165厘米的学生，表达正确。故答案为:D",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·河北邢台·期末"
+    },
+    {
+        "id":  "240",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "有8个大小互不相同的数，其中a最小,b最大,c是这8个数的平均数，则有()。",
+        "options":  [
+                        "A. c\u003ca\u003cb",
+                        "B. a\u003cc\u003cb",
+                        "C. a\u003cb\u003cc",
+                        "D. 不确定"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】平均数反映一组数据的集中情况，平均数应大于这组数据中的最小的数且小于这组数据中最大的数，据此解答.\n【详解】由分析可得：有8个大小互不相同的数，其中a最小,b最大,c是这8个数的平均数，则有a\u003cc\u003cb.\n故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "22-23四年级下·山西·期末"
+    },
+    {
+        "id":  "241",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "在一次跳绳比赛中，明明，萱萱，青青三人的平均成绩是126下，其中明明跳了131下，青青跳了145下，萱萱跳了()下。",
+        "options":  [
+                        "A. 126",
+                        "B. 102",
+                        "C. 150"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】平均数问题的基本数量关系：总数量/份数=平均数;总数量/平均数=份数;平均数×份数=总数量;用126乘3可以计算出三人的总成绩，再依次减去明明的成绩和青青的成绩，计算出萱萱的成绩;据此解答.\n【详解】126×3-131-145\n=378-131-145\n=247-145\n=102(下)\n所以萱萱跳了102下。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法",
+                                 "乘加",
+                                 "乘减混合运算"
+                             ],
+        "source":  "23-24四年级下·河北张家口·期末"
+    },
+    {
+        "id":  "242",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面说法正确的是()。",
+        "options":  [
+                        "A. 从复式条形统计图中很容易比较两种事物数量的多少。",
+                        "B. 三个数的平均数是15，这三个数都大于15。",
+                        "C. 复式条形统计图中的图例颜色可以相同。",
+                        "D. 四1班平均每人捐款4.5元，四1班的丽丽一定捐了4元5角。"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】条形统计图用直条的长短表示数量的多少，从图中直观地看出数量的多少，便于比较，复式条形统计图可以表示多种量的多少，每种事物的图例颜色不相同;一组数据的平均数与其中一个数据的大小没有必然联系，可以大于平均数，可以等于平均数，也可以小于平均数;1元=10角，将1元看作单位\"1\"，平均分成10份，每份是1角，还可以写成0.1元;据此解答.\n【详解】根据分析:A。从复式条形统计图中很容易比较两种事物数量的多少，原题说法正确;B。三个数的平均数是15，这三个数可能有的大于15，有的等于15，有的小于15，不一定都大于15，原题说法错误;C。复式条形统计图中的图例颜色相同的话，无法区分是哪种事物，原题说法错误;D。4元5角为4.5元，那么四1班平均每人捐款4.5元，四1班的丽丽不一定捐了4元5角，原题说法错误。故答案为:A",
+        "knowledge_points":  [
+                                 "平均数的意义及求法",
+                                 "复式条形统计图",
+                                 "元",
+                                 "角",
+                                 "分的认识及换算",
+                                 "根据小数的意义进行单位换算"
+                             ],
+        "source":  "23-24四年级下·河北唐山·期末"
+    },
+    {
+        "id":  "243",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "期末测试中，小明班的平均分是92分，小亮班的平均分是94分，小亮的考试成绩会()。",
+        "options":  [
+                        "A. 比小明分数高",
+                        "B. 比小明的分数低",
+                        "C. 无法比较"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】平均数能代表一组数据的整体水平，一组数据中有一些数比平均数小，有一些数比平均数大。小明班的平均分是92分，小明有可能考的分数比92分低，如86分;也有可能比92分高，如99分。小亮班的平均分是94分，小亮有可能考的分数比94分低，如83分，也有可能比94分高，如100分.\n【详解】期末测试中，小明班的平均分是92分，小亮班的平均分是94分，小亮的考试成绩和小明的考试成绩无法比较。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·河北保定·期末"
+    },
+    {
+        "id":  "244",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "李师傅制作花篮，前3天共制作了54个，后2天共制作了36个，李师傅平均每天制作花篮多少个。列式正确的是()。",
+        "options":  [
+                        "A. 54÷3+36÷2",
+                        "B. (54+36)÷(3+2)",
+                        "C. 54×3+36×2"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】先用后2天制作花篮的个数加上前3天制作花篮的个数，再根据平均数=总数/总份数，进行解答即可.\n【详解】由分析可得：求李师傅平均每天制作花篮多少个，列式为(54+36)÷(3+2).\n故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·河北保定·期末"
+    },
+    {
+        "id":  "245",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "小明家去年12月份用水最多，是15吨;4月份用水最少，是6吨。那么小明家去年一年平均每月的用水量最有可能是()吨。",
+        "options":  [
+                        "A. 16",
+                        "B. 15",
+                        "C. 10",
+                        "D. 5"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】根据平均数是反映一组数据的集中趋势的量，它是这组数据所有数字的和除以这组数据的个数得出的，它比最大数小，比最小数大，据此解答即可.\n【详解】5\u003c6\u003c10\u003c15\u003c16小明家去年12月份用水最多，是15吨;4月份用水最少，是6吨。那么小明家去年一年平均每月的用水量最有可能是10吨。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·河北保定·期末"
+    },
+    {
+        "id":  "246",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "刘宇参加校园歌手比赛,7个评委的打分(单位：分)分别如下:95,94,96,99,93,97,90。如果去掉一个最高分和一个最低分，刘宇的平均得分是()。",
+        "options":  [
+                        "A. 95",
+                        "B. 94",
+                          "C. 96"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】首先确定最高分和最低分：给出的分数为95,94,96,99,93,97,90.99\u003e97\u003e96\u003e95\u003e94\u003e93\u003e90，其中最高分是99分，最低分是90分。然后计算去掉最高分和最低分后的总分：去掉99和90后，剩下的分数为95,94,96,93,97。总分是95+94+96+93+97=475(分)。最后计算平均得分：因为去掉最高分和最低分后还剩下5个评委的分数，所以求平均得分列式为475÷5.\n【详解】(95+94+96+93+97)÷(7-2)\n=475÷5\n=95(分)\n刘宇的平均得分是95分。故答案为:A",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·甘肃天水·期末"
+    },
+    {
+        "id":  "247",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "小雨班的平均身高是1.53m,小慧班的平均身高是1.53m,小雨的身高和小慧的身高相比,()。",
+        "options":  [
+                        "A. 小雨高",
+                        "B. 小慧高",
+                        "C. 一样高",
+                        "D. 无法确定"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】平均数是反映一组数据的平均水平，并不能反映这组数据中各个数据的大小，这组数据的数字分布在平均数的上下，但不可能都比平均数大，也不可能都比平均数小;由此即可进行判断.\n【详解】根据分析可知：不能根据两人所在班的平均身高，确定两人的身高高矮，所以小雨的身高和小慧的身高相比，无法确定谁高谁矮.\n故答案为:D",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·山东济南·期末"
+    },
+    {
+        "id":  "248",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面说法中合理的是()。",
+        "options":  [
+                        "A. 小红从家到学校平均每分钟走80米，用了6分钟。在这6分钟里，她必须每分钟走80米。",
+                        "B. 四(2)班同学的平均身高是143厘米，所以四(2)班的同学身高不可能有145厘米的。",
+                        "C. 冬冬家三口人五一假期自驾游一共花了1200元，平均每个人花了400元。"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】平均数是反映一组数据的平均水平，是一组数据中所有数据之和再除以这组数据的个数，平均数介于一组数据中的最大值与最小值之间，据此即可解答.\n【详解】A。小红从家到学校平均每分钟走80米，用了6分钟。在这6分钟里，她必须每分钟走80米。小红不需要每分钟都走80米，只要走的距离除以时间是80米即可，不符合;B。四(2)班同学的平均身高是143厘米，所以四(2)班的同学身高不可能有145厘米的。平均数介于一组数据中的最大值与最小值之间，四(2)班的同学身高可能有145厘米的，不符合;C。冬冬家三口人五一假期自驾游一共花了1200元，平均每个人花了400元。1200÷3=400(元),即平均每个人花了400元，符合。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·四川德阳·期末"
+    },
+    {
+        "id":  "249",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "五个人踢毽子，丽丽踢了39个，明明踢了28个，华华踢了10个，另外两个人踢的个数比明明少，比华华多，这五个人踢毽子的平均数应是()。",
+        "options":  [
+                        "A. 大于10小于39",
+                        "B. 28",
+                        "C. 大于28小于39"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】根据题意，另外两个人踢的个数比明明少，比华华多，则最大为27个，最少为11个，一组数据的总和除以这组数据个数所得到的商叫这组数据的平均数，把另外两个人踢的个数分别按照最少和最多来计算出平均数，据此选择即可.\n【详解】另外两个都踢11个:(39+28+10+11+11)÷5\n=(67+10+11+11)÷5\n=(77+11+11)÷5\n=(88+11)÷5\n=99÷5\n=19.8(个)\n另外两个人都踢27个:(39+28+10+27+27)÷5\n=(67+10+27+27)÷5\n=(77+27+27)÷5\n=(104+27)÷5\n=131÷5\n=26.2(个)\nA。五个人踢毽子的平均数大于10小于39，符合题意;B。28\u003e26.2，五个人踢毽子的平均数不可能是28，不符合题意;C。五个人踢毽子的平均数不可能大于28，不符合题意。故答案为:A",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·四川乐山·期末"
+    },
+    {
+        "id":  "250",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "三位同学的说法中，正确的是()。",
+        "options":  [
+                        "A. 小明：我身高1.65米，是学校篮球队中最矮的，我们学校篮球队的平均身高一定大于1.65米。",
+                        "B. 小丽：几个不同的物体，不管从什么角度观察，看到的图形一定不同。",
+                        "C. 小磊：用2,4,6和小数点组成的最大小数是6.42。"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】根据平均数的特点：平均数反映一组数据的整体情况，它处于这组数据的中间，比最小的数据大，比最大的数据小;根据从不同方向观察物体所看到的形状，可能相同也可能不同;比较两个小数的大小，先比较整数部分，整数部分大这个数就大，整数部分相同再比较小数部分;要使三个数字组成的小数最大，则应把其中最大的两个数字按从大到小的顺序依次写在整数的十位和个位，最小的一个数字写在第一位小数。据此解答.\n【详解】A。根据平均数的特点，它处在一组数据的中间，比最小的数大，比最大的数小;所以，小明的身高1.65米是篮球队最矮的，则平均身高一定大于1.65米;原题说法正确;B。例如以下两个物体:![](images/2026-03-14%20172623.jpg),它们的形状不同，但左图前面看到的图形和右图左面看到的图形是相同的，都是![](images/2026-03-14%20172641.jpg),所以，几个不同的物体，看到的图形可能相同，也可能不相同;原题说法错误;C。用2,4,6和小数点组成的最大小数是64.2;原题说法错误。所以，三位同学的说法中，正确的是：小明：我身高1.65米，是学校篮球队中最矮的，我们学校篮球队的平均身高一定大于1.65米。故答案为:A",
+        "knowledge_points":  [
+                                 "平均数的意义及求法",
+                                 "多位小数的大小比较",
+                                 "从不同位置观察单个物体"
+                             ],
+        "source":  "23-24四年级下·四川南充·期末"
+    },
+    {
+        "id":  "251",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "这是图图的数学日记，下面说法错误的是()。",
+        "options":  [
+                        "A. 运算定律帮助我，计算37×99+37可以直接用37×100计算",
+                        "B. 三角形的知识我学得好，我明白一个三角形至少有2个锐角",
+                        "C. 平均数知识告诉我，身高1.4米的我到平均水深1.2米的水池不一定安全",
+                        "D. 认识了小数，我知道了4.5和4.50大小相等，计数单位相同"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】两个数的和与一个数相乘，可以用这两个数分别和这个数相乘，再把它们的积相加，这叫乘法分配律.\n直角三角形有1个直角和2个锐角，锐角三角形有3个锐角，钝角三角形有1个钝角和2个锐角，应该三角形最少有2个锐角。平均数能代表一组数据的整体水平，一组数据中有一些数比平均数小，有一些数比平均数大。平均水深1.2米的水池，并不是水池所有地方的水深都是1.2米，有的地方有可能比1.2米深得多，也有可能比1.2米浅得多.\n在小数的末尾添上0或去掉0，小数的大小不变，这叫做小数的性质。一位小数的计数单位是十分之一，两位小数的计数单位是百分之一，三位小数的计数单位是千分之一。分别写作0.1,0.01,0.001。.\n【详解】A。37×99+37=37×(99+1)=37×100=3700，说法正确。B。一个三角形至少有2个锐角。说法正确。C。身高1.4米的我到平均水深1.2米的水池不一定安全。说法正确。D。4.5=4.50,4.5的计数单位是0.1,4.50的计数单位是0.01。说法错误。故答案为:D",
+        "knowledge_points":  [
+                                 "小数的性质",
+                                 "平均数的意义及求法",
+                                 "小数的数位和计数单位的认识",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "23-24四年级下·山西长治·期末"
+    },
+    {
+        "id":  "252",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "学校足球队队员的平均身高是160厘米，现在增加一名身高为170厘米的新队员。现在队员的平均身高()超过160厘米。",
+        "options":  [
+                        "A. 一定",
+                        "B. 可能",
+                        "C. 不可能",
+                        "D. 无法判断"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】平均数是反映一组数据的平均水平，是一组数据中所有数据之和再除以这组数据的个数，平均数介于一组数据中的最大值与最小值之间，据此即可解答.\n【详解】增加的新队员身高是170厘米，新队员的身高大于学校足球队队员的平均身高，所以现在队员的平均身高一定超过160厘米。故答案为:A",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·安徽芜湖·期末"
+    },
+    {
+        "id":  "253",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "四(1)班学生平均身高140厘米，下列说法正确的是()。",
+        "options":  [
+                        "A. 四(1)班每个学生的身高可能都不是140厘米。",
+                        "B. 四(1)班每个学生的身高一定都是140厘米。",
+                        "C. 四(1)班身高最高的学生是小东，他的身高可能比140厘米矮。",
+                        "D. 四(1)班每个学生的身高可能都超过140厘米。"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】平均数是反映一组数据的平均水平，并不能反映这组数据中各个数据的大小，平均数介于一组数据中的最大值与最小值之间，据此即可解答.\n【详解】A。根据分析可知，四(1)班每个学生的身高可能都不是140厘米，原说法正确。B。根据分析可知，四(1)班每个学生的身高可能大于140厘米，小于140厘米，也可能等于140厘米，原说法错误。C。根据分析可知，最高身高大于平均身高，四(1)班身高最高的学生是小东，他的身高比140厘米高，原说法错误.\nD。根据分析可知，四(1)班每个学生的身高可能大于140厘米，小于140厘米，也可能等于140厘米，不可能都超过140厘米，如果都超过140厘米，那么平均数就会比140大，原说法错误.\n故答案为:A",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·安徽宣城·期末"
+    },
+    {
+        "id":  "254",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "四(1)班一组同学进行跳绳测试，成绩最好的同学一分钟跳了132下，这一组7名同学一分钟的平均跳绳成绩可能是()下。",
+        "options":  [
+                        "A. 132",
+                        "B. 107",
+                        "C. 121",
+                        "D. 133"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】平均数：一组数据的总和除以这组数据个数所得到的商叫这组数据的平均数。在一组不同的数据中，平均数应该大于最小数，小于最大数，据此解答即可.\n【详解】107\u003c121\u003c132\u003c133 这一组7名同学一分钟的平均跳绳成绩可能是121下。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·云南昆明·期末"
+    },
+    {
+        "id":  "255",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "四(2)班男生的平均体重是41千克，在这个班的男生中，小君的体重最轻，他的体重不可能是()千克。",
+        "options":  [
+                        "A. 38",
+                        "B. 40",
+                        "C. 42"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】四(2)班男生的平均体重是41千克，平均体重等于所有男生体重之和除以男生的个数;小君的体重最轻，所以体重不可能超过平均体重，据此解答.\n【详解】根据解析可知，小君的体重不可能是42千克。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·云南昭通·期末"
+    },
+    {
+        "id":  "256",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "小嘉同学某次测试的语文，数学和英语三科的平均成绩为85分，已知她的数学和英语的成绩分别是88分和82分，则她的语文成绩是()。",
+        "options":  [
+                        "A. 85分",
+                        "B. 86分",
+                        "C. 88分"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】根据总数/份数=平均数可知：平均数×份数=总数，用三科的平均成绩乘3，即求到三科的总分，再用三科的总分减数学和英语的成绩，剩下的则是语文成绩。据此解答.\n【详解】85×3-88-82=255-88-82=167-82=85(分)所以，她的语文成绩是85分。故答案为:A",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·云南大理·期末"
+    },
+    {
+        "id":  "257",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "丽丽所在班级同学的平均身高是140cm,婷婷所在班级同学的平均身高是143cm,婷婷()比丽丽高。",
+        "options":  [
+                        "A. 一定",
+                        "B. 可能",
+                        "C. 不可能"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】平均数：一组数据的总和除以这组数据个数所得到的商叫这组数据的平均数;一组数据的平均数与其中个数据的大小没有必然联系，可以大于平均数，可以等于平均数，也可以小于平均数;据此解答.\n【详解】根据分析：丽丽的身高可能比140cm高，也可能比140cm矮，也可能正好140cm;婷婷的身高可能比143cm高，也可能比143cm矮，也可能正好143cm,所以婷婷可能比丽丽高.\n故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·吉林通化·期末"
+    },
+    {
+        "id":  "258",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "刘奇和张强两位同学做仰卧起坐，平均每人做了34个，其中张强做了37个，刘奇做了()个。",
+        "options":  [
+                        "A. 31",
+                        "B. 33",
+                        "C. 35"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】先用34×2求出刘奇和张强两位同学共做了多少个仰卧起坐，再减去张强做的个数，即可求出刘奇做了多少个。据此列式解答.\n【详解】34×2-37=68-37=31(个)刘奇做了31个仰卧起坐。故答案为:A",
+        "knowledge_points":  [
+                                 "平均数的意义及求法",
+                                 "乘加",
+                                 "乘减混合运算"
+                             ],
+        "source":  "23-24四年级下·吉林四平·期末"
+    },
+    {
+        "id":  "259",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "某小学图书室周三，周四各借出图书65本，周五上午借出图书47本，下午借出图书45本。这三天平均每天借出图书多少本。正确的算式是()。",
+        "options":  [
+                        "A. (65+47+45)÷3",
+                        "B. (65+65+47+45)÷3",
+                        "C. (65+65+47+45)÷4"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】一组数据的总和除以这组数据个数所得到的商叫这组数据的平均数;先用47+45求出周五借出去的本数，将周三，周四和周五借出去的图书本数相加，再除以3即可求出这三天平均每天借出图书多少本.\n【详解】(65+65+47+45)÷3=(130+47+45)÷3=(177+45)÷3=222÷3=74(本)正确的算式是(65+65+47+45)÷3.\n故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·贵州黔东南·期末"
+    },
+    {
+        "id":  "260",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "明明语文，数学，英语和科学四科成绩的平均分是91分，已知语文93分，英语89分，科学91分，那么他的数学成绩()。",
+        "options":  [
+                        "A. 高于平均分",
+                        "B. 低于平均分",
+                        "C. 等于平均分",
+                        "D. 不确定"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】一组数据的总和除以这组数据的个数所得到的商叫这组数据的平均数，用91×4求出四科的总分是多少，再减去另外三科的分数，即可求出数学的成绩，据此和平均成绩比较选择即可.\n【详解】91×4=364(分)364-93-89-91=271-89-91=182-91=91(分)他的数学成绩等于平均分。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·贵州黔南·期末"
+    },
+    {
+        "id":  "261",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "张华在期末考试时，数学成绩公布前他三门功课的平均分数是94分，数学成绩公布后，他的平均成绩下降了1分。张华的数学考了()分。",
+        "options":  [
+                        "A. 93分",
+                        "B. 92分",
+                        "C. 91分",
+                        "D. 90分"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】平均数=总数/总份数，则总数=平均数×总份数，三门功课的平均分乘3，可以算出三门功课的总分是(94×3)分。数学成绩公布后，他的平均成绩下降了1分，则四门功课的平均分是(94-1=93)分。四门功课的平均分乘4，可以算出四门功课的总分是(93×4)分。四门功课的总分减去三门功课的总分，即可算出张华数学考了多少分.\n【详解】94×3=282(分)(94-1)×4=93×4=372(分) 372-282=90(分)张华的数学考了90分。故答案为:D",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·四川凉山·期末"
+    },
+    {
+        "id":  "262",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "有一个水池，指示牌()能帮助人们根据身高判断下水是否安全。",
+        "options":  [
+                        "A. 最深处154厘米",
+                        "B. 平均水深140厘米",
+                        "C. 此处水深约130厘米",
+                        "D. 最浅处126厘米"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】A。最深处154厘米，说明其他地方的深度都不大于154厘米，人们可以根据身高判断下水是否安全;B。平均水深140厘米，说明有的地方比140厘米浅，有的地方比140厘米深，贸然下水很危险;C。此处水深约130厘米，不知道其他处水有多深，有没有超过下水安全深度，无法帮助人们根据身高判断下水是否安全;D。最浅处126厘米，不知道最深处是多少厘米，贸然下水很危险.\n【详解】根据分析可知，最深处154厘米能帮助人们根据身高判断下水是否安全。故答案为:A",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·四川凉山·期末"
+    },
+    {
+        "id":  "263",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "小天所在班级的平均身高是141cm,小红所在班级的平均身高是142cm,小天和小红的身高相比,()。",
+        "options":  [
+                        "A. 小天高",
+                        "B. 小红高",
+                        "C. 无法确定"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】一组数据的总和除以这组数据个数所得到的商叫这组数据的平均数，平均数只能反映一组数据的平均水平，并不能反应这组数据中所有数据的大小。小天所在班级的平均身高是141cm,但是小天可能高于141cm也可能低于141cm,小红所在班级的平均身高是142cm,小红可能高于142cm也可能低于142cm,无法判断两个人的身高，据此选择即可.\n【详解】由分析得：小天和小红的身高相比，无法确定。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·河南漯河·期末"
+    },
+    {
+        "id":  "264",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面说法正确的是()。",
+        "options":  [
+                        "A. 1.0987精确到百分位是1.1",
+                        "B. 小红身高1.5米，她在平均水深1.2米的水池中游泳没有危险",
+                        "C. 0.03和0.30这两个小数大小不等，计数单位相同"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】A。精确到百分位即保留两位小数，要看小数部分的第三位(千分位)上的数，然后按\"四舍五入\"法确定是\"舍\"还是\"入\";如果被舍去部分首位上的数小于5，就直接舍去;如果被舍去部分首位上的数大于或等于5，就向前一位进1后再舍去;B。平均数是表示一组数据集中趋势的量数，它是反映数据集中趋势的一项指标，在平均水深1.2米的水池中游泳，则某一处的水深可能超过1.2米，或低于1.2米，或等于1.2米，据此判断;C。0.03\u003c0.30，它们的大小不相等,0.03是两位小数，它的计数单位是0.01,0.30是两位小数，它的计数单位是0.01，所以计数单位相同;据此判断即可.\n【详解】A。1.0987≈1.10,1.0987精确到百分位是1.10，故原题说法错误;B。水池平均水深1.2米，并不代表所有的地方的水深都是1.2米，可能有比1.5米深的地方，所以身高1.5米的小红在水池中游泳可能有危险，故原题说法错误;C。0.03和0.30这两个小数大小不等,0.03的计数单位是0.01,0.30的计数单位是0.01，所以计数单位相同，故原题说法正确。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法",
+                                 "小数的数位和计数单位的认识",
+                                 "多位小数的大小比较",
+                                 "小数的近似数"
+                             ],
+        "source":  "23-24四年级下·河南焦作·期末"
+    },
+    {
+        "id":  "265",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "小辉身高1.3米，在平均水深1.1米的河里游泳，他认为没有危险。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数代表的是一组数据的一般水平。由题意得，河里的平均水深为1.1米，那么河里有的区域的水深比1.1米浅，有的区域的水深比1.1米深，所以身高1.3米的小辉去河里游泳，可能会有危险.\n【详解】小辉身高1.3米，在平均水深1.1米的河里游泳，可能会有危险。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·黑龙江佳木斯·期末"
+    },
+    {
+        "id":  "266",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "四(1)班的数学平均成绩是95分，四(2)班的数学平均成绩是92分，四(1)班的王明数学成绩一定比四(2)班的李想数学成绩高。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数是反映一组数据的整体水平，并不能代表其中某个个体的具体数值。四(1)班的平均分高，只能说明该班整体成绩较好，但王明的成绩可能高于，等于或低于该班平均分，同理李想的成绩也可能高于，等于或低于四(2)班的平均分。因此，无法确定两人成绩的高低关系.\n【详解】四(1)班的数学平均成绩是95分，四(2)班的数学平均成绩是92分，四(1)班的王明数学成绩一定比四(2)班的李想数学成绩高。这句话错误，无法确定两人成绩的高低关系。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·甘肃定西·期末"
+    },
+    {
+        "id":  "267",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "四(1)班男生的平均身高是145厘米，女生的平均身高是143厘米，说明该班所有男生都比女生高。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数是整体水平的反映，不能代表每个个体的具体数值。男生的平均身高高于女生，但可能存在个别女生比某些男生高的情况.\n【详解】平均身高表示的是男生和女生身高的整体平均水平，并不能说明每个学生的具体身高。例如：假设男生有3人，身高分别为144厘米,145厘米,146厘米，平均身高为145厘米;女生有3人，身高分别为142厘米,143厘米,144厘米，平均身高为143厘米。此时女生的最高身高(144厘米)与男生的最矮身高(144厘米)相等。若女生中有一人身高为146厘米，而男生中有一人身高为144厘米，则这名女生比该男生高。因此，\"所有男生都比女生高\"的结论不一定成立.\n故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·云南玉溪·期末"
+    },
+    {
+        "id":  "268",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "豆豆身高1.5米，他觉得自己可以去平均深度1.45米的小河边游泳。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数是指在一组数据中所有数据之和再除以数据的个数。该组数据中的数据可以高于平均数，也可以低于平均数，还可以等于平均数，据此判断.\n【详解】平均水深为1.45米的水域，该水域部分水深高于1.45米，豆豆身高1.5米，有可能部分水域的水深超过1.5米。去该水域游泳有可能有危险。即原说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·山东日照·期末"
+    },
+    {
+        "id":  "269",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "平均数一定是这组数据中的一个数。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数是所有数据的总和除以数据的个数;在一组数据中，平均数具有唯一性，是通过计算得到的，它不是数据中的原始数据，只是在平均数计算出来后，有和某个原始数据相等的可能，所以它可能属于原始数据，也可能不属于，需结合具体数据判断.\n【详解】例如:2和4的平均数为(2+4)÷2=6÷2=3，但3不在这组数据中。因此，平均数不一定是这组数据中的一个数，原题表达错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·陕西西安·期末"
+    },
+    {
+        "id":  "270",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "张华所在班级的同学的平均体重是32千克，李东所在班级的同学的平均体重是29千克，则张华的体重一定比李东重。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数是指在一组数据中所有数据之和再除以数据的个数;平均数是表示一组数据集中趋势的量数，它是反映数据集中趋势的一项指标，不能反映每个数据情况，据此判断即可.\n【详解】根据分析可知，张华所在班级的平均体重较高，可能由于部分同学体重较大，但张华本人的体重可能低于班级平均;同理，李东的体重可能高于班级平均，甚至超过张华，因此，无法确定张华一定比李东重，原题表达错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·海南三亚·期末"
+    },
+    {
+        "id":  "271",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "航天实验室工作人员的平均年龄是37岁，其中李教授的年龄可能是41岁。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】平均年龄为37岁，说明所有工作人员的年龄总和除以人数等于37。李教授的年龄41岁高于平均数，但其他工作人员的年龄可能低于平均数，从而平衡总和，因此可能存在这种情况.\n【详解】航天实验室工作人员的平均年龄是37岁，其中李教授的年龄可能是41岁。原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "272",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "篮球队队员的平均身高是163厘米，那么这个篮球队所有队员的身高一定都不低于163厘米。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数是所有数据的总和除以数据的个数，它代表一组数据的平均水平，并不能说明每个数据的具体大小。可能存在部分数据低于平均数的情况.\n【详解】篮球队队员的平均身高是163厘米，说明所有队员的身高的总和除以人数等于163厘米。但队员的身高可能有的高于163厘米，有的等于163厘米，也可能有的低于163厘米。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·陕西·期末"
+    },
+    {
+        "id":  "273",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "有8位同学的平均身高是147cm,其中最高的是139cm。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】一组数据的总和除以这组数据个数所得到的商叫这组数据的平均数。平均数应小于最大值，大于最小值。据此判断.\n【详解】有8位同学的平均身高是147cm,其中最高的应大于平均身高，即大于147cm,不可能是139cm。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·天津滨海新·期末"
+    },
+    {
+        "id":  "274",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "小明的身高1.5米，在平均水深1.1米的河里游泳不会有危险。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均水深1.1米表示整条河水深的平均值，有些地方实际水深可能高于或低于平均值，若某些区域水深超过1.5米，小明游泳时会有危险，据此即可解答.\n【详解】根据分析可知，平均水深1.1米并不能说明河水的每一处深度都是1.1米，可能存在部分区域水深超过1.5米，而小明身高1.5米，存在溺水风险，题中说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·江西抚州·期末"
+    },
+    {
+        "id":  "275",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "复式条形统计图可以直观地比较不同地区同一时间的气温情况。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】复式条形统计图的特点是能够同时显示多组数据，便于比较不同组别在同一项目上的数值。比较不同地区同一时间的气温，符合复式条形统计图的应用场景.\n【详解】复式条形统计图可以直观地比较不同地区同一时间的气温情况。故答案为:√",
+        "knowledge_points":  [
+                                 "复式条形统计图"
+                             ],
+        "source":  "24-25四年级下·河南信阳·期末"
+    },
+    {
+        "id":  "276",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "乐乐所在班级的同学平均每人捐款15元，说明乐乐一定捐了15元。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数是所有数据的总和除以数据的个数，反映的是整体水平，并不代表每个个体都等于平均数。因此，平均每人捐款15元，可能存在有人捐的比15元多，有人捐的比15元少的情况.\n【详解】根据平均数的意义，班级平均每人捐款15元，说明总捐款额等于人数乘15元。但每个同学的捐款金额可能高于，等于或低于15元。例如，若班级有3人，总捐款为45元，可能的捐款情况为10元,15元,20元，此时平均数仍为15元，但乐乐可能捐了10元,15元或20元。因此，乐乐不一定捐了15元。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25三年级下·重庆忠县·期末"
+    },
+    {
+        "id":  "277",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "小强身高1.4m,他在平均水深是1.2m的泳池内游泳，肯定不会有危险。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数只能反映一组数据的平均水平，并不能反应这组数据中所有数据的大小。平均水深1.2m的泳池，说明某些区域的水深可能不超过1.2m,也可能超过1.2m,甚至还可能超过小强的身高1.4m,因此存在危险的可能性.\n【详解】平均水深1.2m并不表示泳池所有区域的水深都是1.2m。若某些区域水深超过1.4m,小强在此处游泳会有危险。因此，题目中\"肯定不会有危险\"的说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "22-23四年级下·青海西宁·期末"
+    },
+    {
+        "id":  "278",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "明明所在班的同学平均身高是160cm,冬冬所在班的同学平均身高是158cm,所以明明一定比冬冬高。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】一组数据的总和除以这组数据个数所得到的商叫这组数据的平均数，明明所在班级同学的平均身高是160cm,但是明明可能低于160cm,冬冬所在班级同学的平均身高是158cm,但是冬冬可能高于158cm,据此判断即可.\n【详解】根据分析可知：明明可能身高158cm(低于班级平均),冬冬可能身高160cm(高于班级平均),因此无法确定明明一定比冬冬高。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级下·天津河东·期末"
+    },
+    {
+        "id":  "279",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "小敏3分钟跳绳180个，她每分钟一定跳60个。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】根据分析可知:180÷3=60(个),题目中给出小敏3分钟跳绳180个，每分钟的平均数量60个。但题目中的\"一定\"表示每分钟都必须恰好跳60个，而平均数并不能保证每一分钟的实际数量都相等，因此结论不一定成立。原题说法错误。故答案为: ×\n【详解】根据题意，小敏3分钟跳绳的总个数为180个，用180除以3，就是平均每分钟跳的个数。平均数仅表示整体的平均速度，并不能说明每分钟都恰好跳的个数。例如，可能存在1分钟跳50个，某分钟跳70个，另一分钟跳60个，但总和仍为180个。因此，题目中\"每分钟一定跳60个\"的结论不成立，应判断为错误。",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "22-23四年级下·安徽淮南·期末"
+    },
+    {
+        "id":  "280",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "小华身高145厘米，所以他去平均水深130厘米的河里游泳不会有危险。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均水深130厘米表示的是河流不同位置水深的平均值，但实际各区域的水深可能高于或低于平均值。小华身高145厘米，若某些区域水深超过145厘米，就会有危险。据此解答.\n【详解】平均水深130厘米的河流中，可能存在比130厘米深或浅的区域。例如，若某处水深150厘米(超过小华身高),则会有危险。由于平均数无法保证所有区域水深均\u003c=145厘米，因此题中说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·云南文山·期末"
+    },
+    {
+        "id":  "281",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "小明所在小组的平均体重是36千克，小明的体重一定是36千克。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数是所有数据的总和除以数据的个数，反映的是整体水平，不能确定每个个体的具体数值。小明的体重可能高于，等于或低于平均体重.\n【详解】根据平均数的定义，小明所在小组的平均体重为36千克，说明全组体重总和除以人数等于36，但小明的体重可能不等于36千克。例如，若小组有3人，体重分别为35千克,36千克,37千克，平均为36千克，但小明可能是其中任意一个数值。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·辽宁盘锦·期末"
+    },
+    {
+        "id":  "282",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "小明所在班级平均身高是135厘米，小刚所在班级平均身高是138厘米，所以小刚一定比小明高。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数是反映一组数据的平均水平，并不能反映这组数据中各个数据的大小，由此即可进行判断.\n【详解】根据分析可得:135厘米,138厘米分别是小明，小刚所在班级的平均身高，并不是他们实际的身高，因此不确定他们两人谁更高，所以原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24三年级下·广东湛江·期末"
+    },
+    {
+        "id":  "283",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "小杰班级数学的平均成绩是85分，所以小杰的数学成绩一定是85分。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】根据平均数的意义，平均分表示的是全班学生成绩的平均水平，平均数表示的是整体水平，不能代表个人成绩。已知小杰班级数学的平均成绩是85分，那么小杰的成绩可能比85高，也可能比85低，也可能是85分;所以无法确定准确的成绩;以此判断.\n【详解】根据分析可知：小杰班级数学的平均成绩是85分，但小杰的数学成绩不一定是85分。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·海南省直辖县级单位·期末"
+    },
+    {
+        "id":  "284",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "小强身高1.4米，肯定能趟过平均水深为1.2米的河，不会有危险。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数是一组数据的总和除以这组数据的个数。平均数是反映一组数据的平均水平，并不能反应这组数据中各个数据的大小，它比最大的数据小，比最小的数据大，据此解答.\n【详解】平均水深1.2米的小河说明此小河水的深度有的地方比1.2米浅，但有的地方会比1.2米深，如果深的地方超过了身高1.4米，那么小强过河就有危险。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24六年级下·江西南昌·期中"
+    },
+    {
+        "id":  "285",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "甜甜4次跳绳的总成绩是400下，她每次跳绳的成绩肯定都是100下。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】总成绩除以跳的次数等于平均每次跳绳的成绩，平均数介于一组数据的最大值和最小值之间，据此解题.\n【详解】400÷4=100(下)\n甜甜4次跳绳的总成绩是400下，她每次跳绳的成绩可能比100下多，也可能比100下少，也可能等于100下。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·甘肃武威·期末"
+    },
+    {
+        "id":  "286",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "张松所在的班级平均身高是1.42米，可以按照平均身高给这个班级的每个同学定制校服。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数是反映一组数据的平均水平，并不能反应这组数据中各个数据的大小。在一组数据中，有的数据可能会大于平均数，有的数据可能会小于平均数，有的数据可能会等于平均数。张松所在的班级平均身高1.42米，不代表每个同学的身高都是1.42米，有的同学的身高可能比1.42米高很多，有的同学的身高可能比1.42米矮，所以不能按照平均身高给这个班级的每个同学定制校服，据此解答即可.\n【详解】由分析可知：张松所在的班级平均身高是1.42米，不可以按照平均身高给这个班级的每个同学定制校服。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·河南焦作·期末"
+    },
+    {
+        "id":  "287",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "张宇所在小组平均每分钟踢毽子个数是30个，那么张宇每分钟踢毽子个数可能大于30个，可能小于30个，也可能刚好30个。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】平均数是表示一组数据的平均值，是指在一组数据中所有数据之和再除以这组数据的个数;其特点是比最大数小，比最小数大;依此判断.\n【详解】根据分析可知，张宇每分钟踢毽子个数可能大于30个，可能小于30个，也可能刚好30个。故答案为:√",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·河北衡水·期末"
+    },
+    {
+        "id":  "288",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "小明班男生平均体重36kg,女生平均体重34kg。那么小明班平均体重是35kg。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】班级平均体重并非简单地将男生平均体重和女生平均体重相加再除以2，而是需要考虑男生和女生的人数，用全班总体重除以全班总人数才能得到准确的班级平均体重。题中未提及人数关系，所以不能直接得出班级平均体重是35kg,据此解答即可.\n【详解】由分析可知，小明班男生平均体重36kg,女生平均体重34kg,无法判断小明班平均体重是不是35kg,原说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·浙江宁波·期末"
+    },
+    {
+        "id":  "289",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "小明的身高是141cm,他到平均水深为120cm的游泳池里游泳，不会有危险。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数表示的是一组数据的平均水平，并不能表示这组数据中各个数据的大小，由此即可进行判断.\n【详解】游泳池的平均水深为120cm,可能有的地方水深超过141cm。小明的身高是141cm,他到平均水深为120cm的游泳池里游泳，可能有危险。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·广西河池·期末"
+    },
+    {
+        "id":  "290",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "小东所在小组同学的平均身高是149cm,小刚所在小组同学的平均身高是145cm,小东一定比小刚高。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数只能反映一组数据的平均水平，并不能反应这组数据中所有数据的大小。小东所在小组同学的平均身高是149cm,小刚所在小组同学的平均身高是145cm。但是小东的身高可能低于149cm,甚至低于145cm。而小刚的身高可能高于145cm,甚至高于149cm。也就是小东的身高可能比小刚的身高矮。也可能两人同样高。据此判断.\n【详解】由分析得：只能明确小东所在小组同学的平均身高比小刚所在小组同学的平均身高要高，但不能明确小东一定比小刚高，还可能小东比小刚矮，甚至两人同样高。原说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·安徽铜陵·期末"
+    },
+    {
+        "id":  "291",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "踢毽子比赛，小红所在的小组平均每人踢36个，小丽所在的小组平均每人踢32个，所以小红踢的比小丽多。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数代表一组数据的一般水平，它一般大于这组数据的最小值，它一般小于这组数据的最大值。由题意得，小红所在的小组平均每人踢36个，小红踢的个数可能比36个少;小丽所在的小组平均每人踢32个，小丽踢的个数可能比32个多。所以小红可能比小丽踢的少.\n【详解】由分析得，小红可能比小丽踢的少。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·山东日照·期末"
+    },
+    {
+        "id":  "292",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "四(1)班男生的平均体重是43kg,女生的平均体重是41kg,所以每个男生都比女生重。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数常用于表示统计对象的一般水平，它代表的不是每个人的具体水平，所以男生的平均体重不代表所有男生的体重，女生的平均体重也不代表所有女生的体重，据此判断.\n【详解】由分析知：平均体重不代表所有人的体重，即可能有的女生要比男生重。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25四年级上·河北张家口·期末"
+    },
+    {
+        "id":  "293",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "小聪身高140厘米，游泳池平均水深110厘米，小聪下水游泳一定不会有危险。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数只能反映一组数据的平均水平，并不能反应这组数据中所有数据的大小，游泳池的平均水深是110厘米，可能有的地方水深超过140厘米，下水游泳可能有危险，据此解答即可.\n【详解】平均水深110厘米的游泳池，并不代表每处的水深都是110厘米，可能比110厘米深，超过140厘米完全有可能，也可能比110厘米浅;所以，小聪在平均水深110厘米的游泳池游泳，可能有危险。因此题干中的结论是错误的。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "2023·广东广州·小升初真题"
+    },
+    {
+        "id":  "294",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "某班男生的平均体重为32kg,女生平均体重为28kg,该班所有男生一定比所有女生重。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】一组数据中所有数据的和除以这组数据中数据的个数，所得的数叫平均数。平均数能较好地反映一组数据的整体情况，不能代表个别情况.\n【详解】某班男生的平均体重为32kg,女生平均体重为28kg,平均体重只能代表整体水平，不能代表个别情况，有可能有个别女生的体重超过男生的体重。所以，某班男生的平均体重为32kg,女生体重为28kg,不能说明该班所有男生一定比所有女生重。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24六年级下·河南漯河·期末"
+    },
+    {
+        "id":  "295",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "池塘的平均水深110厘米，小华身高是145厘米。他下河玩水不会有危险。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】池塘的平均水深为110厘米，表示各个区域水深的平均值，但实际某些区域可能更深。例如，若某处水深达170厘米，身高145厘米的小华在此处会被淹没，存在危险。据此判断即可.\n【详解】由分析可知：池塘的平均水深110厘米，小华身高是145厘米。他下河玩水可能会有危险。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25六年级下·海南海口·单元测试"
+    },
+    {
+        "id":  "296",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "计算全班数学平均分的方法是:(最高分数+最低分数)÷2。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】根据平均数=总数量/总份数，进行分析.\n【详解】计算全班数学平均分的方法是：全班同学数学分数的和/全班人数，原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25六年级下·海南海口·单元测试"
+    },
+    {
+        "id":  "297",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "如果某年的月平均气温最低是3度，最高是33度，那么这一年的平均气温一定是18度。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】月平均气温：指某个月份的平均气温。年平均气温：指一年中12个月的平均气温的平均值。题干中仅给出了最低和最高的月平均气温(3度和33度),但并未提供其他月份的气温数据。年平均气温的计算需要所有12个月的平均气温数据，不能仅凭最低和最高值推断。据此举例验证即可.\n【详解】假设某年12个月的气温分别为:3度,3度,3度,3度,3度,3度,33度,33度,33度,33度,33度,33度，平均气温=(3×6+33×6)÷12=(18+198)÷12=216÷12=18(度)但若某年12个月的气温分别为:3度,3度,3度,3度,3度,3度,3度,3度,3度,3度,3度,33度，平均气温=(3×11+33)÷12=(33+33)÷12=66÷12=5.5(度)由此可见，年平均气温不一定是18度。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "24-25六年级下·海南海口·单元测试"
+    },
+    {
+        "id":  "298",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "脱式计算，能简算的要简算.\n125×88\n656-(164+56)\n288÷12\n28.8+0.56+1.44+12.2",
+        "options":  [
+
+                    ],
+        "answer":  "11000;436;24;43",
+        "solution":  "【分析】(1)125×8=1000，把88分成8×11，利用乘法结合律，使125×8先算。(2)656-56能出整数，一个数减去两个数的和等于一个数连续减去这两个数。(3)多位数与一位数的口算较容易，把12分成4×3，先用288÷4再÷3。(4)28.8+12.2，0.56+1.44能凑成整数，利用加法交换律，结合律使它们先算.\n【详解】125×88\n=125×(8×11)\n=125×8×11\n=1000×11\n=11000\n656-(164+56)\n=656-56-164\n=600-164\n=436\n288÷12\n=288÷(4×3)\n=288÷4÷3\n=72÷3\n=24\n28.8+0.56+1.44+12.2\n=(28.8+12.2)+(0.56+1.44)\n=41+2\n=43",
+        "knowledge_points":  [
+                                 "三位数与两位数的进位乘法",
+                                 "除数是两位数的笔算除法",
+                                 "带有小括号的混合运算",
+                                 "多位小数的进位加法",
+                                 "退位减法"
+                             ],
+        "source":  "20-21四年级下·全国·单元测试"
+    },
+    {
+        "id":  "299",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "甲数和乙数的平均数是94，乙数比甲数多8，甲，乙两数各是多少。",
+        "options":  [
+
+                    ],
+        "answer":  "甲90;乙98",
+        "solution":  "【分析】一组数据的总和除以这组数据个数所得到的商叫这组数据的平均数。据此可知，甲数和乙数的平均数是94，则甲数与乙数的和是94×2=188。乙数比甲数多8，则甲数为(188-8)÷2。再用甲数加上8，即可求出乙数.\n【详解】(188-8)÷2\n=180÷2\n=90\n90+8=98\n答：甲数是90，乙数是98。",
+        "knowledge_points":  [
+                                 "平均数的意义及求法",
+                                 "和差问题"
+                             ],
+        "source":  "13-14三年级下·天津·期中"
+    },
+    {
+        "id":  "300",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "脱式计算，能简算的要简算.\n125×88\n656-(164+56)\n288÷12\n28.8+0.56+1.44+12.2",
+        "options":  [
+
+                    ],
+        "answer":  "11000;436;24;43",
+        "solution":  "【分析】本题考查四则混合运算的简便计算。第一题将88拆分成8×11，利用乘法结合律简算;第二题利用减法的性质去括号后简算;第三题将12拆分成4×3，利用除法的性质简算;第四题利用加法交换律和结合律简算.\n【详解】125×88\n=125×(8×11)\n=125×8×11\n=1000×11\n=11000\n656-(164+56)\n=656-56-164\n=600-164\n=436\n288÷12\n=288÷(4×3)\n=288÷4÷3\n=72÷3\n=24\n28.8+0.56+1.44+12.2\n=(28.8+12.2)+(0.56+1.44)\n=41+2\n=43",
+        "knowledge_points":  [
+                                 "三位数与两位数的进位乘法",
+                                 "除数是两位数的笔算除法",
+                                 "带有小括号的混合运算",
+                                 "多位小数的进位加法",
+                                 "退位减法"
+                             ],
+        "source":  "19-20四年级下·全国·单元测试"
+    },
+    {
+        "id":  "301",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "体质健康测评后，东东算出他和明明，林林三人的平均体重是37kg,加上坤坤的体重40kg。他们四人的平均体重()。",
+        "options":  [
+                        "A. \u003e=37kg",
+                        "B. \u003c37kg",
+                        "C. \u003e37kg",
+                        "D. \u003c=40kg"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】一组数据的总和除以这组数据个数所得到的商叫这组数据的平均数，平均体重是37kg,坤坤的体重大于平均体重，则加上坤坤后，他们四人的平均体重一定大于原来三人的平均体重，据此选择即可.\n【详解】40kg\u003e37kg,他们四人的平均体重\u003e37kg。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "302",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面说法不正确的是()。",
+        "options":  [
+                        "A. 等边三角形一定是等腰三角形",
+                        "B. 一个三角形有三条高",
+                        "C. 小明身高1.48米，他去平均水深为1.4米的水域游泳不会有危险",
+                        "D. 一个三角形有两条边相等并且最大的角是直角，这个三角形是等腰直角三角形"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】等边三角形三条边相等，等腰三角形是两腰相等，所以等边三角形一定是等腰三角形;三角形的高是从一顶点向底边画的垂线，三角形有三个顶点就有三条高;小明身高1.48米，平均水深为1.4米的水域，平均数为1.4米，有可能大于1.4米也有可能小于1.4米，所以也可能会大于1.48米，即可能会有危险;等腰直角三角形，一个角是直角，两个直角边相等，据此解答.\n【详解】A。等边三角形三条边相等，等腰三角形是两腰相等，等边三角形一定是等腰三角形，这句话正确。B。一个三角形有三个顶点，过顶点向底边作垂线即为高，所以一个三角形有三条高这句话正确。C。平均水深为1.4米，水深可能超过1.48米，所以可能会有危险，故小明身高1.48米，他去平均水深为1.4米的水域游泳不会有危险，这句话错误。D。一个三角形有两条边相等并且最大的角是直角，这个三角形是等腰直角三角形。这句话正确。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法",
+                                 "三角形的高及画法",
+                                 "等腰三角形和等边三角形的认识及特征",
+                                 "三角形的分类"
+                             ],
+        "source":  "23-24四年级下·辽宁鞍山·期末"
+    },
+    {
+        "id":  "303",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "在比赛中，计算平均得分时要去掉一个最高分和一个最低分的原因是()。",
+        "options":  [
+                        "A. 避免平均数受极端数据的影响，使比赛更公平",
+                        "B. 因为评委喜欢",
+                        "C. 比较容易看出每个选手得分的情况",
+                        "D. 便于比较"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】根据平均数的含义，平均数来表示一组数据的集中趋势，受到这组数据中每一个值的影响，如果一组数据中出现了极端数据，则得出的平均数不能很好地反映这组数据的总体情况，据此解答即可.\n【详解】由分析得：在比赛中，计算平均得分时要去掉一个最高分和一个最低分的原因是避免平均数受极端数据的影响，使比赛更公平。故答案为:A",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·内蒙古巴彦淖尔·期末"
+    },
+    {
+        "id":  "304",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面说法正确的是()。",
+        "options":  [
+                        "A. 某班级同学体育测试的平均成绩是92分，可能有一个人的成绩是75分。",
+                        "B. 明明所在班级同学的平均体重是36千克，乐乐所在班级同学的平均体重是41千克，乐乐一定比明明重。",
+                        "C. 统计小东家下半年的用电情况适合用复式统计图。"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】(1)平均数介于一组数据的最大值和最小值之间，平均成绩是92分，有的同学成绩小于92分，所以可能有一个人的成绩是75分。(2)平均数是反映一组数据的平均水平，并不能反应这组数据中各个数据的大小，只知道明明班和乐乐班同学的平均体重，不能确定乐乐和明明的体重，所以乐乐不一定比明明重。(3)复式统计图表示两组及以上数据的情况，单式统计图表示一组数据的情况，所以统计小东家下半年的用电情况适合用单式统计图.\n【详解】A。根据分析可知，某班级同学体育测试的平均成绩是92分，可能有一个人的成绩是75分，原说法正确.\nB。根据分析可知，明明所在班级同学的平均体重是36千克，乐乐所在班级同学的平均体重是41千克，但乐乐不一定比明明重，原说法错误。C。根据分析可知，统计小东家下半年的用电情况适合用单式统计图，原说法错误.\n故答案为:A",
+        "knowledge_points":  [
+                                 "1格表示多个单位的单式条形统计图",
+                                 "复式条形统计图",
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·山东临沂·期末"
+    },
+    {
+        "id":  "305",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "小强所在小组同学的平均身高是153厘米，小毅所在小组同学的平均身高是150厘米。下面的说法中，正确的是()。",
+        "options":  [
+                        "A. 小强一定比小毅高",
+                        "B. 小毅有可能比小强高",
+                        "C. 小强所在小组同学的身高都是153厘米",
+                        "D. 小毅所在小组同学的身高都不超过150厘米"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】平均数是表示一组数据一般水平的一个量。小强所在小组同学的平均身高是153厘米，小强的身高有可能比153厘米高，也有可能比153厘米矮，还有可能等于153厘米;小毅所在小组同学的平均身高是150厘米，小毅的身高有可能比150厘米高，也有可能比150厘米矮，还有可能等于150厘米。据此解答.\n【详解】A。小强不一定比小毅高，该选项说法错误。B。小毅有可能比小强高，该选项说法正确。C。小强所在小组有的同学的身高可能是153厘米，可能比153厘米高，也可能比153厘米矮，该选项说法错误。D。小毅所在小组同学的身高有可能超过150厘米，该选项说法错误。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·广东广州·期末"
+    },
+    {
+        "id":  "306",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "甲，乙，丙，丁四个小组比赛跑步，平均成绩依次是17.5秒,18.3秒,16.8秒,17.9秒。小明在甲组，小芳在乙组，小东在丙组，小青在丁组。下面说法正确的是()。",
+        "options":  [
+                        "A. 小芳跑步的速度一定比小明慢。",
+                        "B. 成绩最好的是丙组。",
+                        "C. 小东跑步的速度一定比小青快。",
+                        "D. 成绩最好的是乙组。"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】一组数据的总和除以这组数据个数所得到的商叫这组数据的平均数，平均数只能反映一组数据的平均水平，并不能反应这组数据中所有数据的大小。据此分析每个选项，选出正确的即可.\n【详解】A。小芳所在组的平均成绩是18.3秒，小明所在组的平均成绩是17.5秒,17.5\u003c18.3，则乙组的平均速度比甲组慢，但并不代表小芳的成绩一定比小明慢，选项说法错误;B。16.8\u003c17.5\u003c17.9\u003c18.3，丙组用时最短成绩最好，选项说法正确;C。小东所在组的平均成绩是16.8秒，小青所在组的平均成绩是17.9秒,16.8\u003c17.9，则丙组的平均速度比丁组快，但并不代表小东跑步的速度一定比小青快，选项说法错误;D。成绩最差的是乙组，选项说法错误。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法",
+                                 "一位小数的大小比较"
+                             ],
+        "source":  "23-24四年级下·广东广州·期末"
+    },
+    {
+        "id":  "307",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一条招聘广告上写着\"我公司现招聘员工，月平均工资是3500元.\"关于这则广告，下面叙述中一定正确的是()。",
+        "options":  [
+                        "A. 所有员工的月工资都是3500元",
+                        "B. 至少有一名员工的月工资是3500元",
+                        "C. 所有员工的月工资都高于3500元",
+                        "D. 可能一些员工的月工资会高于3500元，一些员工的月工资会低于3500元"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】平均工资=全公司所有员工的工资/总人数，月平均工资是3500元，说明可能有人的工资高于3500元，可能有人的工资低于3500元，可能有人的工资等于3500元，据此分析每个选项选择即可.\n【详解】A。所有员工的月工资不一定都是3500元，选项说法错误;B。不一定必须有员工的工资正好是3500元，选项说法错误;C。如果所有员工的月工资都高于3500元，则平均工资也一定高于3500元，选项说法错误;D。可能一些员工的月工资会高于3500元，一些员工的月工资会低于3500元，选项说法正确。故答案为:D",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·新疆和田·期末"
+    },
+    {
+        "id":  "308",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "四(1)班全班同学的平均身高是148厘米，最近转来了一名新同学，这名新同学的身高是160厘米，现在四(1)班同学的平均身高()。",
+        "options":  [
+                        "A. 比148厘米高",
+                        "B. 还是148厘米",
+                        "C. 是160厘米",
+                        "D. 比160厘米高"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】一组数据的总和除以这组数据个数所得到的商叫这组数据的平均数。根据题意,160厘米大于平均身高148厘米，则现在四(1)班同学的平均身高一定比原来的平均身高148厘米高，但是一定小于160厘米，据此选择即可.\n【详解】现在四(1)班同学的平均身高比148厘米高。故答案为:A",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·湖北襄阳·期末"
+    },
+    {
+        "id":  "309",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面说法正确的是()。",
+        "options":  [
+                        "A. 两个完全一样的直角三角形可以拼成一个大三角形，这个大三角形的内角和是360°。",
+                        "B. 2.5和2.500的大小相同，计数单位也相同。",
+                        "C. 1-0.2+0.8=1-1=0。",
+                        "D. 复式条形统计图是将两个统计图中的内容合并成一个统计图，便于信息的获取与对比。"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】三角形的内角和都为180°;小数的基本性质：小数的末尾添上\"0\"或去掉\"0\"，小数的大小不变;一位小数的计数单位是0.1，两位小数的计数单位是0.01，三位小数的计数单位是0.001;一个算式中既有减法，又有加法，要按照从左往右的顺序依次计算;条形统计图是用一个单位长度表示一定的数量，根据数量的多少画成长短不同的直条，然后把这些按照一定的顺序排列起来。而复式条形统计图可以用不同的条形表示2种以上的量。条形统计图中很容易看出各种数量的多少。据此解答.\n【详解】A。如图![](images/e76259e2fb7c1939e521383331230c1b167482bcfbff8e782cfae8c7323bb89b.jpg),两个完全一样的直角三角形拼成了一个大的三角形，这个大的三角形的内角和还是180°。该选项说法错误。B。由小数的基本性质可知:2.5=2.500.2.5的计数单位是0.1,2.500的计数单位是0.001，它们的计数单位不一样。该选项说法错误。C。1-0.2+0.8=0.8+0.8=1.6。该选项说法错误。D。复式条形统计图是将两个统计图中的内容合并成一个统计图，便于信息的获取与对比。该选项说法正确。故答案为:D",
+        "knowledge_points":  [
+                                 "小数的数位和计数单位的认识",
+                                 "复式条形统计图",
+                                 "三角形的内角和",
+                                 "小数的性质"
+                             ],
+        "source":  "23-24四年级下·福建龙岩·期末"
+    },
+    {
+        "id":  "310",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "妈妈买了10个大小接近的苹果，拿出其中2个苹果称重，这两个苹果的平均质量是250克，这10个苹果的平均质量，下面说法不正确的是()。",
+        "options":  [
+                        "A. 一定是250克",
+                        "B. 可能是250克",
+                        "C. 可能小于250克",
+                        "D. 可能大于250克"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】一组数据的总和除以这组数据个数所得到的商叫这组数据的平均数。据此选择即可.\n【详解】A。250克是平均质量，并不是苹果的质量一定是250克，原题说法不正确;B。苹果的质量可能是250克，原题说法正确;C。苹果的质量可能小于250克，原题说法正确;D。苹果的质量可能大于250克，原题说法正确。故答案为:A",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·福建漳州·期末"
+    },
+    {
+        "id":  "311",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "四年1班举行魔方速拧比赛，男生的平均成绩是52秒，女生的平均成绩是56秒。下面说法正确的是()。",
+        "options":  [
+                        "A. 每个男生的成绩都是52秒",
+                        "B. 全班的平均成绩一定是54秒",
+                        "C. 男生的整体水平高于女生",
+                        "D. 没有一个女生的成绩超过男生"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】根据平均数的意义和特点，男生和女生的平均成绩分别是用男生和女生的总成绩除以总人数得到的;所以平均数反映的是一组数据的整体情况，平均数的大小与这组数据中每个数据都有关系，它比这组数据中的最大的数小，比最小的数大;据此判断.\n【详解】A。男生的平均成绩是52秒，但每个男生的成绩不一定都是52秒，有大于52秒的也有小于52秒的;原说法错误;B。全班的平均成绩不一定是54秒，因为虽然54是52和56的平均数，但是具体平均成绩还要看男生和女生的人数，例如假设男生20人，女生30人，则平均成绩为(20×52+30×56)÷50=(1040+1680)÷50=2720÷50=54.4(秒),所以全班的平均成绩不一定是54秒;原说法错误;C。在魔方速拧比赛中，时间越少说明水平越高，平均数反映的是一组数据的整体水平,52\u003c56，男生平均成绩小于女生，说明男生的整体水平高于女生;原说法正确;D。虽然女生平均成绩56秒大于男生平均成绩52秒，但也有可能有女生成绩小于56秒甚至小于52秒的，男生中也有可能超过52秒甚至超过56秒的;所以女生的成绩有可能超过男生;原说法错误。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·福建漳州·期末"
+    },
+    {
+        "id":  "312",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "校乒乓球队8名队员的平均体重是43千克，小强的体重是39千克，加入乒乓球队后，现在乒乓球队队员的平均体重与原来比,()。",
+        "options":  [
+                        "A. 比原来重",
+                        "B. 与原来同样重",
+                        "C. 比原来轻",
+                        "D. 无法确定"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】根据题意，用43乘8，求出乒乓球队原来有8个队员的总体重，再加上39千克，即可求出现在乒乓球队队员的总体重，用现在乒乓球队员的总重量除以总人数，求出现在的平均体重，再比较即可解答.\n【详解】(43×8+39)÷(8+1)\n=(344+39)÷9\n=383÷9\n≈42.6(千克)\n42.6\u003c43，现在乒乓球队的平均体重比原来轻。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·福建三明·期末"
+    },
+    {
+        "id":  "313",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面说法正确的是()。",
+        "options":  [
+                        "A. 4.5和4.50大小相同，计数单位也相同",
+                        "B. 三角形的三个角可能分别是锐角，直角和钝角",
+                        "C. 算式\"(405+14)×6\"错算成\"405+14×6\"，结果与正确结果相差2025",
+                        "D. 期末考试中，小新所在的401班数学平均成绩是89.5,402班的小严考了91分，所以小严考得比小新好"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】小数的性质：小数的末尾添上\"0\"或去掉\"0\"，小数的大小不变;首先搞清小数的位数，有一位小数计数单位就是0.1，有两位小数计数单位就是0.01,。,以此类推;锐角三角形：三个角都是锐角的三角形;直角三角形：有一个角是直角的三角形;钝角三角形：有一个角是钝角的三角形;先运用乘法分配律:(a+b)×c=a×c+b×c,将算式(405+14)×6改为405×6+14×6，再减去405+14×6，计算出两个算式之间的差;平均数：一组数据的总和除以这组数据个数所得到的商叫这组数据的平均数;据此解答.\n【详解】根据分析:A。4.5=4.50,4.5的计数单位是0.1,4.50的计数单位是0.01，所以4.5和4.50大小相同，计数单位不相同，原题说法错误;B。三角形的三个角不可能分别是锐角，直角和钝角，因为三角形的内角和是180°,如果有一个直角和一个钝角，两个角之和就已经大于180°,原题说法错误;C。(405+14)×6-(405+14×6)=405×6+14×6-405-14×6=(405×6-405)+(14×6-14×6)=405×6-405=405×(6-1)=405×5=2025;所以结果与正确结果相差2025，原题说法正确;D。平均成绩是一个平均数，所以小新考的有可能比89.5分低，也可能正好89.5分，也可能比89.5分高;而小严考了91分，所以小严不一定考得比小新好，原题说法错误。故答案为:C",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "平均数的意义及求法",
+                                 "三角形的分类",
+                                 "小数的性质"
+                             ],
+        "source":  "23-24四年级下·浙江湖州·期末"
+    },
+    {
+        "id":  "314",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "小明所在的三(1)班同学平均身高是1.38米，小军所在的四(1)班同学的平均身高是1.42米，下列判断正确的是()。",
+        "options":  [
+                        "A. 小明比小军高",
+                        "B. 小军比小明高",
+                        "C. 小军和小明一样高",
+                        "D. A,B,C三种情况都有可能"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】平均数：一组数据的总和除以这组数据个数所得到的商叫这组数据的平均数;小明所在班级同学的平均身高是1.38米，小军所在班级同学的平均身高是1.42米;那么小明可能比1.38米高，也可能比1.38米低，或者正好是1.38米;小军可能比1.42米高，也可能比1.42米低，或者正好是1.42米;所以小明可能比小军高，也可能比小军低，也可能一样高;据此解答.\n【详解】小明所在的三(1)班同学平均身高是1.38米，小军所在的四(1)班同学的平均身高是1.42米，小明有可能比小军高，小军有可能比小明高，小军有可能和小明一样高。故答案为:D",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·浙江绍兴·期末"
+    },
+    {
+        "id":  "315",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "六位同学参加拍球比赛，他们所拍的数量如图所示。下面哪个数值最接近六位同学拍球数量的平均数。()\n![](images/847d28bf2ec4d2f689c7e0c52fd73095a01d477e1b6f116396a26fabe3c074a3.jpg)",
+        "options":  [
+                        "A. 60",
+                        "B. 62",
+                        "C. 71",
+                        "D. 80"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】平均数代表一组数据的平均水平;比这组数据的最小值要大，比这组数据的最大值要小。再根据\"割多补少\"方法解答.\n【详解】观察统计图可知，这组数据最小数大约是62，最大数大约是85，选项A和选项B,不符合题意，选项C,接近六位同学拍球数量的平均数，选项D,比平均数要大得多，所以最接近六位同学拍球数量的平均数是71。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24六年级下·河北保定·期末"
+    },
+    {
+        "id":  "316",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "四(1)班四个同学体重从大到小分别是52千克,a千克,b千克,46千克，他们的平均体重可能是()千克。",
+        "options":  [
+                        "A. 52",
+                        "B. 48",
+                        "C. 46",
+                        "D. 45"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】一组数据的总和除以这组数据个数所得到的商叫这组数据的平均数。据此可知，平均数应大于最小数，小于最大数。据此解答。熟练掌握平均数的意义是解答此题的关键.\n【详解】A。52=52，不符合题意;B。46\u003c48\u003c52，符合题意;C。46=46，不符合题意;D。45\u003c46，不符合题意。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·浙江台州·期末"
+    },
+    {
+        "id":  "317",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面说法错误的有()句。",
+        "options":  [
+                        "A. 1",
+                        "B. 2",
+                        "C. 3",
+                        "D. 4"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】①三角形的内角和等于180°;②被减数=减数+差，被减数+减数+差=150，被减数+被减数=150，被减数*2=150，被减数=150÷2=75;③78+37+122=37+78+122，运用了加法交换律,37+78+122=37+(78+122),运用了加法结合律;④平均数是反映一组数据的平均水平，并不能反应这组数据中各个数据的大小;小红所在班级的同学平均身高是1.55米，但不能确定小红的身高是多少，无法比较小丽和小红的身高.\n【详解】①根据分析可知，把一个大三角形剪成两个小三角形，每个小三角形的内角和都是180°,原说法错误.\n②根据分析可知，被减数+减数+差=150，那么被减数一定是75，原说法正确.\n③根据分析可知，在78+37+122=37+(78+122)中运用了加法交换律和加法结合律，原说法正确.\n④根据分析可知，小红所在班级的同学平均身高是1.55米，这不能确定小红的身高，所以无法比较小丽和小红身高，原说法错误.\n所以说法错误的有2句。故答案为:B",
+        "knowledge_points":  [
+                                 "整数加法结合律",
+                                 "整数加法交换律",
+                                 "平均数的意义及求法",
+                                 "三角形的内角和"
+                             ],
+        "source":  "23-24四年级下·浙江台州·期末"
+    },
+    {
+        "id":  "318",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.94,
+        "question":  "为了改进同学们的伙食，学校食堂打算调整一些菜品，于是进行了调查和相关数据的收集。你认为食堂最需要收集的数据是()。",
+        "options":  [
+                        "A. 全校各年级的人数",
+                        "B. 全校男生和女生的人数",
+                        "C. 全校喜欢吃鸡翅的人数",
+                        "D. 全校喜欢各种菜品的人数"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】为了改进同学们的伙食，学校食堂打算调整一些菜品，那么食堂需要收集的数据应该和菜品有关，调查的重点是同学们喜欢吃什么样的菜.\n【详解】A。调查全校各年级的人数与菜品无关。B。全校男生和女生的人数与菜品无关。C。全校喜欢吃鸡翅的人数，调查的菜品过于单一。D。收集全校喜欢各种菜品的人数，可以根据同学们的喜欢情况制定菜品。故答案为:D",
+        "knowledge_points":  [
+                                 "设计简单的调查表"
+                             ],
+        "source":  "23-24四年级下·浙江宁波·期末"
+    },
+    {
+        "id":  "319",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "小塘，小夏和小沙三人的平均身高是153cm。已知小塘身高是151cm,小夏身高是149cm,那么小沙身高是()。",
+        "options":  [
+                        "A. 等于153cm",
+                        "B. 小于153cm",
+                        "C. 大于153cm",
+                        "D. 无法确定"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】平均数是指在一组数据中所有数据之和再除以数据的个数，平均数介于一组数据的最大数和最小数之间;据此即可解答.\n【详解】平均数介于一组数据的最大数和最小数之间，三人的平均身高是153cm,已知小塘和小夏的身高都小于153cm,所以小沙的身高一定大于153cm。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "22-23四年级下·浙江·期末"
+    },
+    {
+        "id":  "320",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "五一节假日期间，曹村镇田园综合体前3天共接待游客6000人，后2天平均每天接待游客1500人。五一节假日期间共接待游客多少人。解决这个问题需要用到的信息是()。",
+        "options":  [
+                        "A. 6000人,1500人",
+                        "B. 6000人,2天,1500人",
+                        "C. 3天,6000人,1500人",
+                        "D. 3天,6000人,2天,1500人"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据对平均数的认识，先用1500×2求出后两天接待游客的人数，加上6000即可求出五一节假日期间共接待游客多少人，需要先知道后2天平均每天接待游客1500人，再知道前3天共接待游客6000人，据此选择即可.\n【详解】1500×2+6000=3000+6000=9000(人)解决这个问题需要用到的信息是6000人,2天,1500人。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法",
+                                 "\"提问题\"\"填条件\"问题"
+                             ],
+        "source":  "23-24四年级下·浙江温州·期末"
+    },
+    {
+        "id":  "321",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列说法中正确的是()。",
+        "options":  [
+                        "A. 6.895保留两位小数是6.90，小数末尾的0能去掉",
+                        "B. 小红身高1.5米，她在平均水深1.2米的水池中游泳没有危险的",
+                        "C. 2024年2月份有29天",
+                        "D. a/b=5。3，将a和b同时扩大到原来的100倍，则商不变，余数也不变"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】根据小数的近似数，保留两位小数则看千分位上的数，如果小于5则舍去，大于或等于5则向百分位进1;一组数据的总和除以这组数据个数所得到的商叫这组数据的平均数;平年2月有28天，闰年2月有29天;根据商的变化规律，被除数和除数同时扩大到原来的100倍，商不变，但是余数也要扩大到原数的100倍，据此选择即可.\n【详解】A。6.895≈6.90，不能去掉末尾的0，去掉则变成一位小数，选项说法错误;B。平均水深1.2米，可能有低于1.2米的地方，也可能有高于1.2米的地方，身高1.5米，也可能有危险，选项说法错误;C。2024÷4=506,2024年是闰年,2月有29天，选项说法正确;D.a/b=5。3,100a/100b=5。300，选项说法错误。故答案为:C",
+        "knowledge_points":  [
+                                 "商不变的规律及应用",
+                                 "小数的近似数",
+                                 "平均数的意义及求法",
+                                 "平年",
+                                 "闰年的认识及判定方法"
+                             ],
+        "source":  "23-24四年级下·浙江·期末"
+    },
+    {
+        "id":  "322",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "小刚身高140厘米，他到一个平均水深110厘米的池塘里游泳,()危险。",
+        "options":  [
+                        "A. 一定没有",
+                        "B. 可能有",
+                        "C. 一定有"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据题意可知，小刚身高140厘米，他到一个平均水深110厘米的池塘里游泳，平均水深100厘米说明有的地方超过110厘米，有的地方没有110厘米，据此解答即可.\n【详解】由分析可得：小刚身高140厘米，他到一个平均水深110厘米的池塘里游泳，可能有危险。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24五年级下·贵州安顺·期末"
+    },
+    {
+        "id":  "323",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一条长360米的小路，王叔叔去时每分钟走60米，回来时每分钟走90米。王叔叔来回的平均速度是多少。下面列式正确的是()。",
+        "options":  [
+                        "A. (60+90)÷2",
+                        "B. 360÷60+360÷90",
+                        "C. 360÷(360/60+360÷90)",
+                        "D. 360×2÷(360/60+360÷90)"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】根据题意可知王叔叔往返的路程是不变的，都是360米，已知王叔叔去时的速度和回来的速度，根据时间=路程/速度可以算出往返分别所需的时间，最后用往返的路程除以往返的总时间就是往返的平均速度.\n【详解】去的时间是360÷60，回来的时间是360÷90，往返的总时间是360÷60+360÷90，往返的路程是360×2，所以往返的平均速度就是360×2÷(360/60+360÷90)。故答案为:D",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·浙江嘉兴·期末"
+    },
+    {
+        "id":  "324",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "某品牌汽车公布了2023年度各车型销量排行榜。销量最高的为43万辆，各车型平均销量为21万辆。下面的说法正确的是()。",
+        "options":  [
+                        "A. 一定有一种车型销量为21万辆",
+                        "B. 一半的车型销量少于21万辆",
+                        "C. 大部分车型销量为19~23万辆",
+                        "D. 总有一种车型销量少于21万辆"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】平均数是表示一组数据的一般情况，用一组数据的总和除以这组数据的个数，所得的商就是这组数据的平均数。据此可知,2023年度各车型销量排行榜的数据中，各车型的平均销量为21万辆，说明各车型的销量在21万辆上下，销量最高的为43万辆，则一定有车型的销量低于21万辆，据此可以判断.\n【详解】由分析得：一定有车型销量小于21万辆，才能使平均数为21万辆。A。题干只说平均销量为21万辆，不能判断是否一定有一种车型销量为21万辆，此选项错误。B。题干没有给与此选项有关的已知信息，所以不能判断出是否一半的车型销量少于21万辆，此选项错误。C。题干只说平均销量为21万辆，不能判断大部分车型销量为19~23万辆，此选项错误。D。各车型平均销量为21万辆，销量最高的为43万辆，则总有一种车型销量小于21万辆，此选项正确。故答案为:D",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·浙江温州·期末"
+    },
+    {
+        "id":  "325",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "乐乐期末考试语文和数学的平均分为95分，英语低于95分，那么他三科平均分()。",
+        "options":  [
+                        "A. 高于95分",
+                        "B. 低于95分",
+                        "C. 无法判断"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】平均数是表示一组数据的平均值，是指在一组数据中所有数据之和再除以这组数据的个数;由此可知，三科的总分低于(95×3)分，那么三科的平均分低于(95×3÷3)分，依此即可选择.\n【详解】根据分析可知，乐乐期末考试语文和数学的平均分为95分，英语低于95分，那么他三科平均分低于95分。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·浙江杭州·期末"
+    },
+    {
+        "id":  "326",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "乐乐所在的401班的平均身高1.435米，欢欢所在的402班平均身高1.429米，那么()。",
+        "options":  [
+                        "A. 乐乐身高高",
+                        "B. 欢欢身高高",
+                        "C. 无法判断"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】平均数是指在一组数据中所有数据之和再除以数据的个数，所以，它只能代表这组数据的\"平均水平\"，而不能代表每一个具体的数据，这意味着，在401班里，有的同学的身高可能高于1.435米，有的同学的身高可能低于1.435米;同样地，在402班里，也有同学的身高可能高于或低于1.429米，因此，仅凭两个班级的平均身高，不能直接判断两个班级中任意两个同学的身高大小关系，据此作答.\n【详解】根据上述分析可得：乐乐所在的401班的平均身高1.435米，欢欢所在的402班平均身高1.429米，仅凭两个班级的平均身高，无法判断两个班级中任意两个同学的身高大小关系。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·浙江杭州·期末"
+    },
+    {
+        "id":  "327",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "小金统计了1~6月每月的天气情况，下面能代表三月份气温整体情况的是()。",
+        "options":  [
+                        "A. 三月任意一天的温度",
+                        "B. 三月温度的平均数",
+                        "C. 温度最高的那一天",
+                        "D. 温度最低的那一天"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】题干要表示三月份气温整体情况;而平均数是用来表示一组数据集中趋势的数量，是反映数据集中趋势的一项指标;据此逐项分析可解此题.\n【详解】根据分析:A。三月任意一天的温度，只能表示当天温度情况，不能表示三月份整体情况，不符合;B。三月温度的平均数，能代表三月份气温整体情况，符合;C。温度最高的那一天，只能表示当天温度最高，不能表示三月份整体情况，不符合;D。温度最低的那一天，只能表示当天温度最低，不能表示三月份整体情况，不符合。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·江西上饶·期末"
+    },
+    {
+        "id":  "328",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面对平均数理解正确的是()。",
+        "options":  [
+                        "A. 三位同学一分钟跳绳的个数分别是100,90,80，则他们跳绳的平均个数是90个。",
+                        "B. 乐乐一周打篮球4次，平均每次2小时，则乐乐每次打篮球都用了2小时。",
+                        "C. 淘气班同学平均身高148厘米，笑笑班同学平均身高155厘米，则笑笑一定比淘气高。"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】(1)三人跳绳的个数和除以3即等于他们跳绳的平均个数。(2)平均数是反映一组数据的平均水平，并不能反应这组数据中各个数据的大小，它介于一组数据的最大值和最小值之间。(3)平均数是反映一组数据的平均水平，并不能反应这组数据中各个数据的大小.\n【详解】A。(100+90+80)÷3=270÷3=90(个),他们跳绳的平均个数是90个，原说法正确。B。根据分析可知，平均每次2小时，不等于每次打篮球都用了2小时，有时可能多于2小时，有时可能小于2小时，有时可能等于2小时，原说法错误。C。根据分析可知，平均数是反映一组数据的平均水平，并不能反应这组数据中各个数据的大小，笑笑班同学平均身高比淘气班同学平均身高高，并不能说明笑笑一定比淘气高，原说法错误。故答案为:A",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·江西九江·期末"
+    },
+    {
+        "id":  "329",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列说法正确的是()。",
+        "options":  [
+                        "A. 游泳池平均水深1.2米，小明身高1.4米，进入游泳池没有危险。",
+                        "B. 学校根据学生的平均身高为学生定制校服。",
+                        "C. 全班同学的平均体重是32千克，天天的体重可能低于32千克。",
+                        "D. 小小3天共看了12页课外书，她每天看的页数一定都是4页。"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】平均数是表示一组数据的平均值，是指在一组数据中所有数据之和再除以这组数据的个数;其特点是比最大数小，比最小数大;它表示的是一组数据的平均水平，并不能表示这组数据中各个数据的大小，由此即可选择.\n【详解】A。游泳池平均水深1.2米，小明身高1.4米，进入游泳池可能有危险，即原说法错误。B。学校不应该根据学生的平均身高为学生定制校服，即原说法错误。C。全班同学的平均体重是32千克，天天的体重可能低于32千克，即原说法正确。D。12÷3=4(页),小小3天共看了12页课外书，她平均每天看的页数是4页，即原说法错误。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·江西赣州·期末"
+    },
+    {
+        "id":  "330",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "连续3个偶数的平均数是26，这三个偶数中，最大的是()。",
+        "options":  [
+                        "A. 20",
+                        "B. 24",
+                        "C. 28"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】三个连续的偶数的平均数是26，连续的偶数之间都相差2，所以最小的偶数要比中间的偶数小2，最大的偶数要比中间的偶数大2，根据平均数的意义得出中间的偶数是26，最大的偶数就是用26再加2即可.\n【详解】26+2=28，这三个偶数中，最大的是28。故答案为:C",
+        "knowledge_points":  [
+                                 "奇数与偶数的认识",
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·江西九江·期末"
+    },
+    {
+        "id":  "331",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "乐乐前两次数学考试的平均成绩是93分，第三次考试成绩是99分，她这三次考试的平均成绩是()。",
+        "options":  [
+                        "A. 93分",
+                        "B. 94分",
+                        "C. 95分",
+                        "D. 96分"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】平均数=总数/总份数，总数=平均数×总份数。乐乐前两次数学考试的平均成绩乘2，可以算出乐乐前两次数学考试一共考了(93×2)分;乐乐前两次数学考试总分加上第三次考试分数再除以3，即可算出她这三次考试的平均成绩.\n【详解】93×2=186(分)\n(186+99)÷3\n=285÷3\n=95(分)\n乐乐前两次数学考试的平均成绩是93分，第三次考试成绩是99分，她这三次考试的平均成绩是95分。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·湖南湘西·期末"
+    },
+    {
+        "id":  "332",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面说法错误的是()。",
+        "options":  [
+                        "A. 根据小数的性质,5.08可以写成5.80，还可以写成5.0800。",
+                        "B. 一个三角形中至少有两个角是锐角。",
+                        "C. 典典所在班级的同学平均每人捐款15元，典典不一定捐了15元。",
+                        "D. 手机支架，自行车框架等都是三角形的，原因是因为三角形具有稳定性。"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】根据小数的性质改写：小数的末尾添上0或者去掉0，小数的大小不变，这叫做小数的性质。整数改写小数，先在整数后面加小数点，再根据小数位数添0个数。锐角三角形：三个角都是锐角的三角形叫做锐角三角形。直角三角形：有一个角是直角的三角形叫做直角三角形。钝角三角形：有一个角是钝角的三角形叫做钝角三角形。以上三点说明三角形至少有两个角是锐角。平均数表示一组数据的趋势，实际数可以大于平均数，可以等于平均数，也可以小于平均数。三角形具有稳定性，生活中常利用这个原理来建构物体。根据以上分析逐项分析每项.\n【详解】A。根据小数的性质,5.08不可以写成5.80，但可以写成5.0800。原题说法错误。B。任意一个三角形中至少有两个角是锐角。原题说法正确。C。典典所在班级的同学平均每人捐款15元，可能比15元捐的多，可能少，也可能相等。原题说法正确。D。手机支架，自行车框架等都是三角形的，是利用三角形稳定性，原题说法正确.\n故答案为:A",
+        "knowledge_points":  [
+                                 "小数的性质",
+                                 "三角形的分类",
+                                 "平均数的意义及求法",
+                                 "三角形的稳定性及应用"
+                             ],
+        "source":  "23-24四年级下·湖南湘西·期末"
+    },
+    {
+        "id":  "333",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面说法中错误的有()个。",
+        "options":  [
+                        "A. 1",
+                        "B. 2",
+                        "C. 3",
+                        "D. 4"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】①大于0.3而小于0.5的一位小数只有一个是0.4;②计算小数加减法时，要像计算整数加减法一样，把末尾对齐;③小明身高132cm,他去平均水深120cm的河里游泳不会有危险;④0.050和5÷100的大小相等，计数单位也相同.\n【详解】①大于0.3而小于0.5的一位小数只有一个是0.4，原题说法正确;②计算小数加减法时，要像计算整数加减法一样，把末尾对齐，这个说法错误;③小明身高132厘米，他去平均水深120厘米的河里游泳可能有危险，原题说法错误;④0.050和5÷100的大小相等，计数单位不相同，原题干说法错误。故答案为:C",
+        "knowledge_points":  [
+                                 "多位小数的进位加法",
+                                 "退位减法",
+                                 "一位小数的大小比较",
+                                 "小数的数位和计数单位的认识",
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·湖南常德·期末"
+    },
+    {
+        "id":  "334",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "一个三层书架的第一层有52本书，第二层和第三层共有44本书。平均每层有()本书。",
+        "options":  [
+                        "A. 48",
+                        "B. 32",
+                        "C. 52"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据题意，用第一层放书的本数加上第二层，第三层放书的总本数，求这个书架共放书的多少本，再除以书架的层数即可.\n【详解】(52+44)÷3=96÷3=32(本)所以，平均每层有32本书。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·湖南常德·期末"
+    },
+    {
+        "id":  "335",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "小东所在小组同学的平均体重是36kg,小刚所在小组同学平均体重是34kg。下列说法正确的是()。",
+        "options":  [
+                        "A. 小东一定比小刚重",
+                        "B. 小东一定比小刚轻",
+                        "C. 无法确定谁轻谁重"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】平均数反映一组数据的平均情况，并不能反映这组数据中各个数据的大小;在一组数据中，有的数据可能会大于平均数，有的数据可能会小于平均数，有的数据可能会等于平均数;据此解答.\n【详解】由分析可知：知道小东和小刚所在组的平均体重并不能确定小东，小刚的体重是多少，小东可能比小刚重，也可能小东比小刚轻，还可能两人体重相同。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·湖南岳阳·期末"
+    },
+    {
+        "id":  "336",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "优优看一本童话书，前4天每天看了27页，又用5天看了98页，结合以上信息，不能解决()。",
+        "options":  [
+                        "A. 前4天一共看了多少页。",
+                        "B. 一共看了多少页。",
+                        "C. 这9天平均每天看多少页。",
+                        "D. 这本书还要多少天看完。"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】根据题意，求前4天一共看了多少页，用乘法计算即可;求一共看了多少页，先求出前4天看多少页，再加上98页，即可求出一共看了多少页书;求这9天平均每天看多少页书，先求出前4天看多少页，再加上5天看的页数，求出9天看的总页数，然后除以9即可;求这本书还要多少天才能看完，需要知道每天看的页数，题中没有给出，所以不能解决下面这个问题;据此解答.\n【详解】A。27×4=108(页);B。27×4+98=108+98=206(页);C。(27×4+98)÷9=(108+98)÷9=206÷9≈22.9(页);D。求这本书还要多少天才能看完，需要知道每天看的页数，题中没有给出，所以不能解决下面这个问题。故答案为:D",
+        "knowledge_points":  [
+                                 "带有小括号的混合运算",
+                                 "乘加",
+                                 "乘减混合运算",
+                                 "平均数的意义及求法",
+                                 "\"提问题\"\"填条件\"问题"
+                             ],
+        "source":  "23-24四年级下·湖南娄底·期末"
+    },
+    {
+        "id":  "337",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "小明班上最高的是162厘米，最矮的是125厘米，他们的平均身高可能是()。",
+        "options":  [
+                        "A. 120厘米",
+                        "B. 125厘米",
+                        "C. 136厘米",
+                        "D. 162厘米"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】平均数是指在一组数据中所有数据之和再除以数据的个数，平均数介于一组数据的最大值和最小值之间，据此即可解答.\n【详解】根据分析可知，小明班上最高的是162厘米，最矮的是125厘米，故他们的平均身高应大于125厘米小于162厘米，四个选项中只有136厘米符合要求，所以平均身高可能是136厘米。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·湖南·期末"
+    },
+    {
+        "id":  "338",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "四1班男生的平均身高是142厘米，下面说法正确的是().\n①亮亮的身高是151厘米，在四1班男生中个子是偏高的.\n②四1班一定有某位男生的身高正好是142厘米.\n③四1班新转来一位男同学身高是137厘米，现在全班男生的平均身高会比142厘米低.\n④四1班女生的平均身高是144厘米，红红是女生，她一定比本班所有男同学都高。",
+        "options":  [
+                        "A. 只有①",
+                        "B. 只有①③",
+                        "C. 只有①②③",
+                        "D. 只有①③④"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据平均数的定义，平均数是表示一组数据集中趋势的量数，它是反映数据集中趋势的一项指标，进行分析解答即可.\n【详解】①四1班男生的平均身高是142厘米，亮亮的身高是151厘米，所以亮亮在四1班男生中个子是偏高的，原题干说法正确。②身高的平均数是142厘米，不一定就有某位男生的身高正好是142厘米，原题干说法错误。③由于新转来的男同学身高是137厘米，比原来平均身高低，所以现在全班男生的平均身高会比142厘米低，原题干说法正确。④四1班女生的平均身高是144厘米，红红是女生，但她不一定比本班所有男同学都高。原题干说法错误。①和③说法正确。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24六年级下·北京房山·期末"
+    },
+    {
+        "id":  "339",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "同一战队的三个选手参加歌唱比赛,1号选手得了99分,2号选手得了90分,3号选手比2号选手成绩高，但不超过93分。这三个选手的平均成绩可能在()。",
+        "options":  [
+                        "A. 90分以下",
+                        "B. 90分到93分之间",
+                        "C. 93分到95分之间",
+                        "D. 99分以上"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】2号选手得了90分,3号选手比2号选手成绩高，但不超过93分，则3号选手得分91分或92分。三人的平均成绩=三人的总分/3，据此分别求出两种可能的平均成绩即可解答.\n【详解】(99+90+91)÷3\n=280÷3\n≈93.33(分)\n(99+90+92)÷3\n=281÷3\n≈93.67(分)\n两种情况的平均成绩都在93分到95分之间。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24六年级下·北京东城·期末"
+    },
+    {
+        "id":  "340",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列说法中，正确的是().\nA。把0.596保留两位小数约是0.6\nB。小明在计算8×(□+5)时，算成了8*□+5，他算出的结果是正确的\nC。王悦5次跳远的总成绩是10米，她每次跳远的成绩肯定都是2米\nD。一个三位小数，四舍五入后的近似数是3.69，原来的三位小数最小是3.685",
+        "options":  [
+                        "A. 把0.596保留两位小数约是0.6",
+                        "B. 小明在计算8×(□+5)时，算成了8*□+5，他算出的结果是正确的",
+                        "C. 王悦5次跳远的总成绩是10米，她每次跳远的成绩肯定都是2米",
+                        "D. 一个三位小数，四舍五入后的近似数是3.69，原来的三位小数最小是3.685"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】保留两位小数就是精确到百分位，省略百分位后面的尾数，我们看的是千分位，把千分位上的数四舍五入。两个数的和与一个数相乘，可以用这两个数分别和这个数相乘，再把它们的积相加，这叫乘法分配律。平均数能代表一组数据的整体水平，一组数据中有一些数比平均数小，有一些数比平均数大。一个三位小数，四舍五入后的近似数是3.69，用四舍法取近似数，这个三位小数的个位上是3，十分位上是6，百分位上是9，千分位上最大是4。用五入法取近似数，这个三位小数的个位上是3，十分位上是6，百分位上最小是5.\n【详解】A。0.596≈0.60，把0.596保留两位小数约是0.60。原说法错误。B。8×(□+5)=8*□+8×5=8*□+40。小明在计算8×(□+5)时，算成了8*□+5，他算出的结果是错误的。原说法错误。C。10÷5=2(米),王悦跳远的平均成绩是2米，但是她每次跳远的成绩不一定都是2米。原说法错误。D。一个三位小数，四舍五入后的近似数是3.69，原来的三位小数最小是3.685，最大是3.694。原说法正确。故答案为:D",
+        "knowledge_points":  [
+                                 "小数的近似数",
+                                 "整数乘法分配律",
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·湖南怀化·期末"
+    },
+    {
+        "id":  "341",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "小明所在班学生的平均身高是1.4米，小强所在班学生的平均身高是1.5米，小明和小强对比,()。",
+        "options":  [
+                        "A. 小强高",
+                        "B. 小明高",
+                        "C. 一样高",
+                        "D. 无法确定谁高"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】平均数是表示一组数据集中趋势的量数，它是反映数据集中趋势的一项指标，并不能代表具体个体的数据信息;平均数大于数据的最小值，小于数据的最大值，小明所在班级学生平均身高是1.4米，说明小明的身高可能小于1.4米，小强所在班级学生平均身高是1.5米，说明小强的身高可能小于1.5米，无法比较小明和小强的身高;据此解答即可.\n【详解】由分析可知，平均身高只是表示一组数据集中趋势的量数，它是反映数据集中趋势的一项指标，并不能代表具体个体的数据信息，所以无法确定小明和小强谁高。故答案为:D",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·湖南长沙·期末"
+    },
+    {
+        "id":  "342",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "学校舞蹈队队员的平均身高是1.56米，下列说法正确的是()。",
+        "options":  [
+                        "A. 所有队员的身高都是1.56米。",
+                        "B. 高于1.56米的队员人数一定等于低于1.56米的队员人数。",
+                        "C. 再加一个1.6米的队员，队员们的平均身高会增加。",
+                        "D. 从中随意抽取一半的队员，这些队员的平均身高也一定是1.56米。"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】平均身高代表的是这些队员身高的一般水平，有的队员比平均身高高，有的队员比平均身高矮，有的队员的身高就等于平均身高。据此解答.\n【详解】A。舞蹈队队员的平均身高是1.56米，不代表所有队员的身高都是1.56米，该选项说法错误。B。舞蹈队队员的平均身高是1.56米，有的队员会比平均身高高，有的队员比平均身高矮，但他们的具体人数不确定。该选项说法错误。C。再加一个1.6米的队员，这个队员比平均身高高，所以队员们的平均身高会增加。该选项说法正确。D。从中随意抽取一半的队员，抽取的队员身高不确定，抽取出来的队员的平均身高也不确定。该选项说法错误。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·湖南长沙·期末"
+    },
+    {
+        "id":  "343",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列说法，正确的是()。",
+        "options":  [
+                        "A. 小数5.6和5.600大小相等，表示的意义也相同。",
+                        "B. 一个图形无论是向左，右平移还是向上，下平移，它的形状和大小都不变。",
+                        "C. 池塘平均水深1.1米，小明身高1.4米，他下水没有安全隐患。",
+                        "D. 小数3.050中小数点后面的\"0\"，根据小数的性质，都可以去掉。"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】A。根据小数的性质：在小数的末尾添上0或去掉0，小数的大小不变。因此5.6和5.600大小相等,5.6是一位小数，计数单位是0.1,5.600是三位小数，计数单位是0.001，因此它们表示的意义不同。B。平移：在平面内，将一个图形上的所有点都按照某个方向作相同距离移动的图形运动。平移后图形的位置改变，形状，大小，方向不变。C。一组数据的总和除以这组数据的个数，所得到的商叫这组数据的平均数。平均数反映一组数据的平均情况，在一组数据中，有的数据可能会大于平均数，有的数据可能会小于平均数，有的数据可能会等于平均数.\nD。根据小数的性质：在小数的末尾添上0或去掉0，小数的大小不变。因此小数3.050千分位上的0可以去掉，小数的大小不变，而小数3.050的十分位上的0不能去掉，去掉后小数的大小变化了，据此解答.\n【详解】A。5.6的计数单位是0.1,5.600的计数单位是0.001。因此小数5.6和5.600大小相等，表示的意义也不相同;原题说法错误，不符合题意;B。一个图形无论怎样平移，都只是位置发生了变化，它的形状，大小，方向不变。因此一个图形无论是向左，右平移还是向上，下平移，它的形状和大小都不变。原题说法正确，符合题意;C。池塘平均水深1.1米，说明最深的地方会比1.1米深，所以小明身高1.4米，他下水仍然有安全隐患;原题说法错误，不符合题意;D。由分析可知，小数3.050的十分位上的0不能去掉，去掉后小数的大小发生变化，因此原题说法错误，不符合题意.\n故答案为:B",
+        "knowledge_points":  [
+                                 "小数的数位和计数单位的认识",
+                                 "平均数的意义及求法",
+                                 "小数的性质",
+                                 "平移与平移现象"
+                             ],
+        "source":  "23-24四年级下·湖南怀化·期末"
+    },
+    {
+        "id":  "344",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "小刚，小东，小伟三人的数学期末考试平均分是96分，三人的平均分比小熊低4分，他们四人的平均分是()分。",
+        "options":  [
+                        "A. 96",
+                        "B. 98",
+                        "C. 97"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】三人的平均分比小熊低4分，也就是小熊的分数是100分，然后算出四人的总分，最后再根据平均分=总分/人数，就可以求出四人的平均分.\n【详解】96+4=100(分)96×3=288(分) 288+100=388(分) 388÷4=97(分) 所以四个人的平均分是97分。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·湖北黄冈·期末"
+    },
+    {
+        "id":  "345",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "四(1)班和四(2)班同学们积极响应\"健康校园\"的倡议，开展跳绳比赛。如图是这两个班前四次的比赛成绩统计图。四(1)班四次比赛的平均成绩是()下.\n![](images/4a0eab37a93976a2b8f59e22639e543a4aeedef2f39a618aad19dbe37934fe5e.jpg)",
+        "options":  [
+                        "A. 191",
+                        "B. 195",
+                        "C. 193"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】根据对平均数的认识，将四(1)班四次比赛的成绩相加再除以4即可求出平均成绩是多少下，据此选择即可.\n【详解】从统计图中可知，四(1)班四次比赛的成绩分别为185下,164下,206下,209下.\n(185+164+206+209)÷4\n=(349+206+209)÷4\n=(555+209)÷4\n=764÷4\n=191(下)\n四(1)班四次比赛的平均成绩是191下。故答案为:A",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·湖北襄阳·期末"
+    },
+    {
+        "id":  "346",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "4名同学投掷沙包，每人投掷3次，平均成绩大约是9m的是().\n![](images/3cbc8d91b6b956713d01d674dbc3fa8d29b1c945a2545f5b6a9a04dbe7a6beec.jpg)",
+        "options":  [
+                        "A. 华华",
+                        "B. 莉莉",
+                        "C. 津津",
+                        "D. 娟娟"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】平均数：一组数据的总和除以这组数据个数所得到的商叫这组数据的平均数;据此解答.\n【详解】根据分析:A。观察发现华华的3次成绩中有2次都低于9m,并且低于9m的2次中有一次低很多，那么平均数一定低于9m;B。观察发现莉莉的3次成绩中，低于9m和高于9m的两次与9m之间的距离差不多，还有一次在9m附近，那么平均成绩大约是9m;C。观察发现津津的3次成绩都高于9m,那么平均数一定高于9m;D。观察发现娟娟的3次成绩中有2次都低于9m很多，还有一次在9m附近，那么平均数一定低于9m。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "347",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面说法正确的是()。",
+        "options":  [
+                        "A. 某班级同学体育测试的平均成绩是92分，可能有一个人的成绩是75分。",
+                        "B. 明明所在班级同学的平均体重是36kg,乐乐所在班级同学的平均体重是41kg,乐乐一定比明明重。",
+                        "C. 统计小东家下半年的用电情况适合用复式条形统计图。"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】平均数是表示一组数据的平均值，是指在一组数据中所有数据之和再除以这组数据的个数;其特点是比最大数小，比最小数大。下半年为7月到12月，依此再根据复式条形统计图的特点进行判断即可.\n【详解】A。某班级同学体育测试的平均成绩是92分，可能有一个人的成绩是75分，即原说法正确。B。明明所在班级同学的平均体重是36kg,乐乐所在班级同学的平均体重是41kg,乐乐不一定比明明重，即原说法错误。C。统计小东家下半年的用电情况不适合用复式条形统计图，即原说法错误。故答案为:A",
+        "knowledge_points":  [
+                                 "复式条形统计图",
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·山东临沂·期末"
+    },
+    {
+        "id":  "348",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "关于平均数的知识，下面的四句话中表述正确的有()句话.\n①平均数反映了一组数据的整体水平\n②平均数具有敏感性\n③平均数不是真实的数，而是虚拟的数\n④平均数是一种统计量",
+        "options":  [
+                        "A. 1",
+                        "B. 2",
+                        "C. 3",
+                        "D. 4"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】平均数是指在一组数据中所有数据之和再除以数据的个数，表示一组数据集中趋势的量数，它是反映数据集中趋势的一项指标;平均数反映了一组数据的整体水平，当一组数据中的任何一个数值发生变化时，计算出的平均数也会相应地发生变化。平均数是通过计算一组数据的总和然后除以数据的个数得到的，它不是一个实际存在于数据中的具体数值，而是用来表示这组数据的整体水平或趋势的一个统计量。平均数是一个重要的刻画数据集中趋势的统计量，具有代表性.\n【详解】①平均数反映了一组数据的整体水平，说法正确;②当一组数据中的任何一个数值发生变化时，计算出的平均数也会相应地发生变化。所以平均数具有敏感性，说法正确;③平均数是通过计算一组数据的总和然后除以数据的个数得到的，它不是一个实际存在于数据中的具体数值，而是用来表示这组数据的整体水平或趋势的一个统计量。所以平均数是虚拟的数，说法正确;④平均数是一个重要的刻画数据集中趋势的统计量，具有代表性，说法正确。四句话中表述正确的有4句话。故答案为:D",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·山东菏泽·期末"
+    },
+    {
+        "id":  "349",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列说法中，表述正确的是()。",
+        "options":  [
+                        "A. 0.2和0.20的大小相等，计数单位相同。",
+                        "B. 正方形，长方形，平行四边形，等腰三角形都是轴对称图形。",
+                        "C. 用长3厘米,4厘米,5厘米的三根小棒，只能搭出1种三角形。",
+                        "D. 某小队5名队员平均每分钟跳绳185个，队长跳绳个数不可能是185个。"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】(1)根据小数的性质可知,0.2=0.20,0.2的计数单位是0.1,0.20的计数单位是0.01;(2)正方形，长方形和等腰三角形是轴对称图形，平行四边形不是轴对称图形;(3)3厘米+4厘米\u003e5厘米，用长3厘米,4厘米,5厘米的三根小棒能搭出一个三角形;(4)平均数是反映一组数据的平均水平，是一组数据中所有数据之和再除以这组数据的个数，平均数介于一组数据中的最大值与最小值之间，所以队长跳绳个数可能大于185个，也可能小于185个，还可能等于185个.\n【详解】A。根据分析可知,0.2和0.20的大小相等，计数单位不同，原说法错误。B。根据分析可知，正方形，长方形，等腰三角形是轴对称图形，平行四边形不是轴对称图形，原说法错误。C。根据分析可知，用长3厘米,4厘米,5厘米的三根小棒，能搭出1种三角形，原说法正确。D。根据分析可知，某小队5名队员平均每分钟跳绳185个，队长跳绳个数可能是185个，原说法错误。故答案为:C",
+        "knowledge_points":  [
+                                 "三角形三边关系",
+                                 "平均数的意义及求法",
+                                 "小数的性质",
+                                 "轴对称的认识及辨认"
+                             ],
+        "source":  "23-24四年级下·山东菏泽·期末"
+    },
+    {
+        "id":  "350",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "淘气1分钟跳绳成绩分别是165下,185下和186下，他第四次跳了175下。他跳绳的平均成绩会因为最后一次成绩而()。",
+        "options":  [
+                        "A. 上升",
+                        "B. 下降",
+                        "C. 不变",
+                        "D. 无法确定"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据求平均数的方法，用淘气前三次跳的下数和除以3，求出平均数，再和第四次跳的下数比较，如果第四次跳的下数大于前三次的平均数，则他跳绳的平均成绩会因为最后一次成绩而上升，反之，则下降。据此解答.\n【详解】(165+185+186)÷3\n=(350+186)÷3\n=536÷3\n≈178.7(下)\n178.7\u003e175\n所以他跳绳的平均成绩会因为最后一次成绩而下降。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24五年级下·内蒙古包头·期末"
+    },
+    {
+        "id":  "351",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "四(1)班平均每个小组回收废报纸25千克。根据这条信息，下面说法正确的是()。",
+        "options":  [
+                        "A. 所有小组回收报纸都在25千克到26千克之间",
+                        "B. 一半小组回收报纸比25千克少，一半小组比25千克多",
+                        "C. 至少有一个小组回收报纸是25千克",
+                        "D. 有一些小组回收报纸小于或等于25千克"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】本题主要考查平均数的特点，在一组数据中，平均数具有惟一性，是一个\"虚拟\"的数，是通过计算得到的，它不是数据中的原始数据，只是在平均数计算出来后，有和某一个原始数据相等的可能。平均数与每一个数据都有关，其中任何数据的变动都会相应引起平均数的变动。主要缺点是易受极端值的影响，这里的极端值是指偏大或偏小数，当出现偏大数时，平均数将会被抬高，当出现偏小数时，平均数会降低.\n【详解】A。平均每个小组回收废报纸25千克并不代表所有小组回收报纸都在25千克到26千克之间，有可能一个数特别小，一个数特别大也可以使得平均数为25千克，不符合题意。B。平均每个小组回收废报纸25千克并不能说明一半小组回收报纸比25千克少，一半小组比25千克多，不符合题意。C。平均每个小组回收废报纸25千克并不能肯定至少有一个小组回收报纸是25千克，不符合题意。D。四(1)班平均每个小组回收废报纸25千克，不可能每一个小组数的回收报纸数都大于25千克，如果都大于25千克，那么平均数就不可能是25千克，所以有一些小组回收报纸小于或等于25千克，符合题意。故答案为:D",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·天津河西·期末"
+    },
+    {
+        "id":  "352",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "小琪玩拍球游戏，三次的平均成绩是98下。第一次拍了96下，第二次拍了100下。她第三次拍了()下。",
+        "options":  [
+                        "A. 98",
+                        "B. 97",
+                        "C. 96"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】用平均成绩98下乘3求出三次拍球的总数，再减去第一次和第二次的即可。据此代入数值求解.\n【详解】98×3=294(下)294-96-100=98(下) 她第三次拍了98下。 故答案为:A",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·天津蓟州·期末"
+    },
+    {
+        "id":  "353",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "137,138,139,141,145这五个数的平均数是()。",
+        "options":  [
+                        "A. 139",
+                        "B. 140",
+                        "C. 141"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】平均数是指在一组数据中所有数据之和再除以这组数据的个数。据此将5个数字相加，求出5个数字和，再用5个数字和除以5，即可求出这五个数的平均数.\n【详解】(137+138+139+141+145)÷5\n=(275+139+141+145)÷5\n=(414+141+145)÷5\n=(555+145)÷5\n=700÷5\n=140\n137,138,139,141,145这五个数的平均数是140。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法",
+                                 "除加",
+                                 "除减混合运算"
+                             ],
+        "source":  "23-24四年级下·黑龙江佳木斯·期末"
+    },
+    {
+        "id":  "354",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "有6个数的平均数是61，去掉一个数后，余下数的平均数是60，去掉的数是()。",
+        "options":  [
+                        "A. 60",
+                        "B. 61",
+                        "C. 65",
+                        "D. 66"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】首先用平均数61乘6，就是6个数的总和，再用平均数60乘5，就是5个数的总和，最后用6个数的总和减去5个数的总和，即能求出去掉的数。据此解答即可.\n【详解】61×6-60×5=366-300=66去掉的数是66。故答案为:D",
+        "knowledge_points":  [
+                                 "乘加",
+                                 "乘减混合运算",
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·江西南昌·期末"
+    },
+    {
+        "id":  "355",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "四(1)班有50名学生，全班一分钟跳绳测试的平均成绩是116个。小明，小芳，小丽都是这个班的学生，他们的成绩各不相同，下面说法正确的是()。",
+        "options":  [
+                        "A. 如果小明的成绩是三人中最低的，那么他的成绩肯定小于116个。",
+                        "B. 小明，小芳，小丽三人中肯定有人的成绩正好是116个。",
+                        "C. 如果小丽的成绩是全班最高的，那么她的成绩肯定大于116个。",
+                        "D. 这三名同学的平均成绩肯定也是116个。"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】一组数据的总和除以这组数据的个数，所得到的商叫这组数据的平均数。平均数反映一组数据的平均情况，在一组数据中，有的数据可能会大于平均数，有的数据可能会小于平均数，有的数据可能会等于平均数。平均数大于一组数据的最小值，且小于这组数据的最大值.\n【详解】A。全班一分钟跳绳测试的平均成绩是116个，而其中三人的平均成绩不一定是116个;小明的成绩是三人中最低的，也并非全班最低，所以他的成绩不一定小于116个;原题说法错误;B。小明，小芳，小丽三人成绩具有不确定性，可能会大于平均数，可能会小于平均数，也可能会等于平均数;所以原题说法错误;C。小丽的成绩是全班最高的，全班最高的成绩一定大于平均成绩，所以她的成绩大于116个;原题说法正确;D。全班一分钟跳绳测试的平均成绩是116个，而其中这三人的平均成绩不一定是116个;所以原题说法错误。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·湖北武汉·期末"
+    },
+    {
+        "id":  "356",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "四一班学生平均身高为138厘米。下面说法正确的是()。",
+        "options":  [
+                        "A. 四一班一定有学生身高为138厘米。",
+                        "B. 四一班如果转走了两名同学，平均身高一定还是138厘米。",
+                        "C. 四一班可能有比138厘米高的同学，也可能有比138厘米低的同学。"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】平均数是反映一组数据的集中趋势的量，它不能反映具体数据，据此解答即可。熟练掌握平均数的含义和求法，是解答此题的关键.\n【详解】由分析可知:A。四一班可能有学生身高为138厘米，原题说法错误;B。四一班如果转走了两名同学，平均身高可能还是138厘米，但不一定，原说法错误;C。四一班可能有比138厘米高的同学，也可能有比138厘米低的同学，故符合题意。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·河南郑州·期末"
+    },
+    {
+        "id":  "357",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "黄河文化公园位于河南省会郑州市西北20公里处黄河之滨，南依巍巍岳山，北临滔滔黄河。黄河露营地适合团建游玩。周末四二班的学生在此举行了踢毽子比赛，小郑所在的小组平均每人踢36个，小新所在的小组平均每人踢32个。下列说法正确的是()。",
+        "options":  [
+                        "A. 小郑一定比小新踢得多",
+                        "B. 小郑一定比小新踢得少",
+                        "C. 两人踢的一样多",
+                        "D. 无法确定他俩谁踢得多一些"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】根据平均数的意义可知，平均数只是反映的一组数据的集中趋势，不表示这组数据中某一个具体数据。小郑所在的小组平均每人踢36个，小新所在的小组平均每人踢32个,36和32是平均数，所以无法确定小郑和小新谁踢得多，谁踢得少，据此解答即可.\n【详解】由分析可知，无法确定小郑和小新谁踢得多，谁踢得少。故答案为:D",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·河南郑州·期末"
+    },
+    {
+        "id":  "358",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面说法正确的选项是()。",
+        "options":  [
+                        "A. 奇奇5次跳远的总成绩是10米，他每次的跳远成绩肯定都是2米",
+                        "B. 妙妙所在小组同学的平均体重是34千克，淘气所在小组同学的平均体重是36千克，淘气一定比妙妙重",
+                        "C. 一个水库的平均水深138厘米，强强身高1.5米，去游泳没有危险",
+                        "D. 学校足球队队员的平均身高是160厘米，有的队员身高会超过160厘米，有的队员身高不到160厘米"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】平均数是反映一组数据的平均水平，并不能反映这组数据中各个数据的大小，平均数介于这组数据的最大值和最小值之间;一组数据的和除以这组数据的个数，所得的商叫这组数据的平均数，它是描述一组数据集中趋势的统计量;据此逐项分析解答即可.\n【详解】A。奇奇5次跳远的总成绩是10米，他每次的跳远成绩不一定都是2米;总成绩/次数=平均成绩,10÷5=2(米),平均成绩是2米，不是每次肯定2米，故原题说法错误;B。妙妙和淘气所在小组同学的平均体重代表不了她们个人的实际体重，也就是说，妙妙的实际体重可能比淘气重，也可能比淘气轻，还可能与淘气一样重;故原题说法错误;C。1.5米=150厘米，在平均水深138厘米的水库里游泳，并不代表每处的水深都是138厘米，可能比138厘米深，也可能比138厘米浅，还可能等于138厘米，所以强强在平均水深138厘米的水库里游泳，可能有危险，故原题说法错误;D。足球队队员的个人身高会低于或者高于平均身高，故原题说法正确。故答案为:D",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·山东济南·期末"
+    },
+    {
+        "id":  "359",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "丽丽所在小组同学的平均分是87分，明明所在小组同学的平均分是90分，丽丽的个人得分与明明的个人得分相比,()。",
+        "options":  [
+                        "A. 比明明的高",
+                        "B. 比明明的低",
+                        "C. 与明明相等",
+                        "D. 无法判定"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】根据平均数的意义，平均数只能反映一组数据的平均水平，它会受极端数据的影响。只是反应整体水平，不能代表个人的水平，据此解答即可.\n【详解】结合分析，丽丽所在的小组同学的平均分是87分，有的同学的成绩可能比87分高得多，有的同学的成绩可能比87分低得多;同理，明明所在的小组同学的平均分是90分，有的同学的成绩可能比90分高得多，有的同学的成绩可能比90分低得多;所以丽丽的得分与明明的得分无法得出一个具体的数值，也就无法判断。故答案为:D",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·山东济南·期末"
+    },
+    {
+        "id":  "360",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "六年级同学参加趣味知识竞赛，男生组的平均成绩是86分，女生组的平均成绩是83分。男生组的第一名和女生组的第一名相比,()。",
+        "options":  [
+                        "A. 男生的成绩高",
+                        "B. 女生的成绩高",
+                        "C. 一样",
+                        "D. 无法确定谁的成绩高"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】平均数=总数量/总份数，平均成绩只能反映这组的整体情况，不能反映个体的情况;知道男生组和女生组的平均成绩，但不能确定男生组第一名和女生组第一名的成绩是多少，故无法确定谁的成绩高，据此即可解答.\n【详解】根据分析可知，六年级同学参加趣味知识竞赛，男生组的平均成绩是86分，女生组的平均成绩是83分。男生组的第一名和女生组的第一名相比，无法确定谁的成绩高。故答案为:D",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24六年级下·湖南岳阳·期末"
+    },
+    {
+        "id":  "361",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "阳光小学举行\"金话筒\"主持人大赛，在评分时采用去掉一个最高分和一个最低分后再计算选手平均分的评分规则。你认为这种评分规则的合理性在于()。",
+        "options":  [
+                        "A. 可以减少评委统计分数时的工作量",
+                        "B. 平均数具有代表性，可以帮助我们解决问题",
+                        "C. 这样可以提高选手的平均分，能更好地鼓励选手",
+                        "D. 平均数易受较大数和较小数的影响，去掉最高分和最低分后算得的平均分更有代表性。"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】平均数反映一组数据的平均情况。在一组数据中，去掉极端数据再求平均数，更具有代表性。主持人大赛中，因为评委的评分常带有主观性，因此去掉一个最高和最低分，能够使评分更具公平性.\n【详解】由分析结合所给选项可得，评分时采用去掉一个最高分和一个最低分后再计算选手平均分的评分规则。你认为这种评分规则的合理性在于平均数易受较大数和较小数的影响，去掉最高分和最低分后算得的平均分更有代表性。故答案为:D",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24五年级下·广西贺州·期末"
+    },
+    {
+        "id":  "362",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "小明所在小组同学的平均身高是165cm,小华所在小组的同学的平均身高是163cm,下面的说法中，错误的是()。",
+        "options":  [
+                        "A. 小明一定比小华高",
+                        "B. 小华有可能比小明高",
+                        "C. 小华所在小组有的同学的身高会不到163cm",
+                        "D. 小明所在小组有的同学的身高会超过165cm"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】平均身高是代表一个组同学身高的平均水平，可能这个小组的某些同学比平均身高高得多，也可能某些同学比平均身高矮得多，所以，小明所在小组同学的平均身高是165cm,小华所在小组的同学的平均身高是163cm,小明可能比小华高，可能比小华矮，据此解题即可.\n【详解】根据分析可知，小明所在小组同学的平均身高是165cm,小华所在小组的同学的平均身高是163cm,小明可能比小华高，可能比小华矮。A。小明不一定比小华高，故此项说法错误;B。小华有可能比小明高，故此项说法正确;C。小华所在小组有的同学的身高会不到163cm,故此项说法正确;D。小明所在小组有的同学的身高会超过165cm,故此项说法正确。故答案为:A",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·广东广州·期末"
+    },
+    {
+        "id":  "363",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "三个连续自然数的平均数是12，那么三个数中最小的数一定是11。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】由题意可得，三个连续的自然数的平均数是12，说明这3个数分别为:12-1,12,12+1，即这三个连续的自然数就为11,12,13，其中最小的一个自然数是11，据此判断即可.\n【详解】12-1=11,12+1=13，这三个连续自然数分别为11,12,13，所以三个数中最小的数一定是11。故答案为:√",
+        "knowledge_points":  [
+                                 "自然数的认识",
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·河南焦作·期末"
+    },
+    {
+        "id":  "364",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "小东所在小组同学的平均体重是36kg,小刚所在小组同学的平均体重是34.5kg,小东一定比小刚重。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数反应的是一组数的平均水平。小东所在小组同学的平均体重是36kg,小东的体重有可能大于36kg,等于36kg,或者小于36kg,这都是有可能的。小刚所在小组同学的平均体重是34.5kg,小刚的体重有可能大于34.5kg,等于34.5kg,或者小于34.5kg,这都是有可能的，两人的体重无法确定，所以谁的体重更大一些，也无法确定.\n【详解】小东所在小组同学的平均体重是36kg,小刚所在小组同学的平均体重是34.5kg,小东不一定比小刚重，原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·河南省直辖县级单位·期末"
+    },
+    {
+        "id":  "365",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "一条小河平均水深为120cm,小强身高135cm,所以他在这条小河中游泳不会有危险。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数只能反映一组数据的平均水平，并不能反映这组数据的中所有数据的大小，有的数可能比平均数大，有的数可能比平均数小;河水的平均水深是120cm,可能有的地方水深超过120cm,甚至超过135cm,所以小强身高135cm下水游泳可能有危险.\n【详解】根据平均数只能反映一组数据的平均水平，并不能反应一组数据的最高水平可知，一条小河平均水深为120cm,小强身高135cm,所以他在这条小河中游泳不会有危险，这句话是错误的。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·河南·期末"
+    },
+    {
+        "id":  "366",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "河水平均深度是100cm,身高155cm的小明下河游泳不会有危险。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数并不能反应这组数据的中所有数据的大小。河水平均深度是100cm,可能有的地方水深超过100cm很多，甚至超过155cm,所以他淌水过河可能有危险，据此解答即可.\n【详解】由分析可知，河水平均深度是100cm,身高155cm的小明下河游泳可能会有危险。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·河南洛阳·期末"
+    },
+    {
+        "id":  "367",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.94,
+        "question":  "学校篮球队队员的平均身高为172cm,则所有队员的身高不低于172cm。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数是反映一组数据的平均水平，并不能反映这组数据中各个数据的大小，由此即可进行判断.\n【详解】学校篮球队队员的平均身高为172cm,则有的队员的身高可能不到172cm,也可能比172cm高，还可能就是172cm。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·河南商丘·期末"
+    },
+    {
+        "id":  "368",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "小强的身高是132厘米，一条小河的平均水深是120厘米，他蹚水从河岸这一边到对岸，肯定安全。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数反映一组数据的平均水平，并不能反映这组数据里各个数据的大小，平均数会受极端数据的影响，因此平均水深是120厘米，并不能反映出整个小河中每一处的水深大小，有的地方会深一些，有的地方会浅一些，据此判断.\n【详解】根据解析可知，平均水深是120厘米，则有的地方可能比120厘米深，也有可能有的地方可能比120厘米浅，也有可能有的地方可能是120厘米;所以他蹚水从河岸这一边到对岸，可能有危险，原题表达错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·河南安阳·期末"
+    },
+    {
+        "id":  "369",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "三(2)班学生的平均体重是34千克，三(2)班不可能有体重低于32千克的学生。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】根据平均数的意义，一组数据的平均数一定大于这组数据中的最小数，一定小于这组数据中的最大数。平均数是一组数据中所有数据之和除以数据的个数，是表示一组数据集中趋势的量数;三(2)班学生的平均体重是34千克，并不是每个学生的体重都是34千克，有的学生的体重可能比34千克高得多，也有的学生的体重可能比34千克低得多，据此解答即可.\n【详解】三(2)班学生的平均体重是34千克，三(2)班可能有体重低于32千克的学生。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24三年级下·河南·期末"
+    },
+    {
+        "id":  "370",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "小明身高140cm,所以他去平均水深120cm的河里游泳不会有危险。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数反映一组数据的平均情况，在一组数据中，有的数据可能会大于平均数，有的数据可能会小于平均数，有的数据可能会等于平均数.\n【详解】平均水深是120厘米的河里，有的地方的深度可能大于120厘米，有的地方的深度可能小于120厘米，有的地方的深度可能等于120厘米，所以河里某些地方的深度可能会超过140厘米，小明身高140cm,会有危险，故原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·河南信阳·期末"
+    },
+    {
+        "id":  "371",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "小红从家到学校平均用15分钟。一天，她7:45从家出发，那么她8:00前一定能到校。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数代表的是一组数据的平均水平。在这组数据中，有的数会比平均数多，有的数会比平均数少，还有的数可能等于平均数。据此解答.\n【详解】小红从家到学校平均用15分钟，那么她有时候需要的时间会比15分钟短，有的时候需要的时间会比15分钟长。7:45+15分钟=8:00，如果某天小红在路上的用时超过了15分钟，那么她8:00前就不能到校。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "时，分，秒时间的推算",
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·河南信阳·期末"
+    },
+    {
+        "id":  "372",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "复式条形统计图可以直观地比较两种小麦在不同地区的产量。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】条形统计图是用一个单位长度表示一定的数量，根据数量的多少画成长短不同的直条，然后把这些按照一定的顺序排列起来。复式条形统计图可以用不同的条形表示2种及以上的量，复式条形统计图可以直观地比较多种量的多少.\n【详解】由分析可知，\"复式条形统计图可以直观地比较两种小麦在不同地区的产量\"此说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "复式条形统计图"
+                             ],
+        "source":  "23-24四年级下·河南信阳·期末"
+    },
+    {
+        "id":  "373",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "某班体育测试的平均成绩是93分，可能有人得分62分。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】平均数是一组数的总和除以这组数的数量。平均数反映的是一组数据的平均水平，平均数会受偏大或偏小数据的影响.\n【详解】根据解析可知，平均成绩是93分，则有人成绩可能比93分高，也有可能比93分低，也有可能正好是93分;所以原表达正确。故答案为:√",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·河南新乡·期末"
+    },
+    {
+        "id":  "374",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "四(1)班学生平均体重是38千克，有的学生体重会超过38千克。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】平均数反映的一组数据的整体情况，它比最大的数据小，比最小的数据大，处于这组数据的中间;四(1)班学生平均体重是38千克，并不代表四(1)班学生所有学生的体重都是38千克，有的学生的体重可能比38千克重;有的学生体重可能比38千克轻。据此判断.\n【详解】根据分析可知：四(1)班学生平均体重是38千克，有的学生体重会超过38千克。原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·河南新乡·期末"
+    },
+    {
+        "id":  "375",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "王明身高1.5m,在平均水深1.2m的河里游泳没有危险。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数是反映一组数据的平均水平，并不能反应这组数据的中各个数据的大小，由此即可进行判断.\n【详解】平均水深1.2m并不能反映出整条河中每一处的水深都是1.2m,有的地方会比1.2m深一些，有的地方会比1.2m浅一些，也有可能正好是1.2m,所以王明身高1.5m,他在平均水深1.2m的河里游泳可能有危险，原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·河南·期末"
+    },
+    {
+        "id":  "376",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "小明所在小组的平均身高是142厘米，小强所在小组的平均身高是144厘米，小强比小明高。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】一组数据的总和除以这组数据的个数，所得到的商就是这组数据的平均数。平均数反映一组数据的平均情况，在一组数据中，有的数据可能会大于平均数，有的数据可能会小于平均数，有的数据可能会等于平均数，据此判断即可.\n【详解】根据题意,142厘米和144厘米是小明和小强所在小组的平均身高，并不代表小明和小强的实际身高，如果小明的身高是145厘米，小强的身高是144厘米，小明就比小强高，所以小明所在小组的平均身高是142厘米，小强所在小组的平均身高是144厘米，小强比小明高，说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·河南·期末"
+    },
+    {
+        "id":  "377",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "强强身高1.48m,他到平均水深1.05米的池塘中游泳，一定不会有危险。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】一组数据的总和除以这组数据个数所得到的商叫这组数据的平均数，平均水深1.05米，可能有的地方比1.05米浅，也可能有的地方比1.05米深，据此判断即可.\n【详解】强强身高1.48m,他到平均水深1.05米的池塘中游泳，可能会有危险。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·河南·期末"
+    },
+    {
+        "id":  "378",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "池塘的水平均深1米，小强身高1.3米，他下河玩耍可能会有危险。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】平均水深不能代表所有的水深都是1米，所以有的地方可能小于1米，有的地方可能大于1米，据此解答.\n【详解】根据解析可知，池塘的水平均深1米，小强身高1.3米，他下河玩耍可能会有危险，原题表达正确。故答案为:√",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·河南·期末"
+    },
+    {
+        "id":  "379",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "甲所在班级同学平均身高152cm,乙所在班级平均身高150cm,甲一定比乙高。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】根据平均数的意义，平均数是反映一组数据的平均水平，平均数会受极端数据的影响，由此可知，甲所在班级同学的平均身高是152cm,甲的身高可能比152cm高，也可能比152cm矮;乙所在班级的平均身高是150cm,小华的身高可能比150cm高，也可能比150cm矮;所以甲不一定比乙高，据此判断即可.\n【详解】甲所在班级同学平均身高152cm,乙所在班级平均身高150cm,甲不一定比乙高。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·广东汕头·期末"
+    },
+    {
+        "id":  "380",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "四(1)班全体42名同学的平均体重是39千克，这个班有可能出现体重是48千克的情况。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】根据平均数的含义，平均数可以代表一组数据的总体水平或中心位置。四(1)班全体42名同学的平均体重是39千克，反映的是这个班级学生的总体体重情况，可能会出现比39千克体重更高的，也会出现比39千克体重更低的，据此判断即可.\n【详解】结合分析，四(1)班全体42名同学的平均体重是39千克，这个班有可能出现体重是48千克的情况。原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "22-23四年级下·广东汕头·期末"
+    },
+    {
+        "id":  "381",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "一条河的平均水深是1.5m,这条河中一定没有比1.5m深的地方。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数只能反映一组数据的平均水平，并不能反映这组数据的中所有数据的大小，一条河的平均水深是1.5米，可能有的地方水深超过1.5米很多，甚至超过2米，据此判断.\n【详解】因为平均数不能反映这组数据的中所有数据的大小，所以一条河的平均水深是1.5米，这条河中可能有比1.5米深的地方。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·新疆巴音郭楞·期末"
+    },
+    {
+        "id":  "382",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "小东身高1.50米，他不会游泳，要趟过一条平均水深为1.40米的河，不会有任何危险。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】这条河的平均水深是1.40米，那么这条河有的地方可能水深是大于1.40米的，有些地方水深可能是小于1.40米的，并不是处处都是1.40米，所以小东有可能是有危险的.\n【详解】小东身高1.50米，他不会游泳，要趟过一条平均水深为1.40米的河，可能会有危险，所以原题干说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·新疆克孜勒苏·期末"
+    },
+    {
+        "id":  "383",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "学校篮球队队员的平均身高是160厘米，王强是篮球队队员，他的身高不可能比160厘米矮。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数表示一个\"虚拟\"的数，是表示一组数据集中趋势的量数，是指在一组数据中所有数据之和再除以这组数据的个数。它是反映数据集中趋势的一项指标。一组样本中，样本数据可能比平均数大，也可能比平均数小。据此解答.\n【详解】由分析可知，学校篮球队队员的平均身高是160厘米，王强是篮球队队员，他的身高可能比160厘米高，也可能比160厘米矮。题目说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·浙江·期末"
+    },
+    {
+        "id":  "384",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "李强所在的篮球队，队员的平均身高是160厘米，所以李强的身高是160厘米。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数：一组数据的总和除以这组数据个数所得到的商叫这组数据的平均数;平均数只能反映一组数据的平均水平，并不能反应这组数据中所有数据的大小。据此解答.\n【详解】根据分析：篮球队员平均身高160厘米，队员的身高可能大于160厘米，也可能等于160厘米，还可能小于160厘米，所以李强的身高不一定是160厘米，原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·江西宜春·期末"
+    },
+    {
+        "id":  "385",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "四(1)班男生平均体重是32kg,女生平均体重是28kg,这个班的男生李华一定比女生丽丽重。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】结合平均数的含义可知，平均数常用于表示统计对象的一般平均水平，它代表的不是每个人的具体水平，所以男生的平均体重不代表所有男生的体重，女生的平均体重也不代表所有女生的体重，据此判断.\n【详解】结合平均数的含义，四(1)班男生平均体重是32kg,女生平均体重是28kg,这个班的男生李华一定比女生丽丽重的说法有误，李华可能比丽丽重，也可能比丽丽轻。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·江西吉安·期末"
+    },
+    {
+        "id":  "386",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "甲，乙，丙是三个连续的非零自然数，它们的平均数是乙。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】平均数是表示一组数据的平均值，是指在一组数据中所有数据之和再除以这组数据的个数;由此可知，三个连续的非零自然数的平均数是它们从小到大排列的最中间的数，依此判断.\n【详解】甲，乙，丙是三个连续的非零自然数，它们的平均数是乙。比如1,2,3的平均数是2。故答案为:√",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·湖南岳阳·期末"
+    },
+    {
+        "id":  "387",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "军军身高138cm,他要过平均水深1.25m的小河没有任何危险。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】根据1米=100cm,先将138cm的小数点向左移动2位，换算为以m作单位;再根据平均水深1.25m,并不能反映出整条小河中每一处的水深大小，有的地方会深一些，有的地方会浅一些，所以身高1.38m的军军要过河有危险，据此作答.\n【详解】138cm=1.38m,有的地方水深会大于1.25m,有的地方水深甚至大于1.38m,所以过河可能有危险，原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·湖南娄底·期末"
+    },
+    {
+        "id":  "388",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "学校篮球队中男队员的平均身高是160厘米。下列说法是否正确。正确的请画\"√\"，错误的请画\"*\"。",
+        "options":  [
+                        "A. 校篮球男队每个队员的身高一定是160厘米。( )",
+                        "B. 李强是学校篮球队的，他的身高可能是155厘米。( )",
+                        "C. 身高170厘米的小亮新加入了校篮球队，现在篮球男队平均身高会超过160厘米。( )"
+                    ],
+        "answer":  "*;√;√",
+        "solution":  "【分析】根据平均数的认识，平均数是指一组数据的和除以这组数据的个数所得的商，这组数据可能比平均数大，也可能比平均数小.\n【详解】A。平均身高是160厘米，并不意味着每个队员的身高都是160厘米。平均值是所有队员身高的总和除以队员人数的结果，因此这个说法是错误的.\nB。李强的身高可能是155厘米，因为平均身高是160厘米，个别队员的身高可以高于或低于这个值，所以这个说法是正确的.\nC。小亮的身高是170厘米，高于平均身高160厘米，加入后会拉高整体的平均身高，因此这个说法是正确的.\n故答案为: ×;√;√",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·湖南株洲·期末"
+    },
+    {
+        "id":  "389",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "四(1)班同学平均身高为1.37m,那么他们班的王强不可能高1.52m。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数是反映一组数据的平均水平，它介于这组数据的最大值和最小值之间，据此即可解答.\n【详解】根据分析可知，四(1)班同学平均身高1.37m,部分同学的身高大于1.37m,所以王强身高可能是1.52m,原说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·湖南张家界·期末"
+    },
+    {
+        "id":  "390",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "有4个数，它们的平均数是12，若把其中一个数改为3，这4个数的平均数就变为11，则这个被改动的数是7。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】平均数=总数/总份数，则总数=平均数×总份数。这4个数的平均数是12，则这4个数的和是(12×4);改动后的4个数的平均数是11，则改动后4个数的和是(11×4),与改动前相比，和少了(12×4-11×4),也就是改动的数少了(12×4-11×4),(12×4-11×4)加上3，即可算出被改动的数是几.\n【详解】12×4-11×4=48-44=4,4+3=7有4个数，它们的平均数是12，若把其中一个数改为3，这4个数的平均数就变为11，则这个被改动的数是7。故答案为:√",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·湖南张家界·期末"
+    },
+    {
+        "id":  "391",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "小明5次跳远的总成绩是10米，他每次的跳远成绩肯定都是2米。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】根据题意，小明5次跳远的总成绩是10米，平均每次跳远的成绩是10÷5=2米，但这并不意味着他每次跳远的成绩都是2米。平均数只是所有数据的总和除以数据的个数，不能代表每个数据的具体值.\n【详解】根据对平均数意义的认识，平均数只能表示整体水平，不能代表每个数据的具体值。原题说法错误.\n故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·湖南长沙·期末"
+    },
+    {
+        "id":  "392",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "四(2)班同学去植树，平均每人植2棵，四(2)班的王玲一定植了2棵。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均每人植2棵，即这个班植树的平均数是2，平均数反应的是一组数的平均水平，那么王玲植树的数量，有可能比2棵多，也有可能比2棵少，也有可能正好是2棵.\n【详解】四(2)班同学去植树，平均每人植2棵，四(2)班的王玲一定植了2棵，这句话说法不对.\n故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·湖北省直辖县级单位·期末"
+    },
+    {
+        "id":  "393",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "第二小学教师的平均年龄是39岁，第二小学的张老师的年龄可能是40岁。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】\"39\"是这一组数据的平均数，平均数并不是每个教师的实际年龄，而是代表第二小学教师年龄的一种平均水平，可能有的教师年龄比39岁大，也可能有的教师的年龄比39岁小，因此第二小学的张老师的年龄可能是40岁.\n【详解】第二小学教师的平均年龄是39岁，第二小学的张老师的年龄可能是40岁。原题说法正确.\n故答案为:√",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·湖北孝感·期末"
+    },
+    {
+        "id":  "394",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "四(1)班同学平均身高是142cm,小明的身高一定是142cm。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数是表示一组数据集中趋势的量数，它是反映数据集中趋势的一项指标;小明所在班全班同学的平均身高是142cm,只是说明学生身高的平均数是142cm,小明的身高有可能比142cm大，有可能小于142cm,也有可能等于142cm,据此判断即可.\n【详解】小明的身高有可能比142cm大，有可能小于142cm,也有可能等于142cm,所以小明的身高一定是142cm不正确。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·湖北恩施·期末"
+    },
+    {
+        "id":  "395",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "校园歌手大赛,6号选手的得分分别是91分,95分,92分,95分,97分，去掉一个最高分和一个最低分，他的平均分是95分。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】由题目可知，首先找到最高分和最低分，最高分是97分，最低分是91分，不计入计算，将剩下的分数加起来求出和除以个数即为平均数，据此解答即可.\n【详解】由分析可知:(92+95+95)÷3=282÷3=94(分)所以他的平均分是94分，原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·湖北鄂州·期末"
+    },
+    {
+        "id":  "396",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "校篮球队队员的平均身高是155cm,其中篮球队员王亮身高可能是145cm。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】平均数是一组数据的总和除以数据总个数得到的，这组数据中的某个数据可能大于平均数，可能小于平均数，也有可能等于平均数，据此解答即可.\n【详解】校篮球队队员的平均身高是155cm,其中篮球队员王亮身高可能是145cm。原题说法正确.\n故答案为:√",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·湖北黄冈·期末"
+    },
+    {
+        "id":  "397",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "小丽所在的班级同学的平均体重是35kg,小丽的体重不可能超过35kg。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】一组数据的总和除以这组数据个数所得到的商叫这组数据的平均数。平均体重是35kg,小丽的体重可能低于35kg,也可能高于35kg,也可能等于35kg,据此判断即可.\n【详解】小丽所在的班级同学的平均体重是35kg,小丽的体重可能超过35kg,原题说法错误.\n故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·湖北黄冈·期末"
+    },
+    {
+        "id":  "398",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "四(1)班同学的平均身高是131厘米，这个班不可能有身高超过151厘米的同学。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数代表一组数据的平均水平。四(1)班同学的平均身高是131厘米，那么这个班的同学的身高有可能比131厘米高，也有可能比131厘米矮，还有可能等于131厘米.\n【详解】由分析得，四(1)班同学的平均身高是131厘米，这个班有可能有身高超过151厘米的同学。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "23-24四年级下·湖北·期末"
+    },
+    {
+        "id":  "399",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "求下列20个数的平均数:306,312,306,308,314,304,318,311,313,315,314,310,310,320,300,316,320,312,314,315",
+        "options":  [
+
+                    ],
+        "answer":  "311.9",
+        "solution":  "【分析】本题考查平均数的简便计算。当一组数据都在某个基准数附近时，可以先用每个数减去基准数，求出差的平均数，再加上基准数，即可得到原数据的平均数。这里以300为基准数.\n【详解】这些数与300很接近，我们以300作为基准数，即先将每个数减去300，再求这些做差后的数的平均数，再将这个平均数加上300即为所求。有(6+12+6+8+14+4+18+11+13+15+14+10+10+20+0+16+20+12+14+15)÷20=11.9所以题中原来这些数的平均数为311.9。",
+        "knowledge_points":  [
+                                 "平均数问题",
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "2015六年级·全国·竞赛"
+    },
+    {
+        "id":  "400",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "甲数是33.5，乙数与丙数的平均数是30.5，这三个数的平均数是多少。",
+        "options":  [
+
+                    ],
+        "answer":  "31.5",
+        "solution":  "【分析】要求这三个数的平均数是多少，先根据\"平均数×数的个数=总成绩\"求出乙，丙两数的和;然后根据平均数的求法，用三个数的和除以3即可得出结论。此题应根据平均数，数的个数和总数的关系进行分析，进而得出结论.\n【详解】(30.5×2+33.5)÷3\n=94.5÷3\n=31.5\n答：这三个数的平均数是31.5。",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "2015·福建·小升初模拟"
+    },
+    {
+        "id":  "401",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "学校篮球队队员的平均身高是170cm下面两种说法中()是不合理的。",
+        "options":  [
+                        "A. 学校篮球队中有的队员身高会超过170cm。",
+                        "B. 学校篮球队中不会有低于170cm的队员。"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据\"平均数=总身高/总人数\"，平均身高和总身高，总人数有关，并不能代表一个人的身高，由此可得，篮球队队员的身高不确定，可能比平均数高，也可能比平均数低，也可能等于平均数，据此解答即可。此题主要考查对平均数的基础知识的掌握情况，做题时一定要弄清题意，认真审题，然后做出判断.\n【详解】A。学校篮球队中有的队员身高会超过170cm,说法正确;B。学校篮球队中不会有低于170cm的队员，说法错误;故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22四年级下·河北唐山·期末"
+    },
+    {
+        "id":  "402",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "某印刷厂上半年印刷48万册图书，平均每月印多少万册。下面列式正确的是()。",
+        "options":  [
+                        "A. 48÷12",
+                        "B. 48÷2",
+                        "C. 48÷6"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】半年是6个月,6个月刷48万册图书，求平均每月印多少万册，用48除以6计算解答。解答本题的关键是知道半年是6个月，再用平均分问题的解决方法解答.\n【详解】根据分析可知：某印刷厂上半年印刷48万册图书，平均每月印多少万册。列式为:48÷6=8(万册)故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22四年级下·河北唐山·期末"
+    },
+    {
+        "id":  "403",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "四(1)班组织跳绳比赛，第一组4人，平均每人跳了96个;第二组5人，总共跳了485个。()获胜。",
+        "options":  [
+                        "A. 第一组",
+                        "B. 第二组",
+                        "C. 不能确定"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】平均数的计算方法是：一组数据中所有数据之和再除以这组数据的个数。用485÷5，求出第2组平均每人跳多少个，再与第1组平均每人跳了96个，作比较即可。明确平均数的意义，熟练计算三位数除以一位数，是解题关键.\n【详解】485÷5=97(个)96\u003c97 第二组平均每人跳的多。 故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22四年级下·河北沧州·期末"
+    },
+    {
+        "id":  "404",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "莉莉数学，英语，语文三科的平均成绩是95分，其中英语成绩是93分，语文成绩是96，数学成绩是()分。",
+        "options":  [
+                        "A. 95",
+                        "B. 96",
+                        "C. 97",
+                        "D. 98"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】数学，英语，语文三科的平均成绩*3=数学，英语，语文三科的总成绩，数学，英语，语文三科的总成绩-英语成绩-语文成绩=数学成绩，依此计算并选择。此题考查的是平均数的求法，先计算出数学，英语，语文三科的总成绩是解答此题的关键.\n【详解】95×3=285(分)285-93-96=192-96=96(分)即数学成绩是96分。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22四年级下·河北邯郸·期末"
+    },
+    {
+        "id":  "405",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "有5个互不相等的整数，它们的平均数是90，如果去掉其中最大的一个数，剩下的4个数的平均数()。",
+        "options":  [
+                        "A. 大于90",
+                        "B. 等于90",
+                        "C. 小于90",
+                        "D. 无法确定"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】平均数=总数/数据个数，如果去掉其中最大的一个数，由平均数的意义可知，平均数比整体中最大的数据小，所以平均数将会降低，据此解答即可。本题考查学生对平均数的掌握。解答此题的关键是掌握平均数的相关公式.\n【详解】有5个互不相等的整数，它们的平均数是90，如果去掉其中最大的一个数，剩下的4个数的平均数小于90。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22四年级下·湖北武汉·期末"
+    },
+    {
+        "id":  "406",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面关于\"平均数\"的描述正确的是()。",
+        "options":  [
+                        "A. 小明平均每天跑3km,每天一定都跑了3km",
+                        "B. 四(2)班男生的平均体重为30kg,女生的平均体重为28kg,四(2)班的男生张良一定比女生贾灿重",
+                        "C. 现役中国女排队员中，身高最矮的是处于自由人位置的队员林莉，她的身高是1.71m,中国女排的平均身高肯定大于1.71m"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】平均数是一组数据的集中趋势，不能反映具体数据，平均数比最大数小，比最小数大，据此解答即可。本题考查学生对平均数意义的掌握。解答此题时，要注意平均数只能代表整体的平均水平，不能代表整体中的个体.\n【详解】A。小明平均每天跑3km,每天不一定都跑了3km,所以题干说法是错误的;B。四(2)班男生的平均体重为30kg,女生的平均体重为28kg,四(2)班的男生张良无法和女生贾灿比较谁重，所以题干说法是错误的;C。现役中国女排队员中，身高最矮的是处于自由人位置的队员林莉，她的身高是1.71m,中国女排的平均身高肯定大于1.71米，题干说法正确。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22四年级下·河北石家庄·期末"
+    },
+    {
+        "id":  "407",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "\"六一\"儿童节的歌咏比赛中，四(1)班唱的《小小少年》评委最高分是98分，最低分是90分。四(1)这次歌咏比赛的平均分可能是()分。",
+        "options":  [
+                        "A. 98",
+                        "B. 95",
+                        "C. 90"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据平均数是反映一组数据的集中趋势的量，它比最大数小，比最小数大，据此解答即可。平均数是表示一组数据的集中趋势，不是其中每个数据的特征，它是这组数据所有数字的和除以这组数据的个数得出的，它会受偏大偏小数据的影响.\n【详解】\"六一\"儿童节的歌咏比赛中，四(1)班唱的《小小少年》评委最高分是98分，最低分是90分。四(1)这次歌咏比赛的平均分比90分大，比98分小，可能是95分。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22四年级下·四川宜宾·期末"
+    },
+    {
+        "id":  "408",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "关于\"平均数\"，以下说法正确的是()。",
+        "options":  [
+                        "A. 欢欢班同学的平均身高是1.44m,乐乐班同学的平均身高是1.42m,乐乐的身高不可能比欢欢高",
+                        "B. 歌唱比赛时，采取去掉一个最高分和一个最低分再计算参赛选手的平均得分更公平",
+                        "C. 池塘的平均水深1.2米，小明身高1.5米，直立站在池塘里，水不可能淹没头顶",
+                        "D. 第一小组踢毽子，前4人的平均水平是50个，如果第5人踢60个，将会拉低第一小组的平均成绩"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】先对每个选项中的说法进行分析并判断，然后选择正确的一项即可。平均数是反映一组数据的平均水平，并不能反应这组数据中各个数据的大小，由此即可对每项中的说法进行判断。熟练掌握平均数的意义是解答此题的关键.\n【详解】A。欢欢班同学的平均身高是1.44m,乐乐班同学的平均身高是1.42m,乐乐的身高可能比欢欢高，即原说法错误;B。歌唱比赛时，采取去掉一个最高分和一个最低分再计算参赛选手的平均得分更公平，即原说法正确;C。池塘的平均水深1.2米，小明身高1.5米，直立站在池塘里，水可能淹没头顶;D。第一小组踢毽子，前4人的平均水平是50个，如果第5人踢60个，将会拉高第一小组的平均成绩。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22四年级下·四川乐山·期末"
+    },
+    {
+        "id":  "409",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.4,
+        "question":  "五个人折纸鹤，小芳折了29个，小丽折了19个，小华折了9个，另外两人折的个数比小丽少，比小华多。这五个人折纸鹤的平均数应是()。",
+        "options":  [
+                        "A. 小于9",
+                        "B. 大于9小于19",
+                        "C. 19",
+                        "D. 大于19小于29"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】(29+19+9)÷3=57÷3=19(个)通过计算可知：小芳，小丽，小华3人折纸鹤的平均数是19个。如果再添加两个人，两人折的个数都比19小，也就是比3人的平均数小，则5人的平均数应该比3人的平均数19要小。如果两人折的个数都比9大，则5人的平均数应比最小值9要大。此题还可用假设法：假设两人折的个数和小丽同样多，都是19个，计算出5人平均数的最大值;再假设这两人折的个数和小华同样多，都是9个，再计算出5人平均数的最小值，平均数应该在最大值和最小值之间.\n【详解】根据分析可知：这5人折纸鹤的平均数大于9，小于19。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22四年级下·湖北武汉·期末"
+    },
+    {
+        "id":  "410",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面说法错误的有()个。(1)37×99=37×100-1(2)篮球队不同身高队员的平均身高195cm,每个队员的身高一定都是195cm。(3)0.6和0.60的计数单位不同。(4)梯形的腰一定比高长。(5)任意一个三角形至少有2个锐角。",
+        "options":  [
+                        "A. 1",
+                        "B. 2",
+                        "C. 3"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】(1)先把99写成(100-1),再根据乘法分配律计算;(2)平均数反映一组数据的平均情况，在一组数据中，有的数据可能会大于平均数，有的数据可能会小于平均数，有的数据可能会等于平均数;(3)根据小数的性质：在小数的末尾添上\"0\"或去掉\"0\"，小数的大小不变。虽然0.6=0.60，但是它们的计数单位不同;(4)根据梯形的高的含义，在梯形上底上任取一点，过这点向下底作垂线段即为梯形的高。这样的线段可以作无数条，因而一个梯形能画出无数条高，又因为梯形的上底和下底互相平行，因而这些高都相等;腰是从上底到下底的连线，当为直角梯形时，高和一条腰是相等的，所以梯形的高有可能比腰长，也可能与高相等;(5)根据三角形的内角和是180°来分析解决问题。本题考查了乘法分配律的应用，平均数的意义，小数的计数单位，梯形的高的特征以及三角形内角和的定理.\n【详解】(1)37×99=37×(100-1)=37×100-37=3700-37=3663，所以原题说法错误;(2)篮球队不同身高队员的平均身高195cm,每个队员的身高不一定都是195cm,有的可能比平均身高高，有的可能比平均身高矮，有的可能等于平均身高，所以原题说法错误;(3)0.6和0.60的大小相等,0.6表示6个0.1,0.60表示60个0.01，所以原题说法正确;(4)直角梯形的一条腰与梯形的高相等，所以原题说法错误;(5)假设三角形中锐角的个数少于2个，那么三角形中就会出现两个或两个以上的角是钝角或直角，两个钝角或两个直角的和加上第三个角的度数一定大于180°,这就违背了三角形内角和是180°的性质，所以任意一个三角形至少有2个锐角，原题说法正确;所以上面说法错误的有3个。故答案为:C",
+        "knowledge_points":  [
+                                 "小数的数位和计数单位的认识",
+                                 "平均数的意义及求法",
+                                 "梯形的高及画法",
+                                 "三角形的内角和"
+                             ],
+        "source":  "21-22四年级下·四川达州·期末"
+    },
+    {
+        "id":  "411",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面信息中适合用复式条形统计图的是()。",
+        "options":  [
+                        "A. 某超市一周的销售额",
+                        "B. 5名同学的身高",
+                        "C. 四年级4个班男生人数和女生人数"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】复式条形统计图的特点是能清楚地看出数量的多少，便于比较两组或多组数据的多少，依此选择。熟练掌握复式条形统计图的特点是解答此题的关键.\n【详解】A。某超市一周的销售额不适合用复式条形统计图;B。5名同学的身高不适合用复式条形统计图;C。四年级4个班男生人数和女生人数适合用复式条形统计图。故答案为:C",
+        "knowledge_points":  [
+                                 "复式条形统计图"
+                             ],
+        "source":  "21-22四年级下·四川自贡·期末"
+    },
+    {
+        "id":  "412",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "甲乙丙三人分别有卡片2张,4张和6张，丙给甲()张后，三个人的卡片同样多。",
+        "options":  [
+                        "A. 2",
+                        "B. 4",
+                        "C. 6"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】根据平均数的意义及求法，求甲，乙，丙卡片的平均张数，把丙比平均张数多的张数给甲，三个人的卡片同样多，据此解答。本题的关键是根据分析数量关系，求出三人的平均卡片数，再进行解答.\n【详解】(2+4+6)÷3=12÷3=4(张) 6-4=2(张) 丙给甲2张后，三个人的卡片同样多。 故答案为:A",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22四年级下·四川南充·期末"
+    },
+    {
+        "id":  "413",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "王鹏半期考试语文，数学平均分是96分。他语文比数学少4分，数学是()。",
+        "options":  [
+                        "A. 95",
+                        "B. 92",
+                        "C. 98",
+                        "D. 100"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】语文，数学的平均分*2=语文，数学的总成绩，再根据大的数=(和+差)÷2计算出数学的成绩即可。此题考查的是和差问题的计算，先计算出语文，数学的总成绩是解答此题的关键.\n【详解】96×2=192(分) (192+4)÷2=196÷2=98(分) 故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法",
+                                 "和差倍问题"
+                             ],
+        "source":  "21-22四年级下·四川德阳·期末"
+    },
+    {
+        "id":  "414",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "甲，乙，丙三位同学数学单元测试的平均成绩为95分，丁的成绩为99分，他们四个人的平均成绩是()分。",
+        "options":  [
+                        "A. 95",
+                        "B. 96",
+                        "C. 97",
+                        "D. 98"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】甲，乙，丙三位同学的平均成绩*3=甲，乙，丙三位同学的总成绩，甲，乙，丙三位同学的总成绩+丁的成绩=四位同学的总成绩，四位同学的总成绩/4=他们四个人的平均成绩，依此计算此题考查的是平均数的求法，先计算出四位同学的总成绩是解答此题的关键.\n【详解】95×3=285(分) 285+99=384(分) 384÷4=96(分) 故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22四年级下·四川广元·期末"
+    },
+    {
+        "id":  "415",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "妈妈答应朋朋，如果他期末考试语文，数学，英语三科的平均分达到95分，就给他买一双球鞋。朋朋语文和数学分别得了92分和98分，并顺利买到了喜欢的球鞋。他的英语成绩可能是()。",
+        "options":  [
+                        "A. 96分",
+                        "B. 94分",
+                        "C. 90分"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】根据\"平均成绩×科目的数量=总成绩\"算出语文，数学，英语三科的总成绩，再用三科的总成绩减去语文和数学两科的成绩，即可求出英语的成绩。本题考查平均数的意义，关键是求出三科的总成绩.\n【详解】95×3-(92+98)=285-190=95(分) 也就是说，英语成绩要达到95分以上。A。96分\u003e95分 B。94分\u003c95分 C。90分\u003c95分 故答案为:A",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22四年级下·四川广元·期末"
+    },
+    {
+        "id":  "416",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "168,126,157,122,150,185,149，这组数据的平均数是()。",
+        "options":  [
+                        "A. 115",
+                        "B. 147",
+                        "C. 151",
+                        "D. 135"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】要求这组数据的平均数，那需要求出这组数据的总数是多少，再运用平均数=总数/总数量，即可解答。本题考查学生对平均数计算方法的掌握。在求平均数时，要先找到总数和总数量，再运用计算公式即可解答.\n【详解】(168+126+157+122+150+158+149)÷7=1057÷7=151故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22四年级下·四川凉山·期末"
+    },
+    {
+        "id":  "417",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "求4+8+12+16的和，下面算式错误的是()。",
+        "options":  [
+                        "A. (4+16)÷2×4",
+                        "B. 16×4",
+                        "C. (8+12)×2",
+                        "D. 20×2"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】观察4,8,12,16，可以看出这一组数是相差为4的连续的自然数，可知它的特点就是第一个数加第四个数的和等于第二个数加第三个数的和，根据这个特点进行简便运算。此题需要学生找到算式的规律，然后进行简便运算.\n【详解】A。(4+16)÷2可以算出四个数的平均数，再乘4可以算出4,8,12,16的和，所以选项正确;B。16×4是4个16相加的和，不符合题干的算式，所以错误;C。8+12=4+16，所以(8+12)×2就是4+8+12+16的和，所以正确;D。因为8+12=4+16=20，所以20×2就是4+8+12+16的和，所以正确。故答案为:B",
+        "knowledge_points":  [
+                                 "整数加法结合律",
+                                 "整数加法交换律",
+                                 "平均数的意义及求法",
+                                 "数字排列的规律"
+                             ],
+        "source":  "20-21六年级下·云南玉溪·期末"
+    },
+    {
+        "id":  "418",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "18。(21-22四年级下·湖南长沙·期末)小东所在小组同学的平均体重是36千克，小刚所在小组同学的平均体重是34千克。下列说法正确的是()。",
+        "options":  [
+                        "A. 小东的体重比小刚重",
+                        "B. 小刚的体重比小东重",
+                        "C. A,B两种情况都有可能"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】平均数反映一组数据的平均情况，在一组数据中，有的数据可能会大于平均数，有的数据可能会小于平均数，有的数据可能会等于平均数。此题考查了平均数的性质和特点，要熟练掌握.\n【详解】因为小东所在小组同学的平均体重是36千克，小刚所在小组同学的平均体重是34千克;所以小东的体重可能比小刚重，也可能比小刚轻，所以两种情况都有可能。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22四年级下·湖南长沙·期末"
+    },
+    {
+        "id":  "419",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面说法正确的是()。",
+        "options":  [
+                        "A. 4.5和4.50大小相同，计数单位也相同",
+                        "B. 有一个角是锐角的三角形一定是锐角三角形",
+                        "C. 25×32可以用25×4×8进行简便计算",
+                        "D. 条形统计图中1格表示的数是固定的"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】一位小数的计数单位是0.1，两位小数的计数单位是0.01;三角形三个角都是锐角就是锐角三角形;计算25×32时，要给25找\"固定搭配\",32能拆分为4×8;条形统计图中1格可以表示1个单位，也可以表示2个,3个,4个等等。本题考查学生对小数计数单位，三角形分类，乘法结合律，条形统计图的认识。解决此题时，可以运用排除法，将错误答案排除，选出最后的正确答案.\n【详解】A。4.5和4.50的大小相同，但是计数单位不同。所以题目说法不正确;B。钝角三角形，直角三角形都有两个锐角。所以题目说法不正确;C。25×32可以将32拆成4×8,25×32可以用25×4×8来进行简算。题目说法正确;D。条形统计图中的1格可以表示的数是不固定的,1格表示几个单位可以根据实际情况来进行选择，但是一个条形统计图1格表示的数是固定的。题目说法不正确。故答案为:C",
+        "knowledge_points":  [
+                                 "1格表示多个单位的单式条形统计图",
+                                 "整数乘法结合律",
+                                 "小数的数位和计数单位的认识",
+                                 "三角形的分类"
+                             ],
+        "source":  "21-22四年级下·河南三门峡·期末"
+    },
+    {
+        "id":  "420",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "小聪同学周一至周五完成数学作业所花的时间分别是:15分钟,17分钟,14分钟,16分钟和18分钟，她平均每天完成数学作业的时间是()分钟。",
+        "options":  [
+                        "A. 14",
+                        "B. 15",
+                        "C. 16",
+                        "D. 18"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】先算出周一至周五完成数学作业所花的总时间，再除以5即可。根据平均数的含义和求法，解答此题即可.\n【详解】(15+17+14+16+18)÷5=80÷5=16(分钟)所以，她平均每天完成数学作业的时间是16分钟。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22四年级下·湖南长沙·期末"
+    },
+    {
+        "id":  "421",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "小东看一本故事书，前4天平均每天看15页，后3天共看66页，小东平均每天看多少页。正确的算式是()。",
+        "options":  [
+                        "A. 66÷3+15",
+                        "B. (15×4+66)÷(4+3)",
+                        "C. (15+66)÷(4+3)"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】前4天平均每天看15页，根据乘法的意义可知，前4天共看了(15×4)页，又后3天共看66页，则这本故事书共有(15×4+66)页，前后共看了(4+3)天，根据除法的意义可知，平均每天看:(15×4+66)÷(4+3);据此列式解答。本题主要考查平均数的意义及求法，关键是求出前4天看的页数.\n【详解】(15×4+66)÷(4+3)=(60+66)÷7=126÷7=18(页)小东平均每天看18页。所以正确的算式是(15×4+66)÷(4+3)。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22四年级下·湖南怀化·期末"
+    },
+    {
+        "id":  "422",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列说法正确的是()。",
+        "options":  [
+                        "A. 2021年2月份有29天",
+                        "B. 小红身高1.5米，她在平均水深1米的水池中游泳还是有危险的",
+                        "C. a/b=5。3，将a和b同时扩大到原来的100倍，则商不变，余数也不变"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】A。平年的2月有28天，闰年的2月有29天;一般年份中能被4整除的是闰年，不能被4整除的是平年，整百年份能被400整除的数是闰年，不能被400整除的数是平年;B。平均数只能反映一组数据的平均水平，并不能反映这组数据中所有数据的大小，池塘平均水深1米，可能有的地方水深超过1米，所以身高为1.5米的小红到平均水深1米的池中游泳可能会遇到危险，据此解答即可;C。根据商不变的规律，被除数和除数同时乘或除以一个不为0的数，商不变，余数随着被除数和除数的变化而变化。本题考查商不变的规律，明确余数随着被除数和除数的变化而变化是解题的关键.\n【详解】A。2021÷4=505。1，所以2021年是平年,2月有28天，原题干说法错误;B。平均水深1米的水池，可能有的地方水深超过1米，所以可能有危险，原题干说法正确;C.a/b=5。3，将a和b同时扩大到原来的100倍，则商不变，余数也扩大到原来的100倍，原题说法错误。故答案为:B",
+        "knowledge_points":  [
+                                 "平年",
+                                 "闰年的认识及判定方法",
+                                 "平均数的意义及求法",
+                                 "商不变的规律及应用"
+                             ],
+        "source":  "20-21六年级下·湖南邵阳·期末"
+    },
+    {
+        "id":  "423",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面说法正确的是()。",
+        "options":  [
+                        "A. 全班同学的平均身高是150厘米，小亮的身高可能低于150厘米。",
+                        "B. 小明身高140厘米，所以他去平均水深120厘米的河里游泳不会有危险。",
+                        "C. 四年级男生平均体重32千克，女生平均体重28千克，则任意一个男生一定比女生重。",
+                        "D. 东东5次跳远总成绩是10米，他每次的跳远成绩肯定都是2米。"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】平均数是反映一组数据的平均水平，并不能反映这组数据中各个数据的大小，平均数介于这组数据的最大值和最小值之间，据此即可解答。本题主要考查学生对平均数知识的掌握和灵活运用.\n【详解】A。平均身高是150厘米，班上有的同学高于150厘米，有的同学等于150厘米，有的同学低于150厘米，所以小亮的身高可能低于150厘米说法正确。B。平均水深120厘米，河里最深处可能大于140厘米，小明到河里游泳有危险，原说法错误。C。四年级男生平均体重32kg,女生平均体重28kg,但并不能确定任意一个男生一定比女生重，原说法错误。D。东东5次跳远总成绩是10米，他5次的跳远的平均成绩是2米，每次跳远的成绩不一定是2米，原说法错误。故答案为:A",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22四年级下·湖南株洲·期末"
+    },
+    {
+        "id":  "424",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "甲，乙两个数的平均数是16，乙，丙两个数的平均数是14，则甲比丙()。",
+        "options":  [
+                        "A. 多4",
+                        "B. 少4",
+                        "C. 多2"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】根据\"总数=平均数×数量\"，分别求出甲，乙之和，乙，丙之和;因为都有相同数乙，所以(甲+乙)-(乙+丙)就是甲与丙的差，据此解答。掌握平均数的意义及应用是解题的关键.\n【详解】甲，乙之和:16×2=32 乙，丙之和:14×2=28 甲比丙多:32-28=4 故答案为:A",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "20-21六年级下·河南商丘·期末"
+    },
+    {
+        "id":  "425",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "计算一组数据的平均数时，如果该组中一个数据变大了，那么平均数会()。",
+        "options":  [
+                        "A. 变大",
+                        "B. 变小",
+                        "C. 不变",
+                        "D. 无法确定"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】把一组数据中的所有数据加起来，再除以数据个数，即可得出平均数，平均数易受少数几个极端值的影响，据此解答。本题属于基础题，要掌握对平均数的含义的深刻理解.\n【详解】由分析可知：计算一组数据的平均数时，如果该组中一个数据变大了，那么平均数会变大。故答案为:A",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "20-21五年级下·四川成都·期末"
+    },
+    {
+        "id":  "426",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "学校举办\"校园好声音\"歌唱比赛，天天是参赛选手,9位评委给天天打分如下:94分,94分,93分,95分,95分,96分,92分,83分,95分。如果去掉一个最高分和一个最低分再计算平均分，那么她的平均得分是()分。",
+        "options":  [
+                        "A. 93",
+                        "B. 94",
+                        "C. 95",
+                        "D. 96"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】最高分是96分，最低分是83分，去掉之后，根据平均数=总数/个数，代入数据计算即可。解答此题的关键是掌握平均数=总数/个数这个公式.\n【详解】(94+94+93+95+95+92+95)÷7=658÷7=94(分)故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "20-21五年级下·广东深圳·期末"
+    },
+    {
+        "id":  "427",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "汽车公司发布一则汽车促销广告：\"所有汽车平均价格4万\"。关于这则广告，哪个说法一定是正确的。()",
+        "options":  [
+                        "A. 大部分汽车的价格在3万-5万之间",
+                        "B. 至少有一辆汽车的价格是4万",
+                        "C. 有一些汽车的价格不高于4万"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】根据平均数的定义可知，在一组数据中，平均数具有唯一性，是一个\"虚拟\"的数，是通过计算得到的，它不是数据中的原始数据，只是在平均数计算出来后，有和某一个原始数据相等的可能。平均数反映的是一组数据的特征，不是其中每一个数据的特征。逐一判断3个选项里的说法是否正确。此题的解题关键是理解掌握平均数的意义及应用.\n【详解】A。汽车平均价格4万并不能确定大部分汽车的价格范围，有可能是3~5万之间，有可能是2~6万之间，有可能是1~7万之间，无法确定汽车的价格范围;B。根据分析得，平均数并不一定就是某一辆汽车的价格，它可能和某一辆汽车的价格相等，也可能不相等，所以原题说法有误;C。平均数反映的是一组数据的特征，所以有些汽车的价格可能会比平均数高，也可能比平均数低，有一些汽车的价格不高于4万是合理的，所以原题说法正确。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "20-21五年级下·内蒙古包头·期末"
+    },
+    {
+        "id":  "428",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "四(1)班参加五一运动会的四名男队员，身高最高是1.62m,最矮是1.38m,这四位运动员的平均身高可能是()。",
+        "options":  [
+                        "A. 1.38m",
+                        "B. 1.62m",
+                        "C. 1.56m"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】平均数是表示一组数据的平均值，是指在一组数据中所有数据之和再除以这组数据的个数;其特点是比最大数小，比最小数大;依此选择。熟练掌握平均数的特点是解答此题的关键.\n【详解】A。1.38m=1.38m,因此不满足;B。1.62m=1.62m,因此不满足;C。1.38m\u003c1.56m\u003c1.62m,因此满足。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22四年级下·河南安阳·期末"
+    },
+    {
+        "id":  "429",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "解放路小学开展\"爱惜粮食，努力光盘\"的活动，每天记录各班的剩菜剩饭的重量。三(2)班某周的情况如表，按一周五天计算，该班平均每天的剩饭剩菜的质量是()千克左右.\n| 星期 | 星期一 | 星期二 | 星期三 | 星期四 | 星期五 |\n| -------------- | ------ | ------ | ------ | ------ | ------ |\n| 剩饭菜(千克) | 0.15 | 0.20 | 0 | 0 | 0.15 |",
+        "options":  [
+                        "A. 0.17",
+                        "B. 0.15",
+                        "C. 0.10"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】用这组数据的和除以数据的个数，就是平均数，据此解答。掌握平均数的计算方法是解答题目的关键.\n【详解】(0.15+0.20+0+0+0.15)÷5=0.5÷5=0.10(千克)故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "20-21五年级下·浙江宁波·期末"
+    },
+    {
+        "id":  "430",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下图是我国2010年和2020年\"人口普查\"人口地区分布对比图。下列描述与图上信息不符的是().\n![](images/4cde814925299d8cf5c29d2442d0d08496d8c8aff334a0729573c3d1661574a0.jpg)",
+        "options":  [
+                        "A. 东部地区人口上升最多",
+                        "B. 人口下降的只有中部地区",
+                        "C. 我国东部地区人口最多"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】观察复式条形统计图，深色的直条代表4个地区2010年的人口情况，浅色的直条代表4个地区2020年的人口情况，从条形统计图中很容易看出各种数量的多少，直条越高，说明人口越多。据此逐一判断3个选项里说法是否正确。此题考查的目的是理解和掌握复式条形统计图的特点和作用，能够根据它的特点和作用解决有关的实际问题.\n【详解】A。各地区2020年的人口与2010年的人口相比，东部地区人口在上升，中部地区在下降，西部地区基本持平，东北地区在下降，所以\"东部地区人口上升最多\"的说法是正确的;B。各地区2020年的人口与2010年的人口相比，中部地区在下降，东北地区也在下降，所以\"人口下降的只有中部地区\"的说法是错误的;C。无论是2010年，还是2020年，东部地区所代表的直条都是最高的，说明东部地区的人口数都是最多的，所以\"我国东部地区人口最多\"的说法是正确的。故答案为:B",
+        "knowledge_points":  [
+                                 "复式条形统计图"
+                             ],
+        "source":  "20-21五年级下·浙江宁波·期末"
+    },
+    {
+        "id":  "431",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "四(2)班同学数学期末考试，最低分76分，最高分98分，这个班数学的平均成绩可能是()分。",
+        "options":  [
+                        "A. 75",
+                        "B. 90",
+                        "C. 98",
+                        "D. 99"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】平均成绩是整个班的总体水平的反馈。实际成绩最低分是76分，最高分是98分，那么平均成绩就是在这两个分值之间。本题考查学生对于平均数的掌握。平均数是总体成绩的一个反馈，平均数不可能比实际成绩最小值小，也不可能比实际成绩最大值大，只会在这两个数值之间.\n【详解】99\u003e98\u003e90\u003e78\u003e75，平均成绩要在76到95之间，只有90是在这个区间的。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22四年级下·河南洛阳·期末"
+    },
+    {
+        "id":  "432",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "评委给谷爱凌打分为92,94,96,93,93,93，六个数平均分为()。",
+        "options":  [
+                        "A. (94+93+93+93)÷6",
+                        "B. (94+93+93+93)÷5",
+                        "C. (92+94+96+93+93+93)÷6"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】根据平均数=总数量/总份数，代入数据列式即可。此题主要考查了平均数的认识及运算法则.\n【详解】由题可得:(92+94+96+93+93+93)÷6故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22四年级下·河南洛阳·期末"
+    },
+    {
+        "id":  "433",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "四(1)班开展争做环保小能手活动，男生组20人，平均每人收集塑料瓶15个，女生组25人，共收集塑料瓶550个。四(1)班平均每个同学收集多少个塑料瓶。正确算式是()。",
+        "options":  [
+                        "A. (550÷25+15)÷2",
+                        "B. (15×20+550)÷(20+25)",
+                        "C. (15+550)÷(20+25)"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】由题意可知，男生组20人，平均每人收集塑料瓶15个，则男生组一共收集了(15×20)个，再加上女生组收集塑料瓶的个数，可得一共收集了(15×20+550)个，最后再除以男，女生组的总人数(20+25)人，可求出四(1)班平均每个同学收集多少个塑料瓶。熟悉平均数的含义和求法，是解答此题的关键.\n【详解】据分析可知正确算式是:(15×20+550)÷(20+25)。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "434",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列关于平均数的说法不正确的是()。",
+        "options":  [
+                        "A. 平均数易受极端数据影响",
+                        "B. 平均数一定是数据中的一个数",
+                        "C. 去掉一个数这组数的平均数不一定会变化"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据平均数的意义，平均数是反映一组的平均水平，它会受极端数据的影响，也就是平均数会受偏大偏小数据的影响。在一组数据中，有的数据可能会大于平均数，有的数据可能会小于平均数，有的数据可能会等于平均数。平均数=总数/个数，据此解答。解答此题的关键是掌握平均数的意义和平均数的相关公式.\n【详解】A。据分析可知，平均数易受极端数据影响，此说法正确;B。平均数代表的是一组数据的平均水平，不一定等于这组数据中的某个数;所以原说法错误;C。去掉一个数这组数的平均数不一定会变化，此说法正确，例如:5,5,5，去掉其中一个5后，两个5的平均数还是5。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22五年级下·江西九江·期末"
+    },
+    {
+        "id":  "435",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "四(1)班一次数学考试中，班级最高分是96.5分，最低分是88分。那么四(1)班的平均分可能是()分。",
+        "options":  [
+                        "A. 96.5",
+                        "B. 93",
+                        "C. 88",
+                        "D. 都有可能"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】一组数据的总和除以这组数据个数所得到的商叫这组数据的平均数。据此可知，平均分应比最高分小，比最低分大，逐项分析解答。此题考查的目的是理解平均数的意义，一组数据的平均分应大于最小数，小于最大数.\n【详解】A。96.5=96.5，说法错误;B。96.5\u003e93\u003e88，说法正确;C。88=88，说法错误;D。平均分在96.5分和88分之间，则都有可能这种说法错误。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "436",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "小红的体重是32kg,是四(1)班体重最轻的;小刚的体重是41kg,是四(1)班体重最重的。那么四(1)班同学平均体重可能是()kg。",
+        "options":  [
+                        "A. 32",
+                        "B. 36",
+                        "C. 41"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】平均数反映一组数据的平均情况，在一组数据中，有的数据可能会大于平均数，有的数据可能会小于平均数，有的数据可能会等于平均数。平均数大于一组数据的最小值，且小于这组数据的最大值。根据上述分析，对比各选项中的数，即可得到答案。此题考查了平均数的性质和特点，要熟练掌握.\n【详解】选项B只有36kg大于32kg,且小于41kg,则四(1)班同学平均体重可能是36kg。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22四年级下·河南三门峡·期末"
+    },
+    {
+        "id":  "437",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "甜甜的语文，数学，英语三科的平均成绩是94分，其中语文和数学的平均成绩是95分，她的英语成绩是()分。",
+        "options":  [
+                        "A. 92",
+                        "B. 95",
+                        "C. 94",
+                        "D. 93"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】甜甜的语文，数学，英语三科的平均成绩*3=语文，数学，英语三科的总成绩，语文和数学的平均成绩*2=语文和数学的总成绩，语文，数学，英语三科的总成绩-语文和数学的总成绩=英语的成绩，依此计算。此题考查的是平均数的求法，先分别计算出语文，数学，英语三科的总成绩，以及语文和数学的总成绩.\n【详解】94×3=282(分)95×2=190(分)282-190=92(分) 故答案为:A",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22四年级下·河南信阳·期末"
+    },
+    {
+        "id":  "438",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "在朗读比赛中，六位评委给笑笑打的分数分别是:91分,62分,90分,92分,87分,97分。按照去掉一个最高分和一个最低分的方法计算平均分，笑笑的得分是()分。",
+        "options":  [
+                        "A. 90",
+                        "B. 86.5",
+                        "C. 86"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】根据题意，去掉一个最高分97分，去掉一个最低分62分，再把剩余4位评委打的分数加起来，求出总分数，再除以4，即可求出笑笑的得分。此题的解题关键是灵活运用平均数的意义求解.\n【详解】(91+90+92+87)÷4=360÷4=90(分),故答案为:A",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22五年级下·四川成都·期末"
+    },
+    {
+        "id":  "439",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "瑶瑶所在班级学生的平均身高是142厘米，瑶瑶的身高是()。",
+        "options":  [
+                        "A. 比142厘米高",
+                        "B. 比142厘米矮",
+                        "C. 可能比142厘米高，可能比142厘米矮，可能等于142厘米"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】平均数表示一组数据集中趋势，它是反映数据集中趋势的一项指标;瑶瑶所在班级学生的平均身高是142厘米，只是说明学生身高的平均数是142厘米，瑶瑶的身高有可能比142厘米大，有可能小于142厘米，也有可能等于142厘米，据此判断即可。此题主要考查对平均数的含义的理解，做题时应认真分析，想的要周全，不要被数据所迷惑.\n【详解】平均身高为142厘米，并不代表班级里所有同学的身高都是142厘米，瑶瑶的身高可能比142厘米要高的多，也可能比142厘米矮的多，也可能等于142厘米。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22五年级下·四川成都·期末"
+    },
+    {
+        "id":  "440",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "六年级某次考试，一班的平均成绩是89分，二班的平均成绩是90分。根据这个信息，请判断下面说法正确的是()。",
+        "options":  [
+                        "A. 一班最低分一定比二班低",
+                        "B. 二班最高分一定比一班高",
+                        "C. 二班超过90分的人数比一班多",
+                        "D. 这次二班的总体水平比一班略高"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】平均数反映的是一组数据的特征，不是其中每一个数据的特征;一班的平均成绩是89分，二班的平均成绩是90分，只能说明二班的平均成绩大于一班的平均成绩，并不代表二班每个人的成绩都比一班的好;一班最低分不一定比二班低，二班最高分不一定比一班高，二班超过90分的人数不一定比一班多，据此解答即可。本题考查了平均数的意义，掌握平均数的含义是解题的关键.\n【详解】通过分析可知，每个班的平均成绩只能反映每个班的总体水平,90\u003e89，说明这次二班的总体水平比一班略高。故答案为:D",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22六年级下·浙江嘉兴·期末"
+    },
+    {
+        "id":  "441",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "小丽，小红和明明三位选手参加讲故事比赛，小丽以9.9分获得第一名，小红以9.3分获得第三名，明明获得第二名。这三位选手的平均成绩在()。",
+        "options":  [
+                        "A. 9.3分以下",
+                        "B. 9.3分到9.5分之间",
+                        "C. 9.5分到9.7分之间",
+                        "D. 9.8分以上"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】由题意可知，明明的分高于9.3分，小于9.9分，然后根据平均分=总分数/人数，据此解答即可。本题考查平均数，明确平均分=总分数/人数是解题的关键.\n【详解】(9.3+9.3+9.9)÷3=28.5÷3=9.5(分)(9.9+9.9+9.3)÷3=29.1÷3=9.7(分)所以这三位选手的平均成绩在9.5分到9.7分之间。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "2022·浙江温州·小升初真题"
+    },
+    {
+        "id":  "442",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "学校跳绳队队员15人的平均体重是39千克。后来跳绳队又来了一名队员，他的体重是45千克。现在队员的平均体重与原来相比,()。",
+        "options":  [
+                        "A. 比原来轻",
+                        "B. 比原来重",
+                        "C. 与原来相同",
+                        "D. 无法比较"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】平均数介于一组数据的最大值和最小值之间，假设原15人的体重都是39千克，现在增加的一名队员的体重是45千克，增加一名队员后的平均体重就应该在39千克和45千克之间，所以现在队员的平均体重比原来的重，据此即可解答。本题主要考查学生对平均数知识的掌握和灵活运用.\n【详解】根据分析可知，学校跳绳队队员15人的平均体重是39千克。后来跳绳队又来了一名队员，他的体重是45千克。现在队员的平均体重比原来的重。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22四年级下·四川凉山·期末"
+    },
+    {
+        "id":  "443",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一个数比60少2，那么这个数与60的平均数是()。",
+        "options":  [
+                        "A. 58",
+                        "B. 60",
+                        "C. 59"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】一个数比60少2，这个数等于60减2，然后把这个数与60相加，再除以2即可解答。本题主要考查学生求平均数方法的掌握和灵活运用.\n【详解】(60-2+60)÷2=118÷2=59 故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22四年级下·广西玉林·期末"
+    },
+    {
+        "id":  "444",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "小东所在小组同学的平均体重是38kg,小山所在小组同学的平均体重是34kg,下面说法正确的是()。",
+        "options":  [
+                        "A. 小东一定比小山重",
+                        "B. 小山一定比小东重",
+                        "C. 小东和小山不可能一样重",
+                        "D. 小东可能比小山重"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】根据平均数的意义及求法，小东所在小组同学的平均体重是指小东所在的小组的同学体重之和除以人数，同理，小山所在小组同学的平均体重是指小山所在的小组的同学体重之和除以人数;平均体重受偏重或偏轻同学的影响，小东所在的组可能偏重的同学较多，导致平均体重增加，或小山所在的小组体重偏轻的同学多，导致平均体重减少，因此，小东可能比小山重，小山可能比小东重，可能两人一样重。解答本题需要知道平均数受极端数据的影响较大，它不能反映个体的情况.\n【详解】根据分析，小东可能比小山重，小山可能比小东重，可能两人一样重,A,B,C三个选项都过于绝对。故答案为:D",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22四年级下·福建福州·期末"
+    },
+    {
+        "id":  "445",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "甲，乙，丙，丁四个数都是大于0的整数，而且甲\u003e乙\u003e丙\u003e丁，如果甲，乙，丙三个数的平均数是15，那么甲，乙，丙，丁四个数的平均数应该是()。",
+        "options":  [
+                        "A. 等于15",
+                        "B. 比15小",
+                        "C. 比15大",
+                        "D. 无法确定"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据题干甲\u003e乙\u003e丙\u003e丁，如果甲，乙，丙三个数的平均数是15，那么丁一定是小于15的数，所以四个数的平均数小于15，据此即可解答问题。此题考查了平均数的意义及求解方法.\n【详解】根据题干分析可得：甲\u003e乙\u003e丙\u003e丁，如果甲，乙，丙三个数的平均数是15，那么丁一定是小于15的数，所以四个数的平均数小于15。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22六年级下·浙江温州·期末"
+    },
+    {
+        "id":  "446",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "乐乐所在小组的平均身高是1.48m,天天所在小组的平均身高是1.56m。那么乐乐与天天的身高相比()。",
+        "options":  [
+                        "A. 乐乐高一些",
+                        "B. 天天高一些",
+                        "C. 他俩一样高",
+                        "D. 无法确定"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】平均数只能反映一组数据的集中趋势，不能反映具体数据的情况，据此解答即可。理解掌握平均数的含义是解答本题的关键.\n【详解】乐乐所在小组的平均身高是1.48m,天天所在小组的平均身高是1.56m。乐乐与天天的身高无法比较大小。故答案为:D",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22四年级下·山东济南·期末"
+    },
+    {
+        "id":  "447",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "实验小学四年级参加植树活动，第一天植树28棵，第二天植树30棵，第三天和第四天平均每天植树32棵，求平均每天植树多少棵的算式是()。",
+        "options":  [
+                        "A. (28+30+32)÷4",
+                        "B. (28+30+32)÷3",
+                        "C. (28+30+32×2)÷4"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】应先求出4天植树的棵数总和，再去除以4，需要注意的是，第3天和第4天平均每天植树32棵，总共植32×2=64(棵)。平均数的计算方法是：一组数据中所有数据之和再除以这组数据的个数.\n【详解】根据分析可知：求平均每天植树多少棵的算式是(28+30+32×2)÷4。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22四年级下·山西忻州·期末"
+    },
+    {
+        "id":  "448",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列说法中，叙述正确的是()。",
+        "options":  [
+                        "A. 把一个大三角形分成两个小三角形，每个小三角形的内角和是90°",
+                        "B. 钝角三角形中，钝角所对的那条边一定是最长的",
+                        "C. 25×32×125=25×4+125×8",
+                        "D. 在平均水深1.1m的河中，身高1.4m的李兵同学下去游泳是没有危险的"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】先对每个选项中的说法进行分析并判断，然后选择正确的一项即可。A。三角形的内角和为180°,依此判断;B。钝角三角形其中一个角是钝角，其它两个角都是锐角，依此画图判断;C。此题先将32写成4×8，再根据乘法结合律的特点进行计算，乘法结合律的特点是三个数相乘，先把前两个数相乘，或先把后两个数相乘，积不变;依此判断;D。平均数是反映一组数据的平均水平，并不能反映这组数据中各个数据的大小，由此即可进行判断。此题考查的是三角形的内角和，钝角三角形的特点，乘法结合律的特点，以及平均数的意义，应熟练掌握.\n【详解】A。把一个大三角形分成两个小三角形，每个小三角形的内角和是180°,即原说法错误;B。![](images/1f5357a91e3b07325bcc2fdea480932cf13f1444521b3b941fbe58e2e79cc4d3.jpg)钝角三角形中，钝角所对的那条边一定是最长的，即原说法正确;C。25×32×125=25×(4×8)×125=25×4×8×125=(25×4)×(8×125),即原算式错误;D。在平均水深1.1m的河中，身高1.4m的李兵同学下去游泳是有危险的，即原说法错误。故答案为:B",
+        "knowledge_points":  [
+                                 "三角形的分类",
+                                 "三角形的内角和",
+                                 "平均数的意义及求法",
+                                 "整数乘法结合律"
+                             ],
+        "source":  "21-22四年级下·浙江湖州·期末"
+    },
+    {
+        "id":  "449",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面说法错误的是()。",
+        "options":  [
+                        "A. 钝角三角形最多有1个钝角。",
+                        "B. 大于1且小于2的小数有无限个。",
+                    "C. 一组数据的平均数一般比这组数据中最大的数小，比最小的数大。",
+                        "D. 有一个角是45度的等腰三角形一定是锐角三角形"
+                    ],
+        "answer":  "D",
+    "solution":  "【分析】先对每个选项中的说法进行分析并判断，然后选择错误的一项即可。A。钝角三角形其中一个角是钝角，其它两个角都是锐角，依此判断;B。小数比较大小的方法：先看整数部分，整数部分大的，这个小数就大;整数部分相同，就看十分位上的数，十分位上的数大，这个小数就大，十分位上的数相同，再比较百分位上的数，百分位上的数大，这个小数就大，依此判断;C。平均数是表示一组数据的平均值，是指在一组数据中所有数据之和再除以这组数据的个数;平均数通常在这组数据的最大值和最小值之间，依此判断;D。等腰三角形的两腰相等，两个底角都相等;三角形的内角和是180°,三个角都是锐角的三角形是锐角三角形，依此判断。此题考查的是三角形的分类，三角形的内角和，小数的大小比较，以及平均数的特点，应熟练掌握.\n【详解】A。钝角三角形最多有1个钝角，即原说法正确;B。由于没有限制小数的位数，因此大于1且小于2的小数有无限个，即原说法正确;C。一组数据的平均数一般比这组数据中最大的数小，比最小的数大，这种说法正确;D。45°+45°=90°;180°-90°=90°,即有一个角是45度的等腰三角形一定是直角三角形，因此原说法错误。故答案为:D",
+        "knowledge_points":  [
+                                 "多位小数的大小比较",
+                                 "平均数的意义及求法",
+                                 "三角形的内角和",
+                                 "三角形的分类"
+                             ],
+        "source":  "21-22四年级下·山西阳泉·期末"
+    },
+    {
+        "id":  "450",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "欢欢期中检测语文，数学，英语三科的平均分是96分，其中语文92分，英语96分，她的数学是()分。",
+        "options":  [
+                        "A. 90",
+                        "B. 95",
+                        "C. 100",
+                        "D. 98"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】总分数=平均分数×科数，据此求出语文，数学，英语三科的总成绩，然后用总成绩减去语文和英语的成绩即可求出数学成绩。本题考查平均数，明确总分数=平均分数×科数是解题的关键.\n【详解】96×3-92-96=288-92-96=100(分),故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22六年级下·湖南湘潭·期末"
+    },
+    {
+        "id":  "451",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "比赛时往往要去掉评委最高分和最低分，再计算剩余得分的平均数为选手的最终得分。这是因为()。",
+        "options":  [
+                        "A. 去掉后再求平均数更公平",
+                        "B. 平均数具有公平性",
+                        "C. 平均数真的很灵敏",
+                        "D. 这两个数没有作用"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】平均数容易受极大值和极小值的影响，为了能尽量消除评委的个人偏好的极端影响，更好的反映选手的真实水平，所以在一些比赛中，计算选手的最后得分时，往往先去掉一个最高分和一个最低分，再计算。本题主要考查了平均数的含义，要熟练掌握.\n【详解】比赛时往往要去掉评委最高分和最低分，再计算剩余得分的平均数为选手的最终得分。这是因为去掉后再求平均数更公平。故答案为:A",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22五年级下·四川成都·期末"
+    },
+    {
+        "id":  "452",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "班级跳绳比赛中，淘气所在班级的平均跳绳个数是105个，奇思所在班级的平均跳绳个数是112个，淘气与奇思的跳绳个数相比,()。",
+        "options":  [
+                        "A. 淘气跳得多",
+                        "B. 奇思跳得多",
+                        "C. 他俩跳得一样多",
+                        "D. 以上三种情况都有可能"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】平均数是表示一组数据的平均值，是指在一组数据中所有数据之和再除以这组数据的个数，此题依此判断即可。熟练掌握平均数的意义是解答此题的关键.\n【详解】淘气所在班级的平均跳绳个数是105个，奇思所在班级的平均跳绳个数是112个，这两个数分别是淘气和奇思所在班级跳绳的一个平均数，并不代表他们的实际个数，因此三种情况都有可能。故答案为:D",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22五年级下·广东深圳·期末"
+    },
+    {
+        "id":  "453",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一本书共526页，小玉前5天每天看30页，后8天共看了240页，根据以上信息，不能解决的问题的是()。",
+        "options":  [
+                        "A. 这本书已经看了多少页。",
+                        "B. 这本书还剩下多少页没有看。",
+                        "C. 这本书还要多少天才能看完。",
+                        "D. 前13天小玉平均每天看多少页。"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】根据题意对每个选项中的问题进行解答并选择即可;前5天看的页数+后8天看的页数=这本书已经看了的页数;这本书的总页数-已经看了的页数=剩下的页数;剩下的页数/剩下可以看的天数=这本书还要看的天数;前13天小玉看的总页数/13=前13天小玉平均每天看页数，依此选择即可。此题考查的是工程问题的计算，根据题意找到对应的关系是解答此题的关键.\n【详解】A。5×30+240=150+240=390(页),即这本书已经看了390页。B。526-390=136(页),即这本书还剩下136页没有看。C。由于不知道剩下的页数每天看的天数，因此这本书还要多少天才能看完，不能计算。D。8+5=13(天),390÷13=30(页),即前13天小玉平均每天看30页。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法",
+                                 "工程问题"
+                             ],
+        "source":  "21-22四年级下·浙江温州·期末"
+    },
+    {
+        "id":  "454",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面选项正确的是()。",
+        "options":  [
+                        "A. 正方形，圆，长方形，平行四边形都是轴对称图形。",
+                        "B. 0.5和0.50大小相等，计数单位也相同。",
+                        "C. 0.68扩大100倍是68。",
+                        "D. 红红所在的班级的平均身高是140厘米，亮亮所在班级的平均身高是142厘米，所以亮亮肯定比红红高。"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】先对每个选项中的说法进行分析并判断，然后选择正确的一项即可。A。一个图形沿一条直线折叠，直线两旁的部分能够完全重合，这样的图形叫做轴对称图形，依此判断;B。小数的末尾添上\"0\"或去掉\"0\"，小数的大小不变;一位小数的计数单位是0.1，两位小数的计数单位是0.01，依此判断;C。用0.68乘100即可;D。平均数是反映一组数据的平均水平，并不能反映这组数据中各个数据的大小，依此判断。此题考查的是轴对称图形的特点，小数的性质，对小数的计数单位的认识，小数点位置的移动，以及平均数的意义，应熟练掌握.\n【详解】A。正方形，圆，长方形是轴对称图形，平行四边形不是轴对称图形，即原说法错误;B。0.5=0.50,0.5的计数单位是0.1,0.50的计数单位是0.01，即原说法错误;C。0.68×100=68，即0.68扩大100倍是68，因此原说法正确;D。140厘米,142厘米分别是红红，亮亮所在班级的平均身高，并不是他们两人的实际身高，因此他们谁更高无法比较。故答案为:C",
+        "knowledge_points":  [
+                                 "小数的数位和计数单位的认识",
+                                 "平均数的意义及求法",
+                                 "小数点位置的移动",
+                                 "小数的性质"
+                             ],
+        "source":  "21-22四年级下·浙江温州·期末"
+    },
+    {
+        "id":  "455",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "四(1)班同学的平均身高为139厘米，四(2)班同学的平均身高为141厘米，分别从这两个班随机各选一个学生进行比较，那么()。",
+        "options":  [
+                        "A. 四(1)班的高",
+                        "B. 四(2)班的高",
+                        "C. 一样高",
+                        "D. 无法确定"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】根据平均数的意义可知，四(1)班同学的平均身高为139厘米，可能有的同学比139厘米高得多，可能有的同学比139厘米矮得多;同理：四(2)班同学的平均身高为141厘米，可能有的同学比141厘米高得多，可能有的同学比141厘米矮得多，所以，分别从这两个班随机各选一个学生进行比较，是无法比较的。解答此题因根据平均数的意义，进行分析，解答即可，注意平均数只反映一组数据的集中趋势，并不代表其中的哪一个数据.\n【详解】根据分析可知，四(1)班同学的平均身高为139厘米，四(2)班同学的平均身高为141厘米，分别从这两个班随机各选一个学生进行比较，是无法比较的。故答案为:D",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22四年级下·浙江温州·期末"
+    },
+    {
+        "id":  "456",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "某次考试，语文和数学两科平均分小明比小刚高，单看数学成绩小明也比小刚高，那么单看语文成绩()。",
+        "options":  [
+                        "A. 小明比小刚高",
+                        "B. 小明比小刚低",
+                        "C. 无法判断"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】平均数是反映一组数据的平均水平，并不能反映这组数据中各个数据的大小，由此即可进行选择。熟练掌握平均数的意义是解答此题的关键.\n【详解】根据分析可知，小明和小刚的语文成绩谁更高，无法比较。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22四年级下·浙江杭州·期末"
+    },
+    {
+        "id":  "457",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "在两江新区全民健身运动会开幕式中，来自翠云街道的运动员排成4列纵队入场，个子矮的运动员站前面，个子高的运动员站后面。更接近翠云街道全体运动员平均身高的是()。",
+        "options":  [
+                        "A. 第一排运动员的平均身高",
+                        "B. 第一列运动员的平均身高",
+                        "C. 随机挑选1名运动员的身高",
+                        "D. 随机挑选10名运动员的平均身高"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据平均数是反映一组数据的集中趋势的一个量，不能反映具体数据，据此解答即可。根据平均数的含义和求法，解答此题即可.\n【详解】根据分析可知，更接近翠云街道全体运动员平均身高的是第一列运动员的平均身高。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22四年级下·重庆·期末"
+    },
+    {
+        "id":  "458",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "我校四年级数学教师有4人，他们平均年龄是26岁，四人中没有小于23岁的，那么四人中年龄最大的教师最大可能()岁。",
+        "options":  [
+                        "A. 30",
+                        "B. 33",
+                        "C. 35",
+                        "D. 38"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】四个人的平均年龄是26岁，他们的年龄和是26×4岁，要求最大的年龄是多少岁，必须使另外三人年龄最小，又没有小于23岁的，那么其他三人的年龄最小都是23岁，然后再用四人的年龄和减去最小3人的年龄和即可。解答此题的关键是明确，要求四个人中年龄最大的可能是多少岁，可使其他三人年龄尽量小.\n【详解】26×4=104(岁)104-23×3=104-69=35(岁)四人中年龄最大的教师最大可能是35岁。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22四年级下·重庆巴南·期末"
+    },
+    {
+        "id":  "459",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "期中考试中，小明一共考了五门功课，不算语文和数学成绩的话，平均成绩是90分，语文和数学的平均成绩是95分，小明这次期中考试的平均成绩是()分。",
+        "options":  [
+                        "A. 84",
+                        "B. 90",
+                        "C. 92",
+                        "D. 85"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】先计算出不算语文和数学成绩时，另外三科的总成绩，然后再计算出语文和数学的总成绩，再计算出五科的总成绩，最后用五科的总成绩除以5即可，依此计算。熟练掌握平均数的意义及求法是解答此题的关键.\n【详解】90×3=270(分)95×2=190(分)270+190=460(分)460÷5=92(分)故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22四年级下·浙江·期末"
+    },
+    {
+        "id":  "460",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "学校篮球社团的平均身高是155cm,聪聪的身高是1.65m,如果他加入这个篮球社团，他们的平均身高会( )155cm。",
+        "options":  [
+                        "A. 等于",
+                        "B. 高于",
+                        "C. 矮于"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】原来学校篮球社团的平均身高是155cm小于聪聪身高1.65m,根据\"身高和/人数=平均身高\"以及\"移多补少的思想\"可知，聪聪加入篮球社团后，现在球队队员的平均身高跟原来相比高了;据此解答即可。此题应根据平均数的意义，进行分析，解答，身高和/人数=平均身高.\n【详解】据分析可知：学校篮球社团的平均身高是155cm,聪聪的身高是1.65m,如果他加入这个篮球社团，他们的平均身高会高于155cm。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22四年级下·重庆万州·期末"
+    },
+    {
+        "id":  "461",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "阳阳所在班级的平均身高是145厘米，欢欢所在班级的平均身高是143厘米。阳阳与欢欢的身高相比,()。",
+        "options":  [
+                        "A. 阳阳高",
+                        "B. 欢欢高",
+                        "C. 两人一样高",
+                        "D. 无法确定谁高"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】平均数是反映一组数据的平均水平，并不能反映这组数据中各个数据的大小，据此即可解答。本题考查了平均数的意义在实际生活中的灵活应用.\n【详解】班级的平均身高并不能反映单个人的身高，知道阳阳和欢欢班级的平均身高并不能确定阳阳和欢欢的身高，所以无法比较他们的身高。故答案为:D",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22四年级下·重庆渝中·期末"
+    },
+    {
+        "id":  "462",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "学校游泳队平均身高是163cm,下面4名同学中,()加入后，平均身高会减少。",
+        "options":  [
+                        "A. 王东:168cm",
+                        "B. 赵明:164cm",
+                        "C. 李红:163cm",
+                        "D. 陈星:160cm"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】一组数据加入一个大于这组数据平均数的数，则加入后的平均数大于原平均数，反之小于原平均数，如果加入的数等于这组数据的平均数，则加入后的平均数不变;据此即可解答。本题主要考查学生对平均数知识的掌握和灵活运用.\n【详解】根据分析可知，王东，赵明的身高大于163cm,加入后平均身高增加，李红的身高等于163cm,加入后平均身高不变，陈星的身高小于163cm,加入后平均身高会减少。故答案为:D",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22四年级下·重庆沙坪坝·期末"
+    },
+    {
+        "id":  "463",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面说法正确的是()。",
+        "options":  [
+                        "A. 四(一)班数学平均成绩90分，那么这个班一定有学生成绩是90分。",
+                        "B. 把57缩小到它的1÷100是0.57",
+                        "C. 按照\"四舍五入\"法，近似数为5.72的最大的三位小数是5.719。"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】先对每个选项中的说法进行分析并判断，然后选择正确的一项即可。A。平均数是反映一组数据的平均水平，并不能反映这组数据中各个数据的大小，由此即可进行判断。B。把57缩小到它的1÷100，就用57除以100，依此计算并判断。C。保留两位小数时，就把百分位后面的数省略，当千分位上的数等于或大于5时，应向百分位上进1后再省略;因此最大数是通过\"四舍\"得到的，依此判断。此题考查的是平均数的意义及求法，小数点位置的移动，以及小数近似数的计算，应熟练掌握.\n【详解】A。四(一)班数学平均成绩90分，那么这个班不一定有学生成绩是90分，即原说法错误。B。57÷100=0.57，即把57缩小到它的1÷100是0.57，因此原说法正确;C。按照\"四舍五入\"法，近似数为5.72的最大的三位小数是5.724，即原说法错误。故答案为:B",
+        "knowledge_points":  [
+                                 "小数的近似数和改写",
+                                 "平均数的意义及求法",
+                                 "小数点位置的移动"
+                             ],
+        "source":  "21-22四年级下·山西晋中·期末"
+    },
+    {
+        "id":  "464",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "王红调查操场上做游戏的几名同学年龄情况:8岁,12岁,9岁,6岁,10岁。这时另一个同学也加入到游戏的队伍，他年龄是7岁，此时做游戏同学的平均年龄会()。",
+        "options":  [
+                        "A. 变大",
+                        "B. 变小",
+                        "C. 无法确定"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】先计算出开始做游戏的5名同学的平均年龄(用5名同学的总年龄除以5即可),然后与7岁进行比较，根据比较结果进行选择即可。熟练掌握平均数的意义及求法是解答此题的关键.\n【详解】8+12+9+6+10=45(岁)45÷5=9(岁) 7岁\u003c9岁，因此7岁小朋友也加入到游戏的队伍时，此时做游戏同学的平均年龄会变小。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22四年级下·山西晋中·期末"
+    },
+    {
+        "id":  "465",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "有4个同样的杯子，水面的高度分别是9厘米,7厘米,3厘米和5厘米。这四杯水面的平均高度是()厘米。",
+        "options":  [
+                        "A. 8",
+                        "B. 5",
+                        "C. 4",
+                        "D. 6"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】计算平均数就是把给出的几个数相加再除以它的个数即可，在本题中，把水面的几个高度的数据相加，再除以4即可，据此解答。熟练掌握平均数的计算方法并灵活运用是解答本题的关键.\n【详解】(9+7+3+5)÷4=24÷4=6(厘米),故答案为:D",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22四年级下·山东菏泽·期末"
+    },
+    {
+        "id":  "466",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "小玲三天看完一本书，平均每天看30页。她前两天看的页数都超过30页，那么她第三天看的页数()。",
+        "options":  [
+                        "A. 比30页多",
+                        "B. 比30页少",
+                        "C. 还是30页"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据平均数的意义可知，小玲三天看完一本书，平均每天看30页，这本书的总页数是30×3=90页，而她前两天每天看的都超过了30页，说明两天看的页数有比60页多，第三天看的页数比30页少，据此解答。解答本题关键是先求出总的页数，根据前两天每天看的都超过了30页，说明两天看的页数比60页多，然后用总页数减去看的就可以解答.\n【详解】30×3=90(页)她前两天每天看的都超过了30页，说明两天看的页数比60页多，所以第三天看的页数不到30页。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22四年级下·山东菏泽·期末"
+    },
+    {
+        "id":  "467",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "在一分钟跳绳比赛中，第五小组共5人参赛，小组中跳得最多220下，最少的196下。第五小组的平均个数()。",
+        "options":  [
+                        "A. 大于220",
+                        "B. 小于196",
+                        "C. 在196到220之间",
+                        "D. 以上三种情况都有可能"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】平均数反映一组数据的平均情况，在一组数据中，有的数据可能会大于平均数，有的数据可能会小于平均数，有的数据可能会等于平均数。平均数大于一组数据的最小值，且小于这组数据的最大值。此题考查了平均数的性质和特点，关键是明确平均数在最大值和最小值之间.\n【详解】小组中跳得最多220下，最少的196下，则第五小组的平均个数在196到220之间。故答案为:C",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22四年级下·重庆南岸·期末"
+    },
+    {
+        "id":  "468",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "3个连续自然数的平均数是18，其中最小的数是()。",
+        "options":  [
+                        "A. 17",
+                        "B. 18",
+                        "C. 19"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】相邻的两个自然数之间相差为1,3个连续自然数的平均数等于中间一个数，所以平均数减1等于最小的数，据此即可解答。本题主要考查学生对平均数知识的掌握和灵活运用.\n【详解】18-1=17 故答案为:A",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "469",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "奇奇班里同学们的平均身高是136cm,豆豆班里同学们的平均身高是148cm,那么奇奇和豆豆的身高相比,()。",
+        "options":  [
+                        "A. 奇奇高",
+                        "B. 豆豆高",
+                        "C. 一样高",
+                        "D. 无法确定谁高"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】平均数会受极端数据的影响，也就是平均数会受偏大，偏小数据的影响，虽然豆豆班里同学的平均身高比奇奇班里同学的平均身高高，但是他们的身高无法确定，据此解答即可。此题考查的目的是理解平均数的意义，掌握求平均数的方法及应用，明确平均数会受极端数据的影响，也就是平均数会受偏大，偏小数据的影响.\n【详解】奇奇班里同学们的平均身高是136cm,豆豆班里同学们的平均身高是148cm,奇奇和豆豆无法确定谁高。故答案为:D",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "2022·四川广元·小升初真题"
+    },
+    {
+        "id":  "470",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "两个小组做仰卧起坐。第一小组5人，一共做了85个;第二小组4人，一共做了76个。()成绩好。",
+        "options":  [
+                        "A. 第一小组",
+                        "B. 第二小组",
+                        "C. 无法比较"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】85除以5等于第一小组的平均成绩,76除以4等于第二小组的平均成绩，然后进行比较即可解答。本题考查了平均数的应用，通过比较两组平均成绩的大小来进行解答.\n【详解】85÷5=17(个)\n76÷4=19(个)\n19\u003e17，第二小组的成绩好。故答案为:B",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22四年级下·河南郑州·期末"
+    },
+    {
+        "id":  "471",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "()可以很清楚地看出两组数据数量的多少，便于比较。",
+        "options":  [
+                        "A. 条形统计图",
+                        "B. 统计表",
+                        "C. 复式条形统计图"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】条形统计图的特点是能够使人们一眼看出各个数据的大小，易于比较数据之间的差别，能清楚地表示出数量的多少，但是只能表示一个项目的数据;统计表的特点是使统计资料条理化，简明清晰，便于检查数字的完整性和准确性，以及对比分析;复式条形统计图的特点是不仅能更清楚的表示各种数量的多少，还能更直观，形象地比较多种数量之间的关系，可以同时表示多个项目的数据。此题主要考查不同类型的统计图的特征，解答的关键是明确不同类型的统计图的特点.\n【详解】可以很清楚地看出两组数据数量的多少，便于比较的是复式条形统计图。故答案为:C",
+        "knowledge_points":  [
+                                 "单式统计表的特点及填补",
+                                 "复式条形统计图"
+                             ],
+        "source":  "21-22四年级下·河南安阳·期末"
+    },
+    {
+        "id":  "472",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面说法正确的是()。",
+        "options":  [
+                        "A. 有一个角是锐角的三角形就是锐角三角形。",
+                        "B. 算式\"4200÷35\"与算式\"4200÷7×5\"的计算结果相同。",
+                        "C. 平均数是一种能较好地反映一组数据的整体水平的数。"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】A选项中，有可能三角形里两个锐角分别是15°,15°,150°,它是一个钝角三角形;B选项中4200除以35得120,4200除以7再乘5得3000;平均数是一种能较好地反映一组数据的整体水平的数，据此解答。熟练掌握平均数，三角形的分类并灵活运用是解答本题的关键.\n【详解】A。有一个角是锐角的三角形就是锐角三角形，该说法不正确;B。算式\"4200÷35\"与算式\"4200÷7×5\"的计算结果相同，该说法不正确;C。平均数是一种能较好地反映一组数据的整体水平的数，该说法正确.\n故答案为:C",
+        "knowledge_points":  [
+                                 "直角",
+                                 "钝角",
+                                 "锐角的认识及特征",
+                                 "除数是两位数的笔算除法",
+                                 "平均数的意义及求法",
+                                 "三角形的分类"
+                             ],
+        "source":  "21-22四年级下·浙江绍兴·期末"
+    },
+    {
+        "id":  "473",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "四年级(1)班男生的平均身高是148厘米，张建是这个班的男生，他的身高一定是148厘米。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】一组数据的总和除以这组数据个数所得到的商叫这组数据的平均数。根据平均数的意义，这组数据里有些数可能会高于平均数，有些数会低于平均数，据此判断。解决本题的关键是正确理解平均数的意义.\n【详解】根据平均数的意义，张建是这个班的男生，他的身高不一定是148厘米。故原说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22四年级上·贵州毕节·期末"
+    },
+    {
+        "id":  "474",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "小华班里同学的平均身高是1.38m,刘红班里同学的平均身高是1.35m,则小华比刘红高。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数反映的是一组数据的特征，不是其中每个数据的特征，小华所在班级学生平均身高高，并不代表小华的身高就高，他可能比平均身高高，也可能比平均身高矮;刘红所在班级学生平均身高矮，并不代表刘红的身高就矮，可能比平均身高矮，也可能比平均身高高;进而得出结论。本题主要考查了平均数的意义，即平均数反映的是一组数据的特征，不是其中每一个数据的特征，不要被数字迷惑.\n【详解】根据分析得，小华班里同学的平均身高是1.38m,刘红班里同学的平均身高是1.35m,小华和刘红的身高无法比较。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22六年级下·湖北孝感·期末"
+    },
+    {
+        "id":  "475",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "六(1)班为\"希望小学\"捐款，女生平均每人捐7元，男生平均每人捐3元，全班平均每人捐5元。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】知道男生和女生每人捐款的钱数，要求平均每人捐款的钱数，还需要知道男，女生的人数，用总钱数/全班总人数=平均每人捐款钱数。此题应根据总钱数/全班总人数=平均每人捐款钱数进行分析，判断.\n【详解】总钱数/全班总人数=平均每人捐款钱数;若全班男女生人数相等，这题就是对的;若男女生人数不相等就是错的;本题不知道男，女生分别有多少人，所以说是错误的。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22六年级下·湖南永州·期末"
+    },
+    {
+        "id":  "476",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "纵向复式条形统计图比横向复式条形统计图表示的更明白。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】条形统计图分为：单式条形统计图和复式条形统计图，前者只表示1个项目的数据，后者可以同时表示多个项目的数据;可以是纵向的，也可以是横向的。进而判断即可。此题考查了条形统计图的分类和特点.\n【详解】根据条形统计图的特点可知：条形统计图的条形可以表示两种不同的数量，可以是纵向的，也可以是横向的。故答案为: ×",
+        "knowledge_points":  [
+                                 "复式条形统计图"
+                             ],
+        "source":  "21-22四年级下·全国·假期作业"
+    },
+    {
+        "id":  "477",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "可以按全班同学的平均鞋码订购运动鞋。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】一组数据的总和除以这组数据个数所得到的商叫这组数据的平均数，据此判断。解决本题的关键是正确理解平均数的意义.\n【详解】根据平均数的意义，一组数中有些数可能会大于平均数，有些数会小于平均数，所以按全班同学的平均鞋码订购运动鞋，这种说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22四年级下·广东东莞·期末"
+    },
+    {
+        "id":  "478",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "学校排球队队员的平均身高是160厘米，小东是学校排球队队员，他的身高有可能是158厘米。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】平均数反映一组数据的平均情况，在一组数据中，有的数据可能会大于平均数，有的数据可能会小于平均数，有的数据可能会等于平均数;据此判断即可。此题考查了平均数的意义和特点，要熟练掌握.\n【详解】学校排球队队员的平均身高是160厘米，小东是学校排球队队员，他的身高有可能是158厘米，此说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "20-21四年级下·广东肇庆·期末"
+    },
+    {
+        "id":  "479",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "小明的身高是1米45厘米，他在一个平均水深为1米35厘米的游泳池中游泳一定不会有危险。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数是表示一组数据的平均值，是指在一组数据中所有数据之和再除以这组数据的个数;其特点是比最大数小，比最小数大;依此判断。熟练掌握平均数的意义是解答此题的关键.\n【详解】在平均水深1米35厘米的游泳池，并不代表每处的水深都是1米35厘米，可能比1米35厘米深，也可能比1米35厘米浅;所以，他在平均水深1米35厘米的游泳池游泳，会有危险。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22四年级下·全国·期末"
+    },
+    {
+        "id":  "480",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "四(1)班男生平均体重是32kg,女生平均体重是28kg,每个男生的体重一定比每个女生重。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数常用于表示统计对象的一般水平，它代表的不是每个人的具体水平，所以男生的平均体重不代表所有男生的体重，女生的平均体重也不代表所有女生的体重，据此判断。此题考查的是平均数的意义，注意由于平均数受极值的影响，它不代表平均数大的所有个体都比平均数小的所有个体大.\n【详解】平均体重不代表所有人的体重，即可能有的女生要比男生重，所以判断错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "20-21四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "481",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "要反映一组数据的一般水平，通常用这组数据的平均数来表示。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】平均数是表示一组数据的平均值，是指在一组数据中所有数据之和再除以这组数据的个数，依此判断。熟练掌握平均数的意义是解答此题的关键.\n【详解】根据平均数的意义可知：平均数能够较好的反映出一组数据的平均水平。故答案为:√",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22六年级下·贵州遵义·期末"
+    },
+    {
+        "id":  "482",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "王明身高1.5米，他在平均水深1.2米的河里游泳可能有危险。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】平均数是反映一组数据的平均水平，并不能反映这组数据中各个数据的大小，它比最小的数大一些，比最大的数小一些，在它们之间;由此即可进行判断.\n【详解】根据分析可知，平均水深1.2米，并不能反映出整个水域中每一处的水深大小，有的地方会深一些，可能超过1.2米，有的地方会浅一些，所以身高是1.5米的王明去平均水深为1.2的水域游泳，可能有危险。所以原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "19-20四年级·湖南株洲·期末"
+    },
+    {
+        "id":  "483",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "池塘的平均水深是120厘米，小明身高145厘米，去这个池塘游泳很安全。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数是反映一组数据的平均水平，并不能反映这组数据中各个数据的大小，由此即可进行判断。熟练掌握平均数的意义是解答此题的关键.\n【详解】池塘的平均水深是120厘米，并不代表池塘的每一处水深都是120厘米，因此小明身高145厘米，去这个池塘游泳可能会有危险。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "19-20四年级上·安徽芜湖·期末"
+    },
+    {
+        "id":  "484",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "小明所在班级学生的平均身高是1.5米，小红所在班级学生的平均身高是1.48米。小明一定比小红高。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数：一组数据，用这组数据的总和除以总个数，得出的数就是这组数据的平均数。平均数的大小与一组数据里的每个数据都有关系，任何一个数据的变动都会引起平均数的变动，即平均数受较大数和较小数的影响。本题中根据平均数的定义即可得出小明的身高可能高于1.5米，也可能低于1.5米;小红的身高可能高于1.48米，也可能低于1.48米，本题据此判断即可。此题主要考查对平均数的基础知识的掌握情况，做题时一定要弄清题意，认真审题，然后做出判断.\n【详解】小明所在班级学生的平均身高是1.5米，小红所在班级学生的平均身高是1.48米。小明不一定比小红高，所以原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "2022六年级下·全国·专题练习"
+    },
+    {
+        "id":  "485",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "四年级一班同学的平均体重是45千克，赵虎是四年级一班的同学，他的体重可能超过45千克。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】平均数是表示一组数据的平均值，是指在一组数据中所有数据之和再除以这组数据的个数;其特点是比最大数小，比最小数大，依此判断即可。熟练掌握平均数的意义是解答此题的关键.\n【详解】45千克是四年级一班同学的平均体重，并不代表赵虎的实际体重，因此赵虎的体重可能超过45千克。故答案为:√",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "21-22四年级上·河南周口·期末"
+    },
+    {
+        "id":  "486",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "王芳所在班级的数学平均成绩是84分，李明班的数学平均成绩是86分，所以李明的数学成绩一定比王芳的数学成绩高。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数反映一组数据的平均情况，在一组数据中，有的数据可能会大于平均数，有的数据可能会小于平均数，有的数据可能会等于平均数。此题考查了平均数的性质和特点，要熟练掌握.\n【详解】王芳所在班级的数学平均成绩是84分，李明班的数学平均成绩是86分，不能判断李明，王芳的数学成绩之间的大小关系。故原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "2021·宁夏石嘴山·小升初真题"
+    },
+    {
+        "id":  "487",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "在一次数学测试中，全班男生的总分比女生高，说明男生的成绩比女生好。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数是指在一组数据中所有数据之和再除以数据的个数，比较平均数能得出成绩的优差，由于一个班中男女生的人数不一定相等，所以单单比较男女生的总成绩是不能判定成绩的优差。此题考查了平均数的应用，比较平均数能得出成绩的优差.\n【详解】由于一个班中男女生的人数不一定相等，所以单单比较男女生的总成绩是不能判定成绩的优差;所以原说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "20-21四年级下·湖北荆门·期末"
+    },
+    {
+        "id":  "488",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "小红所在班级的数学期中平均分为92分，她的数学期中成绩一定是92分。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】尽管平均数能较好地反映一组数据的总体情况，但不代表这组数据里每一个数都和平均数相等;据此解答。解答本题需要明确平均数的意义，以及准确理解其内在含义.\n【详解】结合平均数的意义可知：小红所在班级的数学期中平均分为92分，她的数学期中成绩不一定就是92分。原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "20-21六年级下·四川宜宾·期末"
+    },
+    {
+        "id":  "489",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "平均数能反映一组数据的整体情况，不是一个真实的数。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【详解】平均数只能反映一组数据的平均水平，并不能反映这组数据中所有数据的大小。平均数=总数量/份数，则平均数能反映一组数据的整体情况，不是一个真实的数。例如小林5次跳绳的总成绩是375下，他5次跳绳的平均成绩是75下，但他每次跳绳的成绩不一定都是75下。故答案为:√",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "20-21四年级下·黑龙江牡丹江·期末"
+    },
+    {
+        "id":  "490",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "在实际比赛中，通常都采取去掉最高分，然后再计算平均数的记分方法，因为去掉最高分后再求平均数就更有代表性了。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】用这组数据的和除以数据的个数，就是平均数，平均数容易受极端数据的影响，所以计算比赛成绩时去掉最高分和最低分。掌握平均数的计算方法是解答题目的关键.\n【详解】在实际比赛中，通常都采取去掉最高分和最低分，然后再计算平均数的记分方法，因为平均数容易受极端数据的影响。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "19-20五年级下·广东韶关·期末"
+    },
+    {
+        "id":  "491",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "复式条形统计图可以很清楚的看出两组数据不同的地方。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【详解】条形统计图上的每个小格表示就是一个长度单位，每个长度单位表示的数量是相同的，由于数据不同，所以画出直条的长短也不同，单式条形统计图中的数据只用一种直条来表示，而复式条形统计图可以同时表示两种或两种以上不同的量。复式条形统计图的优点是：可以直观地看出不同项目的数据是多少，能形象地比较不同数据;所以复式条形统计图可以很清楚的看出两组数据不同的地方，说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "复式条形统计图"
+                             ],
+        "source":  "17-18四年级下·内蒙古巴彦淖尔·期末"
+    },
+    {
+        "id":  "492",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "两组学生做仰卧起坐，一组4人共做了100个，二组5人共做了110个，二组的成绩好些。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】根据题意，用每组做的个数，除以每组的人数，求出平均每人做的个数，然后再比较解答。本题考查平均数的应用，掌握平均数的意义和求法是解题的关键.\n【详解】100÷4=25(个) 110÷5=22(个) 25\u003e22，一组的成绩好些。所以，原题说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "17-18四年级下·内蒙古巴彦淖尔·期末"
+    },
+    {
+        "id":  "493",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "在一组数据中，平均数比该组数据中的数都要小。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】根据平均数的意义，分析判断即可。本题考查了平均数，掌握平均数的概念和求法是解题的关键.\n【详解】平均数是一组数据的总和除以数据的数量的结果，所以这组数据中的数要么比平均数大，要么比平均数小，或者和平均数相等。但是，平均数至少要和这组数据中的数相等，不可能该组数据中的数都要小。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "20-21六年级下·内蒙古通辽·期末"
+    },
+    {
+        "id":  "494",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "复式条形统计图的特点是用直条的长短表示数量的多少，优点是能清楚地看出数量的多少，便于比较两组数据的多少。( )",
+        "options":  [
+
+                    ],
+          "answer":  "√",
+          "solution":  "【分析】条形统计图的特点：能够清楚地看出数量的多少;复式条形统计图：统计两组或两组以上的数据的多少，每组数据用不同的直条表示，能够比较两组或两组以上数据的多少，据此判断。本题考查复式条形统计图的特点，掌握分析中的特点是解题的关键.\n【详解】根据分析可知：复式条形统计图的特点是用直条的长短表示数量的多少，优点是能清楚地看出数量的多少，便于比较两组数据的多少。原题说法正确。故答案为: √",
+        "knowledge_points":  [
+                                 "复式条形统计图"
+                             ],
+        "source":  "20-21四年级下·辽宁朝阳·期末"
+    },
+    {
+        "id":  "495",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "知道小组的平均身高，一定能知道小组中每个人的身高。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【详解】因为平均数反映的是一组数据的特征，不是其中每一个数据的特征，所以知道小组的平均身高，并不能知道每个同学的身高;原题中说一定能知道小组中每个人的身高，所以判断错误。例如:(160+180)÷2=340÷2=170(cm)如果只知道两人的平均身高170cm,不能确定每一个的身高，只能求出两人的身高之和:170×2=340(cm)。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "20-21四年级下·四川雅安·期末"
+    },
+    {
+        "id":  "496",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "小雪语文数学的平均分是96分，她的语文成绩是93分，数学成绩一定高于96分。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】根据小雪语文数学的平均分是96，由此可以求出语文数学的总成绩，然后用总成绩减去语文的成绩即可求出数学的成绩，再与96分比较即可判断。本题考查平均数的意义和求法。平均数问题的基本数量关系：平均数×份数=总数量.\n【详解】96×2-93=192-93=99(分)99分\u003e96分，所以原题说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "20-21四年级下·四川内江·期末"
+    },
+    {
+        "id":  "497",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "篮球队不同身高队员的平均身高189cm,最高队员的身高一定高于189cm。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】平均数是指在一组数据中所有数据之和再除以这组数据的个数，它介于这组数据的最大值与最小值之间，据此即可解答。本题主要考查学生对平均数的意义及特点的掌握和灵活运用.\n【详解】根据分析可知，平均身高189cm介于最高队员身高和最矮队员身高之间，所以最高队员的身高一定高于189cm。故答案为:√",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "20-21四年级下·四川宜宾·期末"
+    },
+    {
+        "id":  "498",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "某公司员工的月平均工资是3800元，李伯伯是这个公司的员工，他这个月的工资可能比3800元低。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】根据平均数的意义可知，平均数一定比最小数大，比最大数小，即这个公司员工的实际工资浮动在3800元上下，据此解题。本题考查了平均数，明确平均数的意义是解题的关键.\n【详解】由分析得：李伯伯作为公司的一员，他的工资可能高于平均工资3800元，也可能低于平均工资3800元。故答案为:√",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "18-19六年级下·浙江杭州·期末"
+    },
+    {
+        "id":  "499",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "小东所在小组同学的平均体重是36千克，小刚所在小组同学的平均体重是35千克，小东一定比小刚重。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】平均数是反映一组数据的平均水平，并不能反映这组数据中各个数据的大小，据此即可解答。本题考查了平均数的意义在实际生活中的灵活应用，平均体重并不能反映某个同学的体重.\n【详解】小东所在小组同学的平均体重是36千克，小刚所在小组同学的平均体重是35千克，并不能确定小东和小刚的体重是多少，故无法比较，原说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数",
+                                 "众数",
+                                 "中位数的异同及应用"
+                             ],
+        "source":  "20-21四年级下·广东珠海·期末"
+    },
+    {
+        "id":  "500",
+        "unit":  "第八单元平均数和条形图",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "期中考试赵平语数外三科的平均成绩是93分，田欣然语数外三科的平均成绩是97分，所以这次期中考试赵平的语文成绩一定比田欣然的语文成绩低。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】根据平均数的意义可知，期中考试赵平语数外三科的平均成绩是93分，田欣然语数外三科的平均成绩是97分，并不代表赵平的语文成绩是93分，他的语文成绩可能比93分高得多，也可能比93分低得多，即赵平的语文成绩可能比田欣然的语文成绩高，所以这次期中考试赵平的语文成绩不一定比田欣然的语文成绩低。解答此题因根据平均数的意义，进行分析，解答即可，注意平均数只反映一组数据的集中趋势，并不代表其中的哪一个数据.\n【详解】根据分析可知，期中考试赵平语数外三科的平均成绩是93分，田欣然语数外三科的平均成绩是97分，所以这次期中考试赵平的语文成绩不一定比田欣然的语文成绩低;故原题干说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "平均数的意义及求法"
+                             ],
+        "source":  "20-21四年级下·湖南张家界·期末"
+    }
+]
diff --git "a/\345\233\233\345\271\264\347\272\247\351\242\230\351\233\206/\345\233\233\345\271\264\347\272\247\344\270\213/\345\233\233\345\271\264\347\272\247\344\270\213\347\254\254\345\205\255\345\215\225\345\205\203\345\260\217\346\225\260\347\232\204\345\212\240\346\263\225\345\222\214\345\207\217\346\263\225.json" "b/\345\233\233\345\271\264\347\272\247\351\242\230\351\233\206/\345\233\233\345\271\264\347\272\247\344\270\213/\345\233\233\345\271\264\347\272\247\344\270\213\347\254\254\345\205\255\345\215\225\345\205\203\345\260\217\346\225\260\347\232\204\345\212\240\346\263\225\345\222\214\345\207\217\346\263\225.json"
new file mode 100644
index 0000000000000000000000000000000000000000..6947fd5f4081327770fdd2b9d65a21a93022c7ab
--- /dev/null
+++ "b/\345\233\233\345\271\264\347\272\247\351\242\230\351\233\206/\345\233\233\345\271\264\347\272\247\344\270\213/\345\233\233\345\271\264\347\272\247\344\270\213\347\254\254\345\205\255\345\215\225\345\205\203\345\260\217\346\225\260\347\232\204\345\212\240\346\263\225\345\222\214\345\207\217\346\263\225.json"
@@ -0,0 +1,8840 @@
+[
+    {
+        "id":  "1",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "有甲，乙，丙三个数，其中甲数比乙数大1.3，乙数比丙数小4.8，那么甲数比丙数()。",
+        "options":  [
+                        "A. 小3.5",
+                        "B. 小6.1",
+                        "C. 大3.5",
+                        "D. 大6.1"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】根据题目所给条件，通过设丙数为一个具体数值，分别求出乙数和甲数，再计算甲数与丙数的差值，即可判断甲数比丙数大还是小以及具体数值.\n【详解】设丙数为10(设为10是为了方便计算，设其他数结果相同)。因为乙数比丙数小4.8，所以乙数=丙数-4.8=10-4.8=5.2。又因为甲数比乙数大1.3，所以甲数=乙数+1.3=5.2+1.3=6.5。则甲数比丙数小:10-6.5=3.5。故答案为A。",
+        "knowledge_points":  [
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "2",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面算式中都有一个数字看不清楚，得数一定大于50的算式是()。",
+        "options":  [
+                        "A. 2![](images/2026-03-12%20112234.jpg)。8+17.1",
+                        "B. 62.5-2![](images/2026-03-12%20112250.jpg)。9",
+                        "C. 7![](images/2026-03-12%20112306.jpg)。5-11.2",
+                        "D. 24.7+![](images/2026-03-12%20112320.jpg)5.6"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】计算小数加，减法要把小数点对齐，从低位算起。加法中\"满十进一\"，减法中，不够减，从前一位\"借一当十\"再减。据此解答.\n【详解】A。二十几加17.1，结果最多是四十几.\nB。62.5减二十几，结果最多是四十几.\nC。七十几减11.2，结果最少是五十几.\nD。24.7加一个数，结果可能大于五十，也可能小于五十.\n故答案为:C",
+        "knowledge_points":  [
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "3",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "乐乐用计算器计算18.39+2.34时，误输了13.89+2.34，下面的方法可以帮助她修正错误的是()。",
+        "options":  [
+                        "A. 加4.5",
+                        "B. 减0.5",
+                        "C. 加0.5",
+                        "D. 减4.5"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】两个算式中,2.34相同，只需比较18.39与13.89的差,18.39与13.89的差值为4.5，即错误算式的结果比正确算式少了4.5，所以要得到正确结果，需在错误结果基础上加4.5.\n【详解】18.39-13.89=4.5，故答案为:A",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "小数的四则运算及法则"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "4",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "某天早晨的气温是22.5(℃),中午升高了8.7(℃),晚上降低了9.3(℃),晚上的气温是()(℃)",
+        "options":  [
+                        "A. 23.1",
+                        "B. 21.9",
+                        "C. 4.5",
+                        "D. 40.5"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据题意，早晨的气温加上中午升高的温度再减去晚上降低的温度，即可求出晚上的气温，据此解答.\n【详解】22.5+8.7-9.3=31.2-9.3=21.9(℃)\n故答案为:B",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算",
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "5",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "□-2.13+1.09=8,□ 里的数是()。",
+        "options":  [
+                        "A. 4.78",
+                        "B. 7.96",
+                        "C. 6.96",
+                        "D. 9.04"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】先用8减去1.09，求出□-2.13的差;再加上2.13，即可求出□里的数，据此解答。\n【详解】8-1.09+2.13=9.04，故答案为:D",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "6",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下列各数中，与880万最接近的是()。",
+        "options":  [
+                        "A. 8801000",
+                        "B. 9000000",
+                        "C. 8891000",
+                        "D. 8008888"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】先把各选项中的整数改写成用\"万\"作单位的数，然后与880万相减，求出差，再比较大小，差值越小的，越接近880万。改写成用\"万\"作单位的数，就是在万位数的右下角点上小数点，然后把小数末尾的0去掉，再在数的后面写上\"万\"字.\n【详解】A。8801000=880.1万,880.1万-880万=0.1万;B。9000000=900万,900万-880万=20万;C。8891000=889.1万,889.1万-880万=9.1万;D。8008888=800.8888万,880万-800.8888万=79.1112万;0.1万\u003c9.1万\u003c20万\u003c79.1112万，所以，与880万最接近的是8801000.\n故答案为:A",
+        "knowledge_points":  [
+                                 "亿以内数的改写",
+                                 "小数的改写",
+                                 "多位小数的进位加法，退位减法",
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "2025·浙江杭州·小升初真题"
+    },
+    {
+        "id":  "7",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面不适合用\"部分量+部分量=总量\"这个关系来表示的是()。",
+        "options":  [
+                        "A. 两艘轮船共行170千米",
+                        "B. 两个工程队共同打通了675米长的隧道",
+                        "C. 买苹果和梨共花了10.4元",
+                        "D. 足球上白色皮比黑色皮的2倍少4块"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】\"部分量+部分量=总量\"关系的理解。该关系表示两个部分相加等于一个整体总量，据此逐项分析，进行解答.\n【详解】A。两艘轮船共行170千米;表示一条轮船行驶距离与另一条轮船行驶距离之和等于170千米，适合用\"部分量+部分量=总量\"关系的理解.\nB。两个工程队共同打通了675米长的隧道;表示一个工程队打通隧道的长度与另一个工程队打通隧道的长度和，适合用\"部分量+部分量=总量\"关系的理解.\nC。买苹果和梨共花了10.4元;表示买苹果花的钱数与买梨花的钱数和，适合用\"部分量+部分量=总量\"关系的理解.\nD。足球上白色皮比黑色皮的2倍少4块;表示足球上白色皮块数=黑色皮块数×2-4，不适合用\"部分量+部分量=总量\"这个关系来表示.\n故答案为:D",
+        "knowledge_points":  [
+                                 "利用小数加减法解决实际问题"
+                             ],
+        "source":  "25-26五年级上·重庆九龙坡·月考"
+    },
+    {
+        "id":  "8",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "黄老师买3本笔记本,2支钢笔和4把直尺一共需要36.2元，买2本笔记本,3支钢笔和1把直尺一共需要43.8元。买1本笔记本,1支钢笔和1把直尺一共需要()元。",
+        "options":  [
+                        "A. 14",
+                        "B. 15",
+                        "C. 16"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】将36.2和43.8求和即可计算出5本笔记本,5支钢笔和5把直尺的总价;再用5本笔记本,5支钢笔和5把直尺的总价除以5即可计算1本笔记本,1支钢笔和1把直尺的总价.\n【详解】3+2=5(本)2+3=5(支)4+1=5(把)\n(36.2+43.8)÷5=80÷5=16(元)\n故答案为:C",
+        "knowledge_points":  [
+                                 "一位小数的进位加法，退位减法",
+                                 "等量代换"
+                             ],
+        "source":  "24-25六年级上·广西梧州·期末"
+    },
+    {
+        "id":  "9",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "一根9.4米长的绳子，第一次剪去3.6米，第二次剪去2.5米，现在比原来短了()。",
+        "options":  [
+                        "A. 5.8米",
+                        "B. 6.1米",
+                        "C. 1.1米"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据第一次剪去3.6米，第二次剪去2.5米，把两次减去的长度求和，求出现在比原来短了多少米即可.\n【详解】3.6+2.5=6.1(米)故答案为:B",
+        "knowledge_points":  [
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "25-26五年级上·湖南邵阳·期中"
+    },
+    {
+        "id":  "10",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "在38.268中，小数点前面的\"8\"和小数点后面的\"8\"相差()。",
+        "options":  [
+                        "A. 0",
+                        "B. 7.92",
+                        "C. 7.992",
+                        "D. 8.008"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】先明确两个\"8\"所表示的数值：小数点前面的\"8\"在个位上，表示8个一，即8;小数点后面的\"8\"在千分位上，表示8个0.001，即0.008，再计算它们的差值即可.\n【详解】根据分析可得:8-0.008=7.992，故答案为:C",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25五年级上·河南南阳·期中"
+    },
+    {
+        "id":  "11",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "轩轩买了五件价格不同的玩具，最便宜的6.2元，最贵的8.3元，她可能一共花了()元。",
+        "options":  [
+                        "A. 31",
+                        "B. 35.6",
+                        "C. 41.5"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】已知最便宜6.2元，最贵8.3元，中间三件价格介于两者之间且互不相同，假设价格均为一位小数，则最小值时中间三件价格尽可能小，为6.3元,6.4元,6.5元;最大值时中间三件价格尽可能大，为8元,8.1元,8.2元。据此分别求出五件玩具总价的最小值和最大值，进而确定总花费的范围。据此解答.\n【详解】6.2+6.3+6.4+6.5+8.3=33.7(元)\n6.2+8+8.1+8.2+8.3=38.8(元)\n所以五件玩具的总价格应大于等于33.7元且小于等于38.8元，选项中35.6符合。故答案为:B",
+        "knowledge_points":  [
+                                 "一位小数的大小比较",
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25五年级上·山东济南·期中"
+    },
+    {
+        "id":  "12",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "淘气用计算器计算4.19-2.83，他错误地输入了4.19-2.33，要修正这个错误，可以接着()。",
+        "options":  [
+                        "A. 减0.5",
+                        "B. 加0.5",
+                        "C. 减5",
+                        "D. 加5"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】因为2.83-2.33=0.5，所以计算4.19-2.83比4.19-2.33少减了0.5，以此做出选择即可.\n【详解】因为2.83-2.33=0.5，所以这两个算式少减了0.5，要修正这个错误，再减0.5即可。故答案为:A",
+        "knowledge_points":  [
+                                 "用计算器计算小数加减法"
+                             ],
+        "source":  "24-25五年级上·湖南邵阳·期中"
+    },
+    {
+        "id":  "13",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.94,
+        "question":  "一根绳子长20米，第一次用去3.15米，第二次用去4.8米，这根绳子和原来相比，短了()米。",
+        "options":  [
+                        "A. 7.95",
+                        "B. 12.05",
+                        "C. 16.85",
+                        "D. 15.2"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】用去多少米就比原来短了多少米，将两次用去的长度相加即可.\n【详解】3.15+4.8=7.95(米)故答案为:A",
+        "knowledge_points":  [
+                                 "多位小数的不进位加法，不退位减法"
+                             ],
+        "source":  "24-25五年级上·湖南邵阳·期中"
+    },
+    {
+        "id":  "14",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一桶油连桶重2.8kg,倒去一半后连桶重1.55kg,油桶的重量为()kg。",
+        "options":  [
+                        "A. 2.6",
+                        "B. 2.5",
+                        "C. 0.3",
+                        "D. 0.2"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】已知\"连桶重2.8kg\"(油+桶的总重),倒去一半油后\"连桶重1.55kg\"(半桶油+桶的总重)。因此，一半油的重量=倒之前总重-倒之后总重，即:2.8-1.55=1.25(kg)。倒去一半油后，\"半桶油+桶=1.55kg\"，其中半桶油重1.25kg,因此：油桶重量=倒之后总重-半桶油重量，即:1.55-1.25=0.3(kg).\n【详解】1.55-(2.8-1.55)=1.55-1.25=0.3(kg)\n故答案为:C",
+        "knowledge_points":  [
+                                 "利用小数加减法解决实际问题",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25五年级上·河南南阳·期中"
+    },
+    {
+        "id":  "15",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面各数中,()最接近5.8。",
+        "options":  [
+                        "A. 5.9",
+                        "B. 5.75",
+                        "C. 5.82"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】用每个选项的数与5.8相减，求出的差越小则越接近5.8，据此选择即可.\n【详解】A。5.9-5.8=0.1\nB。5.8-5.75=0.05\nC。5.82-5.8=0.02\n0.02\u003c0.05\u003c0.1，所以5.82最接近5.8。故答案为:C",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "24-25三年级下·河北唐山·期末"
+    },
+    {
+        "id":  "16",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下列算式中，\"5\"和\"6\"在计算时可以直接相加减的是()。",
+        "options":  [
+                        "A. 607-75",
+                        "B. 0.51+0.68",
+                        "C. 5-0.67",
+                        "D. 12.581-6.781"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据整数加减法和小数加减法的计算法则，只有相同数位上的数字才能直接相加减。需逐一分析各选项中\"5\"和\"6\"所在的数位。以此选择即可.\n【详解】根据分析可知:A。607-75:5在75的个位,6在607的百位，数位不同，不能直接相减。B。0.51+0.68:5在0.51的十分位,6在0.68的十分位，数位相同，可以直接相加。C。5-0.67:5在个位,6在0.67的十分位，数位不同，不能直接相减。D。12.581-6.781:5在12.581的十分位,6在6.781的个位，数位不同，不能直接相减.\n故答案为:B",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·河北廊坊·期末"
+    },
+    {
+        "id":  "17",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "小莫在计算10.54减去一个一位小数时，由于错误地把数的末尾对齐，结果得到错误的答案9.77，正确的得数应该是()。",
+        "options":  [
+                        "A. 0.77",
+                        "B. 2.74",
+                        "C. 2.84"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】由题意可知：用10.54-9.77即可求出减错的那个减数，因为错误把一位小数的末尾对齐，相当于把它当成了两位小数，据此可推断出原来的减数，用10.54减去原来的减数即可求出正确的得数.\n【详解】10.54-9.77=0.77，即原来的减数是7.7,10.54-7.7=2.84故答案为:C",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "加，减法的意义和各部分间的关系"
+                             ],
+        "source":  "24-25四年级下·陕西安康·期末"
+    },
+    {
+        "id":  "18",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "小明的身高是1.4米，与小明的身高最接近的是()。",
+        "options":  [
+                        "A. 淘气身高1.42米",
+                        "B. 笑笑身高1.39米",
+                        "C. 小红身高1.3米"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据题意，先求出每个人与小明的身高差，再比较大小，差值最小的，与小明的身高最接近.\n【详解】A。1.42-1.4=0.02(米)\nB。1.4-1.39=0.01(米)\nC。1.4-1.3=0.1(米)\n因为0.01\u003c0.02\u003c0.1，所以与小明的身高最接近的是笑笑。故答案为:B",
+        "knowledge_points":  [
+                                 "多位小数的大小比较",
+                                 "一位小数的不进位加法，不退位减法",
+                                 "多位小数的不进位加法，不退位减法",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·山西吕梁·期末"
+    },
+    {
+        "id":  "19",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "在15.05中，个位上的\"5\"与百分位上的\"5\"相差()。",
+        "options":  [
+                        "A. 4",
+                        "B. 4.5",
+                        "C. 4.95",
+                        "D. 5.05"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】在15.05中，个位上的\"5\"表示5个一(即5),百分位上的\"5\"表示5个百分之一(即0.05)。用5减去0.05即可求解.\n【详解】5-0.05=4.95，故答案为:C",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·河北唐山·期末"
+    },
+    {
+        "id":  "20",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在()里填上\"\u003e\"\"\u003c\"或\"=\".\n8.5-0.8( )8.5-0.9;10( )5.8+5.9;3.61+0.25( )3.8;6.8( )7.13-0.34;0.7+2.9( )3.6;6.8( )15.06-9.26",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;\u003c;\u003e;\u003e;=;\u003e",
+        "solution":  "【分析】先分别计算出每个算式的结果，再按照小数比较大小的方法进行比较即可。小数比较大小，先比较整数部分，整数部分大的数大;整数部分相同，再比较十分位，十分位大的数大;十分位相同，比较百分位，以此类推.\n【详解】(1)8.5-0.8=7.7,8.5-0.9=7.6,7.7\u003e7.6，所以8.5-0.8\u003e8.5-0.9\n(2)5.8+5.9=11.7,10\u003c11.7，所以10\u003c5.8+5.9\n(3)3.61+0.25=3.86,3.86\u003e3.8，所以3.61+0.25\u003e3.8\n(4)7.13-0.34=6.79,6.8\u003e6.79，所以6.8\u003e7.13-0.34\n(5)0.7+2.9=3.6,3.6=3.6，所以0.7+2.9=3.6\n(6)15.06-9.26=5.8,6.8\u003e5.8，所以6.8\u003e15.06-9.26",
+        "knowledge_points":  [
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "21",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "4吨80千克=( )吨;376公顷=( )平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "4.08;3.76",
+        "solution":  "【分析】根据进率:1吨=1000千克,1平方千米=100公顷;从高级单位向低级单位转换，乘进率;从低级单位向高级单位转换，除以进率;据此解答.\n【详解】(1)80÷1000=0.08(吨),4+0.08=4.08(吨),所以4吨80千克=4.08吨;(2)376÷100=3.76(平方千米),所以376公顷=3.76平方千米。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "公顷，平方千米的进率与换算",
+                                 "吨，千克之间的换算与比较",
+                                 "多位小数的不进位加法，不退位减法"
+                             ],
+        "source":  "25-26五年级上·河北唐山·期末"
+    },
+    {
+        "id":  "22",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小宇在计算a-12.8-2.18时，错算成a-12.8-21.8，这样算得的结果与正确结果相差( )。",
+        "options":  [
+
+                    ],
+        "answer":  "19.62",
+        "solution":  "【分析】正确的算式为a-12.8-2.18，错误的算式为a-12.8-21.8，观察可知计算多减了一部分，实际多减的为21.8-2.18，则结果相差19.62.\n【详解】21.8-2.18=19.62",
+        "knowledge_points":  [
+                                 "小数的四则运算及法则"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "23",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在()里填上\"\u003e\"\"\u003c\"或\"=\".\n3.68+0.63+0.37( )0.63+0.37+3.68;8.72-0.8-0.2( )8.72-(0.8-0.2);7.06-3.7+1.94( )7.06-(3.7-1.94);29.6+8.77-9.6( )29.6-9.6-8.77",
+        "options":  [
+
+                    ],
+        "answer":  "=;\u003c;=;\u003e",
+        "solution":  "【分析】分别计算出左右两边的算式，再比较大小，据此解答.\n【详解】(1)3.68+0.63+0.37=4.68,0.63+0.37+3.68=4.68,4.68=4.68，所以,3.68+0.63+0.37=0.63+0.37+3.68\n(2)8.72-0.8-0.2=7.72,8.72-(0.8-0.2)=8.12,7.72\u003c8.12，所以,8.72-0.8-0.2\u003c8.72-(0.8-0.2)\n(3)7.06-3.7+1.94=5.3,7.06-(3.7-1.94)=5.3,5.3=5.3，所以,7.06-3.7+1.94=7.06-(3.7-1.94)\n(4)29.6+8.77-9.6=28.77,29.6-9.6-8.77=11.23,28.77\u003e11.23，所以,29.6+8.77-9.6\u003e29.6-9.6-8.77",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "小数的四则运算及法则"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "24",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "4.68-(5.1+2.97)与4.68-5.1-2.97的运算顺序( ),计算结果( )。(填\"相同\"或\"不同\")",
+        "options":  [
+
+                    ],
+        "answer":  "不同;相同",
+        "solution":  "【分析】根据整数和小数的四则运算顺序规则：在有括号的算式里，要先算括号里面的，再算括号外面的;在没有括号的算式里，如果只有加，减法或者只有乘，除法，都要从左往右按顺序计算。第一个算式中有括号，应先算括号内的加法，再算括号外的减法;第二个算式中没有括号，只有减法，属于同级运算，应从左往右依次计算，所以两个算式的运算顺序不同。4.68-(5.1+2.97)去括号可得4.68-5.1-2.97，与第二个算式相同，所以计算结果相同.\n【详解】根据分析可知:4.68-(5.1+2.97)与4.68-5.1-2.97的运算顺序(不同),计算结果(相同)。",
+        "knowledge_points":  [
+                                 "无括号的运算顺序",
+                                 "小数的加，减法混合运算",
+                                 "多位小数的进位加法，退位减法",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "25",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "计算6.37+(16.5-8.9)时，应先算( )( )( ),此步结果是( ),再算( )( )( ),最后结果是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "16.5;-;8.9;7.6;6.37;+;7.6;13.97",
+        "solution":  "【分析】根据小数加减混合运算的运算顺序，有括号时先算括号内的运算，再算括号外的运算，先算减法，计算为7.6，再算加法，计算为13.97.\n【详解】计算6.37+(16.5-8.9)时，应先算16.5-8.9，此步结果是7.6，再算6.37+7.6，最后结果是13.97。",
+        "knowledge_points":  [
+                                 "有括号的运算顺序",
+                                 "带有小括号的混合运算",
+                                 "小数的四则运算及法则"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "26",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "中国结象征着人们的美好祝福和心愿，体现着对真善美的追求和渴望。用一根长15米的丝绳编中国结，第一次用去5.6米，第二次用去3.8米，现在剩下的丝绳比原来短了( )米。",
+        "options":  [
+
+                    ],
+        "answer":  "9.4",
+        "solution":  "【分析】丝绳用去多少，就比原来短了多少，剩下的丝绳比原来短的长度=两次一共用去的丝绳长度，可直接通过求两次用去的总和解答.\n【详解】5.6+3.8=9.4(米)",
+        "knowledge_points":  [
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25五年级上·湖南邵阳·期末"
+    },
+    {
+        "id":  "27",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "供电局原来有捆电线长14.85米，工人师傅第一次用去0.75米，第二次又用去1.25米。这根电线比原来短了( )米。",
+        "options":  [
+
+                    ],
+        "answer":  "2",
+        "solution":  "【分析】根据题意，这根电线比原来短的长度即两次一共用去的长度，用第一次用去的0.75米加第二次用去的1.25米即可解答.\n【详解】0.75+1.25=2(米)",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "25-26五年级上·河南洛阳·期中"
+    },
+    {
+        "id":  "28",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "静静有3.6元，给娟娟1.4元后，就和娟娟的钱数相等了，娟娟原来有( )元。",
+        "options":  [
+
+                    ],
+        "answer":  "0.8",
+        "solution":  "【分析】静静给娟娟1.4元后，就和娟娟的钱数相等，说明静静比娟娟多了(1.4+1.4)元，静静的钱数-静静比娟娟多的钱数=娟娟原来的钱数，据此列式计算.\n【详解】3.6-(1.4+1.4)=3.6-2.8=0.8(元)",
+        "knowledge_points":  [
+                                 "利用小数的加，减法混合运算解决实际问题",
+                                 "和差倍问题"
+                             ],
+        "source":  "25-26五年级上·湖南邵阳·期中"
+    },
+    {
+        "id":  "29",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "把0.8的小数点去掉，得到的数比原来多( )。",
+        "options":  [
+
+                    ],
+        "answer":  "7.2",
+        "solution":  "【分析】把0.8的小数点去掉，得到8，两数相减，求出得到的数比原来多几.\n【详解】8-0.8=7.2",
+        "knowledge_points":  [
+                                 "小数点向右移动引起小数大小变化的规律",
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "25-26五年级上·湖南邵阳·期中"
+    },
+    {
+        "id":  "30",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数，保留两位小数后是2.54，这个三位小数最大是( ),最小是( ),相差( )。",
+        "options":  [
+
+                    ],
+        "answer":  "2.544;2.535;0.009",
+        "solution":  "【分析】保留两位小数看千分位，小于5直接舍去，大于或等于5向前一位进一.\"四舍\"得到的近似数比原数小，\"五入\"得到的近似数比原数大，据此分析.\n【详解】2.544-2.535=0.009。一个三位小数，保留两位小数后是2.54，这个三位小数最大是2.544，最小是2.535，相差0.009。",
+        "knowledge_points":  [
+                                 "小数的近似数",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "25-26五年级上·河北邯郸·期中"
+    },
+    {
+        "id":  "31",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "小于1的最大两位小数减去最小的两位小数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "0.98",
+        "solution":  "【分析】小于1的最大两位小数是0.99，最小的两位小数是0.01，根据题意列式计算出结果填空即可.\n【详解】0.99-0.01=0.98",
+        "knowledge_points":  [
+                                 "多位小数的不进位加法，不退位减法",
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "24-25五年级上·河南洛阳·期中"
+    },
+    {
+        "id":  "32",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "小于1的最小三位小数加上小于1的最大四位小数，结果是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "1.0009",
+        "solution":  "【分析】小于1的最小三位小数，说明个位上是0，并且要最小，那么十分位上和百分位上是0，千分位上是1，这个小数是0.001;小于1的最大四位小数，说明个位上是0，并且要最大，那么十分位，百分位，千分位和万分位上都是9，这个小数是0.9999.\n【详解】0.001+0.9999=1.0009",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25五年级上·湖南邵阳·期中"
+    },
+    {
+        "id":  "33",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小于1的最小三位小数加上小于1的最大四位小数，结果是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "1.0009",
+        "solution":  "【分析】小于1的最小三位小数是0.001，小于1的最大四位小数是0.9999，相加即可解答.\n【详解】0.001+0.9999=1.0009",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25五年级上·河北石家庄·期中"
+    },
+    {
+        "id":  "34",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一袋盐重1.7千克，一袋味精重0.9千克，两袋共重( )千克。一袋盐比一袋味精重( )千克。",
+        "options":  [
+
+                    ],
+        "answer":  "2.6;0.8",
+        "solution":  "【分析】求两袋共重多少千克，即为盐的重量加上味精的重量。求一袋盐比一袋味精重多少千克，即为盐的重量减去味精的重量即可.\n【详解】1.7+0.9=2.6(千克);1.7-0.9=0.8(千克)。",
+        "knowledge_points":  [
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25五年级上·河南郑州·期中"
+    },
+    {
+        "id":  "35",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "3个十和3个0.1组成( );( )比0.7少0.25。",
+        "options":  [
+
+                    ],
+        "answer":  "30.3;0.45",
+        "solution":  "【分析】要确定由3个十和3个0.1组成的数，需要明确3个十是30,3个0.1是0.3，将它们相加即可求解;求比0.7少0.25的数，用减法计算即可.\n【详解】30+0.3=30.3;0.7-0.25=0.45。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "一位小数的不进位加法，不退位减法",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25五年级上·湖南邵阳·期中"
+    },
+    {
+        "id":  "36",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.4,
+        "question":  "一个三位小数，用\"四舍五入\"法保留两位小数约是10.01，一个两位小数，用\"四舍五入\"法保留一位小数约是10.1，这两个小数的和最大是( ),和最小是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "20.154;20.055",
+        "solution":  "【分析】对于三位小数保留两位小数约是10.01，要找最大的三位小数需考虑\"四舍\"情况，找最小的三位小数需考虑\"五入\"情况;对于两位小数保留一位小数约是10.1，同样找最大两位小数考虑\"四舍\"，找最小两位小数考虑\"五入\"。然后分别计算最大和最小情况下两个小数的和。关键在于根据\"四舍五入\"规则，准确找出满足条件的最大和最小的小数，再通过小数加法运算得出和的最大值与最小值.\n【详解】用\"四舍\"法得到10.01的三位小数，说明千分位上的数字小于5，那么最大就是4，所以这个三位小数最大是10.014。用\"四舍\"法得到10.1的两位小数，说明百分位上的数字小于5，最大就是4，所以这个两位小数最大是10.14。它们的和最大是:10.014+10.14=20.154。用\"五入\"法得到10.01的三位小数，说明原来百分位是0，千分位上的数字大于或等于5，最小就是5，所以这个三位小数最小是10.005。用\"五入\"法得到10.1的两位小数，说明原来十分位是0，百分位上的数字大于或等于5，最小就是5，所以这个两位小数最小是10.05。它们的和最小是:10.005+10.05=20.055。",
+        "knowledge_points":  [
+                                 "小数的近似数",
+                                 "多位小数的不进位加法，不退位减法"
+                             ],
+        "source":  "24-25五年级上·河南南阳·期中"
+    },
+    {
+        "id":  "37",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "用4,8,7和小数点这四张卡片组成一个两位小数，最大的是( ),最小的是( ),它们的差是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "8.74;4.78;3.96",
+        "solution":  "【分析】4,8,7和小数点这四张卡片组成一个两位小数，如果要最大，个位最大，即为8，百分位最小为4，即可写出这个两位小数的最大数。如果要最小，个位最小，即为4，百分位最大为8，即可写出这个两位小数的最小数。二者做减法即可.\n【详解】用4,8,7和小数点这四张卡片组成一个两位小数，最大的是8.74，最小的是4.78.8.74-4.78=3.96。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "24-25五年级上·湖南邵阳·期中"
+    },
+    {
+        "id":  "38",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "用2,3,5三个数字和小数点组成的两位小数中，最大的数与最小的数相差( )。",
+        "options":  [
+
+                    ],
+        "answer":  "2.97",
+        "solution":  "【分析】把三个数字按从大到小排列，小数点点在第一个数字的右下角组成的两位小数最大，即最大小数为5.32;把三个数字按从小到大排列，小数点点在第一个数字的右下角组成的两位小数最小，即最小小数为2.35;然后用最大的小数减去最小的小数即可解答.\n【详解】5.32-2.35=2.97。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "24-25五年级上·湖南邵阳·期中"
+    },
+    {
+        "id":  "39",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.4,
+        "question":  "淘气在计算15.86减去一个一位小数时，错把数的末位对齐，结果得到14.550，这个一位小数是( ),正确的得数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "13.1;2.76",
+        "solution":  "【分析】根据\"被减数-差=减数\"求出错了的减数，再移动小数点的位置得到这个一位小数，再根据\"被减数-减数=差\"，求出正确结果.\n【详解】15.86-14.550=1.31，这个一位小数是13.1.15.86-13.1=2.76。",
+        "knowledge_points":  [
+                                 "小数点向左移动引起小数大小变化的规律",
+                                 "小数点向右移动引起小数大小变化的规律",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25五年级上·河南周口·期中"
+    },
+    {
+        "id":  "40",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "某超市搞活动，满200元减20元，李阿姨买了两件商品分别是160.8元和80.8元，李阿姨实际应付( )元。",
+        "options":  [
+
+                    ],
+        "answer":  "221.6",
+        "solution":  "【分析】先用加法算出两件商品原价，再和200比较;若满200，则原价减20.\n【详解】160.8+80.8=241.6(元),241.6\u003e200,241.6-20=221.6(元)。",
+        "knowledge_points":  [
+                                 "一位小数的不进位加法，不退位减法",
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25五年级上·湖南邵阳·期中"
+    },
+    {
+        "id":  "41",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "被减数，减数，差的和是10，减数是1.6，差是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "3.4",
+        "solution":  "【分析】设被减数为a,减数为1.6，差为x。由a-1.6=x得a=1.6+x;又a+1.6+x=10，代入得2×(1.6+x)=10，先求1.6+x,再求x.\n【详解】10÷2=5,5-1.6=3.4。",
+        "knowledge_points":  [
+                                 "加，减法的意义和各部分间的关系",
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25五年级上·湖南邵阳·期中"
+    },
+    {
+        "id":  "42",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小乐在计算一道小数减法题时，把被减数百分位上的6看成了1，减数十分位上的7看成了2，他算出的结果是6.45，正确结果是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】被减数百分位6看成1，使差减少0.05;减数十分位7看成2，使差增大0.5。将错误结果按这两部分修正即可.\n【详解】0.06-0.01=0.05,0.7-0.2=0.5,6.45-0.5+0.05=6。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "一位小数的不进位加法，不退位减法",
+                                 "小数的数位和计数单位的认识",
+                                 "和与差的变化规律"
+                             ],
+        "source":  "24-25五年级上·湖南·期中"
+    },
+    {
+        "id":  "43",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "215平方厘米=( )平方分米;9米8厘米=( )米",
+        "options":  [
+
+                    ],
+        "answer":  "2.15;9.08",
+        "solution":  "【分析】低级单位换算高级单位要除以进率。1平方分米=100平方厘米,1米=100厘米.\n【详解】215÷100=2.15(平方分米);8÷100=0.08(米),9+0.08=9.08(米)。",
+        "knowledge_points":  [
+                                 "平方厘米，平方分米，平方米之间的进率与换算",
+                                 "毫米，厘米，分米，米之间的进率与换算",
+                                 "多位小数的不进位加法，不退位减法"
+                             ],
+        "source":  "24-25五年级上·浙江·期中"
+    },
+    {
+        "id":  "44",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "爸爸在家人微信群里发了100元红包，妈妈抢到32.88元，安安抢到45.65元，安安比妈妈多抢到( )元，爸爸发的红包还剩( )元。",
+        "options":  [
+
+                    ],
+        "answer":  "12.77;21.47",
+        "solution":  "【分析】用安安金额减妈妈金额得\"多抢到\";用总红包减两人已抢金额得\"还剩\".\n【详解】45.65-32.88=12.77(元);100-32.88-45.65=21.47(元)。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·陕西安康·期末"
+    },
+    {
+        "id":  "45",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\"\u003e\"，\"\u003c\"或\"=\".\n8.7( )6.35+2.4;5.24-2.24( )1.3+1.07;(36+15)×100( )36+15×100;18×0( )(436-436)÷10",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003e;\u003e;=",
+        "solution":  "【分析】先分别算出两边结果再比较;第三题可用乘法分配律比较;第四题利用0乘法与0除法性质.\n【详解】6.35+2.4=8.75，所以8.7\u003c8.75;5.24-2.24=3,1.3+1.07=2.37，所以3\u003e2.37;(36+15)×100=36×100+15×100\u003e36+15×100;18×0=0,(436-436)÷10=0，所以二者相等。",
+        "knowledge_points":  [
+                                 "多位小数的大小比较",
+                                 "被除数是0的除法运算",
+                                 "多位小数的不进位加法，不退位减法",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25四年级下·山西吕梁·期末"
+    },
+    {
+        "id":  "46",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形的一条边长4cm,另一条边长8.5cm,这个三角形的周长是( )cm。",
+        "options":  [
+
+                    ],
+        "answer":  "21",
+        "solution":  "【分析】等腰三角形需两边相等，并满足三角形两边之和大于第三边。若腰为4，则4+4\u003c8.5不成立;故两腰应为8.5，底为4.\n【详解】周长=8.5+8.5+4=21(cm)。",
+        "knowledge_points":  [
+                                 "利用小数加减法解决实际问题",
+                                 "三角形的周长",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·河北唐山·期末"
+    },
+    {
+        "id":  "47",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "( )+1.4=2;7.05×( )=7050",
+        "options":  [
+
+                    ],
+        "answer":  "0.6;1000",
+        "solution":  "【分析】第一空用2减1.4;第二空比较7.05到7050小数点右移3位，相当于乘1000.\n【详解】2-1.4=0.6;7.05×1000=7050。",
+        "knowledge_points":  [
+                                 "一位小数的进位加法，退位减法",
+                                 "小数点向右移动引起小数大小变化的规律"
+                             ],
+        "source":  "24-25四年级下·山西晋中·期末"
+    },
+    {
+        "id":  "48",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "判断题",
+        "difficulty":  0.94,
+        "question":  "计算小数加减法时，要把小数点对齐。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】小数加减法要把相同数位对齐，实质就是小数点对齐.\n【详解】题干说法正确，故答案为:√。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "2025·广东揭阳·小升初模拟"
+    },
+    {
+        "id":  "49",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "判断题",
+        "difficulty":  0.94,
+        "question":  "5.6+(0.7+0.8)=5.6+0.7+5.6+0.8。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】分别计算等式两边并比较.\n【详解】左边:5.6+(0.7+0.8)=5.6+1.5=7.1;右边:5.6+0.7+5.6+0.8=12.7。7.1不等于12.7，故答案为: ×。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算",
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25五年级上·广东江门·期中"
+    },
+    {
+        "id":  "50",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "判断题",
+        "difficulty":  0.94,
+        "question":  "0.1+0.9+0.1+0.9=1。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】先分组相加计算结果再判断.\n【详解】(0.1+0.9)+(0.1+0.9)=1+1=2，不等于1，故答案为: ×。",
+        "knowledge_points":  [
+                                 "一位小数的进位加法，退位减法",
+                                 "整数加法运算律推广到小数"
+                             ],
+        "source":  "24-25五年级上·湖南株洲·期中"
+    },
+    {
+        "id":  "51",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "判断题",
+        "difficulty":  0.94,
+        "question":  "1.5+0.7-1.5+0.7=0。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】利用运算律化简后计算.\n【详解】(1.5-1.5)+(0.7+0.7)=0+1.4=1.4，不等于0，故答案为: ×。",
+        "knowledge_points":  [
+                                 "整数加法运算律推广到小数",
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25五年级上·河南周口·期中"
+    },
+    {
+        "id":  "52",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "和是小数，两个加数一定都是小数。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】和是小数，说明至少有一个加数是小数，另一个可以是整数.\n【详解】如3.5+2=5.5，和是小数但第二个加数是整数，故题干错误，答案为:×。",
+        "knowledge_points":  [
+                                 "小数的初步认识",
+                                 "一位小数的不进位加法，不退位减法"
+                             ],
+        "source":  "24-25五年级上·河南周口·期中"
+    },
+    {
+        "id":  "53",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "计算题",
+        "difficulty":  0.85,
+        "question":  "列竖式计算，并验算.\n24.08+4.33= 20-10.99=",
+        "options":  [
+
+                    ],
+        "answer":  "28.41;9.01",
+        "solution":  "【分析】小数加减法计算时小数点对齐，按整数加减法法则计算，最后点上小数点.\n【详解】24.08+4.33=28.41\n![](images/2026-03-12%20114842.jpg) 验算:![](images/2026-03-12%20114847.jpg)\n20-10.99=9.01\n![](images/2026-03-12%20114853.jpg) 验算:![](images/2026-03-12%20114859.jpg)",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "54",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "脱式计算，能简算的要用简便方法.\n45.55-(5.55+15.5)\n26-4.2-5.8\n3.72-0.94+5.28-2.06",
+        "options":  [
+
+                    ],
+        "answer":  "24.5;16;6",
+        "solution":  "【分析】第一题去括号后连续减法;第二题用减法性质先把后两数相加;第三题交换结合后凑整计算.\n【详解】45.55-(5.55+15.5)=45.55-5.55-15.5=40-15.5=24.5\n26-4.2-5.8=26-(4.2+5.8)=26-10=16\n3.72-0.94+5.28-2.06=3.72+5.28-0.94-2.06=(3.72+5.28)-(0.94+2.06)=9-3=6",
+        "knowledge_points":  [
+                                 "与小数减法相关的简便运算",
+                                 "整数加法运算律推广到小数"
+                             ],
+        "source":  "25-26五年级上·湖南邵阳·期中"
+    },
+    {
+        "id":  "55",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "计算题",
+        "difficulty":  0.85,
+        "question":  "用竖式计算.\n0.86+3.74= 8-2.35= 19.87-6.9=",
+        "options":  [
+
+                    ],
+        "answer":  "4.6;5.65;12.97",
+        "solution":  "【分析】小数加减法：小数点对齐，从低位算起，按整数法则计算，结果小数点对齐.\n【详解】0.86+3.74=4.6\n![](images/2026-03-12%20114602.jpg)\n8-2.35=5.65\n![](images/2026-03-12%20114608.jpg)\n19.87-6.9=12.97\n![](images/2026-03-12%20114613.jpg)",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "25-26五年级上·湖南邵阳·期中"
+    },
+    {
+        "id":  "56",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "计算下面各题，能简算的要简算.\n49.6-12.8+0.2\n17.4-2.25-7.75",
+        "options":  [
+
+                    ],
+        "answer":  "37;7.4",
+        "solution":  "【分析】第一题按顺序算;第二题用减法性质转化为减去和.\n【详解】49.6-12.8+0.2=36.8+0.2=37\n17.4-2.25-7.75=17.4-(2.25+7.75)=17.4-10=7.4",
+        "knowledge_points":  [
+                                 "与小数减法相关的简便运算",
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "25-26五年级上·湖南邵阳·期中"
+    },
+    {
+        "id":  "57",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "列式计算.\n37.46减去5.25与7.85的和，差是多少。",
+        "options":  [
+
+                    ],
+        "answer":  "37.46-(5.25+7.85)=24.36",
+        "solution":  "【分析】先算5.25与7.85的和，再用37.46去减.\n【详解】37.46-(5.25+7.85)=37.46-13.1=24.36。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·河南驻马店·期末"
+    },
+    {
+        "id":  "58",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "列式计算.\n比4.75小1.8的数，再加上7.253，结果是多少。",
+        "options":  [
+
+                    ],
+        "answer":  "10.203",
+        "solution":  "【分析】\"比4.75小1.8\"先做减法，再把结果加上7.253.\n【详解】4.75-1.8+7.253=2.95+7.253=10.203。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·河南驻马店·期末"
+    },
+    {
+        "id":  "59",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "列竖式计算.\n99.89-17.97= 26.75+8.77=",
+        "options":  [
+
+                    ],
+        "answer":  "81.92;35.52",
+        "solution":  "【分析】小数加减法先小数点对齐，再按整数法则计算，最后在得数中点小数点.\n【详解】99.89-17.97=81.92\n![](images/2026-03-12%20114354.jpg)\n26.75+8.77=35.52\n![](images/2026-03-12%20114401.jpg)",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·陕西安康·期末"
+    },
+    {
+        "id":  "60",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "列竖式计算下面各题.\n8.3-2.57= 15.98+16.79=",
+        "options":  [
+
+                    ],
+        "answer":  "5.73;32.77",
+        "solution":  "【分析】计算小数加减法时小数点对齐，位数不足可在末尾补0，再按整数法则计算.\n【详解】8.3-2.57=5.73\n![](images/2026-03-12%20114128.jpg)\n15.98+16.79=32.77\n![](images/2026-03-12%20114135.jpg)",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·河北邢台·期末"
+    },
+    {
+        "id":  "61",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "一条地铁线分高架线和地下线两部分，其中高架线16.5千米，地下线比高架线短2.8千米，地下线长多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "13.7千米",
+        "solution":  "【分析】地下线比高架线短2.8千米，所以地下线=16.5-2.8.\n【详解】16.5-2.8=13.7(千米)。答：地下线长13.7千米。",
+        "knowledge_points":  [
+                                 "利用小数加减法解决实际问题",
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "25-26五年级上·海南省直辖县级单位·期末"
+    },
+    {
+        "id":  "62",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "乐乐在计算一个数减0.69时，不小心把0.69写成了0.96，得数是3.99。这道题的正确结果是多少。",
+        "options":  [
+
+                    ],
+        "answer":  "4.26",
+        "solution":  "【分析】先由错误结果求被减数，再用被减数减去正确减数0.69.\n【详解】3.99+0.96-0.69=4.95-0.69=4.26。答：正确结果是4.26。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "63",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "乐乐想买一本26.48元的作文选集，他现在的钱再添0.67元就正好可以买一本，结果乐乐只买了一本12.33元的故事书。乐乐还剩下多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "13.48元",
+        "solution":  "【分析】先算乐乐原有钱数:26.48-0.67;再减去故事书价钱12.33.\n【详解】26.48-0.67-12.33=26.48-(0.67+12.33)=26.48-13=13.48(元)。答：乐乐还剩13.48元。",
+        "knowledge_points":  [
+                                 "利用小数加减法解决实际问题"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "64",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.4,
+        "question":  "把一根竹竿的一端竖直插入水中，浸湿部分长1.35m,反过来把另一端竖直插入水中，这时竹竿还有比全长的一半多0.45m是干的。这根竹竿的全长是多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "6.3米",
+        "solution":  "【分析】两次插入后浸湿总长为1.35×2=2.7米。已知干的部分比全长一半多0.45米，则浸湿部分比全长一半少0.45米，因此全长一半=2.7+0.45.\n【详解】1.35×2=2.7(米)\n(2.7+0.45)×2=3.15×2=6.3(米)\n答：这根竹竿全长6.3米。",
+        "knowledge_points":  [
+                                 "利用小数加减法解决实际问题",
+                                 "倒推问题"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "65",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "乐乐家，园园家和图书馆同在一条东西走向的笔直的路上。乐乐家到园园家有多远。\n![](images/2026-03-12%20113551.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "5.29km或2.31km",
+        "solution":  "【分析】若两家分居图书馆两侧，则两家距离为到图书馆距离之和;若同侧，则两家距离为两段距离之差.\n【详解】1.49+3.8=5.29(km)\n3.8-1.49=2.31(km)\n答：乐乐家到园园家可能是5.29km或2.31km。",
+        "knowledge_points":  [
+                                 "两，三位数的一次退位减法",
+                                 "两，三位数的一次进位加法",
+                                 "图文问题"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "66",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "在跳远比赛中，小宇跳了2.1m,比小恒多跳了0.25m,比海海少跳了0.18m。小恒，海海各跳了多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "小恒:1.85米;海海:2.28米",
+        "solution":  "【分析】小恒比小宇少0.25米，用减法;海海比小宇多0.18米，用加法.\n【详解】小恒:2.1-0.25=1.85(m)\n海海:2.1+0.18=2.28(m)\n答：小恒跳了1.85米，海海跳了2.28米。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "一位小数的不进位加法，不退位减法",
+                                 "和差倍问题"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "67",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "2024年全国低碳日为5月15日，小恒从自身做起，他今天少看了一小时电视减少碳排放0.096kg,少用了一小时电脑减少碳排放190g。小恒今天共减少了多少千克的碳排放量。",
+        "options":  [
+
+                    ],
+        "answer":  "0.286千克",
+        "solution":  "【分析】先把190g换算成千克:190g=0.19kg,再与0.096kg相加.\n【详解】190g=0.19kg,0.096+0.19=0.286(kg)\n答：小恒今天共减少了0.286千克的碳排放量。",
+        "knowledge_points":  [
+                                 "克，千克之间的换算与比较",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "68",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "哥哥身高1.38m,比小恒高0.09m,小恒比妈妈矮0.3m。妈妈身高多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "1.59米",
+        "solution":  "【分析】先求小恒身高:1.38-0.09，再加0.3求妈妈身高.\n【详解】1.38-0.09+0.3=1.29+0.3=1.59(米)\n答：妈妈身高1.59米。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "69",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "A,B两地之间是山路，相距60千米，其中一部分是上坡路，其余是下坡路，某人骑电动车从A地到B地，再沿原路返回，去时用了4.5小时，返回时用了3.5小时。已知下坡路每小时行20千米，那么上坡路每小时行多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "12千米",
+        "solution":  "【分析】往返中，上下坡路程各总计60千米。已知下坡速度20千米/小时，可求下坡总时间;再由往返总时间减去下坡时间，得到上坡总时间，最后用路程除以时间求上坡速度.\n【详解】下坡总时间:60÷20=3(小时)\n往返总时间:4.5+3.5=8(小时)\n上坡总时间:8-3=5(小时)\n上坡速度:60÷5=12(千米/小时)\n答：上坡路每小时行12千米。",
+        "knowledge_points":  [
+                                 "基础行程问题",
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "2025·湖南长沙·小升初真题"
+    },
+    {
+        "id":  "70",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "园林局准备在一段公路旁种花卉来装扮城市，第一天种了74.8米，第二天比第一天多种8.2米，第三天比第二天少种了11.6米，第三天种了多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "71.4米",
+        "solution":  "【分析】先求第二天:74.8+8.2;再求第三天：第二天-11.6.\n【详解】74.8+8.2-11.6=83-11.6=71.4(米)。答：第三天种了71.4米。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算",
+                                 "利用小数的加，减法混合运算解决实际问题"
+                             ],
+        "source":  "25-26五年级上·湖南邵阳·期中"
+    },
+    {
+        "id":  "71",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.4,
+        "question":  "乐乐和壮壮去买一本科幻小说，到书店看了价格后发现，乐乐还差5.7元，壮壮还缺2.3元，若两人合买一本，则多10.8元。这本科幻小说多少元。乐乐和壮壮各带了多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "科幻小说18.8元，乐乐带了13.1元，壮壮带了16.5元",
+        "solution":  "【分析】设书价为x。乐乐有x-5.7元，壮壮有x-2.3元;两人合起来比书价多10.8元，即(x-5.7)+(x-2.3)=x+10.8，据此可求x.\n【详解】x-5.7+x-2.3=x+10.8\n2x-8=x+10.8\nx=18.8(元)\n乐乐:18.8-5.7=13.1(元)\n壮壮:18.8-2.3=16.5(元)\n答：科幻小说18.8元，乐乐带了13.1元，壮壮带了16.5元。",
+        "knowledge_points":  [
+                                 "利用小数的加，减法混合运算解决实际问题"
+                             ],
+        "source":  "25-26五年级上·山西太原·期中"
+    },
+    {
+        "id":  "72",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "一块抹茶蛋糕20.8元，一杯奶茶比这块蛋糕便宜5.9元，小星买一块抹茶蛋糕和一杯奶茶一共要付多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "35.7元",
+        "solution":  "【分析】奶茶价=20.8-5.9，再加上蛋糕价20.8.\n【详解】20.8-5.9+20.8=14.9+20.8=35.7(元)。答：一共要付35.7元。",
+        "knowledge_points":  [
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25五年级上·湖南邵阳·期末"
+    },
+    {
+        "id":  "73",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "校园文化节，一场爱心与快乐同行的义卖活动在学校拉开帷幕，下表是部分义卖商品价格.\n| 商品 | 盆栽 | 科普书 | 笔记本 |\n|------|------|--------|--------|\n| 价格 | 8.45元 | 7.85元 | 。元 |\n(1)聪聪买了一盆盆栽和一本科普书，一共花了多少钱。\n(2)明明用10元买了一本笔记本后，还剩4.3元，这本笔记本的单价是多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "16.3元;5.7元",
+        "solution":  "【分析】(1)总价=8.45+7.85;(2)笔记本单价=10-4.3.\n【详解】(1)8.45+7.85=16.3(元)\n(2)10-4.3=5.7(元)\n答:(1)一共花16.3元;(2)笔记本单价5.7元。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "25-26五年级上·海南省直辖县级单位·期中"
+    },
+    {
+        "id":  "74",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "粮站新收购大豆2.56万吨，收购的玉米比大豆多0.75万吨，粮站收购大豆和玉米共多少万吨。",
+        "options":  [
+
+                    ],
+        "answer":  "5.87万吨",
+        "solution":  "【分析】先求玉米:2.56+0.75，再与大豆2.56相加.\n【详解】2.56+0.75+2.56=3.31+2.56=5.87(万吨)。答：共5.87万吨。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "25-26五年级上·湖南邵阳·期中"
+    },
+    {
+        "id":  "75",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "王小胜原来有35元钱，买钢笔用去6.7元，买文具盒用去9.8元，还剩下多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "18.5元",
+        "solution":  "【分析】剩余钱数=35-6.7-9.8.\n【详解】35-6.7-9.8=28.3-9.8=18.5(元)。答：还剩18.5元。",
+        "knowledge_points":  [
+                                 "一位小数的进位加法，退位减法",
+                                 "经济问题"
+                             ],
+        "source":  "25-26五年级上·湖南邵阳·期中"
+    },
+    {
+        "id":  "76",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "小明家距离学校1.45千米，小华家距离学校2.6千米。两家最多相距多少千米。最少相距多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "4.05千米;1.15千米",
+        "solution":  "【分析】两家在学校两侧时最远，为两段和;同侧时最近，为两段差.\n【详解】最多:1.45+2.6=4.05(千米)\n最少:2.6-1.45=1.15(千米)\n答：最多4.05千米，最少1.15千米。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25五年级上·湖南邵阳·期中"
+    },
+    {
+        "id":  "77",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "学校食堂五月份烧煤9.33吨，比四月份节约1.37吨，这两个月共烧煤多少吨。",
+        "options":  [
+
+                    ],
+        "answer":  "20.03吨",
+        "solution":  "【分析】五月比四月少1.37吨，所以四月=9.33+1.37，再与五月相加.\n【详解】9.33+1.37=10.7(吨)\n10.7+9.33=20.03(吨)\n答：两个月共烧煤20.03吨。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25五年级上·湖南邵阳·期中"
+    },
+    {
+        "id":  "78",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一个三角形三条边的长度分别为5.2厘米,6.68厘米,7.12厘米，这个三角形的周长是多少厘米。",
+        "options":  [
+
+                    ],
+        "answer":  "19厘米",
+        "solution":  "【分析】三角形周长等于三边长度之和.\n【详解】5.2+6.68+7.12=11.88+7.12=19(厘米)。答：周长是19厘米。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算",
+                                 "三角形的周长"
+                             ],
+        "source":  "24-25五年级上·河南商丘·期中"
+    },
+    {
+        "id":  "79",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "小明家距离学校1.45千米，小华家距离学校2.6千米。两家最多相距多少千米。最少相距多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "最多4.05千米;最少1.15千米",
+        "solution":  "【分析】若在学校两侧，距离为和;若在同侧，距离为差.\n【详解】1.45+2.6=4.05(千米)\n2.6-1.45=1.15(千米)\n答：两家最多相距4.05千米，最少相距1.15千米。",
+        "knowledge_points":  [
+                                 "利用小数加减法解决实际问题",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25五年级上·河北石家庄·期中"
+    },
+    {
+        "id":  "80",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "学校食堂五月份烧煤9.33吨，比四月份节约1.37吨，这两个月共烧煤多少吨。",
+        "options":  [
+
+                    ],
+        "answer":  "20.03吨",
+        "solution":  "【分析】先算四月份:9.33+1.37，再加五月份9.33.\n【详解】9.33+1.37+9.33=10.7+9.33=20.03(吨)。答：两个月共烧煤20.03吨。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25五年级上·河北石家庄·期中"
+    },
+    {
+        "id":  "81",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "一台拖拉机上午耕地3.96公顷，比下午多耕地0.78公顷，这一天一共耕地多少公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "7.14公顷",
+        "solution":  "【分析】下午=3.96-0.78，再与上午相加.\n【详解】3.96-0.78+3.96=3.18+3.96=7.14(公顷)。答：这一天一共耕地7.14公顷。",
+        "knowledge_points":  [
+                                 "利用小数的加，减法混合运算解决实际问题"
+                             ],
+        "source":  "24-25五年级上·湖南邵阳·期中"
+    },
+    {
+        "id":  "82",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "把一根4米长的竹竿垂直插入水池中，竹竿入泥部分是0.28米，露出水面的部分是0.85米，池中水深多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "2.87米",
+        "solution":  "【分析】水深=竹竿总长-入泥部分-露出水面部分.\n【详解】4-0.28-0.85=3.72-0.85=2.87(米)。答：池中水深2.87米。",
+        "knowledge_points":  [
+                                 "利用小数的加，减法混合运算解决实际问题"
+                             ],
+        "source":  "24-25五年级上·河南周口·期中"
+    },
+    {
+        "id":  "83",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "爸爸身高1.85米，乐乐站在0.25米高的餐凳上比爸爸矮0.15米，乐乐身高多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "1.45米",
+        "solution":  "【分析】乐乐站在凳上身高=1.85-0.15;再减凳高0.25得乐乐实际身高.\n【详解】1.85-0.25-0.15=1.6-0.15=1.45(米)。答：乐乐身高1.45米。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25五年级上·河南周口·期中"
+    },
+    {
+        "id":  "84",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "乐乐付了100元买了一本45.6元的《汉语大词典》和一本20.5元的《格林童话》,营业员应找给他多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "33.9元",
+        "solution":  "【分析】找零=100-45.6-20.5.\n【详解】100-45.6-20.5=54.4-20.5=33.9(元)。答：应找回33.9元。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算",
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25五年级上·河南周口·期中"
+    },
+    {
+        "id":  "85",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "两根绳子的长分别是1.25米和2.39米，爸爸把两根绳子接在一起，接头处用去0.25米。接好后的绳子长多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "3.39米",
+        "solution":  "【分析】接后长度=两根长度和-接头损耗.\n【详解】1.25+2.39-0.25=3.64-0.25=3.39(米)。答：接好后的绳子长3.39米。",
+        "knowledge_points":  [
+                                 "利用小数加减法解决实际问题",
+                                 "小数的加，减法混合运算",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25五年级上·河南周口·期中"
+    },
+    {
+        "id":  "86",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "李乐想买一本定价为20.5元的《趣味数学》,可是他带的钱还差4元，于是他买了一本定价为12.00元的《数学思维》,他还剩多少元钱。",
+        "options":  [
+
+                    ],
+        "answer":  "4.5元",
+        "solution":  "【分析】先求带的钱:20.5-4;再减去实际买书12元.\n【详解】20.5-4-12=16.5-12=4.5(元)。答：他还剩4.5元。",
+        "knowledge_points":  [
+                                 "利用小数的加，减法混合运算解决实际问题"
+                             ],
+        "source":  "24-25五年级上·湖南·期中"
+    },
+    {
+        "id":  "87",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "两根绳子分别长1.07米和1.2米，亮亮把两根绳子接在一起，接头处用去0.26米，接好后的绳子是几米。",
+        "options":  [
+
+                    ],
+        "answer":  "2.01米",
+        "solution":  "【分析】接后长度=1.07+1.2-0.26.\n【详解】1.07+1.2-0.26=2.27-0.26=2.01(米)。答：接好后的绳子是2.01米。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算",
+                                 "多位小数的不进位加法，不退位减法"
+                             ],
+        "source":  "24-25四年级下·河北唐山·期末"
+    },
+    {
+        "id":  "88",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "某玩具店10周年店庆促销，购物每满100元减20元。王叔叔买了一个机器人和一辆遥控汽车，他实际应付多少元。\n![](images/575515eff325461090d9e5e1fa111c22.png)",
+        "options":  [
+
+                    ],
+        "answer":  "212.78元",
+        "solution":  "【分析】机器人156.98元，遥控汽车95.80元，总价252.78元。每满100减20，满2个100，共减40元.\n【详解】156.98+95.80=252.78(元)\n20×2=40(元)\n252.78-40=212.78(元)\n答：实际应付212.78元。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "89",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "下面是聪聪的购物小票的一部分，请你算一算聪聪应该付多少钱。\n| 货品名称 | 数量 | 单价/元 | 合计/元 |\n|----------|------|----------|----------|\n| 提拉米苏 | 1块 | 24.75 | 24.75 |\n| 果汁 | 1瓶 | 12.99 | 12.99 |\n| 饼干 | 1盒 | 17.25 | 17.25 |",
+        "options":  [
+
+                    ],
+        "answer":  "54.99元",
+        "solution":  "【分析】三种商品数量都为1，合计等于三项单价之和.\n【详解】24.75+12.99+17.25=37.74+17.25=54.99(元)。答：聪聪应付54.99元。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算",
+                                 "利用小数的加，减法混合运算解决实际问题"
+                             ],
+        "source":  "24-25四年级下·河北廊坊·期末"
+    },
+    {
+        "id":  "90",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "有两张纸条，一张长0.36米，另一张长0.58米，粘贴成一张后，这张纸条的长是0.9米，粘贴部分用去了多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "0.04米",
+        "solution":  "【分析】两张原长和减去粘后总长，即重叠(粘贴)部分.\n【详解】0.36+0.58-0.9=0.94-0.9=0.04(米)。答：粘贴部分用去了0.04米。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·河南驻马店·期末"
+    },
+    {
+        "id":  "91",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "李阿姨在超市买了一盒鸡蛋25.8元，一包火锅底料19.75元，她使用支付宝余额付款，支付宝余额中原有50.23元，支付完这两件商品后，还剩多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "4.68元",
+        "solution":  "【分析】先求总花费，再用原余额减总花费.\n【详解】50.23-(25.8+19.75)=50.23-45.55=4.68(元)。答：还剩4.68元。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·陕西安康·期末"
+    },
+    {
+        "id":  "92",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "2023年6月5日是第50个世界环境日。为打好绿水青山就是金山银山的环境保卫战，大力弘扬生态文明，推进美丽校园建设，某小学四(1)班的同学们发起了收集废纸的活动。最后，第一小组收集了40.08千克，第二小组收集了36.2千克.\n(1)两个小组共收废纸多少千克。\n(2)哪个小组收集的废纸多。多多少千克。",
+        "options":  [
+
+                    ],
+        "answer":  "76.28千克;第一小组;3.88千克",
+        "solution":  "【分析】(1)总量=40.08+36.2;(2)比较两组大小，再用较大减较小求相差量.\n【详解】(1)40.08+36.2=76.28(千克)\n(2)40.08\u003e36.2,40.08-36.2=3.88(千克)\n答：两个小组共收废纸76.28千克;第一小组多，多3.88千克。",
+        "knowledge_points":  [
+                                 "多位小数的大小比较",
+                                 "多位小数的不进位加法，不退位减法",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·山西吕梁·期末"
+    },
+    {
+        "id":  "93",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "二氧化碳排放量是可以计算出来的。李明开车15.6千米，排放二氧化碳3.6千克;李明乘坐电梯回家比开车15.6千米少排放二氧化碳3.38千克。李明开车15.6千米和乘坐电梯回家，一共排放了多少千克二氧化碳。",
+        "options":  [
+
+                    ],
+        "answer":  "3.82千克",
+        "solution":  "【分析】先求乘电梯排放量:3.6-3.38，再与开车排放量3.6相加.\n【详解】3.6-3.38+3.6=0.22+3.6=3.82(千克)。答：一共排放3.82千克二氧化碳。",
+        "knowledge_points":  [
+                                 "利用小数的加，减法混合运算解决实际问题",
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·河北邢台·期末"
+    },
+    {
+        "id":  "94",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一台电风扇69.8元，比一个电饭锅便宜79.2元，张叔叔要买一个电饭锅，付给售货员200元，应找回多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "51元",
+        "solution":  "【分析】电饭锅价=69.8+79.2，再用200减去电饭锅价得找零.\n【详解】69.8+79.2=149(元)\n200-149=51(元)\n答：应找回51元。",
+        "knowledge_points":  [
+                                 "利用小数的加，减法混合运算解决实际问题",
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·河北唐山·期末"
+    },
+    {
+        "id":  "95",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "二氧化碳排放量是可以计算出来的。李明开车15.6千米，排放二氧化碳3.6千克;李明乘坐电梯回家比开车15.6千米少排放二氧化碳3.38千克。李明开车15.6千米和乘坐电梯回家，一共排放了多少千克二氧化碳。",
+        "options":  [
+
+                    ],
+        "answer":  "3.82千克",
+        "solution":  "【分析】先算乘电梯排放量，再与开车排放量相加.\n【详解】3.6+(3.6-3.38)=3.6+0.22=3.82(千克)\n答：一共排放了3.82千克二氧化碳。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·河北邢台·期末"
+    },
+    {
+        "id":  "96",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "妈妈到超市买了一袋售价为108.3元的大米和一桶售价为48.2元的花生油，付给收银员160元，收银员说：\"您再给我1元5角，我找您5元.\"收银员的说法正确吗。",
+        "options":  [
+
+                    ],
+        "answer":  "正确",
+        "solution":  "【分析】先算应付总额和160元应找零，再看若多给1.5元是否会找回5元.\n【详解】160-(108.3+48.2)=160-156.5=3.5(元)\n1元5角=1.5元,3.5+1.5=5(元)\n答：收银员的说法正确。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算",
+                                 "元，角，分的认识及换算"
+                             ],
+        "source":  "24-25四年级下·湖南长沙·期末"
+    },
+    {
+        "id":  "97",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "学校举行歌手大赛。小明得了9.43分，比小红高0.4分，小华比小红高0.27分。小华得了多少分。",
+        "options":  [
+
+                    ],
+        "answer":  "9.3分",
+        "solution":  "【分析】先由小明分数求小红:9.43-0.4;再求小华：小红+0.27.\n【详解】9.43-0.4+0.27=9.03+0.27=9.3(分)\n答：小华得了9.3分。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·陕西安康·期末"
+    },
+    {
+        "id":  "98",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "骑行是一种健康自然的运动旅游方式，一辆单车，一个背包即可出行，简单又环保。文叔叔周末常常去骑行，本周末的骑行目标是55千米。第一小时骑行了22.38千米，第二小时骑行了23.62千米，他还要骑行多少千米才能完成预定目标。",
+        "options":  [
+
+                    ],
+        "answer":  "9千米",
+        "solution":  "【分析】剩余路程=总目标-前两小时路程和.\n【详解】55-(22.38+23.62)=55-46=9(千米)\n答：还要骑行9千米。",
+        "knowledge_points":  [
+                                 "整数加法运算律推广到小数"
+                             ],
+        "source":  "24-25四年级下·河北邢台·期末"
+    },
+    {
+        "id":  "99",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "文叔叔要乘坐飞机去西安开会，航空公司规定免费托运的行李限重20千克，文叔叔的行李箱自重2.56千克，里面放的生活用品重14.75千克，文件资料包重3.44千克，想免费托运的话超重了吗。",
+        "options":  [
+
+                    ],
+        "answer":  "超重了",
+        "solution":  "【分析】把三部分重量相加后与20千克比较.\n【详解】2.56+14.75+3.44=2.56+3.44+14.75=6+14.75=20.75(千克)\n20.75\u003e20\n答：想免费托运的话超重了。",
+        "knowledge_points":  [
+                                 "整数加法运算律推广到小数",
+                                 "多位小数的进位加法，退位减法",
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "24-25四年级下·河北邢台·期末"
+    },
+    {
+        "id":  "100",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "妈妈给小明买了一件上衣和一条裤子，上衣每件89.5元，一件上衣比一条裤子便宜5.49元。买一件上衣和一条裤子一共多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "184.49元",
+        "solution":  "【分析】裤子价=89.5+5.49，再与上衣价89.5相加.\n【详解】89.5+5.49+89.5=94.99+89.5=184.49(元)\n答：一共184.49元。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·安徽淮北·期末"
+    },
+    {
+        "id":  "101",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "如图，在加法竖式计算中，\"8\"和\"8\"相加得().\n![](images/2026-03-12%20121813.jpg)",
+        "options":  [
+                        "A. 16个一",
+                        "B. 6个0.1",
+                        "C. 16个0.1",
+                        "D. 6个十分之一"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】小数加减法先小数点对齐，再按整数加减法计算，最后点上小数点.\n【详解】4.8+2.85=7.65。两个\"8\"都在十分位，表示8个0.1,8+8=16，表示16个0.1。故答案为:C",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·四川乐山·期末"
+    },
+    {
+        "id":  "102",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面小数中最接近20的是()。",
+        "options":  [
+                        "A. 20.01",
+                        "B. 19.998",
+                        "C. 19.9"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】分别算各数与20的差，差最小的最接近20.\n【详解】A:20.01-20=0.01;B:20-19.998=0.002;C:20-19.9=0.1，且0.002\u003c0.01\u003c0.1，所以最接近20的是19.998。故答案为:B",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "24-25四年级下·湖南衡阳·期末"
+    },
+    {
+        "id":  "103",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "一个两位小数\"0.4□\"与另一个两位小数\"0.0□\"相加，它们的和可能是()。",
+        "options":  [
+                        "A. 0.39",
+                        "B. 0.5",
+                        "C. 0.4"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】最小和为0.40+0.01=0.41，最大和为0.49+0.09=0.58，故和在0.41到0.58之间.\n【详解】A 0.39\u003c0.41，不可能;B 0.5在区间内，可能;C 0.4\u003c0.41，不可能。故答案为:B",
+        "knowledge_points":  [
+                                 "多位小数的大小比较",
+                                 "多位小数的不进位加法，不退位减法",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·河北衡水·期末"
+    },
+    {
+        "id":  "104",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "3□.5+8.□6的结果不可能是()。",
+        "options":  [
+                        "A. 47.66",
+                        "B. 38.56",
+                        "C. 38.61",
+                        "D. 43.76"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】最大:39.5+8.96=48.46;最小:30.5+8.06=38.56，且百分位必是6.\n【详解】A在范围内可能;B是最小值可能;C百分位是1，不可能;D在范围内可能。故答案为:C",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·内蒙古呼和浩特·期末"
+    },
+    {
+        "id":  "105",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一个笔记本4.5元，比一个文具盒便宜26.3元。明明买一个这样的文具盒和一个笔记本一共需要()元钱。",
+        "options":  [
+                        "A. 21.8",
+                        "B. 26.3",
+                        "C. 30.8",
+                        "D. 35.3"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】文具盒价=4.5+26.3，再加笔记本4.5.\n【详解】4.5+26.3+4.5=35.3(元)。故答案为:D",
+        "knowledge_points":  [
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·四川南充·期末"
+    },
+    {
+        "id":  "106",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列算式中，\"2\"和\"8\"不能直接相加减的算式是()。",
+        "options":  [
+                        "A. 20+580",
+                        "B. 132-98",
+                        "C. 7.25+1.8",
+                        "D. 6.2-4.08"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】能否直接算要看是否在相同数位.\n【详解】A十位对十位可直接算;B个位对个位可直接算;C十分位对十分位可直接算;D里2在十分位,8在百分位，不能直接算。故答案为:D",
+        "knowledge_points":  [
+                                 "两，三位数的一次进位加法",
+                                 "两，三位数的连续退位减法",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·浙江杭州·期末"
+    },
+    {
+        "id":  "107",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "一根绳子长15米，第一次用去4.6米，第二次用去5.1米，算式4.6+5.1不能解决的问题是这根绳子()。",
+        "options":  [
+                        "A. 两次一共用去多少米",
+                        "B. 还剩多少米",
+                        "C. 比原来短了多少米"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】4.6+5.1只能求\"共用去\"，不能直接求\"还剩\".\n【详解】A,C都等于\"共用去\";B还需再算15-(4.6+5.1)。故答案为:B",
+        "knowledge_points":  [
+                                 "\"提问题\"\"填条件\"问题",
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·河南驻马店·期末"
+    },
+    {
+        "id":  "108",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列运算过程体现了运算律和运算方法之间的联系，其中没有体现加法交换律和结合律的是()。",
+        "options":  [
+                        "A. 5.2+3.6=(5+0.2)+(3+0.6)=(5+3)+(0.2+0.6)=8+0.8=8.8",
+                        "B. 52×14=52×(10+4)=52×10+52×4=520+208=728",
+                        "C. 260+332=(200+60)+(300+30+2)=(200+300)+(60+30)+2=592"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】A,C体现加法交换律和结合律;B体现乘法分配律.\n【详解】只有B没有体现加法交换律和结合律。故答案为:B",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "整数加法运算律推广到小数"
+                             ],
+        "source":  "24-25四年级下·河北廊坊·期末"
+    },
+    {
+        "id":  "109",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "钢笔和圆规两样物品的总价保留一位小数后是30.5元，钢笔的价格是13.59元，圆规的价格不可能是()元。",
+        "options":  [
+                        "A. 16.97",
+                        "B. 16.92",
+                        "C. 16.87",
+                        "D. 16.94"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】分别与13.59相加并保留一位小数，找不等于30.5的.\n【详解】A:16.97+13.59=30.56≈30.6(不符);B:30.51≈30.5;C:30.46≈30.5;D:30.53≈30.5。故答案为:A",
+        "knowledge_points":  [
+                                 "小数的近似数",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·湖北十堰·期末"
+    },
+    {
+        "id":  "110",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面的小数中，最接近1的是()。",
+        "options":  [
+                        "A. 1.001",
+                        "B. 0.98",
+                        "C. 1.01",
+                        "D. 0.99"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】比较各数与1的差.\n【详解】A差0.001,B差0.02,C差0.01,D差0.01，最小是0.001。故答案为:A",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "多位小数的不进位加法，不退位减法",
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "24-25四年级下·河南濮阳·期末"
+    },
+    {
+        "id":  "111",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "将一根长5.6米的竹竿插入河水中，插入淤泥部分0.5米，露出水面1.2米，水深多少米。下面列式正确的是()。",
+        "options":  [
+                        "A. 5.6+0.5+1.2",
+                        "B. 5.6-0.5+1.2",
+                        "C. 5.6-0.5-1.2"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】水深=竹竿总长-入泥-露出水面.\n【详解】5.6-0.5-1.2=3.9(米)。列式应为C。故答案为:C",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算",
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25三年级下·河南洛阳·期末"
+    },
+    {
+        "id":  "112",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面的数，最接近10的是()。",
+        "options":  [
+                        "A. 10.01",
+                        "B. 9.9",
+                        "C. 10.1",
+                        "D. 11"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】比较各数与10的差.\n【详解】A差0.01,B差0.1,C差0.1,D差1，最小是0.01。故答案为:A",
+        "knowledge_points":  [
+                                 "一位小数的不进位加法，不退位减法",
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "24-25四年级下·内蒙古包头·期末"
+    },
+    {
+        "id":  "113",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "小虎在计算12.5-2.7+1.3时，是这样计算的:12.5-2.7+1.3=12.5-(2.7+1.3)=12.5-4=8.5。小虎的计算结果与原结果相比()。",
+        "options":  [
+                        "A. 没有变化",
+                        "B. 减少了1.3",
+                        "C. 减少了2.6",
+                        "D. 增加了2.7"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】分别算原式与错误式，再比较差值.\n【详解】原式:12.5-2.7+1.3=11.1;错算得8.5;11.1-8.5=2.6，错算结果减少2.6。故答案为:C",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算",
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·广西南宁·期末"
+    },
+    {
+        "id":  "114",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "乐乐在玩积木时，把两块长度为7.5cm的积木拼接在一起，作为机器人的一只手臂(如图)。这只手臂长()cm.\n![](images/39a0a3eff7076462586c942fcf181769f5742e1d0a96ee5f4cd4de5aa0ac9497.jpg)",
+        "options":  [
+                        "A. 10",
+                        "B. 12.5",
+                        "C. 15"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】两块长度相加，再减去重叠接头长度.\n【详解】7.5+7.5-2.5=12.5(cm)。故答案为:B",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·江西吉安·期末"
+    },
+    {
+        "id":  "115",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "两个小数相加，一个加数增加3.4，另一个加数减少了1.5，和()。",
+        "options":  [
+                        "A. 增加了1.9",
+                        "B. 增加了4.9",
+                        "C. 减少了1.9",
+                        "D. 减少了4.9"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】和的变化量=+3.4-1.5.\n【详解】3.4-1.5=1.9，和增加1.9。故答案为:A",
+        "knowledge_points":  [
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·河北邢台·期末"
+    },
+    {
+        "id":  "116",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面各数中与\"10\"最接近的数是()。",
+        "options":  [
+                        "A. 10.4",
+                        "B. 10.01",
+                        "C. 9.98",
+                        "D. 9.9"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】比较与10的差值，差最小最接近.\n【详解】A差0.4,B差0.01,C差0.02,D差0.1，最小是0.01。故答案为:B",
+        "knowledge_points":  [
+                                 "多位小数的大小比较",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·福建漳州·期末"
+    },
+    {
+        "id":  "117",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "在下图的加法竖式中，\"9\"和\"6\"相加得().\n![](images/2026-03-12%20122114.jpg)",
+        "options":  [
+                        "A. 15个1",
+                        "B. 15个0.1",
+                        "C. 15个0.01"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】看数位:9和6都在十分位，表示十分之一.\n【详解】9个0.1+6个0.1=15个0.1。故答案为:B",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·江西吉安·期末"
+    },
+    {
+        "id":  "118",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "妈妈扫码领红包得5.61元，比爸爸多领1.7元，爸爸和妈妈共领了()元。",
+        "options":  [
+                        "A. 9.52",
+                        "B. 11.05",
+                        "C. 12.92",
+                        "D. 8.64"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】爸爸=5.61-1.7，再与妈妈金额相加.\n【详解】5.61-1.7+5.61=9.52(元)。故答案为:A",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·福建漳州·期末"
+    },
+    {
+        "id":  "119",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "根据下面的线段图,1号井采煤()万吨.\n![](images/2026-03-12%20122233.jpg)",
+        "options":  [
+                        "A. 37.8",
+                        "B. 24",
+                        "C. 48",
+                        "D. 42.9"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】先由线段关系求总量，再除以2求1号井.\n【详解】42.9+5.1=48(万吨),48÷2=24(万吨)。故答案为:B",
+        "knowledge_points":  [
+                                 "用画图法解决问题",
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·河南开封·期末"
+    },
+    {
+        "id":  "120",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "星期天小明从家出发，先去体育馆打羽毛球，再到超市购物，最后回家。小明走的路线近似于一个三角形(如下图),超市到小明家的距离可能是()千米.\n![](images/d194c8096ee15831ac5b05e51b6e776b37a2318effc73194b725c69d7813ac99.jpg)",
+        "options":  [
+                        "A. 0.2",
+                        "B. 0.3",
+                        "C. 0.7",
+                        "D. 2.2"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】第三边应满足两边之差\u003c第三边\u003c两边之和.\n【详解】已知两边0.9和1.2，则0.3\u003c第三边\u003c2.1，选项中只有0.7符合。故答案为:C",
+        "knowledge_points":  [
+                                 "一位小数的进位加法，退位减法",
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·贵州遵义·期末"
+    },
+    {
+        "id":  "121",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面各数中与10最接近的数是()。",
+        "options":  [
+                        "A. 9.98",
+                        "B. 10.101",
+                        "C. 10.05",
+                        "D. 10.01"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】比较各数与10的差值.\n【详解】A差0.02,B差0.101,C差0.05,D差0.01，最小是0.01。故答案为:D",
+        "knowledge_points":  [
+                                 "多位小数的大小比较",
+                                 "多位小数的不进位加法，不退位减法",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·河北石家庄·期末"
+    },
+    {
+        "id":  "122",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面式子中的\"7\"和\"2\"，可以直接相减的是()。",
+        "options":  [
+                        "A. 37-21",
+                        "B. 3.7-1.2",
+                        "C. 7÷8-2÷3"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】能直接减要在同数位(或分数同分母同单位).\n【详解】A中7在个位,2在十位;C需先通分;B中7和2都在十分位，可直接相减。故答案为:B",
+        "knowledge_points":  [
+                                 "两位数与两位数的不退位减法",
+                                 "一位小数的不进位加法，不退位减法",
+                                 "分母在10以内的同分母分数加，减法"
+                             ],
+        "source":  "24-25四年级下·河南省直辖县级单位·期末"
+    },
+    {
+        "id":  "123",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "妈妈发了10元的红包(如图),小红抢到的红包金额最少，爸爸手气最佳，他抢到的红包金额最多是()元.\n![](images/2026-03-12%20122518.jpg)",
+        "options":  [
+                        "A. 7.57",
+                        "B. 8.57",
+                        "C. 4.69",
+                        "D. 5.13"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】已知小红2.43元，总额10元。逐项判断是否满足总额，三人都抢到且爸爸最多，小红最少.\n【详解】A两人已到10元，妈妈无红包，不符;B两人超10元，不符;C可行但爸爸4.69;D也可行且5.13\u003e4.69，为可行中的最大。故答案为:D",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "24-25四年级下·福建漳州·期末"
+    },
+    {
+        "id":  "124",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "乐乐买了一个杯子和一本故事书，她可能一共花了()元.\n![](images/2026-03-12%20122645.jpg)",
+        "options":  [
+                        "A. 19.25",
+                        "B. 26.45",
+                        "C. 30.35",
+                        "D. 27.68"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】根据图中\"□\"取0到9可得总价范围，并注意百分位固定特征.\n【详解】最小19.35，最大29.25，且百分位应为5。只有26.45符合。故答案为:B",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·浙江台州·期末"
+    },
+    {
+        "id":  "125",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "小芳在计算小数减法时，错把减数54.2看成5.42得到的差是70.3，正确的差应该是()。",
+        "options":  [
+                        "A. 75.72",
+                        "B. 21.52",
+                        "C. 48.78"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】先用\"差+减数\"求被减数，再减正确减数.\n【详解】被减数=70.3+5.42=75.72;正确差=75.72-54.2=21.52。故答案为:B",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·江西萍乡·期末"
+    },
+    {
+        "id":  "126",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "由2,6,3和小数点组成最大的一位小数是(),最小的两位小数是(),这两个数的和是(),差是()。",
+        "options":  [
+
+                    ],
+        "answer":  "63.2;2.36;65.56;60.84",
+        "solution":  "【分析】最大一位小数应把较大数字放整数部分高位;最小两位小数应把最小数字作整数部分，再算和与差.\n【详解】最大一位小数63.2，最小两位小数2.36;63.2+2.36=65.56,63.2-2.36=60.84。",
+        "knowledge_points":  [
+                                 "多位小数的大小比较",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·安徽淮北·期末"
+    },
+    {
+        "id":  "127",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈在水电一卡通里充了100元钱用来缴水电费，该月水费20.92元，电费36.4元，一共要缴()元，缴完水电费后一卡通里还剩下()元。",
+        "options":  [
+
+                    ],
+        "answer":  "57.32;42.68",
+        "solution":  "【分析】先求总缴费，再用100减去总缴费.\n【详解】20.92+36.4=57.32(元),100-57.32=42.68(元)。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·河北衡水·期末"
+    },
+    {
+        "id":  "128",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "28.37比71.8小();比10.8大4.65的数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "43.43;15.45",
+        "solution":  "【分析】\"比71.8小多少\"用减法;\"比10.8大4.65\"用加法.\n【详解】71.8-28.37=43.43;10.8+4.65=15.45。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·河北衡水·期末"
+    },
+    {
+        "id":  "129",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一根铁丝长20分米，剪成3段，第一段长10.6分米，第二段长4.4分米。第三段长()分米，这三段首尾相接后()围成一个三角形(填\"能\"或\"不能\")。",
+        "options":  [
+
+                    ],
+        "answer":  "5;不能",
+        "solution":  "【分析】第三段=总长-前两段;再判断三角形两短边和是否大于最长边.\n【详解】20-10.6-4.4=5(分米);5+4.4\u003c10.6，不能围成三角形。",
+        "knowledge_points":  [
+                                 "与小数减法相关的简便运算",
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·山东日照·期末"
+    },
+    {
+        "id":  "130",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "10.01整数部分的1比小数部分的1大()。",
+        "options":  [
+
+                    ],
+        "answer":  "9.99",
+        "solution":  "【分析】整数部分的1表示10，小数部分的1表示0.01.\n【详解】10-0.01=9.99。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·内蒙古呼和浩特·期末"
+    },
+    {
+        "id":  "131",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个两位数，在它的十位和个位数字之间加上一个小数点，再和这个两位数相加，得数是30.8，这个两位数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "28",
+        "solution":  "【分析】设两位数为ab,则小数为a.b,满足ab+a.b=30.8。试值可得28+2.8=30.8.\n【详解】该两位数是28。",
+        "knowledge_points":  [
+                                 "一位小数的不进位加法，不退位减法"
+                             ],
+        "source":  "24-25四年级下·河南驻马店·期末"
+    },
+    {
+        "id":  "132",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈买来一桶蜂蜜，连桶共重4.5kg,用去一半蜂蜜后连桶共重2.53kg,这桶蜂蜜原来重()kg,桶重()kg。",
+        "options":  [
+
+                    ],
+        "answer":  "3.94;0.56",
+        "solution":  "【分析】前后差值是半桶蜂蜜重，乘2得蜂蜜总重，再用总重减蜂蜜重得桶重.\n【详解】4.5-2.53=1.97(kg),蜂蜜重=1.97×2=3.94(kg),桶重=4.5-3.94=0.56(kg)。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·四川广元·期末"
+    },
+    {
+        "id":  "133",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "9.8比12.92少(),比5.63多1.86的数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "3.12;7.49",
+        "solution":  "【分析】前空用12.92-9.8;后空用5.63+1.86.\n【详解】12.92-9.8=3.12;5.63+1.86=7.49。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "134",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "人口老龄化是21世纪很多国家都面临的问题。截至2014年底，央视网报道我国65岁及以上人口约有1.38亿人，到2024年底，此年龄段人口已增至约2.2亿人，十年间65岁及以上人口增加了()亿。",
+        "options":  [
+
+                    ],
+        "answer":  "0.82",
+        "solution":  "【分析】用2024年底65岁及以上人口数减去2014年底65岁及以上人口数即可.\n【详解】2.2-1.38=0.82(亿)。十年间65岁及以上人口增加了0.82亿。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·山西长治·期末"
+    },
+    {
+        "id":  "135",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个数由3个百和5个百分之一组成，这个数写作(),读作(),再添上6个十分之一是()。",
+        "options":  [
+
+                    ],
+        "answer":  "300.05;三百点零五;300.65",
+        "solution":  "【分析】3个百是300,5个百分之一是0.05，合起来是300.05;再加6个十分之一即加0.6.\n【详解】300.05+0.6=300.65。这个数写作300.05，读作三百点零五，再添上6个十分之一是300.65。",
+        "knowledge_points":  [
+                                 "多位小数的不进位加法，不退位减法",
+                                 "小数的数位和计数单位的认识",
+                                 "小数的读法和写法"
+                             ],
+        "source":  "24-25四年级下·四川南充·期末"
+    },
+    {
+        "id":  "136",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "再填上两个数，使得这六个数的大小排列有规律。3.5,(),3.9,4.1,(),4.5。",
+        "options":  [
+
+                    ],
+        "answer":  "3.7;4.3",
+        "solution":  "【分析】相邻已知数4.1-3.9=0.2，可知每次增加0.2.\n【详解】3.5+0.2=3.7,4.1+0.2=4.3，所以依次是3.5,3.7,3.9,4.1,4.3,4.5。",
+        "knowledge_points":  [
+                                 "一位小数的进位加法，退位减法",
+                                 "小数排列的规律"
+                             ],
+        "source":  "24-25四年级下·浙江杭州·期末"
+    },
+    {
+        "id":  "137",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "今年笑笑的身高是1.38米，比淘气矮0.09米，小华比淘气矮0.18米，今年小华的身高是()米。",
+        "options":  [
+
+                    ],
+        "answer":  "1.29",
+        "solution":  "【分析】先求淘气身高:1.38+0.09，再求小华身高：淘气-0.18.\n【详解】1.38+0.09-0.18=1.47-0.18=1.29(米)。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·陕西宝鸡·期末"
+    },
+    {
+        "id":  "138",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一根绳子长10米，第一次剪去2.6米，第二次剪去3.8米，两次共剪了()米，还剩下()米。",
+        "options":  [
+
+                    ],
+        "answer":  "6.4;3.6",
+        "solution":  "【分析】两次共剪长度=2.6+3.8;剩余长度=10-共剪长度.\n【详解】2.6+3.8=6.4(米),10-6.4=3.6(米)。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25三年级下·河北邢台·期末"
+    },
+    {
+        "id":  "139",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "比12.25多6.75的数是();()比31.08少4.65。",
+        "options":  [
+
+                    ],
+        "answer":  "19;26.43",
+        "solution":  "【分析】第1空用加法，第2空用减法.\n【详解】12.25+6.75=19;31.08-4.65=26.43。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·河南漯河·期末"
+    },
+    {
+        "id":  "140",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小明去新华书店买书，扫码支付了2.85元，售货员说：\"小朋友，刚才我在收银机上把小数点输错了，应该是28.5元.\"小明还需要再支付()元。",
+        "options":  [
+
+                    ],
+        "answer":  "25.65",
+        "solution":  "【分析】应付金额减去已付金额就是还需支付金额.\n【详解】28.5-2.85=25.65(元)。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·河南焦作·期末"
+    },
+    {
+        "id":  "141",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把4.5改写成两位小数是(),改写后的这个两位小数里面有()个0.01，再加上()个这样的计数单位就是整数5。",
+        "options":  [
+
+                    ],
+        "answer":  "4.50;450;50",
+        "solution":  "【分析】4.5改写成两位小数是4.50;4.50里有450个0.01;到5还差0.50即50个0.01.\n【详解】4.5=4.50,5-4.50=0.50，所以还需50个0.01。",
+        "knowledge_points":  [
+                                 "小数的性质",
+                                 "小数的数位和计数单位的认识",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·河南许昌·期末"
+    },
+    {
+        "id":  "142",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在()里填上\"\u003e\"、\"\u003c\"或\"=\"，在□里填上合适的数字.\n3.07吨( )3吨700千克\n8.89+0.01( )8.99\n87×99( )87×100-99\n21600平方米\u003e2.□6公顷",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003c;\u003e;0",
+        "solution":  "【分析】先统一单位并计算左右两边结果，再比较大小;最后一空先将21600平方米化为2.16公顷再判断□.\n【详解】3吨700千克=3.7吨，所以3.07\u003c3.7;8.89+0.01=8.9,8.9\u003c8.99;87×99=8613,87×100-99=8601,8613\u003e8601;21600平方米=2.16公顷，要使2.16\u003e2.□6,□应为0。",
+        "knowledge_points":  [
+                                 "多位小数的大小比较",
+                                 "多位小数的进位加法，退位减法",
+                                 "整数乘法分配律",
+                                 "运用小数点移动解决小数的单位换算问题"
+                             ],
+        "source":  "24-25四年级下·河南许昌·期末"
+    },
+    {
+        "id":  "143",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "5.46的计数单位是(),它有()个这样的计数单位，再加()个这样的计数单位就是6。",
+        "options":  [
+
+                    ],
+        "answer":  "0.01;546;54",
+        "solution":  "【分析】两位小数计数单位是0.01;5.46有546个0.01;到6还差0.54即54个0.01.\n【详解】6-5.46=0.54，所以再加54个0.01就是6。",
+        "knowledge_points":  [
+                                 "小数的数位和计数单位的认识",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·四川乐山·期末"
+    },
+    {
+        "id":  "144",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小红在计算一道减法时，将减数4.8误看成0.48，计算结果与正确结果相差()。",
+        "options":  [
+
+                    ],
+        "answer":  "4.32",
+        "solution":  "【分析】减数少看了多少，差就会多多少.\n【详解】4.8-0.48=4.32，所以计算结果与正确结果相差4.32。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·河南安阳·期末"
+    },
+    {
+        "id":  "145",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数的十位上是6，百分位上是7，其他数位上都是0，这个数是(),这个数再加上()是100。",
+        "options":  [
+
+                    ],
+        "answer":  "60.070;39.93",
+        "solution":  "【分析】按数位写出该三位小数，再用100减去它.\n【详解】该数是60.070;100-60.070=39.93。",
+        "knowledge_points":  [
+                                 "小数的读法和写法",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·四川乐山·期末"
+    },
+    {
+        "id":  "146",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "比1.2多0.09的数是(),2.3比()少1.7。",
+        "options":  [
+
+                    ],
+        "answer":  "1.29;4",
+        "solution":  "【分析】\"比1.2多0.09\"用加法;\"2.3比某数少1.7\"即该数=2.3+1.7.\n【详解】1.2+0.09=1.29;2.3+1.7=4。",
+        "knowledge_points":  [
+                                 "多位小数的不进位加法，不退位减法",
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25三年级下·福建漳州·期末"
+    },
+    {
+        "id":  "147",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "比0.58多4.5的数是(),()比4.5少0.65;0.96的计数单位是(),它有()个这样的计数单位，再添上()个这样的计数单位就是1。",
+        "options":  [
+
+                    ],
+        "answer":  "5.08;3.85;0.01;96;4",
+        "solution":  "【分析】前两空分别用加法和减法;0.96是两位小数，计数单位是0.01，有96个0.01，到1还差4个0.01.\n【详解】0.58+4.5=5.08;4.5-0.65=3.85;1-0.96=0.04=4个0.01。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·河北石家庄·期末"
+    },
+    {
+        "id":  "148",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "2025年春晚的《秧BOT》震撼亮相，机器人身着花坎肩，手持手绢，随着欢快的节奏扭起秧歌。据悉，机器人身高约1.80米，轻量版网上售价为8.8万元.\n(1)1.80的计数单位是(),再加上()个这样的计数单位就是2.\n(2)8.8小数点左边的\"8\"表示8个(),右面的8表示8个()。",
+        "options":  [
+
+                    ],
+        "answer":  "0.01或百分之一;20;一;0.1或十分之一",
+        "solution":  "【分析】1.80是两位小数，计数单位是0.01，到2还差0.20即20个0.01;8.8中左边8在个位，右边8在十分位.\n【详解】(1)2-1.80=0.20，所以再加20个0.01.\n(2)左边\"8\"表示8个一，右边\"8\"表示8个0.1。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "小数的数位和计数单位的认识",
+                                 "小数的意义"
+                             ],
+        "source":  "24-25四年级下·河南郑州·期末"
+    },
+    {
+        "id":  "149",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "0.016千米=()米;3吨26千克=()吨",
+        "options":  [
+
+                    ],
+        "answer":  "16;3.026",
+        "solution":  "【分析】1千米=1000米,1吨=1000千克.\n【详解】0.016×1000=16(米);26÷1000=0.026(吨),3+0.026=3.026(吨)。",
+        "knowledge_points":  [
+                                 "吨，千克之间的换算与比较",
+                                 "千米和米之间的进率与换算",
+                                 "运用小数点移动解决小数的单位换算问题"
+                             ],
+        "source":  "24-25六年级下·云南昆明·期末"
+    },
+    {
+        "id":  "150",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小云在超市购物结账时，账单显示花费135.60元，这个小数读作()元，这个数中\"6\"表示(),若使用微信支付，满100元减10元，实际需支付()元。",
+        "options":  [
+
+                    ],
+        "answer":  "一百三十五点六零;6角;125.60或125.6",
+        "solution":  "【分析】按小数读法读取;\"6\"在十分位表示0.6元即6角;满减后用135.60-10.\n【详解】135.60读作一百三十五点六零;\"6\"表示6角;135.60-10=125.60(元)。",
+        "knowledge_points":  [
+                                 "多位小数的不进位加法，不退位减法",
+                                 "小数的数位和计数单位的认识",
+                                 "小数的读法和写法"
+                             ],
+        "source":  "24-25四年级下·江西上饶·期末"
+    },
+    {
+        "id":  "151",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一根长10.4m的绳子，第一次剪去3.5m,第二次剪去2.5m。现在比原来短了()m。",
+        "options":  [
+
+                    ],
+        "answer":  "6",
+        "solution":  "【分析】比原来短的长度等于两次剪去长度之和.\n【详解】3.5+2.5=6(m)。",
+        "knowledge_points":  [
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·四川乐山·期末"
+    },
+    {
+        "id":  "152",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "算式108.8+6.56+3.44=108.8+(6.56+3.44)运用了加法交换律和结合律。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】该变形只改变了运算顺序，没有交换加数位置.\n【详解】原式到等式右边只体现加法结合律，不体现交换律，所以说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "整数加法运算律推广到小数"
+                             ],
+        "source":  "24-25四年级下·四川乐山·期末"
+    },
+    {
+        "id":  "153",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "判断题",
+        "difficulty":  0.94,
+        "question":  "用竖式计算小数加，减法时，首先要把各数的末尾对齐。( )",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】小数加减法应先把小数点对齐(相同数位对齐).\n【详解】题干说\"末尾对齐\"错误，应是\"小数点对齐\"。故答案为: ×",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·河南驻马店·期末"
+    },
+    {
+        "id":  "154",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "计算小数加减法时相同数位要对齐。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】小数加减法规则就是先小数点对齐，即相同数位对齐.\n【详解】题干说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25三年级下·河北邯郸·期末"
+    },
+    {
+        "id":  "155",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "两位小数加一位小数，和是两位小数。( )",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】按小数加法，和的小数位数不超过加数中最多的小数位数，这里最多是两位.\n【详解】如0.55+0.5=1.05，和是两位小数。题干说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "多位小数的不进位加法，不退位减法"
+                             ],
+        "source":  "24-25四年级下·河南·期末"
+    },
+    {
+        "id":  "156",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "计算题",
+        "difficulty":  0.85,
+        "question":  "列式计算。90减去27.25与16.84的和，差是多少。",
+        "options":  [
+
+                    ],
+        "answer":  "45.91",
+        "solution":  "【分析】先求27.25与16.84的和，再用90减去该和.\n【详解】90-(27.25+16.84)=90-44.09=45.91。",
+        "knowledge_points":  [
+                                 "小数的四则运算及法则",
+                                 "带有小括号的混合运算",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·辽宁盘锦·期末"
+    },
+    {
+        "id":  "157",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "和我一起来巧算.\n(1)85+85+73+85+85\n(2)33.2-13.4-12.6\n(3)38×12\n(4)56×39",
+        "options":  [
+
+                    ],
+        "answer":  "413;7.2;456;2184",
+        "solution":  "【分析】(1)将相同加数合并;(2)用减法性质;(3)(4)用乘法分配律.\n【详解】(1)85×4+73=413\n(2)33.2-(13.4+12.6)=7.2\n(3)38×(10+2)=456\n(4)56×(40-1)=2184",
+        "knowledge_points":  [
+                                 "整数加法运算律推广到小数",
+                                 "整数乘法分配律"
+                             ],
+        "source":  "24-25三年级下·河南周口·期末"
+    },
+    {
+        "id":  "158",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "计算题",
+        "difficulty":  0.94,
+        "question":  "直接写出得数.\n1000×0.05= 2.5+0.8= 0÷59= 0.97-0.66=\n31.8÷10= 439×0= 4×25= 81÷9/9=",
+        "options":  [
+
+                    ],
+        "answer":  "50;3.3;0;0.31;3.18;0;100;1",
+        "solution":  "【分析】根据小数点移动规律,0的运算性质和同级运算顺序直接口算.\n【详解】1000×0.05=50;2.5+0.8=3.3;0÷59=0;0.97-0.66=0.31;31.8÷10=3.18;439×0=0;4×25=100;81÷9/9=1。",
+        "knowledge_points":  [
+                                 "一位小数的进位加法，退位减法",
+                                 "多位小数的进位加法，退位减法",
+                                 "小数点向右移动引起小数大小变化的规律",
+                                 "小数点向左移动引起小数大小变化的规律"
+                             ],
+        "source":  "24-25四年级下·安徽淮北·期末"
+    },
+    {
+        "id":  "159",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "计算题",
+        "difficulty":  0.85,
+        "question":  "直接写出得数.\n1.23+0.64= 2.7×100= 45×8×0= 70.6÷100=\n125×4= 7-3.92= 320-160÷4= 1-0.43=",
+        "options":  [
+
+                    ],
+        "answer":  "1.87;270;0;0.706;500;3.08;280;0.57",
+        "solution":  "【分析】按小数运算，整数运算和混合运算顺序直接计算.\n【详解】1.23+0.64=1.87;2.7×100=270;45×8×0=0;70.6÷100=0.706;125×4=500;7-3.92=3.08;320-160÷4=280;1-0.43=0.57。",
+        "knowledge_points":  [
+                                 "多位小数的不进位加法，不退位减法",
+                                 "多位小数的进位加法，退位减法",
+                                 "小数点向右移动引起小数大小变化的规律",
+                                 "小数点向左移动引起小数大小变化的规律"
+                             ],
+        "source":  "24-25四年级下·四川南充·期末"
+    },
+    {
+        "id":  "160",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "列竖式计算.\n8.42+9.48= 42.56+3.124= 91.26-30.35=",
+        "options":  [
+
+                    ],
+        "answer":  "17.9;45.684;60.91",
+        "solution":  "【分析】小数加减法都要小数点对齐，再按整数法则计算.\n【详解】8.42+9.48=17.9;42.56+3.124=45.684;91.26-30.35=60.91。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "161",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "某网购平台为了获取流量，在端午节期间推出了4天的促销活动。一名推销粽子的主播在第1天卖出了3.8万个粽子，以后每一天都比前一天多卖出约1.5万个，这个主播在这4天活动期间共约卖出多少万个粽子。",
+        "options":  [
+
+                    ],
+        "answer":  "24.2万个",
+        "solution":  "【分析】逐天递推：第2天=第1天+1.5，第3天=第2天+1.5，第4天=第3天+1.5，再求4天总和.\n【详解】第2天5.3万个，第3天6.8万个，第4天8.3万个;总数=3.8+5.3+6.8+8.3=24.2(万个)。答：共约卖出24.2万个粽子。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算",
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·安徽宣城·期末"
+    },
+    {
+        "id":  "162",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "学校图书馆原有3.47万册图书，上个月借出0.7万册，本月又新购买了0.53万册。现在学校图书馆有多少万册图书。",
+        "options":  [
+
+                    ],
+        "answer":  "3.3万册",
+        "solution":  "【分析】现有=原有-借出+新购.\n【详解】3.47-0.7+0.53=3.47+0.53-0.7=4-0.7=3.3(万册)。答：现在有3.3万册图书。",
+        "knowledge_points":  [
+                                 "整数加法运算律推广到小数",
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·安徽宣城·期末"
+    },
+    {
+        "id":  "163",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "![](images/06733fad76d30ade2f558e4e239ce842.png)\n(1)一个篮球比一副羽毛球拍贵多少元。(2)万老师有150元，买了一个排球，一副羽毛球拍和一根跳绳，还剩多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "52.05元;21.5元",
+        "solution":  "【分析】(1)篮球价减羽毛球拍价;(2)150元减去三样商品总价.\n【详解】(1)130.55-78.5=52.05(元).\n(2)150-34.3-78.5-15.7=21.5(元).\n答：一个篮球贵52.05元;还剩21.5元。",
+        "knowledge_points":  [
+                                 "利用小数的加，减法混合运算解决实际问题",
+                                 "小数的加，减法混合运算",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·河南濮阳·期末"
+    },
+    {
+        "id":  "164",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "![](images/e50c40de310c8747ec35a15124718981.png)\n小英的妈妈带了50元准备买其中三种食品，最多能找回多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "25.4元",
+        "solution":  "【分析】要找回最多就要花最少，选三种最便宜食品后用50减总价.\n【详解】三种最低价为11.50元,9.60元,3.50元.\n50-(11.50+9.60+3.50)=50-24.60=25.4(元).\n答：最多找回25.4元。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·河南洛阳·期末"
+    },
+    {
+        "id":  "165",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "端午节是我国重要的传统节日之一。端午节这天，聪聪和妈妈去超市购买了三样物品，下面是超市的购物小票.\n![](images/2026-03-12%20123312.jpg)\n(1)妈妈一共花了多少钱。\n(2)应找回多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "50.83元;49.17元",
+        "solution":  "【分析】(1)三样物品总价相加;(2)用100元减总价.\n【详解】(1)5.80+29.98+15.05=50.83(元).\n(2)100-50.83=49.17(元).\n答：一共花50.83元，应找回49.17元。",
+        "knowledge_points":  [
+                                 "利用小数的加，减法混合运算解决实际问题",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·河南郑州·期末"
+    },
+    {
+        "id":  "166",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "哲学家培根说过：\"读史使人明智，读诗使人灵秀\"。每一部名著都是时代的缩影，每一部流传到现在的名著，都意味着时代的筛选。近日四年级开启四大名著阅读热潮，方方在买书的时候遇到了书店促销活动,《水浒传》原价27.68元，现价24.29元,《红楼梦》原价25.98元，现价20.56元，各买一本共便宜了多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "8.81元",
+        "solution":  "【分析】先求两本书原价总和与现价总和，再相减得优惠金额.\n【详解】原价:27.68+25.98=53.66(元);现价:24.29+20.56=44.85(元);便宜:53.66-44.85=8.81(元)。答：共便宜8.81元。",
+        "knowledge_points":  [
+                                 "利用小数加减法解决实际问题",
+                                 "经济问题"
+                             ],
+        "source":  "24-25四年级下·河南郑州·期末"
+    },
+    {
+        "id":  "167",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "小亮家七月份电费是58.6元，比六月份多18.8元，小亮家六、七月份共交电费多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "98.4元",
+        "solution":  "【分析】六月份=58.6-18.8，再与七月份相加.\n【详解】58.6-18.8+58.6=39.8+58.6=98.4(元)。答：六、七月份共交98.4元。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·湖南衡阳·期末"
+    },
+    {
+        "id":  "168",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "![](images/b01b20f01abe13dd77fa314aba5a1362.png)",
+        "options":  [
+
+                    ],
+        "answer":  "14元",
+        "solution":  "【分析】先算总花费55.50+30.50=86，再用100减去86.\n【详解】100-55.50-30.50=100-(55.50+30.50)=100-86=14(元)。答：应找回14元。",
+        "knowledge_points":  [
+                                 "与小数减法相关的简便运算"
+                             ],
+        "source":  "24-25四年级下·福建福州·期末"
+    },
+    {
+        "id":  "169",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "在跳远考试中，明明跳了1.83米，比苗苗远0.17米，萌萌比苗苗远0.33米。萌萌的跳远成绩是多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "1.99米",
+        "solution":  "【分析】先求苗苗成绩:1.83-0.17，再求萌萌成绩：苗苗+0.33.\n【详解】1.83-0.17+0.33=1.66+0.33=1.99(米)。答：萌萌的跳远成绩是1.99米。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "小数的加，减法混合运算",
+                                 "多位小数的不进位加法，不退位减法"
+                             ],
+        "source":  "24-25四年级下·四川广元·期末"
+    },
+    {
+        "id":  "170",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "某学校幼儿园第一天为学生准备了14.35千克苹果，第二天准备的苹果比第一天多2.3千克。这两天一共准备了多少千克苹果。",
+        "options":  [
+
+                    ],
+        "answer":  "31千克",
+        "solution":  "【分析】先求第二天准备了多少千克，再把两天的苹果质量相加.\n【详解】14.35+2.3=16.65(千克)\n14.35+16.65=31(千克)\n答：这两天一共准备了31千克苹果。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "171",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "仓库中有30吨粮食运往灾区，第一次运了8.25吨，第二次运的比第一次少1.2吨，第二次运了多少吨。还剩下多少吨粮食。",
+        "options":  [
+
+                    ],
+        "answer":  "7.05吨;14.7吨",
+        "solution":  "【分析】第二次比第一次少1.2吨，用第一次减1.2求第二次。再用总吨数减去两次运走的吨数求剩余.\n【详解】8.25-1.2=7.05(吨)\n30-8.25-7.05=21.75-7.05=14.7(吨)\n答：第二次运了7.05吨，还剩下14.7吨粮食。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "多位小数的不进位加法，不退位减法"
+                             ],
+        "source":  "24-25四年级下·四川南充·期末"
+    },
+    {
+        "id":  "172",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "为参加此次研学活动，小明在妈妈的陪同下精心准备出行物资，他选购了一个旅行包72.5元，一双运动鞋187.5元。妈妈付给售货员阿姨300元，应找回多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "40元",
+        "solution":  "【分析】先求两件物品总价，再用300元减去总价求找回的钱.\n【详解】300-72.5-187.5=300-260=40(元)\n答：应找回40元。",
+        "knowledge_points":  [
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·山西晋中·期末"
+    },
+    {
+        "id":  "173",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "李叔叔家5月份电费83.54元，比水费多用40.6元。李叔叔家5月份电费和水费一共多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "126.48元",
+        "solution":  "【分析】先用电费减去40.6求水费，再把电费和水费相加.\n【详解】83.54-40.6=42.94(元)\n83.54+42.94=126.48(元)\n答：李叔叔家5月份电费和水费一共126.48元。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·浙江杭州·期末"
+    },
+    {
+        "id":  "174",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "世界上最快的无人机\"X-43\"每秒可飞3千米，我国神舟16号飞船的速度比它的2倍还要多1.8千米，神舟16号飞船的速度是每秒多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "7.8千米",
+        "solution":  "【分析】先求3千米的2倍，再加上1.8千米.\n【详解】3×2+1.8=6+1.8=7.8(千米)\n答：神舟16号飞船的速度是每秒7.8千米。",
+        "knowledge_points":  [
+                                 "求一个数的几倍是多少",
+                                 "小数的四则运算及法则"
+                             ],
+        "source":  "2025·重庆綦江·小升初真题"
+    },
+    {
+        "id":  "175",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "妈妈到超市买水果，买苹果用去18.75元，买桃用去9.6元。妈妈付给营业员100元，营业员应找回多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "71.65元",
+        "solution":  "【分析】用100元减去苹果和桃一共花的钱.\n【详解】100-18.75-9.6=81.25-9.6=71.65(元)\n答：营业员应找回71.65元。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·河南驻马店·期末"
+    },
+    {
+        "id":  "176",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一筐苹果连筐共重24.64千克，卖掉一半苹果后，连筐共重12.64千克。苹果和筐各有多少千克。",
+        "options":  [
+
+                    ],
+        "answer":  "苹果:24千克;筐:0.64千克",
+        "solution":  "【分析】前后相差的是一半苹果的质量。先求一半苹果，再求苹果总质量，最后求筐重.\n【详解】24.64-12.64=12(千克)\n苹果:12×2=24(千克)\n筐:24.64-24=0.64(千克)\n答：苹果重24千克，筐重0.64千克。",
+        "knowledge_points":  [
+                                 "多位小数的不进位加法，不退位减法"
+                             ],
+        "source":  "24-25四年级下·湖南永州·期末"
+    },
+    {
+        "id":  "177",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "为欢庆六一儿童节，四年级各班用彩带布置教室。一卷彩带50米，四(1)班用去16.75米，四(2)班用去了15.25米。还剩多少米彩带。",
+        "options":  [
+
+                    ],
+        "answer":  "18米",
+        "solution":  "【分析】用总长度减去两个班用去的长度，可用减法性质简算.\n【详解】50-16.75-15.25=50-(16.75+15.25)=50-32=18(米)\n答：还剩18米彩带。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "与小数减法相关的简便运算"
+                             ],
+        "source":  "24-25四年级下·河南省直辖县级单位·期末"
+    },
+    {
+        "id":  "178",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "每年4月是全民读书月。小明妈妈买了一套《徐霞客游记》,花了131.34元;小刚妈妈买了一套《唐诗选》,比小明妈妈多花28.26元;小红妈妈买了一套《红楼梦》,比小刚妈妈少花了8.9元。一套《红楼梦》多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "150.7元",
+        "solution":  "【分析】先求《唐诗选》价格，再减去8.9元得到《红楼梦》价格.\n【详解】131.34+28.26-8.9=159.6-8.9=150.7(元)\n答：一套《红楼梦》150.7元。",
+        "knowledge_points":  [
+                                 "利用小数的加，减法混合运算解决实际问题"
+                             ],
+        "source":  "24-25四年级下·河南许昌·期末"
+    },
+    {
+        "id":  "179",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "在一次跳远比赛中，聪聪跳了1.53米，比明明多跳了0.09米，奇奇比明明多跳了0.18米，奇奇跳了多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "1.62米",
+        "solution":  "【分析】先求明明跳远成绩，再加0.18求奇奇成绩.\n【详解】1.53-0.09+0.18=1.44+0.18=1.62(米)\n答：奇奇跳了1.62米。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·河南漯河·期末"
+    },
+    {
+        "id":  "180",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "随着移动支付的普及，微信支付已成为我们日常生活中主要的支付方式之一。妈妈的微信钱包里有50元，买水果用去38.6元，剩下的钱刚好买蔬菜用完。买蔬菜比买水果少花多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "27.2元",
+        "solution":  "【分析】先求买蔬菜花了多少钱，再与买水果的钱比较.\n【详解】50-38.6=11.4(元)\n38.6-11.4=27.2(元)\n答：买蔬菜比买水果少花27.2元。",
+        "knowledge_points":  [
+                                 "一位小数的进位加法，退位减法",
+                                 "经济问题"
+                             ],
+        "source":  "24-25四年级下·河南漯河·期末"
+    },
+    {
+        "id":  "181",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "新华路小学开设了科技运动项目\"纸桥承重\"。在一次测试活动中，聪聪小组搭建的纸桥一共可以承载质量为7.56千克。如果在该纸桥上放一个质量为2.57千克的柚子和一个质量为4.38千克的西瓜，还能再放一个质量为0.65千克的苹果吗。",
+        "options":  [
+
+                    ],
+        "answer":  "不能",
+        "solution":  "【分析】先算三种物体总质量，再与纸桥承载上限比较.\n【详解】2.57+4.38+0.65=6.95+0.65=7.60(千克)\n7.60\u003e7.56，所以不能再放.\n答：不能。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "24-25四年级下·河南郑州·期末"
+    },
+    {
+        "id":  "182",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "我国自主研发的人工智能绘画软件，第一次创作一幅水墨画耗时12.35分钟，经过系统升级后，第二次创作同样风格的水墨画比第一次少用了3.6分钟。第三次更新算法后，创作时间又比第二次缩短了1.25分钟。第三次创作这幅水墨画需要多少分钟。",
+        "options":  [
+
+                    ],
+        "answer":  "7.5分钟",
+        "solution":  "【分析】第三次时间=第一次时间-第二次减少时间-第三次再缩短时间.\n【详解】12.35-3.6-1.25=8.75-1.25=7.5(分钟)\n答：第三次创作这幅水墨画需要7.5分钟。",
+        "knowledge_points":  [
+                                 "利用小数的加，减法混合运算解决实际问题",
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·河南郑州·期末"
+    },
+    {
+        "id":  "183",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "端午节是中国首个入选人类非物质文化遗产的节日，丰富多彩的文旅活动拉动了节日消费。2025年端午节假期，只有河南戏剧幻城的旅游收入是0.26亿元，比郑州方特欢乐世界的旅游收入高0.06亿元，开封清明上河园比郑州方特欢乐世界的旅游收入高0.15亿元。开封清明上河园的旅游收入是多少亿元。",
+        "options":  [
+
+                    ],
+        "answer":  "0.35亿元",
+        "solution":  "【分析】先求郑州方特欢乐世界收入，再加上0.15亿元求开封清明上河园收入.\n【详解】0.26-0.06+0.15=0.20+0.15=0.35(亿元)\n答：开封清明上河园的旅游收入是0.35亿元。",
+        "knowledge_points":  [
+                                 "利用小数的加，减法混合运算解决实际问题"
+                             ],
+        "source":  "24-25四年级下·河南郑州·期末"
+    },
+    {
+        "id":  "184",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "妈妈在商场买了一套睡衣，花了99.8元;买了一盒发圈，花了9.9元，买完后又从商场领了3.5元的红包。妈妈买这些东西实际花了多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "106.2元",
+        "solution":  "【分析】先求睡衣和发圈总价，再减去红包金额.\n【详解】99.8+9.9-3.5=109.7-3.5=106.2(元)\n答：妈妈买这些东西实际花了106.2元。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算",
+                                 "利用小数的加，减法混合运算解决实际问题"
+                             ],
+        "source":  "24-25四年级下·河南安阳·期末"
+    },
+    {
+        "id":  "185",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "陈路在淘宝网买了一本价格为23.71元的《格林童话》,还买了一本比《格林童话》便宜14.27元的笔记本。他购买这两种物品一共用去多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "33.15元",
+        "solution":  "【分析】先求笔记本价格，再与《格林童话》价格相加.\n【详解】23.71-14.27=9.44(元)\n9.44+23.71=33.15(元)\n答：一共用去33.15元。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·四川乐山·期末"
+    },
+    {
+        "id":  "186",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一筐水果连筐共重18.2千克，卖去水果的一半后，连筐共重9.2千克，水果和筐分别重多少千克。",
+        "options":  [
+
+                    ],
+        "answer":  "水果:18千克;筐:0.2千克",
+        "solution":  "【分析】前后重量差是卖去的一半水果质量。先求一半水果，再求筐重和水果总重.\n【详解】18.2-9.2=9(千克)\n筐重:9.2-9=0.2(千克)\n水果重:18.2-0.2=18(千克)\n答：水果重18千克，筐重0.2千克。",
+        "knowledge_points":  [
+                                 "利用小数的加，减法混合运算解决实际问题",
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·四川凉山·期末"
+    },
+    {
+        "id":  "187",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "4月23日为\"世界读书日\"，这天书店推出满100减15元的优惠活动。优优将下列三种课外读物各买了一本，一共需要支付多少钱。\n![](images/b5e27a742a3df4f0acb13c0c57510902.png)",
+        "options":  [
+
+                    ],
+        "answer":  "120.4元",
+        "solution":  "【分析】先求三本书原价总和，再按\"满100减15\"计算实付.\n【详解】52.6+58.9+23.9=135.4(元)\n135.4-15=120.4(元)\n答：一共需要支付120.4元。",
+        "knowledge_points":  [
+                                 "利用小数加减法解决实际问题",
+                                 "一位小数的进位加法，退位减法",
+                                 "经济问题"
+                             ],
+        "source":  "24-25四年级下·广西南宁·期末"
+    },
+    {
+        "id":  "188",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "壮壮和爸爸利用周末时间开展骑行运动，计划从家出发，途经学校到青秀山。路线图如下，最近的骑行路程是多少千米。\n![](images/950cb59464d0afc03a98ad24133428572f33aa682bb5b44eae57df4cacdd7546.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "4.05千米",
+        "solution":  "【分析】先把1200米化成1.2千米，再在每段路线中选较短的一条，最后相加.\n【详解】1200米=1.2千米，且1.2\u003c1.7,2.85\u003c3.3\u003c3.54.\n最近路程=1.2+2.85=4.05(千米)\n答：最近的骑行路程是4.05千米。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "多位小数的进位加法，退位减法",
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "24-25四年级下·广西南宁·期末"
+    },
+    {
+        "id":  "189",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "在一次跳高比赛中，天天跳了1.2米，欢欢比天天跳得低0.13米，乐乐比欢欢跳得高0.07米，乐乐跳了多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "1.14米",
+        "solution":  "【分析】先求欢欢成绩，再加0.07求乐乐成绩.\n【详解】1.2-0.13=1.07(米)\n1.07+0.07=1.14(米)\n答：乐乐跳了1.14米。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·江西吉安·期末"
+    },
+    {
+        "id":  "190",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "民生商场正在举办\"购物狂欢节\"抽奖活动，顾客消费满50元即可参与抽奖。妈妈在购物后抽到了两个现金红包，分别是2.57元和1.12元，她购买的商品总价为90元，且活动规则允许红包叠加使用，使用这两个红包后，妈妈实际需要支付多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "86.31元",
+        "solution":  "【分析】实际支付=商品总价-两个红包金额.\n【详解】90-2.57-1.12=87.43-1.12=86.31(元)\n答：妈妈实际需要支付86.31元。",
+        "knowledge_points":  [
+                                 "利用小数的加，减法混合运算解决实际问题",
+                                 "多位小数的进位加法，退位减法",
+                                 "多位小数的不进位加法，不退位减法"
+                             ],
+        "source":  "24-25四年级下·陕西商洛·期末"
+    },
+    {
+        "id":  "191",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "阅读材料，回答问题.\n铁路运营里程，是指能运行客车，货车，帮我们运送人和东西的铁路线加起来的总长度。1949年，中国铁路运营里程是2.18万千米;1976年是4.63万千米。2006年，中国建成世界海拔最高的铁路--青藏铁路，全国铁路运营里程为7.71万千米;2024年，中国铁路运营里程达到16.2万千米，其中高铁运营里程超过4.6万千米，高铁运营里程位居世界第一.\n(1)从1949年到2006年，中国铁路运营里程增加了多少万千米。\n(2)2024年，除了高铁运营里程以外，中国其它铁路运营里程大约是多少万千米。",
+        "options":  [
+
+                    ],
+        "answer":  "5.53万千米;11.6万千米",
+        "solution":  "【分析】(1)用2006年里程减去1949年里程;(2)用2024年总里程减去高铁里程.\n【详解】(1)7.71-2.18=5.53(万千米)\n(2)16.2-4.6=11.6(万千米)\n答:(1)增加了5.53万千米;(2)约11.6万千米。",
+        "knowledge_points":  [
+                                 "两，三位数的一次退位减法"
+                             ],
+        "source":  "24-25四年级下·河北廊坊·期末"
+    },
+    {
+        "id":  "192",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "妙妙在景区买了一瓶蜂蜜柚子茶8.5元，爸爸买了一瓶矿泉水1.95元和一袋面包6.55元，两人一共花了多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "17元",
+        "solution":  "【分析】把三项费用相加.\n【详解】1.95+6.55+8.5=8.5+8.5=17(元)\n答：两人一共花了17元。",
+        "knowledge_points":  [
+                                 "一位小数的进位加法，退位减法",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·福建漳州·期末"
+    },
+    {
+        "id":  "193",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "智能垃圾分类机器人是融合人工智能，图像识别的环保设备。为降低成本和碳排放量,AI工程师从电子废弃物中提取铜等金属原料。现有甲，乙两种电路板，分别重100克和250克，已知1千克电路板可提取32克铜.\n(1)AI工程师用100克甲电路板能提取多少克铜。\n(2)AI工程师又用250克乙电路板提取出8克铜，提取过程中甲和乙电路板铜氧化共损耗了2.53克，最终可用的铜有多少克。",
+        "options":  [
+
+                    ],
+        "answer":  "3.2克;8.67克",
+        "solution":  "【分析】(1)1千克=1000克,100克是1千克的1÷10，所以可提取铜是32÷10。(2)最终可用铜=甲提取铜+乙提取铜-损耗.\n【详解】(1)32÷10=3.2(克)\n(2)3.2+8-2.53=11.2-2.53=8.67(克)\n答:(1)能提取3.2克铜;(2)最终可用铜有8.67克。",
+        "knowledge_points":  [
+                                 "小数点向左移动引起数大小变化的规律",
+                                 "克，千克之间的换算与比较",
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·河南郑州·期末"
+    },
+    {
+        "id":  "194",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "厨余垃圾处理设备可将厨余垃圾转化为有机肥料，用于小区绿地养护。某次设备产出22.4千克的肥料，小区内绿地养护使用了6.28千克，小区周边绿地养护使用了12.72千克，还剩多少千克肥料。",
+        "options":  [
+
+                    ],
+        "answer":  "3.4千克",
+        "solution":  "【分析】剩余肥料=总产出-两处使用量，可先把两处使用量相加.\n【详解】22.4-6.28-12.72=22.4-(6.28+12.72)=22.4-19=3.4(千克)\n答：还剩3.4千克肥料。",
+        "knowledge_points":  [
+                                 "与小数减法相关的简便运算"
+                             ],
+        "source":  "24-25四年级下·河北石家庄·期末"
+    },
+    {
+        "id":  "195",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "在4×100m的接力赛中，四(3)班的成绩是57.36秒，其中第一棒用了14.08秒，第二棒用了15.23秒，第三棒用了13.21秒。请你算一算第四棒用了多长时间。",
+        "options":  [
+
+                    ],
+        "answer":  "14.84秒",
+        "solution":  "【分析】第四棒时间=总时间-前三棒时间和.\n【详解】57.36-14.08-15.23-13.21=43.28-15.23-13.21=28.05-13.21=14.84(秒)\n答：第四棒用了14.84秒。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·河北邯郸·期末"
+    },
+    {
+        "id":  "196",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "李阿姨买苹果花了35.5元，买苹果比买香蕉少花了7.8元。李阿姨买水果一共花了多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "78.8元",
+        "solution":  "【分析】先求香蕉花费，再与苹果花费相加.\n【详解】35.5+7.8+35.5=43.3+35.5=78.8(元)\n答：李阿姨买水果一共花了78.8元。",
+        "knowledge_points":  [
+                                 "利用小数的加，减法混合运算解决实际问题"
+                             ],
+        "source":  "24-25四年级下·江西宜春·期末"
+    },
+    {
+        "id":  "197",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "体育课上，同学们在进行立定跳远练习。小余跳了1.7米，小王比小余少跳0.3米，小智比小王多跳0.15米。小智跳了多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "1.55米",
+        "solution":  "【分析】先求小王成绩，再求小智成绩.\n【详解】1.7-0.3+0.15=1.4+0.15=1.55(米)\n答：小智跳了1.55米。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算",
+                                 "多位小数的不进位加法，不退位减法"
+                             ],
+        "source":  "24-25四年级下·江西上饶·期末"
+    },
+    {
+        "id":  "198",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "交通费用问题.\n同学们乘坐大巴前往农场。实际支付时每人享受2项优惠。那么，每位同学实际需支付多少钱购买车票。\n![](images/2026-03-12%20123704.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "13元",
+        "solution":  "【分析】每人实付=原价-第一项优惠-第二项优惠.\n【详解】18.5-3-2.5=15.5-2.5=13(元)\n答：每位同学实际需支付13元。",
+        "knowledge_points":  [
+                                 "一位小数的不进位加法，不退位减法",
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·广西百色·期末"
+    },
+    {
+        "id":  "199",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "此次主题共发布了6条主题旅游线路。第一条线路全长38.5千米，第二条线路比第一条长12.7千米。第二条线路长多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "51.2千米",
+        "solution":  "【分析】第二条线路比第一条长12.7千米，用38.5+12.7计算.\n【详解】38.5+12.7=51.2(千米)\n答：第二条线路长51.2千米。",
+        "knowledge_points":  [
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·河南洛阳·期末"
+    },
+    {
+        "id":  "200",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一堆沙子76吨，第一次运走15.36吨，第二次运走14.64吨。还剩多少吨没有运完。",
+        "options":  [
+
+                    ],
+        "answer":  "46吨",
+        "solution":  "【分析】剩余=总吨数-两次运走吨数和，可先算15.36+14.64=30.\n【详解】76-15.36-14.64=76-(15.36+14.64)=76-30=46(吨)\n答：还剩46吨没有运完。",
+        "knowledge_points":  [
+                                 "整数加法运算律推广到小数"
+                             ],
+        "source":  "24-25四年级下·江西上饶·期末"
+    },
+    {
+        "id":  "201",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "在小数70.07中，整数部分的\"7\"与小数部分的\"7\"相差()。",
+        "options":  [
+                        "A. 69",
+                        "B. 6.993",
+                        "C. 69.93",
+                        "D. 6993"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】整数部分\"7\"在十位，表示70;小数部分\"7\"在百分位，表示0.07，用70减0.07即可.\n【详解】70-0.07=69.93，所以相差69.93。故答案为:C",
+        "knowledge_points":  [
+                                 "小数的数位和计数单位的认识",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·浙江宁波·期末"
+    },
+    {
+        "id":  "202",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面各式中，\"6\"和\"2\"不能直接相加减的算式是()。",
+        "options":  [
+                        "A. 420+60",
+                        "B. 8.61-5.2",
+                        "C. 6150-2099",
+                        "D. 4.86+0.2"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】直接相加减要求在相同数位.\n【详解】A,B,C中\"6\"和\"2\"在相同数位;D中\"6\"在百分位，\"2\"在十分位，不能直接相加减。故答案为:D",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·浙江杭州·期末"
+    },
+    {
+        "id":  "203",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面的算式中，\"2\"和\"3\"不能直接相加减的是()。",
+        "options":  [
+                        "A. 265+347",
+                        "B. 5.29-1.3",
+                        "C. 11.02+4.93",
+                        "D. 50.62-7.3"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】是否能直接相加减取决于是否在同一数位.\n【详解】A中都在百位,B中都在十分位,C中都在百分位;D中\"2\"在百分位，\"3\"在十分位，不能直接相加减。故答案为:D",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "一位小数的进位加法，退位减法",
+                                 "两，三位数的连续进位加法"
+                             ],
+        "source":  "24-25四年级下·贵州黔西·期末"
+    },
+    {
+        "id":  "204",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "如图所示的方框中，百分位不够减，需要向\"5\"借\"1\"，这个\"1\"表示().\n![](images/2026-03-12%20130439.jpg)",
+        "options":  [
+                        "A. 10个一",
+                        "B. 10个十分之一",
+                        "C. 10个百分之一",
+                        "D. 10个千分之一"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】\"5\"在十分位，向前一位借1表示1个0.1，换到百分位是10个0.01.\n【详解】这个\"1\"表示10个百分之一。故答案为:C",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·河南漯河·期末"
+    },
+    {
+        "id":  "205",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面关于5.88元的说法错误的是()。",
+        "options":  [
+                        "A. 两个数字\"8\"表示的意义不同",
+                        "B. 5.88元就是588分",
+                        "C. 比5.80元多0.8分"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】5.88中两个\"8\"分别在十分位和百分位;1元=100分.\n【详解】5.88-5.80=0.08元=8分，不是0.8分。错误的是C。故答案为:C",
+        "knowledge_points":  [
+                                 "元，角，分的认识及换算",
+                                 "多位小数的不进位加法，不退位减法",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·江西吉安·期末"
+    },
+    {
+        "id":  "206",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列运算过程错误的选项是()。",
+        "options":  [
+                        "A. a-9.9=a-10-0.1",
+                        "B. a÷11÷8=a÷(11×8)",
+                        "C. a×99+a=a×100"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】9.9=10-0.1，所以a-9.9=a-(10-0.1)=a-10+0.1.\n【详解】A错误;B,C正确。故答案为:A",
+        "knowledge_points":  [
+                                 "整数除法的性质",
+                                 "整数乘法分配律",
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·江西南昌·期末"
+    },
+    {
+        "id":  "207",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面()和5最接近。",
+        "options":  [
+                        "A. 4.9",
+                        "B. 5.01",
+                        "C. 4.995"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】比较各数与5的差，差越小越接近.\n【详解】5-4.9=0.1,5.01-5=0.01,5-4.995=0.005;最小是0.005。故答案为:C",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·河南新乡·期末"
+    },
+    {
+        "id":  "208",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "在下面的算式中，\"6\"和\"1\"不能直接相加减的是()。",
+        "options":  [
+                        "A. 43.65+9.15",
+                        "B. 7.6+7.1",
+                        "C. 9.96-9.1"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】同数位才能直接运算.\n【详解】A,B里\"6\"和\"1\"都在十分位;C里\"6\"在百分位，\"1\"在十分位，不能直接相加减。故答案为:C",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "多位小数的不进位加法，不退位减法"
+                             ],
+        "source":  "24-25四年级下·河南新乡·期末"
+    },
+    {
+        "id":  "209",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "爸爸通过扫码领红包得了4.8元红包，比妈妈多领了1.9元，妈妈领了()元红包。",
+        "options":  [
+                        "A. 2.9",
+                        "B. 6.7",
+                        "C. 7.7"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】妈妈金额=爸爸金额-多领部分.\n【详解】4.8-1.9=2.9(元)。故答案为:A",
+        "knowledge_points":  [
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·陕西西安·期末"
+    },
+    {
+        "id":  "210",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面各数中，最接近10的是()。",
+        "options":  [
+                        "A. 9.9",
+                        "B. 9.98",
+                        "C. 10.1",
+                        "D. 10.101"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】比较与10的差值.\n【详解】A差0.1,B差0.02,C差0.1,D差0.101，最小是0.02。故答案为:B",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "多位小数的不进位加法，不退位减法",
+                                 "一位小数的不进位加法，不退位减法",
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "24-25四年级下·贵州遵义·期末"
+    },
+    {
+        "id":  "211",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "小丽用10元钱买一块0.45元的橡皮擦，应找回()。",
+        "options":  [
+                        "A. 5.5元",
+                        "B. 0.55元",
+                        "C. 9.55元",
+                        "D. 9.45元"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】找回金额=10-0.45.\n【详解】10-0.45=9.55(元)。故答案为:C",
+        "knowledge_points":  [
+                                 "利用小数加减法解决实际问题",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25三年级下·河南开封·期末"
+    },
+    {
+        "id":  "212",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "a,b,c,d四个数在数轴上的位置如下图。得数最接近1的算式是().\n![](images/d1060fb11eb9279bd707aa0a946aae5df5349128c92aaeb9f0c339ff5bad3cc2.jpg)",
+        "options":  [
+                        "A. b+c",
+                        "B. a+d",
+                        "C. d-a",
+                        "D. d-b"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】由数轴可估算a,b,c,d,再比较四个式子的结果与1的差.\n【详解】估算得d-b约为1.01，与1差约0.01，最接近1。故答案为:D",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "多位小数的不进位加法，不退位减法",
+                                 "多位小数的大小比较",
+                                 "小数的意义"
+                             ],
+        "source":  "24-25四年级下·浙江嘉兴·期末"
+    },
+    {
+        "id":  "213",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列式子中，\"3\"和\"2\"可以直接相加减的是()。",
+        "options":  [
+                        "A. 380+22",
+                        "B. 6.35+2.7",
+                        "C. 5.3-1.21",
+                        "D. 13.7-10.2"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】要看\"3\"和\"2\"是否在同一数位.\n【详解】只有C中\"3\"和\"2\"都在十分位，可以直接相减。故答案为:C",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "一位小数的不进位加法，不退位减法"
+                             ],
+        "source":  "24-25四年级下·浙江嘉兴·期末"
+    },
+    {
+        "id":  "214",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "若△-☆=9.9，则12.38-△+☆=()。",
+        "options":  [
+                        "A. 2.48",
+                        "B. 3.48",
+                        "C. 7.43",
+                        "D. 17.33"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】12.38-△+☆=12.38-(△-☆),代入△-☆=9.9.\n【详解】12.38-9.9=2.48。故答案为:A",
+        "knowledge_points":  [
+                                 "整数加法运算律推广到小数",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·河南开封·期末"
+    },
+    {
+        "id":  "215",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "小丽在列竖式计算3.54加一个一位小数时，错误地把数的末位对齐了，算得结果为4.03。正确的结果应是()。",
+        "options":  [
+                        "A. 0.49",
+                        "B. 7.57",
+                        "C. 8.45",
+                        "D. 8.44"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】先求错误加数:4.03-3.54=0.49。末位对齐相当于把正确一位小数小数点左移一位，所以正确加数是4.9.\n【详解】3.54+4.9=8.44。故答案为:D",
+        "knowledge_points":  [
+                                 "小数点向左移动引起小数大小变化的规律",
+                                 "小数点向右移动引起小数大小变化的规律",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·福建漳州·期末"
+    },
+    {
+        "id":  "216",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面算式中，\"3\"和\"5\"不能直接相加的是()。",
+        "options":  [
+                        "A. 1.73+0.05",
+                        "B. 0.3+0.05",
+                        "C. 136+52",
+                        "D. 300+1500"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】\"3\"和\"5\"是否能直接相加，取决于是否同数位.\n【详解】A,C,D同数位可直接加;B里\"3\"在十分位，\"5\"在百分位，不能直接相加。故答案为:B",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·浙江杭州·期末"
+    },
+    {
+        "id":  "217",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "两个数相减的差是6.5，如果被减数减少2.4，减数增加1.5，那么差是()。",
+        "options":  [
+                        "A. 5.6",
+                        "B. 2.6",
+                        "C. 10.4",
+                        "D. 7.4"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】被减数减少会使差减少，减数增加也会使差减少.\n【详解】新差=6.5-2.4-1.5=2.6。故答案为:B",
+        "knowledge_points":  [
+                                 "加，减法的意义和各部分间的关系",
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·贵州铜仁·期末"
+    },
+    {
+        "id":  "218",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列算式中\"6\"和\"2\"能直接相加等于\"8\"的是()。",
+        "options":  [
+                        "A. 640+25",
+                        "B. 14.6+7.24",
+                        "C. 4.6+5.32",
+                        "D. 18.6-8.2"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】要求\"6\"和\"2\"直接相加，必须在同数位，且是加法算式.\n【详解】B中\"6\"和\"2\"都在十分位，能直接相加得8(十分位);其余不符合。故答案为:B",
+        "knowledge_points":  [
+                                 "多位小数的不进位加法，不退位减法",
+                                 "一位小数的不进位加法，不退位减法"
+                             ],
+        "source":  "24-25四年级下·浙江·期末"
+    },
+    {
+        "id":  "219",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "小虎在计算小数加法时，把其中一个加数5.8看成了58，结果得72.6，正确的得数是()。",
+        "options":  [
+                        "A. 14.6",
+                        "B. 20.4",
+                        "C. 62.2",
+                        "D. 72.5"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】先求另一个加数，再换回正确加数5.8计算.\n【详解】另一个加数=72.6-58=14.6;正确和=14.6+5.8=20.4。故答案为:B",
+        "knowledge_points":  [
+                                 "利用小数的加，减法混合运算解决实际问题"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "220",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "根据人民币面值的特点，购买下图文具盒，下列现金支付方法不合理的是()。铅笔盒:27.3元",
+        "options":  [
+                        "A. 付27.5元，找回0.2元",
+                        "B. 付27.8元，找回0.5元",
+                        "C. 付28元，找回0.7元",
+                        "D. 付30元，找回2.7元"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】按人民币常用面值判断支付与找零是否合理.\n【详解】A,C,D都可由常见面值组合实现;B\"付27.8元找0.5元\"对应支付结构不合理。故答案为:B",
+        "knowledge_points":  [
+                                 "十元以上人民币的认识与计算",
+                                 "元，角，分的认识及换算",
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·浙江台州·期末"
+    },
+    {
+        "id":  "221",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "小辰在计算小数减法时，错把减数10.2看成了1.02，得到的差是17.6，那么正确的结果是()。",
+        "options":  [
+                        "A. 6.82",
+                        "B. 8.42",
+                        "C. 8.24"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】先由\"被减数=差+减数\"求被减数，再减去正确减数10.2.\n【详解】被减数=17.6+1.02=18.62;正确差=18.62-10.2=8.42。故答案为:B",
+        "knowledge_points":  [
+                                 "加，减法的意义和各部分间的关系",
+                                 "多位小数的不进位加法，不退位减法"
+                             ],
+        "source":  "24-25四年级下·陕西安康·期末"
+    },
+    {
+        "id":  "222",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "如图所示的算式，方框中百分位上的\"0\"不够减\"8\"，需要向\"3\"借\"1\"，这个\"1\"表示().\n![](images/2026-03-12%20200637.jpg)",
+        "options":  [
+                        "A. 10个1",
+                        "B. 10个0.1",
+                        "C. 10个0.01",
+                        "D. 10个0.001"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】从十分位借1到百分位，相当于10个百分之一.\n【详解】这个\"1\"表示10个0.01。故答案为:C",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·浙江·期末"
+    },
+    {
+        "id":  "223",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列各数与6最接近的是()。",
+        "options":  [
+                        "A. 5.89",
+                        "B. 5.98",
+                        "C. 6.01",
+                        "D. 6.1"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】比较各数与6的差，差最小者最接近6.\n【详解】A差0.11,B差0.02,C差0.01,D差0.1，最小是0.01。故答案为:C",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "多位小数的不进位加法，不退位减法",
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "24-25四年级下·云南昭通·期中"
+    },
+    {
+        "id":  "224",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列说法正确的是()。",
+        "options":  [
+                        "A. 0除以任何数都得0",
+                        "B. 列竖式计算3.27-1.7时，两个\"7\"要对齐",
+                        "C. 减数与差的和加上被减数等于0",
+                        "D. 6.05与6.059保留一位小数都约是6.1"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】逐项判断:0不能作除数;小数减法应小数点对齐;减数+差=被减数;按四舍五入保留一位小数.\n【详解】A错(0不能作除数);B错(应小数点对齐，不是\"7\"对齐);C错;D对(6.05≈6.1,6.059≈6.1)。故答案为:D",
+        "knowledge_points":  [
+                                 "小数的近似数",
+                                 "加，减法的意义和各部分间的关系",
+                                 "多位小数的进位加法，退位减法",
+                                 "被除数是0的除法运算"
+                             ],
+        "source":  "24-25四年级下·福建漳州·期末"
+    },
+    {
+        "id":  "225",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下边的竖式中，百分位不够减，向前一位借\"1\"，这个\"1\"表示().\n![](images/2026-03-12%20200958.jpg)",
+        "options":  [
+                        "A. 10个0.001",
+                        "B. 10个0.01",
+                        "C. 10个0.1",
+                        "D. 10个1"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】从十分位借1到百分位,1个0.1等于10个0.01.\n【详解】这个\"1\"表示10个0.01。故答案为:B",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·浙江杭州·期末"
+    },
+    {
+        "id":  "226",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面小数中最接近于10的数是()。",
+        "options":  [
+                        "A. 10.1",
+                        "B. 9.99",
+                        "C. 10.001"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】求与10的差，差最小最接近.\n【详解】A差0.1,B差0.01,C差0.001，最小是0.001。故答案为:C",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "227",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列各数中，最接近5.06的是()。",
+        "options":  [
+                        "A. 5.061",
+                        "B. 5.059",
+                        "C. 5.0601",
+                        "D. 5.0611"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】分别与5.06作差比较大小.\n【详解】A差0.001,B差0.001,C差0.0001,D差0.0011，最小是0.0001。故答案为:C",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "多位小数的不进位加法，不退位减法",
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "24-25四年级下·辽宁鞍山·期末"
+    },
+    {
+        "id":  "228",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一只蚂蚁外出寻找食物。如图，它沿着墙边直线爬行，蚂蚁第一次找到食物爬行的距离比第二次找到食物爬行的距离多了()dm.\n![](images/bf56688782a6992cab8f3841ccf91bf4c84522eee05dd4b3f7e453d5448d777e.jpg)",
+        "options":  [
+                        "A. 1.1",
+                        "B. 0.7",
+                        "C. 0.4",
+                        "D. 0.3"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】每小格0.1dm,第一次约0.7dm,第二次约0.4dm,作差即可.\n【详解】0.7-0.4=0.3(dm)。故答案为:D",
+        "knowledge_points":  [
+                                 "一位小数的不进位加法，不退位减法"
+                             ],
+        "source":  "24-25四年级下·广东珠海·期末"
+    },
+    {
+        "id":  "229",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "7个0.01加上()就等于2;()米比28.63米少3.19米。",
+        "options":  [
+
+                    ],
+        "answer":  "1.93;25.44",
+        "solution":  "【分析】7个0.01是0.07，第一空用2-0.07;第二空用28.63-3.19.\n【详解】2-0.07=1.93;28.63-3.19=25.44(米)。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·江西吉安·期末"
+    },
+    {
+        "id":  "230",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "乒乓球是我国的国球,2000年2月23日，国际乒联特别大会和代表大会在吉隆坡通过40mm大球改革方案，决定从2000年10月1日起，也就是在悉尼奥运会之后，乒乓球比赛将使用直径40mm,质量2.7±0.18g的大球，改革后的球的最大质量是()克，最小质量是()克。",
+        "options":  [
+
+                    ],
+        "answer":  "2.88;2.52",
+        "solution":  "【分析】最大质量=2.7+0.18;最小质量=2.7-0.18.\n【详解】2.7+0.18=2.88(克);2.7-0.18=2.52(克)。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25六年级下·河北石家庄·期末"
+    },
+    {
+        "id":  "231",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "要将3km450m改写成用km作单位表示时，小明想到了1km=()m,450m=()km,所以3km450m=()km。",
+        "options":  [
+
+                    ],
+        "answer":  "1000;0.45;3.45",
+        "solution":  "【分析】1km=1000m,450m换算成千米是450÷1000=0.45km,再与3km相加.\n【详解】3km450m=3+0.45=3.45km。",
+        "knowledge_points":  [
+                                 "千米和米之间的进率与换算",
+                                 "多位小数的不进位加法，不退位减法"
+                             ],
+        "source":  "24-25四年级下·河南新乡·期末"
+    },
+    {
+        "id":  "232",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "两个数相减的差是38.4，如果被减数增加2.56，减数减少2.7，此时两个数的差是()。",
+        "options":  [
+
+                    ],
+        "answer":  "43.66",
+        "solution":  "【分析】被减数增加会使差增加，减数减少也会使差增加.\n【详解】新差=38.4+2.56+2.7=43.66。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·河南开封·期末"
+    },
+    {
+        "id":  "233",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形两条边的长分别是1.3cm和2.8cm,这个等腰三角形的周长是()cm。",
+        "options":  [
+
+                    ],
+        "answer":  "6.9",
+        "solution":  "【分析】等腰三角形第三边只能等于1.3或2.8。若第三边是1.3，则1.3+1.3\u003c2.8不成立;故第三边是2.8.\n【详解】周长=1.3+2.8+2.8=6.9(cm)。",
+        "knowledge_points":  [
+                                 "三角形三边关系",
+                                 "小数的加，减法混合运算",
+                                 "三角形的周长",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·贵州遵义·期末"
+    },
+    {
+        "id":  "234",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一根6.3米长的红绳，编中国结用去了1.5米，编手链用去了2.9米。一共用去了()米，还剩下()米。",
+        "options":  [
+
+                    ],
+        "answer":  "4.4;1.9",
+        "solution":  "【分析】先把两次用去的长度相加，求出一共用去多少米;再用总长度减去用去的长度，求剩下多少米.\n【详解】1.5+2.9=4.4(米)\n6.3-4.4=1.9(米)\n所以一共用去了4.4米，还剩下1.9米。",
+        "knowledge_points":  [
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·江西上饶·期末"
+    },
+    {
+        "id":  "235",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "修一段公路，第一天修了7.1千米，比第二天多修了1.35千米，第二天修了()千米，两天一共修了()千米。",
+        "options":  [
+
+                    ],
+        "answer":  "5.75;12.85",
+        "solution":  "【分析】第一天比第二天多1.35千米，所以第二天=7.1-1.35;再把两天长度相加求总长度.\n【详解】7.1-1.35=5.75(千米)\n5.75+7.1=12.85(千米)\n所以第二天修了5.75千米，两天一共修了12.85千米。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "236",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在潍坊风筝节上，有一个等腰三角形的风筝，它的两条边分别是0.8米和0.4米，那么这个等腰三角形的周长是()米。",
+        "options":  [
+
+                    ],
+        "answer":  "2",
+        "solution":  "【分析】等腰三角形有两条边相等。若相等边是0.4米，则0.4+0.4=0.8，不能组成三角形;所以相等边应是0.8米.\n【详解】周长=0.8+0.8+0.4=2(米)。",
+        "knowledge_points":  [
+                                 "三角形三边关系",
+                                 "三角形的周长",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "237",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "爸爸在家人微信群里发了100元红包，妈妈抢到25.65元，晓晓抢到32.54元，晓晓比妈妈多抢到()元，爸爸发的红包还剩()元。",
+        "options":  [
+
+                    ],
+        "answer":  "6.89;41.81",
+        "solution":  "【分析】晓晓比妈妈多抢到的钱数=32.54-25.65;剩余红包=100-(32.54+25.65).\n【详解】32.54-25.65=6.89(元)\n32.54+25.65=58.19(元)\n100-58.19=41.81(元)\n晓晓比妈妈多抢到6.89元，红包还剩41.81元。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·河南漯河·期末"
+    },
+    {
+        "id":  "238",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "7.91读作(),它的计数单位是(),它里面含有()个这样的计数单位，再加上()个这样的计数单位就是8。",
+        "options":  [
+
+                    ],
+        "answer":  "七点九一;0.01或百分之一或1÷100;791;9",
+        "solution":  "【分析】7.91是两位小数，计数单位是0.01;去掉小数点得到791，表示有791个0.01;8-7.91=0.09，即9个0.01.\n【详解】7.91读作七点九一;计数单位是0.01;含有791个这样的计数单位;再加9个这样的计数单位就是8。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "小数的数位和计数单位的认识",
+                                 "小数的读法和写法"
+                             ],
+        "source":  "24-25四年级下·贵州黔南·期末"
+    },
+    {
+        "id":  "239",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "天宫空间站的主体是由天和核心舱，问天实验舱和梦天实验舱组成。天和核心舱的舱体长16.6m,问天实验舱舱体长17.9m,梦天实验舱舱体长17.88m。()的舱体长度最短，问天实验舱舱体比梦天实验舱舱体长()m。",
+        "options":  [
+
+                    ],
+        "answer":  "天和核心舱;0.02",
+        "solution":  "【分析】先比较16.6,17.9,17.88大小，最小者最短;再用17.9-17.88求长度差.\n【详解】16.6\u003c17.88\u003c17.9，所以天和核心舱最短.\n17.9-17.88=0.02(m)。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "24-25四年级下·山西晋中·期末"
+    },
+    {
+        "id":  "240",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一本故事书定价12.85元，小明付给售货员20元，应找回()元。",
+        "options":  [
+
+                    ],
+        "answer":  "7.15",
+        "solution":  "【分析】找回金额=20-12.85.\n【详解】20-12.85=7.15(元)。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·云南玉溪·期末"
+    },
+    {
+        "id":  "241",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "比4.3多0.25的数是(),比12.4少2.8的数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "4.55;9.6",
+        "solution":  "【分析】\"比4.3多0.25\"用加法，\"比12.4少2.8\"用减法.\n【详解】4.3+0.25=4.55;12.4-2.8=9.6。",
+        "knowledge_points":  [
+                                 "多位小数的不进位加法，不退位减法",
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "242",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "30平方厘米=()平方分米;5.2米=()分米;4吨+25千克=()吨",
+        "options":  [
+
+                    ],
+        "answer":  "0.3;52;4.025",
+        "solution":  "【分析】1平方分米=100平方厘米,1米=10分米,1吨=1000千克，按进率换算.\n【详解】30÷100=0.3(平方分米)\n5.2×10=52(分米)\n25千克=0.025吨,4+0.025=4.025(吨)。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "平方厘米，平方分米，平方米之间的进率与换算",
+                                 "毫米，厘米，分米，米之间的进率与换算",
+                                 "吨，千克之间的换算与比较"
+                             ],
+        "source":  "24-25四年级下·广东云浮·期末"
+    },
+    {
+        "id":  "243",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把0.86加上()个0.01是1;()个0.001加上0.624是0.9。",
+        "options":  [
+
+                    ],
+        "answer":  "14;276",
+        "solution":  "【分析】第一空先算1-0.86=0.14,0.14里有14个0.01;第二空算0.9-0.624=0.276,0.276里有276个0.001.\n【详解】1-0.86=0.14，所以是14个0.01;0.9-0.624=0.276，所以是276个0.001。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·辽宁鞍山·期末"
+    },
+    {
+        "id":  "244",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "根据以下信息，判断这辆货车()从公路限高杆下通过。(填\"能\"或\"不能\")\n![](images/4dc88f5efd076d3819c0709fd1330e5a84c6454179fa3edabf6b462d738b3633.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "能",
+        "solution":  "【分析】货车总高=1.6+0.9，再与限高3.5米比较.\n【详解】1.6+0.9=2.5(米),2.5\u003c3.5，所以能通过。",
+        "knowledge_points":  [
+                                 "一位小数的进位加法，退位减法",
+                                 "一位小数的大小比较"
+                             ],
+        "source":  "24-25四年级下·云南昆明·期末"
+    },
+    {
+        "id":  "245",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一堆煤重50.5吨，卖出21.5吨，这堆煤还剩()吨。",
+        "options":  [
+
+                    ],
+        "answer":  "29",
+        "solution":  "【分析】剩余=总量-卖出量.\n【详解】50.5-21.5=29(吨)。",
+        "knowledge_points":  [
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·广东云浮·期末"
+    },
+    {
+        "id":  "246",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "比5.24多1.03的数是(),比1.65多()的数是6.78。",
+        "options":  [
+
+                    ],
+        "answer":  "6.27;5.13",
+        "solution":  "【分析】第一空用加法，第二空用减法.\n【详解】5.24+1.03=6.27;6.78-1.65=5.13。",
+        "knowledge_points":  [
+                                 "多位小数的不进位加法，不退位减法"
+                             ],
+        "source":  "24-25四年级下·广东肇庆·期末"
+    },
+    {
+        "id":  "247",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "列竖式计算小数加减法时，小数点对齐就是相同数位对齐。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】小数点对齐的本质就是让相同数位对齐，才能正确相加减.\n【详解】题干说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·广西贵港·期末"
+    },
+    {
+        "id":  "248",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "小数加减法计算时，小数点对齐就是小数末位对齐。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】小数加减法要求小数点对齐(相同数位对齐),并不一定是末位对齐.\n【详解】如3.4与2.65相加，先写成3.40与2.65。可见关键是小数点对齐，不是末位对齐。故答案为: ×",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "多位小数的不进位加法，不退位减法"
+                             ],
+        "source":  "24-25四年级下·贵州·期末"
+    },
+    {
+        "id":  "249",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "两个小数相减，如果被减数和减数都增加1.9，则差不变。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】被减数和减数同时增加同一个数，差不变.\n【详解】例如2.56-2.5=(2.56+1.9)-(2.5+1.9),差相同。故答案为:√",
+        "knowledge_points":  [
+                                 "和与差的变化规律"
+                             ],
+        "source":  "24-25四年级下·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "250",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "在小数加减法的竖式中，要把各数的小数点对齐，其目的就是要把相同数位上的数对齐。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】小数点对齐后，个位对个位，十分位对十分位，保证同数位运算.\n【详解】题干说法正确。故答案为:√",
+        "knowledge_points":  [
+                                 "多位小数的不进位加法，不退位减法"
+                             ],
+        "source":  "24-25四年级下·贵州黔东南·期末"
+    },
+    {
+        "id":  "251",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "37.6+23.6-37.6+23.6=0。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】把相反项与同类项合并计算.\n【详解】(37.6-37.6)+(23.6+23.6)=0+47.2=47.2，不等于0。故答案为: ×",
+        "knowledge_points":  [
+                                 "整数加法运算律推广到小数"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "252",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "2.5+2.98+7.5=2.98+(2.5+7.5)。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】运用了加法交换律与结合律，和不变.\n【详解】左边=2.5+2.98+7.5=12.98;右边=2.98+(2.5+7.5)=12.98。故答案为:√",
+        "knowledge_points":  [
+                                 "整数加法运算律推广到小数"
+                             ],
+        "source":  "24-25四年级下·安徽安庆·期末"
+    },
+    {
+        "id":  "253",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "列竖式计算2.38+3.5时，要把末位对齐。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】小数加法应小数点对齐，不是末位对齐.\n【详解】2.38与3.5应写成2.38和3.50进行计算。题干错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "多位小数的不进位加法，不退位减法"
+                             ],
+        "source":  "23-24三年级下·云南昭通·期末"
+    },
+    {
+        "id":  "254",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "20个0.01减去2个0.1等于18个0.001。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】20个0.01=0.2,2个0.1=0.2，差为0;18个0.001=0.018.\n【详解】0不等于0.018，所以说法错误。故答案为: ×",
+        "knowledge_points":  [
+                                 "一位小数的不进位加法，不退位减法",
+                                 "小数的意义"
+                             ],
+        "source":  "24-25四年级下·贵州·期末"
+    },
+    {
+        "id":  "255",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "用竖式计算下面各题.\n360×24= 508×18= 22.38+17.42= 52.2-23.06=",
+        "options":  [
+
+                    ],
+        "answer":  "8640;9144;39.8;29.14",
+        "solution":  "【分析】整数乘法按三位数乘两位数笔算;小数加减法先小数点对齐再计算.\n【详解】360×24=8640;508×18=9144;22.38+17.42=39.8;52.2-23.06=29.14。",
+        "knowledge_points":  [
+                                 "三位数乘两位数，乘数末尾有0",
+                                 "三位数乘两位数，三位数中间有0",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·广西贵港·期末"
+    },
+    {
+        "id":  "256",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "用竖式计算.\n92-15.48= 1215÷45=",
+        "options":  [
+
+                    ],
+        "answer":  "76.52;27",
+        "solution":  "【分析】小数减法小数点对齐;除法按除数是两位数的笔算方法.\n【详解】92-15.48=76.52;1215÷45=27。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "除数是两位数的笔算除法"
+                             ],
+        "source":  "24-25四年级下·贵州铜仁·期末"
+    },
+    {
+        "id":  "257",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "从102.58里减去12.58与3.42的和，差是多少。",
+        "options":  [
+
+                    ],
+        "answer":  "86.58",
+        "solution":  "【分析】先求12.58与3.42的和，再用102.58减去这个和.\n【详解】102.58-(12.58+3.42)=102.58-16=86.58。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算",
+                                 "带有小括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·河南开封·期末"
+    },
+    {
+        "id":  "258",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "一箱牛奶38.5元，一袋面包比牛奶便宜12.8元，买一箱牛奶和一袋面包一共需要多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "64.2元",
+        "solution":  "【分析】先求面包价格:38.5-12.8，再与牛奶价格相加.\n【详解】38.5-12.8+38.5=25.7+38.5=64.2(元)\n答：一共需要64.2元。",
+        "knowledge_points":  [
+                                 "一位小数的进位加法，退位减法",
+                                 "利用小数的加，减法混合运算解决实际问题"
+                             ],
+        "source":  "24-25四年级下·广西贵港·期末"
+    },
+    {
+        "id":  "259",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "宁海开游节，天明湖集市的龙虾摊位第一天卖出龙虾65.7千克，比第二天多卖12.5千克，第三天卖出龙虾比前两天卖出的总和少36.1千克。第三天卖出龙虾多少千克。",
+        "options":  [
+
+                    ],
+        "answer":  "82.8千克",
+        "solution":  "【分析】第二天=65.7-12.5;前两天和=第一天+第二天;第三天=前两天和-36.1.\n【详解】65.7+(65.7-12.5)-36.1=65.7+53.2-36.1=118.9-36.1=82.8(千克)\n答：第三天卖出82.8千克。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·浙江宁波·期末"
+    },
+    {
+        "id":  "260",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "李老师从家出发，经过学校到公园。最近的路和最远的路相差多少千米。\n![](images/f2080f76bc0f79a9ab43723da40af76f8afbbc0bfd140ba3f5e1787d6af0d851.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "0.71千米",
+        "solution":  "【分析】家到学校选最短1.55千米，最长1.80千米;学校到公园选最短2.30千米，最长2.76千米。分别求总长再作差.\n【详解】最远:1.80+2.76=4.56(千米)\n最近:1.55+2.30=3.85(千米)\n差:4.56-3.85=0.71(千米)\n答：相差0.71千米。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·浙江杭州·期末"
+    },
+    {
+        "id":  "261",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "地球表面的总面积是5.1亿平方千米，其中陆地面积是1.49亿平方千米。陆地面积比海洋面积少多少亿平方千米。",
+        "options":  [
+
+                    ],
+        "answer":  "2.12亿平方千米",
+        "solution":  "【分析】先求海洋面积:5.1-1.49，再求海洋面积与陆地面积之差.\n【详解】5.1-1.49=3.61(亿平方千米)\n3.61-1.49=2.12(亿平方千米)\n答：陆地面积比海洋面积少2.12亿平方千米。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "2015四年级下·全国·专题练习"
+    },
+    {
+        "id":  "262",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "为了丰富同学们的数学知识，学校准备开展\"数学趣味周\"活动。张老师来到书店，打算购买一本《数学乐园》和一本《同步练习》作为活动参考资料。书店里，一本《同步练习》的价钱是18.62元，比一本《数学乐园》便宜7.9元，张老师付给售货员50元，应找回多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "4.86元",
+        "solution":  "【分析】《数学乐园》价钱=18.62+7.9;找回=50-两本总价.\n【详解】18.62+7.9=26.52(元)\n50-18.62-26.52=31.38-26.52=4.86(元)\n答：应找回4.86元。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·贵州铜仁·期末"
+    },
+    {
+        "id":  "263",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一捆电线，剪下了28.6米，剩下的比剪下的多12.24米，这捆电线原来有多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "69.44米",
+        "solution":  "【分析】剩下长度=28.6+12.24，再与剪下长度相加得原长.\n【详解】28.6+12.24+28.6=40.84+28.6=69.44(米)\n答：这捆电线原来有69.44米。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·江西吉安·期末"
+    },
+    {
+        "id":  "264",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "把8米长的竹竿插入水池中，竹竿入泥部分是1.3米，露出水面部分是3.28米，水池中水深多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "3.42米",
+        "solution":  "【分析】水深=竹竿总长-入泥部分-露出水面部分.\n【详解】8-1.3-3.28=6.7-3.28=3.42(米)\n答：水池中水深3.42米。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·江西吉安·期末"
+    },
+    {
+        "id":  "265",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "修路队修一条长9千米的公路，第一天修了1.45千米，第二天比第一天多修了0.3千米，还剩多少千米路没修。",
+        "options":  [
+
+                    ],
+        "answer":  "5.8千米",
+        "solution":  "【分析】第二天=1.45+0.3;已修总长=第一天+第二天;剩余=9-已修总长.\n【详解】1.45+0.3=1.75(千米)\n9-(1.45+1.75)=9-3.2=5.8(千米)\n答：还剩5.8千米路没修。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·江西吉安·期末"
+    },
+    {
+        "id":  "266",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "修路队修一条长9千米的公路，第一天修了1.45千米，第二天比第一天多修了0.3千米，还剩多少千米路没修。",
+        "options":  [
+
+                    ],
+        "answer":  "5.8千米",
+        "solution":  "【分析】先求第二天修路长度，再求两天共修长度，最后用总长度减去已修长度.\n【详解】1.45+0.3=1.75(千米)\n9-(1.45+1.75)=9-3.2=5.8(千米)\n答：还剩5.8千米路没修。",
+        "knowledge_points":  [
+                                 "利用小数的加，减法混合运算解决实际问题"
+                             ],
+        "source":  "24-25四年级下·江西吉安·期末"
+    },
+    {
+        "id":  "267",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一支钢笔9.7元，比一支圆珠笔贵6.3元。小华买了一支圆珠笔和一支钢笔，共用了多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "13.1元",
+        "solution":  "【分析】先求圆珠笔价格:9.7-6.3，再加钢笔价格9.7.\n【详解】9.7-6.3+9.7=3.4+9.7=13.1(元)\n答：共用了13.1元。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算",
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·河南新乡·期末"
+    },
+    {
+        "id":  "268",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "妈妈带60元钱去菜市场买菜，买水果用去28.70元，买蔬菜用去21.30元，还剩多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "10元",
+        "solution":  "【分析】剩余钱数=60-(水果钱+蔬菜钱).\n【详解】60-(28.7+21.3)=60-50=10(元)\n答：还剩10元。",
+        "knowledge_points":  [
+                                 "经济问题",
+                                 "一位小数的进位加法，退位减法",
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·江西南昌·期末"
+    },
+    {
+        "id":  "269",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "端午节的前一天，小辰跟随妈妈去超市购买了一些物品.\n![](images/e591697df30ef2a9cdaea9a65dc769a2.png)\n(1)小辰买了1个水杯和1盒牛奶，一共花了多少钱。\n(2)妈妈买了1个防晒帽子与1袋面包，付给售货员50元，应找回多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "25.25元;3.95元",
+        "solution":  "【分析】(1)水杯和牛奶价格相加;(2)帽子和面包价格相加后，用50元减去总价.\n【详解】(1)18.9+6.35=25.25(元)\n(2)36.5+9.55=46.05(元)\n50-46.05=3.95(元)\n答:(1)一共花了25.25元;(2)应找回3.95元。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "经济问题"
+                             ],
+        "source":  "24-25四年级下·陕西安康·期末"
+    },
+    {
+        "id":  "270",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "河南是孕育文明的摇篮，也是中国唯一地跨长江，黄河，淮河，海河四大流域的省份。长江在河南的流域面积是2.72万平方千米，比黄河的省内流域面积少0.9万平方千米，淮河的省内流域面积比黄河多5.21万平方千米。淮河在河南省内的流域面积是多少万平方千米。",
+        "options":  [
+
+                    ],
+        "answer":  "8.83万平方千米",
+        "solution":  "【分析】先求黄河在河南省内的流域面积:2.72+0.9;再加上5.21求淮河在河南省内的流域面积.\n【详解】2.72+0.9+5.21=3.62+5.21=8.83(万平方千米)\n答：淮河在河南省内的流域面积是8.83万平方千米。",
+        "knowledge_points":  [
+                                 "利用小数的加，减法混合运算解决实际问题",
+                                 "多位小数的进位加法，退位减法",
+                                 "多位小数的不进位加法，不退位减法"
+                             ],
+        "source":  "24-25四年级下·河南周口·期末"
+    },
+    {
+        "id":  "271",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "小玲每天早上要吃一个面包，喝一盒牛奶。她有30元钱，买一个星期的牛奶和面包够吗。\n![](images/2c9f9c394a3d046d04574274362ebe95.png)",
+        "options":  [
+
+                    ],
+        "answer":  "够",
+        "solution":  "【分析】先算每天花费，再乘7求一个星期总花费，最后与30元比较.\n【详解】1.8+2.2=4(元)\n4×7=28(元)\n28\u003c30\n答：买一个星期的牛奶和面包够。",
+        "knowledge_points":  [
+                                 "一位小数的进位加法，退位减法",
+                                 "经济问题"
+                             ],
+        "source":  "24-25四年级下·吉林·期末"
+    },
+    {
+        "id":  "272",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "乐乐身高1.36米，他站在0.45米的床上时，还比爸爸矮0.07米，爸爸身高多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "1.88米",
+        "solution":  "【分析】先求乐乐站在床上的总高度，再加上0.07米得到爸爸身高.\n【详解】1.36+0.45=1.81(米)\n1.81+0.07=1.88(米)\n答：爸爸身高是1.88米。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "多位小数的不进位加法，不退位减法"
+                             ],
+        "source":  "24-25四年级下·河南新乡·期末"
+    },
+    {
+        "id":  "273",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "李叔叔在体育用品店买了一副羽毛球拍和一个足球，他使用微信钱包付款，微信钱包中原有248.26元，支付完这两件商品后，还剩多少钱。\n![](images/c285c5d6d45b2e5e43e1c936a84b130c.png)",
+        "options":  [
+
+                    ],
+        "answer":  "47.96元",
+        "solution":  "【分析】用原有金额减去羽毛球拍和足球的总价.\n【详解】248.26-117.80-82.50=130.46-82.50=47.96(元)\n答：还剩47.96元。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·内蒙古鄂尔多斯·期末"
+    },
+    {
+        "id":  "274",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "修路队修一条公路，第一天修了2.8千米，比第二天多修了0.6千米。两天一共修了多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "5千米",
+        "solution":  "【分析】第二天修的长度=2.8-0.6，再与第一天相加.\n【详解】2.8-0.6+2.8=2.2+2.8=5(千米)\n答：两天一共修了5千米。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·浙江嘉兴·期末"
+    },
+    {
+        "id":  "275",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "学校进行身高测量，强强的身高是1.53米，比乐乐矮0.04米。笑笑比乐乐高0.11米，笑笑的身高是多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "1.68米",
+        "solution":  "【分析】先求乐乐身高，再加0.11求笑笑身高.\n【详解】1.53+0.04+0.11=1.57+0.11=1.68(米)\n答：笑笑的身高是1.68米。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算",
+                                 "多位小数的不进位加法，不退位减法"
+                             ],
+        "source":  "24-25四年级下·浙江嘉兴·期末"
+    },
+    {
+        "id":  "276",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "为了培养同学们旧物利用，勤俭节约的意识，学校开展了\"跳蚤市场\"活动，\n![](images/701ef81d192ea52cc9e44136c0e26e36b3c2b5de0f28ce163c92b500dad35d61.jpg)\n(1)在售卖活动中，林林收入多少钱。\n(2)辰辰利用旧物制作手工作品，售卖时开展\"满五送一\"的促销活动，每个手工作品0.25元，小泽想买12个，需要多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "30.5元;2.5元",
+        "solution":  "【分析】(1)先算苗苗收入:37.5-4.75，再减2.25得到林林收入。(2)\"满五送一\"即买5得6,12个相当于买10个，按0.25元/个计算.\n【详解】(1)37.5-4.75-2.25=37.5-(4.75+2.25)=37.5-7=30.5(元)\n答：林林收入30.5元.\n(2)12÷(5+1)=2(组)\n5×2×0.25=10×0.25=2.5(元)\n答：需要2.5元。",
+        "knowledge_points":  [
+                                 "小数与整数的乘法",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·河北石家庄·期末"
+    },
+    {
+        "id":  "277",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.94,
+        "question":  "体育课上，同学们进行立定跳远测试。龙一鸣跳了1.45米，淘淘比龙一鸣少跳了0.07米，壮壮比淘淘多跳0.13米，壮壮跳了多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "1.51米",
+        "solution":  "【分析】先求淘淘成绩，再加0.13求壮壮成绩.\n【详解】1.45-0.07+0.13=1.38+0.13=1.51(米)\n答：壮壮跳了1.51米。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·江西宜春·期末"
+    },
+    {
+        "id":  "278",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "小红用智能体重秤记录体重变化：上周体重是34.2千克，这周坚持锻炼后体重下降了0.8千克，周末户外野餐后体重又增加了0.35千克。小红现在的体重是多少千克。",
+        "options":  [
+
+                    ],
+        "answer":  "33.75千克",
+        "solution":  "【分析】先减去下降的0.8千克，再加上增加的0.35千克.\n【详解】34.2-0.8=33.4(千克)\n33.4+0.35=33.75(千克)\n答：小红现在的体重是33.75千克。",
+        "knowledge_points":  [
+                                 "多位小数的不进位加法，不退位减法",
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·吉林·期末"
+    },
+    {
+        "id":  "279",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "四年级3个班进行垃圾分类比赛，四(1)班收集塑料瓶12.55千克，四(2)班收集塑料瓶15.4千克，四(3)班收集塑料瓶20.45千克。三个班共收集塑料瓶多少千克。",
+        "options":  [
+
+                    ],
+        "answer":  "48.4千克",
+        "solution":  "【分析】将三个班收集量相加，可先用加法交换律凑整.\n【详解】12.55+15.4+20.45=12.55+20.45+15.4=33+15.4=48.4(千克)\n答：三个班共收集塑料瓶48.4千克。",
+        "knowledge_points":  [
+                                 "整数加法运算律推广到小数",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·浙江·期末"
+    },
+    {
+        "id":  "280",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "明明到商店里买钢笔，买10元一支的差0.4元，她只好买了一支9元的钢笔，她还剩多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "0.6元",
+        "solution":  "【分析】\"买10元一支的差0.4元\"说明带了10-0.4元，再减去9元.\n【详解】10-0.4-9=9.6-9=0.6(元)\n答：她还剩0.6元。",
+        "knowledge_points":  [
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·河南开封·期末"
+    },
+    {
+        "id":  "281",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "为了打造城市品牌，增强人民体质，相关部门多次在新蒲新区湿地公园举办\"筑梦健康·酷跑遵义\"系列活动。湿地公园跑道一圈长约6.8千米，沿途绿树成荫，天鹅成群，景色宜人，是理想的锻炼身体，休闲放松之地。星期六早上，王叔叔在湿地公园晨跑锻炼，围绕跑道正好跑了一圈半，王叔叔跑了多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "10.2千米",
+        "solution":  "【分析】一圈半=1圈+0.5圈，先求半圈长度，再与一圈相加.\n【详解】6.8÷2=3.4(千米)\n6.8+3.4=10.2(千米)\n答：王叔叔跑了10.2千米。",
+        "knowledge_points":  [
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·贵州遵义·期末"
+    },
+    {
+        "id":  "282",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一个油桶装有一些汽油，如果先倒入26.57千克汽油，再倒出48.26千克汽油，此时桶内恰好有90千克汽油，这个油桶原来有多少千克汽油。",
+        "options":  [
+
+                    ],
+        "answer":  "111.69千克",
+        "solution":  "【分析】用倒推法：原有=最后90千克+倒出48.26千克-倒入26.57千克.\n【详解】90+48.26-26.57=138.26-26.57=111.69(千克)\n答：这个油桶原来有111.69千克汽油。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·河南漯河·期末"
+    },
+    {
+        "id":  "283",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "在2023年世界泳联跳水世界杯蒙特利尔站女双10米台的决赛中，中国组合陈芋汐/全红婵发挥出色，以378.60分的成绩夺得冠军，领先亚军多达61.92分。而亚军比季军多了14.10分。季军得了多少分。",
+        "options":  [
+
+                    ],
+        "answer":  "302.58分",
+        "solution":  "【分析】先求亚军分数，再减去14.10求季军分数.\n【详解】378.60-61.92-14.10=316.68-14.10=302.58(分)\n答：季军得了302.58分。",
+        "knowledge_points":  [
+                                 "利用小数的加，减法混合运算解决实际问题",
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "23-24四年级下·河南濮阳·期末"
+    },
+    {
+        "id":  "284",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "妈妈的微信钱包里原来有83.05元，下图是妈妈某日的微信账单(单位：元)。这天妈妈的微信钱包里还剩多少钱。\n![](images/2026-03-12%20202245.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "31.83元",
+        "solution":  "【分析】先算总支出，再从原金额中减去总支出，最后加上收到的红包.\n【详解】83.05-(19.83+32.17)+0.78=83.05-52+0.78=31.05+0.78=31.83(元)\n答：这天妈妈的微信钱包里还剩31.83元。",
+        "knowledge_points":  [
+                                 "利用小数的加，减法混合运算解决实际问题",
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·辽宁抚顺·期末"
+    },
+    {
+        "id":  "285",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "修一条长7.56千米的公路，第一周修了1.9千米，第二周比第一周多修了0.28千米，还剩多少千米没有修。",
+        "options":  [
+
+                    ],
+        "answer":  "3.48千米",
+        "solution":  "【分析】第二周=1.9+0.28，已修总长=第一周+第二周，剩余=7.56-已修总长.\n【详解】1.9+0.28=2.18(千米)\n7.56-(1.9+2.18)=7.56-4.08=3.48(千米)\n答：还剩3.48千米没有修。",
+        "knowledge_points":  [
+                                 "利用小数的加，减法混合运算解决实际问题"
+                             ],
+        "source":  "24-25四年级下·江西宜春·期末"
+    },
+    {
+        "id":  "286",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "有甲，乙两箱货物，共重24.72kg,甲箱重13.3kg。甲箱比乙箱重多少千克。",
+        "options":  [
+
+                    ],
+        "answer":  "1.88千克",
+        "solution":  "【分析】先求乙箱重量，再求甲箱比乙箱重多少.\n【详解】24.72-13.3=11.42(千克)\n13.3-11.42=1.88(千克)\n答：甲箱比乙箱重1.88千克。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·辽宁抚顺·期末"
+    },
+    {
+        "id":  "287",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "聪聪养成了良好的读书习惯，每个周末都坚持读书。妈妈为了奖励他到书店买一本《十万个为什么》和一本《宝葫芦的秘密》。一本《宝葫芦的秘密》18.62元，一本《十万个为什么》17.9元。妈妈付给售货员50元，应找回多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "13.48元",
+        "solution":  "【分析】先求两本书总价，再用50减去总价.\n【详解】50-(18.62+17.9)=50-36.52=13.48(元)\n答：应找回13.48元。",
+        "knowledge_points":  [
+                                 "利用小数的加，减法混合运算解决实际问题",
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·辽宁葫芦岛·期末"
+    },
+    {
+        "id":  "288",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "一条公路，已经修好了85.65千米，比剩下的多13.8千米。这条公路全长多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "157.5千米",
+        "solution":  "【分析】剩下长度=85.65-13.8，全长=已修+剩下.\n【详解】85.65-13.8+85.65=71.85+85.65=157.5(千米)\n答：这条公路全长157.5千米。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·湖南岳阳·期末"
+    },
+    {
+        "id":  "289",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "2008年奥运会男子单人3米跳板比赛中，何冲以领先第二名德斯帕蒂耶斯32.45分的优势进入最后一跳，秦凯落后德斯帕蒂耶斯7.65分。最后一跳前，秦凯落后何冲多少分。",
+        "options":  [
+
+                    ],
+        "answer":  "40.1分",
+        "solution":  "【分析】秦凯落后何冲=何冲领先德斯帕蒂耶斯+德斯帕蒂耶斯领先秦凯.\n【详解】32.45+7.65=40.1(分)\n答：最后一跳前，秦凯落后何冲40.1分。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·湖北省直辖县级单位·期末"
+    },
+    {
+        "id":  "290",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "一根绳子长10米，第一次用去3.6米，第二次用去2.5米，还剩下多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "3.9米",
+        "solution":  "【分析】剩余长度=总长度-(两次用去长度之和).\n【详解】10-(3.6+2.5)=10-6.1=3.9(米)\n答：还剩下3.9米。",
+        "knowledge_points":  [
+                                 "利用小数的加，减法混合运算解决实际问题",
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·浙江温州·期末"
+    },
+    {
+        "id":  "291",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "工程队要修一条长12.5千米的公路，第一个月修了3.5千米，第二个月修了4.7千米，两个月修了多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "8.2千米",
+        "solution":  "【分析】两个月修路总长度=第一个月+第二个月.\n【详解】3.5+4.7=8.2(千米)\n答：两个月修了8.2千米。",
+        "knowledge_points":  [
+                                 "利用小数加减法解决实际问题",
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·浙江宁波·期末"
+    },
+    {
+        "id":  "292",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "为了更好地了解神舟二十号，参观科技馆前李老师买了《你好!空间站》和《飞船升空了》两本有关航天科技的书籍，其中《你好!空间站》花了12.36元,《飞船升空了》比《你好!空间站》每本便宜1.25元，李老师买两本书应付多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "23.47元",
+        "solution":  "【分析】先求《飞船升空了》的价格，再与《你好!空间站》价格相加.\n【详解】12.36-1.25=11.11(元)\n11.11+12.36=23.47(元)\n答：应付23.47元。",
+        "knowledge_points":  [
+                                 "多位小数的不进位加法，不退位减法"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "293",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一桶油连桶重8.7千克，用去一半后连桶重4.6千克。油重多少千克。",
+        "options":  [
+
+                    ],
+        "answer":  "8.2千克",
+        "solution":  "【分析】前后差是半桶油重，再乘2得到整桶油重.\n【详解】8.7-4.6=4.1(千克)\n4.1+4.1=8.2(千克)\n答：油重8.2千克。",
+        "knowledge_points":  [
+                                 "一位小数的不进位加法，不退位减法"
+                             ],
+        "source":  "24-25四年级下·四川德阳·期末"
+    },
+    {
+        "id":  "294",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "兰兰去书店买了2本书，其中故事书的价格是28.70元，另一本漫画书的价格是36.23元。她付给售货员100元，应找回多少元钱。",
+        "options":  [
+
+                    ],
+        "answer":  "35.07元",
+        "solution":  "【分析】找回金额=100-故事书价-漫画书价.\n【详解】100-28.70-36.23=71.3-36.23=35.07(元)\n答：应找回35.07元。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·浙江温州·期末"
+    },
+    {
+        "id":  "295",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "小辉跳远跳了1.9米，小东跳得比小辉多0.07米，小乐跳得比小东少0.18米.\n(1)小东跳了多少米。\n(2)小乐跳了多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "1.97米;1.79米",
+        "solution":  "【分析】(1)小东=1.9+0.07;(2)小乐=小东-0.18.\n【详解】(1)1.9+0.07=1.97(米)\n(2)1.97-0.18=1.79(米)\n答:(1)小东跳了1.97米;(2)小乐跳了1.79米。",
+        "knowledge_points":  [
+                                 "多位小数的不进位加法，不退位减法",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·广东云浮·期末"
+    },
+    {
+        "id":  "296",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "张英，李明和黎红参加跳高比赛，张英跳了1.1米，比李明低了0.15米。黎红比李明低了0.09米，黎红跳了多高。",
+        "options":  [
+
+                    ],
+        "answer":  "1.16米",
+        "solution":  "【分析】先求李明成绩，再减去0.09求黎红成绩.\n【详解】1.1+0.15-0.09=1.25-0.09=1.16(米)\n答：黎红跳了1.16米。",
+        "knowledge_points":  [
+                                 "利用小数的加，减法混合运算解决实际问题",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·广东肇庆·期末"
+    },
+    {
+        "id":  "297",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "据统计,2024年春节假期，潮州市牌坊街共接待游客165.06万人，国庆期间接待的游客数量比春节期间少69.52万人。这两个节日共接待游客多少万人。",
+        "options":  [
+
+                    ],
+        "answer":  "260.6万人",
+        "solution":  "【分析】先求国庆接待人数，再与春节接待人数相加.\n【详解】165.06-69.52+165.06=95.54+165.06=260.6(万人)\n答：这两个节日共接待游客260.6万人。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·广东潮州·期末"
+    },
+    {
+        "id":  "298",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "在社团活动中，老师将一条长15米的彩带分给3个小组做创意编织，第一小组分得5.2米彩带，比第二小组长0.25米，剩下的分给第三小组。第三小组分得多少米彩带。",
+        "options":  [
+
+                    ],
+        "answer":  "4.85米",
+        "solution":  "【分析】第二小组=5.2-0.25，第三小组=15-(第一+第二).\n【详解】15-(5.2-0.25+5.2)=15-(4.95+5.2)=15-10.15=4.85(米)\n答：第三小组分得4.85米彩带。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "299",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一套运动衣49元，一套速干衣44.5元。买这两套衣服，一共需要多少钱。如果付100元，要找回多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "93.5元;6.5元",
+        "solution":  "【分析】先求两套衣服总价，再用100减去总价求找回金额.\n【详解】49+44.5=93.5(元)\n100-93.5=6.5(元)\n答：一共需要93.5元;要找回6.5元。",
+        "knowledge_points":  [
+                                 "一位小数的不进位加法，不退位减法",
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·吉林白城·期末"
+    },
+    {
+        "id":  "300",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "李老师去超市购物，买肉类用去76.8元，又买了2千克苹果，一千克苹果5元钱，她带了100元，还剩下多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "13.2元",
+        "solution":  "【分析】先求苹果总价，再用100减去买肉和买苹果的总花费.\n【详解】100-76.8-5×2=23.2-10=13.2(元)\n答：还剩下13.2元。",
+        "knowledge_points":  [
+                                 "乘加，乘减混合运算",
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "301",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面各数中，最接近8.08万的是()。",
+        "options":  [
+                        "A. 8.081万",
+                        "B. 8.0801万",
+                        "C. 8.018万"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】分别计算各选项与8.08万的差，差最小的最接近.\n【详解】A:8.081-8.08=0.001(万);B:8.0801-8.08=0.0001(万);C:8.08-8.018=0.062(万).\n0.0001\u003c0.001\u003c0.062，所以最接近8.08万的是8.0801万。故答案为:B",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "多位小数的不进位加法，不退位减法",
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "24-25四年级下·广东潮州·期末"
+    },
+    {
+        "id":  "302",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列算式中，\"3\"和\"5\"不能直接相加的是()。",
+        "options":  [
+                        "A. 326+569",
+                        "B. 4.23+6.75",
+                        "C. 52.8+132.2",
+                        "D. 463+16.5"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】直接相加需要在相同数位.\n【详解】A中\"3\"和\"5\"都在百位;B都在百分位;C都在十位;D中\"3\"在个位，\"5\"在十分位，不能直接相加。故答案为:D",
+        "knowledge_points":  [
+                                 "多位小数的不进位加法，不退位减法",
+                                 "一位小数的进位加法，退位减法",
+                                 "一位小数的不进位加法，不退位减法"
+                             ],
+        "source":  "24-25四年级下·浙江杭州·期末"
+    },
+    {
+        "id":  "303",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "0.54扩大10倍后与原数相比，增加了()。",
+        "options":  [
+                        "A. 4.84",
+                        "B. 4.86",
+                        "C. 5.4",
+                        "D. 4.94"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】先求0.54扩大10倍后的数，再减去原数.\n【详解】0.54×10=5.4,5.4-0.54=4.86。故答案为:B",
+        "knowledge_points":  [
+                                 "小数点向右移动引起小数大小变化的规律",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·湖南岳阳·期末"
+    },
+    {
+        "id":  "304",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "如图，要求3号彩带的长度列式正确的是().\n![](images/be8bd43bff585641ad9d6a0968eee2d796c72a7ebbf41bc3dbfe58a20a7d8b66.jpg)",
+        "options":  [
+                        "A. 2.4+0.8+0.4",
+                        "B. 2.4+(0.8-0.4)",
+                        "C. 2.4-0.8-0.4",
+                        "D. 2.4-(0.8-0.4)"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】3号比1号多(0.8-0.4)dm,所以3号=2.4+(0.8-0.4).\n【详解】列式应为2.4+(0.8-0.4)。故答案为:B",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·山东临沂·期末"
+    },
+    {
+        "id":  "305",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面各数中，与168最接近的数是()。",
+        "options":  [
+                        "A. 168.15",
+                        "B. 168.015",
+                        "C. 167.99",
+                        "D. 167.998"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】比较各选项与168的差值.\n【详解】A差0.15,B差0.015,C差0.01,D差0.002，最小是0.002。故答案为:D",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "多位小数的不进位加法，不退位减法",
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "24-25四年级下·广东江门·期末"
+    },
+    {
+        "id":  "306",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列数中与10最接近的数是()。",
+        "options":  [
+                        "A. 9.9",
+                        "B. 10.01",
+                        "C. 10.1"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】计算各数与10的差，差最小最接近.\n【详解】A差0.1,B差0.01,C差0.1，最小是0.01。故答案为:B",
+        "knowledge_points":  [
+                                 "多位小数的不进位加法，不退位减法",
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "24-25四年级下·广东汕头·期末"
+    },
+    {
+        "id":  "307",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "如图，要求3号彩带的长度列式正确的是().\n![](images/89ed3c742e22e16e2edf6af140888d2030682d16a4353c882a07f3bf0e7dd4a5.jpg)",
+        "options":  [
+                        "A. 2.4+0.8+0.4",
+                        "B. 2.4+(0.8-0.4)",
+                        "C. 2.4-0.8-0.4",
+                        "D. 2.4-(0.8-0.4)"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】3号比1号多0.8-0.4，所以3号长度=2.4+(0.8-0.4).\n【详解】列式正确的是2.4+(0.8-0.4)。故答案为:B",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·山东临沂·期末"
+    },
+    {
+        "id":  "308",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "关于290.090，下面说法正确的是()。",
+        "options":  [
+                        "A. 这个数中的三个零都要读出来",
+                        "B. 它与290.09大小相同，计数单位也相同",
+                        "C. 这个数再加上0.1就是291",
+                        "D. 从左边数第一个9表示的是第二个9的1000倍"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】按小数读法，计数单位和小数性质逐项判断.\n【详解】A错;B大小相同但计数单位不同;C错,290.090+0.1=290.190;D对:90是0.09的1000倍。故答案为:D",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "小数的数位和计数单位的认识",
+                                 "小数的读法和写法",
+                                 "小数的性质"
+                             ],
+        "source":  "24-25四年级下·山东临沂·期末"
+    },
+    {
+        "id":  "309",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "一道减法算式的被减数增加4.8，减数减少1.4，则差()。",
+        "options":  [
+                        "A. 增加3.4",
+                        "B. 减少3.4",
+                        "C. 增加6.2",
+                        "D. 减少6.2"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】被减数增加，差同增;减数减少，差也增.\n【详解】差共增加4.8+1.4=6.2。故答案为:C",
+        "knowledge_points":  [
+                                 "一位小数的进位加法，退位减法",
+                                 "和与差的变化规律"
+                             ],
+        "source":  "24-25四年级下·广东汕头·期末"
+    },
+    {
+        "id":  "310",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "小文在做加法计算时，把其中一个加数8.7看成了7.8，得出的结果是9.15，如果她没有看错，得出的结果应该是()。",
+        "options":  [
+                        "A. 10.05",
+                        "B. 9.24",
+                        "C. 8.25",
+                        "D. 7.15"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】先求另一个加数，再与8.7相加.\n【详解】9.15-7.8=1.35,1.35+8.7=10.05。故答案为:A",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·山东临沂·期末"
+    },
+    {
+        "id":  "311",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "小文在做加法计算时，把其中一个加数8.7看成了7.8，得出的结果是9.15，如果她没有看错，得出的结果应该是()。",
+        "options":  [
+                        "A. 10.05",
+                        "B. 9.24",
+                        "C. 8.25",
+                        "D. 7.15"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】错误和减去错看加数，得到另一加数;再与正确加数相加.\n【详解】9.15-7.8=1.35,1.35+8.7=10.05。故答案为:A",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·山东临沂·期末"
+    },
+    {
+        "id":  "312",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面的算式中，\"4\"和\"5\"不能直接相加减的是()。",
+        "options":  [
+                        "A. 408+502",
+                        "B. 0.47+1.5",
+                        "C. 3.84+10.52",
+                        "D. 75.2-24.16"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】判断\"4\"和\"5\"是否在同数位.\n【详解】A百位对百位,B十分位对十分位,D个位对个位，都可以;C中\"4\"在百分位，\"5\"在十分位，不能直接相加减。故答案为:C",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "两，三位数的一次进位加法"
+                             ],
+        "source":  "24-25四年级下·山东临沂·期末"
+    },
+    {
+        "id":  "313",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "小文在做加法计算时，把其中一个加数8.7看成了7.8，得出的结果是9.15，如果她没有看错，得出的结果应该是()。",
+        "options":  [
+                        "A. 10.05",
+                        "B. 9.24",
+                        "C. 8.25",
+                        "D. 7.15"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】正确结果=错误结果-错看加数+正确加数.\n【详解】9.15-7.8+8.7=1.35+8.7=10.05。故答案为:A",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·山东临沂·期末"
+    },
+    {
+        "id":  "314",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "笑笑用计算器计算13.79+2.34时，她错误的输入了13.19+2.34。要修正这个错误，应该()。",
+        "options":  [
+                        "A. 加60",
+                        "B. 减0.6",
+                        "C. 减60",
+                        "D. 加0.6"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】错误少加了13.79-13.19=0.6，所以应补加0.6.\n【详解】应加0.6。故答案为:D",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "多位小数的不进位加法，不退位减法"
+                             ],
+        "source":  "24-25四年级下·内蒙古赤峰·期末"
+    },
+    {
+        "id":  "315",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列各数中，最接近2.9的是()。",
+        "options":  [
+                        "A. 2.89",
+                        "B. 2.99",
+                        "C. 3.0"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】比较各数与2.9的差值.\n【详解】A差0.01,B差0.09,C差0.1，最小是0.01。故答案为:A",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "多位小数的不进位加法，不退位减法",
+                                 "一位小数的进位加法，退位减法",
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "24-25四年级下·内蒙古包头·期末"
+    },
+    {
+        "id":  "316",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面的算式中\"3\"和\"7\"可以直接相加减的是()。",
+        "options":  [
+                        "A. 567+358",
+                        "B. 5.97+2.3",
+                        "C. 5.79-2.3"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】同数位才能直接运算.\n【详解】A中\"3\"在百位，\"7\"在个位;B中\"3\"在十分位，\"7\"在百分位;C中二者都在十分位，可以直接相减。故答案为:C",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "多位小数的不进位加法，不退位减法",
+                                 "两，三位数的连续进位加法"
+                             ],
+        "source":  "24-25四年级下·湖北孝感·期末"
+    },
+    {
+        "id":  "317",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "王老师到新华书店买三本价格不同的书，其中最贵的一本是32.60元，最便宜的一本是20.40元，这三本书的总价可能是()元。",
+        "options":  [
+                        "A. 65.60",
+                        "B. 73.40",
+                        "C. 75.60",
+                        "D. 86.40"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】第三本价格在20.40与32.60之间且不同。总价范围:20.40+20.41+32.60=73.41至20.40+32.59+32.60=85.59.\n【详解】只有75.60落在73.41到85.59之间。故答案为:C",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·广东汕尾·期末"
+    },
+    {
+        "id":  "318",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面小数中，与7最接近的是()。",
+        "options":  [
+                        "A. 6.99",
+                        "B. 7.05",
+                        "C. 7.1",
+                        "D. 8"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】比较各数与7的差值，最小者最接近.\n【详解】A差0.01,B差0.05,C差0.1,D差1，最小是0.01。故答案为:A",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "多位小数的不进位加法，不退位减法",
+                                 "一位小数的不进位加法，不退位减法",
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "24-25四年级下·江西南昌·期末"
+    },
+    {
+        "id":  "319",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列各数中与\"1\"最接近的数是()。",
+        "options":  [
+                        "A. 0.99",
+                        "B. 1.002",
+                        "C. 1.01",
+                        "D. 1.1"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】分别求与1的差.\n【详解】A差0.01,B差0.002,C差0.01,D差0.1，最小是0.002。故答案为:B",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "多位小数的不进位加法，不退位减法",
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "24-25四年级下·内蒙古赤峰·期末"
+    },
+    {
+        "id":  "320",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "文文在书店买了两本书，一本16.45元，另一本18.3元，一共花了多少钱。解决这个问题，列式为16.45+18.3，如图竖式虚线框中的数表示的是().\n![](images/2d2a76695a9bda78b9159ebccc98e7a8568dc797fb773002e0a16d51e1e9b2b7.jpg)",
+        "options":  [
+                        "A. 4分加3分等于7分",
+                        "B. 4个0.1加3个0.1等于7个0.1",
+                        "C. 4元加3元等于7元",
+                        "D. 4个一加3个一等于7个一"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】16.45中的4和18.3中的3都在十分位，表示十分之一.\n【详解】虚线框表示4个0.1+3个0.1=7个0.1。故答案为:B",
+        "knowledge_points":  [
+                                 "根据小数的意义进行单位换算",
+                                 "元，角，分的认识及换算",
+                                 "多位小数的进位加法，退位减法",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·贵州遵义·期末"
+    },
+    {
+        "id":  "321",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列小数中，与10最接近的是()。",
+        "options":  [
+                        "A. 10.001",
+                        "B. 10.1",
+                        "C. 9.9",
+                        "D. 10.01"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】比较各数与10的差值.\n【详解】A差0.001,B差0.1,C差0.1,D差0.01，最小是0.001。故答案为:A",
+        "knowledge_points":  [
+                                 "多位小数的不进位加法，不退位减法",
+                                 "一位小数的进位加法，退位减法",
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "24-25四年级下·江西南昌·期末"
+    },
+    {
+        "id":  "322",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "要计算7.9-2.85，以下计算过程错误的是()。",
+        "options":  [
+                        "A. 原式=7.90-2.85",
+                        "B. 原式=7.9-3+0.15",
+                        "C. 原式=7.9-3-0.15",
+                        "D. 原式=8.0-2.85-0.1"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】把2.85看成3-0.15，则7.9-2.85=7.9-(3-0.15)=7.9-3+0.15.\n【详解】A,B,D都成立;C变成了7.9-3.15，与原式不等。故答案为:C",
+        "knowledge_points":  [
+                                 "与小数减法相关的简便运算",
+                                 "整数加法运算律推广到小数",
+                                 "小数的性质"
+                             ],
+        "source":  "24-25四年级下·广东东莞·期末"
+    },
+    {
+        "id":  "323",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面各小数最接近1的是()。",
+        "options":  [
+                        "A. 1.1",
+                        "B. 1.01",
+                        "C. 0.98"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】求与1的差，差最小最接近.\n【详解】A差0.1,B差0.01,C差0.02，最小是0.01。故答案为:B",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "多位小数的不进位加法，不退位减法",
+                                 "一位小数的不进位加法，不退位减法"
+                             ],
+        "source":  "24-25三年级下·山东济宁·期末"
+    },
+    {
+        "id":  "324",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "列竖式计算小数加减法时，需要小数点对齐，目的是为了()。",
+        "options":  [
+                        "A. 计算方便",
+                        "B. 末尾对齐",
+                        "C. 相同计数单位的个数相加减",
+                        "D. 前三种都是"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】小数点对齐本质是相同数位(相同计数单位)对齐，便于正确相加减.\n【详解】目的是\"相同计数单位的个数相加减\"。故答案为:C",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·浙江嘉兴·期末"
+    },
+    {
+        "id":  "325",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "知夏家，豆豆家和学校的位置在同一条直线上，知夏家在距学校1.25千米处，豆豆家在学校2.4千米处，知夏家到豆豆家有多远。()",
+        "options":  [
+                        "A. 3.65千米",
+                        "B. 1.15千米",
+                        "C. 3.65或1.15千米",
+                        "D. 无法判断"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】两家可能在学校两侧或同侧.\n【详解】两侧时:1.25+2.4=3.65(千米);同侧时:2.4-1.25=1.15(千米).\n所以距离为3.65或1.15千米。故答案为:C",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "多位小数的不进位加法，不退位减法"
+                             ],
+        "source":  "24-25四年级下·山东日照·期末"
+    },
+    {
+        "id":  "326",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "把0.56的小数点向右移动一位后，所得的数比原数大()。",
+        "options":  [
+                        "A. 5",
+                        "B. 5.01",
+                        "C. 5.02",
+                        "D. 5.04"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】0.56小数点右移一位得5.6，再减原数.\n【详解】5.6-0.56=5.04。故答案为:D",
+        "knowledge_points":  [
+                                 "小数点向右移动引起小数大小变化的规律",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·甘肃定西·期末"
+    },
+    {
+        "id":  "327",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "1.2的计数单位比0.38的计数单位大()。",
+        "options":  [
+                        "A. 0.09",
+                        "B. 0.82",
+                        "C. 1.58"
+                    ],
+        "answer":  "A",
+        "solution":  "【分析】1.2是一位小数，计数单位0.1;0.38是两位小数，计数单位0.01.\n【详解】0.1-0.01=0.09。故答案为:A",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·云南楚雄·期末"
+    },
+    {
+        "id":  "328",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面各数中最接近6.92的是()。",
+        "options":  [
+                        "A. 7.16",
+                        "B. 7.3",
+                        "C. 6.8",
+                        "D. 6.88"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】比较各数与6.92的差值.\n【详解】A差0.24,B差0.38,C差0.12,D差0.04，最小是0.04。故答案为:D",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "24-25四年级下·云南大理·期末"
+    },
+    {
+        "id":  "329",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列各数，最接近10的是()。",
+        "options":  [
+                        "A. 9.1",
+                        "B. 9.01",
+                        "C. 10.001",
+                        "D. 10.01"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】分别与10作差，差最小最接近.\n【详解】A差0.9,B差0.99,C差0.001,D差0.01。最小是0.001。故答案为:C",
+        "knowledge_points":  [
+                                 "多位小数的大小比较",
+                                 "一位小数的进位加法，退位减法",
+                                 "多位小数的不进位加法，不退位减法",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·云南昭通·期末"
+    },
+    {
+        "id":  "330",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面各数中和5最接近的数是()。",
+        "options":  [
+                        "A. 5.02",
+                        "B. 4.99",
+                        "C. 5.001"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】比较与5的差值.\n【详解】A差0.02,B差0.01,C差0.001，最小是0.001。故答案为:C",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "多位小数的不进位加法，不退位减法",
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "24-25四年级下·甘肃陇南·期末"
+    },
+    {
+        "id":  "331",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "在立定跳远比赛中，蓝蓝跳了1.45m,同同比蓝蓝多跳了0.29m,华华比蓝蓝少跳了0.33m,同同比华华多跳了()m。",
+        "options":  [
+                        "A. 0.04",
+                        "B. 0.62",
+                        "C. 0.52"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】同同比蓝蓝多0.29m,华华比蓝蓝少0.33m,所以同同比华华多0.29+0.33.\n【详解】0.29+0.33=0.62(m)。故答案为:B",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·甘肃定西·期末"
+    },
+    {
+        "id":  "332",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "明明在计算7.65-(b+0.65)时漏掉了括号，这样算得的结果比正确结果()。",
+        "options":  [
+                        "A. 大0.65",
+                        "B. 相等",
+                        "C. 小0.65",
+                        "D. 大1.3"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】正确式:7.65-b-0.65;漏括号后:7.65-b+0.65。后者比前者多0.65+0.65=1.3.\n【详解】漏括号后的结果比正确结果大1.3。故答案为:D",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算",
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·贵州黔西·期末"
+    },
+    {
+        "id":  "333",
+        "unit":  "第六单元 小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面各数中，与10最接近的数是()。",
+        "options":  [
+                        "A. 10.02",
+                        "B. 9.99",
+                        "C. 10.002"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】分别计算各数与10的差，差最小最接近.\n【详解】A差0.02,B差0.01,C差0.002，最小是0.002。故答案为:C",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "多位小数的不进位加法，不退位减法",
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "24-25四年级下·山东济南·期末"
+    },
+    {
+        "id":  "334",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小数0.99的计数单位是(),0.99再加上6个这样的计数单位是(),得到的小数读作()。",
+        "options":  [
+
+                    ],
+        "answer":  "0.01或百分之一或1÷100;1.05;一点零五",
+        "solution":  "【分析】两位小数的计数单位是0.01(百分之一或1÷100)，6个这样的计数单位就是0.06，用0.99加0.06即可。小数的读法：整数部分按整数的读法来读，如果是0读作零，小数点读作点。小数部分依次读出每位上的数字;小数部分有几个0就读出几个零.\n【详解】0.99是两位小数，它的计数单位是0.01(百分之一或1÷100),6个这样的计数单位是0.06,0.99+0.06=1.05,0.99再加上6个这样的计数单位是1.05，得到的小数读作一点零五。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法",
+                                 "退位减法",
+                                 "小数的数位和计数单位的认识",
+                                 "小数的读法和写法"
+                             ],
+        "source":  "24-25四年级下·广东潮州·期末"
+    },
+    {
+        "id":  "335",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把1.08的小数点去掉，得到的数比原来多了()倍，多了()。",
+        "options":  [
+
+                    ],
+        "answer":  "99;106.92",
+        "solution":  "【分析】小数点向右移动一位，两位，三位，四位，小数就扩大到原数的10倍,100倍,1000倍,10000倍。由题意得，把1.08的小数点去掉，原数变为108，相当于把小数点向右移动了两位，这个小数变为原来的100倍，比原来多了(100-1)倍。求多了多少，直接用108减去1.08即可算出多了多少.\n【详解】100-1=99;108-1.08=106.92。把1.08的小数点去掉，得到的数比原来多了99倍，多了106.92。",
+        "knowledge_points":  [
+                                 "小数点向右移动引起小数大小变化的规律",
+                                 "多位小数的进位加法",
+                                 "退位减法"
+                             ],
+        "source":  "24-25四年级下·广东潮州·期末"
+    },
+    {
+        "id":  "336",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在一个直角三角形中，如果有一个锐角是32.5°,则另一个锐角是()°。",
+        "options":  [
+
+                    ],
+        "answer":  "57.5",
+        "solution":  "【分析】有一个角是直角的三角形是直角三角形。三角形内角和是180°,180°减去90°再减去32.5°,即可算出另一个锐角的度数.\n【详解】180°-90°-32.5°=90°-32.5°=57.5°。在一个直角三角形中，如果有一个锐角是32.5°,则另一个锐角是57.5°。",
+        "knowledge_points":  [
+                                 "三角形的内角和",
+                                 "三角形的分类"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "337",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "比7.2多8的数是();15.78比()少2.89。",
+        "options":  [
+
+                    ],
+        "answer":  "15.2;18.67",
+        "solution":  "【分析】求比7.2多8的数：比7.2多8，就是在7.2的基础上加上8，用加法计算;求15.78比哪个数少2.89:15.78比一个数少2.89，那么这个数就比15.78多2.89，用加法计算.\n【详解】7.2+8=15.2;15.78+2.89=18.67;所以，比7.2多8的数是15.2;15.78比18.67少2.89。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法",
+                                 "退位减法"
+                             ],
+        "source":  "24-25四年级下·安徽安庆·期末"
+    },
+    {
+        "id":  "338",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "红领巾是少先队员的标志，分为大号和小号两种规格。红领巾按角分是()三角形，最小的角是30°,其他两个角分别是()°和()°;按边分是()三角形，大号红领巾的两条边分别是1.2m和72cm,大号红领巾的周长是()m。![](images/55e3b10be349af36bea986aaac95456392589b826003b0c5175cc045e3d09786.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "钝角;30;120;等腰;2.64",
+        "solution":  "【分析】首先红领巾是等腰三角形，然后等腰三角形的特征：两底角相等;再根据三角形内角和是180°和一个角是30°,先求得两个底角的度数和，进而求得它的顶角的度数。根据1m=100cm,72cm等于0.72m,根据红领巾标准尺寸，大号红领巾底边为1.2m,腰长为0.72m,等腰三角形周长为两腰加底边即可.\n【详解】180°-30°×2=180°-60°=120°,72cm=0.72m,0.72+0.72+1.2=1.44+1.2=2.64(m)。红领巾是少先队员的标志，分为大号和小号两种规格。红领巾按角分是钝角三角形，最小的角是30°,其他两个角分别是30°和120°;按边分是等腰三角形，大号红领巾的两条边分别是1.2m和72cm,大号红领巾的周长是2.64m。",
+        "knowledge_points":  [
+                                 "多位小数的不进位加法",
+                                 "不退位减法",
+                                 "三角形的周长",
+                                 "三角形的内角和",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·山西忻州·期末"
+    },
+    {
+        "id":  "339",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "4.8km=()m;2公顷80平方米=()公顷",
+        "options":  [
+
+                    ],
+        "answer":  "4800;2.008",
+        "solution":  "【分析】1km=1000m,1公顷=10000平方米，高级单位到低级单位，乘进率;低级单位到高级单位，除以进率。据此解答.\n【详解】(1)1km=1000m,4.8×1000=4800,4.8km=4800m;(2)1公顷=10000平方米,80÷10000=0.008,80平方米=0.008公顷,2公顷80平方米=2公顷+0.008公顷=2.008公顷。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "面积单位间的进率及换算",
+                                 "千米和米之间的进率与换算",
+                                 "多位小数的不进位加法",
+                                 "不退位减法"
+                             ],
+        "source":  "24-25四年级下·广东汕头·期末"
+    },
+    {
+        "id":  "340",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "1.68里面有()个0.01，再添上()个0.01就是2。",
+        "options":  [
+
+                    ],
+        "answer":  "168;32",
+        "solution":  "【分析】由数位表可知小数点后边的第一位是十分位，十分位的计数单位是1÷10(或0.1),第二位是百分位，百分位的计数单位是1÷100(或0.01)。1.68里面有8个0.01,6个0.1和1个1，相邻计数单位之间的进率都是10,1个1相当于100个0.01，所以1.68里面有168个0.01;求1.68添加多少个0.01是2，用2-1.68=0.32,0.32里面有32个0.01，据此解答即可.\n【详解】1.68里面有168个0.01，再添上32个0.01就是2。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法",
+                                 "退位减法",
+                                 "小数的数位和计数单位的认识",
+                                 "小数的意义"
+                             ],
+        "source":  "24-25四年级下·山东临沂·期末"
+    },
+    {
+        "id":  "341",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "十分位上的3比十位上的3少(),0.9与最小的两位小数之和是()。",
+        "options":  [
+
+                    ],
+        "answer":  "29.7;0.91",
+        "solution":  "【分析】十分位上的3表示3个0.1，即0.3，十位上的3表示3个十，即30，两者相差30-0.3=29.7;最小的两位小数，整数部分和十分位都应该最小，为0，百分位上为1，即0.01，用0.9+0.01可算出两数之和.\n【详解】根据分析:30-0.3=29.7,0.9+0.01=0.91，所以十分位上的3比十位上的3少29.7,0.9与最小的两位小数之和是0.91。",
+        "knowledge_points":  [
+                                 "多位小数的不进位加法",
+                                 "不退位减法",
+                                 "一位小数的进位加法",
+                                 "退位减法",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·广东汕头·期末"
+    },
+    {
+        "id":  "342",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "比2.9多()的数是8;()比3.24多6.78。",
+        "options":  [
+
+                    ],
+        "answer":  "5.1;10.02",
+        "solution":  "【分析】求比2.9多多少等于8，就是求8减去2.9的差，用减法计算;求比3.24多6.78的数，就是求3.24与6.78的和，用加法计算.\n【详解】8-2.9=5.1,3.24+6.78=10.02，即比2.9多5.1的数是8,10.02比3.24多6.78。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法",
+                                 "退位减法",
+                                 "一位小数的进位加法",
+                                 "退位减法"
+                             ],
+        "source":  "24-25四年级下·广东汕头·期末"
+    },
+    {
+        "id":  "343",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小马虎在计算一道小数加法题时，误将其中一个加数3.3的小数点看漏了，结果得50.82，正确的结果是()。",
+        "options":  [
+
+                    ],
+        "answer":  "21.12",
+        "solution":  "【分析】根据题意，小马虎将加数3.3的小数点看漏，导致该数被误认为33。根据错误结果50.82，用50.82减去33，可先求出另一个正确的加数，再加上3.3，计算正确结果。以此答题即可.\n【详解】50.82-33=17.82;17.82+3.3=21.12。小马虎在计算一道小数加法题时，误将其中一个加数3.3的小数点看漏了，结果得50.82，正确的结果是21.12。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法",
+                                 "退位减法",
+                                 "多位小数的不进位加法",
+                                 "不退位减法"
+                             ],
+        "source":  "24-25四年级下·山东临沂·期末"
+    },
+    {
+        "id":  "344",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "陈阿姨到超市买菜，买西红柿要付14.23元，买牛肉要付45.77元。陈阿姨共要付()元，她给营业员100元，应找回()元。",
+        "options":  [
+
+                    ],
+        "answer":  "60;40",
+        "solution":  "【分析】求一共付多少元，需将所购买东西的价钱相加得到总和;求找回多少元，将所付钱数减去应付的钱数即可;据此作答.\n【详解】一共要付的钱数:14.23+45.77=60(元);找回的钱数:100-60=40(元);因此陈阿姨共要付60元，应找回40元。",
+        "knowledge_points":  [
+                                 "十元以上人民币的认识与计算",
+                                 "多位小数的进位加法",
+                                 "退位减法"
+                             ],
+        "source":  "24-25四年级下·广东佛山·期末"
+    },
+    {
+        "id":  "345",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "有三袋小麦，甲袋和乙袋共重18kg,乙袋和丙袋共重18.3kg,甲袋和丙袋共重17.7kg。甲袋小麦重()kg,乙袋小麦重()kg,丙袋小麦重()kg。",
+        "options":  [
+
+                    ],
+        "answer":  "8.7;9.3;9",
+        "solution":  "【分析】根据题意，有三袋小麦，甲袋和乙袋共重18kg,乙袋和丙袋共重18.3kg,甲袋和丙袋共重17.7kg。把三个数相加，求出和，就是三袋总重量的2倍，用总和除以2，求出甲，乙，丙三袋的重量;再减去18，求出丙袋的重量;用总和减去18.3，求出甲袋的重量;用总和减去17.7，求出乙袋的重量，列式计算即可.\n【详解】(18+18.3+17.7)÷2=(36.3+17.7)÷2=54÷2=27(千克);27-18=9(千克);27-18.3=8.7(千克);27-17.7=9.3(千克)。有三袋小麦，甲袋小麦重8.7kg,乙袋小麦重9.3kg,丙袋小麦重9kg。",
+        "knowledge_points":  [
+                                 "一位小数的进位加法",
+                                 "退位减法"
+                             ],
+        "source":  "24-25四年级下·广东佛山·期末"
+    },
+    {
+        "id":  "346",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小丽的妈妈用传统蜡染技艺制作一条长3.6米的布料，第一次剪下0.93米做杯垫，第二次剪下1.56米做桌布，剩下的布料还有()米。",
+        "options":  [
+
+                    ],
+        "answer":  "1.11",
+        "solution":  "【分析】根据题意，小丽的妈妈用传统蜡染技艺制作一条长3.6米的布料，第一次剪下0.93米做杯垫，第二次剪下1.56米做桌布，用3.6减去0.93，再减去1.56，就是剩下的布料.\n【详解】3.6-0.93-1.56=2.67-1.56=1.11(米)。剩下的布料还有1.11米。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法",
+                                 "退位减法"
+                             ],
+        "source":  "24-25四年级下·贵州黔东南·期末"
+    },
+    {
+        "id":  "347",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "4个一,4个百分之一,5个千分之一组成的数是(),它的计数单位是(),至少减去()个这样的计数单位，这个数就成了自然数，组成的数中两个\"4\"相差()。",
+        "options":  [
+
+                    ],
+        "answer":  "4.045;0.001;45;3.96",
+        "solution":  "【分析】根据小数的组成和计数单位定义,4个一对应个位4,4个百分之一对应百分位4,5个千分之一对应千分位5，组合成4.045。计数单位由最低位千分位决定，为0.001。将4.045变为自然数需减去小数部分0.045，即45个0.001。两个\"4\"分别位于个位和百分位，数值差为4-0.04=3.96.\n【详解】4-0.04=3.96;根据分析:4个一,4个百分之一,5个千分之一组成的数是4.045，它的计数单位是0.001，至少减去45个这样的计数单位，这个数就成了自然数，组成的数中两个\"4\"相差3.96。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法",
+                                 "退位减法",
+                                 "小数的数位和计数单位的认识",
+                                 "小数的意义"
+                             ],
+        "source":  "24-25四年级下·湖北孝感·期末"
+    },
+    {
+        "id":  "348",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "1.9里面有()个0.1，如果把1.9的小数点去掉，得到的数就()(填\"扩大\"或\"缩小\")到原来的()倍，与原数相差()。",
+        "options":  [
+
+                    ],
+        "answer":  "19;扩大;10;17.1",
+        "solution":  "【分析】1.9由1个1和9个0.1组成,1个1等于10个0.1，因此共有10+9=19个0.1。去掉1.9的小数点后变为19，相当于将原数的小数点向右移动一位，即扩大到原来的10倍，然后计算19与1.9的差即可.\n【详解】19-1.9=17.1,1.9里面有19个0.1，如果把1.9的小数点去掉，得到的数就扩大到原来的10倍，与原数相差17.1。",
+        "knowledge_points":  [
+                                 "小数点向右移动引起小数大小变化的规律",
+                                 "一位小数的进位加法",
+                                 "退位减法",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·广东东莞·期末"
+    },
+    {
+        "id":  "349",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "东门路的美食街上，有很多传统美食，比如:(1)肉圆粿12.50元/份，读作()元/份。(2)肉圆粿和铅山烫粉的价格相比,()更贵，每份贵()元。(3)小丽买了1斤艾米果和一些天桂梨，在计算价钱时误把12.7看成了127，算得的结果是167元。她买了()元天桂梨，正确的总价应是()元。",
+        "options":  [
+
+                    ],
+        "answer":  "十二点五零;肉圆粿;1.7;40;52.7",
+        "solution":  "【分析】(1)小数的读法是整数部分按照整数的读法来读，小数点读作\"点\"，小数部分依次读出每一位上的数字。根据小数读法读出小数。(2)根据小数大小的比较方法比较，再用减法算出贵多少元。(3)小丽误把艾米果价格12.7看成127，算得结果是167元。那么她买天桂梨花费的钱数为167-127=40元，正确的总价应该是12.7+40.\n【详解】(1)12.50读作十二点五零。(2)12.50-10.80=1.7(元),所以肉圆粿更贵，每份贵1.7元。(3)167-127=40(元),12.7+40=52.7(元),所以她买了40元天桂梨，正确总价是52.7元。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法",
+                                 "退位减法",
+                                 "多位小数的大小比较",
+                                 "小数的读法和写法"
+                             ],
+        "source":  "24-25四年级下·江西上饶·期末"
+    },
+    {
+        "id":  "350",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小雨在计算一道小数减法题时，把减数2.13错看成了2.73，结果得16.48，这道题的正确结果是()。",
+        "options":  [
+
+                    ],
+        "answer":  "17.08",
+        "solution":  "【分析】先根据错误的减数和错误结果求出被减数，再用被减数减去正确减数即可。被减数=差+减数.\n【详解】16.48+2.73=19.21,19.21-2.13=17.08。这道题的正确结果是17.08。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法",
+                                 "退位减法"
+                             ],
+        "source":  "24-25四年级下·陕西宝鸡·期末"
+    },
+    {
+        "id":  "351",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "妈妈每年都会在小颖生日那天给她测量身高，去年测得她的身高是1m36cm,今年是142cm,这一年小颖长高了()m。",
+        "options":  [
+
+                    ],
+        "answer":  "0.06",
+        "solution":  "【分析】1m=100cm。把去年身高1m36cm化成米是1.36m,把今年142cm化成米是1.42m,再相减.\n【详解】36cm=0.36m,所以1m36cm=1.36m;142cm=1.42m;1.42-1.36=0.06(m)。这一年小颖长高了0.06m。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "多位小数的进位加法",
+                                 "退位减法"
+                             ],
+        "source":  "24-25四年级下·贵州遵义·期末"
+    },
+    {
+        "id":  "352",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一桶油，连桶重15.5千克。用去一半油后，连桶重8.3千克。原来桶里有油()千克，桶重()千克。",
+        "options":  [
+
+                    ],
+        "answer":  "14.4;1.1",
+        "solution":  "【分析】连桶重15.5千克，用去一半后连桶8.3千克，所以喝掉的一半油重15.5-8.3=7.2千克;整桶油重7.2×2=14.4千克;桶重=15.5-14.4.\n【详解】15.5-8.3=7.2(千克);7.2+7.2=14.4(千克);15.5-14.4=1.1(千克)。原来桶里有油14.4千克，桶重1.1千克。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·浙江杭州·期末"
+    },
+    {
+        "id":  "353",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "列竖式计算小数加减法时,()要对齐，也就是()数位要对齐。",
+        "options":  [
+
+                    ],
+        "answer":  "小数点;相同",
+        "solution":  "【分析】小数加减法竖式计算的关键是先把小数点对齐，即相同数位对齐，再按整数加减法法则计算.\n【详解】例如:15.21+2.9=18.11![](images/2026-03-12%20205911.jpg);25.41-2.6=22.81![](images/2026-03-12%20205916.jpg)。列竖式计算小数加减法时，小数点要对齐，也就是相同数位要对齐。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法",
+                                 "退位减法"
+                             ],
+        "source":  "24-25四年级下·陕西西安·期末"
+    },
+    {
+        "id":  "354",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "0.80的计数单位是(),再加上()个这样的计数单位就得到1。",
+        "options":  [
+
+                    ],
+        "answer":  "0.01(或百分之一或1÷100);20",
+        "solution":  "【分析】0.80是两位小数，计数单位是0.01(百分之一)。先算1-0.80=0.20，再看0.20里有多少个0.01.\n【详解】0.80是两位小数，它的计数单位是0.01(或百分之一或1÷100);1-0.80=0.20,0.20里面有20个0.01，所以再加20个这样的计数单位就得到1。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法",
+                                 "退位减法",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "23-24四年级下·云南昭通·期末"
+    },
+    {
+        "id":  "355",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "29km80m=()km;560kg=()t",
+        "options":  [
+
+                    ],
+        "answer":  "29.08;0.56",
+        "solution":  "【分析】1km=1000m,80m=0.08km,所以29km80m=29.08km;1t=1000kg,560kg=0.56t.\n【详解】80m=0.08km,29km80m=29+0.08=29.08km;560kg=0.56t。",
+        "knowledge_points":  [
+                                 "小数点向左移动引起小数大小变化的规律",
+                                 "千米和米之间的进率与换算",
+                                 "吨，千克之间的换算与比较",
+                                 "多位小数的不进位加法",
+                                 "不退位减法"
+                             ],
+        "source":  "23-24四年级下·云南昭通·期末"
+    },
+    {
+        "id":  "356",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一箱苹果连箱共重30.3千克，倒出一半苹果后连箱共重16.3千克，全部苹果重()千克，箱子重()千克。",
+        "options":  [
+
+                    ],
+        "answer":  "28;2.3",
+        "solution":  "【分析】连箱质量差30.3-16.3是一半苹果重，再乘2得全部苹果重;连箱重减苹果重得箱重.\n【详解】30.3-16.3=14(千克);14×2=28(千克);30.3-28=2.3(千克)。全部苹果重28千克，箱子重2.3千克。",
+        "knowledge_points":  [
+                                 "一位小数的不进位加法",
+                                 "不退位减法",
+                                 "两位数与一位数的乘法口算"
+                             ],
+        "source":  "24-25四年级下·贵州遵义·期末"
+    },
+    {
+        "id":  "357",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "360平方千米=()公顷;3吨54千克=()吨;3.2千米=()米;10元8角=()元;4.07平方米=()平方分米=()平方厘米",
+        "options":  [
+
+                    ],
+        "answer":  "36000;3.054;3200;10.8;407;40700",
+        "solution":  "【分析】1平方千米=100公顷,1000千克=1吨,1千米=1000米,10角=1元,1平方米=100平方分米=10000平方厘米，大单位到小单位乘进率，小单位到大单位除以进率.\n【详解】360平方千米=360×100=36000公顷;3吨54千克=3+54÷1000=3.054吨;3.2千米=3.2×1000=3200米;10元8角=10+8÷10=10.8元;4.07平方米=4.07×100=407平方分米=407×100=40700平方厘米。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "平方厘米，平方分米，平方米之间的进率与换算",
+                                 "吨，千克之间的换算与比较",
+                                 "多位小数的不进位加法",
+                                 "不退位减法"
+                             ],
+        "source":  "24-25四年级下·甘肃平凉·期末"
+    },
+    {
+        "id":  "358",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "找规律填数。(1)3.31,3.86,4.41,4.96,(),()。(2)8.799,8.792,8.785,8.778,(),()。",
+        "options":  [
+
+                    ],
+        "answer":  "5.51;6.06;8.771;8.764",
+        "solution":  "【分析】(1)相邻两项都增加0.55;(2)相邻两项都减少0.007.\n【详解】(1)4.96+0.55=5.51,5.51+0.55=6.06;(2)8.778-0.007=8.771,8.771-0.007=8.764。",
+        "knowledge_points":  [
+                                 "小数排列的规律",
+                                 "多位小数的进位加法",
+                                 "退位减法"
+                             ],
+        "source":  "24-25四年级下·甘肃定西·期末"
+    },
+    {
+        "id":  "359",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "马拉松是一项风靡全球的长跑比赛项目，它代表毅力，勇气和超越自我。马拉松全程是42195m,用km作单位是()km;鸵鸟是世界上最大的鸟，鸵鸟蛋可达1kg450g,也就是()kg。",
+        "options":  [
+
+                    ],
+        "answer":  "42.195;1.45",
+        "solution":  "【分析】1000m=1km,1000g=1kg,小单位换算成大单位除以进率.\n【详解】42195÷1000=42.195(km);450÷1000=0.45(kg);1kg450g=1+0.45=1.45kg。马拉松全程是42.195km;鸵鸟蛋是1.45kg。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "千米和米之间的进率与换算",
+                                 "克，千克之间的换算与比较",
+                                 "多位小数的不进位加法",
+                                 "不退位减法"
+                             ],
+        "source":  "24-25四年级下·辽宁盘锦·期末"
+    },
+    {
+        "id":  "360",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "64缩小到它的1÷1000是();0.78扩大到原数的100倍是(),扩大后的数比原数增加了()。",
+        "options":  [
+
+                    ],
+        "answer":  "0.064;78;77.22",
+        "solution":  "【分析】小数点向左移动三位是缩小到1÷1000，向右移动两位是扩大到100倍。最后用扩大后的数减去原数.\n【详解】64缩小到它的1÷1000是0.064;0.78扩大到原数的100倍是78;78-0.78=77.22，所以扩大后的数比原数增加了77.22。",
+        "knowledge_points":  [
+                                 "小数点向左移动引起小数大小变化的规律",
+                                 "小数点向右移动引起小数大小变化的规律",
+                                 "多位小数的进位加法",
+                                 "退位减法"
+                             ],
+        "source":  "24-25四年级下·甘肃兰州·期末"
+    },
+    {
+        "id":  "361",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "已知甲数是3.27，比乙数多1.78，甲，乙两数的和是()。",
+        "options":  [
+
+                    ],
+        "answer":  "4.76",
+        "solution":  "【分析】先用甲数减去1.78求乙数，再把甲乙两数相加.\n【详解】3.27-1.78+3.27=1.49+3.27=4.76，所以甲，乙两数的和是4.76。",
+        "knowledge_points":  [
+                                 "利用小数的加，减法混合运算解决实际问题",
+                                 "多位小数的进位加法",
+                                 "退位减法"
+                             ],
+        "source":  "24-25四年级下·山东济南·期末"
+    },
+    {
+        "id":  "362",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "先列出综合算式，再计算。从106.5里减去12.58与3.42的和，差是多少。",
+        "options":  [
+
+                    ],
+        "answer":  "106.5-(12.58+3.42);90.5",
+        "solution":  "【分析】先算12.58与3.42的和，再用106.5减去这个和.\n【详解】106.5-(12.58+3.42)=106.5-16=90.5，所以差是90.5。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算",
+                                 "多位小数的进位加法",
+                                 "退位减法"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "363",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "竖式计算。34.7+45.38=();23.68+5.62=();20-9.46=();30.8-7.23=()。",
+        "options":  [
+
+                    ],
+        "answer":  "80.08;29.3;10.54;23.57",
+        "solution":  "【分析】小数加减法先把小数点对齐，再按整数加减法计算，最后点上小数点.\n【详解】34.7+45.38=80.08![](images/2026-03-12%20210351.jpg);23.68+5.62=29.3![](images/2026-03-12%20210357.jpg);20-9.46=10.54![](images/2026-03-12%20210403.jpg);30.8-7.23=23.57![](images/2026-03-12%20210410.jpg)。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法",
+                                 "退位减法"
+                             ],
+        "source":  "24-25四年级下·吉林·期末"
+    },
+    {
+        "id":  "364",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "列竖式计算。30.2-9.89=();20-10.87=();14.56+9.6=()。",
+        "options":  [
+
+                    ],
+        "answer":  "20.31;9.13;24.16",
+        "solution":  "【分析】小数加减法先把小数点对齐，再按整数法则计算，最后在得数里点小数点.\n【详解】30.2-9.89=20.31![](images/2026-03-12%20210616.jpg);20-10.87=9.13![](images/2026-03-12%20210625.jpg);14.56+9.6=24.16![](images/2026-03-12%20210642.jpg)。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法",
+                                 "退位减法"
+                             ],
+        "source":  "24-25四年级下·湖南岳阳·期末"
+    },
+    {
+        "id":  "365",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "计算下面各题，能简算的要简算。78.12-0.87-0.13;102×45;3600÷45÷2;158×[(27+153)÷9]。",
+        "options":  [
+
+                    ],
+        "answer":  "77.12;4590;40;3160",
+        "solution":  "【分析】78.12-0.87-0.13用减法性质a-b-c=a-(b+c);102×45用乘法分配律;3600÷45÷2用除法性质a÷b÷c=a÷(b×c);最后一题按运算顺序先括号后乘法.\n【详解】78.12-0.87-0.13=78.12-(0.87+0.13)=78.12-1=77.12;102×45=(100+2)×45=100×45+2×45=4500+90=4590;3600÷45÷2=3600÷(45×2)=3600÷90=40;158×[(27+153)÷9]=158×[180÷9]=158×20=3160。",
+        "knowledge_points":  [
+                                 "与小数减法相关的简便运算",
+                                 "整数除法的性质",
+                                 "整数乘法分配律",
+                                 "带有中括号的混合运算"
+                             ],
+        "source":  "24-25四年级下·山东临沂·期末"
+    },
+    {
+        "id":  "366",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "计算题",
+        "difficulty":  0.65,
+        "question":  "比3.7多2.05的数减去1.8，差是多少。",
+        "options":  [
+
+                    ],
+        "answer":  "3.95",
+        "solution":  "【分析】先求比3.7多2.05的数，再减去1.8.\n【详解】3.7+2.05-1.8=5.75-1.8=3.95，所以差是3.95。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算",
+                                 "多位小数的进位加法",
+                                 "退位减法"
+                             ],
+        "source":  "24-25三年级下·云南昭通·期末"
+    },
+    {
+        "id":  "367",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "爸爸买了一条鱼和一个西瓜，一条鱼23.12元，一个西瓜24.88元。他付给售货员100元，应找回多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "52元",
+        "solution":  "【分析】先算鱼和西瓜总价，再用100元减去总价.\n【详解】100-(23.12+24.88)=100-48=52(元)。答：他付给售货员100元，应找回52元。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算",
+                                 "多位小数的进位加法",
+                                 "退位减法"
+                             ],
+        "source":  "24-25四年级下·广东肇庆·期末"
+    },
+    {
+        "id":  "368",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "王阿姨用一卷长20米的彩带编织中国结，第一次用去6.35米，第二次用去5.85米，还剩下多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "7.8米",
+        "solution":  "【分析】用总长度依次减去两次用去的长度.\n【详解】20-6.35-5.85=13.65-5.85=7.8(米)。答：还剩下7.8米。",
+        "knowledge_points":  [
+                                 "利用小数的加，减法混合运算解决实际问题",
+                                 "多位小数的进位加法",
+                                 "退位减法"
+                             ],
+        "source":  "24-25四年级下·广东潮州·期末"
+    },
+    {
+        "id":  "369",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "小聪要给\"手拉手\"的小伙伴准备礼物，他带了50元去买文具。买笔记本花了25.6元，买水笔花了18.4元，他还剩下多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "6元",
+        "solution":  "【分析】总钱数依次减去两项花费，也可用减法性质先把两项花费相加再减.\n【详解】50-25.6-18.4=50-(25.6+18.4)=50-44=6(元)。答：他还剩下6元。",
+        "knowledge_points":  [
+                                 "与小数减法相关的简便运算"
+                             ],
+        "source":  "24-25四年级下·广东潮州·期末"
+    },
+    {
+        "id":  "370",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一本《儿童文学》21.9元，一本《科技博览》比一本《儿童文学》便宜3.4元。用42元钱买这两本书，够吗。",
+        "options":  [
+
+                    ],
+        "answer":  "够了",
+        "solution":  "【分析】由题意得，一本《儿童文学》21.9元，一本《科技博览》比一本《儿童文学》便宜3.4元，直接用21.9减去3.4可以算出一本《科技博览》需要多少元，然后加上一本《儿童文学》的价钱即可算出两本书一共需要多少钱。最后再与42元比较大小即可.\n【详解】21.9-3.4+21.9=18.5+21.9=40.4(元),40.4\u003c42。答：用42元钱买这两本书，够了。",
+        "knowledge_points":  [
+                                 "利用小数的加，减法混合运算解决实际问题"
+                             ],
+        "source":  "24-25四年级下·广东江门·期末"
+    },
+    {
+        "id":  "371",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "惠民超市去年全年收入49.27万元，其中上半年收入27.8万元，下半年比上半年少收入多少万元。",
+        "options":  [
+
+                    ],
+        "answer":  "6.33万元",
+        "solution":  "【分析】已知惠民超市去年全年收入49.27万元，其中上半年收入27.8万元，用减法先算出下半年收入是多少万元，再与上半年收入作差，即可求出下半年比上半年少收入多少万元.\n【详解】49.27-27.8=21.47(万元),27.8-21.47=6.33(万元)。答：下半年比上半年少收入6.33万元。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·安徽安庆·期末"
+    },
+    {
+        "id":  "372",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一本书原价45元，第一次降价4.2元，第二次降价1.25元。这本书现在卖价多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "39.55元",
+        "solution":  "【分析】用原价依次减去两次降价的钱即为这本书现在卖价多少元.\n【详解】45-4.2-1.25=40.8-1.25=39.55(元)。答：这本书现在卖价39.55元。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·海南省直辖县级单位·期末"
+    },
+    {
+        "id":  "373",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "同学们体育测试跳远项目，王平跳了1.63米，李强比王平远0.19米，张萌跳得比李强近0.25米，张萌跳了多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "1.57米",
+        "solution":  "【分析】李强比王平远0.19米，张萌跳得比李强近0.25米，所以王平跳的距离加0.19米等于李强跳的距离，李强跳的距离减去0.25米等于张萌跳的距离，据此即可解答.\n【详解】李强:1.63+0.19=1.82(米);张萌:1.82-0.25=1.57(米)。答：张萌跳了1.57米。",
+        "knowledge_points":  [
+                                 "利用小数加减法解决实际问题",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·广东佛山·期末"
+    },
+    {
+        "id":  "374",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "妈妈的电子钱包里面有60元，去超市买了一盏台灯和一个垃圾桶，用电子钱包付款后，电子钱包里面还剩下多少钱。\n![](images/2026-03-12%20210905.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "21.3元",
+        "solution":  "【分析】根据题图可知，一盏台灯的价钱是27.8元，一个垃圾桶的价钱是12.9元，将买的一盏台灯和一个垃圾桶的价钱相加，先算出购买一盏台灯和一个垃圾桶需要的钱数;又已知满30元减2元，则用求出的购买一盏台灯和一个垃圾桶需要的钱数与30作比较，如果大于等于30元，则用求出的购买一盏台灯和一个垃圾桶需要的钱数减去2，再与电子钱包里原来有的钱数作差，即可求出电子钱包里面还剩下多少钱;如果小于30元，则用求出的购买一盏台灯和一个垃圾桶需要的钱数与电子钱包里原来有的钱数作差，即可求出电子钱包里面还剩下多少钱.\n【详解】27.8+12.9=40.7(元),40.7\u003e30,60-(40.7-2)=60-38.7=21.3(元)。答：电子钱包里面还剩下21.3元。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算",
+                                 "一位小数的进位加法，退位减法",
+                                 "一位小数的不进位加法，不退位减法",
+                                 "一位小数的大小比较"
+                             ],
+        "source":  "24-25四年级下·湖北孝感·期末"
+    },
+    {
+        "id":  "375",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "修路队修补一段公路，已经修补好了9.25千米，比未修补的多1.8千米。这段公路全长是多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "16.7千米",
+        "solution":  "【分析】由题意可知：用已经修补好的千米数减去1.8，求出未修补的千米数，再加上已经修补好的千米数，即可解答.\n【详解】9.25-1.8+9.25=7.45+9.25=16.7(千米)。答：这段公路全长是16.7千米。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·广东汕头·期末"
+    },
+    {
+        "id":  "376",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一根绳子长15.28米，第一次剪下3.5米，第二次剪下5.5米，余下的绳子长多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "6.28米",
+        "solution":  "【分析】由题意得，一根绳子长15.28米，第一次剪下3.5米，第二次剪下5.5米。求余下的绳子长多少米，直接用绳子的总长度减去第一次剪下的长度再减去第二次剪下的长度即可算出余下的绳子长多少米。计算时，利用减法的性质可使计算简便.\n【详解】15.28-(3.5+5.5)=15.28-9=6.28(米)。答：余下的绳子长6.28米。",
+        "knowledge_points":  [
+                                 "与小数减法相关的简便运算"
+                             ],
+        "source":  "24-25四年级下·广东汕头·期末"
+    },
+    {
+        "id":  "377",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "把一根长3米的竹竿垂直放入水池中，竹竿入泥部分是0.36米，露出水面0.7米。水池中的水深多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "1.94米",
+        "solution":  "【分析】竹竿的总长度=入泥部分的深度+水深+露出水面的长度，所以水深=竹竿的总长度-入泥部分的深度-露出水面的长度，代入数据即可.\n【详解】3-0.36-0.7=2.64-0.7=1.94(米)。答：水池中的水深1.94米。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·内蒙古赤峰·期末"
+    },
+    {
+        "id":  "378",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "化肥厂去年十二月份计划生产化肥80吨，结果上旬生产了25.8吨，比中旬少生产了2.8吨，下旬再生产多少吨就完成任务。",
+        "options":  [
+
+                    ],
+        "answer":  "25.6吨",
+        "solution":  "【分析】由题意可知：用25.8+2.8求出中旬生产的数量，再用计划生产的总数量分别减去上旬和中旬生产的数量，即可求解.\n【详解】25.8+2.8=28.6(吨),80-28.6-25.8=51.4-25.8=25.6(吨)。答：下旬再生产25.6吨就完成任务。",
+        "knowledge_points":  [
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·广东汕头·期末"
+    },
+    {
+        "id":  "379",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "微信是一款在我国使用非常广泛的社交软件，使用微信支付既简单又便捷。妈妈的微信钱包里有300元钱，某电商平台周年庆促销期间，妈妈领取了一张\"满200元减16.8元\"的优惠券，并购买了一台电风扇和一盏台灯，妈妈用微信付完款后，微信钱包里还剩多少钱。\n![](images/2d9dab78a58cc9d6d4ae21b20aae4001.png)",
+        "options":  [
+
+                    ],
+        "answer":  "116.8元",
+        "solution":  "【分析】根据题意，已知一台电风扇的单价是175.5元，一盏台灯的单价是24.5元，把两数相加，就是商品的总价;已知妈妈领取了一张\"满200元减16.8元\"的优惠券，求出的总价如果大于等于200，就减去16.8元，求出实际的付款数，最后用300减去付款数，就是剩余的金额，列式计算即可.\n【详解】300-(175.5+24.5-16.8)=300-(200-16.8)=300-183.2=116.8(元)。答：微信钱包里还剩116.8元钱。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·贵州遵义·期末"
+    },
+    {
+        "id":  "380",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "截至2025年2月20日21时10分，全球动画电影票房榜第一名为《哪吒之魔童闹海》,总票房为126.93亿元人民币，第二名为《头脑特工队2》,票房为123.19亿元人民币，第一名票房比第二名票房多多少亿元人民币。",
+        "options":  [
+
+                    ],
+        "answer":  "3.74亿元",
+        "solution":  "【分析】根据题意，用第一名的总票房减去第二名的总票房，即可求出第一名票房比第二名票房多多少亿元人民币.\n【详解】126.93-123.19=3.74(亿元)。答：第一名票房比第二名票房多3.74亿元人民币。",
+        "knowledge_points":  [
+                                 "利用小数加减法解决实际问题",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·江西南昌·期末"
+    },
+    {
+        "id":  "381",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "地球像一颗巨大的蓝色水球。它表面七成是海洋，三成是陆地。海洋是天气的\"空调\"和鱼儿的家，陆地是我们的家园。爱护地球很简单：不往海里扔垃圾，随手关灯关水。现在，来算算它们有多大吧!根据最新数据，地球的海洋面积约是3.61亿平方千米，比陆地面积约多2.12亿平方千米，地球总面积约是多少亿平方千米。\n![](images/13722b91f34cd49f77bc524ab3057a26f5bb8efdd9ed0b3af1c1b590f32d7030.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "5.1亿平方千米",
+        "solution":  "【分析】根据题意，用海洋的面积减去海洋比陆地多的面积就是陆地的面积，再用陆地的面积加上海洋的面积就是地球的总面积.\n【详解】3.61-2.12+3.61=1.49+3.61=5.1(亿平方千米)。答：地球总面积约是5.1亿平方千米。",
+        "knowledge_points":  [
+                                 "利用小数的加，减法混合运算解决实际问题",
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "382",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一本《中华上下五千年》24.8元，比一本《科学家》(少年版)便宜3.6元。小赵用50元买这两本书，够吗。",
+        "options":  [
+
+                    ],
+        "answer":  "不够",
+        "solution":  "【分析】由题意得，一本《中华上下五千年》24.8元，比一本《科学家》(少年版)便宜3.6元，那么一本《科学家》(少年版)比一本《中华上下五千年》贵3.6元。可以先用24.8加上3.6算出一本《科学家》需要多少钱，然后再加上一本《中华上下五千年》的钱数算出两本书一共需要多少钱。最后再把得数与50元作比较即可.\n【详解】24.8+3.6+24.8=28.4+24.8=53.2(元),53.2\u003e50，不够。答：小赵用50元买这两本书，不够。",
+        "knowledge_points":  [
+                                 "利用小数的加，减法混合运算解决实际问题"
+                             ],
+        "source":  "24-25四年级下·贵州黔东南·期末"
+    },
+    {
+        "id":  "383",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "某网店3周年店庆促销，购物每满100元减20元.\n![](images/b69019a9fbe69b90d6e5b5e4bb8f42b4.png)\n(1)爸爸准备买一个电风扇和一个电饭煲。他应付多少钱。\n(2)王阿姨买了其中两个电器，实际付的钱没有超过100元。她可能买了哪两个电器。",
+        "options":  [
+
+                    ],
+        "answer":  "172.99元;可能买了一个电风扇和一个吹风机，也可能买了一个吹风机和一个烧水壶",
+        "solution":  "【分析】(1)由题意得，一个电风扇73.99元，一个电饭煲139.00元。爸爸准备买一个电风扇和一个电饭煲，直接用加法即可算出一共需要多少钱。该网店每满100元减20元，可以看前面的得数里面有几个100，那么就可以减几个20元，据此算出爸爸应该付多少钱。(2)由题意得，王阿姨买了其中两个电器，实际付的钱没有超过100元，那么这两件商品的价格之和没有超过100元或者减去优惠的钱后没有超过100元，据此找出王阿姨可能买的两件电器即可.\n【详解】(1)73.99+139=212.99(元),212.99元里面有2个100元，所以应该优惠2个20元,20×2=40(元),212.99-40=172.99(元)。答：爸爸应付172.99元。(2)①买一个电风扇和一个吹风机:73.99+42.5=116.49(元),116.49-20=96.49(元),96.49元\u003c100元，满足题意。②买一个吹风机和一个烧水壶:42.50+56.50=99(元),99元\u003c100元，满足题意。答：她可能买了一个电风扇和一个吹风机，也可能买了一个吹风机和一个烧水壶。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "24-25四年级下·云南昭通·期末"
+    },
+    {
+        "id":  "384",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "妈妈买水果花了25.8元，买蔬菜比买水果少花6.5元。妈妈买蔬菜和水果一共花了多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "45.1元",
+        "solution":  "【分析】根据题意，用25.8-6.5，求出买蔬菜的钱数，再用买蔬菜的钱数加上买水果的钱数，即可求出妈妈买蔬菜和水果一共花了多少钱.\n【详解】25.8+(25.8-6.5)=25.8+19.3=45.1(元)。答：妈妈买蔬菜和水果一共花了45.1元钱。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·河北张家口·期末"
+    },
+    {
+        "id":  "385",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "过春节时，妈妈在一家三口的家庭群里发了一个88元的红包，爸爸抢到33.25元，华华抢到19.67元，妈妈会抢到多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "35.08元",
+        "solution":  "【分析】用红包的总金额依次减去爸爸和华华抢到的金额，剩下的就是妈妈抢到的金额.\n【详解】88-33.25-19.67=54.75-19.67=35.08(元)。答：妈妈会抢到35.08元。",
+        "knowledge_points":  [
+                                 "利用小数的加，减法混合运算解决实际问题"
+                             ],
+        "source":  "24-25四年级下·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "386",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "小明到书店买了一本《新华字典》和一本《少儿百科全书》,《新华字典》16.5元，小明付了30元，营业员找他2.8元。《少儿百科全书》的单价是多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "10.7元",
+        "solution":  "【分析】由题意得，小明付了30元，营业员找他2.8元，可以先用30减去2.8算出一本《新华字典》和一本《少儿百科全书》一共花了多少钱。一本《新华字典》需要16.5元，再用前面的得数减去16.5即可算出《少儿百科全书》的单价是多少元.\n【详解】30-2.8-16.5=27.2-16.5=10.7(元)。答:《少儿百科全书》的单价是10.7元。",
+        "knowledge_points":  [
+                                 "利用小数的加，减法混合运算解决实际问题"
+                             ],
+        "source":  "24-25四年级下·湖北武汉·期末"
+    },
+    {
+        "id":  "387",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "把一根长4米的竹竿垂直放入水池中，竹竿入泥部分是0.36米，露出水面0.89米，水池中的水深多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "2.75米",
+        "solution":  "【分析】竹竿的总长度减去入泥部分的长度和露出水面部分的长度，即为水池中的水深，据此作答.\n【详解】4-0.36-0.89=3.64-0.89=2.75(米)。答：水池中的水深2.75米。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·广东广州·期末"
+    },
+    {
+        "id":  "388",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "明珠小学迎六一，组织四年级学生进行自制动力小船竞速赛。比赛中，乐乐成绩是15.34秒，欢欢比乐乐慢0.76秒，明明比欢欢快1.2秒。明明的成绩是多少秒。",
+        "options":  [
+
+                    ],
+        "answer":  "14.9秒",
+        "solution":  "【分析】在小船竞速赛中，时间越少，成绩越好。先用乐乐的成绩加上0.76秒，求出欢欢的成绩，再用欢欢的成绩减去1.2秒，即可求出明明的成绩。据此作答.\n【详解】15.34+0.76-1.2=16.1-1.2=14.9(秒)。答：明明的成绩是14.9秒。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算",
+                                 "时间快慢的比较"
+                             ],
+        "source":  "24-25四年级下·浙江杭州·期末"
+    },
+    {
+        "id":  "389",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "周一午餐后，明珠小学对各班泔水进行称重。四年级泔水总重14.9千克，其中401班泔水重量为3.55千克,402班泔水重量为4千克,403班泔水比402班少0.85千克。那么404班泔水重量是多少千克。",
+        "options":  [
+
+                    ],
+        "answer":  "4.2千克",
+        "solution":  "【分析】首先计算出403班泔水的重量，就用402班的泔水重量减去0.85;再用四年级总的泔水重量分别减去401班,402班,403班的泔水重量，就是404班的泔水重量.\n【详解】4-0.85=3.15(千克),14.9-3.55-4-3.15=4.2(千克)。答:404班泔水重量是4.2千克。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·浙江杭州·期末"
+    },
+    {
+        "id":  "390",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "在学校庆六一演出中，摄影师操控无人机从76米的高度先下降18.63米进行拍摄，又下降了10.37米进行再次拍摄，拍摄完成后返航。航拍无人机是从多少米的高度返航的。",
+        "options":  [
+
+                    ],
+        "answer":  "47米",
+        "solution":  "【分析】无人机初始高度为76米，先下降18.63米，之后又下降10.37米。用初始高度减去两次下降的高度即等于返航时的高度，计算时可以利用减法性质进行简算.\n【详解】76-18.63-10.37=76-(18.63+10.37)=76-29=47(米)。答：航拍无人机是从47米的高度返航的。",
+        "knowledge_points":  [
+                                 "整数加法运算律推广到小数",
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·山东日照·期末"
+    },
+    {
+        "id":  "391",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一瓶饮料连瓶共重1.05千克，喝了一半后，连瓶共重0.55千克。(1)这瓶饮料喝了多少千克。(2)这个瓶子重多少千克。",
+        "options":  [
+
+                    ],
+        "answer":  "0.5千克;0.05千克",
+        "solution":  "【分析】(1)用饮料连瓶共重的千克数减去剩余的连瓶共重的千克数就是喝了的数量。(2)喝了一半，说明饮料剩下的也是一半，所以用剩下的饮料连瓶共重的千克数减去喝了的饮料即为瓶的重量.\n【详解】(1)1.05-0.55=0.5(千克),答：这瓶饮料喝了0.5千克。(2)0.55-0.5=0.05(千克),答：这个瓶子重0.05千克。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "多位小数的不进位加法，不退位减法"
+                             ],
+        "source":  "24-25四年级下·广东东莞·期末"
+    },
+    {
+        "id":  "392",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "陈阿姨在超市买了一些西兰花和一盒鸡蛋，已知西兰花要付7.2元，鸡蛋要付26.8元。她使用手机钱包付款，手机钱包中原有70.26元，支付完这两件商品后，还剩多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "36.26元",
+        "solution":  "【分析】由题意得，西兰花要付7.2元，鸡蛋要付26.8元，可以先用加法算出西兰花和鸡蛋一共要多少元。手机钱包中原有70.26元，支付完这两件商品后，直接用70.26减去前面的得数即可算出还剩多少钱.\n【详解】70.26-(7.2+26.8)=70.26-34=36.26(元)。答：支付完这两件商品后，还剩36.26元。",
+        "knowledge_points":  [
+                                 "利用小数的加，减法混合运算解决实际问题"
+                             ],
+        "source":  "23-24四年级下·云南昭通·期末"
+    },
+    {
+        "id":  "393",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "妈妈在超市买了一些水果和蔬菜，买水果用去53.4元，买蔬菜用去34.95元，超市促销：满60元减7.9元，妈妈实际支付多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "80.45元",
+        "solution":  "【分析】根据题意，用买水果的钱数加上买蔬菜的钱数，求出一共需要付多少钱，超市促销：满60元减7.9元，如果需要付的钱数大于或等于60元，则减去7.9后是实际需要支付的钱数，如果小于60元，则直接付，据此解答即可.\n【详解】53.4+34.95=88.35(元),88.35\u003e60,88.35-7.9=80.45(元)。答：妈妈实际支付80.45元。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "24-25四年级下·甘肃陇南·期末"
+    },
+    {
+        "id":  "394",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "水果批发市场运来橘子2.43吨，运来的苹果比橘子少0.7吨，比香蕉多0.57吨。水果批发市场运来香蕉多少吨。",
+        "options":  [
+
+                    ],
+        "answer":  "1.16吨",
+        "solution":  "【分析】根据题意，用橘子的吨数-苹果比橘子少的吨数=苹果的吨数。再用苹果的吨数-苹果比香蕉多的吨数=香蕉的吨数.\n【详解】2.43-0.7-0.57=1.73-0.57=1.16(吨)。答：水果批发市场运来香蕉1.16吨。",
+        "knowledge_points":  [
+                                 "利用小数加减法解决实际问题",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·云南昭通·期末"
+    },
+    {
+        "id":  "395",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "王奶奶去医院看病，挂号和病历本共花费5.74元，买了两盒药，一盒24.38元，一盒18.26元，王奶奶这次看病一共花了多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "48.38元",
+        "solution":  "【分析】用挂号和病历本的费用再加上两盒药的价格即可算出.\n【详解】5.74+18.26+24.38=24+24.38=48.38(元)。答：王奶奶这次看病一共花了48.38元。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·云南昭通·期末"
+    },
+    {
+        "id":  "396",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "中国温度最高的地方是新疆吐鲁番的火焰山，夏季最高气温可达47.8摄氏度，而位于黑龙江的漠河夏季的最高气温一般比火焰山低22.2摄氏度。漠河夏季的最高气温一般是多少摄氏度。",
+        "options":  [
+
+                    ],
+        "answer":  "25.6摄氏度",
+        "solution":  "【分析】根据题意，已知夏季最高气温可达47.8摄氏度，而位于黑龙江的漠河夏季的最高气温一般比火焰山低22.2摄氏度。用47.8减去22.2，求出漠河夏季的最高气温，列式计算即可.\n【详解】47.8-22.2=25.6(摄氏度)。答：漠河夏季的最高气温一般是25.6摄氏度。",
+        "knowledge_points":  [
+                                 "多位小数的不进位加法，不退位减法"
+                             ],
+        "source":  "24-25四年级下·云南楚雄·期末"
+    },
+    {
+        "id":  "397",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "工程队挖一条水渠，第一天挖了0.84千米，比第二天少挖0.16千米。工程队两天一共挖了多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "1.84千米",
+        "solution":  "【分析】已知第一天挖了0.84千米，且第一天比第二天少挖0.16千米，这意味着第二天挖的长度比第一天多0.16千米。所以第二天挖的长度为第一天挖的长度加上0.16千米;两天一共挖的长度就是将这两天挖的长度相加.\n【详解】0.84+0.16+0.84=1+0.84=1.84(千米)。答：工程队两天一共挖了1.84千米。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·云南大理·期末"
+    },
+    {
+        "id":  "398",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "上海市星空之境海绵公园占地面积约54.47公顷，其中水域面积是16.17公顷，剩下的是陆域面积，陆域面积比水域面积大多少公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "22.13公顷",
+        "solution":  "【分析】由题意得，可以先用海绵公园的总面积减去水域面积算出陆域面积，接着再用陆域面积减去水域面积即可算出陆域面积比水域面积大多少公顷.\n【详解】54.47-16.17-16.17=38.3-16.17=22.13(公顷)。答：陆域面积比水域面积大22.13公顷。",
+        "knowledge_points":  [
+                                 "利用小数的加，减法混合运算解决实际问题"
+                             ],
+        "source":  "24-25四年级下·甘肃定西·期末"
+    },
+    {
+        "id":  "399",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "王老师买了两本参考书,《小学数学》22.36元,《数学手册》比《小学数学》多4.25元，王老师给了售货员50元，应该找回多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "1.03元",
+        "solution":  "【分析】根据题意,《数学手册》比《小学数学》多4.25元，那么先算出《数学手册》要多少钱，用《小学数学》的钱数+多出的钱数，即22.36+4.25=26.61元，因为王老师买了《数学手册》和《小学数学》两本参考书，所以先算出这两本参考书一共要多少钱，即22.36+26.61=48.97元，再用50元减去买这两本书的钱，就是应该要找回的钱数.\n【详解】50-[(22.36+4.25)+22.36]=50-[26.61+22.36]=50-48.97=1.03(元)。答：应该找回1.03元。",
+        "knowledge_points":  [
+                                 "多位小数的不进位加法，不退位减法",
+                                 "多位小数的进位加法，退位减法",
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·辽宁盘锦·期末"
+    },
+    {
+        "id":  "400",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一根绳子长5米，第一次剪去2.3米，第二次剪去的比第一次短0.6米，两次一共剪去多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "4米",
+        "solution":  "【分析】用第一次剪去的长度减去0.6米求出第二次剪去的长度，然后把两次剪去的长度相加求出两次一共剪去的长度.\n【详解】2.3-0.6+2.3=1.7+2.3=4(米)。答：两次一共剪去4米。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算",
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·浙江绍兴·期末"
+    },
+    {
+        "id":  "401",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "计算5.74吨比3.6吨多多少吨。对如图竖式中描述错误的是().\n![](images/2026-03-12%20220649.jpg)",
+        "options":  [
+                        "A. 2.14中的\"1\"表示0.1吨",
+                        "B. 3.6中的\"6\"表示600千克",
+                        "C. 5.74中的\"5\"表示5吨",
+                        "D. 最后结果表示2吨14千克"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】小数的数位从左到右依次为十分位，百分位等，对应的计数单位分别是0.1,0.01等。在小数减法竖式计算中，相同数位要对齐，然后按照计数单位进行减法运算。我们需要分析竖式中每个数字所在的数位以及对应的计数单位，来判断每个选项描述是否正确.\n【详解】A。2.14中的\"1\"在十分位上，表示0.1吨，所以A选项描述正确。B。3.6中的\"6\"在十分位上，表示0.6吨，根据1吨=1000千克，即0.6吨=600千克，所以B选项描述正确。C。5.74中的\"5\"在个位上，表示5吨，所以C选项描述正确。D。最后结果是2.14吨，根据1吨=1000千克，即2.14吨=2吨140千克，所以D选项描述错误。故答案为:D",
+        "knowledge_points":  [
+                                 "吨，千克之间的换算与比较",
+                                 "多位小数的不进位加法，不退位减法",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "402",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下图竖式中虚线圈起来的部分表示().\n![](images/2026-03-12%20220825.jpg)",
+        "options":  [
+                        "A. 4+1=5",
+                        "B. 0.4+0.1=0.5",
+                        "C. 0.04+0.01=0.05"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】2.40中的4在十分位上，表示0.4;5.16中的1在十分位上，表示0.1;7.56中的5在十分位上，表示0.5，据此解答即可.\n【详解】根据分析可知：下图竖式中虚线圈起来的部分表示0.4+0.1=0.5。故答案为:B",
+        "knowledge_points":  [
+                                 "多位小数的不进位加法，不退位减法",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·广东广州·期末"
+    },
+    {
+        "id":  "403",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "晨晨在用竖式计算13.7减去一个两位数时，把减号写成了加号，算出的结果是22.33，正确的结果应该是()。",
+        "options":  [
+                        "A. 8.63",
+                        "B. 5.13",
+                        "C. 5.07"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】根据晨晨的错误算法,13.7是加数，这个两位数是加数,22.33是和，根据\"加数=和-另一个加数\"，用22.33-13.7=8.63，即可求出另一个加数，也就是正确算式的减数，然后用13.7减去这个减数，即可求出正确结果.\n【详解】22.33-13.7=8.63,13.7-8.63=5.07。故答案为:C",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "404",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列算式的结果最接近1的是()。",
+        "options":  [
+                        "A. 9÷10-11÷100",
+                        "B. 65÷100+74÷100",
+                        "C. 6.57-5.56"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】由题意得，可以先把十分之几或百分之几的分数转化为一位小数或两位小数，然后分别计算出它们的结果。最后把各个算式的结果分别与1作差，最后找出差值最小的即可.\n【详解】A。9÷10-11÷100=0.9-0.11=0.79,1-0.79=0.21。B。65÷100+74÷100=0.65+0.74=1.39,1.39-1=0.39。C。6.57-5.56=1.01,1.01-1=0.01。0.01\u003c0.21\u003c0.39，所以算式6.57-5.56的结果最接近1。故答案为:C",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "多位小数的不进位加法，不退位减法",
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "24-25四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "405",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "解决题目\"小林在书店买了两本书，一本16.45元，一本18.3元，一共花了多少钱。\"的算式是16.45+18.3。下面竖式中的\"4+3\"表示的是().\n![](images/2026-03-12%20220943.jpg)",
+        "options":  [
+                        "A. 4个一加3个一等于7个一",
+                        "B. 4元加3元等于7元",
+                        "C. 4个0.1加3个0.1等于7个0.1",
+                        "D. 4分加3分等于7分"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】小数点右边第一位是十分位，计数单位是0.1;第二位是百分位，计数单位是0.01。;人民币用小数表示时，整数部分表示\"元\"，十分位表示\"角\"，百分位表示\"分\"。当小数位数不同时，先在位数较少的小数的末尾添上\"0\"，变成位数相同的小数加减法。然后小数点对齐，也就是相同数位对齐，按照整数加减法的法则进行计算。算出得数的小数点要与横线上面的小数点对齐.\n【详解】A。4和3都在十分位上，表示4个0.1和3个0.1，不是4个一和3个一。B。4和3都在十分位上，表示4角和3角，不是4元和3元。C。4和3都在十分位上，表示4个0.1和3个0.1，所以该表述正确。D。4和3都在十分位上，表示4角和3角，不是4分和3分。故答案为:C",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·河南信阳·期末"
+    },
+    {
+        "id":  "406",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下面小数中最接近6的是()。",
+        "options":  [
+                        "A. 6.01",
+                        "B. 5.998",
+                        "C. 5.9",
+                        "D. 6.1"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】分别计算每个选项与6的差值，差值最小的就是最接近6的数.\n【详解】A。6.01-6=0.01;B。6-5.998=0.002;C。6-5.9=0.1;D。6.1-6=0.1。比较可得0.002\u003c0.01\u003c0.1，所以最接近6的是5.998。故答案为:B",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "24-25四年级下·山东济南·期末"
+    },
+    {
+        "id":  "407",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "如下图，方框中的2减8不够减，需要向\"5\"借\"1\"，这个\"1\"表示().\n![](images/2026-03-12%20221046.jpg)",
+        "options":  [
+                        "A. 1个10",
+                        "B. 10个0.1",
+                        "C. 1个0.01",
+                        "D. 10个0.01"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】百分位不够减，需要向十分位借\"1\"。十分位上的1借到百分位，等于10个百分之一，即10个0.01.\n【详解】这个\"1\"表示10个0.01。故答案为:D",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·浙江温州·期末"
+    },
+    {
+        "id":  "408",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "\"2□.4-7.□8\"是一道小数减法，计算结果不可能是()。",
+        "options":  [
+                        "A. 12.52",
+                        "B. 16.22",
+                        "C. 20.02",
+                        "D. 13.56"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】小数减法中，百分位0减8不够减，要向十分位借1，百分位结果必为10-8=2，所以结果末尾应是2.\n【详解】A,B,C末尾均为2,D末尾为6，不可能。故答案为:D",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·广东广州·期末"
+    },
+    {
+        "id":  "409",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下列算式中，\"8\"和\"6\"能直接相加减的算式是()。",
+        "options":  [
+                        "A. 10.8-6.57",
+                        "B. 28+260",
+                        "C. 7.81-0.6",
+                        "D. 9.83-2.06"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】只有相同数位上的数字才能直接相加减.\n【详解】A中8在十分位,6在个位;B中8在个位,6在十位;C中8和6都在十分位;D中8在十分位,6在百分位。只有C可直接相减。故答案为:C",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·浙江·期末"
+    },
+    {
+        "id":  "410",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列问题用加法解决的是()。",
+        "options":  [
+                        "A. 魏老师花36元买了4个同样的羽毛球，每个羽毛球多少钱。",
+                        "B. 四(1)班有学生55人，其中女生有28人，男生有多少人。",
+                        "C. 一辆汽车平均每小时行驶75千米,3小时行驶了多少千米。",
+                        "D. 一根绳子，先用去3.2米，再用去2.6米，两次用去多少米。"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】A是总价/数量(除法);B是总人数-女生人数(减法);C是速度×时间(乘法);D是两次用去长度相加(加法).\n【详解】用加法解决的是D。故答案为:D",
+        "knowledge_points":  [
+                                 "一位小数的不进位加法，不退位减法",
+                                 "两，三位数与一位数连续进位的乘法",
+                                 "两位数与两位数的退位减法",
+                                 "经济问题"
+                             ],
+        "source":  "24-25四年级下·云南曲靖·期末"
+    },
+    {
+        "id":  "411",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "小明买售价1.4元的圆珠笔一支，根据你的生活经验，结合人民币币值特点，下列付钱方法不合理的是()。",
+        "options":  [
+                        "A. 付出1.5元，找回0.1元",
+                        "B. 付出1.7元，找回0.3元",
+                        "C. 付出2.0元，找回0.6元"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】找回钱数=付出钱数-1.4元，并结合常见币值判断是否合理.\n【详解】A:1.5-1.4=0.1，合理。B:1.7-1.4=0.3，虽然数值对，但可直接付1.5元更合理,1.7元找0.3元不合理。C:2.0-1.4=0.6，合理。故答案为:B",
+        "knowledge_points":  [
+                                 "一位小数的进位加法，退位减法",
+                                 "十元以下人民币的认识与计算"
+                             ],
+        "source":  "24-25四年级下·河南信阳·期末"
+    },
+    {
+        "id":  "412",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "林林用计算器计算\"15.79+2.45\"时，他错误地输入\"15.19+2.45\"，为了输出正确的结果，下面的操作正确的是()。",
+        "options":  [
+                        "A. 继续输入\"-0.6\"",
+                        "B. 继续输入\"+0.6\"",
+                        "C. 继续输入\"-0.06\""
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】把15.79错输成15.19，少了0.6，所以应补加0.6.\n【详解】15.79-15.19=0.6，应继续输入\"+0.6\"。故答案为:B",
+        "knowledge_points":  [
+                                 "多位小数的不进位加法，不退位减法"
+                             ],
+        "source":  "24-25四年级下·广西玉林·期末"
+    },
+    {
+        "id":  "413",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "用计算器计算13.49+28.37时，按成了13.49+28.73，要修正错误，可接着输入()。",
+        "options":  [
+                        "A. +0.36",
+                        "B. -0.63",
+                        "C. +0.36",
+                        "D. -0.36"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】第二个加数由28.37变为28.73，多了0.36，和也多0.36，所以减去0.36修正.\n【详解】28.73-28.37=0.36，应输入-0.36。故答案为:D",
+        "knowledge_points":  [
+                                 "计算器的初步认识与使用",
+                                 "多位小数的进位加法，退位减法",
+                                 "和与差的变化规律"
+                             ],
+        "source":  "24-25四年级下·浙江宁波·期末"
+    },
+    {
+        "id":  "414",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列算式中，\"7\"和\"5\"不能直接相加减的算式是()。",
+        "options":  [
+                        "A. 70+250",
+                        "B. 7-5.08",
+                        "C. 4.7米-4.5米",
+                        "D. 1.87-0.5"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】只有相同数位上的数才能直接加减.\n【详解】A十位对十位可直接算;B个位对个位可直接算;C十分位对十分位可直接算;D中7在百分位,5在十分位，不能直接算。故答案为:D",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·浙江宁波·期末"
+    },
+    {
+        "id":  "415",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "如图所示的框中，百分位不够减，需要向\"2\"借\"1\"，这个\"1\"表示().\n![](images/2026-03-12%20221151.jpg)",
+        "options":  [
+                        "A. 10个十分之一",
+                        "B. 10个百分之一",
+                        "C. 10个千分之一"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】向十分位借1，相当于在百分位得到10个0.01.\n【详解】这个\"1\"表示10个百分之一。故答案为:B",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·青海果洛·期末"
+    },
+    {
+        "id":  "416",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下列各数中与1最接近的是()。",
+        "options":  [
+                        "A. 0.9",
+                        "B. 1.001",
+                        "C. 0.99",
+                        "D. 1.1"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】比较各数与1的差，差最小的最接近1.\n【详解】A差0.1,B差0.001,C差0.01,D差0.1。最小是0.001，所以最接近1的是1.001。故答案为:B",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "多位小数的不进位加法，不退位减法",
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "24-25四年级下·江西吉安·期末"
+    },
+    {
+        "id":  "417",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面与10最接近的数是()。",
+        "options":  [
+                        "A. 9.99",
+                        "B. 9.999",
+                        "C. 10.01",
+                        "D. 10.1"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】比较各选项与10的差值.\n【详解】A差0.01,B差0.001,C差0.01,D差0.1。最小是0.001，所以最接近10的是9.999。故答案为:B",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "一位小数的不进位加法，不退位减法",
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "24-25四年级下·湖北孝感·期末"
+    },
+    {
+        "id":  "418",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面算式中，\"7\"和\"8\"可以直接相加或相减的是()。",
+        "options":  [
+                        "A. 702-48",
+                        "B. 7+8.5",
+                        "C. 0.57+0.8",
+                        "D. 0.79-0.38"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】判断7与8是否在同一数位.\n【详解】A百位与个位，不行;B个位与个位，可以;C百分位与十分位，不行;D十分位与百分位，不行。故答案为:B",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "多位小数的不进位加法，不退位减法",
+                                 "一位小数的不进位加法，不退位减法",
+                                 "两，三位数的连续退位减法"
+                             ],
+        "source":  "24-25四年级下·湖南株洲·期末"
+    },
+    {
+        "id":  "419",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "小青用计算器计算13.79+2.34，她错误地输入了13.19+2.34，要修正这个错误，你同意()的说法。",
+        "options":  [
+                        "A. 加60",
+                        "B. 加0.6",
+                        "C. 减0.6"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】13.19比13.79少0.6，故和也少0.6，应补加0.6.\n【详解】13.79-13.19=0.6，所以应再加0.6。故答案为:B",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·湖南常德·期末"
+    },
+    {
+        "id":  "420",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "下面各数中，和\"8\"最接近的数是()。",
+        "options":  [
+                        "A. 8.1",
+                        "B. 8.02",
+                        "C. 7.98",
+                        "D. 7.987"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】分别求与8的差，差最小者最接近.\n【详解】A差0.1,B差0.02,C差0.02,D差0.013.0.013最小，所以最接近8的是7.987。故答案为:D",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "多位小数的不进位加法，不退位减法",
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "24-25四年级下·湖北武汉·期末"
+    },
+    {
+        "id":  "421",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "用计算器计算\"18.64+15.68\"时，错误地输入为\"18.64+15.8\"，要修正这个错误，需要再()。",
+        "options":  [
+                        "A. 加0.6",
+                        "B. 加0.12",
+                        "C. 减0.12"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】把15.68错输成15.8，输入值大了0.12，结果应减去0.12.\n【详解】15.8-15.68=0.12，所以需要再减0.12。故答案为:C",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·湖北武汉·期末"
+    },
+    {
+        "id":  "422",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.65,
+        "question":  "张叔叔家的太阳能热水器支架损坏了(如图),如果更换钢条需要更换钢条的长度可能是().\n![](images/12d9b5c2a9af105a14c0e6db9efb24c2a2b88647ba3a45584dc2ca10a7a30d9f.jpg)",
+        "options":  [
+                        "A. 3.6m",
+                        "B. 3.2m",
+                        "C. 0.2m",
+                        "D. 2.7m"
+                    ],
+        "answer":  "D",
+        "solution":  "【分析】根据三角形三边关系：第三边应满足两边之差\u003c第三边\u003c两边之和.\n【详解】已知两边1.7m和1.5m,和为3.2m,差为0.2m,所以第三边需满足0.2m\u003c第三边\u003c3.2m。仅2.7m符合。故答案为:D",
+        "knowledge_points":  [
+                                 "三角形三边关系",
+                                 "一位小数的不进位加法，不退位减法",
+                                 "一位小数的大小比较"
+                             ],
+        "source":  "24-25四年级下·湖北荆州·期末"
+    },
+    {
+        "id":  "423",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "小林用计算器计算\"18.64+15.18\"时，她错误地输入\"18.64+15.68\"，要修正这个错误，她需要再()。",
+        "options":  [
+                        "A. 加0.5",
+                        "B. 减0.5",
+                        "C. 加5",
+                        "D. 减5"
+                    ],
+        "answer":  "B",
+        "solution":  "【分析】把15.18错输为15.68，多了0.5，应减去0.5修正.\n【详解】15.68-15.18=0.5，所以需再减0.5。故答案为:B",
+        "knowledge_points":  [
+                                 "多位小数的不进位加法，不退位减法",
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "24-25四年级下·湖北荆州·期末"
+    },
+    {
+        "id":  "424",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "选择题",
+        "difficulty":  0.85,
+        "question":  "下列算式中\"2\"和\"4\"能直接相加的算式是()。",
+        "options":  [
+                        "A. 520+134",
+                        "B. 182+7.4",
+                        "C. 206+7458"
+                    ],
+        "answer":  "C",
+        "solution":  "【分析】直接相加要求两数在同一数位.\n【详解】A中2在十位,4在个位;B中2在个位,4在十分位;C中2和4都在百位，可直接相加。故答案为:C",
+        "knowledge_points":  [
+                                 "两，三位数的不进位加法",
+                                 "一位小数的不进位加法，不退位减法"
+                             ],
+        "source":  "24-25四年级下·湖北恩施·期末"
+    },
+    {
+        "id":  "425",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "将0.58的小数点向右移动两位，得到的数比原数大()。",
+        "options":  [
+
+                    ],
+        "answer":  "57.42",
+        "solution":  "【分析】小数点向右移动两位，原数扩大100倍,0.58变为58，再求58与0.58的差.\n【详解】58-0.58=57.42，所以得到的数比原数大57.42。",
+        "knowledge_points":  [
+                                 "小数点向右移动引起小数大小变化的规律",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·河南三门峡·期末"
+    },
+    {
+        "id":  "426",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "2千克450克-1千克400克=()千克;4千米800米+3千米50米=()千米",
+        "options":  [
+
+                    ],
+        "answer":  "1.05;7.85",
+        "solution":  "【分析】先统一单位再计算：克转千克除以1000，米转千米除以1000.\n【详解】2千克450克=2.45千克,1千克400克=1.4千克,2.45-1.4=1.05千克;4千米800米=4.8千米,3千米50米=3.05千米,4.8+3.05=7.85千米。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "多位小数的不进位加法，不退位减法"
+                             ],
+        "source":  "24-25四年级下·广东广州·期末"
+    },
+    {
+        "id":  "427",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "5.12比0.49多();0.91比3.25少()。",
+        "options":  [
+
+                    ],
+        "answer":  "4.63;2.34",
+        "solution":  "【分析】求\"多/少多少\"都用较大数减较小数.\n【详解】5.12-0.49=4.63,3.25-0.91=2.34。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·甘肃天水·期末"
+    },
+    {
+        "id":  "428",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\"\u003e\"\"\u003c\"或\"=\".\n120分()1小时20分 8+0.1()8+0.012分米()0.3米 4年()50个月 32×34()33×336平方米()600平方分米",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;\u003e;\u003c;\u003c;\u003c;=",
+        "solution":  "【分析】统一单位或先算结果后比较大小.\n【详解】1小时20分=80分，所以120分\u003e1小时20分;8+0.1=8.1,8+0.01=8.01，所以前者大;0.3米=3分米，所以2分米\u003c0.3米;4年=48个月\u003c50个月;32×34=1088,33×33=1089，所以前者小;6平方米=600平方分米，所以相等。",
+        "knowledge_points":  [
+                                 "多位小数的不进位加法，不退位减法",
+                                 "两位数乘两位数的进位乘法",
+                                 "多位小数的大小比较",
+                                 "年，月，日的认识及换算"
+                             ],
+        "source":  "24-25三年级下·浙江宁波·期末"
+    },
+    {
+        "id":  "429",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "小马虎在写1.2时忘记点小数点了，写出来的数比原来大()。",
+        "options":  [
+
+                    ],
+        "answer":  "10.8",
+        "solution":  "【分析】忘记小数点后1.2变成12，求12比1.2大多少.\n【详解】12-1.2=10.8，所以比原来大10.8。",
+        "knowledge_points":  [
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·贵州黔西·期末"
+    },
+    {
+        "id":  "430",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个小数由2个一和2个百分之一组成，这个小数是(),这两个2相差()。",
+        "options":  [
+
+                    ],
+        "answer":  "2.02;1.98",
+        "solution":  "【分析】2个一是2,2个百分之一是0.02，合成2.02;两个2分别表示2和0.02，求差即可.\n【详解】2-0.02=1.98，所以这个小数是2.02，这两个2相差1.98。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·吉林四平·期末"
+    },
+    {
+        "id":  "431",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\"\u003e\"\"\u003c\"或\"=\".\n7.4吨()7.38吨 60÷(30×2)()60÷30×26米8厘米()6.8米 320÷4/8()320÷8/4",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;\u003c;\u003c;=",
+        "solution":  "【分析】分别比较小数大小，运算结果和单位换算后的数值.\n【详解】7.4\u003e7.38;60÷(30×2)=1,60÷30×2=4，所以\u003c;6米8厘米=6.08米\u003c6.8米;320÷4/8=10,320÷8/4=10，所以=。",
+        "knowledge_points":  [
+                                 "整数除法的性质",
+                                 "多位小数的不进位加法，不退位减法",
+                                 "带有小括号的混合运算",
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "24-25四年级下·新疆喀什·期末"
+    },
+    {
+        "id":  "432",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上合适的数.\n5.06元=()元()分,5.59吨=()千克,12平方米8平方分米=()平方米",
+        "options":  [
+
+                    ],
+        "answer":  "5;6;5590;12.08",
+        "solution":  "【分析】1元=100分,1吨=1000千克,1平方米=100平方分米，按进率换算.\n【详解】5.06元=5元6分;5.59吨=5590千克;12平方米8平方分米=12+0.08=12.08平方米。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "平方厘米，平方分米，平方米之间的进率与换算",
+                                 "吨，千克之间的换算与比较",
+                                 "多位小数的不进位加法，不退位减法"
+                             ],
+        "source":  "24-25四年级下·新疆喀什·期末"
+    },
+    {
+        "id":  "433",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上\"\u003e\"\"\u003c\"或\"=\".\n1.137()1.139;6.7亿()6700万;3kg30g()3.03kg;3.84+4.3()10.64-3.5",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003e;=;\u003e",
+        "solution":  "【分析】按小数大小比较，单位统一和算式计算后比较.\n【详解】1.137\u003c1.139;6.7亿=67000万\u003e6700万;3kg30g=3.03kg;3.84+4.3=8.14,10.64-3.5=7.14，故8.14\u003e7.14。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "多位小数的进位加法，退位减法",
+                                 "多位小数的大小比较",
+                                 "大数的比较"
+                             ],
+        "source":  "24-25四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "434",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "广州地铁1号线是广州市第一条建成运营的地铁线路，全长18.50千米，其中地面段长2.05千米，剩余部分都是地下段，地下段长()千米。",
+        "options":  [
+
+                    ],
+        "answer":  "16.45",
+        "solution":  "【分析】用地铁全长减去地面段长等于地下段长，代入数据即可.\n【详解】18.50-2.05=16.45(千米)因此地下段长16.45千米。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·广东广州·期末"
+    },
+    {
+        "id":  "435",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "2.085吨=()吨()千克;9平方千米-7公顷=()平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "2;85;8.93",
+        "solution":  "【分析】(1)2.085吨=()吨()千克，只需要把0.085吨转化为多少千克。大单位转化为小单位，需要乘上它们之间的进率1000，相当于把小数点向右移动三位。(2)先把7公顷转化为平方千米，再做减法.\n【详解】(1)0.085×1000=85，所以0.085吨=85千克,2.085吨=2吨85千克。(2)7÷100=0.07，所以7公顷=0.07平方千米。9平方千米-0.07平方千米=8.93平方千米。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "436",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把5.04扩大到它的()倍是50400,3.31加上()个1÷100等于4.1。",
+        "options":  [
+
+                    ],
+        "answer":  "10000;79",
+        "solution":  "【分析】根据小数点移动规律，求5.04需扩大多少倍到50400;再用4.1-3.31得到需要补上的数量，并换算成1÷100的个数.\n【详解】5.04×10000=50400;4.1-3.31=0.79=79个1÷100。",
+        "knowledge_points":  [
+                                 "小数点向右移动引起小数大小变化的规律",
+                                 "多位小数的进位加法，退位减法",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "437",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "去年陕西新能源汽车产量大幅增长，某工厂上半年生产新能源汽车23.56万辆，下半年生产34.18万辆，全年共生产()万辆，下半年比上半年多生产()万辆。",
+        "options":  [
+
+                    ],
+        "answer":  "57.74;10.62",
+        "solution":  "【分析】全年产量=上半年+下半年;下半年比上半年多=下半年-上半年.\n【详解】23.56+34.18=57.74(万辆);34.18-23.56=10.62(万辆)。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·陕西延安·期末"
+    },
+    {
+        "id":  "438",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个等腰三角形中，有两条边分别是7.85cm和3.6cm,它的周长是()cm。",
+        "options":  [
+
+                    ],
+        "answer":  "19.3",
+        "solution":  "【分析】等腰三角形两腰相等，并满足三角形三边关系。若腰为3.6，则3.6+3.6\u003c7.85，不成立;故腰为7.85.\n【详解】周长=7.85+7.85+3.6=19.3(cm)。",
+        "knowledge_points":  [
+                                 "三角形三边关系",
+                                 "一位小数的进位加法，退位减法",
+                                 "三角形的周长",
+                                 "等腰三角形和等边三角形的认识及特征"
+                             ],
+        "source":  "24-25四年级下·辽宁鞍山·期末"
+    },
+    {
+        "id":  "439",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小文从水培植物风信子出芽后第3周的第一天开始记录生长高度，到第12周的第一天(如图所示),风信子增长了()厘米.\n![](images/138032d97f00a8664715f134be121be0e1844f948e89085edf774d737dd055d5.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "9.25",
+        "solution":  "【分析】增长高度=第12周高度-第3周高度.\n【详解】11.35-2.1=9.25(厘米)。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·云南文山·期末"
+    },
+    {
+        "id":  "440",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "爷爷通过扫二维码领了5.6元红包，比奶奶多领了1.03元。爷爷和奶奶一共领了红包()元。",
+        "options":  [
+
+                    ],
+        "answer":  "10.17",
+        "solution":  "【分析】奶奶红包=5.6-1.03;总额=爷爷+奶奶.\n【详解】5.6-1.03+5.6=4.57+5.6=10.17(元)。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·浙江·期末"
+    },
+    {
+        "id":  "441",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在下面括号里填上\"\u003e\"、\"\u003c\"或\"=\".\n6.8+(4.7-2.2)()6.8+4.7-2.2\n7.69千克()7千克69克",
+        "options":  [
+
+                    ],
+        "answer":  "=;\u003e",
+        "solution":  "【分析】第一空分别计算两边;第二空统一单位比较.\n【详解】6.8+(4.7-2.2)=9.3,6.8+4.7-2.2=9.3，所以\"=\"。69克=0.069千克,7千克69克=7.069千克,7.69\u003e7.069，所以\"\u003e\"。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "小数的加，减法混合运算",
+                                 "多位小数的不进位加法，不退位减法",
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "24-25四年级下·浙江·期末"
+    },
+    {
+        "id":  "442",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "6.4公顷=()平方米;16.80元=()元()角;9吨30千克=()吨;1.56千米=()千米()米",
+        "options":  [
+
+                    ],
+        "answer":  "64000;16;8;9.03;1;560",
+        "solution":  "【分析】依据进率:1公顷=10000平方米,1元=10角,1吨=1000千克,1千米=1000米进行单位换算.\n【详解】6.4公顷=64000平方米;16.80元=16元8角;9吨30千克=9.03吨;1.56千米=1千米560米。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "公顷，平方千米的进率与换算",
+                                 "千米和米之间的进率与换算",
+                                 "多位小数的不进位加法，不退位减法"
+                             ],
+        "source":  "24-25四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "443",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "奥运会的男子跳高纪录是2.39米，这个小数的计数单位是(),再添加()个这样的计数单位就是3。",
+        "options":  [
+
+                    ],
+        "answer":  "0.01或百分之一或1÷100;61",
+        "solution":  "【分析】2.39是两位小数，计数单位是0.01;补到3需要3-2.39=0.61，即61个0.01.\n【详解】计数单位是0.01，再添加61个这样的计数单位就是3。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·湖北十堰·期末"
+    },
+    {
+        "id":  "444",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用6,2,7三个数字组成小数部分是两位的小数，其中组成的最小的小数和最大的小数相差()。",
+        "options":  [
+
+                    ],
+        "answer":  "4.95",
+        "solution":  "【分析】最大数取7.62，最小数取2.67，求差即可.\n【详解】7.62-2.67=4.95。",
+        "knowledge_points":  [
+                                 "多位小数的大小比较",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·江西吉安·期末"
+    },
+    {
+        "id":  "445",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在下面括号里填上适当的数.\n3吨60千克=()吨;0.5小时=()分钟",
+        "options":  [
+
+                    ],
+        "answer":  "3.06;30",
+        "solution":  "【分析】60千克=0.06吨，所以3吨60千克=3.06吨;1小时=60分,0.5小时是其一半.\n【详解】3吨60千克=3.06吨;0.5小时=30分钟。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "吨，千克之间的换算与比较",
+                                 "时，分，秒之间的换算与比较",
+                                 "一位小数的不进位加法，不退位减法"
+                             ],
+        "source":  "24-25四年级下·湖南永州·期末"
+    },
+    {
+        "id":  "446",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在1.1,1.95和2.01这三个数中，最接近整数1的是(),最接近整数2的是()。",
+        "options":  [
+
+                    ],
+        "answer":  "1.1;2.01",
+        "solution":  "【分析】分别求每个数与1，与2的差，差最小者最接近.\n【详解】与1比较:0.1,0.95,1.01，最小是0.1，对应1.1;与2比较:0.9,0.05,0.01，最小是0.01，对应2.01。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·湖南长沙·期末"
+    },
+    {
+        "id":  "447",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小红计算某数加3.6时，把3.6误看成了36，算得的结果是48.8，这道题的正确结果是()。",
+        "options":  [
+
+                    ],
+        "answer":  "16.4",
+        "solution":  "【分析】先由错误算式求原数，再代入正确加数.\n【详解】原数=48.8-36=12.8，正确结果=12.8+3.6=16.4。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·湖北恩施·期末"
+    },
+    {
+        "id":  "448",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "明明用AI大模型DeepSeek预测了2025年武汉黄鹤楼\"十一\"黄金周的接待量下限为43.2万人次，上限为50.4万人次，上限和下限差是()万人次，改写成\"亿\"为单位写作()亿。",
+        "options":  [
+
+                    ],
+        "answer":  "7.2;0.00072",
+        "solution":  "【分析】先求差:50.4-43.2;再把\"万\"改写为\"亿\"，除以10000.\n【详解】50.4-43.2=7.2(万人次);7.2万=0.00072亿。",
+        "knowledge_points":  [
+                                 "亿以上数的改写",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·湖北武汉·期末"
+    },
+    {
+        "id":  "449",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个数由8个一和8个百分之一组成，这个数是(),两个8相差()。",
+        "options":  [
+
+                    ],
+        "answer":  "8.08;7.92",
+        "solution":  "【分析】8个一是8,8个百分之一是0.08，组成8.08;两个8分别表示8和0.08，作差.\n【详解】8-0.08=7.92。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "小数的数位和计数单位的认识",
+                                 "小数的读法和写法"
+                             ],
+        "source":  "24-25四年级下·湖北武汉·期末"
+    },
+    {
+        "id":  "450",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "0.53的计数单位是(),再加上()个这样的计数单位就是1。",
+        "options":  [
+
+                    ],
+        "answer":  "0.01;47",
+        "solution":  "【分析】0.53是两位小数，计数单位是0.01;补到1需1-0.53=0.47，即47个0.01.\n【详解】计数单位为0.01，再加47个这样的计数单位就是1。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·河北保定·期末"
+    },
+    {
+        "id":  "451",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在一次跳高比赛中，小明跳了2.48米，小亮比小明少跳了0.09米，小军比小亮多跳了0.25米。小军跳了()米。",
+        "options":  [
+
+                    ],
+        "answer":  "2.64",
+        "solution":  "【分析】小亮=2.48-0.09;小军=小亮+0.25.\n【详解】2.48-0.09+0.25=2.39+0.25=2.64(米)。",
+        "knowledge_points":  [
+                                 "利用小数的加，减法混合运算解决实际问题"
+                             ],
+        "source":  "24-25四年级下·湖南永州·期末"
+    },
+    {
+        "id":  "452",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "比3.25多0.8的数是(),3.25比()少0.8。",
+        "options":  [
+
+                    ],
+        "answer":  "4.05;4.05",
+        "solution":  "【分析】两空本质相同，都是求3.25+0.8.\n【详解】3.25+0.8=4.05。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·湖南永州·期末"
+    },
+    {
+        "id":  "453",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "爸爸身高1.86米，比妈妈高27厘米，妈妈身高()m。",
+        "options":  [
+
+                    ],
+        "answer":  "1.59",
+        "solution":  "【分析】先把27厘米化成米:0.27米，再用爸爸身高减去0.27.\n【详解】1.86-0.27=1.59(米)。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·湖北孝感·期末"
+    },
+    {
+        "id":  "454",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在2024年巴黎奥运会男子100米自由泳决赛中，中国选手潘展乐以46.4秒的成绩打破世界纪录夺得冠军，领先第二名1.08秒，第二名的成绩是()秒。",
+        "options":  [
+
+                    ],
+        "answer":  "47.48",
+        "solution":  "【分析】\"领先1.08秒\"表示第二名用时更长1.08秒.\n【详解】46.4+1.08=47.48(秒)。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·湖北武汉·期末"
+    },
+    {
+        "id":  "455",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小数2.76读作(),计数单位是(),至少再添上()个这样的计数单位就能得到一个整数。",
+        "options":  [
+
+                    ],
+        "answer":  "二点七六;0.01或百分之一;24",
+        "solution":  "【分析】2.76读作\"二点七六\";两位小数计数单位0.01;补到3需3-2.76=0.24，即24个0.01.\n【详解】2.76读作二点七六，计数单位是0.01，至少再添上24个这样的计数单位就能得到整数3。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "小数的数位和计数单位的认识",
+                                 "小数的读法和写法"
+                             ],
+        "source":  "24-25四年级下·湖北武汉·期末"
+    },
+    {
+        "id":  "456",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "妈妈腿长0.98米，她说自己走一步最长2.2米。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】最大步长通常不超过腿长两倍，先算0.98×2.\n【详解】0.98+0.98=1.96(米),1.96\u003c2.2，所以说法错误。故答案为: ×。",
+        "knowledge_points":  [
+                                 "多位小数的大小比较",
+                                 "多位小数的进位加法，退位减法",
+                                 "三角形三边关系"
+                             ],
+        "source":  "24-25四年级下·辽宁盘锦·期末"
+    },
+    {
+        "id":  "457",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "62.8-1.6+8.4=62.8-(1.6+8.4)。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】分别计算左右两边.\n【详解】左边=62.8-1.6+8.4=69.6;右边=62.8-(1.6+8.4)=52.8，二者不等，说法错误。故答案为: ×。",
+        "knowledge_points":  [
+                                 "与小数减法相关的简便运算",
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "458",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "计算小数加减法时要把末位对齐，实质上就是把相同数位对齐。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】小数加减法应小数点对齐，即相同数位对齐，不是末位对齐.\n【详解】如3.25+1.3，若末位对齐会错位;应小数点对齐。故题目说法错误。故答案为: ×。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "459",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "42+3.25+6.75=42+(3.25+6.75)运用了加法交换律。()",
+        "options":  [
+
+                    ],
+        "answer":  "×",
+        "solution":  "【分析】该式改变的是结合方式，不是交换位置.\n【详解】这是加法结合律(a+b)+c=a+(b+c),不是交换律a+b=b+a,故说法错误。故答案为: ×。",
+        "knowledge_points":  [
+                                 "整数加法运算律推广到小数"
+                             ],
+        "source":  "24-25四年级下·广东广州·期末"
+    },
+    {
+        "id":  "460",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "26.98-12.56-7.44=26.98-(12.56+7.44),运用了减法的性质进行简便计算。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】连续减两个数可写成减去这两个数的和.\n【详解】26.98-12.56-7.44=26.98-(12.56+7.44),运用的是减法性质，故说法正确。故答案为:√。",
+        "knowledge_points":  [
+                                 "与小数减法相关的简便运算"
+                             ],
+        "source":  "24-25四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "461",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "判断题",
+        "difficulty":  0.65,
+        "question":  "整数加减法的运算定律对小数加减法也同样适用。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】加法交换律，结合律和减法性质在小数运算中同样成立.\n【详解】可用小数举例验证结果一致，因此说法正确。故答案为:√。",
+        "knowledge_points":  [
+                                 "与小数减法相关的简便运算",
+                                 "整数加法运算律推广到小数"
+                             ],
+        "source":  "24-25四年级下·贵州铜仁·期末"
+    },
+    {
+        "id":  "462",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "判断题",
+        "difficulty":  0.85,
+        "question":  "在计算12.58-3.5时，算式中的\"5\"数位相同，可以直接相减。()",
+        "options":  [
+
+                    ],
+        "answer":  "√",
+        "solution":  "【分析】12.58中的5和3.5中的5都在十分位.\n【详解】同为十分位，计数单位相同，可以直接相减，故说法正确。故答案为:√。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·吉林四平·期末"
+    },
+    {
+        "id":  "463",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "计算题",
+        "difficulty":  0.85,
+        "question":  "直接写得数.\n12.3-2.4=( );0.8+0.22=( );2000÷4/25=( );8×10÷10×8=( );7-0.77=( );0.609×100=( );50.5×1000÷10=( );3.6÷100×10=( );",
+        "options":  [
+
+                    ],
+        "answer":  "9.9;1.02;20;64;6.23;60.9;5050;0.36",
+        "solution":  "【分析】按小数加减法，四则运算顺序以及小数点移动规律计算.\n【详解】12.3-2.4=9.9;0.8+0.22=1.02;2000÷4/25=20;8×10÷10×8=64;7-0.77=6.23;0.609×100=60.9;50.5×1000÷10=5050;3.6÷100×10=0.36。",
+        "knowledge_points":  [
+                                 "小数点向左移动引起小数大小变化的规律",
+                                 "小数点向右移动引起小数大小变化的规律",
+                                 "整数除法的性质",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "464",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "王奶奶家响应节水号召，更换节水器具前,6月份用水量是30.2吨，更换后7月份用水量减少了2.4吨，王奶奶家这两个月一共用水多少吨。",
+        "options":  [
+
+                    ],
+        "answer":  "58吨",
+        "solution":  "【分析】7月用水=30.2-2.4，再与6月相加.\n【详解】30.2-2.4+30.2=27.8+30.2=58(吨)。答：这两个月一共用水58吨。",
+        "knowledge_points":  [
+                                 "利用小数的加，减法混合运算解决实际问题"
+                             ],
+        "source":  "24-25四年级下·河南三门峡·期末"
+    },
+    {
+        "id":  "465",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "明明在参观\"人与健康\"馆时，利用里面的方法估算了他平均每天摄入的脂肪量如下表。明明平均每天通过这三类食物共摄入多少克脂肪。\n| 食物类型 | 牛奶 | 肉类 | 蛋类 |\n|----------|------|------|------|\n| 脂肪摄入量 | 10.8g | 3.53g | 4.2g |",
+        "options":  [
+
+                    ],
+        "answer":  "18.53克",
+        "solution":  "【分析】把三类食物的脂肪摄入量相加.\n【详解】10.8+3.53+4.2=10.8+4.2+3.53=15+3.53=18.53(克)。答：共摄入18.53克脂肪。",
+        "knowledge_points":  [
+                                 "整数加法运算律推广到小数",
+                                 "小数的四则运算及法则"
+                             ],
+        "source":  "24-25四年级下·广东广州·期末"
+    },
+    {
+        "id":  "466",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "张亮家5月份缴水费33.5元,6月份比5月份多缴了6.5元。张亮家这两个月一共缴了多少元水费。",
+        "options":  [
+
+                    ],
+        "answer":  "73.5元",
+        "solution":  "【分析】先求6月:33.5+6.5，再与5月相加.\n【详解】33.5+6.5+33.5=40+33.5=73.5(元)。答：这两个月一共缴了73.5元水费。",
+        "knowledge_points":  [
+                                 "利用小数的加，减法混合运算解决实际问题"
+                             ],
+        "source":  "24-25四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "467",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "超市运来香蕉18.5千克，运来苹果比香蕉少3.6千克，两种水果一共运来多少千克。",
+        "options":  [
+
+                    ],
+        "answer":  "33.4千克",
+        "solution":  "【分析】苹果=18.5-3.6，再与香蕉相加.\n【详解】18.5-3.6+18.5=14.9+18.5=33.4(千克)。答：两种水果一共运来33.4千克。",
+        "knowledge_points":  [
+                                 "利用小数的加，减法混合运算解决实际问题"
+                             ],
+        "source":  "24-25四年级下·甘肃天水·期末"
+    },
+    {
+        "id":  "468",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一支钢笔38.5元，一个书包比一支钢笔贵25.9元。一个书包比一个文具盒贵35.5元，钢笔和文具盒哪个贵。",
+        "options":  [
+
+                    ],
+        "answer":  "钢笔贵",
+        "solution":  "【分析】先求书包价格，再求文具盒价格，最后比较文具盒与钢笔.\n【详解】38.5+25.9=64.4(元),64.4-35.5=28.9(元),38.5\u003e28.9。答：钢笔更贵。",
+        "knowledge_points":  [
+                                 "一位小数的进位加法，退位减法",
+                                 "一位小数的大小比较"
+                             ],
+        "source":  "24-25四年级下·甘肃兰州·期末"
+    },
+    {
+        "id":  "469",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "一本《童话世界》售价是13.8元，一本《西游记》售价比《童话世界》少2.9元，这两本书一共要多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "24.7元",
+        "solution":  "【分析】先求《西游记》售价，再求总价.\n【详解】13.8-2.9=10.9(元),13.8+10.9=24.7(元)。答：这两本书一共要24.7元。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·重庆·期末"
+    },
+    {
+        "id":  "470",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "小明家七月份的电费是56.8元，比八月份少26.6元，小明家七月和八月共交电费多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "140.2元",
+        "solution":  "【分析】先用56.8+26.6求出八月份电费，再加上七月份电费.\n【详解】56.8+26.6+56.8=83.4+56.8=140.2(元)。答：小明家七月和八月共交电费140.2元。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·新疆喀什·期末"
+    },
+    {
+        "id":  "471",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "2025年钻石联赛厦门站，我国运动员张溟鲲在男子跳远中以8.18米的成绩获得冠军。比美国选手多跳了0.08米，另一位中国选手舒衡比美国选手少跳了0.02米。舒衡跳了多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "8.08米",
+        "solution":  "【分析】先用8.18-0.08求美国选手成绩，再减0.02求舒衡成绩.\n【详解】8.18-0.08-0.02=8.18-(0.08+0.02)=8.18-0.1=8.08(米)。答：舒衡跳了8.08米。",
+        "knowledge_points":  [
+                                 "与小数减法相关的简便运算",
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "472",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "摄影师操控航拍无人机从距地面89米的高度先下降18.63米进行拍摄，再下降10.37米进行拍摄，拍摄完成后返航。航拍无人机从距地面多少米的高度返航。",
+        "options":  [
+
+                    ],
+        "answer":  "60米",
+        "solution":  "【分析】返航高度=初始高度-两次下降高度.\n【详解】89-18.63-10.37=89-(18.63+10.37)=89-29=60(米)。答：航拍无人机从距地面60米的高度返航。",
+        "knowledge_points":  [
+                                 "整数加法运算律推广到小数",
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "473",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "星期日，小红和妈妈一起去逛超市，妈妈实际花了多少钱。\n![](images/2026-03-12%20221538.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "45.77元",
+        "solution":  "【分析】买肉和蔬菜的钱相加，再减去优惠金额.\n【详解】38.35+8.7-1.28=47.05-1.28=45.77(元)。答：妈妈实际花了45.77元。",
+        "knowledge_points":  [
+                                 "图文问题",
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·河南信阳·期末"
+    },
+    {
+        "id":  "474",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "某校举办\"阳光校园，乐享运动\"的春季体育节，四年级的张英，王强和肖红三名同学参加了跳高比赛。肖红跳了1.16米，肖红比王强跳得低0.09米，张英比王强跳得低0.15米。张英跳了多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "1.1米",
+        "solution":  "【分析】先求王强高度，再减去0.15米求张英高度.\n【详解】1.16+0.09=1.25(米),1.25-0.15=1.1(米)。答：张英跳了1.1米。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·河南南阳·期末"
+    },
+    {
+        "id":  "475",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "2025年国家补贴政策期间，小明购买了一台原价为5799元的笔记本电脑，在享受补贴1154元和会员折扣28.99元后，他实际支付多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "4616.01元",
+        "solution":  "【分析】实际支付=原价-补贴-会员折扣.\n【详解】5799-1154-28.99=4645-28.99=4616.01(元)。答：他实际支付4616.01元。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·广东广州·期末"
+    },
+    {
+        "id":  "476",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "今年五一假期，陕西某著名旅游景点接待游客数量如下：第一天接待游客12.38万人次，第二天接待游客14.53万人次，第三天接待游客比前两天人数总和少0.89万人次。第三天接待游客多少万人次。",
+        "options":  [
+
+                    ],
+        "answer":  "26.02万人次",
+        "solution":  "【分析】先求前两天总和，再减去0.89.\n【详解】12.38+14.53-0.89=26.91-0.89=26.02(万人次)。答：第三天接待游客26.02万人次。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法",
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·陕西延安·期末"
+    },
+    {
+        "id":  "477",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "在跳远测试中，明明跳了1.73米，比苗苗远0.15米，萌萌比苗苗远0.22米。萌萌的跳远成绩是多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "1.8米",
+        "solution":  "【分析】先求苗苗成绩，再加0.22求萌萌成绩.\n【详解】1.73-0.15+0.22=1.58+0.22=1.8(米)。答：萌萌的跳远成绩是1.8米。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·辽宁鞍山·期末"
+    },
+    {
+        "id":  "478",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "用一根长2米的丝带装饰乌牛早茶礼盒，打包礼盒用去9.56分米，扎蝴蝶结用去5.44分米，这根丝带一共用去了多少分米。",
+        "options":  [
+
+                    ],
+        "answer":  "15分米",
+        "solution":  "【分析】总用去长度=两部分长度之和.\n【详解】9.56+5.44=15(分米)。答：这根丝带一共用去了15分米。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·浙江温州·期末"
+    },
+    {
+        "id":  "479",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "修路队修一条公路，第一天修了3.4千米，第二天比第一天多修了0.6千米，两天共修了多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "7.4千米",
+        "solution":  "【分析】先求第二天修路长度，再与第一天相加.\n【详解】3.4+0.6+3.4=4+3.4=7.4(千米)。答：两天共修了7.4千米。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算",
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·河南信阳·期末"
+    },
+    {
+        "id":  "480",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "\"活化石\"是指在地质历史时期，其身体结构变化很小，保留了很多原始特征的现存生物。下面是我国现有动物\"活化石\"中的3种，扬子鳄重多少吨。\n![](images/d2ff959de087f0bcd3a0cd6cc72ea405.png)",
+        "options":  [
+
+                    ],
+        "answer":  "0.03吨",
+        "solution":  "【分析】先由白鳍豚求中华鲟，再由中华鲟求扬子鳄.\n【详解】0.23-0.03-0.17=0.2-0.17=0.03(吨)。答：扬子鳄重0.03吨。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算",
+                                 "多位小数的进位加法，退位减法",
+                                 "多位小数的不进位加法，不退位减法"
+                             ],
+        "source":  "24-25四年级下·贵州六盘水·期末"
+    },
+    {
+        "id":  "481",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "妈妈买了一件上衣和一条裤子，上衣每件85元，一件上衣比一条裤子便宜5.38元。她付给售货员200元，应找回多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "24.62元",
+        "solution":  "【分析】先求裤子价格，再求总价，最后用200减总价.\n【详解】85+5.38=90.38(元),200-(90.38+85)=200-175.38=24.62(元)。答：应找回24.62元。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算",
+                                 "多位小数的进位加法，退位减法",
+                                 "经济问题"
+                             ],
+        "source":  "24-25四年级下·内蒙古呼伦贝尔·期末"
+    },
+    {
+        "id":  "482",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "在献爱心活动中，新村小学共捐款3.21万元，其中0.87万元用于帮助困难学生,1.13万元捐给红十字会，剩下的作为学校慈善基金会的启动资金，启动资金有多少万元。",
+        "options":  [
+
+                    ],
+        "answer":  "1.21万元",
+        "solution":  "【分析】总捐款减去两项支出，利用减法性质简算.\n【详解】3.21-0.87-1.13=3.21-(0.87+1.13)=3.21-2=1.21(万元)。答：启动资金有1.21万元。",
+        "knowledge_points":  [
+                                 "与小数减法相关的简便运算"
+                             ],
+        "source":  "24-25四年级下·重庆黔江·期末"
+    },
+    {
+        "id":  "483",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "小红身高1.48米，她站在0.4米高的凳子上，就和爸爸身高一样高了，爸爸身高多少。",
+        "options":  [
+
+                    ],
+        "answer":  "1.88米",
+        "solution":  "【分析】爸爸身高=小红身高+凳子高度.\n【详解】1.48+0.4=1.88(米)。答：爸爸身高1.88米。",
+        "knowledge_points":  [
+                                 "多位小数的不进位加法，不退位减法"
+                             ],
+        "source":  "24-25四年级下·浙江宁波·期末"
+    },
+    {
+        "id":  "484",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一堆煤由甲，乙两个运输队负责运输，甲队运了97.45吨，比乙队少运4.82吨，这堆煤共有多少吨。",
+        "options":  [
+
+                    ],
+        "answer":  "199.72吨",
+        "solution":  "【分析】先求乙队运量，再与甲队运量相加.\n【详解】97.45+4.82+97.45=102.27+97.45=199.72(吨)。答：这堆煤共有199.72吨。",
+        "knowledge_points":  [
+                                 "利用小数的加，减法混合运算解决实际问题"
+                             ],
+        "source":  "24-25四年级下·湖南衡阳·期末"
+    },
+    {
+        "id":  "485",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "把一根长3米的竹竿垂直地插入水池中，竹竿入泥部分是0.36米，露出水面的部分是0.75米。水池中的水深多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "1.89米",
+        "solution":  "【分析】水深=竹竿总长-入泥部分-露出水面部分.\n【详解】3-0.36-0.75=2.64-0.75=1.89(米)。答：水池的水深1.89米。",
+        "knowledge_points":  [
+                                 "利用小数的加，减法混合运算解决实际问题"
+                             ],
+        "source":  "24-25四年级下·湖南岳阳·期末"
+    },
+    {
+        "id":  "486",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "李老师用数字人民币在智能书店购书，系统自动生成账单：数学参考书24.28元，语文参考书23.76元。若用50元数字钱包支付，找回多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "1.96元",
+        "solution":  "【分析】找回金额=支付金额-两本书总价.\n【详解】50-24.28-23.76=25.72-23.76=1.96(元)。答：找回1.96元。",
+        "knowledge_points":  [
+                                 "利用小数的加，减法混合运算解决实际问题"
+                             ],
+        "source":  "24-25四年级下·江西吉安·期末"
+    },
+    {
+        "id":  "487",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "把一根长2.6米的竹竿垂直插入池塘中，竹竿入泥部分是0.25米，露出水面0.85米，池塘中的水深多少米。",
+        "options":  [
+
+                    ],
+        "answer":  "1.5米",
+        "solution":  "【分析】水深=竹竿长度-(入泥部分+露出水面部分).\n【详解】2.6-(0.25+0.85)=2.6-1.1=1.5(米)。答：池塘中的水深1.5米。",
+        "knowledge_points":  [
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·湖南长沙·期末"
+    },
+    {
+        "id":  "488",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "美术社团制作宣传海报，海报是长方形，长3.6米，长比宽长1.2米。东东要在海报四周贴装饰条，每米装饰条8元，贴装饰条一共需要多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "96元",
+        "solution":  "【分析】先求宽，再求长方形周长，最后乘单价.\n【详解】(3.6-1.2+3.6)×2×8=6×2×8=96(元)。答：贴装饰条一共需要96元。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算",
+                                 "长方形的周长",
+                                 "经济问题"
+                             ],
+        "source":  "24-25四年级下·湖南株洲·期末"
+    },
+    {
+        "id":  "489",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "\"六一\"儿童节，妈妈带小明去超市购物。他们买了一箱牛奶花了48.5元，一袋水果花了36.8元，妈妈付给收银员100元，应找回多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "14.7元",
+        "solution":  "【分析】先求总花费，再用100减去总花费.\n【详解】48.5+36.8=85.3(元),100-85.3=14.7(元)。答：应找回14.7元。",
+        "knowledge_points":  [
+                                 "利用小数加减法解决实际问题",
+                                 "经济问题"
+                             ],
+        "source":  "24-25四年级下·湖南株洲·期末"
+    },
+    {
+        "id":  "490",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "修路队修一条公路，第一天修了34千米，第一天比第二天多修了0.6千米，两天共修了多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "67.4千米",
+        "solution":  "【分析】先求第二天修路长度，再求两天和.\n【详解】34-0.6+34=33.4+34=67.4(千米)。答：两天共修了67.4千米。",
+        "knowledge_points":  [
+                                 "利用小数的加，减法混合运算解决实际问题"
+                             ],
+        "source":  "24-25四年级下·湖北恩施·期末"
+    },
+    {
+        "id":  "491",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "修路队修一条公路，第一天修了3.25千米，第二天比第一天多修0.8千米，两天共修多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "7.3千米",
+        "solution":  "【分析】第二天=第一天+0.8，两天总和=第一天+第二天.\n【详解】3.25+0.8+3.25=3.25+3.25+0.8=6.5+0.8=7.3(千米)。答：两天共修7.3千米。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·湖南株洲·期末"
+    },
+    {
+        "id":  "492",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "小华从家出发去图书馆，家到图书馆的距离是2.3千米。他走了0.45千米后发现借书证忘带了，于是立即返回家拿借书证，然后再次前往图书馆。小华这次去图书馆一共走了多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "3.2千米",
+        "solution":  "【分析】总路程=正常去图书馆路程+多走的往返0.45千米.\n【详解】0.45+0.45+2.3=0.9+2.3=3.2(千米)。答：小华这次去图书馆一共走了3.2千米。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·湖北武汉·期末"
+    },
+    {
+        "id":  "493",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.85,
+        "question":  "黄鹤楼客流\"五一\"期间，黄鹤楼景区迎来旅游高峰。5月1日上午售出门票3.25万张，下午售出门票1.75万张。5月2日售出的门票总数比5月1日多0.8万张,5月2日售出多少万张门票。",
+        "options":  [
+
+                    ],
+        "answer":  "5.8万张",
+        "solution":  "【分析】先求5月1日总量，再加0.8万张.\n【详解】3.25+1.75+0.8=5+0.8=5.8(万张)。答:5月2日售出5.8万张门票。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·湖北武汉·期末"
+    },
+    {
+        "id":  "494",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "小明酷爱体育运动，他前往体育用品店购买了一个篮球，花了49.5元。随后他又购买了一个足球，价格比篮球贵12.6元。请问小明购买足球和篮球共花了多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "111.6元",
+        "solution":  "【分析】先求足球价格，再与篮球价格相加.\n【详解】49.5+12.6+49.5=62.1+49.5=111.6(元)。答：小明购买足球和篮球共花了111.6元。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算"
+                             ],
+        "source":  "24-25四年级下·河北保定·期末"
+    },
+    {
+        "id":  "495",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "李阿姨带100元去超市购物，买水果用了35.8元，买蔬菜用了24.6元，剩下的钱想买一箱牛奶，牛奶的价格是42.5元。钱够吗。",
+        "options":  [
+
+                    ],
+        "answer":  "不够",
+        "solution":  "【分析】先算剩余钱数，再与42.5元比较.\n【详解】100-35.8-24.6=64.2-24.6=39.6(元),39.6\u003c42.5。答：剩下的钱不够买一箱牛奶。",
+        "knowledge_points":  [
+                                 "利用小数的加，减法混合运算解决实际问题",
+                                 "一位小数的大小比较"
+                             ],
+        "source":  "24-25四年级下·湖南永州·期末"
+    },
+    {
+        "id":  "496",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "妈妈带50元买水果，买苹果用去18.5元，买香蕉用去12.7元，应找回多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "18.8元",
+        "solution":  "【分析】找回金额=50-18.5-12.7.\n【详解】50-18.5-12.7=31.5-12.7=18.8(元)。答：应找回18.8元。",
+        "knowledge_points":  [
+                                 "小数的加，减法混合运算",
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·湖北黄石·期末"
+    },
+    {
+        "id":  "497",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "一本故事书18.5元，一本科技书22.9元，小明买这两本书，付给营业员50元，应找回多少元。",
+        "options":  [
+
+                    ],
+        "answer":  "8.6元",
+        "solution":  "【分析】先求两本书总价，再用50元减去总价.\n【详解】18.5+22.9=41.4(元),50-41.4=8.6(元)。答：应找回8.6元。",
+        "knowledge_points":  [
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·河北沧州·期末"
+    },
+    {
+        "id":  "498",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "某市地铁1号线全长是42.08千米,2号线的全长比1号线短10.32千米,5号线的全长比2号线长9.75千米。5号线的全长是多少千米。",
+        "options":  [
+
+                    ],
+        "answer":  "41.51千米",
+        "solution":  "【分析】先求2号线全长，再加9.75求5号线.\n【详解】42.08-10.32+9.75=31.76+9.75=41.51(千米)。答:5号线的全长是41.51千米。",
+        "knowledge_points":  [
+                                 "利用小数的加，减法混合运算解决实际问题",
+                                 "多位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25四年级下·河南许昌·期末"
+    },
+    {
+        "id":  "499",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "某平台电器商城开展\"6·18\"年中促销活动，购物满500元立减88元，妈妈购买了一台空气炸锅262.8元和一台破壁机318.6元，妈妈实际支付了多少钱。",
+        "options":  [
+
+                    ],
+        "answer":  "493.4元",
+        "solution":  "【分析】先求总价并判断满足满减，再减去88元.\n【详解】262.8+318.6=581.4(元),581.4\u003e500,581.4-88=493.4(元)。答：妈妈实际支付了493.4元钱。",
+        "knowledge_points":  [
+                                 "利用小数加减法解决实际问题"
+                             ],
+        "source":  "24-25四年级下·湖北十堰·期末"
+    },
+    {
+        "id":  "500",
+        "unit":  "第六单元小数的加法和减法",
+        "question_type":  "解答题",
+        "difficulty":  0.65,
+        "question":  "父亲节那天，爸爸在家人微信群发了一个120元的红包，妈妈抢到了34.08元，姐姐抢到了11.92元，还剩下多少钱没有被领取。",
+        "options":  [
+
+                    ],
+        "answer":  "74元",
+        "solution":  "【分析】总金额减去妈妈和姐姐领取金额，可用减法性质简算.\n【详解】120-34.08-11.92=120-(34.08+11.92)=120-46=74(元)。答：还剩下74元没有被领取。",
+        "knowledge_points":  [
+                                 "与小数减法相关的简便运算"
+                             ],
+        "source":  "24-25四年级下·湖北荆州·期末"
+    }
+]
diff --git "a/\345\233\233\345\271\264\347\272\247\351\242\230\351\233\206/\345\233\233\345\271\264\347\272\247\344\270\213/\345\233\233\345\271\264\347\272\247\344\270\213\347\254\254\345\233\233\345\215\225\345\205\203\345\260\217\346\225\260\347\232\204\346\204\217\344\271\211.json" "b/\345\233\233\345\271\264\347\272\247\351\242\230\351\233\206/\345\233\233\345\271\264\347\272\247\344\270\213/\345\233\233\345\271\264\347\272\247\344\270\213\347\254\254\345\233\233\345\215\225\345\205\203\345\260\217\346\225\260\347\232\204\346\204\217\344\271\211.json"
new file mode 100644
index 0000000000000000000000000000000000000000..c59c53bdc56bb0903ceacf851a74cc7b1252bf0d
--- /dev/null
+++ "b/\345\233\233\345\271\264\347\272\247\351\242\230\351\233\206/\345\233\233\345\271\264\347\272\247\344\270\213/\345\233\233\345\271\264\347\272\247\344\270\213\347\254\254\345\233\233\345\215\225\345\205\203\345\260\217\346\225\260\347\232\204\346\204\217\344\271\211.json"
@@ -0,0 +1,8467 @@
+[
+    {
+        "id":  "1",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "2024年国庆期间，苏州的旅游景点和商圈也迎来了大量游客，全市累计接待游客约(12856000)人次。把括号内的数改写成用\"亿\"作单位的数，是()亿，把改写后的数保留两位小数是()亿。",
+        "options":  [
+
+                    ],
+        "answer":  "0.12856;0.13",
+        "solution":  "【分析】要把12856000改写成用\"亿\"作单位的数,1亿等于100000000，得到0.12856亿;对0.12856亿保留两位小数，需要看千分位上的数字，这里千分位是8，根据\"四舍五入\"的规则,8大于5，要向百分位进1，百分位的2加1变成3，所以保留两位小数的结果是0.13亿.\n【详解】1亿=100000000,12856000=0.12856亿，将12856000改写成用\"亿\"作单位的数是0.12856亿，保留两位小数后为0.13亿。",
+        "knowledge_points":  [
+                                 "小数的改写",
+                                 "小数的近似数",
+                                 "亿以上数的改写"
+                             ],
+        "source":  "24-25五年级上·江苏苏州·期末"
+    },
+    {
+        "id":  "2",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在()里填上\"\u003e\"、\"\u003c\"或\"=\".\n800分()8元;3日()70时;5L400mL()5.05L\n3.3时()3时30分;8500g()8.05kg;5t60kg()5.6t",
+        "options":  [
+
+                    ],
+        "answer":  "=;\u003e;\u003e;\u003c;\u003e;\u003c",
+        "solution":  "【分析】带单位比大小，若单位相同，数字大的数就大;若单位不同，先把单位换成相同的，再比较数的大小.\n【详解】(1)8元=800分,800分=800分，所以800分(=)8元。(2)3日=72时,72时\u003e70时，所以3日(\u003e)70时。(3)5L400mL=5.4L,5.4L\u003e5.05L,所以5L400mL(\u003e)5.05L。(4)3.3时=3时18分,3时18分\u003c3时30分，所以3.3时(\u003c)3时30分。(5)8.05kg=8050g,8500g\u003e8050g,所以8500g(\u003e)8.05kg。(6)5t60kg=5.06,5.06t\u003c5.6t,所以5t60kg(\u003c)5.6t。",
+        "knowledge_points":  [
+                                 "吨，千克之间的换算与比较",
+                                 "克，千克之间的换算与比较",
+                                 "元，角，分的认识及换算",
+                                 "时，分，秒之间的换算与比较"
+                             ],
+        "source":  "25-26六年级·全国·随堂练习"
+    },
+    {
+        "id":  "3",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "张大爷绕一个长1000米的长方形果园走一圈，一共走了3600米。这个果园的面积是()平方米，合()平方千米。",
+        "options":  [
+
+                    ],
+        "answer":  "800000;0.8",
+        "solution":  "【分析】长方形的周长公式为：周长=(长+宽)×2，先计算长与宽的和，即周长除以2;再用长与宽的和减去长，得到宽;再根据长方形的面积公式为：面积=长×宽，代入数据计算得出长方形果园的面积。注意单位的换算:1公顷=10000平方米,1平方千米=100公顷，所以1平方千米=1000000平方米，从平方米换算成平方千米，除以进率.\n【详解】3600÷2-1000=1800-1000=800(米);1000×800=800000(平方米);800000÷1000000=0.8(平方千米)。因此，张大爷绕一个长1000米的长方形果园走一圈，一共走了3600米。这个果园的面积是800000平方米，合0.8平方千米。",
+        "knowledge_points":  [
+                                 "长方形的周长",
+                                 "长方形的面积",
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26五年级上·湖南邵阳·期末"
+    },
+    {
+        "id":  "4",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "2024年国庆期间，江苏全省共接待游客(38837400)人次，把括号内的数改写成用\"万\"作单位的数是()万，改写成用\"亿\"作单位并保留两位小数约是()亿。",
+        "options":  [
+
+                    ],
+        "answer":  "3883.74;0.39",
+        "solution":  "【分析】改写成用\"万\"作单位的数，在万位数的右下角点上小数点，把小数末尾的0去掉，再在数的后面写上\"万\"字。改写成用\"亿\"作单位的数，在亿位数的右下角点上小数点，把小数末尾的0去掉，再在数的后面写上\"亿\"字;保留两位小数，就是精确到百分位，需看千分位上的数字，根据\"四舍五入\"法取近似值.\n【详解】38837400=3883.74万,38837400=0.388374亿≈0.39亿，把括号内的数改写成用\"万\"作单位的数是3883.74万，改写成用\"亿\"作单位并保留两位小数约是0.39亿。",
+        "knowledge_points":  [
+                                 "亿以内数的组成",
+                                 "小数的改写",
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·湖南邵阳·期末"
+    },
+    {
+        "id":  "5",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上合适的数.\n3.026km=()km()m;4kg-490g=()kg",
+        "options":  [
+
+                    ],
+        "answer":  "3;26;3.51",
+        "solution":  "【分析】因为1km=1000m,3.026km的整数部分3表示3km,小数部分0.026km换算为米为26m,所以3.026km=3km26m。先统一单位，因为1kg=1000g,所以490g=0.49kg,则4kg-490g=3.51kg.\n【详解】0.026×1000=26m;3.026km=3km26m;490g=490÷1000=0.49kg;4kg-490g=4kg-0.49kg=3.51kg;",
+        "knowledge_points":  [
+                                 "千米和米之间的进率与换算",
+                                 "克，千克之间的换算与比较"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "6",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "截至2025年5月18日，据网络平台数据显示，国产动画电影《哪吒之魔童闹海》总票房达到(一百五十八亿五千五百二十六万)元，登顶全球动画电影票房榜首!\n(1)括号内的数字写作()元，省略亿位后面的尾数约是()亿元.\n(2)电影播放总时长144分钟，合()小时。",
+        "options":  [
+
+                    ],
+        "answer":  "15855260000;159;2.4",
+        "solution":  "【分析】(1)写数时，从高位写起，先写亿级，再写万级，最后写个级，哪个数位上一个单位也没有，就在那个数位上写0;省略\"亿\"后面的尾数就是四舍五入到亿位，就是把亿位后的千万位上的数进行四舍五入，再在数的后面写上\"亿\"字;(2)1小时=60分，低级单位换算成高级单位，用144÷60即可解答.\n【详解】(1)一百五十八亿五千五百二十六万写作:15855260000,15855260000≈159亿，一百五十八亿五千五百二十六万元写作15855260000元，省略亿位后面的尾数约是159亿元.\n(2)1小时=60分,144分=2小时24分,2小时24分=2.4小时，所以电影播放总时长144分钟，合2.4小时。",
+        "knowledge_points":  [
+                                 "亿以上数的读，写法",
+                                 "时，分的认识",
+                                 "根据小数的意义进行单位换算",
+                                 "亿以上数的近似数"
+                             ],
+        "source":  "25-26四年级上·黑龙江鸡西·期末"
+    },
+    {
+        "id":  "7",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数四舍五入到百分位后是2.40，这个三位小数最小是(),最大是()。",
+        "options":  [
+
+                    ],
+        "answer":  "2.395;2.404",
+        "solution":  "【分析】小数的近似数：根据\"四舍五入\"法求近似数的方法，找到要求保留的数位，看下一位;如果下一位的数字大于或等于5，要往前进一;如果下一位的数字小于5，要舍去;据此解答.\n【详解】一个三位小数四舍五入到百分位是2.40，这个最小的三位小数用\"五入\"法得近似数2.40，它的千分位上的数要大于或等于5，大于5或等于5的数有5~9，最小是5，所以这个三位小数最小是2.395。这个最大的三位小数用\"四舍\"法得近似数2.40，它的千分位上的数要小于5，小于5的数有0~4，最大是4，所以这个三位小数最大是2.404",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·广东揭阳·期末"
+    },
+    {
+        "id":  "8",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "28.28读作(),20.005读作()。",
+        "options":  [
+
+                    ],
+        "answer":  "二十八点二八;二十点零零五",
+        "solution":  "【分析】小数的读法：整数部分按整数的读法来读，小数点读作点，小数部分要依次读出每个数字，据此解答.\n【详解】28.28读作二十八点二八,20.005读作二十点零零五。",
+        "knowledge_points":  [
+                                 "小数的读法和写法"
+                             ],
+        "source":  "25-26四年级上·新疆省直辖县级单位·期末"
+    },
+    {
+        "id":  "9",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "540厘米=()米;5200平方米=()公顷;3.2吨=()千克;400000平方米=()平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "5.4;0.52;3200;0.4",
+        "solution":  "【分析】因为1米=100厘米，厘米换算为米，是小单位换算为大单位，要除以进率100;因为1公顷=10000平方米，平方米换算为公顷，是小单位换算为大单位，要除以进率10000;因为1吨=1000千克，吨换算为千克，是大单位换算为小单位，要乘进率1000;因为1平方千米=100公顷=1000000平方米，平方米换算为平方千米，是小单位换算为大单位，要除以进率1000000.\n【详解】540÷100=5.4，所以540厘米=5.4米;5200÷10000=0.52，所以5200平方米=0.52公顷;3.2×1000=3200，所以3.2吨=3200千克;400000÷1000000=0.4，所以400000平方米=0.4平方千米。",
+        "knowledge_points":  [
+                                 "吨，千克之间的换算与比较",
+                                 "面积单位间的进率及换算",
+                                 "运用小数点移动解决小数的单位换算问题"
+                             ],
+        "source":  "25-26五年级上·吉林辽源·期末"
+    },
+    {
+        "id":  "10",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "2025年福安国民生产总值(17670114000)元，把括号内的数改写成用\"亿元\"作单位的数是()亿元，把改写后的数保留两位小数写出近似数是()亿元。",
+        "options":  [
+
+                    ],
+        "answer":  "176.70114;176.70",
+        "solution":  "【分析】改写成用\"亿元\"作单位，需要将原数(17670114000)的小数点向左移动8位，得到176.70114亿元;保留两位小数时，需看小数点后第三位数字(即千分位上的数),根据\"四舍五入\"法写出近似数.\n【详解】根据分析可知:17670114000=176.70114亿;176.70114千分位上是1,1\u003c5，舍去，所以176.70114≈176.70。因此,2025年福安国民生产总值17670114000元，把括号内的数改写成用\"亿元\"作单位的数是176.70114亿元，把改写后的数保留两位小数写出近似数是176.70亿元。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·吉林辽源·期末"
+    },
+    {
+        "id":  "11",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "2025年人口普查结果，福安人口为(563640)人。把括号内的数改写成用\"万人\"作单位的数是()万人，把改写后的数保留一位小数写出近似数是()万人。",
+        "options":  [
+
+                    ],
+        "answer":  "56.364;56.4",
+        "solution":  "【分析】将原数563640改写成用\"万人\"作单位的数，需要在万位后面点上小数点，再加上\"万\"字，得到56.364万人;将56.364保留一位小数，需看百分位上的数字，根据\"四舍五入\"法写出近似数.\n【详解】563640=56.364万,56.364的百分位上是6,6\u003e5，向前一位进1,3+1=4，所以56.364≈56.4，因此,2025年人口普查结果，福安人口为563640人。把括号内的数改写成用\"万人\"作单位的数是56.364万人，把改写后的数保留一位小数写出近似数是56.4万人。",
+        "knowledge_points":  [
+                                 "小数的改写",
+                                 "小数的近似数",
+                                 "亿以内数的改写"
+                             ],
+        "source":  "25-26五年级上·吉林辽源·期末"
+    },
+    {
+        "id":  "12",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "把3.67扩大100倍是(),缩小到它的1÷10是()。",
+        "options":  [
+
+                    ],
+        "answer":  "367;0.367",
+        "solution":  "【分析】一个数扩大100倍是将小数点向右移动两位，缩小到它的1÷10是将小数点向左移动一位.\n【详解】根据分析,3.67小数点向右移动两位是367，小数点向左移动一位是0.367;即把3.67扩大100倍是367，缩小到它的1÷10是0.367",
+        "knowledge_points":  [
+                                 "小数点向左移动引起小数大小变化的规律",
+                                 "小数点向右移动引起小数大小变化的规律"
+                             ],
+        "source":  "25-26五年级上·新疆和田·期末"
+    },
+    {
+        "id":  "13",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "40.5平方千米=()公顷;60800平方米=()公顷;5700000平方米=( )平方千米;3.4公顷=( )平方米",
+        "options":  [
+
+                    ],
+        "answer":  "4050;6.08;5.7;34000",
+        "solution":  "【分析】1平方千米=100公顷，高级单位换算低级单位乘进率，即40.5×100;1公顷=10000平方米，低级单位换算高级单位除以进率，即60800÷10000;1平方千米=1000000平方米，低级单位换算高级单位除以进率，即5700000÷1000000;1公顷=10000平方米，高级单位换算低级单位乘进率，即3.4×10000.\n【详解】40.5平方千米=40.5×100=4050公顷;60800平方米=60800÷10000=6.08公顷;5700000平方米=5700000÷1000000=5.7平方千米\n3.4公顷=3.4×10000=34000平方米;所以,40.5平方千米=4050公顷,60800平方米=6.08公顷,5700000平方米=5.7平方千米,3.4公顷=34000平方米。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "面积单位间的进率及换算",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26五年级上·天津北辰·期末"
+    },
+    {
+        "id":  "14",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "第19届亚运会在中国杭州举行。作为亚运会主场馆的杭州奥体博览城核心区占地1543700平方米，合()公顷。建筑面积约2720000平方米，把这个数改写成用\"万平方米\"作单位的数是()万平方米。亚运会开幕式电视端累计观众收视人次达199086782次，这个数省略亿位后面的尾数约是()次。",
+        "options":  [
+
+                    ],
+        "answer":  "154.37;272;2亿",
+        "solution":  "【分析】把高级单位的名数换算成低级单位的名数，用高级单位的数乘进率;把低级单位的名数换算成高级单位的名数，用低级单位的数除以进率.\n【详解】1公顷=10000平方米,1543700÷10000=154.37(公顷),即作为亚运会主场馆的杭州奥体博览城核心区占地1543700平方米，合(154.37)公顷。2720000平方米=272万平方米，建筑面积约2720000平方米，把这个数改写成用\"万平方米\"作单位的数是(272)万平方米。199086782≈2亿，这个数省略亿位后面的尾数约是(2亿)次。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "亿以上数的近似数"
+                             ],
+        "source":  "2025·浙江·小升初模拟"
+    },
+    {
+        "id":  "15",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "()平方米=0.39公顷;6.08公顷=()公顷()平方米。",
+        "options":  [
+
+                    ],
+        "answer":  "3900;6;800",
+        "solution":  "【分析】根据进率:1公顷=10000平方米;从高级单位向低级单位转换，乘进率;从低级单位向高级单位转换，除以进率;据此解答.\n【详解】(1)0.39×10000=3900(平方米),所以3900平方米=0.39公顷;(2)6.08公顷=6公顷+0.08公顷,0.08×10000=800(平方米),所以6.08公顷=6公顷800平方米。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "25-26五年级上·河北唐山·期末"
+    },
+    {
+        "id":  "16",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把2.9635保留一位小数是(),保留三位小数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "3.0;2.964",
+        "solution":  "【分析】根据题意，保留一位小数需看百分位上的数字，保留三位小数需看万分位上的数字，再按照四舍五入的规则进行取舍，据此解答.\n【详解】保留一位小数:2.9635百分位为6，向十分位进1，再向个位进1，得3.0。保留三位小数:2.9635万分位为5，向千分位进1，得2.964综上所述可得，把2.9635保留一位小数是3.0，保留三位小数是2.964",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·河北唐山·期末"
+    },
+    {
+        "id":  "17",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上合适的数.\n0.08平方千米=()公顷;8.06千米=()米;78平方分米=()平方米;12元5分=()元;2公顷=()平方千米=()平方米",
+        "options":  [
+
+                    ],
+        "answer":  "8;8060;0.78;12.05;0.02;20000",
+        "solution":  "【分析】(1)1平方千米=100公顷，大单位换算成小单位时乘进率;(2)1千米=1000米，大单位换算成小单位时乘进率;(3)1平方米=100平方分米，小单位换算成大单位时除以进率;(4)1元=100分，小单位换算成大单位时除以进率;(5)1平方千米=100公顷,1公顷=10000平方米，公顷化平方千米，小单位换算成大单位时除以进率;公顷化平方米，大单位换算成小单位时乘进率.\n【详解】(1)0.08×100=8(公顷),所以0.08平方千米=8公顷。(2)8.06×1000=8060(米),所以8.06千米=8060米。(3)78÷100=0.78(平方米),所以78平方分米=0.78平方米。(4)5÷100=0.05(元),12+0.05=12.05(元),所以12元5分=12.05元。(5)2÷100=0.02(平方千米),2×10000=20000(平方米),所以2公顷=0.02平方千米=20000平方米。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "公顷，平方千米的进率与换算",
+                                 "平方厘米，平方分米，平方米之间的进率与换算",
+                                 "千米和米之间的进率与换算"
+                             ],
+        "source":  "25-26五年级上·河北邢台·期末"
+    },
+    {
+        "id":  "18",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "由3个亿,5个百万,6个千组成的数，写作(),将它改写成用\"亿\"为单位的数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "305006000;3.05006亿",
+        "solution":  "【分析】整数的数位顺序表如下:\n| 数级 | 。 | 亿级 | 万级 | 个级 | | | | | | | | | |\n| -------- | ---- | ------ | ------ | ------ | ---- | ------ | ------ | ------ | ---- | ---- | ---- | ---- | ---- |\n| 数位 | 。 | 千亿位 | 百亿位 | 十亿位 | 亿位 | 千万位 | 百万位 | 十万位 | 万位 | 千位 | 百位 | 十位 | 个位 |\n| 计数单位 | 。 | 千亿 | 百亿 | 十亿 | 亿 | 千万 | 百万 | 十万 | 万 | 千 | 百 | 十 | 个 |\n把一个非整亿数改写成用\"亿\"作单位的数，只需要找到亿位，在亿位数的右下角点上小数点，然后把小数末尾的0去掉，最后在数的后面写上\"亿\"字。据此解答.\n【详解】由题意得，一个数由3个亿,5个百万和6个千组成，所以这个数的亿位上是3，百万位上是5，千位上是6，其余数位上是0，所以这个数是305006000.305006000=3.05006亿。由3个亿,5个百万,6个千组成的数，写作305006000，将它改写成用\"亿\"为单位的数是3.05006亿。",
+        "knowledge_points":  [
+                                 "亿以上数的组成",
+                                 "小数的改写"
+                             ],
+        "source":  "25-26四年级上·甘肃天水·期末"
+    },
+    {
+        "id":  "19",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个两位小数，保留一位小数后是1.4，这个两位小数最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "1.35",
+        "solution":  "【分析】要考虑1.4是一个两位小数的近似数，有两种情况：\"四舍\"得到的1.4，有1.41,1.42,1.43,1.44，其中最大是1.44;\"五入\"得到的1.4，有1.35,1.36,1.37,1.38,1.39，其中最小是1.35.\n【详解】一个两位小数，保留一位小数后是1.4，这个两位小数最小是(1.35)。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·河北唐山·期末"
+    },
+    {
+        "id":  "20",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "世界第一大洋太平洋面积是(一亿八千一百三十四万四千)平方千米，括号内的数写作()平方千米，改写成以\"亿\"作单位保留一位小数约是()亿平方千米。",
+        "options":  [
+
+                    ],
+        "answer":  "181344000;1.8",
+        "solution":  "【分析】写数时，先把\"一亿八千一百三十四万四千\"按亿级，万级，个级分级，依次写出每一级的数字，缺位补0，得到对应的整数;改写并保留一位小数时，先将这个数换算成以\"亿\"为单位的小数，再根据四舍五入法，看小数百分位上的数字，判断是否向十分位进1，最终得到保留一位小数的结果.\n【详解】181344000=1.81344亿,1.81344≈1.8，世界第一大洋太平洋面积是一亿八千一百三十四万四千平方千米，横线上的数写作181344000平方千米，改写成以\"亿\"作单位保留一位小数约是1.8亿平方千米。",
+        "knowledge_points":  [
+                                 "亿以上数的读，写法",
+                                 "小数的近似数"
+                             ],
+        "source":  "2025·广东汕头·小升初模拟"
+    },
+    {
+        "id":  "21",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上合适的数.\n3公顷50平方米=()平方米;0.3平方千米=()公顷;650千克=()吨",
+        "options":  [
+
+                    ],
+        "answer":  "30050;30;0.65",
+        "solution":  "【分析】1公顷=10000平方米，先用乘法把3公顷转化为30000平方米，再加上50平方米;1平方千米=100公顷，用题目中的数据乘进率，把高级单位转化为低级单位，即0.3×100;1吨=1000千克，用题目中的数据除以进率，把低级单位转化为高级单位，即650÷1000，据此解答.\n【详解】3公顷50平方米=(3×10000)平方米+50平方米=30000平方米+50平方米=30050平方米,0.3平方千米=0.3×100=30公顷,650千克=650÷1000=0.65吨，所以,3公顷50平方米=30050平方米,0.3平方千米=30公顷,650千克=0.65吨。",
+        "knowledge_points":  [
+                                 "吨，千克之间的换算与比较",
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算",
+                                 "运用小数点移动解决小数的单位换算问题"
+                             ],
+        "source":  "25-26五年级上·海南省直辖县级单位·期末"
+    },
+    {
+        "id":  "22",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "200.68读作(),它由2个百,()个十分之一和()个百分之一组成。",
+        "options":  [
+
+                    ],
+        "answer":  "二百点六八;6;8",
+        "solution":  "【分析】小数的读法：整数部分按照整数的读法来读，小数点读作\"点\"，小数部分按顺序读出每一个数位上面的数字;小数点左边第一位是个位，计数单位是一(个);小数点左边第二位是十位，计数单位是十;小数点左边第三位是百位，计数单位是百;小数点左边第四位是千位，计数单位是千。小数点右边第一位是十分位，计数单位是十分之一(0.1);小数点右边第二位是百分位，计数单位是百分之一(0.01);小数点右边第三位是千分位，计单位是千分之一(0.001)。每相邻两个计数单位之间的进率都是10，据此解答.\n【详解】分析可知,200.68读作二百点六八，它由2个百,6个十分之一和8个百分之一组成。",
+        "knowledge_points":  [
+                                 "小数的读法和写法",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "25-26五年级上·海南省直辖县级单位·期末"
+    },
+    {
+        "id":  "23",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "118900改写成用\"万\"作单位的数是()万;3067000000改写成用\"亿\"作单位的数是()亿，精确到十分位约是()亿。",
+        "options":  [
+
+                    ],
+        "answer":  "11.89;30.67;30.7",
+        "solution":  "【分析】改写成用\"万\"作单位的数，就是在万位数的右下角点上小数点，然后把小数末尾的0去掉，再在数的后面写上\"万\"字;改写成用\"亿\"作单位的数，就是在亿位数的右下角点上小数点，然后把小数末尾的0去掉，再在数的后面写上\"亿\"字;小数取近似数时，观察保留位数后面的数字，满5向保留位数进一，不满5直接舍去，据此解答.\n【详解】分析可知,118900改写成用\"万\"作单位的数是11.89万;3067000000改写成用\"亿\"作单位的数是30.67亿，百分位上的数字是7，需要向前一位进一，所以精确到十分位约是30.7亿。",
+        "knowledge_points":  [
+                                 "小数的改写",
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·海南省直辖县级单位·期末"
+    },
+    {
+        "id":  "24",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "武冈是\"铜鹅之乡\"，\"卤菜之都\"。其中武冈豆腐品牌价值达(六十三亿零三百万)元，括号内的数写作(),改写成用亿做单位的数是()亿。",
+        "options":  [
+
+                    ],
+        "answer":  "6303000000;63.03",
+        "solution":  "【分析】(1)根据整数的写法，从高位到低位，一级一级地写，哪一个数位上一个单位也没有，就在那个数位上写0。(2)改写成用\"亿\"作单位的数，就是在亿位数的右下角点上小数点，然后把小数末尾的0去掉，在数的后面带上\"亿\"字.\n【详解】亿级：六十三亿，写作63;万级：零三百万，写作0300;个级：没有具体的数字，全部补0，即0000。因此，六十三亿零三百万写作:6303000000。(2)6303000000的亿位是3，在亿位右下角点上小数点，为63.03000000，去掉小数末尾的0，加上单位\"亿\"。即6303000000改写成用亿做单位的数是63.03亿。",
+        "knowledge_points":  [
+                                 "小数的改写",
+                                 "亿以上数的读，写法"
+                             ],
+        "source":  "2025·湖南邵阳·小升初模拟"
+    },
+    {
+        "id":  "25",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数四舍五入保留两位小数是9.28，这个数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "9.284;9.275",
+        "solution":  "【分析】要考虑9.28是一个三位小数的近似数，有两种情况：\"四舍\"得到的9.28，最大是9.284，\"五入\"得到的9.28，最小是9.275，由此填空即可.\n【详解】一个三位小数四舍五入保留两位小数是9.28，这个数最大是9.284，最小是9.275。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·辽宁盘锦·期末"
+    },
+    {
+        "id":  "26",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数\"四舍五入\"后是10.01，这个数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "10.014;10.005",
+        "solution":  "【分析】根据题意，根据分析可知：一个三位小数\"四舍五入\"后是10.01，表示这个数精确到百分位后是10.01。根据四舍五入规则，千分位上的数字决定是否进位：如果千分位小于5，则舍去，原数的百分位为1;如果千分位大于或等于5，则进位，原数的百分位为0。分别找出所有可能值后，比较得出最大和最小数。这个数最大是10.014，最小是10.005。以此答题即可.\n【详解】根据分析可知：一个三位小数\"四舍五入\"后是10.01，这个数最大是10.014，最小是10.005。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·浙江杭州·期末"
+    },
+    {
+        "id":  "27",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "958000000读作(),改写成以\"亿\"作单位是()亿。",
+        "options":  [
+
+                    ],
+        "answer":  "九亿五千八百万;9.58",
+        "solution":  "【分析】整数的读法：从高位到低位，一级一级地读，每级末尾的0都不读出来，其他数位连续几个0都只读一个\"零\"。改写成用\"亿\"作单位的数，就是在亿位数的右下角点上小数点，然后把小数末尾的0去掉，再在数的后面写上\"亿\"字.\n【详解】958000000=9.58亿,958000000读作(九亿五千八百万),改写成以\"亿\"作单位是(9.58)亿。",
+        "knowledge_points":  [
+                                 "亿以上数的改写",
+                                 "小数的改写",
+                                 "亿以上数的读，写法"
+                             ],
+        "source":  "2025·广东汕头·小升初模拟"
+    },
+    {
+        "id":  "28",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "地球表面陆地面积是149000000平方千米，把这个数改写成用\"亿\"作单位的数是()亿平方千米;海洋面积是361000000平方千米，把这个数省略\"亿\"后面的尾数约是()亿平方千米。",
+        "options":  [
+
+                    ],
+        "answer":  "1.49;4",
+        "solution":  "【分析】把149000000平方千米改写成用\"亿平方千米\"作单位的数，在亿位的右下角点上小数点，把末尾的0去掉同时在后面写上\"亿\"字;把361000000平方千米省略亿位后面的尾数求近似数约是多少亿平方千米，因为361000000平方千米的千万位上的数字是6大于5，所以用\"五入\"法保留近似数.\n【详解】149000000平方千米=1.49亿平方千米;361000000平方千米≈4亿平方千米。所以地球表面陆地面积是149000000平方千米，把这个数改写成用\"亿\"作单位的数是1.49亿平方千米;海洋面积是361000000平方千米，把这个数省略\"亿\"后面的尾数约是4亿平方千米。",
+        "knowledge_points":  [
+                                 "亿以上数的近似数",
+                                 "小数的改写"
+                             ],
+        "source":  "2025·四川绵阳·小升初真题"
+    },
+    {
+        "id":  "29",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "地球上最大的海洋是太平洋，它的面积是(179968000)平方千米，括号内的数读作(),改写成\"亿\"作单位并保留一位小数约是()亿平方千米。",
+        "options":  [
+
+                    ],
+        "answer":  "一亿七千九百九十六万八千;1.8",
+        "solution":  "【分析】整数的读法：从高位到低位，一级一级地读，每级都按照个级的读法来读，读完亿级读一个亿字，读完万级读一个万字，每一级末尾的0都不读出来，其他数位连续几个0都只读一个\"零\"。改写成\"亿\"作单位的数，在亿位的右下角点个小数点，并去掉末尾的0，同时末尾加个亿字，保留一位小数，看百分位上的数，百分位上的数大于或等于5，则进一，小于5，则舍去.\n【详解】179968000读作一亿七千九百九十六万八千;179968000≈1.8亿;即地球上最大的海洋是太平洋，它的面积是(179968000)平方千米，括号内的数读作一亿七千九百九十六万八千，改写成\"亿\"作单位并保留一位小数约是1.8亿平方千米。",
+        "knowledge_points":  [
+                                 "亿以上数的改写",
+                                 "小数的近似数",
+                                 "亿以上数的读，写法"
+                             ],
+        "source":  "2025·浙江宁波·小升初模拟"
+    },
+    {
+        "id":  "30",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "五一节期间，杭州西湖接待游客约563400人次，改写成用\"万\"作单位的数是()万人次，省略\"万\"后面的尾数约是()万人次。",
+        "options":  [
+
+                    ],
+        "answer":  "56.34;56",
+        "solution":  "【分析】根据题意，改写成用万作单位的数，就是在万位数的右下角点上小数点，然后把小数末尾的0去掉，再在数的后面写上\"万\"字，据此改写。省略\"万\"后面的尾数求它的近似数，要看万位的下一位千位上的数进行四舍五入，再在数的后面带上\"万\"字，据此解答.\n【详解】所以,563400改写成用\"万\"作单位的数是56.34万人次。563400的千位上是3，需要舍去，省略\"万\"后面的尾数约是56万人次。",
+        "knowledge_points":  [
+                                 "亿以内数的近似数",
+                                 "小数的改写"
+                             ],
+        "source":  "2025·浙江杭州·小升初模拟"
+    },
+    {
+        "id":  "31",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "将最小的三位数与最大的两位数的乘积缩小到原来的1÷10000是()。",
+        "options":  [
+
+                    ],
+        "answer":  "0.99",
+        "solution":  "【分析】小数点向左移动四位缩小到原来的1÷10000。最小的三位数是100，最大的两位数是99，求出100和99的积，再将这个积的小数点向左移动四位即可.\n【详解】100×99=9900,9900小数点向左移动四位是0.99，将最小的三位数与最大的两位数的乘积缩小到原来的1÷10000是0.99。",
+        "knowledge_points":  [
+                                 "小数点向左移动引起小数大小变化的规律",
+                                 "三位数乘两位数，乘数末尾有0"
+                             ],
+        "source":  "25-26五年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "32",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数\"四舍五入\"后是2.56，这个三位小数最大可能是(),最小可能是()。",
+        "options":  [
+
+                    ],
+        "answer":  "2.564;2.555",
+        "solution":  "【分析】根据题意可得：一个三位小数\"四舍五入\"后是2.56，则这个三位小数最大的情况：百分位上是6，则千分位上的数要小于5，即最大为4;三位小数最小的情况：百分位上是5，千分位上的数要大于或等于5，最小为5。据此可得出答案.\n【详解】这个三位小数最大可能是2.564，最小可能是2.555。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·甘肃武威·期末"
+    },
+    {
+        "id":  "33",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "中央大街是哈尔滨的标志性街道，全长约1.45千米，保留一位小数约是()千米。",
+        "options":  [
+
+                    ],
+        "answer":  "1.5",
+        "solution":  "【分析】结果保留一位小数，需要看小数点后第二位，根据\"四舍五入\"原则，小于5则直接舍去，大于等于5则需要向前进1.\n【详解】1.45小数点后第二位为5，满5需要进1,4+1=5，则中央大街是哈尔滨的标志性街道，全长约1.45千米，保留一位小数约是1.5千米。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "34",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "60000平方米=()公顷;3006000平方米=()平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "6;3.006",
+        "solution":  "【分析】根据1公顷=10000平方米,1平方千米=1000000平方米，单位小变大除以进率，进行换算即可.\n【详解】60000÷10000=6(公顷),60000平方米=6公顷;3006000÷1000000=3.006(平方千米),3006000平方米=3.006平方千米。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "25-26五年级上·吉林辽源·期末"
+    },
+    {
+        "id":  "35",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "成语\"百尺竿头，更进一步\"比喻到了极高的境地，仍需继续努力，争取更高的进步。其中\"百尺\"大约为现在的33米，那么\"一尺\"大约相当于现在的()米。",
+        "options":  [
+
+                    ],
+        "answer":  "0.33",
+        "solution":  "【分析】已知\"百尺\"等于33米，求一尺对应的米数，即要将其缩小为原来的1÷100，根据小数点位置的移动改变小数大小的规律，要将小数点向左移动两位，据此作答.\n【详解】根据分析,33的小数点向左移动两位变为0.33。因此一尺相当于现在的0.33米。",
+        "knowledge_points":  [
+                                 "小数点向左移动引起小数大小变化的规律"
+                             ],
+        "source":  "24-25五年级上·江苏盐城·期末"
+    },
+    {
+        "id":  "36",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数四舍五入后是1.60，这个三位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "1.604;1.595",
+        "solution":  "【分析】分析题目，这个三位小数如果是\"四舍\"得到1.60，则原来的小数比1.60大,1.60是把原来小数的千分位的数字直接去掉得到的，所以这个小数的整数部分还是1，十分位还是6，百分位的数字还是0，千分位的数字最大是4;这个三位小数如果是\"五入\"得到1.60，则原来的小数比1.60小,1.60是给原来的小数的百分位数字加1之后得到的，所以这个小数原来的百分位数字是9，十分位的数字是5，原来的整数部分是1，千分位的数字最小是5，据此解答.\n【详解】一个三位小数四舍五入后是1.60，这个三位小数最大是1.604，最小是1.595。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·河北邢台·期中"
+    },
+    {
+        "id":  "37",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数\"四舍五入\"后是3.06，这个数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "3.064;3.055",
+        "solution":  "【分析】先明确四舍五入到百分位，即看千分位的数，再分别考虑\"四舍\"得到最大数和\"五入\"得到最小数的情况.\n【详解】(1)要使三位小数四舍五入后近似数为3.06且最大，需满足\"四舍\"条件，即千分位上的数字小于5，此时千分位最大为4，整数部分，十分位，百分位分别为3,0,6，这个数最大是3.064。(2)要使三位小数四舍五入后近似数为3.06且最小，需满足\"五入\"条件，即千分位上的数字大于或等于5，此时千分位最小为5，整数部分，十分位为3,0，百分位为5(进位后变为6),这个数最小是3.055。因此，一个三位小数\"四舍五入\"后是3.06，这个数最大是3.064，最小是3.055。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·山西晋中·期中"
+    },
+    {
+        "id":  "38",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数，四舍五入后是4.68，这个三位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "4.684;4.675",
+        "solution":  "【分析】\"四舍\"法取近似值时，原数大于近似数，千分位上的数字最大是4，其它数位上面的数字不变，此时就是最大值的原数，即4.684;\"五入\"法取近似值时，原数小于近似数，千分位上的数字最小是5，百分位上的数字最小是7，其它数位上面的数字不变，此时就是最小值的原数，即4.675，据此解答.\n【详解】分析可知，一个三位小数，四舍五入后是4.68，这个三位小数最大是4.684，最小是4.675。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·甘肃天水·期末"
+    },
+    {
+        "id":  "39",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数，保留两位小数是5.76，这个数最小是(),最大是()。",
+        "options":  [
+
+                    ],
+        "answer":  "5.755;5.764",
+        "solution":  "【分析】取一个小数的近似数，有两种情况：\"四舍\"得到的近似数比原数小，\"五入\"得到的近似数比原数大;要考虑5.76是一个三位小数的近似数，有两种情况：\"四舍\"得到的5.76最大是5.764，\"五入\"得到的5.76最小是5.755，据此解答.\n【详解】根据分析可知，一个三位小数，保留两位小数是5.76，这个数最小是5.755，最大是5.764。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·辽宁鞍山·期末"
+    },
+    {
+        "id":  "40",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数\"四舍五入\"精确到百分位是6.00，这个三位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "6.004;5.995",
+        "solution":  "【分析】一个三位小数\"四舍五入\"精确到百分位是6.00，\"四舍\"得到的近似数比原数小，此时原数的千分位最大是4，其他数位上的数不变，即原数最大是6.004;\"五入\"得到的近似数比原数大，此时千分位最小是5，百分位和十分位最小是9，个位最小是5，即原数最小是5.995，据此解答.\n【详解】分析可知，一个三位小数\"四舍五入\"精确到百分位是6.00，这个三位小数最大是6.004，最小是5.995。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·浙江温州·期末"
+    },
+    {
+        "id":  "41",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个两位小数，用\"四舍五入\"法保留一位小数，得到的近似值是6.8，这个小数最小是(),最大是()。",
+        "options":  [
+
+                    ],
+        "answer":  "6.75;6.84",
+        "solution":  "【分析】取一个数的近似数，有两种情况：\"四舍\"得到的近似数比原数小，\"五入\"得到的近似数比原数大。要考虑6.8是一个两位小数\"四舍五入\"得到的最小是6.75，最大是6.84，据此解答.\n【详解】根据分析可知，一个两位小数，用\"四舍五入\"法保留一位小数，得到的近似值是6.8，这个小数最小是6.75，最大是6.84",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·重庆沙坪坝·期末"
+    },
+    {
+        "id":  "42",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "()dm²=50cm²;0.036km²=()m²;0.08hm²=()m²",
+        "options":  [
+
+                    ],
+        "answer":  "0.5;36000;800",
+        "solution":  "【分析】1dm²=100cm²;1km²=1000000m²;1hm²=10000m²;高级单位换算低级单位，乘进率;低级单位换算高级单位，除以进率，据此解答.\n【详解】50÷100=0.5(dm²),所以0.5dm²=50cm².\n0.036×1000000=36000(m²),所以0.036km²=36000m².\n0.08×10000=800(m²),所以0.08hm²=800m²。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "面积单位间的进率及换算",
+                                 "平方厘米，平方分米，平方米之间的进率与换算"
+                             ],
+        "source":  "24-25五年级上·重庆沙坪坝·期末"
+    },
+    {
+        "id":  "43",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数，四舍五入后是7.64，这个三位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "7.644;7.635",
+        "solution":  "【分析】先明确\"四舍五入\"到百分位的规则，即看千分位上的数;再分别考虑\"四舍\"得到最大值，\"五入\"得到最小值的情况.\n【详解】(1)要使三位小数\"四舍五入\"后是7.64且最大，需考虑\"四舍\"的情况。即千分位上的数字小于5时直接舍去，此时原数的前两位小数为7.64，千分位最大可取4(因为千分位若为5及以上则会进位，不符合\"四舍\"),因此，这个三位小数最大是7.644。(2)要使三位小数\"四舍五入\"后是7.64且最小，需考虑\"五入\"的情况。即千分位上的数字大于或等于5时向百分位进1，此时原数的前两位小数为7.63，千分位最小可取5(因为千分位为5时向百分位进1，百分位3加1变为4，得到7.64),因此，这个三位小数最小是7.635。所以，一个三位小数，四舍五入后是7.64，这个三位小数最大是7.644，最小是7.635。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·贵州铜仁·期末"
+    },
+    {
+        "id":  "44",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个两位小数，四舍五入后得到的近似数为7.0，这个两位小数最小是(),最大是()。",
+        "options":  [
+
+                    ],
+        "answer":  "6.95;7.04",
+        "solution":  "【分析】一个两位小数，四舍五入后得到的近似数为7.0，可以是四舍得到的，那么整数部分和十分位上是7.0，百分位上的数要舍去;百分位上的数可以是1,2,3,4;也可以是五入得到的，那么整数部分和十分位上是6.9，百分位上的数要进一，百分位上的数要大于等于5，百分位上的数可以是5,6,7,8,9;据此解答.\n【详解】由分析可知，一个两位小数，四舍五入后得到的近似数为7.0，这个两位小数最小是6.95，最大是7.04",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·浙江台州·期末"
+    },
+    {
+        "id":  "45",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数取近似值后是5.60，这个三位小数最小可能是(),最大可能是()。",
+        "options":  [
+
+                    ],
+        "answer":  "5.595;5.604",
+        "solution":  "【分析】保留两位小数看千分位，小于5直接舍去，大于或等于5向前一位进一，由此可知，四舍得到的近似数，原数\u003e近似数;五入得到的近似数，原数\u003c近似数，据此分析.\n【详解】一个三位小数取近似值后是5.60，考虑五入的情况，这个三位小数最小可能是5.595，考虑四舍的情况，最大可能是5.604。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·河南三门峡·期末"
+    },
+    {
+        "id":  "46",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "近似数是7.86的最大三位小数是(),最小三位小数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "7.864;7.855",
+        "solution":  "【分析】一个三位小数的近似数是7.86，有两种情况:①原数通过\"四舍\"得到7.86，则原数的个位上的数是7，十分位上的数是8，百分位上的数是6，千分位上的数可能是1,2,3,4，即原数可能是7.861,7.862,7.863,7.864，最大是7.864。②原数通过\"五入\"得到7.86，则原数的个位上的数是7，十分位上的数是8，百分位上的数是5，千分位上的数可能是5,6,7,8,9，即原数可能是7.855,7.856,7.857,7.858,7.859，最小是7.855.\n【详解】根据分析得出：四舍得到7.86，可能是7.861,7.862,7.863,7.864，最大是7.864;五入得到7.86，可能是7.855,7.856,7.857,7.858,7.859，最小是7.855。综上可知，近似数是7.86的最大三位小数是7.864，最小三位小数是7.855。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·江西宜春·期中"
+    },
+    {
+        "id":  "47",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "6050千克=()吨;2.5公顷=()平方米;2时15分=()时;3千米80米=()千米",
+        "options":  [
+
+                    ],
+        "answer":  "6.05;25000;2.25;3.08",
+        "solution":  "【分析】从高级单位到低级单位转化需要乘进率，从低级单位到高级单位转化需要除以进率。质量单位中,1吨=1000千克，因此6050千克转换为\"吨\"作单位需要除以进率1000;面积单位中,1公顷=10000平方米，因此2.5公顷转换为\"平方米\"作单位需要乘进率10000;时间单位中,1时=60分，因此15分转换为\"时\"作单位时需要除以进率60;长度单位中1千米=1000米，因此80米转换为\"千米\"作单位需要除以进率1000;据此解答.\n【详解】6050÷1000=6.05，所以6050千克=6.05吨;2.5×10000=25000，所以2.5公顷=25000平方米;15÷60=0.25，所以2时15分=2时+0.25时=2.25时，即2时15分=2.25时;80÷1000=0.08，所以3千米80米=3千米+0.08千米=3.08千米，即3千米80米=3.08千米。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "公顷，平方千米的进率与换算",
+                                 "千米和米之间的进率与换算",
+                                 "吨，千克之间的换算与比较"
+                             ],
+        "source":  "25-26五年级上·江西宜春·期中"
+    },
+    {
+        "id":  "48",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个两位小数\"四舍五入\"保留一位小数是9.9，这个两位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "9.94;9.85",
+        "solution":  "【分析】根据题意，先明确\"四舍五入\"保留一位小数的规则：看百分位上的数，若百分位上的数\u003e=5，则向十分位进1;若百分位上的数\u003c5，则舍去百分位及后面的数。要找这个两位小数的最大值，需考虑\"四舍\"的情况，即十分位原本是9，百分位最大取4(因为4\u003c5会被舍去),此时这个两位小数最大;要找最小值，需考虑\"五入\"的情况，即十分位原本是8，百分位最小取5(因为5\u003e=5会向十分位进1，使十分位的8变成9),此时这个两位小数最小，据此解答.\n【详解】(1)求最大的两位小数：\"四舍\"时，十分位是9，百分位最大为4，所以这个两位小数最大是9.94。(2)求最小的两位小数：\"五入\"时，十分位是8，百分位最小为5，所以这个两位小数最小是9.85。综上所述可得，这个两位小数最大是9.94，最小是9.85。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·浙江台州·期末"
+    },
+    {
+        "id":  "49",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个数是三位小数，将它四舍五入到百分位是3.98，这个三位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "3.984;3.975",
+        "solution":  "【分析】一个三位小数四舍五入到百分位是3.98，说明原数可能通过舍去千分位或向百分位进一得到3.98。因此，原数的范围是从3.975到3.984，其中最大是3.984，最小是3.975.\n【详解】当千分位\u003c5时，四舍五入后百分位不变，所以原数为3.981,3.982,3.983,3.984，此时原数最大为3.984。当千分位\u003e=5时，四舍五入向百分位进一，所以百分位原来是7，进一后变为8，原数为3.975,3.976,3.977,3.978,3.979，此时原数最小为3.975。因此，这个三位小数最大是3.984，最小是3.975。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·广东汕头·期中"
+    },
+    {
+        "id":  "50",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "把5.6789保留三位小数约是()。",
+        "options":  [
+
+                    ],
+        "answer":  "5.679",
+        "solution":  "【分析】根据\"四舍五入\"求近似数的方法，找到要求保留的数位，看下一位;如果下一位的数字大于或等于5，要往前进一;如果下一位的数字小于5，要舍去.\n【详解】5.6789≈5.679，把5.6789保留三位小数约是5.679。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·内蒙古通辽·期末"
+    },
+    {
+        "id":  "51",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "32.5厘米=()米;2时45分=()分;5千米800米=()千米",
+        "options":  [
+
+                    ],
+        "answer":  "0.325;165;5.8",
+        "solution":  "【分析】(1)1米=100厘米，低级单位换算高级单位除以进率，即32.5÷100;(2)1小时=60分，先用乘法把2时转化为120分，再加上45分;(3)1千米=1000米，先用除法把800米转化为0.8千米，再加上5千米，据此解答.\n【详解】32.5厘米=32.5÷100=0.325米;2时45分=2时+45分=(2×60)分+45分=120分+45分=165分;5千米800米=5千米+800米=5千米+(800÷1000)千米=5千米+0.8千米=5.8千米.\n综上所述,32.5厘米=0.325米,2时45分=165分,5千米800米=5.8千米。",
+        "knowledge_points":  [
+                                 "时，分的认识",
+                                 "厘米和米之间的进率与换算",
+                                 "千米和米之间的进率与换算",
+                                 "运用小数点移动解决小数的单位换算问题"
+                             ],
+        "source":  "25-26五年级上·江西宜春·期中"
+    },
+    {
+        "id":  "52",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个两位小数\"四舍五入\"后是4.6，这个两位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "4.64;4.55",
+        "solution":  "【分析】一个两位小数\"四舍五入\"后是4.6，若百分位小于5，则舍去，十分位不变，结果为4.6(例如4.60至4.64);若百分位大于等于5，则进位，但进位后十分位变为6，因此原始数的十分位必须为5且百分位大于等于5(例如4.55至4.59)。据此解答.\n【详解】一个两位小数\"四舍五入\"后是4.6，\"四舍\"得到的4.6比原数小，所以这个两位小数最大是4.64;\"五入\"得到的4.6比原数大，所以这个两位小数最小是4.55。因此这个两位小数最大是4.64，最小是4.55。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·天津滨海新·期末"
+    },
+    {
+        "id":  "53",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "青海湖是我国最大的内陆湖泊，湖水面积(4557.95)平方千米，蓄水量达1050亿立方米。括号内的数读作(),保留整数后约是()平方千米.\n![](images/063b0ea9a0c977a3e6d9180b2710c2e4f812fe06564e15ca55c3e9d5dbab75ee.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "四千五百五十七点九五;4558",
+        "solution":  "【分析】按照小数的读法规则，整数部分按照整数的读法来读，即4557读作四千五百五十七，小数点读作点，小数部分依次读出每一位上的数字,95读作九五，所以4557.95读作四千五百五十七点九五。根据四舍五入法，保留整数时看十分位上的数字,4557.95十分位是9,9大于5，则向个位进1，所以4557.95≈4558.\n【详解】青海湖是我国最大的内陆湖泊，湖水面积4557.95平方千米，蓄水量达1050亿立方米。括号内的数读作四千五百五十七点九五，保留整数后约是4558平方千米。",
+        "knowledge_points":  [
+                                 "小数的近似数",
+                                 "小数的读法和写法"
+                             ],
+        "source":  "24-25五年级上·重庆九龙坡·期末"
+    },
+    {
+        "id":  "54",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "3.7m²=()dm²;6km230m=()km",
+        "options":  [
+
+                    ],
+        "answer":  "370;6.23",
+        "solution":  "【分析】1m²=100dm²,把高级单位换算成低级单位要乘进率。1km=1000m,把低级单位换算成高级单位要除以进率.\n【详解】3.7×100=370(dm²),所以3.7m²=370dm²。6km230m-6km=230m,230÷1000=0.23(km),6+0.23=6.23(km),所以,6km230m=6.23km",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "平方厘米，平方分米，平方米之间的进率与换算",
+                                 "千米和米之间的进率与换算"
+                             ],
+        "source":  "24-25五年级上·广东云浮·期末"
+    },
+    {
+        "id":  "55",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数保留两位小数四舍五入后是2.40，那么这个三位小数最大可能是(),最小可能是()。",
+        "options":  [
+
+                    ],
+        "answer":  "2.404;2.395",
+        "solution":  "【分析】求保留两位小数后是2.40的三位小数的最大值时，需遵循\"四舍\"原则，即第三位小数要小于5，所以取最大的4，得到最大数2.404;求最小值时，需遵循\"五入\"原则，原数的前两位应为2.39，第三位小数要大于或等于5，取最小的5，得到最小数2.395.\n【详解】根据分析：一个三位小数保留两位小数四舍五入后是2.40，那么这个三位小数最大可能是2.404，最小可能是2.395。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·河北沧州·期中"
+    },
+    {
+        "id":  "56",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把4.7的小数点向右移动三位是();把60缩小到它的千分之一是()。",
+        "options":  [
+
+                    ],
+        "answer":  "4700;0.06",
+        "solution":  "【分析】先根据小数点位置移动引起数的大小变化规律，小数点向右移动一位，原数扩大到原来的10倍;向右移动两位，原数扩大到原来的100倍;向右移动三位，原数扩大到原来的1000倍。再将4.7的小数点向右移动三位。然后缩小到原数的千分之一，就是原数除以1000，相当于把小数点向左移动三位即可.\n【详解】根据小数点位置移动引起数的大小变化规律，小数点向右移动三位，原数扩大到原来的1000倍。所以4.7的小数点向右移动三位，就是4.7×1000=4700。缩小到原数的千分之一，就是除以1000，即60÷1000=0.06。把4.7的小数点向右移动三位是4700;把60缩小到它的千分之一是0.06。",
+        "knowledge_points":  [
+                                 "小数点向左移动引起小数大小变化的规律",
+                                 "小数点向右移动引起小数大小变化的规律"
+                             ],
+        "source":  "25-26五年级上·河南许昌·期中"
+    },
+    {
+        "id":  "57",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "把3.05改写成三位小数是();把4.500末尾的0去掉，小数的大小()。",
+        "options":  [
+
+                    ],
+        "answer":  "3.050;不变",
+        "solution":  "【分析】小数的性质：小数的末尾添上\"0\"或去掉\"0\"，小数的大小不变;改写指定位数小数时，仅在末尾补0，不改变小数大小.\n【详解】将3.05改写成三位小数根据小数的性质，在3.05的末尾添上1个0，得到三位小数3.050。去掉4.500末尾的0，小数大小变化情况去掉4.500末尾的0后变为4.5，由小数的性质可知，小数的大小不变。",
+        "knowledge_points":  [
+                                 "小数的性质"
+                             ],
+        "source":  "25-26六年级·全国·随堂练习"
+    },
+    {
+        "id":  "58",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "第六次人口普查中统计显示：诸暨市共计41个民族。总计人口为(1157938)人，其中汉族人口为1133923人。把括号内的数改写成以\"万\"为单位的数是(),把汉族人口数用\"四舍五入\"法省略万后面的尾数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "115.7938万;113万",
+        "solution":  "【分析】根据整数的改写方法，如果是整万的数，只要省略万位后面的0，并加一个\"万\"字。如果不是整万的数，要在万位的后边，点上小数点，去掉小数点末尾的0，并加上一个\"万\"字;整数的近似数，通过四舍五入法求整数的近似数，要对省略的尾数部分的最高位上的数进行四舍五入，若小于5则直接舍去，若大于或等于5，则向前进一位，并加上\"万\"字.\n【详解】1157938=115.7938万,1133923的千位上的数字是3，小于5，所以要舍去千位及后面的尾数,1133923≈113万。所以，把括号内的数改写成以\"万\"为单位的数是115.7938万，把汉族人口数用\"四舍五入\"法省略万后面的尾数是113万。",
+        "knowledge_points":  [
+                                 "小数点向左移动引起小数大小变化的规律",
+                                 "亿以内数的改写",
+                                 "亿以内数的近似数"
+                             ],
+        "source":  "2025·浙江温州·小升初真题"
+    },
+    {
+        "id":  "59",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "近似数是7.85的最大三位小数是(),最小三位小数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "7.854;7.845",
+        "solution":  "【分析】一个三位小数的近似数是7.85，有两种情况:①原数通过\"四舍\"得到7.85，则原数的个位上的数是7，十分位上的数是8，百分位上的数是5，千分位上的数可能是1,2,3,4，即原数可能是7.851,7.852,7.853,7.854;②原数通过\"五入\"得到7.85，则原数的个位上的数是7，十分位上的数是8，百分位上的数是4，千分位上的数可能是5,6,7,8,9，即原数可能是7.845,7.846,7.847,7.848,7.849;【详解】根据分析：四舍得到7.85，可能是7.851,7.852,7.853,7.854;最大是7.854;五入得到7.85，可能是7.845,7.846,7.847,7.848,7.849;最小是7.845。综上可知，近似数是7.85的最大三位小数是7.854，最小三位小数是7.845。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·广东江门·期中"
+    },
+    {
+        "id":  "60",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "去掉4.58的小数点，等于把它的小数点向右移动了()位，它就扩大到原来的()倍。",
+        "options":  [
+
+                    ],
+        "answer":  "两;100",
+        "solution":  "【分析】小数点位置移动引起数的大小变化规律：一个数的小数点向右移动一位、两位、三位，相当于把原数乘10、100、1000，这个数就扩大到原来的10倍、100倍、1000倍。据此解答.\n【详解】4.58去掉小数点后变为458，原小数点在\"4\"和\"5\"之间，去掉后小数点相当于向右移动到了\"8\"的右边，共移动了两位。根据小数点移动规律，小数点向右移动了两位，小数扩大到原来的100倍。",
+        "knowledge_points":  [
+                                 "小数点向右移动引起小数大小变化的规律"
+                             ],
+        "source":  "25-26五年级上·广东江门·期中"
+    },
+    {
+        "id":  "61",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数，它保留两位小数是30.00。这个小数最小是(),最大是()。",
+        "options":  [
+
+                    ],
+        "answer":  "29.995;30.004",
+        "solution":  "【分析】根据题意，取一个小数的近似数，有两种情况：\"四舍\"得到的近似数比原数小，\"五入\"得到的近似数比原数大，一个三位小数，如果它保留两位小数是30.00，要看千分位的数，如果千分位上的数大于或等于5，就要向百分位进一;如果小于5，则舍去，据此解答.\n【详解】根据分析可知，一个三位小数，它保留两位小数是30.00，这个数可能是29.995,29.996,29.997,29.998,29.999,30.001,30.002,30.003,30.004。所以这个小数最小是29.995，最大是30.004",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26六年级·全国·随堂练习"
+    },
+    {
+        "id":  "62",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "说出下面各数中的\"2\"表示的含义。25中的\"2\"表示(),0.12中的\"2\"表示()。",
+        "options":  [
+
+                    ],
+        "answer":  "2个十;2个0.01",
+        "solution":  "【分析】根据数位表的顺序判断，小数点左起依次是：个位，十位，百位。,计数单位对应是：个，十，百。;小数点右起依次是：十分位，百分位，千分位。,计数单位对应是:0.1,0.01,0.001。.\n【详解】25中的\"2\"在十位上，表示2个十;0.12中的\"2\"在百分位上，表示2个0.01.25中的\"2\"表示2个十,0.12中的\"2\"表示2个0.01。",
+        "knowledge_points":  [
+                                 "100以内数的组成",
+                                 "1000以内数的组成",
+                                 "整数的数级，数位和计数单位的认识",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "25-26六年级·全国·随堂练习"
+    },
+    {
+        "id":  "63",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "去掉3.14的小数点，等于把小数点向右移动了()位，它就扩大到原来的()倍。",
+        "options":  [
+
+                    ],
+        "answer":  "两;100",
+        "solution":  "【分析】去掉3.14的小数点后，数字变为314，这相当于将小数点向右移动两位。根据小数点移动的规律，向右移动两位，数字就扩大到原来的100倍.\n【详解】3.14去掉小数点后变为314，小数点向右移动两位。去掉3.14的小数点，等于把小数点向右移动了两位，它就扩大到原来的100倍。",
+        "knowledge_points":  [
+                                 "小数点向右移动引起小数大小变化的规律"
+                             ],
+        "source":  "25-26五年级上·黑龙江齐齐哈尔·期中"
+    },
+    {
+        "id":  "64",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "青海湖是我国最大的内陆湖泊，湖水面积(4557.95)平方千米，蓄水量达1050亿立方米。括号内的数读作(),保留整数后约是()平方千米。",
+        "options":  [
+
+                    ],
+        "answer":  "四千五百五十七点九五;4558",
+        "solution":  "【分析】按照小数的读法规则，整数部分按照整数的读法来读，即4557读作四千五百五十七，小数点读作点，小数部分依次读出每一位上的数字,95读作九五，所以4557.95读作四千五百五十七点九五。根据四舍五入法，保留整数时看十分位上的数字,4557.95十分位是9,9大于5，则向个位进1，所以4557.95≈4558.\n【详解】青海湖是我国最大的内陆湖泊，湖水面积4557.95平方千米，蓄水量达1050亿立方米。括号内的数读作四千五百五十七点九五，保留整数后约是4558平方千米。",
+        "knowledge_points":  [
+                                 "小数的读法和写法",
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·重庆九龙坡·月考"
+    },
+    {
+        "id":  "65",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数\"四舍五入\"后是8.50，这个小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "8.504;8.495",
+        "solution":  "【分析】取一个小数的近似数，有两种情况：\"四舍\"得到的近似数比原数小，\"五入\"得到的近似数比原数大，要考虑8.50是一个三位小数的近似数，有两种情况：\"四舍\"得到的8.50最大是8.504，\"五入\"得到的8.50最小是8.495，据此解答.\n【详解】由分析可知：一个三位小数\"四舍五入\"后是8.50，这个小数最大是8.504，最小是8.495。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·浙江杭州·期中"
+    },
+    {
+        "id":  "66",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "信阳位于河南省南部，总面积约(一万八千九百一十六)平方千米，括号内的数写作(),改写成用\"万\"作单位的数是()万。",
+        "options":  [
+
+                    ],
+        "answer":  "18916;1.8916",
+        "solution":  "【分析】①从高位到低位，一级一级地写，哪个数位上一个单位也没有，就在那个数位上写0。②把一些较大的数改写成用\"万\"作单位的数时，如果是整万的数，只要省略万位后面的0，并加一个\"万\"字;如果不是整万数，要在万位右下角点上小数点，去掉小数末尾的0，并加上一个\"万\"字.\n【详解】①一万八千九百一十六的万级为1，个级为8916，即写作18916;②18916改写成用\"万\"作单位的数是1.8916万。",
+        "knowledge_points":  [
+                                 "亿以内数的读，写法",
+                                 "亿以内数的改写"
+                             ],
+        "source":  "25-26四年级上·河南信阳·期中"
+    },
+    {
+        "id":  "67",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "()里可以填哪些数字。\n(1)3.()8\u003e3.28;(2)0.94\u003e0.9();(3)45.3=45.3();(4)0.55\u003e0.()3",
+        "options":  [
+
+                    ],
+        "answer":  "3、4、5、6、7、8、9;0、1、2、3;0;0、1、2、3、4、5",
+        "solution":  "【分析】3.□8\u003e3.28，整数部分都是3，比较小数部分，从十分位开始比较□\u003e2，即□可以填3,4,5,6,7,8,9;0.94\u003e0.9□,整数部分和十分位都相同，比较百分位4\u003e□,即□可以填0,1,2,3;45.3=45.3，整数部分和十分位相同，根据小数的性质，小数的末尾添0大小不变，所以□填0;0.55\u003e0.□3，整数部分相同，比较十分位5\u003e□,填0,1,2,3,4时成立，百分位5\u003e3当□=5时，依旧成立，即□填0,1,2,3,4,5.\n【详解】3.()8\u003e3.28，括号可以填3,4,5,6,7,8,9;0.94\u003e0.9(),括号可以填0,1,2,3;45.3=45.3(),括号填0;0.55\u003e0.()3，括号可以填0,1,2,3,4,5。",
+        "knowledge_points":  [
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "68",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个小数，小数点右边第一位上是6，左边第一位上是8，左边第二位上是4。这个数是(),它里面有()个0.1。",
+        "options":  [
+
+                    ],
+        "answer":  "48.6;486",
+        "solution":  "【分析】小数分为整数部分和小数部分，小数点右边第一个数位是十分位，十分位上是几就表示几个十分之一;小数点左边第一个数位是个位，个位上是几就表示几个一;小数点左边第二个数位是十位，十位上是几就表示几个十，据此解答.\n【详解】因此，一个小数，小数点右边第一位上是6，左边第一位上是8，左边第二位上是4。这个数是48.6，它里面有486个0.1。",
+        "knowledge_points":  [
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "69",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "由8个一,4个百分之一,2个千分之一组成的数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "8.042",
+        "solution":  "【分析】8个一表示个位上是8,4个百分之一表示百分位上是4,2个千分之一表示千分位上是2，十分位上没有计数单位用0补足，即可写出这个数.\n【详解】由8个一,4个百分之一,2个千分之一组成的数是8.042。",
+        "knowledge_points":  [
+                                 "小数的意义",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "70",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "6.503中的\"6\"在()位上，表示6个();\"5\"在()位上，表示5个();\"3\"在()位上，表示3个()。",
+        "options":  [
+
+                    ],
+        "answer":  "个;一;十分;十分之一;千分;千分之一",
+        "solution":  "【分析】小数分为整数部分和小数部分，小数点左边第个数位是个位，个位上是几就表示几个一;小数点右边第一个数位是十分位，十分位上是几就表示几个十分之一;小数点右边第二个数位是百分位，百分位上是几就表示几个百分之几;小数点右边第三个数位是千分位，千分位上是几就表示几个千分之几，据此解答.\n【详解】因此,6.503中的\"6\"在个位上，表示6个一;\"5\"在十分位上，表示5个十分之一;\"3\"在千分位上，表示3个千分之一。",
+        "knowledge_points":  [
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "71",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数，保留两位小数的近似数是20.00。这个小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "20.004;19.995",
+        "solution":  "【分析】\"四舍\"法取近似值时，原数大于近似数，近似数的小数点后面第三位数字最大并且不能向前一位进一，此时就是最大值的原数，即20.004;\"五入\"法取近似值时，原数小于近似数，近似数的小数点后面第二位数字减1(不够减时向前一位借\"1\"),近似数的小数点后面第三位数字最小并且向前一位进一，此时就是最小值的原数，即19.995，据此解答.\n【详解】由分析可得：一个三位小数，保留两位小数的近似数是20.00。这个小数最大是20.004，最小是19.995。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "72",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在()里填上\"\u003e\"\"\u003c\"或\"=\".\n70g()0.7kg;2.4t()2400kg;4km5m()4.5km;23dm()0.3m;3.4m²()3m²4dm²;23t500kg()3.05t",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;=;\u003c;\u003e;\u003e;\u003e",
+        "solution":  "【分析】根据1kg=1000g,1t=1000kg,1km=1000m,1m=10dm,1m²=100dm²,高级单位换算成低级单位乘进率，此时小数点向右移动，低级单位换算成高级单位除以进率，此时小数点向左移动，解答此题即可.\n【详解】0.7×1000=700,70\u003c700，所以70g\u003c0.7kg;2.4×1000=2400,2400=2400，所以2.4t=2400kg;5÷1000=0.005，即4km5m=4.005km,4.005\u003c4.5，所以4km5m\u003c4.5km;0.3×10=3,23\u003e3，所以23dm\u003e0.3m;4÷100=0.04，即3m²4dm²=3.04m²,3.4\u003e3.04，所以3.4m²\u003e3m²4dm²;500÷1000=0.5，即3t500kg=3.5t,3.5\u003e3.05，所以3t500kg\u003e3.05t。",
+        "knowledge_points":  [
+                                 "克，千克之间的换算与比较",
+                                 "吨，千克之间的换算与比较",
+                                 "平方厘米，平方分米，平方米之间的进率与换算",
+                                 "运用小数点移动解决小数的单位换算问题"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "73",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "把20.86的小数点向左移动一位，再向右移动两位后是();10.25的小数点向()移动()位后是0.1025。",
+        "options":  [
+
+                    ],
+        "answer":  "208.6;左;两",
+        "solution":  "【分析】根据小数点位置移动引起数的大小变化规律可知：个数的小数点向左移动一位，再向右移动两位，实际相当于小数点向右移动了一位，即这个数扩大了10倍，是208.6;10.25到0.1025小数点向左移动了两位;据此解答即可.\n【详解】20.86÷10×100=208.6,10.25÷100=0.1025，所以把20.86的小数点向左移动一位，再向右移动两位后是208.6;10.25的小数点向左移动两位后是0.1025。",
+        "knowledge_points":  [
+                                 "小数点向右移动引起小数大小变化的规律",
+                                 "小数点向左移动引起小数大小变化的规律"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "74",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "由8个十,6个一,9个十分之一和2个百分之一组成的数是(),读作(),保留整数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "86.92;八十六点九二;87",
+        "solution":  "【分析】根据数的组成写出每个数位上的数，哪个数位上个数也没有就写0，读小数，整数部分按照整数的读法来读，小数点读作点，小数部分顺次读出每一个数位上的数;保留整数就是看十分位上的数进行四舍五入，据此解答即可.\n【详解】由分析可得：由8个十,6个一,9个十分之一和2个百分之一组成的数是86.92，读作八十六点九二，保留整数是87。",
+        "knowledge_points":  [
+                                 "小数的读法和写法",
+                                 "小数的数位和计数单位的认识",
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "75",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "0.36去掉小数点后是(),相当于扩大到原来的()倍，也就是把原数的小数点向()移动()位。",
+        "options":  [
+
+                    ],
+        "answer":  "36;100;右;两",
+        "solution":  "【分析】本题主要考查小数点位置移动引起数的大小变化规律：一个数的小数点向右移动一位、两位、三位，这个数就扩大到原来的10倍、100倍、1000倍，反之也成立。把一个数的小数点向左移动一位、两位、三位，就是将这个数缩小到它的1÷10、1÷100、1÷1000，反之也成立，由此解答即可.\n【详解】由分析可知,0.36去掉小数点后是36，相当于扩大到原来的100倍，也就是把原数的小数点向右移动两位。",
+        "knowledge_points":  [
+                                 "小数的性质",
+                                 "小数点向右移动引起小数大小变化的规律"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "76",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "把25.79的小数点向()移动()位，它就变成了0.2579，相当于缩小到原来的()。",
+        "options":  [
+
+                    ],
+        "answer":  "左;两;1÷100",
+        "solution":  "【分析】根据小数点的移动规律，小数点向左移动一位，两位，三位。,小数就缩小到原来的1÷10,1÷100,1÷1000。,据此解答.\n【详解】把25.79的小数点向左移动两位，它就变成了0.2579，相当于缩小到原来的1÷100。",
+        "knowledge_points":  [
+                                 "小数点向左移动引起小数大小变化的规律"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "77",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用4,0,9,6这几个数字和小数点\".\"按要求写出下面各数。(每个数字都要用上并且只能用一次)。(1)小于1且小数部分是三位的小数:()。(2)大于9且小数部分是三位的小数:()。(3)0不读出来且小数部分是两位的小数:()。",
+        "options":  [
+
+                    ],
+        "answer":  "0.469,0.496,0.649,0.694,0.946,0.964;9.046,9.064,9.406,9.460,9.604,9.640;40.69,40.96,60.49,60.94,90.46,90.64",
+        "solution":  "【分析】根据题意写出符合题意的小数。小于1的三位小数，即整数部分是0且小数部分有三位，从小到大依次列举出来即可;大于9的三位小数，即整数部分是9且小数部分有三位，从小到大依次列举出来即可;0不读出来的两位小数，即0需要放在整数的末尾，即保证整数部分有2位，小数部分也有两位，依次从小到大列举出来即可;据此解答.\n【详解】(1)0.469,0.496,0.649,0.694,0.946,0.964;(2)9.046,9.064,9.406,9.460,9.604,9.640;(3)40.69,40.96,60.49,60.94,90.46,90.64",
+        "knowledge_points":  [
+                                 "小数的读法和写法",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "78",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "下面的小数各在哪两个相邻的整数之间。它们各近似于哪个整数。\n(1)()\u003c3.49\u003c(),近似于()。(2)()\u003c16.84\u003c(),近似于()。(3)()\u003c9.61\u003c(),近似于().\n(4)()\u003c9.09\u003c(),近似于()。",
+        "options":  [
+
+                    ],
+        "answer":  "3;4;3;16;17;17;9;10;10;9;10;9",
+        "solution":  "【分析】小数与相邻整数的关系及小数的近似数(四舍五入法)。相邻整数：小数的整数部分是较小的整数，整数部分+1是较大的整数;近似整数：看小数的十分位,\u003e=5则向整数部分进1,\u003c5则保留整数部分.\n【详解】(1)由分析可知，相邻整数:3\u003c3.49\u003c4近似整数：十分位是4(\u003c5),近似于3;(2)相邻整数:16\u003c16.84\u003c17近似整数：十分位是8(\u003e=5),近似于17;(3)相邻整数:9\u003c9.61\u003c10近似整数：十分位是6(\u003e=5),近似于10;(4)相邻整数:9\u003c9.09\u003c10近似整数：十分位是0(\u003c5),近似于9。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "79",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "世界上最大的青花瓷碗名为\"盛虹之巨\"，重0.875t,约是()t。(精确到百分位)。",
+        "options":  [
+
+                    ],
+        "answer":  "0.88",
+        "solution":  "【分析】求小数的近似数，精确到百分位时，要看它的千分位，若千分位上的数字大于等于5，就向百分位进1;若千分位上的数字小于5，就舍去千分位及其后面数位上的数，据此解答.\n【详解】0.875的千分位上是5，\"五入\"后保留两位小数是0.88。因此世界上最大的青花瓷碗名为\"盛虹之巨\"，重0.875t,约是0.88t。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "80",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个快递包裹的质量是1.438kg,约是()kg。(保留一位小数)。",
+        "options":  [
+
+                    ],
+        "answer":  "1.4",
+        "solution":  "【分析】要保留到某一位小数，就看这一位的下一位数字，根据\"四舍五入\"规则处理(下一位\u003e=5则进1,\u003c5则舍去)。本题要求\"保留一位小数\"，即保留到十分位，所以对百分位上的数进行四舍五入.\n【详解】因为百分位上的数字3\u003c5，根据四舍五入法，舍去百分位及后面的数，所以1.438保留一位小数约是1.4。综上可知，一个快递包裹的质量是1.438kg,约是1.4kg(保留一位小数)。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "81",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "南京长江大桥公路桥长4.589km,约是()km。(保留两位小数)。",
+        "options":  [
+
+                    ],
+        "answer":  "4.59",
+        "solution":  "【分析】求小数的近似数，保留两位小数时，要看它的千分位，若千分位上的数字大于等于5，就向百分位进1;若千分位上的数字小于5，就舍去千分位及其后面数位上的数，据此解答.\n【详解】4.589的千分位是9，\"五入\"后保留两位小数是4.59。因此，南京长江大桥公路桥长4.589km,约是4.59km。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "82",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "1里面有()个0.01,()个0.001是0.86。",
+        "options":  [
+
+                    ],
+        "answer":  "100;860",
+        "solution":  "【分析】根据小数的意义,0.01是百分之一，即1个0.01表示百分之一。1可以看作是1.00,1.00的计数单位是0.01,1.00里面有100个0.01.0.001是千分之一,0.86表示86个0.01;又因为1个0.01等于10个0.001，所以0.86=860个0.001.\n【详解】根据分析可得:1里面有100个0.01,860个0.001是0.86。",
+        "knowledge_points":  [
+                                 "小数的意义",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "83",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "0.6的计数单位是(),它有()个这样的计数单位;1.21的计数单位是(),它有()个这样的计数单位。",
+        "options":  [
+
+                    ],
+        "answer":  "0.1;6;0.01;121",
+        "solution":  "【分析】小数的计数单位要看有几位小数，一位小数计数单位就是0.1(或十分之一),两位小数计数单位就是0.01(或百分之一),三位小数计数单位就是0.001(或千分之一);小数去掉小数点后得到的整数是几，就有几个该小数的计数单位。据此解答.\n【详解】0.6是一位小数，计数单位是0.1，因此0.6里面有6个0.1;1.21是两位小数，它的计数单位是0.01，去掉小数点后是121，因此有121个这样的单位。0.6的计数单位是0.1，它有6个这样的计数单位;1.21的计数单位是0.01，它有121个这样的计数单位。",
+        "knowledge_points":  [
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "84",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "8cm写成分数是()m,写成小数是()m。",
+        "options":  [
+
+                    ],
+        "answer":  "8÷100;0.08",
+        "solution":  "【分析】低级单位的数换算成高级单位的数，用低级单位的数除以进率，结果可以写成分数也可以写成小数，根据此解答.\n【详解】1米=100厘米;8cm=(8÷100)m=8÷100m=0.08m,8cm写成分数是8÷100m,写成小数是0.08m。",
+        "knowledge_points":  [
+                                 "小数的意义",
+                                 "分数与除法的关系",
+                                 "毫米，厘米，分米，米之间的进率与换算"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "85",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在○里填上\"*\"或\"/\"，在□内填上合适的数.\n7.8○□=0.78;0.05○□=50;23.4○□=0.234;3.65○□=365;9.02○□=90.2;58.1○□=0.0581",
+        "options":  [
+
+                    ],
+        "answer":  "/;10;*;1000;/;100;*;100;*;10;/;1000",
+        "solution":  "【分析】小数点移动规律：一个数乘以10,100,1000，小数点分别向右移动一位，两位，三位;一个数除以10,100,1000，小数点分别向左移动一位，两位，三位。据此进行分析.\n【详解】观察原数7.8与结果0.78,7.8的小数点向左移动了一位得到0.78。所以括号里填\"/\"，横线上应填10;观察原数0.05与结果50,0.05的小数点向右移动了三位得到50。所以括号里填\"*\"，横线上应填1000;观察原数23.4与结果0.234,23.4的小数点向左移动了两位得到0.234。所以括号里填\"/\"，横线上应填100;观察原数3.65与结果365,3.65的小数点向右移动了两位得到365。所以括号里填\"*\"，横线上应填100;观察原数9.02与结果90.2,9.02的小数点向右移动了一位得到90.2。所以括号里填\"*\"，横线上应填10;观察原数58.1与结果0.0581,58.1的小数点向左移动了三位得到0.0581。所以括号里填\"/\"，横线上应填1000。",
+        "knowledge_points":  [
+                                 "小数点向右移动引起小数大小变化的规律",
+                                 "小数点向左移动引起小数大小变化的规律"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "86",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "西沙群岛的第二大岛--东岛，是我国最著名的鸟岛之一，面积约(170公顷)。括号内的数量用\"平方米\"作单位是()。",
+        "options":  [
+
+                    ],
+        "answer":  "1700000平方米",
+        "solution":  "【分析】高级单位的数换算成低级单位的数，用高级单位上的数乘进率，据此解答.\n【详解】1公顷=10000平方米,170公顷=(170×10000)平方米=1700000平方米，西沙群岛的第二大岛--东岛，是我国最著名的鸟岛之一，面积约(170公顷)。括号内的数量用\"平方米\"作单位是1700000平方米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的认识",
+                                 "面积单位的选择"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "87",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "王老师在超市买了一些葡萄，总价四舍五入后约是20.0元。葡萄的单价和质量都是一位小数，并且两个小数末位的数字都是9，葡萄的总价四舍五入前是()元。",
+        "options":  [
+
+                    ],
+        "answer":  "20.01",
+        "solution":  "【分析】葡萄的单价和质量都是一位小数，则积是两位小数，两个小数末位的数字都是9,9×9=81，那么总价末尾的数字是1，即百分位上数字是1,1\u003c5，不向十分位进位，四舍五入保留一位小数是20.0元，说明葡萄总价是20.01元.\n【详解】葡萄的单价和质量都是一位小数，并且两个小数末位的数字都是9，说明总价是两位小数，且百分位数字是1，葡萄的总价四舍五入前是20.01元。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·福建漳州·期中"
+    },
+    {
+        "id":  "88",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "把下面各数中的小数点都移到最高位的左边，得到的数分别是原数的几分之一。63.78;409.7;5.00;364;12.01();();();();()",
+        "options":  [
+
+                    ],
+        "answer":  "1÷100;1÷1000;1÷10;1÷1000;1÷100",
+        "solution":  "【分析】根据小数点移动规律，小数点向左移动一位，得到的数是原数的1÷10;小数点向左移动两位，得到的数是原数的1÷100;小数点向左移动三位，得到的数是原数的1÷1000。据此进行分析.\n【详解】根据分析得出:63.78小数点都移到最高位的左边，即小数点向左移动了两位，得到的数是原数的1÷100;409.7小数点都移到最高位的左边，即小数点向左移动了三位，得到的数是原数的1÷1000;5.00小数点都移到最高位的左边，即小数点向左移动了一位，得到的数是原数的1÷10;364小数点都移到最高位的左边，即小数点向左移动了三位，得到的数是原数的1÷1000;12.01小数点都移到最高位的左边，即小数点向左移动了两位，得到的数是原数的1÷100。",
+        "knowledge_points":  [
+                                 "小数点向左移动引起小数大小变化的规律"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "89",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "把下面各数中的小数点都去掉，得到的数分别扩大到原数的多少倍。\n0.8;0.385;2.57;6.001;4.0439 ();();();();()",
+        "options":  [
+
+                    ],
+        "answer":  "10倍;1000倍;100倍;1000倍;10000倍",
+        "solution":  "【分析】根据小数点移动规律，小数点向右移动1位，小数就扩大到原数的10倍;小数点向右移动2位，小数就扩大到原数的100倍;小数点向右移动3位，小数就扩大到原数的1000倍;小数点向右移动4位，小数就扩大到原数的10000倍。先确定每个数的小数位数，再根据小数点向右移动的位数确定扩大到原数的多少倍.\n【详解】0.8是一位小数，去掉小数点后，相当于把小数点向右移动了1位，扩大到原数的10倍;0.385是三位小数，去掉小数点后，相当于把小数点向右移动了3位，扩大到原数的1000倍;2.57是两位小数，去掉小数点后，相当于把小数点向右移动了2位，扩大到原数的100倍;6.001是三位小数，去掉小数点后，相当于把小数点向右移动了3位，扩大到原数的1000倍;4.0439是四位小数，去掉小数点后，相当于把小数点向右移动了4位，扩大到原数的10000倍。",
+        "knowledge_points":  [
+                                 "小数点向右移动引起小数大小变化的规律"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "90",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在0.4的末尾添上两个\"0\"，它的大小(),计数单位由()变为()。",
+        "options":  [
+
+                    ],
+        "answer":  "不变;0.1;0.001",
+        "solution":  "【分析】根据小数的基本性质，在小数的末尾添上或去掉0，小数的大小不变，可以知道0.4和0.400大小相等，再由小数的数位顺序表知，小数的计数单位从小数点向右依次是十分位，百分位，千分位。,它们间的进率是10，据此分析解答.\n【详解】根据分析得出：在0.4的末尾添上两个\"0\"，它的大小不变;0.4的计数单位是十分之一，写作0.1;0.400的计数单位是千分之一，写作0.001。计数单位由0.1变为0.001。",
+        "knowledge_points":  [
+                                 "小数的性质",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "91",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "57610=()万;834590000=()亿;345760=()万;637930000=()亿;784630=()万;976540=()万",
+        "options":  [
+
+                    ],
+        "answer":  "5.761;8.3459;34.576;6.3793;78.463;97.654",
+        "solution":  "【分析】改写成用\"万\"作单位的数，就是在万位数的右下角点上小数点，然后把小数末尾的0去掉，再在数的后面写上\"万\"字;改写成用\"亿\"作单位的数，就是在亿位数的右下角点上小数点，然后把小数末尾的0去掉，再在数的后面写上\"亿\"字.\n【详解】由分析可得:57610=5.761万;834590000=8.3459亿;345760=34.576万;637930000=6.3793亿;784630=78.463万;976540=97.654万。",
+        "knowledge_points":  [
+                                 "小数的改写",
+                                 "亿以内数的改写",
+                                 "亿以上数的改写"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "92",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在()里填\"\u003e\"\"\u003c\"或\"=\".\n4km6dm()4.6km;348g()0.348kg;2.08元()2元8角;0.05km²()10公顷;10.5t()1050kg;6kg30g()6.3kg",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;=;\u003c;\u003c;\u003e;\u003c",
+        "solution":  "【分析】根据1km=10000dm,1kg=1000g,1元=10角,1km²=100公顷,1t=1000kg。高级单位换成低级单位乘进率，低级单位换算成高级单位除以进率，解答此题即可.\n【详解】6÷10000=0.0006，即4km6dm=4.0006km,4.0006\u003c4.6，所以4km6dm\u003c4.6km;348÷1000=0.348,0.348=0.348，所以348g=0.348kg;8÷10=0.8，即2元8角=2.8元，所以2.08\u003c2.8;0.05×100=5,5\u003c10，所以0.05km²\u003c10公顷;1050÷1000=1.05,10.5\u003e1.05，所以10.5t\u003e1050kg;30÷1000=0.03，即6kg30g=6.03kg,6.03\u003c6.3，所以6kg30g\u003c6.3kg。",
+        "knowledge_points":  [
+                                 "元，角，分的认识及换算",
+                                 "克，千克之间的换算与比较",
+                                 "公顷，平方千米的进率与换算",
+                                 "运用小数点移动解决小数的单位换算问题"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "93",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "写出下面括号内的数.\n(1)一只北极熊的体重是(零点七五)吨。写作:()\n(2)成年人的头发直径约是(零点零七)毫米。写作:()\n(3)马里亚纳海沟最深处约为(十一点零三四)千米。写作:()",
+        "options":  [
+
+                    ],
+        "answer":  "0.75;0.07;11.034",
+        "solution":  "【分析】小数的写法：整数部分按照整数的写法来写，小数点写在个位右下角，小数部分依次写出每一位上的数字;据此解答.\n【详解】(1)一只北极熊的体重是零点七五吨。写作:0.75;(2)成年人的头发直径约是零点零七毫米。写作:0.07;(3)马里亚纳海沟最深处约为十一点零三四千米。写作:11.034。",
+        "knowledge_points":  [
+                                 "小数的读法和写法"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "94",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "我会读小数.\n2.3读作:();7.42读作:()\n0.0601读作:();100.06读作:()\n5.005读作:();20.202读作:()",
+        "options":  [
+
+                    ],
+        "answer":  "二点三;七点四二;零点零六零一;一百点零六;五点零零五;二十点二零二",
+        "solution":  "【分析】小数的读法：整数部分按照整数的读法来读(整数部分是0的读\"零\"),小数点读作\"点\"，小数部分依次读出每一位上的数字.\n【详解】2.3读作：二点三;7.42读作：七点四二;0.0601读作：零点零六零一;100.06读作：一百点零六;5.005读作：五点零零五;20.202读作：二十点二零二。",
+        "knowledge_points":  [
+                                 "小数的读法和写法"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "95",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "白河国家城市湿地公园面积24.5平方千米，合()公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "2450",
+        "solution":  "【分析】1平方千米=100公顷，从低级单位换算成高级单位，除以单位间的进率;从高级单位换算成低级单位，乘单位间的进率，据此解答.\n【详解】1平方千米=100公顷,24.5的小数点向右移动两位。所以白河国家城市湿地公园面积24.5平方千米，合2450公顷。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "运用小数点移动解决小数的单位换算问题"
+                             ],
+        "source":  "25-26四年级上·河南南阳·期中"
+    },
+    {
+        "id":  "96",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数，保留两位小数\"四舍五入\"后是8.51，这个三位小数最大可能是(),最小可能是()。",
+        "options":  [
+
+                    ],
+        "answer":  "8.514;8.505",
+        "solution":  "【分析】取一个小数的近似数，有两种情况：\"四舍\"得到的近似数比原数小，\"五入\"得到的近似数比原数大。要考虑8.51是一个三位小数的近似数，有两种情况：\"四舍\"得到的8.51最大是8.514，\"五入\"得到的8.51最小是8.505，由此解答问题即可.\n【详解】由分析可知：一个三位小数，保留两位小数\"四舍五入\"后是8.51，这个三位小数最大可能是8.514，最小可能是8.505。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·河南南阳·期中"
+    },
+    {
+        "id":  "97",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数保留两位小数\"四舍五入\"后是6.00，这个三位小数最小是(),最大是()。",
+        "options":  [
+
+                    ],
+        "answer":  "5.995;6.004",
+        "solution":  "【分析】三位小数保留两位小数后是6.00，需要考虑\"四舍\"和\"五入\"两种情况，通过\"五入\"确定最小，通过\"四舍\"确定最大。当三位小数通过\"五入\"保留两位小数得到6.00时，意味着千分位上的数字大于或等于5，且\"五入\"前的两位小数是5.99。 当三位小数通过\"四舍\"保留两位小数得到6.00时，意味着千分位上的数字小于5，且前两位小数保持6.00不变.\n【详解】求最小：根据\"五入\"的情况可以得到以下几个小数:5.995,5.996,5.997,5.998,5.999，其中最小的是5.995。求最大：根据\"四舍\"的情况可以得到以下几个小数:6.000,6.001,6.002,6.003,6.004，其中最大的是6.004。所以这个三位小数最小是5.995，最大是6.004。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·湖北黄冈·期中"
+    },
+    {
+        "id":  "98",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把98990000改写成用\"亿\"作单位的数且保留一位小数是()亿;128000省略\"万\"位后面的尾数约是()万。",
+        "options":  [
+
+                    ],
+        "answer":  "1.0;13",
+        "solution":  "【分析】把一些较大的数改写成用\"万\"或\"亿\"作单位的数时，如果不是整万或整亿的数，要在万位或亿位的后边点上小数点，去掉小数点末尾的0，并加上一个\"万\"或\"亿\"字。保留一位小数，即精确到十分位，要看百分位上的数。根据四舍五入法，若百分位上的数字大于或等于5，就向十分位进1;若小于5，就舍去百分位及其后面数位上的数。求整数的近似数，要对省略的尾数部分的最高位上的数进行四舍五入.\n【详解】98990000=0.9899亿≈1.0亿,128000=12.8万≈13万，所以把98990000改写成用\"亿\"作单位的数且保留一位小数是1.0亿;128000省略\"万\"位后面的尾数约是13万。",
+        "knowledge_points":  [
+                                 "亿以内数的改写",
+                                 "亿以上数的改写",
+                                 "亿以上数的近似数"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "99",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数\"四舍五入\"后得到的近似数是4.00，这个三位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "4.004;3.995",
+        "solution":  "【分析】根据\"四舍五入\"求小数近似数的法则，当\"四舍\"时，三位小数的千分位最大为4，所以这个三位小数最大是4.004;当\"五入\"时，三位小数的千分位最小为5，且百分位为9，十分位为9，所以这个三位小数最小是3.995.\n【详解】根据分析可知，一个三位小数，四舍五入后是4.00，所以这个三位小数最大是4.004，最小是3.995。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·湖北襄阳·期中"
+    },
+    {
+        "id":  "100",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "先把下面各数改写成用\"亿\"作单位的数，再保留两位小数.\n376200000=()亿≈()亿;99860000=()亿≈()亿",
+        "options":  [
+
+                    ],
+        "answer":  "3.762;3.76;0.9986;1.00",
+        "solution":  "【分析】把一个数改写成用\"亿\"作单位的数，在亿位的右下角点上小数点，把末尾的0去掉同时在后面写上\"亿\"字;保留小数点后面第二位，看小数点后面第三位，利用\"四舍五入\"法解答即可.\n【详解】376200000=3.762亿≈3.76亿;99860000=0.9986亿≈1.00亿。",
+        "knowledge_points":  [
+                                 "亿以上数的改写",
+                                 "亿以上数的近似数"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "101",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在()里填上\"\u003e\"、\"\u003c\"或\"=\".\n314000()3.14万;1206000()120.6万;254700()25.5万;1072860000()107.286亿;90600000()9.06亿;0.651亿()6510万",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;=;\u003c;\u003c;\u003c;=",
+        "solution":  "【分析】(1)将314000改写成以\"万\"为单位的数，要先找到万位，再在万位的右下角点上一个小数点，然后比较大小即可。(2)将1206000改写成以\"万\"为单位的数，要先找到万位，再在万位的右下角点上一个小数点，然后比较大小即可。(3)将254700改写成以\"万\"为单位的数，要先找到万位，再在万位的右下角点上一个小数点，然后比较大小即可。(4)将1072860000改写成以\"亿\"为单位的数，要先找到亿位，再在亿位的右下角点上一个小数点，然后比较大小即可。(5)将90600000改写成以\"亿\"为单位的数，要先找到亿位，再在亿位的右下角点上一个小数点，然后比较大小即可。(6)将0.651亿改写成以\"万\"为单位的数，然后比较大小即可。多位数的大小比较，数位不同时，位数多的就大。数位相同时，从最高位比起，如果最高位上的数字相同，就比较下一位，依次类推，直到比较出大小。小数大小的比较：先看小数的整数部分，整数部分大的这个数就大，整数部分相同的就看十分位，十分位大的这个数就大，十分位相同的，再看百分位，百分位大的这个数就大。,据此比较大小.\n【详解】(1)314000=31.4万,31.4\u003e3.14，所以314000\u003e3.14万。(2)1206000=120.6万,120.6=120.6，所以1206000=120.6万。(3)254700=25.47万,25.47\u003c25.5，所以254700\u003c25.5万。(4)1072860000=10.7286亿,10.7286\u003c107.286，所以1072860000\u003c107.286亿。(5)90600000=0.906亿,0.906\u003c9.06，所以90600000\u003c9.06亿。(6)0.651亿=6510万,6510=6510，所以0.651亿=6510万。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "亿以内数的改写",
+                                 "亿以上数的改写"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "102",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "先把下面各数改写成用\"亿\"作单位的数，再保留两位小数.\n376200000=()亿≈()亿;99860000=()亿≈()亿",
+        "options":  [
+
+                    ],
+        "answer":  "3.762;3.76;0.9986;1.00",
+        "solution":  "【分析】把一个数改写成用\"亿\"作单位的数，在亿位的右下角点上小数点，把末尾的0去掉同时在后面写上\"亿\"字;保留小数点后面第二位，看小数点后面第三位，利用\"四舍五入\"法解答即可.\n【详解】376200000=3.762亿≈3.76亿;99860000=0.9986亿≈1.00亿",
+        "knowledge_points":  [
+                                 "亿以上数的改写",
+                                 "亿以上数的近似数"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "103",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "先把下面各数改写成用\"万\"作单位的数，再保留一位小数.\n364210=()万≈()万;504900=()万≈()万;99820=()万≈()万;6793=()万≈()万",
+        "options":  [
+
+                    ],
+        "answer":  "36.421;36.4;50.49;50.5;9.982;10.0;0.6793;0.7",
+        "solution":  "【分析】改写成用\"万\"作单位的数，就是在万位的右下角点上小数点，然后把小数末尾的0去掉，再在数的后面写上\"万\"字即可;保留一位小数就是精确到十分位，把百分位上的数进行\"四舍五入\"，据此解答.\n【详解】由分析可得:364210=36.421万≈36.4万;504900=50.49万≈50.5万;99820=9.982万≈10.0万;6793=0.6793万≈0.7万",
+        "knowledge_points":  [
+                                 "亿以内数的改写",
+                                 "亿以内数的近似数"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "104",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "19.973保留两位小数是(),精确到十分位是()。",
+        "options":  [
+
+                    ],
+        "answer":  "19.97;20.0",
+        "solution":  "【分析】保留两位小数看千分位，精确到十分位看百分位，小于5直接舍去，大于或等于5向前一位进一.\n【详解】19.973≈19.97;19.973≈20.0,19.973保留两位小数是19.97，精确到十分位是20.0",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·浙江杭州·期末"
+    },
+    {
+        "id":  "105",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "()中最大能填几。\n221mm\u003e2.2()dm;5()g\u003c0.056kg;5.4()m\u003c548cm\n2.()t\u003c2760kg;8.()6km\u003e8580m;4500g\u003e4.()kg\n7.4cm\u003e()cm5mm;0.()9m²\u003c50dm²;7600cm²\u003e0.()m²",
+        "options":  [
+
+                    ],
+        "answer":  "0;5;7;7;9;4;6;4;7",
+        "solution":  "【分析】常用计量单位的进率：不同单位(长度，质量，面积)之间的换算关系，是统一单位的基础，从高级单位转换为低级单位，需用高级单位的数值乘进率。从低级单位转换为高级单位，需用低级单位的数值除以进率。小数/整数的数位大小规则：相同数位上的数字越大，数的整体值越大(需结合不等式方向判断最大可填数)整数比较大小步骤：数位越多，数越大。当数位相同，从高位到低位依次比先看最高位(左起第一位),最高位数字大的数大;若最高位相同，就比下一位，以此类推。小数比较大小步骤：先比整数部分，整数部分大的小数就大。当整数部分相同，比小数部分从小数部分的十分位开始比，十分位数字大的数大;若十分位相同，比百分位，以此类推.\n【详解】221mm=2.21dm,2.21\u003e2.2(),括号内填0;0.056kg=56g,5()\u003c56，括号位于个位，需满足\"5(括号内的数)\u003c56\"，因此括号内最大填5(55\u003c56);548cm=5.48m,5.4()\u003c5.48，括号位于百分位，需满足\"5.4(括号内的数)\u003c5.48\"，因此括号内最大填7(5.47\u003c5.48);2760kg=2.76t,2.()\u003c2.76，括号位于十分位，需满足\"2.(括号内的数)\u003c2.76\"，因此括号内最大填7(2.7\u003c2.76);8580m=8.58km。8.()6\u003e8.58，括号在十分位，最大填9(8.96\u003e8.58);4500g=4.5kg。4.5\u003e4.()，括号在十分位，最大填4(4.5\u003e4.4);5mm=0.5cm,即\"()cm5mm=()+0.5cm\"。7.4\u003e()+0.5，得()\u003c6.9，最大填6(6+0.5=6.5,6.5\u003c7.4);50dm²=0.5m²。0.()9\u003c0.5，括号在十分位，最大填4(0.49\u003c0.5);7600cm²=0.76m²。0.76\u003e0.()，括号在十分位，最大填7(0.76\u003e0.7)。",
+        "knowledge_points":  [
+                                 "克，千克之间的换算与比较",
+                                 "吨，千克之间的换算与比较",
+                                 "毫米，厘米，分米，米之间的进率与换算",
+                                 "平方厘米，平方分米，平方米之间的进率与换算"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "106",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "马拉松比赛路程全长(42km195m)。括号内的数量用\"米\"作单位是()。",
+        "options":  [
+
+                    ],
+        "answer":  "42195m",
+        "solution":  "【分析】长度单位的换算，需明确\"千米(km)\"和\"米(m)\"之间的进率:1km=1000m。题目中给出的路程是42km195m,要转换为以米为单位，需先将千米部分换算成米，从高级单位转换为低级单位，需用高级单位的数值乘进率1000，再与原有的米数相加.\n【详解】根据进率1km=1000m,可得42×1000=42000(m)。合并米数：将换算后的米数与原有的195m相加，即42000+195=42195(m)。括号内的数量用\"米\"作单位是42195m。",
+        "knowledge_points":  [
+                                 "千米和米之间的进率与换算"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "107",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "一部电梯上标明承载量为(1.2t)。括号内的数量用\"千克\"作单位是()。",
+        "options":  [
+
+                    ],
+        "answer":  "1200kg",
+        "solution":  "【分析】确定单位进率，明确要换算的两个单位之间的数量关系，题中\"吨(t)\"和\"千克(kg)\"的进率是1吨=1000千克。从\"吨\"换算到\"千克\"是从高级单位转换为低级单位，需用高级单位的数值乘进率.\n【详解】1.2×1000=1200(千克),一部电梯上标明承载量为(1.2t)。括号内的数量用\"千克\"作单位是1200千克。",
+        "knowledge_points":  [
+                                 "吨，千克之间的换算与比较"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "108",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在()里填上合适的单位.\n590()=0.59();125()=1.25();8750()=8.75();3452()=34.52()",
+        "options":  [
+
+                    ],
+        "answer":  "kg或千克;t或吨;cm或厘米;m或米;m或米;km或千米;cm²或平方厘米;dm²或平方分米",
+        "solution":  "【分析】题目是将较小数值的单位转化为较大数值的单位(低级单位转化为高级单位),需根据\"低级单位数/进率=高级单位数\"，结合数值的变化确定进率，再选择对应的常用计量单位.\n【详解】590÷1000=0.59，选择进率为1000的单位(如kg和t),1t=1000kg,所以590kg=0.59t;125÷100=1.25，选择进率为100的单位(如cm和m),1m=100cm,所以125cm=1.25m;8750÷1000=8.75，选择进率为1000的单位(如m和km),1km=1000m,所以8750m=8.75km;3452÷100=34.52，选择进率为100的单位(如cm²和dm²),1dm²=100cm²,所以3452cm²=34.52dm²。(答案不唯一)",
+        "knowledge_points":  [
+                                 "吨，千克之间的换算与比较",
+                                 "厘米和米之间的进率与换算",
+                                 "千米和米之间的进率与换算",
+                                 "平方厘米，平方分米，平方米之间的进率与换算"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "109",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个两位小数用四舍五入法保留一位小数后得到5.0，这个数最小是(),最大是()。",
+        "options":  [
+
+                    ],
+        "answer":  "4.95;5.04",
+        "solution":  "【分析】用四舍五入法保留一位小数，需看小数点后第二位，大于等于5则进一，小于5则舍去。因此，\"四舍\"得到的近似数比原数小，\"五入\"得到的近似数比原数大，所以保留一位小数后得到5.0的两位小数，最小的应该是\"五入\"到5.0的，而最大的应该是\"四舍\"到5.0的.\n【详解】\"五入\"得到5.0的最小两位小数是4.95，所以这个数最小是4.95.\"四舍\"得到5.0的最大两位小数是5.04，所以这个数最大是5.04。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·河南驻马店·月考"
+    },
+    {
+        "id":  "110",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果一个三位小数，保留两位小数后是5.00，那么这个三位小数最小是(),最大是()。",
+        "options":  [
+
+                    ],
+        "answer":  "4.995;5.004",
+        "solution":  "【分析】要确定这个三位小数的最大值和最小值，需根据\"四舍五入\"保留两位小数的规则，最小是通过\"五入\"得到的，最大是通过\"四舍\"得到的.\n【详解】根据分析，要通过\"五入\"得到5.00，那么原来的两位小数是4.99，此时小数点后第三位数字需要大于或等于5才能向前进1，所以第三位最小数字是5，即这三位小数最小是4.995;另外要通过\"四舍\"得到5.00，那么原来的两位小数就是5.00，此时小数点后第三位数字需小于5才能舍去，所以第三位最大数字是4，即这个三位小数最大是5.004。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·河北唐山·期末"
+    },
+    {
+        "id":  "111",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "6.08平方千米=()平方米;3公顷90平方米=()公顷",
+        "options":  [
+
+                    ],
+        "answer":  "6080000;3.009",
+        "solution":  "【分析】根据1平方千米=100公顷=1000000平方米，用6.08乘1000000相当于小数点向右移动六位换算成\"平方米\"作单位。1公顷=10000平方米，用90除以10000相当于小数点向左移动四位，换算单位后加3公顷即可.\n【详解】6.08×1000000=6080000(平方米),故6.08平方千米=6080000平方米。90÷10000=0.009(公顷),3+0.009=3.009(公顷),故3公顷90平方米=3.009公顷。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25五年级上·河北唐山·期末"
+    },
+    {
+        "id":  "112",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "29.999精确到十分位约是(),省略百分位后面的尾数约是(),保留整数约是()。",
+        "options":  [
+
+                    ],
+        "answer":  "30.0;30.00;30",
+        "solution":  "【分析】根据\"四舍五入\"求近似数，找到要求保留的数位，看下一位。如果下一位的数字大于或等于5，要向前位进1;如果下一位的数字小于5，要舍去。据此解答.\n【详解】29.999百分位上是9,9\u003e5，向前一位进1,29.999≈30.0;29.999千分位上是9,9\u003e5，向前一位进1,29.999≈30.00;29.999十分位上是9,9\u003e5，向前一位进1,29.999≈30。因此,29.999精确到十分位约是30.0，省略百分位后面的尾数约是30.00，保留整数约是30。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·江西赣州·期末"
+    },
+    {
+        "id":  "113",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "结果保留两位小数.\n976120000≈()亿;8943570≈()万;437200000≈()亿;590043≈()万;130640000≈()亿;23456789≈()万",
+        "options":  [
+
+                    ],
+        "answer":  "9.76;894.36;4.37;59.00;1.31;2345.68",
+        "solution":  "【分析】将一个数改写成以万为单位的数，将小数点向左移动4位，在得到的数字后面加上\"万\";将一个数改写成以\"亿\"为单位的数，将小数点向左移动8位，在数字后面加上\"亿\";要保留两位小数，需要看小数点后第三位小数，根据\"四舍五入\"法则计算得出答案.\n【详解】976120000=9.7612亿,9.7612亿的小数部分第三位是1，小于5，所以976120000≈9.76亿;8943570=894.357万,894.357万的小数部分第三位是7，大于5，所以8943570≈894.36万;437200000=4.372亿,4.372亿的小数部分第三位是2，小于5，所以437200000≈4.37亿;590043=59.0043万,59.0043万的小数部分第三位是4，小于5，所以590043≈59.00万;130640000=1.3064亿,1.3064亿的小数部分第三位是6，大于5，所以130640000≈1.31亿;23456789=2345.6789万,2345.6789万的小数部分第三位是8，大于5，所以23456789≈2345.68万。",
+        "knowledge_points":  [
+                                 "亿以内数的改写",
+                                 "亿以上数的改写",
+                                 "亿以内数的近似数",
+                                 "亿以上数的近似数"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "114",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在()里填上\"\u003e\"、\"\u003c\"或\"=\".\n11.8()10.8;9()9.00;3.80()0.38;2.78()2.87;4.71()4.709;0.6500()0.65;10.60()10.06;2.555()2.56;1.67()1.607;1.723()1.73;3.75m()3.57m;4.56元()45.6元",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;=;\u003e;\u003c;\u003e;=;\u003e;\u003c;\u003e;\u003c;\u003e;\u003c",
+        "solution":  "【分析】比较两个小数的大小：看它们的整数部分，整数部分大的那个数就大;整数部分相同的，十分位上的数大的那个数就大;十分位上的数也相同的，百分位上的数大的那个数就大，依次往右进行比较，直到比出大小为止;根据小数的性质，在小数的末尾添上0或去掉0，小数的大小不变.\n【详解】11.8的整数部分是11,10.8的整数部分是10,11\u003e10，所以11.8\u003e10.8;9.00去掉小数末尾的0，小数的大小不变，所以9=9.00;3.80的个位是3,0.38个位是0,3\u003e0，所以3.80\u003e0.38;2.78和2.87的整数部分都是2,2.78的十分位是7,2.87十分位是8,7\u003c8，所以2.78\u003c2.87;4.71和4.709的整数部分都是4,4.71的百分位是1,4.709百分位是0,1\u003e0，所以4.71\u003e4.709;0.6500去掉小数末尾的0，小数的大小不变，所以0.6500=0.65;10.60和10.06的整数部分都是10,10.60的十分位是6,10.06十分位是0,6\u003e0，所以10.60\u003e10.06;2.555和2.56的整数部分都是2,2.555的百分位是5,2.56百分位是6,5\u003c6，所以2.555\u003c2.56;1.67和1.607的整数部分都是1,1.67的百分位是7,1.607百分位是0,7\u003e0，所以1.67\u003e1.607;1.723和1.73的整数部分都是1,1.723的百分位是2,1.73百分位是3,2\u003c3，所以1.723\u003c1.73;3.75m和3.57m的整数部分都是3且单位相同,3.75的十分位是7,3.57十分位是5,7\u003e5，所以3.75m\u003e3.57m;4.56元和45.6元单位相同,4.56的整数部分是4,45.6元的整数部分是45,4\u003c45，所以4.56元\u003c45.6元。",
+        "knowledge_points":  [
+                                 "多位小数的大小比较",
+                                 "小数的性质"
+                             ],
+        "source":  "24-25四年级下·全国·课后作业"
+    },
+    {
+        "id":  "115",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个两位小数，保留一位小数约是4.7，这个两位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "4.74;4.65",
+        "solution":  "【分析】\"四舍\"法取近似值时，原数大于近似数，小数点后面第二位上面数字最大并且不能向前一位进一，原数取最大值;\"五入\"法取近似值时，原数小于近似数，近似值的十分位上面数字减1，后面百分位上面数字最小并且向前一位进一，原数取最小值，据此解答.\n【详解】根据四舍五入法，\"四舍\"得到的4.7，说明原两位小数的百分位上的数字小于5直接舍去了，此时这个两位小数最大，即百分位最大是4，所以这个两位小数最大是4.74;\"五入\"得到的4.7，说明原两位小数的十分位原来是6，百分位上的数字大于或等于5向十分位进1得到4.7，此时这个两位小数最小，即百分位最小是5，所以这个两位小数最小是4.65。因此，这个两位小数最大是4.74，最小是4.65。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·湖南株洲·期末"
+    },
+    {
+        "id":  "116",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数，经四舍五入后，保留两位小数约是7.45，这个三位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "7.454;7.445",
+        "solution":  "【分析】四舍五入保留两位小数得7.45，若千分位小于5，则舍去，百分位不变，结果为7.45(例如7.450至7.454);若千分位大于等于5，则进位，但进位后百分位变为5，因此原始数的百分位必须为4且千分位大于等于5(例如7.445至7.449)。据此解答.\n【详解】一个三位小数\"四舍五入\"保留两位小数约是7.45，\"四舍\"得到的7.45比原数小，所以这个三位小数最大是7.454;\"五入\"得到的7.45比原数大，所以这个三位小数最小是7.445。因此这个小数最大是7.454，最小是7.445。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·新疆哈密·期末"
+    },
+    {
+        "id":  "117",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数\"四舍五入\"保留两位小数后是10.00，这个三位小数最小是(),最大是()。",
+        "options":  [
+
+                    ],
+        "answer":  "9.995;10.004",
+        "solution":  "【分析】根据\"四舍五入\"求小数近似数的法则，当\"四舍\"时，三位小数的千分位最大为4，所以这个三位小数最大是10.004。当\"五入\"时，三位小数的千分位最小为5，且百分位为9，十分位为9，所以这个三位小数最小是9.995.\n【详解】一个三位小数\"四舍五入\"保留两位小数后是10.00，这个三位小数最小是9.995，最大是10.004。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·湖北黄冈·期中"
+    },
+    {
+        "id":  "118",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "2.15km²=()hm²;65000m²=()hm²;5.02km²=()m²;253hm²=()km²",
+        "options":  [
+
+                    ],
+        "answer":  "215;6.5;5020000;2.53",
+        "solution":  "【分析】根据1km²=100hm²,1hm²=10000m²,1km²=1000000m²,高级单位化低级单位乘进率，低级单位化高级单位除以进率，据此解答.\n【详解】因为2.15×100=215(hm²),所以2.15km²=215hm²。因为65000÷10000=6.5(hm²),所以65000m²=6.5hm²。因为5.02×1000000=5020000(m²),所以5.02km²=5020000m²。因为253÷100=2.53(km²),所以253hm²=2.53km²",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "面积单位间的进率及换算",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25五年级上·重庆合川·期末"
+    },
+    {
+        "id":  "119",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "300公顷=()平方千米;0.9公顷=()平方米",
+        "options":  [
+
+                    ],
+        "answer":  "3;9000",
+        "solution":  "【分析】(1)1平方千米=100公顷，小单位换算成大单位除以进率;(2)1公顷=10000平方米，大单位换算成小单位乘进率.\n【详解】(1)300÷100=3(平方千米)所以300公顷=3平方千米。(2)0.9×10000=9000(平方米)所以0.9公顷=9000平方米。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25五年级上·天津河西·期末"
+    },
+    {
+        "id":  "120",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "55000平方米=()公顷;0.23平方千米=()平方米",
+        "options":  [
+
+                    ],
+        "answer":  "5.5;230000",
+        "solution":  "【分析】1公顷=10000平方米，将平方米换算为公顷，需要除以进率10000.1平方千米=1000000平方米，将平方千米换算为平方米，需要乘进率1000000.\n【详解】①55000÷10000=5.5,55000平方米=5.5公顷。②0.23×1000000=230000,0.23平方千米=230000平方米",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "面积单位间的进率及换算",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25五年级上·河北邯郸·期末"
+    },
+    {
+        "id":  "121",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数，用\"四舍五入\"法保留两位小数，得到的近似数是3.00，这个三位小数最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "2.995",
+        "solution":  "【分析】一个三位小数四舍五入后是3.00，有\"四舍\"和\"五入\"两种情况：\"四舍\"的情况：当用\"四舍\"法得到3.00时，说明这个三位小数的十分位和百分位是0，千分位上的数字小于5直接舍去了，此时这个三位小数最大为3.004;当用\"五入\"法得到3.00时，说明这个三位小数的十分位和百分位是9，千分位上的数字大于或等于5，向百分位进1后得到3.00，此时这个三位小数最小为2.995.\n【详解】根据分析可知：一个三位小数，用\"四舍五入\"法保留两位小数，得到的近似数是3.00，这个三位小数最小是2.995。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·河南南阳·期末"
+    },
+    {
+        "id":  "122",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "2023年底统计邵阳市常住人口总数达6358800人，生产总值达273142000000元。把人口总数改写成用\"万\"作单位的数是()万人。把生产总值省略\"亿\"后面的尾数约是()亿元。",
+        "options":  [
+
+                    ],
+        "answer":  "635.88;2731",
+        "solution":  "【分析】(1)改写成用\"万\"作单位的数，就是在万位数的右下角点上小数点，然后把小数末尾的0去掉，再在数的后面写上\"万\"字;(2)省略\"亿\"后面的尾数就是先根据千万位上的数字进行\"四舍五入\"，再把亿位后面的数字去掉，在数的后面加上一个\"亿\"字，据此解答.\n【详解】6358800=635.88万,273142000000≈2731亿,2023年底统计邵阳市常驻人口总数达6358800人，生产总值达273142000000元。把人口总数改写成用\"万\"作单位的数是635.88万人。把生产总值省略\"亿\"后面的尾数约是2731亿元。",
+        "knowledge_points":  [
+                                 "亿以上数的近似数",
+                                 "小数的改写"
+                             ],
+        "source":  "24-25六年级上·湖南邵阳·期末"
+    },
+    {
+        "id":  "123",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "去掉3.41的小数点，等于把它的小数点向右移动()位，它的值就扩大到原来的()倍。",
+        "options":  [
+
+                    ],
+        "answer":  "两;100",
+        "solution":  "【分析】小数点位置移动引起数的大小变化规律：一个数的小数点向右移动一位，两位，三位。,相当于把原数乘10,100,1000。,这个数就扩大到原来的10倍,100倍,1000倍。据此解答.\n【详解】去掉3.41的小数点后是341，等于把它的小数点向右移动两位，它的值就扩大到原来的100倍。",
+        "knowledge_points":  [
+                                 "小数点向右移动引起小数大小变化的规律"
+                             ],
+        "source":  "24-25五年级上·山东菏泽·期末"
+    },
+    {
+        "id":  "124",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "7吨80千克=()吨;5000cm²=()m²",
+        "options":  [
+
+                    ],
+        "answer":  "7.08;0.5",
+        "solution":  "【分析】高级单位换算低级单位乘进率，低级单位换算高级单位除以进率，根据1吨=1000千克,1m²=10000cm²,据此进行计算即可.\n【详解】因为80÷1000=0.08,7+0.08=7.08，所以7吨80千克=7.08吨。因为5000÷10000=0.5，所以5000cm²=0.5m²",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "平方厘米，平方分米，平方米之间的进率与换算",
+                                 "吨，千克之间的换算与比较"
+                             ],
+        "source":  "24-25五年级上·黑龙江哈尔滨·期末"
+    },
+    {
+        "id":  "125",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "1.27的计数单位是(),它有()个这样的计数单位，再添上()个这样的计数单位是2。",
+        "options":  [
+
+                    ],
+        "answer":  "0.01;127;73",
+        "solution":  "【分析】根据小数的组成及计数单位,1.27是一个两位小数，它的计数单位是0.01;1.27里含有127个0.01,2里面含有200个0.01，所以要再添上(200-127)个0.01就是2，据此解答.\n【详解】1.27的计数单位是0.01，它有127个这样的计数单位;200-127=73(个),再添上73个这样的计数单位是2。",
+        "knowledge_points":  [
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25五年级上·河南平顶山·期末"
+    },
+    {
+        "id":  "126",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "请用改写或求近似数的知识把下面的数改成以\"万\"为单位的数.\n53150≈()万;160000=()万;377440=()万;181700≈()万",
+        "options":  [
+
+                    ],
+        "answer":  "16;37.744;18",
+        "solution":  "【分析】改写成用\"万\"作单位的数，就是在万位数的右下角点上小数点，然后把小数末尾的0去掉，再在数的后面写上\"万\"字;通过四舍五入法求整数的近似数，要对省略的尾数部分的最高位上的数进行四舍五入，若小于5则直接舍去，若大于或等于5，则向前进一位，并加上\"万\"字;。据此解答.\n【详解】根据分析:53150≈5万;160000=16万;377440=37.744万;181700≈18万。",
+        "knowledge_points":  [
+                                 "小数的改写",
+                                 "亿以内数的近似数"
+                             ],
+        "source":  "25-26四年级上·江西赣州·期中"
+    },
+    {
+        "id":  "127",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个两位小数保留一位小数后是2.0，这个两位小数最小是(),最大是()。",
+        "options":  [
+
+                    ],
+        "answer":  "1.95;2.04",
+        "solution":  "【分析】保留一位小数看百分位，小于5直接舍去，大于或等于5向前一位进一，由此可知，四舍得到的近似数，原数\u003e近似数;五入得到的近似数，原数\u003c近似数，据此分析.\n【详解】一个两位小数保留一位小数后是2.0，考虑五入的情况，这个两位小数最小是1.95，考虑四舍的情况，最大是2.04。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·内蒙古通辽·期末"
+    },
+    {
+        "id":  "128",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个数的十位，百分位和千分位都是5，其余各位都是0，这个数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "50.055",
+        "solution":  "【分析】根据数位表，十位的数是小数点左边的第二位，百分位的数是小数点右边的第二位，千分位是小数点右边的第三位，根据题意，十位，百分位和千分位都是5，其余各位上都是0，由此写出这个数是:50.055，据此解答.\n【详解】根据分析可知，一个数的十位，百分位和千分位都是5，其余各位都是0，这个数是50.055。",
+        "knowledge_points":  [
+                                 "小数的数位和计数单位的认识",
+                                 "小数的读法和写法"
+                             ],
+        "source":  "24-25五年级上·湖南邵阳·期末"
+    },
+    {
+        "id":  "129",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "我国是农业大国，据国家统计局发布的2022年全国及各省粮食产量数据显示，江苏省2022年粮食总产量约75380000000斤，改写成用\"亿\"作单位的数是()亿，省略\"亿\"后面的尾数大约是()亿。",
+        "options":  [
+
+                    ],
+        "answer":  "753.8;754",
+        "solution":  "【分析】改写时，如果不是整亿的数，要在亿位的后边，点上小数点，去掉小数点末尾的0，并加上一个\"亿\"字。通过四舍五入法求整数的近似数，要对省略的尾数部分的最高位上的数进行四舍五入，若小于5则直接舍去，若大于或等于5，则向前进一位，并加上\"亿\".\n【详解】75380000000=753.8亿;75380000000≈754亿。江苏省2022年粮食总产量约75380000000斤，改写成用\"亿\"作单位的数是753.8亿，省略\"亿\"后面的尾数大约是754亿。",
+        "knowledge_points":  [
+                                 "亿以上数的近似数",
+                                 "亿以上数的改写",
+                                 "小数的改写"
+                             ],
+        "source":  "24-25五年级上·湖南邵阳·期末"
+    },
+    {
+        "id":  "130",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "200cm²=()m²;3km²=()公顷",
+        "options":  [
+
+                    ],
+        "answer":  "0.02;300",
+        "solution":  "【分析】1m²=10000cm²,小单位换算成大单位除以进率;1km²=100公顷，大单位换算成小单位乘进率。据此解答.\n【详解】200÷10000=0.02，所以200cm²=0.02m²;3×100=300，所以3km²=300公顷。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "公顷，平方千米的进率与换算",
+                                 "平方厘米，平方分米，平方米之间的进率与换算"
+                             ],
+        "source":  "24-25五年级上·广东河源·期末"
+    },
+    {
+        "id":  "131",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "2025年中秋国庆叠加，新疆迎来8天超长假期，全区文旅市场与交通系统协同发力，呈现供需两旺，安全有序的繁荣景象。据大数据平台和抽样调查综合测算,10月1日至10月8日，全区接待游客(一千八百六十三万八千四百)人次，括号内的数写作(),游客花费(250.96)亿元，括号内的数读作()。省略亿位后面的尾数约是()亿。",
+        "options":  [
+
+                    ],
+        "answer":  "18638400;二百五十点九六;251",
+        "solution":  "【分析】根据题意，整数的读法：读数之前先分级，从最高级读起，一级一级往下读，先读亿级，再读万级，最后读个级;读亿级和万级时按读个级的方法来读，读完亿级后加上一个\"亿\"字，读完万级后加上一个\"万\"字;每级末尾不管有几个0都不读，每级中间和前面有一个或连续几个0，都只读一个0。数字\"250.96\"的读法：整数部分250读作\"二百五十\"，小数点读作\"点\"，小数部分96读作\"九六\"。故括号内的数读作二百五十点九六。用\"四舍五入\"法省略亿位后面的尾数求近似数，先看千万位上的数.\n【详解】根据分析可知：全区接待游客(一千八百六十三万八千四百)人次，括号内的数写作18638400;游客花费(250.96)亿元，括号内的数读作二百五十点九六。省略亿位后面的尾数约是251亿。",
+        "knowledge_points":  [
+                                 "小数的近似数",
+                                 "小数的读法和写法",
+                                 "亿以内数的读，写法"
+                             ],
+        "source":  "25-26四年级上·新疆乌鲁木齐·期中"
+    },
+    {
+        "id":  "132",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "9000平方米=()公顷;1.5平方千米=()平方米",
+        "options":  [
+
+                    ],
+        "answer":  "0.9;1500000",
+        "solution":  "【分析】低级单位换算高级单位除以进率，高级单位换算低级单位乘进率，根据1公顷=10000平方米,1平方千米=1000000平方米进行填空.\n【详解】因为9000÷10000=0.9，所以9000平方米=0.9公顷。因为1.5×1000000=1500000，所以1.5平方千米=1500000平方米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算",
+                                 "运用小数点移动解决小数的单位换算问题"
+                             ],
+        "source":  "24-25五年级上·北京房山·期末"
+    },
+    {
+        "id":  "133",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数\"四舍五入\"到百分位是6.45，这个三位小数在()与()之间。",
+        "options":  [
+
+                    ],
+        "answer":  "6.445;6.454",
+        "solution":  "【分析】小数的近似数：根据\"四舍五入\"法求近似数的方法，找到要求保留的数位，看下一位;如果下一位的数字大于或等于5，要往前进一;如果下一位的数字小于5，要舍去;据此解答.\n【详解】四舍五入到百分位是6.45，说明这个三位小数的百分位可能为4或5。若百分位是4，则千分位必须大于或等于5，才能进一得到6.45，此时最小值为6.445;若百分位是5，则千分位必须小于5，才能舍去得到6.45，此时最大值为6.454。因此，这个三位小数在6.445与6.454之间。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·内蒙古兴安盟·期中"
+    },
+    {
+        "id":  "134",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个两位小数按\"四舍五入\"法保留一位小数后是6.0，这个两位小数最大可能是(),最小可能是()。",
+        "options":  [
+
+                    ],
+        "answer":  "6.04;5.95",
+        "solution":  "【分析】(1)保留一位小数后是6.0，要使原两位小数最大，需通过\"四舍\"得到。保留一位小数看百分位，\"四舍\"时百分位上的数字应小于5，此时十分位不变仍为0，个位为6，则百分位最大取4;(2)保留一位小数后是6.0，要使原两位小数最小，需通过\"五入\"得到.\"五入\"时百分位上的数字应大于或等于5，此时十分位原来是9进位后变为0，个位原来是5，进位后变为6，则百分位最小取5.\n【详解】根据分析可得：通过\"四舍\"得到这个两位小数最大可能值，为6.04;通过\"五入\"得到这个两位小数的最小可能值，为5.95。因此，一个两位小数按\"四舍五入\"法保留一位小数后是6.0，这个两位小数最大可能是6.04，最小可能是5.95。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·湖北随州·期末"
+    },
+    {
+        "id":  "135",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "甲数的小数点向右移动一位后就与乙数相等，乙数为5，甲，乙两数的差是()。",
+        "options":  [
+
+                    ],
+        "answer":  "4.5",
+        "solution":  "【分析】甲数的小数点向右移动一位，相当于把甲数扩大到原来的10倍，也就是甲数乘10等于乙数，已知乙数是5，用乙数除以10即可求出甲数，再用减法求出甲乙两数的差.\n【详解】5÷10=0.5,5-0.5=4.5，甲数的小数点向右移动一位后就与乙数相等，乙数为5，甲，乙两数的差是4.5。",
+        "knowledge_points":  [
+                                 "小数点向右移动引起小数大小变化的规律",
+                                 "一位小数的进位加法，退位减法"
+                             ],
+        "source":  "24-25五年级上·河北保定·期末"
+    },
+    {
+        "id":  "136",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.4,
+        "question":  "把一个小数扩大到原来的100倍后，小数点向左移动一位，得0.06，那么原来这个小数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "0.006",
+        "solution":  "【分析】可以用倒推法来解决。先看最后一步：小数点向左移动一位后得0.06，倒推的话就要把0.06的小数点向右移动一位，得到0.6。这一步的结果是\"小数扩大到原来的100倍\"之后的数，所以再倒推，把0.6缩小到原来的1÷100(也就是小数点向左移动两位),得到0.006.\n【详解】根据倒推法，将最终的结果0.06的小数点向右移动一位得到0.6;再把0.6缩小到原来的1÷100，也就是小数点向左移动两位，得到0.006。因此，把一个小数扩大到原来的100倍后，小数点向左移动一位，得0.06，那么原来这个小数是0.006。",
+        "knowledge_points":  [
+                                 "小数点向左移动引起小数大小变化的规律",
+                                 "小数点向右移动引起小数大小变化的规律"
+                             ],
+        "source":  "24-25五年级上·河北保定·期末"
+    },
+    {
+        "id":  "137",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "0.48平方千米=()公顷;1.205平方米=()平方厘米;1米5分米=()米;3.04吨=()吨()千克",
+        "options":  [
+
+                    ],
+        "answer":  "48;12050;1.5;3;40",
+        "solution":  "【分析】根据1平方千米=100公顷,1平方米=10000平方厘米,1米=10分米,1吨=1000千克;高级单位化低级单位要乘进率，低级单位化高级单位要除以进率;据此解答.\n【详解】0.48×100=48(公顷),所以0.48平方千米=48公顷;1.205×10000=12050(平方厘米),所以1.205平方米=12050平方厘米;5÷10=0.5(米),所以1米5分米=1.5米。0.04×1000=40(千克),所以3.04吨=3吨40千克。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "公顷，平方千米的进率与换算",
+                                 "平方厘米，平方分米，平方米之间的进率与换算",
+                                 "吨，千克之间的换算与比较"
+                             ],
+        "source":  "24-25五年级上·河北保定·期末"
+    },
+    {
+        "id":  "138",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "定州众春园的面积是305500平方米，合()公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "30.55",
+        "solution":  "【分析】1公顷=10000平方米，根据高级单位化低级单位要乘进率，低级单位化高级单位要除以进率，所以平方米化为公顷要除以10000。据此解答.\n【详解】305500÷10000=30.55(公顷),定州众春园的面积是305500平方米，合30.55公顷。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "运用小数点移动解决小数的单位换算问题"
+                             ],
+        "source":  "24-25五年级上·河北保定·期末"
+    },
+    {
+        "id":  "139",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一幅装饰画的面积是0.3825平方米，保留两位小数是()平方米，精确到千分位约是()平方米。",
+        "options":  [
+
+                    ],
+        "answer":  "0.38;0.383",
+        "solution":  "【分析】保留两位小数，看小数的千分位，千分位上的数字大于或等于5，就向前一位进1，千分位上的数字小于5，就直接舍去;精确到千分位就是看万分位上的数字，根据\"四舍五入\"进行保留.\n【详解】0.3825的千分位是2，小于5，直接舍去，则0.3825保留两位数字是0.38;0.3825的万分位是5，等于5，向前一位进1，则0.3825精确到千分位约是0.383;所以一幅装饰画的面积是0.3825平方米，保留两位小数是0.38平方米，精确到千分位约是0.383平方米。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·广东潮州·期中"
+    },
+    {
+        "id":  "140",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数，如果它的近似值保留两位小数是3.68，这个三位小数最大(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "3.684;3.675",
+        "solution":  "【分析】根据题意，取一个小数的近似数，有两种情况：\"四舍\"得到的近似数比原数小，\"五入\"得到的近似数比原数大，一个三位小数，如果它的近似值保留两位小数是3.68，要看千分位的数，如果千分位上的数大于等于5，就要向百分位进一;如果小于5，则舍去，据此解答.\n【详解】根据分析可知，一个三位小数，如果它的近似值保留两位小数是3.68，这个三位小数可能是3.675,3.676,3.677,3.678,3.679,3.681,3.682,3.683,3.684，所以这个三位小数最大是3.684，最小是3.675。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·浙江·期中"
+    },
+    {
+        "id":  "141",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个两位小数精确到十分位是5.0，这个数最小是()。最大是()。",
+        "options":  [
+
+                    ],
+        "answer":  "4.95;5.04",
+        "solution":  "【分析】精确到十分位看百分位，小于5直接舍去，大于或等于5向前一位进一，由此可知，四舍得到的近似数，原数\u003e近似数;五入得到的近似数，原数\u003c近似数.\n【详解】一个两位小数精确到十分位是5.0，根据分析，这个数最小是4.95，最大是5.04。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "2025·江西上饶·小升初真题"
+    },
+    {
+        "id":  "142",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个数的十位，十分位和千分位上都是6，其余各位上都是0，这个数是(),精确到百分位是()。",
+        "options":  [
+
+                    ],
+        "answer":  "60.606;60.61",
+        "solution":  "【分析】一个数的十位上是6，个位上是0，十分位上是6，百分位上是0，千分位上是6，这个数是60.606;要求精确到百分位，需看千分位上的数，千分位上是6，要向百分位进1,60.606精确到百分位是60.61.\n【详解】根据分析可知：这个数是60.606;60.606≈60.61，精确到百分位是60.61。",
+        "knowledge_points":  [
+                                 "小数的近似数",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25五年级上·湖南邵阳·期末"
+    },
+    {
+        "id":  "143",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上合适的面积单位。我家住房的面积大约是120();我们学校操场的占地面积大约是1();北京故宫的占地面积约是0.72()。",
+        "options":  [
+
+                    ],
+        "answer":  "平方米/m²;公顷/hm²;平方千米/km²",
+        "solution":  "【分析】常用面积单位中，平方厘米，平方分米用于度量较小的面积，如书本封面，课桌面;平方米用于度量较大的日常空间面积，如房间，操场;公顷，平方千米用于度量超大的土地面积，如校园，城市建筑，景区。据此解答.\n【详解】(1)1平方米大约是一块边长1米的地砖面积，用于度量房间等日常空间;住房面积约120块边长1米的地砖面积，所以我家住房的面积大约是120平方米。(2)1公顷=10000平方米，相当于边长100米的正方形区域，用于度量校园，操场等较大土地面积;操场面积约1个\"10000平方米\"的区域规模，所以我们学校操场的占地面积大约是1公顷。(3)1平方千米=100公顷，相当于边长1000米的正方形区域，用于度量大型建筑群，景区等超大土地面积;故宫面积约0.72个\"100公顷\"的区域规模，所以北京故宫的占地面积约是0.72平方千米。",
+        "knowledge_points":  [
+                                 "面积单位的选择",
+                                 "公顷，平方千米的认识",
+                                 "平方厘米，平方分米，平方米的认识"
+                             ],
+        "source":  "24-25五年级上·湖南邵阳·期末"
+    },
+    {
+        "id":  "144",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "2024年10月1日，电影《志愿军：存亡之战》票房收入约(153229000)元，把括号内的数改写成用\"亿\"作单位的数是()亿，省略\"亿\"后面的尾数约是()亿。",
+        "options":  [
+
+                    ],
+        "answer":  "1.53229;2",
+        "solution":  "【分析】改写成用亿作单位的数，就是在亿位的右下角点上小数点，然后把小数末尾的0去掉，再在数的后面写上亿字;省略\"亿\"后面的尾数就是四舍五入到亿位，把亿位后的千万位上的数进行四舍五入.\n【详解】153229000=1.53229亿,153229000≈2亿,153229000改写成用\"亿\"作单位的数是1.53229亿，省略\"亿\"后面的尾数约是2亿。",
+        "knowledge_points":  [
+                                 "亿以上数的近似数",
+                                 "亿以上数的改写",
+                                 "小数的改写"
+                             ],
+        "source":  "24-25五年级上·湖南邵阳·期末"
+    },
+    {
+        "id":  "145",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数\"四舍五入\"后得到的近似数是5.98，这个三位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "5.984;5.975",
+        "solution":  "【分析】取一个数的近似数，有两种情况：\"四舍\"得到的近似数比原数小，\"五入\"得到的近似数比原数大。要考虑三位小数的近似数是5.98，最大是5.984，最小是5.975，据此解答.\n【详解】根据分析可知，一个三位小数\"四舍五入\"后得到的近似数是5.98，这个三位小数最大是5.984，最小是5.975。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·河南南阳·期末"
+    },
+    {
+        "id":  "146",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.4,
+        "question":  "一个三位小数精确到百分位是0.80，这个数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "0.804;0.795",
+        "solution":  "【分析】根据求小数的近似数的方法：利用\"四舍五入法\"，一个三位小数四舍五入到百分位是0.80，需要看千分位数字。有两种情况：根据千分位数字，判断为\"四舍\"得到的0.80，则原来的三位小数有:0.801,0.802,0.803,0.804，最大是0.804;根据千分位数字，判断为\"五入\"得到的0.80，则原来的三位小数有:0.795,0.796,0.797,0.798,0.799，最小是0.795;由此解答问题即可.\n【详解】一个三位小数精确到百分位是0.80，千分位\"四舍\"得到的0.80，则原来的三位小数有:0.801,0.802,0.803,0.804，这个数最大是0.804;千分位\"五入\"得到的0.80，则原来的三位小数有:0.795,0.796,0.797,0.798,0.799，这个数最小是0.795。",
+        "knowledge_points":  [
+                                 "多位小数的大小比较",
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·黑龙江鸡西·期中"
+    },
+    {
+        "id":  "147",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数精确到十分位是8.8，这个三位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "8.849;8.750",
+        "solution":  "【分析】精确到十分位看的是百分位上的数，小于5直接舍去，大于或等于5向前一位进一，由此可知，四舍得到的近似数，原数\u003e近似数;五入得到的近似数，原数\u003c近似数，据此分析.\n【详解】要找这个三位小数的最大值：精确到十分位是8.8，说明\"四舍\"的时候得到这个结果。十分位已经是8，百分位最大取4(因为4及以下会被舍去),千分位取最大的数字9，所以最大值是8.849。要找这个三位小数的最小值：精确到十分位是8.8，说明\"五入\"的时候得到这个结果。十分位原本是7，百分位最小取5(因为5及以上会进1),千分位取最小的数字0，所以最小值是8.750。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "2025五年级上·广东·专题练习"
+    },
+    {
+        "id":  "148",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "3.899保留一位小数约是(),保留两位小数约是(),保留整数约是()。",
+        "options":  [
+
+                    ],
+        "answer":  "3.9;3.90;4",
+        "solution":  "【分析】保留整数时，看十分位;保留一位小数时，看百分位;保留两位小数时，看千分位;按\"四舍五入\"取近似值。表示近似数时，小数末尾的0不能去掉.\n【详解】3.899≈3.9,3.899≈3.90,3.899≈4,3.899保留一位小数约是3.9，保留两位小数约是3.90，保留整数约是4。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "2025五年级上·广东·专题练习"
+    },
+    {
+        "id":  "149",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个两位小数\"四舍五入\"保留一位小数后是8.0，这个两位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "8.04;7.95",
+        "solution":  "【分析】一个两位小数，用\"四舍五入\"法保留一位小数是8.0。最大的两位小数用\"四舍\"得到，百分位最大是4，得到8.04;最小的两位小数用\"五入\"得到，百分位最小是5，且十分位应为9进位，得到7.95.\n【详解】根据分析，一个两位小数\"四舍五入\"保留一位小数后是8.0，这个两位小数最大是8.04，最小是7.95。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·湖北黄冈·期中"
+    },
+    {
+        "id":  "150",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数精确到百分位后约是2.16，这个三位小数最小是(),最大是()。",
+        "options":  [
+
+                    ],
+        "answer":  "2.155;2.164",
+        "solution":  "【分析】取一个小数的近似数，有两种情况：\"四舍\"得到的近似数比原数小，\"五入\"得到的近似数比原数大。一个三位小数精确到百分位后约是2.16，要看千分位的数，千分位大于等于5则向百分位进一，小于5则舍去.\n【详解】根据分析可知：这个三位小数可能是2.155,2.156,2.157,2.158,2.159,2.160,2.161,2.162,2.163,2.164。最小是2.155，最大是2.164。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "2025五年级上·广东·专题练习"
+    },
+    {
+        "id":  "151",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数四舍五入保留两位小数后是9.87，这个三位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "9.874;9.865",
+        "solution":  "【分析】依据\"四舍五入\"保留两位小数的规则：如果是\"四舍\"得到9.87，那原数的前两位就是9.87，千分位要取不超过4的最大数，也就是4，所以最大的三位小数是9.874;如果是\"五入\"得到9.87，那原数的百分位为6，千分位取不小于5的最小数，也就是5，所以最小的三位小数是9.865.\n【详解】依据四舍五入规则，\"四舍\"时这个三位小数最大为9.874，\"五入\"时最小为9.865。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·广东潮州·期中"
+    },
+    {
+        "id":  "152",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "0.56公顷=()平方米;75cm²=()dm²",
+        "options":  [
+
+                    ],
+        "answer":  "5600;0.75",
+        "solution":  "【分析】因为1公顷=10000平方米，公顷换算为平方米，是大单位换算为小单位，要乘进率10000;因为1dm²=100cm²,cm²换算为dm²,是小单位换算为大单位，要除以进率100.\n【详解】0.56×10000=5600，所以0.56公顷=5600平方米;75÷100=0.75，所以75cm²=0.75dm²。",
+        "knowledge_points":  [
+                                 "平方厘米，平方分米，平方米之间的进率与换算",
+                                 "公顷，平方千米的进率与换算",
+                                 "运用小数点移动解决小数的单位换算问题"
+                             ],
+        "source":  "24-25五年级上·湖南永州·期末"
+    },
+    {
+        "id":  "153",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数用\"四舍五入\"法取近似值是7.66，这个数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "7.664;7.655",
+        "solution":  "【分析】要考虑7.66是一个三位小数的近似数，有两种情况：\"四舍\"得到的7.66，则原数可能是7.661,7.662,7.663,7.664，即最大是7.664;\"五入\"得到的7.66，则原数可能是7.655,7.656,7.657,7.658,7.659，即最小是7.655，据此分析即可.\n【详解】结合分析知：一个三位小数用\"四舍五入\"法取近似值是7.66，这个数最大是7.664，最小是7.655。",
+        "knowledge_points":  [
+                                 "多位小数的大小比较",
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·山东日照·期中"
+    },
+    {
+        "id":  "154",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "三位同学比身高，小利比小芳高0.05米，小芳比小菲矮0.06米。三个同学中，最高的是(),最矮的是()。",
+        "options":  [
+
+                    ],
+        "answer":  "小菲;小芳",
+        "solution":  "【分析】小利比小芳高0.05米，小芳比小菲矮0.06米，即小菲比小芳高0.06米,0.06\u003e0.05，则三个同学中，最高的是小菲，最矮的是小芳.\n【详解】0.06\u003e0.05，三个同学中，最高的是小菲，最矮的是小芳。",
+        "knowledge_points":  [
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "25-26五年级上·河南洛阳·期中"
+    },
+    {
+        "id":  "155",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数\"四舍五入\"到百分位是2.75，这个三位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "2.754;2.745",
+        "solution":  "【分析】取一个数的近似数，有两种情况：\"四舍\"得到的近似数比原数小，\"五入\"得到的近似数比原数大。要考虑一个三位小数到百分位是2.75.\n【详解】根据分析可知，一个三位小数\"四舍五入\"到百分位是2.75，这个三位小数最大是2.754，最小是2.745。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·内蒙古鄂尔多斯·期中"
+    },
+    {
+        "id":  "156",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个十位数，最高位上的数是8，从右边起第八位数是5，第五位数是9，最低位上的数是2，其余各位上的数是0，这个数写作()。把这个数改写成用万作单位的数是()。这个数\"四舍五入\"到亿位是()。",
+        "options":  [
+
+                    ],
+        "answer":  "8050090002;805009.0002万;81亿",
+        "solution":  "【分析】整数的写法：从高位到低位，一级一级地写，哪个数位上一个单位也没有，就在那个数位上写0。这个数是十位数，最高位是十亿，所以十亿位上是8，从右边起第八位是千万位，所以千万位上是5，从右边起第五位是万位，所以万位上是9，最低位是个位，所以个位上是2，其余数位上是0。据此写出;把这个数改写成用万作单位的数是在万位右边点上小数点，再在末尾加上\"万\"字即可;\"四舍五入\"到亿位，要看千万位上的数字.\n【详解】十亿位上是8，千万位上是5，万位上是9，个位上是2，其余数位上是0，这个数是8050090002;8050090002=805009.0002万;8050090002≈81亿。",
+        "knowledge_points":  [
+                                 "亿以上数的近似数",
+                                 "小数的改写",
+                                 "亿以上数的读，写法"
+                             ],
+        "source":  "25-26四年级上·湖南张家界·期中"
+    },
+    {
+        "id":  "157",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个两位小数用\"四舍五入\"法保留一位小数后是4.0，这个数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "4.04;3.95",
+        "solution":  "【分析】取一个数的近似数，有两种情况：\"四舍\"得到的近似数比原数小，\"五入\"得到的近似数比原数大，所以\"四舍\"可求最大的数，\"五入\"可求最小的数。本题两位小数保留一位小数即精确到十分位是4.0，要看百分位.\n【详解】根据分析可知：一个两位小数用\"四舍五入\"法保留一位小数后是4.0，这个数最大是4.04，最小是3.95。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·湖南常德·期中"
+    },
+    {
+        "id":  "158",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数取近似值是22.10，这个数最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "22.095",
+        "solution":  "【分析】三位小数取近似值为22.10，需分\"四舍\"和\"五入\"两种情况：\"四舍\"时，原数的千分位\u003c=4，此时原数最大;\"五入\"时，原数的千分位\u003e=5，且百分位会因\"五入\"进1后变为0，十分位进1后变为1，因此原数的十分位是0，百分位是9，千分位最小取5，此时原数最小.\n【详解】原数的十分位是0，百分位是9，千分位最小取5。这个三位小数最小是22.095。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·山东临沂·期中"
+    },
+    {
+        "id":  "159",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数，保留两位小数是8.90，这个三位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "8.904;8.895",
+        "solution":  "【分析】根据\"四舍五入\"原则，则\"四舍\"得到8.90的原数比8.90大，\"五入\"得到8.90的原数比8.90小.\n【详解】\"四舍\"得到8.90的原数:8.901,8.902,8.903,8.904;\"五入\"得到8.90的原数:8.895,8.896,8.897,8.898,8.899;即这个三位小数最大是8.904，最小是8.895。",
+        "knowledge_points":  [
+                                 "小数的近似数",
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "25-26五年级上·河南商丘·期中"
+    },
+    {
+        "id":  "160",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "把4.5扩大到原来的()倍是450，把9.6缩小到原来的()是0.096。",
+        "options":  [
+
+                    ],
+        "answer":  "100;1÷100",
+        "solution":  "【分析】把一个数扩大到原来的10倍,100倍,1000倍。只要把这个数的小数点向右移动一位，两位，三位。;把一个数缩小到原来的1÷10,1÷100,1÷1000。只要把这个数的小数点向左移动一位，两位，三位。由4.5变成450，是小数点向右移动了两位;由9.6变成0.096，是小数点向左移动了两位.\n【详解】由分析可知：把4.5扩大到原来的100倍是450，把9.6缩小到原来的1÷100是0.096。",
+        "knowledge_points":  [
+                                 "小数点向左移动引起小数大小变化的规律",
+                                 "小数点向右移动引起小数大小变化的规律"
+                             ],
+        "source":  "25-26五年级上·河北唐山·期中"
+    },
+    {
+        "id":  "161",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数\"四舍五入\"到百分位是4.50，这个三位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "4.504;4.495",
+        "solution":  "【分析】根据\"四舍五入\"求近似数的方法，已知近似数时，用\"四舍\"法得到最大的原数，用\"五入\"法得到最小的原数.\n【详解】要考虑4.50是一个三位小数的近似数，有两种情况：\"四舍\"得到的4.50，有4.501,4.502,4.503,4.504，其中最大是4.504;\"五入\"得到的4.50，有4.495,4.496,4.497,4.498,4.499，其中最小是4.495。这个三位小数最大是4.504，最小是4.495。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·江西鹰潭·期中"
+    },
+    {
+        "id":  "162",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个两位小数四舍五入后是8.7，那么这个两位小数最大可能是(),最小可能是()。",
+        "options":  [
+
+                    ],
+        "answer":  "8.74;8.65",
+        "solution":  "【分析】取一个小数的近似数，有两种情况：\"四舍\"得到的近似数比原数小，\"五入\"得到的近似数比原数大。要考虑8.7是一个两位小数的近似数.\n【详解】由分析可知：一个两位小数四舍五入后是8.7，那么这个两位小数最大可能是8.74，最小可能是8.65。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·福建福州·期中"
+    },
+    {
+        "id":  "163",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "要使47.4□≈47.4,□里最大填()。要使28.9□≈29.0,□里最小填()。",
+        "options":  [
+
+                    ],
+        "answer":  "4;5",
+        "solution":  "【分析】保留一位小数看百分位，小于5直接舍去，大于或等于5向前一位进一，据此填空.\n【详解】根据四舍五入法保留近似数的方法，要使47.4□≈47.4,□里最大填4。要使28.9□≈29.0,□里最小填5。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·湖南郴州·期中"
+    },
+    {
+        "id":  "164",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数，精确到0.01后近似数是8.74，这个三位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "8.744;8.735",
+        "solution":  "【分析】一个三位小数精确到0.01，即四舍五入到百分位。近似数为8.74，则原数可能通过舍去或进一得到。最大时，原数千分位舍去后仍为8.74，故千分位最大为4;最小时，原数千分位进一后为8.74，故百分位为3，千分位最小为5.\n【详解】这个三位小数最大时，精确到0.01，舍去了千分位上的数，千分位最大是4，这个小数最大是8.744;这个三位小数最小时，精确到0.01，从千分位进一，此时百分位上比4小1，是3，千分位上最小是5，这个小数最小是8.735。这个三位小数最大是8.744，最小是8.735。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·四川绵阳·期中"
+    },
+    {
+        "id":  "165",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数精确到百分位后，得到的近似数是5.90，这个三位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "5.904;5.895",
+        "solution":  "【分析】精确到百分位看千分位，小于5直接舍去，大于或等于5向前一位进一，由此可知，四舍得到的近似数，原数\u003e近似数;五入得到的近似数，原数\u003c近似数，据此分析.\n【详解】一个三位小数精确到百分位后，得到的近似数是5.90，考虑四舍的情况，这个三位小数最大是5.904，考虑五入的情况，最小是5.895。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·湖南邵阳·期中"
+    },
+    {
+        "id":  "166",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "由5个百,5个一,5个十分之一和5个千分之一组成的数是(),它精确到百分位是(),保留整数部分是()。",
+        "options":  [
+
+                    ],
+        "answer":  "505.505;505.51;506",
+        "solution":  "【分析】百位的计数单位是百，个位的计数单位是一，十分位的计数单位是十分之一，千分位的计数单位是千分之一，哪一个数位上有几个该数位的计数单位就在这个数位上写几，十位和百分位一个计数单位也没有用0补足，据此写出这个小数。精确到百分位看千分位，保留整数部分看十分位，小于5直接舍去，大于或等于5向前一位进一.\n【详解】505.505≈505.51;505.505≈506。由5个百,5个一,5个十分之一和5个千分之一组成的数是505.505，它精确到百分位是505.51，保留整数部分是506。",
+        "knowledge_points":  [
+                                 "小数的近似数",
+                                 "小数的数位和计数单位的认识",
+                                 "小数的读法和写法"
+                             ],
+        "source":  "25-26五年级上·湖南邵阳·期中"
+    },
+    {
+        "id":  "167",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\"\u003e\"、\"\u003c\"或\"=\".\n3.7000()3.7;1.1()110÷1000;314000()31.4万;7.02()7.028;326.8万()0.03亿;4.5平方千米()450公顷",
+        "options":  [
+
+                    ],
+        "answer":  "=;\u003e;=;\u003c;\u003e;=",
+        "solution":  "【分析】小数的性质：小数的末尾添上0或去掉0，小数的大小不变;分母是10,100,1000。的分数化成小数;改写成\"万\"或\"亿\"作单位时在相应数位点小数点并加单位。比较小数时先比整数部分，再比十分位，百分位。;面积换算中1平方千米=100公顷.\n【详解】(1)3.7000=3.7;(2)110÷1000=0.11,1.1\u003e0.11;(3)314000=31.4万;(4)7.02\u003c7.028;(5)0.03亿=300万,326.8万\u003e300万;(6)4.5平方千米=450公顷。",
+        "knowledge_points":  [
+                                 "小数的改写",
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "多位小数的大小比较",
+                                 "小数的性质"
+                             ],
+        "source":  "25-26五年级上·湖南邵阳·期中"
+    },
+    {
+        "id":  "168",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "要使1.9☐≈2.0，☐里最小填()。要使98.05≈98.05☐，☐里能填()。",
+        "options":  [
+
+                    ],
+        "answer":  "5;0、1、2、3、4",
+        "solution":  "【分析】保留一位小数看百分位，小于5直接舍去，大于或等于5向前一位进一。第一空要进一，最小填5;第二空保持不变，末位应是可舍去的数字0~4.\n【详解】要使1.9☐≈2.0，☐里最小填5。要使98.05≈98.05☐，☐里能填0,1,2,3,4。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·湖南邵阳·期中"
+    },
+    {
+        "id":  "169",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "剪纸艺术作为一种民俗文化，在历史的发展过程中具有深刻的人文底蕴，其文化价值也越来越受重视。某县的剪纸工坊全年创作了(969808000)件作品，通过举办剪纸艺术节，全县文化产业增加值为6.9亿元。(1)括号内的数读作(),改写成用\"万\"作单位是()万，四舍五入到亿位约是()亿。(2)将6.9的小数点向右移动两位后得到的数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "九亿六千九百八十万八千;96980.8;10;690",
+        "solution":  "【分析】(1)读数时，先分级，从高位读起;改写成\"万\"作单位时在万位右下角点小数点并加\"万\";四舍五入到亿位看千万位。(2)将6.9的小数点向右移动两位就是将原数扩大到原来的100倍.\n【详解】969808000读作：九亿六千九百八十万八千;改写成用\"万\"作单位是96980.8万;千万位上的数是6，大于5，四舍五入到亿位约是10亿。将6.9的小数点向右移动两位后得到的数是690。",
+        "knowledge_points":  [
+                                 "亿以上数的读，写法",
+                                 "小数点向右移动引起小数大小变化的规律",
+                                 "小数的改写",
+                                 "小数的近似数"
+                             ],
+        "source":  "2025·江西吉安·小升初真题"
+    },
+    {
+        "id":  "170",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个两位小数四舍五入后是6.3，这个两位小数最大可能是(),最小可能是()。",
+        "options":  [
+
+                    ],
+        "answer":  "6.34;6.25",
+        "solution":  "【分析】根据题意，四舍五入后是6.3的两位小数：最大的情况是\"四舍\"，即百分位上的数\u003c=4，最大取4，所以是6.34;最小的情况是\"五入\"，即百分位上的数\u003e=5，最小取5，十分位上的数减1(3-1=2),所以是6.25.\n【详解】最大:6.34≈6.3;最小:6.25≈6.3。综上，一个两位小数四舍五入后是6.3，这个两位小数最大可能是6.34，最小可能是6.25。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·河南南阳·月考"
+    },
+    {
+        "id":  "171",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "小数2.945四舍五入保留一位小数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "2.9",
+        "solution":  "【分析】四舍五入保留一位小数，也就是去掉十分位后面的尾数，对百分位上的数进行四舍五入.\n【详解】小数2.945四舍五入保留一位小数是2.9。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·河南南阳·月考"
+    },
+    {
+        "id":  "172",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数，用\"四舍五入\"法保留两位小数约是4.90，这个三位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "4.904;4.895",
+        "solution":  "【分析】根据\"四舍五入\"的基本规则：当需要保留位数的下一位数字小于或等于4时舍去，大于或等于5时向前位进1。当三位小数保留两位后为4.90，最大值需满足原数的前两位为4.90，且第三位尽可能大但舍去后不影响结果，那么第三位最大值为4;最小值需满足原数保留两位后进位为4.90，那么第三位最小值为5，且前两位应为4.89。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·海南省直辖县级单位·期中"
+    },
+    {
+        "id":  "173",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "3.987保留一位小数约是(),保留两位小数约是()。",
+        "options":  [
+
+                    ],
+        "answer":  "4.0;3.99",
+        "solution":  "【分析】运用\"四舍五入\"法取近似值：要看精确到哪一位，从它的下一位运用\"四舍五入\"取值.\n【详解】3.987保留一位小数，看小数点后第二位是8，五入法向十分位进一约是4.0;保留两位小数，看小数点后第三位是7，五入法向百分位进一约是3.99。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·河北石家庄·期中"
+    },
+    {
+        "id":  "174",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "把1.2扩大到原来的1000倍是(),把()缩小到原来的1÷100是0.058。",
+        "options":  [
+
+                    ],
+        "answer":  "1200;5.8",
+        "solution":  "【分析】求把1.2扩大到原来的1000倍，就是将1.2的小数点向右移动三位。把一个数缩小到原来的1÷100是0.058，反推原数应将0.058的小数点向右移动两位.\n【详解】1.2×1000=1200,0.058×100=5.8，所以把1.2扩大到原来的1000倍是1200，把5.8缩小到原来的1÷100是0.058。",
+        "knowledge_points":  [
+                                 "小数点向左移动引起小数大小变化的规律",
+                                 "小数点向右移动引起小数大小变化的规律"
+                             ],
+        "source":  "25-26五年级上·河北唐山·期中"
+    },
+    {
+        "id":  "175",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "我国的铁矿石储存量非常庞大，约有4958000000吨，改写成以\"亿\"做单位的数是()亿吨，保留整数约是()亿吨。",
+        "options":  [
+
+                    ],
+        "answer":  "49.58;50",
+        "solution":  "【分析】改写成用\"亿\"作单位的数，就是在亿位数的右下角点上小数点，再在数的后面写上\"亿\"字;保留整数就是精确到个位，看小数点后第一位，用\"四舍五入\"法取近似数.\n【详解】4958000000=49.58亿≈50亿，所以改写成以\"亿\"作单位是49.58亿吨，保留整数约是50亿吨。",
+        "knowledge_points":  [
+                                 "小数的改写",
+                                 "小数的近似数",
+                                 "亿以上数的改写"
+                             ],
+        "source":  "25-26五年级上·湖南邵阳·期中"
+    },
+    {
+        "id":  "176",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "5个十,7个十分之一,2个千分之一组成的数是();0.8里有()个0.1;28个0.01组成的数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "50.702;8;0.28",
+        "solution":  "【分析】5个十表示十位是5;7个十分之一表示十分位是7;2个千分之一表示千分位是2，其它数位补0，得50.702.0.8的计数单位是0.1，有8个0.1.28个0.01即0.28。",
+        "knowledge_points":  [
+                                 "小数的意义",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "25-26五年级上·湖南邵阳·期中"
+    },
+    {
+        "id":  "177",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数的近似数是8.40，那么这个三位小数最小是(),最大是()。",
+        "options":  [
+
+                    ],
+        "answer":  "8.395;8.404",
+        "solution":  "【分析】一个三位小数保留两位小数得到8.40。最大值来自\"四舍\"，即8.40后千分位最大取4;最小值来自\"五入\"，即应由8.39进一得到8.40，千分位最小取5.\n【详解】因此最小是8.395，最大是8.404。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·山东临沂·期中"
+    },
+    {
+        "id":  "178",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把一个两位小数，保留一位小数后是1.6，这个两位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "1.64;1.55",
+        "solution":  "【分析】要考虑1.6是一个两位小数的近似数，有两种情况：\"四舍\"得到的1.6最大是1.64，\"五入\"得到的1.6最小是1.55.\n【详解】把一个两位小数，保留一位小数后是1.6，这个两位小数最大是1.64，最小是1.55。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·重庆潼南·期中"
+    },
+    {
+        "id":  "179",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "0.09公顷=()平方米;2时45分=()时(填小数);40g=()kg(填小数);6km40m=()km(填小数)",
+        "options":  [
+
+                    ],
+        "answer":  "900;2.75;0.04;6.04",
+        "solution":  "【分析】①1公顷=10000平方米;②1时=60分;③1kg=1000g;④1km=1000m。按进率进行单位换算后计算.\n【详解】0.09×10000=900(平方米);45÷60=0.75(时),2+0.75=2.75(时);40÷1000=0.04(kg);40÷1000=0.04(km),6+0.04=6.04(km)。",
+        "knowledge_points":  [
+                                 "时，分的认识",
+                                 "千米和米之间的进率与换算",
+                                 "面积单位间的进率及换算",
+                                 "运用小数点移动解决小数的单位换算问题"
+                             ],
+        "source":  "25-26五年级上·甘肃天水·期中"
+    },
+    {
+        "id":  "180",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "0.742精确到十分位约是(),8.6358省略百分位后面的尾数约是()。",
+        "options":  [
+
+                    ],
+        "answer":  "0.7;8.64",
+        "solution":  "【分析】精确到十分位，要看百分位上的数;省略百分位后面的尾数，要看千分位上的数.\n【详解】0.742的百分位上是4，舍去，约是0.7;8.6358的千分位上是5，进一，约是8.64。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·广东韶关·期中"
+    },
+    {
+        "id":  "181",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数精确到百分位是0.70，这个数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "0.704;0.695",
+        "solution":  "【分析】精确到百分位看千分位。千分位小于5时\"四舍\"，可得最大值0.704;千分位大于等于5时\"五入\"，可得最小值0.695.\n【详解】一个三位小数精确到百分位是0.70，这个数最大是0.704，最小是0.695。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·河南三门峡·期中"
+    },
+    {
+        "id":  "182",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个两位小数四舍五入后得到的近似数是3.7，这个两位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "3.74;3.65",
+        "solution":  "【分析】近似数为3.7(一位小数),原数是两位小数。百分位\u003c5时舍去，百分位\u003e=5时进一。故原数范围是3.65到3.74.\n【详解】一个两位小数四舍五入后得到的近似数是3.7，这个两位小数最大是3.74，最小是3.65。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·浙江·期中"
+    },
+    {
+        "id":  "183",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "请把以下的小数保留两位小数.\n0.854≈( );3.2876≈( );4.242≈( );9.895≈( )",
+        "options":  [
+
+                    ],
+        "answer":  "0.85;3.29;4.24;9.90",
+        "solution":  "【分析】根据\"四舍五入\"法，看小数的千分位数字：若千分位\u003e=5则向百分位进1;若千分位\u003c5则舍去千分位及后面的数.\n【详解】0.854≈0.85;3.2876≈3.29;4.242≈4.24;9.895≈9.90。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·广东揭阳·期中"
+    },
+    {
+        "id":  "184",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数四舍五入后是5.20，这个三位小数最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "5.195",
+        "solution":  "【分析】一个三位小数四舍五入后是5.20，有\"四舍\"和\"五入\"两种情况。最小值来自\"五入\"：原数应为5.19□,且千分位最小取5.\n【详解】因此，一个三位小数四舍五入后是5.20，这个三位小数最小是5.195。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·福建莆田·期中"
+    },
+    {
+        "id":  "185",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数，用四舍五入法保留两位小数后是6.32，这个三位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "6.324;6.315",
+        "solution":  "【分析】\"四舍五入\"保留两位小数时，需看千分位。若千分位\u003c=4，直接舍去(\"四舍\"),此时原数较大;若千分位\u003e=5，向百分位进1(\"五入\"),此时原数较小.\n【详解】保留两位小数后是6.32，\"四舍\"时千分位最大为4，所以最大是6.324;\"五入\"时原数应为6.31□,千分位最小为5，所以最小是6.315。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·广东东莞·期中"
+    },
+    {
+        "id":  "186",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "国务院新闻办公室举行新闻发布会，公布了第七次全国人口普查结果：全国人口共(1411780000)人。括号内的数读作(),改写成用\"万\"作单位是(),改写成用\"亿\"作单位并保留一位小数约是()亿。",
+        "options":  [
+
+                    ],
+        "answer":  "十四亿一千一百七十八万;141178万;14.1",
+        "solution":  "【分析】亿以上数先分级再读;改写成\"万\"作单位在万位后点小数点并加\"万\";改写成\"亿\"并保留一位小数，先改写再按\"四舍五入\"看百分位.\n【详解】1411780000读作十四亿一千一百七十八万;1411780000=141178万;1411780000=14.1178亿，保留一位小数约是14.1亿。",
+        "knowledge_points":  [
+                                 "亿以上数的读，写法",
+                                 "小数的近似数",
+                                 "亿以上数的改写"
+                             ],
+        "source":  "25-26四年级上·广东潮州·期中"
+    },
+    {
+        "id":  "187",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "截至2025年5月13日某时，电影《哪吒之魔童闹海》的全球票房已达(15844000000)元人民币。将电影票房改写成用亿作单位的数是()亿，省略亿位后面的尾数是()亿。",
+        "options":  [
+
+                    ],
+        "answer":  "158.44;158",
+        "solution":  "【分析】改写时，如果不是整亿的数，要在亿位后点小数点并加\"亿\";省略亿位后面的尾数看千万位，按四舍五入取整亿.\n【详解】15844000000=158.44亿,15844000000≈158亿。",
+        "knowledge_points":  [
+                                 "亿以上数的近似数",
+                                 "小数的改写"
+                             ],
+        "source":  "2025·云南昆明·小升初真题"
+    },
+    {
+        "id":  "188",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "二十点零六写作(),它的计数单位是(),它里面有()个这样的计数单位。",
+        "options":  [
+
+                    ],
+        "answer":  "20.06;0.01;2006",
+        "solution":  "【分析】写小数时整数部分按整数写法，小数点写在个位右下角，小数部分依次写数字;两位小数的计数单位是0.01;去掉小数点后是几，就有几个该计数单位.\n【详解】二十点零六写作20.06，它的计数单位是0.01，它里面有2006个这样的计数单位。",
+        "knowledge_points":  [
+                                 "小数的数位和计数单位的认识",
+                                 "小数的读法和写法"
+                             ],
+        "source":  "2025·重庆梁平·小升初真题"
+    },
+    {
+        "id":  "189",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "2.256先扩大到原来的100倍，再缩小到原来的1÷1000是()。",
+        "options":  [
+
+                    ],
+        "answer":  "0.2256",
+        "solution":  "【分析】先乘100(小数点右移两位),再除以1000(小数点左移三位),合起来相当于小数点左移一位.\n【详解】2.256×100÷1000=0.2256。",
+        "knowledge_points":  [
+                                 "小数点向右移动引起小数大小变化的规律",
+                                 "小数点向左移动引起小数大小变化的规律"
+                             ],
+        "source":  "2025·湖南邵阳·小升初真题"
+    },
+    {
+        "id":  "190",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数，四舍五入后得到的近似数是3.56，这个三位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "3.564;3.555",
+        "solution":  "【分析】若通过\"四舍\"得到3.56，则原三位小数为3.56□且千分位1~4;若通过\"五入\"得到3.56，则原三位小数为3.55□且千分位5~9.\n【详解】满足条件的三位小数中，最大是3.564，最小是3.555。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "2025·湖南邵阳·小升初真题"
+    },
+    {
+        "id":  "191",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个数亿位上是9，十万位上是5，千位上是8，百位上是6，其余各位上都是0，这个数写作(),读作(),改写成以\"万\"作单位的数是(),省略亿后面的尾数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "900508600;九亿零五十万八千六百;90050.86万;9亿",
+        "solution":  "【分析】按数位写数：亿位9，十万位5，千位8，百位6，其余补0。读数先分级;改写\"万\"在万位点小数点并加\"万\";省略亿后尾数看千万位.\n【详解】写作900508600，读作九亿零五十万八千六百，改写成以\"万\"作单位是90050.86万，省略亿后面的尾数是9亿。",
+        "knowledge_points":  [
+                                 "亿以上数的读，写法",
+                                 "小数的改写",
+                                 "亿以上数的近似数"
+                             ],
+        "source":  "2025·河南南阳·小升初模拟"
+    },
+    {
+        "id":  "192",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数，精确到百分位是1.30，这个三位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "1.304;1.295",
+        "solution":  "【分析】精确到百分位看千分位。最大值来自\"四舍\":1.30□且千分位最大4;最小值来自\"五入\"：应由1.29□进一到1.30，千分位最小5.\n【详解】这个三位小数最大是1.304，最小是1.295。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·海南省直辖县级单位·期中"
+    },
+    {
+        "id":  "193",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "由4个一,5个十分之一和9个千分之一组成的数是(),这个数精确到百分位是()。",
+        "options":  [
+
+                    ],
+        "answer":  "4.509;4.51",
+        "solution":  "【分析】4个一在个位,5个十分之一在十分位,9个千分之一在千分位，得4.509。精确到百分位看千分位,9\u003e5，向前进一.\n【详解】由4个一,5个十分之一和9个千分之一组成的数是4.509，这个数精确到百分位是4.51。",
+        "knowledge_points":  [
+                                 "小数的数位和计数单位的认识",
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·海南省直辖县级单位·期中"
+    },
+    {
+        "id":  "194",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "2025年国庆中秋假期，海南全省接待游客总人数达到(4805800)人次。把括号内的数改写成用\"万\"作单位的数是()万，保留一位小数是()万。",
+        "options":  [
+
+                    ],
+        "answer":  "480.58;480.6",
+        "solution":  "【分析】改写成\"万\"为单位，在万位点小数点并加\"万\";保留一位小数看百分位，按四舍五入取近似值.\n【详解】4805800=480.58万,480.58万≈480.6万。",
+        "knowledge_points":  [
+                                 "小数的改写",
+                                 "小数的近似数",
+                                 "亿以内数的改写"
+                             ],
+        "source":  "25-26五年级上·海南省直辖县级单位·期中"
+    },
+    {
+        "id":  "195",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数用\"四舍五入\"法精确到百分位是6.20，这个三位小数最小是(),最大是()。",
+        "options":  [
+
+                    ],
+        "answer":  "6.195;6.204",
+        "solution":  "【分析】\"四舍\"得到6.20时原数为6.20□且千分位1~4，最大6.204;\"五入\"得到6.20时原数为6.19□且千分位5~9，最小6.195.\n【详解】这个三位小数最小是6.195，最大是6.204。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·广东汕尾·期中"
+    },
+    {
+        "id":  "196",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数，\"四舍五入\"后保留两位小数是8.21，这个数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "8.214;8.205",
+        "solution":  "【分析】\"四舍\"时前两位是8.21，千分位最大4，得最大8.214;\"五入\"时应由8.20□进一为8.21，千分位最小5，得最小8.205.\n【详解】一个三位小数\"四舍五入\"后保留两位小数是8.21，这个数最大是8.214，最小是8.205。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·山东济南·期中"
+    },
+    {
+        "id":  "197",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上合适的数.\n3.2公顷=()平方米;4500平方米=()公顷;2.08平方千米=()公顷;6千米50米=()千米",
+        "options":  [
+
+                    ],
+        "answer":  "32000;0.45;208;6.05",
+        "solution":  "【分析】根据进率:1公顷=10000平方米,1平方千米=100公顷,1千米=1000米;从高级单位到低级单位乘进率，从低级单位到高级单位除以进率.\n【详解】3.2×10000=32000(平方米);4500÷10000=0.45(公顷);2.08×100=208(公顷);50÷1000=0.05(千米),6+0.05=6.05(千米)。",
+        "knowledge_points":  [
+                                 "千米和米之间的进率与换算",
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算",
+                                 "运用小数点移动解决小数的单位换算问题"
+                             ],
+        "source":  "24-25五年级上·浙江宁波·期中"
+    },
+    {
+        "id":  "198",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "2025年国庆浙江文旅综合收入达546.8亿元，改写成用\"万\"作单位的数是()万元;接待游客3760.3万人次，保留整数约是()万人次。",
+        "options":  [
+
+                    ],
+        "answer":  "5468000;3760",
+        "solution":  "【分析】把546.8亿改写成\"万\"为单位，相当于乘10000;把3760.3保留整数，精确到个位，看十分位按四舍五入取值.\n【详解】546.8亿=5468000万,3760.3≈3760。",
+        "knowledge_points":  [
+                                 "小数的改写",
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·浙江宁波·期中"
+    },
+    {
+        "id":  "199",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "嘉善县缪家村打造了13.6千米的甜蜜花海风景线，其中玫瑰花田占0.35千米,0.35中的\"3\"在()位上，表示3个(),\"5\"在()位上，表示5个()。",
+        "options":  [
+
+                    ],
+        "answer":  "十分;0.1;百分;0.01",
+        "solution":  "【分析】小数点右边第一位是十分位，计数单位0.1;第二位是百分位，计数单位0.01。数位上是几就表示几个该计数单位.\n【详解】0.35中的\"3\"在十分位，表示3个0.1;\"5\"在百分位，表示5个0.01。",
+        "knowledge_points":  [
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25五年级上·浙江宁波·期中"
+    },
+    {
+        "id":  "200",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个两位小数四舍五入后是9.6，这个两位小数最大可能是(),最小可能是()。",
+        "options":  [
+
+                    ],
+        "answer":  "9.64;9.55",
+        "solution":  "【分析】一个两位小数四舍五入后是9.6，有两种情况:①\"四舍\"得到9.6，百分位最大是4，得最大9.64;②\"五入\"得到9.6，百分位最小是5且十分位原来应是5，得最小9.55.\n【详解】9.64≈9.6,9.55≈9.6，所以这个两位小数最大可能是9.64，最小可能是9.55。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·内蒙古巴彦淖尔·期中"
+    },
+    {
+        "id":  "201",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "小数2.855四舍五入保留整数是(),保留一位小数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "3;2.9",
+        "solution":  "【分析】保留整数时，根据\"四舍五入\"法，需要看十分位上的数字;保留一位小数时，需要看百分位上的数字.\n【详解】2.855保留整数：十分位是8，向整数部分进1，得3;2.855保留一位小数：百分位是5，向前一位进1，得2.9。即2.855≈3,2.855≈2.9。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·广东江门·期中"
+    },
+    {
+        "id":  "202",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "永州市土地面积是22441平方千米，划线数字改成用万作单位的数是()平方千米，第七次全国人口普查永州常住人口为5289824人，划线数字用四舍五入法省略\"万\"后面尾数，约是()人。",
+        "options":  [
+
+                    ],
+        "answer":  "2.2441万;529万",
+        "solution":  "【分析】改写成\"万\"为单位：在万位右下角点小数点;省略\"万\"后尾数：看千位，按四舍五入处理.\n【详解】22441=2.2441万;5289824≈529万。",
+        "knowledge_points":  [
+                                 "亿以内数的近似数",
+                                 "小数的改写"
+                             ],
+        "source":  "25-26四年级上·湖南永州·期中"
+    },
+    {
+        "id":  "203",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "4千米50米=()千米;3.08吨=()吨()千克",
+        "options":  [
+
+                    ],
+        "answer":  "4.05;3;80",
+        "solution":  "【分析】1千米=1000米,1吨=1000千克，单位换算按进率计算.\n【详解】50米=0.05千米，所以4千米50米=4.05千米;3.08吨=3吨+0.08吨,0.08吨=80千克，所以3.08吨=3吨80千克。",
+        "knowledge_points":  [
+                                 "吨，千克之间的换算与比较",
+                                 "千米和米之间的进率与换算",
+                                 "运用小数点移动解决小数的单位换算问题"
+                             ],
+        "source":  "25-26五年级上·河北承德·期中"
+    },
+    {
+        "id":  "204",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个四位小数用\"四舍五入\"法保留三位小数是2.345，那么这个四位小数可能在()和()之间。",
+        "options":  [
+
+                    ],
+        "answer":  "2.3445;2.3454",
+        "solution":  "【分析】保留三位小数看万分位。最大值来自\"四舍\"，最小值来自\"五入\".\n【详解】\"四舍\"可得最大2.3454;\"五入\"可得最小2.3445，所以这个四位小数在2.3445和2.3454之间。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·河北石家庄·期中"
+    },
+    {
+        "id":  "205",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数，\"四舍五入\"保留两位小数约是4.50，那么这个三位小数最小是(),最大是()。",
+        "options":  [
+
+                    ],
+        "answer":  "4.495;4.504",
+        "solution":  "【分析】保留两位小数看千分位。五入得到最小，四舍得到最大.\n【详解】最小是4.495，最大是4.504。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·广东广州·期中"
+    },
+    {
+        "id":  "206",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.4,
+        "question":  "一个三位小数，精确到百分位是3.50，这个数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "3.504;3.495",
+        "solution":  "【分析】精确到百分位看千分位。千分位\u003c5时舍去,\u003e=5时进一.\n【详解】最大是3.504(四舍);最小是3.495(五入)。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·广西河池·期中"
+    },
+    {
+        "id":  "207",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上\"\u003e\"、\"\u003c\"或\"=\".\n2.56()2.65;32万()0.32亿;293000()29.3万",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003c;=",
+        "solution":  "【分析】先比较小数大小;再统一单位比较大数。0.32亿=3200万,29.3万=293000.\n【详解】2.56\u003c2.65;32万\u003c3200万;293000=29.3万。",
+        "knowledge_points":  [
+                                 "小数的改写",
+                                 "多位小数的大小比较",
+                                 "大数的比较"
+                             ],
+        "source":  "24-25五年级上·湖南邵阳·期中"
+    },
+    {
+        "id":  "208",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果一个三位小数取近似值是0.70，那么这个数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "0.704;0.695",
+        "solution":  "【分析】0.70是三位小数保留两位小数后的近似值，分\"四舍\"和\"五入\"两种情况.\n【详解】最大0.704(四舍),最小0.695(五入)。",
+        "knowledge_points":  [
+                                 "多位小数的大小比较",
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·天津河西·期中"
+    },
+    {
+        "id":  "209",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数，其保留两位小数后的近似数是3.17，那么这个三位小数最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "3.165",
+        "solution":  "【分析】保留两位小数看千分位。最小值来自\"五入\"到3.17的下界.\n【详解】原数范围含3.165~3.169与3.171~3.174，最小是3.165。",
+        "knowledge_points":  [
+                                 "小数的近似数",
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "2025·河南郑州·小升初真题"
+    },
+    {
+        "id":  "210",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个数的小数点向右移动两位后是15.6，这个数原来是();一个数的小数点向左移动一位后是0.08，这个数原来是()。",
+        "options":  [
+
+                    ],
+        "answer":  "0.156;0.8",
+        "solution":  "【分析】反向移动小数点即可求原数.\n【详解】15.6向左移两位得0.156;0.08向右移一位得0.8。",
+        "knowledge_points":  [
+                                 "小数点向左移动引起小数大小变化的规律",
+                                 "小数点向右移动引起小数大小变化的规律"
+                             ],
+        "source":  "25-26五年级上·山东菏泽·期中"
+    },
+    {
+        "id":  "211",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个两位小数\"四舍五入\"到十分位后是6.0，这个两位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "6.04;5.95",
+        "solution":  "【分析】6.0作为一位小数近似值，最大来自\"四舍\"，最小来自\"五入\".\n【详解】最大6.04，最小5.95。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·山东菏泽·期中"
+    },
+    {
+        "id":  "212",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数\"四舍五入\"后约是2.65，这个三位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "2.654;2.645",
+        "solution":  "【分析】保留两位小数近似为2.65，最大值由四舍得，最小值由五入得.\n【详解】最大2.654，最小2.645。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·山东济宁·月考"
+    },
+    {
+        "id":  "213",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "\"九天揽月探索不止\"，中国从未停止过对月球的探索。月球到地球的平均距离是(384000000)米,2024年嫦娥六号从月球背面采集样品1935.3克。括号内数字改写成以亿为单位的数是(),省略亿位后面的尾数约是()亿。",
+        "options":  [
+
+                    ],
+        "answer":  "3.84亿;4",
+        "solution":  "【分析】改写\"亿\"为单位在亿位点小数点;省略亿后尾数看千万位四舍五入.\n【详解】384000000=3.84亿,384000000≈4亿。",
+        "knowledge_points":  [
+                                 "小数的改写",
+                                 "亿以上数的近似数"
+                             ],
+        "source":  "24-25四年级上·山西长治·期中"
+    },
+    {
+        "id":  "214",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "2小时=()分;7.29千米=()米;350克=()千克;8.12吨=()吨()千克;1.2平方米=()平方分米",
+        "options":  [
+
+                    ],
+        "answer":  "120;7290;0.35;8;120;120",
+        "solution":  "【分析】利用1小时=60分,1千米=1000米,1千克=1000克,1吨=1000千克,1平方米=100平方分米进行换算.\n【详解】2×60=120;7.29×1000=7290;350÷1000=0.35;8.12吨=8吨+0.12吨=8吨120千克;1.2×100=120。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "千米和米之间的进率与换算",
+                                 "克，千克之间的换算与比较",
+                                 "时，分的认识"
+                             ],
+        "source":  "25-26五年级上·河北邢台·月考"
+    },
+    {
+        "id":  "215",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个两位小数四舍五入后是6.0，这个两位小数最大可能是(),最小可能是()。",
+        "options":  [
+
+                    ],
+        "answer":  "6.04;5.95",
+        "solution":  "【分析】四舍得最大，五入得最小.\n【详解】最大6.04，最小5.95。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·海南省直辖县级单位·期中"
+    },
+    {
+        "id":  "216",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "7.169保留两位小数约是(),4.98精确到十分位约是()。",
+        "options":  [
+
+                    ],
+        "answer":  "7.17;5.0",
+        "solution":  "【分析】保留两位小数看千分位;精确到十分位看百分位.\n【详解】7.169≈7.17;4.98≈5.0。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·海南省直辖县级单位·期中"
+    },
+    {
+        "id":  "217",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "把1.4545保留整数约是(),精确到千分位约是()。",
+        "options":  [
+
+                    ],
+        "answer":  "1;1.455",
+        "solution":  "【分析】保留整数看十分位;精确到千分位看万分位.\n【详解】1.4545十分位4\u003c5，约是1;万分位5，千分位进1，约是1.455。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·山东济宁·月考"
+    },
+    {
+        "id":  "218",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "有一个三位小数，\"四舍五入\"后取近似数为9.30，原三位小数最小是(),最大是()。",
+        "options":  [
+
+                    ],
+        "answer":  "9.295;9.304",
+        "solution":  "【分析】保留到百分位得9.30，最小来自\"五入\"，最大来自\"四舍\".\n【详解】最小9.295，最大9.304。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·浙江杭州·期中"
+    },
+    {
+        "id":  "219",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果A×1000=B/1000，那么将A的小数点向()移动()位就等于B。",
+        "options":  [
+
+                    ],
+        "answer":  "右;6",
+        "solution":  "【分析】由A×1000=B/1000，可两边同乘1000，得A×1000000=B.\n【详解】A变为B需乘1000000，即小数点向右移动6位。",
+        "knowledge_points":  [
+                                 "小数点向右移动引起小数大小变化的规律"
+                             ],
+        "source":  "25-26五年级上·云南楚雄·期中"
+    },
+    {
+        "id":  "220",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数用\"四舍五入\"法，保留两位小数约是4.00，这个三位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "4.004;3.995",
+        "solution":  "【分析】保留两位小数近似为4.00，最大值来自四舍，最小值来自五入.\n【详解】最大4.004，最小3.995。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·河南信阳·期中"
+    },
+    {
+        "id":  "221",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "7.9504保留一位小数约是(),精确到百分位约是()。",
+        "options":  [
+
+                    ],
+        "answer":  "8.0;7.95",
+        "solution":  "【分析】保留一位小数看百分位;精确到百分位看千分位.\n【详解】7.9504保留一位小数时百分位5，进一得8.0;精确到百分位时千分位0，舍去得7.95。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·广东汕头·期中"
+    },
+    {
+        "id":  "222",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "5.2895保留三位小数是(),保留两位小数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "5.290;5.29",
+        "solution":  "【分析】保留三位小数看第四位，保留两位小数看第三位.\n【详解】5.2895保留三位小数是5.290，保留两位小数是5.29。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·辽宁鞍山·期中"
+    },
+    {
+        "id":  "223",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "10.84保留一位小数约是(),7.96保留一位小数约是()。",
+        "options":  [
+
+                    ],
+        "answer":  "10.8;8.0",
+        "solution":  "【分析】保留一位小数看百分位，按四舍五入.\n【详解】10.84百分位4，约10.8;7.96百分位6，约8.0。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·内蒙古兴安盟·月考"
+    },
+    {
+        "id":  "224",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "把3.14159四舍五入保留两位小数后约是()。",
+        "options":  [
+
+                    ],
+        "answer":  "3.14",
+        "solution":  "【分析】保留两位小数看第三位(千分位).\n【详解】3.14159的千分位是1，小于5，舍去后为3.14。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·内蒙古通辽·月考"
+    },
+    {
+        "id":  "225",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数用\"四舍五入\"法保留两位小数是7.05，这个三位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "7.054;7.045",
+        "solution":  "【分析】保留两位小数看千分位，最大来自四舍，最小来自五入.\n【详解】最大7.054，最小7.045。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·湖北襄阳·期中"
+    },
+    {
+        "id":  "226",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "三峡大坝旅游区占地15.28平方千米，改写成用\"公顷\"作单位的数是()公顷;合()平方米。",
+        "options":  [
+
+                    ],
+        "answer":  "1528;15280000",
+        "solution":  "【分析】1平方千米=100公顷=1000000平方米.\n【详解】15.28×100=1528(公顷);15.28×1000000=15280000(平方米)。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "面积单位间的进率及换算",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25五年级上·湖北襄阳·期中"
+    },
+    {
+        "id":  "227",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "2.6先扩大到原来的10倍，再缩小到新数的1÷1000，结果是()。",
+        "options":  [
+
+                    ],
+        "answer":  "0.026",
+        "solution":  "【分析】先乘10，再除以1000，整体相当于除以100.\n【详解】2.6×10÷1000=0.026。",
+        "knowledge_points":  [
+                                 "小数点向左移动引起小数大小变化的规律",
+                                 "小数点向右移动引起小数大小变化的规律"
+                             ],
+        "source":  "25-26五年级上·辽宁鞍山·月考"
+    },
+    {
+        "id":  "228",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "把6.35扩大到原来的()倍是635;4.6缩小到原来的()是0.46。",
+        "options":  [
+
+                    ],
+        "answer":  "100;1÷10",
+        "solution":  "【分析】6.35到635小数点右移两位;4.6到0.46小数点左移一位.\n【详解】6.35扩大到原来的100倍是635;4.6缩小到原来的1÷10是0.46。",
+        "knowledge_points":  [
+                                 "小数点向左移动引起小数大小变化的规律",
+                                 "小数点向右移动引起小数大小变化的规律"
+                             ],
+        "source":  "25-26五年级上·湖南永州·月考"
+    },
+    {
+        "id":  "229",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个两位小数保留一位小数是5.0，这个数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "5.04;4.95",
+        "solution":  "【分析】5.0作为一位小数近似值，最大由四舍得到，最小由五入得到.\n【详解】最大5.04，最小4.95。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·湖南永州·期中"
+    },
+    {
+        "id":  "230",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数用\"四舍五入\"法保留两位小数后是6.40，这个三位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "6.404;6.395",
+        "solution":  "【分析】保留两位小数看千分位，最大来自四舍，最小来自五入并考虑进位.\n【详解】最大6.404，最小6.395。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·河北唐山·期中"
+    },
+    {
+        "id":  "231",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数\"四舍五入\"后是2.68，这个三位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "2.684;2.675",
+        "solution":  "【分析】保留两位小数近似为2.68，千分位\u003c5时四舍,\u003e=5时五入.\n【详解】最大2.684，最小2.675。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·湖北襄阳·期中"
+    },
+    {
+        "id":  "232",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "武汉地铁16号线全长32.46千米，读作(),其中\"4\"在()位上，表示4个()。",
+        "options":  [
+
+                    ],
+        "answer":  "三十二点四六;十分;0.1",
+        "solution":  "【分析】小数读法：整数部分按整数读，小数点读作\"点\"，小数部分依次读出。小数点后第一位是十分位，计数单位是0.1.\n【详解】32.46读作三十二点四六，其中\"4\"在十分位上，表示4个0.1。",
+        "knowledge_points":  [
+                                 "小数的数位和计数单位的认识",
+                                 "小数的读法和写法"
+                             ],
+        "source":  "24-25五年级上·湖北襄阳·期中"
+    },
+    {
+        "id":  "233",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数\"四舍五入\"后是2.60，这个三位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "2.604;2.595",
+        "solution":  "【分析】2.60是保留两位小数后的近似值，最大来自四舍，最小来自五入.\n【详解】最大2.604，最小2.595。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·黑龙江佳木斯·期中"
+    },
+    {
+        "id":  "234",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数四舍五入后是3.78，这个三位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "3.784;3.775",
+        "solution":  "【分析】要考虑3.78是一个三位小数的近似数，有两种情况：\"四舍\"得到的3.78最大是3.784，\"五入\"得到的3.78最小是3.775.\n【详解】一个三位小数四舍五入后是3.78，这个三位小数最大是3.784，最小是3.775。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·河南商丘·期中"
+    },
+    {
+        "id":  "235",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "两个数相除的商是5.36，将被除数的小数点先向左移动两位，再向右移动一位，得到的商是()。",
+        "options":  [
+
+                    ],
+        "answer":  "0.536",
+        "solution":  "【分析】被除数先除以100再乘10，相当于除以10;商也相应除以10.\n【详解】5.36÷10=0.536，所以得到的商是0.536。",
+        "knowledge_points":  [
+                                 "小数点向左移动引起小数大小变化的规律",
+                                 "小数点向右移动引起小数大小变化的规律",
+                                 "商的变化规律及应用"
+                             ],
+        "source":  "24-25五年级上·河南商丘·期中"
+    },
+    {
+        "id":  "236",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "中国古代的\"算筹\"是早期的计算工具，小明用算筹摆出的小数6.08，它的计数单位是(),它有()个这样的计数单位。",
+        "options":  [
+
+                    ],
+        "answer":  "0.01;608",
+        "solution":  "【分析】6.08是两位小数，计数单位是0.01;把小数点去掉得到608，表示有608个0.01.\n【详解】小数6.08的计数单位是0.01，它有608个这样的计数单位。",
+        "knowledge_points":  [
+                                 "小数的数位和计数单位的认识",
+                                 "小数的意义"
+                             ],
+        "source":  "25-26五年级上·黑龙江佳木斯·期中"
+    },
+    {
+        "id":  "237",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "把5.098的小数点向右移动两位是(),它是原数的();如果把原数小数点向左移动一位它是原数的()。",
+        "options":  [
+
+                    ],
+        "answer":  "509.8;100倍;1÷10",
+        "solution":  "【分析】小数点右移两位扩大100倍;左移一位缩小到1÷10.\n【详解】5.098右移两位是509.8，是原数的100倍;左移一位后是原数的1÷10。",
+        "knowledge_points":  [
+                                 "小数点向左移动引起小数大小变化的规律",
+                                 "小数点向右移动引起小数大小变化的规律"
+                             ],
+        "source":  "25-26五年级上·河北唐山·月考"
+    },
+    {
+        "id":  "238",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "7.9837保留整数约是(),精确到十分位约是(),保留两位小数约是(),精确到千分位约是()。",
+        "options":  [
+
+                    ],
+        "answer":  "8;8.0;7.98;7.984",
+        "solution":  "【分析】按\"四舍五入\"分别看个位，十分位，百分位，千分位后一位数字.\n【详解】7.9837≈8;≈8.0;≈7.98;≈7.984。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·河北唐山·月考"
+    },
+    {
+        "id":  "239",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个两位小数，用\"四舍五入\"法保留一位小数是5.6，这个小数最小是(),最大是()。",
+        "options":  [
+
+                    ],
+        "answer":  "5.55;5.64",
+        "solution":  "【分析】保留一位小数看百分位;\"五入\"得最小，\"四舍\"得最大.\n【详解】最小是5.55，最大是5.64。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·甘肃武威·期中"
+    },
+    {
+        "id":  "240",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "0.78扩大到原来的100倍是();56缩小到原来的1÷10是()。",
+        "options":  [
+
+                    ],
+        "answer":  "78;5.6",
+        "solution":  "【分析】扩大100倍即乘100;缩小到1÷10即除以10.\n【详解】0.78×100=78;56÷10=5.6。",
+        "knowledge_points":  [
+                                 "小数点向左移动引起小数大小变化的规律",
+                                 "小数点向右移动引起小数大小变化的规律"
+                             ],
+        "source":  "25-26五年级上·河北邢台·月考"
+    },
+    {
+        "id":  "241",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个数的小数点向右移动一位后是7.8，原数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "0.78",
+        "solution":  "【分析】已知右移一位后的数，求原数应左移一位.\n【详解】7.8÷10=0.78，原数是0.78。",
+        "knowledge_points":  [
+                                 "小数点向左移动引起小数大小变化的规律",
+                                 "小数点向右移动引起小数大小变化的规律"
+                             ],
+        "source":  "25-26五年级上·黑龙江佳木斯·期中"
+    },
+    {
+        "id":  "242",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个两位小数用\"四舍五入\"法取近似数得2，这个两位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "2.49;1.50",
+        "solution":  "【分析】四舍五入到整数为2的两位小数范围是[1.50,2.49].\n【详解】最大2.49，最小1.50。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·河南南阳·月考"
+    },
+    {
+        "id":  "243",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "保留两位小数，写出下面小数的近似数.\n16.275≈();3.8961≈();2.83≈();29.777≈()",
+        "options":  [
+
+                    ],
+        "answer":  "16.28;3.90;2.83;29.78",
+        "solution":  "【分析】保留两位小数看千分位，按四舍五入处理.\n【详解】16.275≈16.28;3.8961≈3.90;2.83≈2.83;29.777≈29.78。",
+        "knowledge_points":  [
+                                 "小数的近似数",
+                                 "循环小数的认识与简写"
+                             ],
+        "source":  "24-25五年级上·陕西安康·期中"
+    },
+    {
+        "id":  "244",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\"\u003e\"、\"\u003c\"或\"=\".\n3.3亿()310000000;5340100()5314000;32平方千米()32000公顷;7公顷()700平方米",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;\u003e;\u003c;\u003e",
+        "solution":  "【分析】统一单位后比较。3.3亿=330000000;1平方千米=100公顷;1公顷=10000平方米.\n【详解】3.3亿\u003e310000000;5340100\u003e5314000;32平方千米=3200公顷\u003c32000公顷;7公顷=70000平方米\u003e700平方米。",
+        "knowledge_points":  [
+                                 "小数的改写",
+                                 "公顷，平方千米的进率与换算",
+                                 "大数的比较"
+                             ],
+        "source":  "24-25四年级上·江西赣州·期中"
+    },
+    {
+        "id":  "245",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "去掉0.48的小数点，就是把这个数扩大到原来的()倍;把40.8的小数点向左移动两位，就是把它缩小到原来的()。",
+        "options":  [
+
+                    ],
+        "answer":  "100;1÷100",
+        "solution":  "【分析】小数点右移两位扩大100倍;左移两位缩小到1÷100.\n【详解】去掉0.48的小数点是48，相当于扩大100倍;40.8左移两位相当于缩小到1÷100。",
+        "knowledge_points":  [
+                                 "小数点向左移动引起小数大小变化的规律",
+                                 "小数点向右移动引起小数大小变化的规律"
+                             ],
+        "source":  "25-26五年级上·河北保定·月考"
+    },
+    {
+        "id":  "246",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个两位小数\"四舍五入\"保留一位小数后是3.5，这个两位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "3.54;3.45",
+        "solution":  "【分析】保留一位小数为3.5的两位小数，四舍部分最大，五入部分最小.\n【详解】最大3.54，最小3.45。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·云南曲靖·期中"
+    },
+    {
+        "id":  "247",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数\"四舍五入\"后是2.40，这个三位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "2.404;2.395",
+        "solution":  "【分析】最大来自\"四舍\":2.40后千分位最大4;最小来自\"五入\"：由2.39进一，千分位最小5.\n【详解】最大2.404，最小2.395。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·甘肃天水·月考"
+    },
+    {
+        "id":  "248",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数的近似数是5.60，这个三位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "5.604;5.595",
+        "solution":  "【分析】近似数5.60对应保留两位小数;最大四舍，最小五入.\n【详解】最大5.604，最小5.595。",
+        "knowledge_points":  [
+                                 "还原小数近似数的问题"
+                             ],
+        "source":  "24-25五年级上·海南省直辖县级单位·期中"
+    },
+    {
+        "id":  "249",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数，用\"四舍五入\"法取近似数是4.60，这个三位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "4.604;4.595",
+        "solution":  "【分析】保留两位小数为4.60，四舍时最大，五入时最小.\n【详解】最大4.604，最小4.595。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·山西晋中·期中"
+    },
+    {
+        "id":  "250",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数精确到百分位后是0.90，这个三位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "0.904;0.895",
+        "solution":  "【分析】精确到百分位看千分位。四舍得最大，五入得最小.\n【详解】最大0.904，最小0.895。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·江西赣州·期中"
+    },
+    {
+        "id":  "251",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "单位换算.\n935dm²=()m²;20.24t=()kg",
+        "options":  [
+
+                    ],
+        "answer":  "9.35;20240",
+        "solution":  "【分析】1m²=100dm²,1t=1000kg.\n【详解】935÷100=9.35，所以935dm²=9.35m²;20.24×1000=20240，所以20.24t=20240kg。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "平方厘米，平方分米，平方米之间的进率与换算",
+                                 "吨，千克之间的换算与比较"
+                             ],
+        "source":  "24-25五年级上·湖北孝感·期中"
+    },
+    {
+        "id":  "252",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.4,
+        "question":  "一个三位小数，用\"四舍五入\"法保留两位小数后是6.00，这个三位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "6.004;5.995",
+        "solution":  "【分析】保留两位小数看千分位;四舍取最大，五入取最小.\n【详解】最大6.004，最小5.995。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·湖南永州·期中"
+    },
+    {
+        "id":  "253",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "\"浙BA\"是全省规模最大，规格最高的群众性篮球赛事，覆盖全省11个设区市,90个县(市，区),截至10月9日，浙BA共举办赛事339场，观众场场爆满，累计观赛人数1169873人次，全网传播量21638080000次,864个话题登上各大平台热搜热榜，带动文商旅体消费2200000000元.\n(1)1169873读作(),改写成用\"万\"作单位并保留整数约是()万.\n(2)把21638080000读作(),省略亿位后面的尾数是()亿.\n(3)把2200000000元改写成用\"万\"作单位是()万元。",
+        "options":  [
+
+                    ],
+        "answer":  "一百一十六万九千八百七十三;117;二百一十六亿三千八百零八万;216;220000",
+        "solution":  "【分析】按大数读法分级读数;改写\"万\"在万位点小数点;省略亿后尾数看千万位;整数改写\"万\"去末尾4个0并加\"万\".\n【详解】1169873读作一百一十六万九千八百七十三，改写为116.9873万，保留整数约117万;21638080000读作二百一十六亿三千八百零八万，省略亿后尾数为216亿;2200000000元=220000万元。",
+        "knowledge_points":  [
+                                 "亿以上数的近似数",
+                                 "小数的近似数",
+                                 "亿以上数的读，写法",
+                                 "亿以内数的读，写法"
+                             ],
+        "source":  "25-26四年级上·浙江杭州·阶段练习"
+    },
+    {
+        "id":  "254",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "9.598用\"四舍五入\"法保留整数是();精确到百分位是()。",
+        "options":  [
+
+                    ],
+        "answer":  "10;9.60",
+        "solution":  "【分析】保留整数看十分位;精确到百分位看千分位.\n【详解】9.598≈10;9.598≈9.60。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·山东济宁·期中"
+    },
+    {
+        "id":  "255",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "1234567890=()万≈()万(保留一位小数);9865432100=()亿≈()亿(保留两位小数)",
+        "options":  [
+
+                    ],
+        "answer":  "123456.789;123456.8;98.654321;98.65",
+        "solution":  "【分析】改写\"万\"左移4位，改写\"亿\"左移8位;再按四舍五入保留位数.\n【详解】1234567890=123456.789万≈123456.8万;9865432100=98.654321亿≈98.65亿。",
+        "knowledge_points":  [
+                                 "亿以上数的近似数",
+                                 "小数的近似数",
+                                 "小数的改写"
+                             ],
+        "source":  "24-25五年级上·河北石家庄·期中"
+    },
+    {
+        "id":  "256",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个数由5个10,3个0.01和8个1÷1000组成，这个数是(),它读作()。",
+        "options":  [
+
+                    ],
+        "answer":  "50.038;五十点零三八",
+        "solution":  "【分析】5个10在十位,3个0.01在百分位,8个0.001在千分位，其余位补0.\n【详解】这个数是50.038，读作五十点零三八。",
+        "knowledge_points":  [
+                                 "小数的数位和计数单位的认识",
+                                 "小数的读法和写法",
+                                 "小数的意义"
+                             ],
+        "source":  "24-25五年级上·河北石家庄·期中"
+    },
+    {
+        "id":  "257",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小强在商店读某商品的标价时，看漏了小数点，误读成了四千零八元。已知该商品的标价低于50元，高于10元，则该商品的标价应该是()元。",
+        "options":  [
+
+                    ],
+        "answer":  "40.08",
+        "solution":  "【分析】误读数是4008，实际价格在10~50之间，应将4008小数点左移两位.\n【详解】4008→40.08，符合范围，所以标价是40.08元。",
+        "knowledge_points":  [
+                                 "小数点向左移动引起小数大小变化的规律"
+                             ],
+        "source":  "24-25五年级上·河南洛阳·期中"
+    },
+    {
+        "id":  "258",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "现在，天安门城楼上的中华人民共和国新国徽，经过精准测量后，高度确定为2.67米，宽度确定为二点四八米.\n①2.67读作(),它的整数部分是(),小数部分的7表示().\n②二点四八写作(),它是由()个1,()个0.1和()个0.01组成的。",
+        "options":  [
+
+                    ],
+        "answer":  "二点六七;2;7个0.01;2.48;2;4;8",
+        "solution":  "【分析】按小数读写法和数位计数单位进行判断.\n【详解】2.67读作二点六七，整数部分是2,7表示7个0.01;二点四八写作2.48，由2个1,4个0.1和8个0.01组成。",
+        "knowledge_points":  [
+                                 "小数的数位和计数单位的认识",
+                                 "小数的读法和写法",
+                                 "小数的意义"
+                             ],
+        "source":  "24-25五年级上·河南洛阳·期中"
+    },
+    {
+        "id":  "259",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "把3.0800化简后是(),把4改写成用千分之一作计数单位的数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "3.08;4.000",
+        "solution":  "【分析】小数末尾添0或去0大小不变;千分之一计数单位表示保留到千分位.\n【详解】3.0800化简是3.08;4写成千分之一作单位是4.000。",
+        "knowledge_points":  [
+                                 "小数的性质"
+                             ],
+        "source":  "24-25五年级上·河南洛阳·期中"
+    },
+    {
+        "id":  "260",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个数由5个10,3个0.01和8个1÷1000组成，这个数是(),它读作()。",
+        "options":  [
+
+                    ],
+        "answer":  "50.038;五十点零三八",
+        "solution":  "【分析】十位是5，百分位是3，千分位是8，其余位补0.\n【详解】这个数是50.038，读作五十点零三八。",
+        "knowledge_points":  [
+                                 "小数的数位和计数单位的认识",
+                                 "小数的读法和写法"
+                             ],
+        "source":  "24-25五年级上·湖南邵阳·期中"
+    },
+    {
+        "id":  "261",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个两位小数，用\"四舍五入\"法取近似值是5.8。这个小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "5.84;5.75",
+        "solution":  "【分析】5.8为一位小数近似值。四舍得到最大，五入得到最小.\n【详解】最大5.84，最小5.75。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·山西忻州·期中"
+    },
+    {
+        "id":  "262",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "1234567890=()万≈()万(保留一位小数);9865432100=()亿≈()亿(保留两位小数)",
+        "options":  [
+
+                    ],
+        "answer":  "123456.789;123456.8;98.654321;98.65",
+        "solution":  "【分析】改写单位后，再按四舍五入保留对应位数.\n【详解】1234567890=123456.789万≈123456.8万;9865432100=98.654321亿≈98.65亿。",
+        "knowledge_points":  [
+                                 "亿以上数的改写",
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·湖南邵阳·期中"
+    },
+    {
+        "id":  "263",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "7.034×100就是把7.034的小数点向()移动()位,4.92÷1000就是把4.92的小数点向()移动()位。",
+        "options":  [
+
+                    ],
+        "answer":  "右;两;左;三",
+        "solution":  "【分析】乘100右移两位，除以1000左移三位.\n【详解】7.034×100小数点向右两位;4.92÷1000小数点向左三位。",
+        "knowledge_points":  [
+                                 "小数点向左移动引起小数大小变化的规律",
+                                 "小数点向右移动引起小数大小变化的规律"
+                             ],
+        "source":  "24-25五年级上·山西忻州·期中"
+    },
+    {
+        "id":  "264",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "经文化和旅游部数据中心测算,2024年国庆节假期国内出游有七亿六千五百万人次，横线上的数写作(),改写成以\"亿\"作单位的数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "765000000;7.65亿",
+        "solution":  "【分析】按整数写法写出\"七亿六千五百万\";改写成\"亿\"为单位在亿位点小数点.\n【详解】七亿六千五百万写作765000000,765000000=7.65亿。",
+        "knowledge_points":  [
+                                 "小数的改写",
+                                 "亿以上数的读，写法"
+                             ],
+        "source":  "24-25四年级上·广西柳州·期中"
+    },
+    {
+        "id":  "265",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数的近似值是2.90，这个三位小数最大可能是(),最小可能是()。",
+        "options":  [
+
+                    ],
+        "answer":  "2.904;2.895",
+        "solution":  "【分析】保留两位小数近似为2.90，最大来自四舍，最小来自五入.\n【详解】最大2.904，最小2.895。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·河南新乡·期中"
+    },
+    {
+        "id":  "266",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个两位小数\"四舍五入\"后是9.0，这个两位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "9.04;8.95",
+        "solution":  "【分析】保留一位小数看百分位;四舍得最大，五入得最小.\n【详解】9.04≈9.0,8.95≈9.0，所以最大9.04，最小8.95。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·甘肃平凉·期中"
+    },
+    {
+        "id":  "267",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "把53095000改写成以\"万\"为单位的数是()万。",
+        "options":  [
+
+                    ],
+        "answer":  "5309.5",
+        "solution":  "【分析】改写成\"万\"为单位，在万位后点小数点并去掉末尾无效0.\n【详解】53095000=5309.5万。",
+        "knowledge_points":  [
+                                 "小数的改写"
+                             ],
+        "source":  "24-25六年级上·贵州安顺·期中"
+    },
+    {
+        "id":  "268",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数用\"四舍五入\"法保留两位小数是6.80，这个三位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "6.804;6.795",
+        "solution":  "【分析】保留两位小数为6.80，四舍时最大，五入时最小.\n【详解】最大6.804，最小6.795。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·湖北襄阳·期中"
+    },
+    {
+        "id":  "269",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "武汉市生态环境局排查整治1842个长江入河排污口，改写成用\"千\"作单位的数是()个，保留一位小数约是()个。",
+        "options":  [
+
+                    ],
+        "answer":  "1.842千;1.8千",
+        "solution":  "【分析】改写成\"千\"为单位，在千位后点小数点;保留一位小数看百分位并四舍五入.\n【详解】1842=1.842千≈1.8千。",
+        "knowledge_points":  [
+                                 "小数的近似数",
+                                 "小数的改写"
+                             ],
+        "source":  "24-25五年级上·湖北襄阳·期中"
+    },
+    {
+        "id":  "270",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个两位小数\"四舍五入\"后得到的近似数是8.0，这个两位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "8.04;7.95",
+        "solution":  "【分析】四舍五入到十分位看百分位.\"四舍\"得到8.0时原数最大;\"五入\"得到8.0时原数最小.\n【详解】最大是8.04，最小是7.95。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·新疆阿克苏·期中"
+    },
+    {
+        "id":  "271",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数用\"四舍五入\"法保留两位小数后得到的近似数是8.00，那么这个三位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "8.004;7.995",
+        "solution":  "【分析】保留两位小数看千分位。四舍可取到最大，五入可取到最小.\n【详解】最大是8.004，最小是7.995。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·河南郑州·阶段练习"
+    },
+    {
+        "id":  "272",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上\"\u003e\"\"\u003c\"或\"=\".\n2.56()2.65;32万()0.32亿;293000()29.3万",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003c;=",
+        "solution":  "【分析】比较小数先比整数部分，再比分位;把0.32亿换算成万再比较;把29.3万换算成整数再比较.\n【详解】2.56\u003c2.65;0.32亿=3200万，所以32万\u003c0.32亿;29.3万=293000，所以293000=29.3万。",
+        "knowledge_points":  [
+                                 "亿以上数的改写",
+                                 "小数的改写",
+                                 "多位小数的大小比较",
+                                 "大数的比较"
+                             ],
+        "source":  "24-25五年级上·湖南邵阳·期中"
+    },
+    {
+        "id":  "273",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "1.96里有()个千分之一，保留一位小数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "1960;2.0",
+        "solution":  "【分析】1.96=1960个0.001;保留一位小数看百分位，按四舍五入处理.\n【详解】1.96里有1960个千分之一，保留一位小数是2.0。",
+        "knowledge_points":  [
+                                 "小数的近似数",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25五年级上·湖南邵阳·期中"
+    },
+    {
+        "id":  "274",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "2.153保留一位小数约是(),精确到百分位约是()。",
+        "options":  [
+
+                    ],
+        "answer":  "2.2;2.15",
+        "solution":  "【分析】保留一位小数看百分位;精确到百分位看千分位.\n【详解】2.153百分位是5，进一得2.2;千分位是3，舍去得2.15。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·河南漯河·期中"
+    },
+    {
+        "id":  "275",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数，保留两位小数是4.10，那这个数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "4.104;4.095",
+        "solution":  "【分析】保留两位小数看千分位。四舍可得最大，五入可得最小.\n【详解】最大是4.104，最小是4.095。",
+        "knowledge_points":  [
+                                 "小数的近似数",
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "24-25五年级上·西藏日喀则·期中"
+    },
+    {
+        "id":  "276",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "2.45扩大到原来的()倍是2450;2.4缩小到原来的()是0.024。",
+        "options":  [
+
+                    ],
+        "answer":  "1000;1÷100",
+        "solution":  "【分析】小数点右移三位扩大1000倍;左移两位缩小到1÷100.\n【详解】2.45到2450是扩大1000倍;2.4到0.024是缩小到1÷100。",
+        "knowledge_points":  [
+                                 "小数点向左移动引起小数大小变化的规律",
+                                 "小数点向右移动引起小数大小变化的规律"
+                             ],
+        "source":  "24-25五年级上·西藏日喀则·期中"
+    },
+    {
+        "id":  "277",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数\"四舍五入\"保留两位小数约是8.50，这个三位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "8.504;8.495",
+        "solution":  "【分析】保留两位小数时看千分位，四舍得最大，五入得最小.\n【详解】最大是8.504，最小是8.495。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·山东济宁·期中"
+    },
+    {
+        "id":  "278",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "3.959四舍五入保留整数是(),保留一位小数是(),保留两位小数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "4;4.0;3.96",
+        "solution":  "【分析】保留到哪一位就看下一位，按四舍五入处理.\n【详解】3.959≈4;3.959≈4.0;3.959≈3.96。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·山东济宁·期中"
+    },
+    {
+        "id":  "279",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数\"四舍五入\"到百分位约是2.78，这个三位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "2.784;2.775",
+        "solution":  "【分析】保留两位小数看千分位。四舍时原数较大，五入时原数较小.\n【详解】最大是2.784，最小是2.775。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·河北邢台·月考"
+    },
+    {
+        "id":  "280",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个两位小数，将它四舍五入到十分位约是4.6，这个数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "4.64;4.55",
+        "solution":  "【分析】四舍五入到十分位看百分位。四舍得到最大，五入得到最小.\n【详解】最大是4.64，最小是4.55。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·海南省直辖县级单位·阶段练习"
+    },
+    {
+        "id":  "281",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数，保留两位小数约是5.86，这个三位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "5.864;5.855",
+        "solution":  "【分析】保留两位小数看千分位，四舍得最大，五入得最小.\n【详解】最大是5.864，最小是5.855。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·广东揭阳·月考"
+    },
+    {
+        "id":  "282",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果将一个两位小数\"四舍五入\"精确到十分位后大约是9.0，那么这个两位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "9.04;8.95",
+        "solution":  "【分析】精确到十分位看百分位，范围是8.95到9.04.\n【详解】最大是9.04，最小是8.95。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·广西玉林·期中"
+    },
+    {
+        "id":  "283",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "地球的表面积为510083042平方千米，这个数读作(),改写成用\"万\"作单位的数是(),四舍五入到亿位约是()。",
+        "options":  [
+
+                    ],
+        "answer":  "五亿一千零八万三千零四十二;51008.3042万;5亿",
+        "solution":  "【分析】先按大数读法分级读数;改写成万在万位点小数点;四舍五入到亿位看千万位.\n【详解】510083042读作五亿一千零八万三千零四十二;改写成万是51008.3042万;到亿位约是5亿。",
+        "knowledge_points":  [
+                                 "亿以上数的读，写法",
+                                 "小数的改写",
+                                 "亿以内数的改写"
+                             ],
+        "source":  "25-26四年级上·广东佛山·月考"
+    },
+    {
+        "id":  "284",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "李小刚在做求一个三位小数的近似数时，求得的近似数是2.56，这个小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "2.564;2.555",
+        "solution":  "【分析】2.56是三位小数保留两位小数后的近似值。四舍得最大，五入得最小.\n【详解】最大是2.564，最小是2.555。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·贵州六盘水·期中"
+    },
+    {
+        "id":  "285",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个两位小数\"四舍五入\"后约是5.0，这个小数最大是()。",
+        "options":  [
+
+                    ],
+        "answer":  "5.04",
+        "solution":  "【分析】约是5.0时，两位小数最大值对应\"四舍\"情形.\n【详解】百分位最大取4，所以最大是5.04。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·河南南阳·期中"
+    },
+    {
+        "id":  "286",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "0.96的计数单位是(),它有()个这样的计数单位，那么再添上()个这样的计数单位结果等于1。",
+        "options":  [
+
+                    ],
+        "answer":  "0.01;96;4",
+        "solution":  "【分析】0.96是两位小数，计数单位是0.01;0.96有96个0.01;到1还差4个0.01.\n【详解】计数单位是0.01，有96个，再添4个等于1。",
+        "knowledge_points":  [
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25五年级上·湖南邵阳·期中"
+    },
+    {
+        "id":  "287",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "3公顷50平方米=()公顷;300平方厘米=()平方米",
+        "options":  [
+
+                    ],
+        "answer":  "3.005;0.03",
+        "solution":  "【分析】1公顷=10000平方米,1平方米=10000平方厘米.\n【详解】50平方米=0.005公顷，所以3公顷50平方米=3.005公顷;300平方厘米=300÷10000=0.03平方米。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "面积单位间的进率及换算",
+                                 "平方厘米，平方分米，平方米之间的进率与换算"
+                             ],
+        "source":  "24-25五年级上·湖南邵阳·期中"
+    },
+    {
+        "id":  "288",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "把996800000改写成用\"万\"作单位的数是()万，改写成用\"亿\"作单位的数并保留一位小数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "99680;10.0亿",
+        "solution":  "【分析】改写成\"万\"去掉末4位并加万;改写成\"亿\"在亿位点小数点，再保留一位小数.\n【详解】996800000=99680万;996800000=9.968亿≈10.0亿。",
+        "knowledge_points":  [
+                                 "亿以上数的改写",
+                                 "小数的近似数",
+                                 "小数的改写"
+                             ],
+        "source":  "24-25五年级上·湖南邵阳·期中"
+    },
+    {
+        "id":  "289",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "由9个十,8个0.1和5个百分之一组成的数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "90.85",
+        "solution":  "【分析】9个十在十位,8个0.1在十分位,5个百分之一在百分位.\n【详解】组成的数是90.85。",
+        "knowledge_points":  [
+                                 "小数的数位和计数单位的认识",
+                                 "小数的读法和写法"
+                             ],
+        "source":  "24-25五年级上·湖南邵阳·期中"
+    },
+    {
+        "id":  "290",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数保留两位小数后是8.00，这个三位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "8.004;7.995",
+        "solution":  "【分析】保留两位小数看千分位。四舍情形取最大，五入情形取最小.\n【详解】最大是8.004，最小是7.995。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·河南商丘·阶段练习"
+    },
+    {
+        "id":  "291",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "根据新闻报道，我国每年消耗一次性筷子45000000000双，这就需要砍伐25000000棵树，减少200万平方米森林面积。因此，人人要节约资源，还自然一片绿色。45000000000改写成用\"亿\"作单位的数是()亿,25000000精确到十分位是()亿。",
+        "options":  [
+
+                    ],
+        "answer":  "450;0.3",
+        "solution":  "【分析】改写成\"亿\"看亿级;25000000=0.25亿，精确到十分位看百分位.\n【详解】45000000000=450亿;25000000=0.25亿≈0.3亿。",
+        "knowledge_points":  [
+                                 "亿以上数的改写",
+                                 "亿以上数的近似数"
+                             ],
+        "source":  "24-25五年级上·河南洛阳·期中"
+    },
+    {
+        "id":  "292",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "45厘米=()米;28.08千米=()千米()米",
+        "options":  [
+
+                    ],
+        "answer":  "0.45;28;80",
+        "solution":  "【分析】1米=100厘米,1千米=1000米.\n【详解】45÷100=0.45米;28.08千米=28千米+0.08千米,0.08千米=80米。",
+        "knowledge_points":  [
+                                 "厘米和米之间的进率与换算",
+                                 "千米和米之间的进率与换算"
+                             ],
+        "source":  "24-25五年级上·湖南长沙·期中"
+    },
+    {
+        "id":  "293",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "下面各数分别是由三位小数四舍五入得到的近似数，请分别写出每个三位小数的最小值和最大值.\n()\u003c9.0\u003c();()\u003c0.10\u003c()",
+        "options":  [
+
+                    ],
+        "answer":  "8.950;9.049;0.095;0.104",
+        "solution":  "【分析】9.0和0.10都看作保留一位/两位小数后的近似值，分别从五入最小值和四舍最大值确定区间.\n【详解】8.950\u003c9.0\u003c9.049;0.095\u003c0.10\u003c0.104。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·河北沧州·期中"
+    },
+    {
+        "id":  "294",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "如果一个三位小数取近似值是4.80，那么它最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "4.804;4.795",
+        "solution":  "【分析】近似值4.80对应保留两位小数，最大来自四舍，最小来自五入.\n【详解】最大是4.804，最小是4.795。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·山东日照·期中"
+    },
+    {
+        "id":  "295",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "0.5054,0.505,0.5056,0.50555中，最大的是(),最小的是()。",
+        "options":  [
+
+                    ],
+        "answer":  "0.5056;0.505",
+        "solution":  "【分析】整数部分和前几位相同，继续比较后续位数，位数不足补0再比较.\n【详解】0.505=0.5050，比较得0.505最小,0.5056最大。",
+        "knowledge_points":  [
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "24-25五年级上·天津滨海新·期中"
+    },
+    {
+        "id":  "296",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "5时=()分;2.05米=()米()厘米",
+        "options":  [
+
+                    ],
+        "answer":  "300;2;5",
+        "solution":  "【分析】1时=60分;1米=100厘米。单名数改复名数时整数部分作高级单位，小数部分换算成低级单位.\n【详解】5×60=300分;2.05米=2米+0.05米=2米5厘米。",
+        "knowledge_points":  [
+                                 "时，分的认识",
+                                 "厘米和米之间的进率与换算"
+                             ],
+        "source":  "24-25五年级上·山东济宁·期中"
+    },
+    {
+        "id":  "297",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把2.44扩大到它的()倍是244，把23缩小到它的1÷1000是()。",
+        "options":  [
+
+                    ],
+        "answer":  "100;0.023",
+        "solution":  "【分析】2.44到244小数点右移两位;23缩小到1÷1000即除以1000.\n【详解】2.44扩大100倍得244;23÷1000=0.023。",
+        "knowledge_points":  [
+                                 "小数点向左移动引起小数大小变化的规律",
+                                 "小数点向右移动引起小数大小变化的规律"
+                             ],
+        "source":  "24-25四年级下·河北唐山·期末"
+    },
+    {
+        "id":  "298",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "8公顷=()平方米;32平方千米=()公顷;450dm²=()m²;40dm²=()cm²",
+        "options":  [
+
+                    ],
+        "answer":  "80000;3200;4.5;4000",
+        "solution":  "【分析】1公顷=10000平方米,1平方千米=100公顷,1平方米=100平方分米,1平方分米=100平方厘米.\n【详解】8×10000=80000;32×100=3200;450÷100=4.5;40×100=4000。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题"
+                             ],
+        "source":  "24-25五年级上·河南周口·期中"
+    },
+    {
+        "id":  "299",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在63079500这个数中，\"3\"在()位上，万位上的数是(),改写成用\"万\"作单位的数是(),省略万位后面的尾数约是()。",
+        "options":  [
+
+                    ],
+        "answer":  "百万;7;6307.95万;6308万",
+        "solution":  "【分析】按数位顺序判断\"3\"的位置和万位数字;改写成万在万位点小数点;省略万后尾数看千位四舍五入.\n【详解】\"3\"在百万位，万位是7;63079500=6307.95万;约是6308万。",
+        "knowledge_points":  [
+                                 "亿以内数的近似数",
+                                 "小数的改写",
+                                 "整数的数级，数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级上·河南信阳·期中"
+    },
+    {
+        "id":  "300",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "由3个十,4个十分之一,6个百分之一组成的数是(),它的计数单位是(),将这个数精确到十分位是()。",
+        "options":  [
+
+                    ],
+        "answer":  "30.46;0.01;30.5",
+        "solution":  "【分析】3个十在十位,4个十分之一在十分位,6个百分之一在百分位，得30.46。最低数位是百分位，所以计数单位是0.01。精确到十分位看百分位,6\u003e=5进一.\n【详解】这个数是30.46，计数单位是0.01，精确到十分位是30.5。",
+        "knowledge_points":  [
+                                 "小数的近似数",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·河北唐山·期末"
+    },
+    {
+        "id":  "301",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "把8.88扩大到它的100倍是(),然后缩小到该数的1÷1000是()。",
+        "options":  [
+
+                    ],
+        "answer":  "888;0.888",
+        "solution":  "【分析】把8.88扩大到100倍，小数点向右移动两位;再缩小到1÷1000，小数点向左移动三位.\n【详解】把8.88扩大到它的100倍是888，然后缩小到该数的1÷1000是0.888。",
+        "knowledge_points":  [
+                                 "小数点向右移动引起小数大小变化的规律",
+                                 "小数点向左移动引起小数大小变化的规律"
+                             ],
+        "source":  "24-25四年级下·安徽阜阳·期末"
+    },
+    {
+        "id":  "302",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "1.5小时=()分;0.5公顷=()平方米;3吨50克=()吨;3.8米=()米()厘米",
+        "options":  [
+
+                    ],
+        "answer":  "90;5000;3.00005;3;80",
+        "solution":  "【分析】根据1小时=60分,1公顷=10000平方米,1吨=1000000克,1米=100厘米进行换算.\n【详解】1.5小时=90分;0.5公顷=5000平方米;3吨50克=3.00005吨;3.8米=3米80厘米。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "公顷，平方千米的进率与换算",
+                                 "吨，千克之间的换算与比较",
+                                 "时，分的认识"
+                             ],
+        "source":  "24-25四年级下·河北唐山·期末"
+    },
+    {
+        "id":  "303",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把5.248扩大到它的100倍是(),再缩小到它的1÷1000是()。",
+        "options":  [
+
+                    ],
+        "answer":  "524.8;0.5248",
+        "solution":  "【分析】扩大100倍小数点右移两位;缩小到1÷1000小数点左移三位.\n【详解】把5.248扩大到它的100倍是524.8，再缩小到它的1÷1000是0.5248。",
+        "knowledge_points":  [
+                                 "小数点向左移动引起小数大小变化的规律",
+                                 "小数点向右移动引起小数大小变化的规律"
+                             ],
+        "source":  "24-25四年级下·河北邢台·期末"
+    },
+    {
+        "id":  "304",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个数由3个十亿,9个千万和8个十万组成，这个数写作(),读作(),改写成用\"亿\"作单位的数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "3090800000;三十亿九千零八十万;30.908亿",
+        "solution":  "【分析】十亿位是3，千万位是9，十万位是8，其余位补0;再按亿级，万级读数并改写成\"亿\".\n【详解】写作3090800000，读作三十亿九千零八十万，改写成用\"亿\"作单位是30.908亿。",
+        "knowledge_points":  [
+                                 "小数的改写",
+                                 "整数的数级，数位和计数单位的认识",
+                                 "亿以上数的读，写法"
+                             ],
+        "source":  "24-25四年级上·广东珠海·期中"
+    },
+    {
+        "id":  "305",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数保留两位小数后是7.60，这个数最小是(),最大是()。",
+        "options":  [
+
+                    ],
+        "answer":  "7.595;7.604",
+        "solution":  "【分析】保留两位小数看千分位。最小来自\"五入\"，最大来自\"四舍\".\n【详解】最小是7.595，最大是7.604。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·广西玉林·期中"
+    },
+    {
+        "id":  "306",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个两位小数，用\"四舍五入法\"得到它的近似值为8.4，这样的两位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "8.44;8.35",
+        "solution":  "【分析】保留一位小数看百分位，四舍时可取最大，五入时可取最小.\n【详解】最大是8.44，最小是8.35。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·广东湛江·期中"
+    },
+    {
+        "id":  "307",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "6平方千米=()公顷;3000平方米=()公顷;200公顷=()平方千米=()平方米",
+        "options":  [
+
+                    ],
+        "answer":  "600;0.3;2;2000000",
+        "solution":  "【分析】1平方千米=100公顷,1公顷=10000平方米，按进率换算.\n【详解】6平方千米=600公顷;3000平方米=0.3公顷;200公顷=2平方千米=2000000平方米。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "公顷，平方千米的进率与换算",
+                                 "公顷，平方千米的认识"
+                             ],
+        "source":  "24-25四年级上·安徽阜阳·期中"
+    },
+    {
+        "id":  "308",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数，保留两位小数后是1.35，这个三位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "1.354;1.345",
+        "solution":  "【分析】保留两位小数看千分位。最大对应四舍(千分位4),最小对应五入(千分位5且原百分位4).\n【详解】最大是1.354，最小是1.345。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·河南焦作·期中"
+    },
+    {
+        "id":  "309",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "某两位小数，四舍五入后的近似值是2.4，原数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "2.44;2.35",
+        "solution":  "【分析】保留一位小数看百分位。四舍取最大，五入取最小.\n【详解】原数最大2.44，最小2.35。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·河北唐山·期中"
+    },
+    {
+        "id":  "310",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "300公顷=()平方米;250公顷=()平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "3000000;2.5",
+        "solution":  "【分析】1公顷=10000平方米,1平方千米=100公顷.\n【详解】300公顷=3000000平方米;250公顷=2.5平方千米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "运用小数点移动解决小数的单位换算问题"
+                             ],
+        "source":  "24-25四年级上·湖南永州·期中"
+    },
+    {
+        "id":  "311",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "用6,5,2,0这四个数字和小数点按要求组数，组成最大的三位小数是(),最小的三位小数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "6.520;0.256",
+        "solution":  "【分析】要最大：整数部分取最大6，小数部分按5,2,0排;要最小：整数部分取最小0，小数部分按2,5,6排.\n【详解】最大的三位小数是6.520，最小的是0.256。",
+        "knowledge_points":  [
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "24-25五年级上·河南周口·期中"
+    },
+    {
+        "id":  "312",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个数的十位上是6，百分位上是7，其他各位上都是0，这个数是(),读作()。",
+        "options":  [
+
+                    ],
+        "answer":  "60.07;六十点零七",
+        "solution":  "【分析】十位是6，百分位是7，其余位补0即可写数并读数.\n【详解】这个数是60.07，读作六十点零七。",
+        "knowledge_points":  [
+                                 "小数的数位和计数单位的认识",
+                                 "小数的读法和写法"
+                             ],
+        "source":  "24-25五年级上·河南周口·期中"
+    },
+    {
+        "id":  "313",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "250个0.001是(),化简后是(),1.5里有()个0.01。",
+        "options":  [
+
+                    ],
+        "answer":  "0.250;0.25;150",
+        "solution":  "【分析】250个0.001=0.250;小数末尾0可去掉;1.5=150个0.01.\n【详解】250个0.001是0.250，化简后是0.25,1.5里有150个0.01。",
+        "knowledge_points":  [
+                                 "小数的数位和计数单位的认识",
+                                 "小数的性质"
+                             ],
+        "source":  "24-25五年级上·河南周口·期中"
+    },
+    {
+        "id":  "314",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数，精确到百分位后约是5.62，这个三位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "5.624;5.615",
+        "solution":  "【分析】精确到百分位看千分位。四舍得最大，五入得最小.\n【详解】最大是5.624，最小是5.615。",
+        "knowledge_points":  [
+                                 "小数的近似数",
+                                 "小数的改写"
+                             ],
+        "source":  "24-25五年级上·河南周口·期中"
+    },
+    {
+        "id":  "315",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "439029万≈()亿,49.3亿=()万。",
+        "options":  [
+
+                    ],
+        "answer":  "44;493000",
+        "solution":  "【分析】万改亿要除以10000并按亿位取近似;亿改万要乘10000.\n【详解】439029万≈44亿;49.3亿=493000万。",
+        "knowledge_points":  [
+                                 "小数点向右移动引起小数大小变化的规律",
+                                 "小数的改写",
+                                 "亿以上数的近似数"
+                             ],
+        "source":  "24-25四年级上·福建龙岩·期中"
+    },
+    {
+        "id":  "316",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如果一个三位小数取近似值是0.35，那么这个三位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "0.354;0.345",
+        "solution":  "【分析】0.35是保留两位小数后的近似值。四舍得最大，五入得最小.\n【详解】最大0.354，最小0.345。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·广东·期中"
+    },
+    {
+        "id":  "317",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个三位小数四舍五入后是4.52，这个三位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "4.524;4.515",
+        "solution":  "【分析】保留两位小数为4.52，四舍时最大，五入时最小.\n【详解】最大是4.524，最小是4.515。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·广东·期中"
+    },
+    {
+        "id":  "318",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "23.47扩大到原数的()倍是2347;9.8缩小到原数的()是0.098。",
+        "options":  [
+
+                    ],
+        "answer":  "100;1÷100",
+        "solution":  "【分析】23.47到2347小数点右移2位;9.8到0.098小数点左移2位.\n【详解】分别是扩大100倍，缩小到1÷100。",
+        "knowledge_points":  [
+                                 "小数点向左移动引起小数大小变化的规律",
+                                 "小数点向右移动引起小数大小变化的规律"
+                             ],
+        "source":  "24-25五年级上·广东·期中"
+    },
+    {
+        "id":  "319",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个三位小数精确到十分位是4.0，这个三位小数最小是(),最大是()。",
+        "options":  [
+
+                    ],
+        "answer":  "3.950;4.049",
+        "solution":  "【分析】精确到十分位看百分位。最小来自五入到4.0，最大来自四舍到4.0.\n【详解】最小是3.950，最大是4.049。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·湖南·期中"
+    },
+    {
+        "id":  "320",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "9.963保留一位小数是(),精确到百分位是()。",
+        "options":  [
+
+                    ],
+        "answer":  "10.0;9.96",
+        "solution":  "【分析】保留一位小数看百分位;精确到百分位看千分位.\n【详解】9.963≈10.0,9.963≈9.96。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·湖南·期中"
+    },
+    {
+        "id":  "321",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\"\u003e\"、\"\u003c\"或\"=\"。\n3.606()3.060;763万()0.763亿;3米5厘米()35分米",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;\u003c;\u003c",
+        "solution":  "【分析】小数比较看位值;0.763亿=7630万;3米5厘米=30.5分米.\n【详解】3.606\u003e3.060;763万\u003c0.763亿;3米5厘米\u003c35分米。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "多位小数的大小比较",
+                                 "毫米，厘米，分米，米之间的进率与换算",
+                                 "亿以内数的改写"
+                             ],
+        "source":  "24-25五年级上·湖南·期中"
+    },
+    {
+        "id":  "322",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "由2个一,3个十分之一和5个千分之一组成的数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "2.305",
+        "solution":  "【分析】个位2，十分位3，千分位5，其余位补0.\n【详解】组成的数是2.305。",
+        "knowledge_points":  [
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25五年级上·湖南·期中"
+    },
+    {
+        "id":  "323",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上合适的小数.\n8角5分=()元;5米6厘米=()米",
+        "options":  [
+
+                    ],
+        "answer":  "0.85;5.06",
+        "solution":  "【分析】1元=10角=100分;1米=100厘米.\n【详解】8角=0.8元,5分=0.05元，合0.85元;6厘米=0.06米，所以5米6厘米=5.06米。",
+        "knowledge_points":  [
+                                 "元，角，分的认识及换算",
+                                 "厘米和米之间的进率与换算",
+                                 "运用小数点移动解决小数的单位换算问题"
+                             ],
+        "source":  "24-25五年级上·湖南·期中"
+    },
+    {
+        "id":  "324",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "800000m²=()公顷;50公顷=()平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "80;0.5",
+        "solution":  "【分析】1公顷=10000平方米,1平方千米=100公顷.\n【详解】800000m²=80公顷;50公顷=0.5平方千米。",
+        "knowledge_points":  [
+                                 "公顷，平方千米的进率与换算",
+                                 "面积单位间的进率及换算"
+                             ],
+        "source":  "24-25五年级上·湖南·期中"
+    },
+    {
+        "id":  "325",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个两位小数保留一位小数后约是6.3，这个小数最大可能是(),最小可能是()。",
+        "options":  [
+
+                    ],
+        "answer":  "6.34;6.25",
+        "solution":  "【分析】保留一位小数看百分位，四舍取最大，五入取最小.\n【详解】最大是6.34，最小是6.25。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·湖南·期中"
+    },
+    {
+        "id":  "326",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数，\"四舍五入\"保留两位小数约是3.40，这个三位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "3.404;3.395",
+        "solution":  "【分析】3.40是三位小数保留两位后的近似值。四舍得最大，五入得最小.\n【详解】最大是3.404，最小是3.395。",
+        "knowledge_points":  [
+                                 "小数的近似数",
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "24-25五年级上·山东·期中"
+    },
+    {
+        "id":  "327",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数四舍五入保留两位小数是6.87，这个三位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "6.874;6.865",
+        "solution":  "【分析】保留两位小数看千分位。四舍得到最大，五入得到最小.\n【详解】最大是6.874，最小是6.865。",
+        "knowledge_points":  [
+                                 "小数的近似数",
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "24-25五年级上·湖南·期中"
+    },
+    {
+        "id":  "328",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\"\u003e\"、\"\u003c\"或\"=\"。\n90万()899999;320个万+320个一()320320;707070()770070;700公顷()7平方千米",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;\u003e;\u003c;=",
+        "solution":  "【分析】先统一单位再比较大小.\n【详解】90万=900000\u003e899999;320个万+320个一=3200320\u003e320320;707070\u003c770070;700公顷=7平方千米。",
+        "knowledge_points":  [
+                                 "大数的比较",
+                                 "公顷，平方千米的进率与换算"
+                             ],
+        "source":  "24-25四年级上·山西长治·期中"
+    },
+    {
+        "id":  "329",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上适当的数。9平方米8平方分米=()平方米;8.06吨=()吨()千克",
+        "options":  [
+
+                    ],
+        "answer":  "9.08;8;60",
+        "solution":  "【分析】1平方米=100平方分米,1吨=1000千克.\n【详解】9平方米8平方分米=9+8÷100=9.08平方米;8.06吨=8吨+0.06吨=8吨60千克。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "平方厘米，平方分米，平方米之间的进率与换算",
+                                 "吨，千克之间的换算与比较"
+                             ],
+        "source":  "24-25五年级上·广西·期中"
+    },
+    {
+        "id":  "330",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数\"四舍五入\"后的近似数为5.70，这个小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "5.704;5.695",
+        "solution":  "【分析】保留两位小数看千分位。最大来自四舍，最小来自五入.\n【详解】最大5.704，最小5.695。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·广东·期中"
+    },
+    {
+        "id":  "331",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "马拉松全长是42千米195米，合()千米。",
+        "options":  [
+
+                    ],
+        "answer":  "42.195",
+        "solution":  "【分析】1千米=1000米,195米=0.195千米.\n【详解】42千米195米=42.195千米。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "千米和米之间的进率与换算"
+                             ],
+        "source":  "24-25五年级上·广西贵港·期中"
+    },
+    {
+        "id":  "332",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "奥运会指定项目滑板运动起源于海上的冲浪运动。航航喜欢滑板运动，他的滑板质量是一个三位小数，将这个三位小数保留两位小数得到的近似数是3.80。这个滑板的质量最重是()kg,最轻是()kg。",
+        "options":  [
+
+                    ],
+        "answer":  "3.804;3.795",
+        "solution":  "【分析】保留两位小数看千分位，四舍取最大，五入取最小.\n【详解】最重是3.804kg,最轻是3.795kg。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·福建莆田·阶段练习"
+    },
+    {
+        "id":  "333",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "2025年\"五一\"假期，国内旅游出游(295000000)人次。把括号内的数改写成用\"亿\"作单位的数是()亿，用\"四舍五入\"法保留整数是()亿。",
+        "options":  [
+
+                    ],
+        "answer":  "2.95;3",
+        "solution":  "【分析】改写成\"亿\"在亿位后点小数点;保留整数看十分位四舍五入.\n【详解】295000000=2.95亿,2.95亿≈3亿。",
+        "knowledge_points":  [
+                                 "亿以上数的改写",
+                                 "小数的近似数"
+                             ],
+        "source":  "25-26五年级上·福建莆田·阶段练习"
+    },
+    {
+        "id":  "334",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "天坛祈年殿从台明到宝顶全高为31.78米，合清代营造尺九丈九尺九寸。横线上的数读作(),改写成三位小数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "三十一点七八;31.780",
+        "solution":  "【分析】小数读法：整数部分按整数读，小数点读作\"点\"，小数部分依次读出各位数字。根据小数性质，在末尾添0大小不变，可改写为三位小数.\n【详解】31.78读作三十一点七八;改写成三位小数是31.780。",
+        "knowledge_points":  [
+                                 "小数的读法和写法",
+                                 "小数的性质"
+                             ],
+        "source":  "24-25四年级下·河南郑州·期末"
+    },
+    {
+        "id":  "335",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "宁波天一阁是中国现存最早的私家藏书楼，也是亚洲现有最古老的图书馆，世界最早的三大家族图书馆之一，其中的宝书楼面积约2.6万平方米，通高8.5米,8.5米是()米()分米。",
+        "options":  [
+
+                    ],
+        "answer":  "8;5",
+        "solution":  "【分析】1米=10分米,8.5米=8米+0.5米,0.5米=5分米.\n【详解】8.5米是8米5分米。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "毫米，厘米，分米，米之间的进率与换算"
+                             ],
+        "source":  "24-25四年级下·河南郑州·期末"
+    },
+    {
+        "id":  "336",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把9.9的小数点向左移动一位，这个数的计数单位从十分之一变为(),也就是缩小到原来的()(填分数)。",
+        "options":  [
+
+                    ],
+        "answer":  "百分之一;1÷10",
+        "solution":  "【分析】小数点向左移动一位，数值缩小到原来的1÷10;9.9变成0.99，最低位由十分位变成百分位.\n【详解】计数单位从十分之一变为百分之一，数缩小到原来的1÷10。",
+        "knowledge_points":  [
+                                 "小数的数位和计数单位的认识",
+                                 "小数点向左移动引起小数大小变化的规律"
+                             ],
+        "source":  "24-25四年级下·河北廊坊·期末"
+    },
+    {
+        "id":  "337",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "卢沟桥全长二百六十六点五米，宽(七点五)米，桥两侧雁翅桥面呈喇叭口状。括号内的数写作(),它是由()个1和()个0.1组成的。",
+        "options":  [
+
+                    ],
+        "answer":  "7.5;7;5",
+        "solution":  "【分析】\"七点五\"写作7.5;7.5由7个1和5个0.1组成.\n【详解】写作7.5，由7个1和5个0.1组成。",
+        "knowledge_points":  [
+                                 "小数的数位和计数单位的认识",
+                                 "小数的读法和写法"
+                             ],
+        "source":  "24-25四年级下·河南郑州·期末"
+    },
+    {
+        "id":  "338",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数四舍五入后是5.70，原来这个三位小数最大是()，最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "5.704;5.695",
+        "solution":  "【分析】保留两位小数看千分位。四舍时取最大为4;五入时取最小为5并由5.69进到5.70.\n【详解】最大是5.704，最小是5.695。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·河南新乡·期中"
+    },
+    {
+        "id":  "339",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "9.045保留一位小数是(),保留两位小数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "9.0;9.05",
+        "solution":  "【分析】保留一位小数看百分位;保留两位小数看千分位.\n【详解】9.045保留一位小数是9.0，保留两位小数是9.05。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·广东河源·期中"
+    },
+    {
+        "id":  "340",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个两位小数，\"四舍五入\"到十分位是7.8，那么这个两位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "7.84;7.75",
+        "solution":  "【分析】保留一位小数看百分位。四舍得到最大，五入得到最小.\n【详解】最大7.84，最小7.75。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·河北张家口·期中"
+    },
+    {
+        "id":  "341",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "4.3867保留一位小数约是(),保留两位小数约是(),保留整数约是()。",
+        "options":  [
+
+                    ],
+        "answer":  "4.4;4.39;4",
+        "solution":  "【分析】保留一位看百分位8，进一;保留两位看千分位6，进一;保留整数看十分位3，不进一.\n【详解】结果分别是4.4,4.39,4。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·广东韶关·期中"
+    },
+    {
+        "id":  "342",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "由4个十,7个十分之一和5个百分之一组成的小数是(),精确到十分位是()。",
+        "options":  [
+
+                    ],
+        "answer":  "40.75;40.8",
+        "solution":  "【分析】4个十是40,7个十分之一是0.7,5个百分之一是0.05，合起来40.75。精确到十分位看百分位5，向前进一.\n【详解】小数是40.75，精确到十分位是40.8。",
+        "knowledge_points":  [
+                                 "小数的近似数",
+                                 "小数的数位和计数单位的认识",
+                                 "小数的读法和写法"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "343",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "3.05千克=()千克()克;5平方米6平方厘米=()平方米",
+        "options":  [
+
+                    ],
+        "answer":  "3;50;5.0006",
+        "solution":  "【分析】1千克=1000克,0.05千克=50克;1平方米=10000平方厘米,6平方厘米=0.0006平方米.\n【详解】3.05千克=3千克50克;5平方米6平方厘米=5.0006平方米。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "平方厘米，平方分米，平方米之间的进率与换算",
+                                 "克，千克之间的换算与比较"
+                             ],
+        "source":  "24-25五年级上·河北邢台·期中"
+    },
+    {
+        "id":  "344",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "近似数是0.98的小数有()个，近似数是0.98的三位小数有()个。",
+        "options":  [
+
+                    ],
+        "answer":  "无数;10",
+        "solution":  "【分析】近似到百分位为0.98的小数可以有无限多位，所以无数个。三位小数则是有限枚举:0.975~0.979与0.980~0.984，共10个.\n【详解】近似数是0.98的小数有无数个，三位小数有10个。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·广东汕头·期中"
+    },
+    {
+        "id":  "345",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数，\"四舍五入\"后是5.74，这个小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "5.744;5.735",
+        "solution":  "【分析】保留两位小数为5.74。四舍时千分位最大4，得最大;五入时从5.73进到5.74，千分位最小5，得最小.\n【详解】最大5.744，最小5.735。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25四年级上·山东菏泽·期中"
+    },
+    {
+        "id":  "346",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "把9改写成以\"千分之一\"为计数单位的数是();把7.2改写成计数单位是0.01的数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "9.000;7.20",
+        "solution":  "【分析】千分之一计数单位对应三位小数;0.01计数单位对应两位小数.\n【详解】9写成9.000;7.2写成7.20。",
+        "knowledge_points":  [
+                                 "小数的性质",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级上·山东菏泽·期中"
+    },
+    {
+        "id":  "347",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "把14698400改写成用\"万\"作单位的数是(),保留整数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "1469.84万;1470万",
+        "solution":  "【分析】改写\"万\"在万位后点小数点;保留整数看十分位8，进一.\n【详解】14698400=1469.84万≈1470万。",
+        "knowledge_points":  [
+                                 "小数的改写",
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25四年级上·山东菏泽·期中"
+    },
+    {
+        "id":  "348",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "同学们参加赛跑，小敏每秒跑5.2米，小红跑48米要10秒，小花10秒跑50米，则()跑得最慢,()跑得最快。",
+        "options":  [
+
+                    ],
+        "answer":  "小红;小敏",
+        "solution":  "【分析】求速度：小红48÷10=4.8米/秒，小花50÷10=5米/秒，小敏5.2米/秒，比较4.8\u003c5\u003c5.2.\n【详解】小红最慢，小敏最快。",
+        "knowledge_points":  [
+                                 "一位小数的大小比较",
+                                 "基础行程问题",
+                                 "小数点向左移动引起小数大小变化的规律"
+                             ],
+        "source":  "24-25四年级上·山东菏泽·期中"
+    },
+    {
+        "id":  "349",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "整数部分是14个一，小数部分是1个百分之一,4个千分之一，组成的数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "14.014",
+        "solution":  "【分析】整数部分14个一是14，小数部分0.01和0.004相加是0.014.\n【详解】组成的数是14.014。",
+        "knowledge_points":  [
+                                 "小数的读法和写法",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级上·山东菏泽·期中"
+    },
+    {
+        "id":  "350",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上合适的数.\n5吨70千克=()吨;9平方米=()平方分米;3元5角7分=()元;5.045千米=()千米()米",
+        "options":  [
+
+                    ],
+        "answer":  "5.07;900;3.57;5;45",
+        "solution":  "【分析】1吨=1000千克,1平方米=100平方分米,1元=10角=100分,1千米=1000米.\n【详解】5吨70千克=5.07吨;9平方米=900平方分米;3元5角7分=3.57元;5.045千米=5千米45米。",
+        "knowledge_points":  [
+                                 "元，角，分的认识及换算",
+                                 "吨，千克之间的换算与比较",
+                                 "平方厘米，平方分米，平方米之间的进率与换算",
+                                 "运用小数点移动解决小数的单位换算问题"
+                             ],
+        "source":  "24-25四年级上·山东菏泽·期中"
+    },
+    {
+        "id":  "351",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把2.57扩大到它原来的1000倍，只要把小数点向()移动()位。",
+        "options":  [
+
+                    ],
+        "answer":  "右;三",
+        "solution":  "【分析】扩大1000倍即小数点向右移动三位.\n【详解】向右移动三位。",
+        "knowledge_points":  [
+                                 "小数点向右移动引起小数大小变化的规律"
+                             ],
+        "source":  "24-25四年级上·山东菏泽·期中"
+    },
+    {
+        "id":  "352",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数\"四舍五入\"保留两位小数约是8.54，这个三位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "8.544;8.535",
+        "solution":  "【分析】保留两位小数看千分位，四舍取最大，五入取最小.\n【详解】最大是8.544，最小是8.535。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·新疆喀什·期中"
+    },
+    {
+        "id":  "353",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "10÷11=0.9090...，保留两位小数是(),保留三位小数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "0.91;0.909",
+        "solution":  "【分析】保留两位看千分位9进一;保留三位看万分位0舍去.\n【详解】0.9090...≈0.91(两位),≈0.909(三位)。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·河北石家庄·期中"
+    },
+    {
+        "id":  "354",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数保留两位小数后得到11.60，这个三位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "11.604;11.595",
+        "solution":  "【分析】11.60是保留两位后的近似值，四舍得最大，五入得最小.\n【详解】最大11.604，最小11.595。",
+        "knowledge_points":  [
+                                 "小数的近似数",
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "25-26五年级上·广东汕头·阶段练习"
+    },
+    {
+        "id":  "355",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数保留两位小数后是6.60，这个三位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "6.604;6.595",
+        "solution":  "【分析】保留两位小数看千分位。四舍对应最大，五入对应最小.\n【详解】最大6.604，最小6.595。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·河南南阳·期中"
+    },
+    {
+        "id":  "356",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数\"四舍五入\"到百分位约是3.58，这个三位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "3.584;3.575",
+        "solution":  "【分析】约到百分位看千分位。四舍时最大，五入时最小.\n【详解】最大3.584，最小3.575。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·河南安阳·期中"
+    },
+    {
+        "id":  "357",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数保留两位小数的近似数是7.20，这个三位小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "7.204;7.195",
+        "solution":  "【分析】保留两位小数为7.20，四舍得最大，五入得最小.\n【详解】最大7.204，最小7.195。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·天津西青·期中"
+    },
+    {
+        "id":  "358",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个两位小数的近似值是10.0，这个小数在取近似值以前最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "10.04;9.95",
+        "solution":  "【分析】精确到十分位为10.0，范围是9.95到10.04.\n【详解】最大10.04，最小9.95。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·山东临沂·期中"
+    },
+    {
+        "id":  "359",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个三位小数\"四舍五入\"后的近似数为5.70，这个小数最大是()。",
+        "options":  [
+
+                    ],
+        "answer":  "5.704",
+        "solution":  "【分析】求最大值应取\"四舍\"情形，千分位取可舍去的最大值4.\n【详解】最大是5.704。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·江西赣州·期中"
+    },
+    {
+        "id":  "360",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "一个数扩大到原来的100倍，小数点向()移动()位;一个数缩小到原来的1÷100，小数点向()移动()位。",
+        "options":  [
+
+                    ],
+        "answer":  "右;两;左;两",
+        "solution":  "【分析】扩大100倍小数点右移两位;缩小到1÷100小数点左移两位.\n【详解】应填\"右，两;左，两\"。",
+        "knowledge_points":  [
+                                 "小数点向左移动引起小数大小变化的规律",
+                                 "小数点向右移动引起小数大小变化的规律"
+                             ],
+        "source":  "25-26五年级上·河北邢台·阶段练习"
+    },
+    {
+        "id":  "361",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个两位小数，四舍五入后是0.8，这个小数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "0.84;0.75",
+        "solution":  "【分析】保留一位小数为0.8，四舍得最大0.84，五入得最小0.75.\n【详解】最大0.84，最小0.75。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·浙江宁波·期中"
+    },
+    {
+        "id":  "362",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个数由6个一,5个十分之一，和8个百分之一组成，这个数写作(),读作(),它也可以看成是由()个0.01组成的。",
+        "options":  [
+
+                    ],
+        "answer":  "6.58;六点五八;658",
+        "solution":  "【分析】6个一是6,5个十分之一是0.5,8个百分之一是0.08，合成6.58;6.58÷0.01=658.\n【详解】写作6.58，读作六点五八，可看成658个0.01。",
+        "knowledge_points":  [
+                                 "小数的数位和计数单位的认识",
+                                 "小数的读法和写法"
+                             ],
+        "source":  "24-25五年级上·江苏淮安·期中"
+    },
+    {
+        "id":  "363",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "下面的□里可以填哪些数字。\n(1)46□391≈46万,□里可以填().\n(2)260□348731≈26.1亿,□里最小应该填().\n(3)95.□54亿≈95亿,□里最大应该填()。",
+        "options":  [
+
+                    ],
+        "answer":  "0,1,2,3,4;5;4",
+        "solution":  "【分析】按四舍五入分别看对应保留位的下一位数字.\n【详解】(1)千位需\u003c5，可填0~4;(2)百万位需\u003e=5，最小填5;(3)千万位需\u003c5，最大填4。",
+        "knowledge_points":  [
+                                 "小数的近似数",
+                                 "亿以内数的近似数",
+                                 "亿以上数的近似数"
+                             ],
+        "source":  "25-26四年级上·贵州黔南·阶段练习"
+    },
+    {
+        "id":  "364",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "九千七百四十万九千五百这个数写作(),改写成以\"亿\"作单位保留一位小数约是()亿。",
+        "options":  [
+
+                    ],
+        "answer":  "97409500;1.0",
+        "solution":  "【分析】按数位写出整数;改写成亿并保留一位小数按四舍五入处理.\n【详解】写作97409500，改写为0.974095亿，保留一位小数约1.0亿。",
+        "knowledge_points":  [
+                                 "亿以内数的读，写法",
+                                 "小数的近似数",
+                                 "亿以上数的改写"
+                             ],
+        "source":  "25-26四年级上·贵州黔南·阶段练习"
+    },
+    {
+        "id":  "365",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "2024年年末我国总人口数约为1408280000人，比2023年减少了139万人。1408280000读作()。把1408280000改写成用\"亿\"作单位的数为()亿,1408280000四舍五入到亿位的近似数记作()亿。",
+        "options":  [
+
+                    ],
+        "answer":  "十四亿零八百二十八万;14.0828;14",
+        "solution":  "【分析】大数分级读写;改写成亿是小数点左移8位;四舍五入到亿位看千万位.\n【详解】1408280000读作十四亿零八百二十八万;改写为14.0828亿;四舍五入到亿位是14亿。",
+        "knowledge_points":  [
+                                 "亿以上数的读，写法",
+                                 "小数的改写",
+                                 "亿以上数的改写",
+                                 "亿以上数的近似数"
+                             ],
+        "source":  "25-26四年级上·贵州黔南·阶段练习"
+    },
+    {
+        "id":  "366",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里面填上\"\u003e\"、\"\u003c\"或\"=\"。\n2.5小时()2小时30分钟;3652千克()3.6吨;38.72平方分米()3.872平方米;0.76千克()760克;8千米()8000米;2公顷()2万平方米",
+        "options":  [
+
+                    ],
+        "answer":  "=;\u003e;\u003c;=;=;=",
+        "solution":  "【分析】统一单位后比较:2.5小时=2小时30分;3652千克=3.652吨;3.872平方米=387.2平方分米;0.76千克=760克;8千米=8000米;2公顷=20000平方米.\n【详解】依次填\"=,\u003e,\u003c,=,=,=\"。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "千米和米之间的进率与换算",
+                                 "时，分的认识",
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "24-25四年级上·山东德州·期中"
+    },
+    {
+        "id":  "367",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "1亿粒大米约重2500千克，合()吨;如果每人每天吃500克大米,1亿粒大米大约够()人吃一天。",
+        "options":  [
+
+                    ],
+        "answer":  "2.5;5000",
+        "solution":  "【分析】2500千克换吨：除以1000。再换成克后除以每人500克求人数.\n【详解】2500千克=2.5吨;2500千克=2500000克,2500000÷500=5000人。",
+        "knowledge_points":  [
+                                 "一亿有多大",
+                                 "克，千克之间的换算与比较",
+                                 "吨，千克之间的换算与比较",
+                                 "运用小数点移动解决小数的单位换算问题"
+                             ],
+        "source":  "25-26四年级上·甘肃天水·阶段练习"
+    },
+    {
+        "id":  "368",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "7005000=()万;3040000000=()亿",
+        "options":  [
+
+                    ],
+        "answer":  "700.5;30.4",
+        "solution":  "【分析】改写\"万\"在万位后点小数点;改写\"亿\"在亿位后点小数点.\n【详解】7005000=700.5万;3040000000=30.4亿。",
+        "knowledge_points":  [
+                                 "小数的改写"
+                             ],
+        "source":  "25-26四年级上·甘肃天水·阶段练习"
+    },
+    {
+        "id":  "369",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "晋中，东依太行山，西临汾河水，全市总面积约1.64万平方千米。1.64读作(),它是由()个一,()个十分之一和()个百分之一组成的，如果把这个数表示在下面的直线上，它在点()的位置.\n![](images/ef95f3297f0743cf5a5a9f60b89a5a934a8da1aaeda743305d8653a6a447fb21.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "一点六四;1;6;4;B",
+        "solution":  "【分析】1.64读作一点六四。其数位含义为1个一,6个十分之一,4个百分之一。结合数轴位置判断在B点.\n【详解】填：一点六四;1;6;4;B。",
+        "knowledge_points":  [
+                                 "小数的数位和计数单位的认识",
+                                 "多位小数的大小比较",
+                                 "小数的读法和写法"
+                             ],
+        "source":  "24-25四年级下·山西晋中·期末"
+    },
+    {
+        "id":  "370",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "2.15厘米=()米;2时45分=()分;5千米800米=()千米",
+        "options":  [
+
+                    ],
+        "answer":  "0.0215;165;5.8",
+        "solution":  "【分析】因为1米=100厘米，厘米换算为米要除以进率。因为1时=60分，时换算为分要乘进率。因为1千米=1000米，米换算为千米要除以进率.\n【详解】1米=100厘米,2.15÷100=0.0215(米),2.15厘米=0.0215米;1时=60分,2×60=120(分),120+45=165(分),2时45分=165分;1千米=1000米,800÷1000=0.8(千米),5+0.8=5.8(千米),5千米800米=5.8千米。",
+        "knowledge_points":  [
+                                 "时，分，秒之间的换算与比较",
+                                 "毫米，厘米，分米，米之间的进率与换算",
+                                 "千米和米之间的进率与换算",
+                                 "运用小数点移动解决小数的单位换算问题"
+                             ],
+        "source":  "24-25五年级上·江西宜春·期中"
+    },
+    {
+        "id":  "371",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "5G是第五代移动通信技术,5G网络理论下载速度可以达到每秒1300000KB以上，用科学记数法表示1300000是()。",
+        "options":  [
+
+                    ],
+        "answer":  "1.3×106",
+        "solution":  "【分析】根据科学记数法的定义，把一个数表示成a×10n的形式，其中1\u003c=a\u003c10,n为整数。要将1300000用科学记数法表示，原数1300000变为1.3时小数点向左移动了6位，所以n=6.\n【详解】由分析可知，用科学记数法表示1300000是1.3×106。",
+        "knowledge_points":  [
+                                 "十进制计数法",
+                                 "小数点向右移动引起小数大小变化的规律",
+                                 "小数的改写"
+                             ],
+        "source":  "24-25六年级上·山东济宁·期中"
+    },
+    {
+        "id":  "372",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数的近似数是8.00，这个三位小数最小是(),最大是()。",
+        "options":  [
+
+                    ],
+        "answer":  "7.995;8.004",
+        "solution":  "【分析】如果通过\"四舍\"得到8.00，则原三位小数可在8.000~8.004之间，最大是8.004;如果通过\"五入\"得到8.00，则原三位小数可在7.995~7.999之间，最小是7.995.\n【详解】通过\"四舍\"得到近似数8.00，千分位小于5，最大的三位小数是8.004;通过\"五入\"得到近似数8.00，千分位大于等于5，最小的三位小数是7.995。所以最小是7.995，最大是8.004。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·江西宜春·期中"
+    },
+    {
+        "id":  "373",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.94,
+        "question":  "2,2.8,2.89,2.895中，最大的数是(),最小的数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "2.895;2",
+        "solution":  "【分析】四个数的整数部分都为2，继续比较小数部分:2.0\u003c2.8\u003c2.89\u003c2.895.\n【详解】比较可得：最大的数是2.895，最小的数是2。",
+        "knowledge_points":  [
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "24-25五年级上·山东济宁·期中"
+    },
+    {
+        "id":  "374",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "4.44平方米=()平方厘米;5.06吨=()吨()千克;3平方米600平方厘米=()平方米;36.06平方米=()平方米()平方分米",
+        "options":  [
+
+                    ],
+        "answer":  "44400;5;60;3.06;36;6",
+        "solution":  "【分析】根据1平方米=100平方分米=10000平方厘米,1吨=1000千克进行换算：高级单位到低级单位乘进率，低级单位到高级单位除以进率.\n【详解】4.44×10000=44400，所以4.44平方米=44400平方厘米;0.06吨=60千克，所以5.06吨=5吨60千克;600÷10000=0.06，所以3平方米600平方厘米=3.06平方米;0.06平方米=6平方分米，所以36.06平方米=36平方米6平方分米。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "平方厘米，平方分米，平方米之间的进率与换算",
+                                 "吨，千克之间的换算与比较"
+                             ],
+        "source":  "24-25四年级下·河南驻马店·期末"
+    },
+    {
+        "id":  "375",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "5吨70千克=()吨;2.31米=()厘米",
+        "options":  [
+
+                    ],
+        "answer":  "5.07;231",
+        "solution":  "【分析】依据1吨=1000千克,1米=100厘米。低级单位化高级单位除以进率，高级单位化低级单位乘以进率.\n【详解】70千克=70÷1000=0.07吨，所以5吨70千克=5.07吨;2.31米=2.31×100=231厘米。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "厘米和米之间的进率与换算",
+                                 "吨，千克之间的换算与比较"
+                             ],
+        "source":  "24-25四年级下·广东汕头·期末"
+    },
+    {
+        "id":  "376",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "6公顷=()平方米;5.08吨=()吨()千克",
+        "options":  [
+
+                    ],
+        "answer":  "60000;5;80",
+        "solution":  "【分析】1公顷=10000平方米，公顷化平方米乘10000;1吨=1000千克，吨的小数部分化千克乘1000.\n【详解】6×10000=60000，所以6公顷=60000平方米;0.08×1000=80，所以5.08吨=5吨80千克。",
+        "knowledge_points":  [
+                                 "吨，千克之间的换算与比较",
+                                 "面积单位间的进率及换算",
+                                 "运用小数点移动解决小数的单位换算问题"
+                             ],
+        "source":  "2025·湖北武汉·小升初真题"
+    },
+    {
+        "id":  "377",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "2024年某国人口普查数据显示，全国总人口约为十四亿三千九百六十八万二千五百人，这个数写作();若将其改写成用\"亿\"作单位的近似数(保留两位小数),约是()亿。",
+        "options":  [
+
+                    ],
+        "answer":  "1439682500;14.40",
+        "solution":  "【分析】整数写法按数位顺序写出;改写成\"亿\"作单位时在亿位后点小数点;保留两位小数看千分位并四舍五入.\n【详解】十四亿三千九百六十八万二千五百写作1439682500;1439682500=14.396825亿≈14.40亿。",
+        "knowledge_points":  [
+                                 "亿以上数的读，写法",
+                                 "小数的近似数",
+                                 "亿以上数的改写"
+                             ],
+        "source":  "24-25六年级下·广东东莞·期末"
+    },
+    {
+        "id":  "378",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "\"沉睡数千年，一醒惊天下\"，中央广播电视总台《三星堆新发现》直播特别节目播出后，三星堆相关话题全网阅读量约(30089600000)次。括号内的数读作:(),改写成用\"亿\"作单位是()亿，精确到百分位是()亿。",
+        "options":  [
+
+                    ],
+        "answer":  "三百亿八千九百六十万;300.896;300.90",
+        "solution":  "【分析】先按亿级，万级分级读数;改写成\"亿\"作单位在亿位后点小数点;精确到百分位即保留两位小数，看千分位四舍五入.\n【详解】30089600000读作三百亿八千九百六十万;30089600000=300.896亿;保留两位小数得300.90亿。",
+        "knowledge_points":  [
+                                 "亿以上数的读，写法",
+                                 "小数的改写",
+                                 "小数的近似数"
+                             ],
+        "source":  "2025·四川内江·小升初真题"
+    },
+    {
+        "id":  "379",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "7.5m²=()dm²;720cm²=()dm²",
+        "options":  [
+
+                    ],
+        "answer":  "750;7.2",
+        "solution":  "【分析】1m²=100dm²,1dm²=100cm²。高级单位到低级单位乘进率，低级单位到高级单位除进率.\n【详解】7.5×100=750，所以7.5m²=750dm²;720÷100=7.2，所以720cm²=7.2dm²。",
+        "knowledge_points":  [
+                                 "平方厘米，平方分米，平方米之间的进率与换算",
+                                 "运用小数点移动解决小数的单位换算问题"
+                             ],
+        "source":  "24-25五年级上·甘肃兰州·期中"
+    },
+    {
+        "id":  "380",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把一个三位小数精确到百分位是5.20，这个小数最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "5.195",
+        "solution":  "【分析】精确到百分位得到5.20，则原三位小数范围是5.195~5.204，最小取下界.\n【详解】这个三位小数最小是5.195。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "2025·四川南充·小升初真题"
+    },
+    {
+        "id":  "381",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个四位小数四舍五入到百分位后是7.20，原数最大是(),最小是()。",
+        "options":  [
+
+                    ],
+        "answer":  "7.2049;7.1950",
+        "solution":  "【分析】保留两位小数：最大值来自\"四舍\"边界7.2049;最小值来自\"五入\"边界7.1950.\n【详解】原数最大是7.2049，最小是7.1950。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "2025·江西抚州·小升初真题"
+    },
+    {
+        "id":  "382",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "2025年\"五一\"假期期间，新疆自治区累计接待游客919.55万人次，游客消费金额(8924000000)元，括号内的数读作(),改写成用亿作单位的数，并保留一位小数是()亿。",
+        "options":  [
+
+                    ],
+        "answer":  "八十九亿二千四百万;89.2",
+        "solution":  "【分析】先分级读数;改写为\"亿\"作单位时在亿位后点小数点;保留一位小数看百分位四舍五入.\n【详解】8924000000读作八十九亿二千四百万;改写后约为89.2亿。",
+        "knowledge_points":  [
+                                 "亿以上数的读，写法",
+                                 "小数的改写",
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25六年级下·甘肃临夏·期末"
+    },
+    {
+        "id":  "383",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个两位小数用四舍五入法精确到十分位是10.0，这个小数最小是(),最大是()。",
+        "options":  [
+
+                    ],
+        "answer":  "9.95;10.04",
+        "solution":  "【分析】精确到十分位为10.0的两位小数范围是9.95~10.04.\n【详解】最小是9.95，最大是10.04。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25六年级下·重庆巴南·期末"
+    },
+    {
+        "id":  "384",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "巴南区2023~2024学年全区各级各类在校生(299567)人，括号内的数读作(),改写成用\"万\"作单位并保留一位小数是()万。",
+        "options":  [
+
+                    ],
+        "answer":  "二十九万九千五百六十七;30.0",
+        "solution":  "【分析】先按万级，个级读数;改写成\"万\"作单位在万位后点小数点;保留一位小数看百分位四舍五入.\n【详解】299567读作二十九万九千五百六十七;299567=29.9567万≈30.0万。",
+        "knowledge_points":  [
+                                 "亿以内数的读，写法",
+                                 "小数的改写"
+                             ],
+        "source":  "24-25六年级下·重庆巴南·期末"
+    },
+    {
+        "id":  "385",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "0.15平方米=()平方分米;3.05千克=()千克()克;70公顷=()平方千米;16厘米=()分米",
+        "options":  [
+
+                    ],
+        "answer":  "15;3;50;0.7;1.6",
+        "solution":  "【分析】用进率换算:1平方米=100平方分米,1千克=1000克,1平方千米=100公顷,1分米=10厘米.\n【详解】0.15×100=15;3.05千克=3千克+0.05千克=3千克50克;70÷100=0.7;16÷10=1.6。",
+        "knowledge_points":  [
+                                 "克，千克之间的换算与比较",
+                                 "平方厘米，平方分米，平方米之间的进率与换算",
+                                 "公顷，平方千米的进率与换算",
+                                 "运用小数点移动解决小数的单位换算问题"
+                             ],
+        "source":  "24-25四年级下·河北廊坊·期末"
+    },
+    {
+        "id":  "386",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "宫面是藁城特产之一，属河北省传统名产名优食品，亦称贡面。李叔叔吃一碗宫面花了13元7角，用小数表示是()元，读作()元。",
+        "options":  [
+
+                    ],
+        "answer":  "13.7;十三点七",
+        "solution":  "【分析】1元=10角,7角=0.7元;小数读法为整数部分按整数读，小数点读\"点\"，小数部分依次读数字.\n【详解】13元7角=13.7元，读作十三点七元。",
+        "knowledge_points":  [
+                                 "小数的读法和写法",
+                                 "运用小数点移动解决小数的单位换算问题"
+                             ],
+        "source":  "24-25四年级下·河北廊坊·期末"
+    },
+    {
+        "id":  "387",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "近年来，国内新能源汽车产业蓬勃发展，某品牌新能源汽车5月交付新车10321台，把横线上的数改写成用\"万\"作单位的数是()万台，保留一位小数约是()万台。",
+        "options":  [
+
+                    ],
+        "answer":  "1.0321;1.0",
+        "solution":  "【分析】10321改写成\"万\"为1.0321万;保留一位小数看百分位,3\u003c5，舍去得到1.0万.\n【详解】10321=1.0321万≈1.0万。",
+        "knowledge_points":  [
+                                 "小数的改写",
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25四年级下·河北廊坊·期末"
+    },
+    {
+        "id":  "388",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "我们的伟大祖国幅员辽阔，国土面积为九百五十九万六千九百六十平方千米，在各国领土面积排行榜中排名第三;最高峰珠穆朗玛峰的海拔高度是8848.86米，是世界第一高峰。\"九百五十九万六千九百六十\"写作(),改写成用\"万\"作单位的数是()万，省略万位后面的尾数后约是()万;\"8848.86\"读作()。",
+        "options":  [
+
+                    ],
+        "answer":  "9596960;959.696;960;八千八百四十八点八六",
+        "solution":  "【分析】整数按数位写出;改写成\"万\"作单位在万位后点小数点;省略万位后尾数即四舍五入到万位;小数读法按\"点\"后逐位读.\n【详解】九百五十九万六千九百六十写作9596960;9596960=959.696万≈960万;8848.86读作八千八百四十八点八六。",
+        "knowledge_points":  [
+                                 "亿以内数的读，写法",
+                                 "小数的读法和写法",
+                                 "小数的近似数",
+                                 "亿以内数的改写"
+                             ],
+        "source":  "2025·河南信阳·小升初真题"
+    },
+    {
+        "id":  "389",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "要使1□0.□7小于110且最大，这个数是();要使□0.□7最接近70，这个数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "100.97;70.07",
+        "solution":  "【分析】比较小数大小先比整数部分。1□0.□7要小于110且最大，百位后空格取0，十分位取9;□0.□7最接近70，则十位取7，十分位取0.\n【详解】满足条件的两个数分别是100.97和70.07。",
+        "knowledge_points":  [
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "24-25四年级下·浙江宁波·期末"
+    },
+    {
+        "id":  "390",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "制作100千克的牡丹花酱需要25千克的牡丹花瓣，那么制作1千克的牡丹花酱需要牡丹花瓣()千克，制作1吨的牡丹花酱需要牡丹花瓣()千克。",
+        "options":  [
+
+                    ],
+        "answer":  "0.25;250",
+        "solution":  "【分析】先求每千克花酱需要花瓣:25÷100=0.25千克;1吨=1000千克，再算1000千克花酱所需花瓣:0.25×1000.\n【详解】制作1千克牡丹花酱需要0.25千克花瓣;制作1吨牡丹花酱需要250千克花瓣。",
+        "knowledge_points":  [
+                                 "吨，千克之间的换算与比较",
+                                 "小数点向右移动引起小数大小变化的规律",
+                                 "小数点向左移动引起小数大小变化的规律"
+                             ],
+        "source":  "24-25四年级下·浙江宁波·期末"
+    },
+    {
+        "id":  "391",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上合适的数。20千克89克=()千克;1.38千米=()米",
+        "options":  [
+
+                    ],
+        "answer":  "20.089;1380",
+        "solution":  "【分析】1千克=1000克,89克=0.089千克;1千米=1000米，千米化米乘1000.\n【详解】20千克89克=20.089千克;1.38千米=1380米。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题"
+                             ],
+        "source":  "24-25四年级下·浙江宁波·期末"
+    },
+    {
+        "id":  "392",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "10.6里面有()个十分之一;0.75里面有()个0.01。",
+        "options":  [
+
+                    ],
+        "answer":  "106;75",
+        "solution":  "【分析】10.6=106×0.1;0.75=75×0.01.\n【详解】10.6里面有106个十分之一;0.75里面有75个0.01。",
+        "knowledge_points":  [
+                                 "小数的意义",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·浙江宁波·期末"
+    },
+    {
+        "id":  "393",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "太平洋是世界上最大，最深的大洋，它的面积约是(181344000)平方千米。将括号内的数改写成用\"亿\"作单位的数是()亿，保留两位小数约是()亿。",
+        "options":  [
+
+                    ],
+        "answer":  "1.81344;1.81",
+        "solution":  "【分析】改写成\"亿\"作单位时在亿位后点小数点;保留两位小数看千分位,3\u003c5，舍去.\n【详解】181344000=1.81344亿，保留两位小数约是1.81亿。",
+        "knowledge_points":  [
+                                 "小数的改写",
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25四年级下·浙江宁波·期末"
+    },
+    {
+        "id":  "394",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "()的小数点先向右移动两位，再向左移动三位是14.32。",
+        "options":  [
+
+                    ],
+        "answer":  "143.2",
+        "solution":  "【分析】先右移两位再左移三位，净效果是左移一位。反推原数可将14.32先右移三位再左移两位.\n【详解】14.32→14320(右移三位)→143.2(左移两位),所以原数是143.2。",
+        "knowledge_points":  [
+                                 "小数点向右移动引起小数大小变化的规律",
+                                 "小数点向左移动引起小数大小变化的规律"
+                             ],
+        "source":  "24-25四年级下·河南驻马店·期末"
+    },
+    {
+        "id":  "395",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "8t70kg=()t;8500m²=()公顷;5.06km=()km()m;4.5m=()cm",
+        "options":  [
+
+                    ],
+        "answer":  "8.07;0.85;5;60;450",
+        "solution":  "【分析】1t=1000kg,1公顷=10000m²,1km=1000m,1m=100cm。按进率进行单位换算.\n【详解】70kg=0.07t,所以8t70kg=8.07t;8500m²=0.85公顷;5.06km=5km60m;4.5m=450cm。",
+        "knowledge_points":  [
+                                 "吨，千克之间的换算与比较",
+                                 "千米和米之间的进率与换算",
+                                 "面积单位间的进率及换算",
+                                 "根据小数的意义进行单位换算"
+                             ],
+        "source":  "24-25四年级下·河南驻马店·期末"
+    },
+    {
+        "id":  "396",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "不改变3.6的大小，把它改写成小数部分是三位的小数是()。",
+        "options":  [
+
+                    ],
+        "answer":  "3.600",
+        "solution":  "【分析】小数性质：末尾添0或去0，小数大小不变.\n【详解】3.6改写成小数部分三位为3.600。",
+        "knowledge_points":  [
+                                 "小数的性质"
+                             ],
+        "source":  "24-25四年级下·河南驻马店·期末"
+    },
+    {
+        "id":  "397",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "5.4里面有()个0.01,358个0.001是()。",
+        "options":  [
+
+                    ],
+        "answer":  "540;0.358",
+        "solution":  "【分析】5.4=540×0.01;358个0.001就是0.358.\n【详解】5.4里面有540个0.01,358个0.001是0.358。",
+        "knowledge_points":  [
+                                 "小数的意义"
+                             ],
+        "source":  "24-25四年级下·河南驻马店·期末"
+    },
+    {
+        "id":  "398",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "1.54km=()m;3200g=()kg;26.54m=()m()cm",
+        "options":  [
+
+                    ],
+        "answer":  "1540;3.2;26;54",
+        "solution":  "【分析】1km=1000m,1kg=1000g,1m=100cm;高级到低级乘进率，低级到高级除进率.\n【详解】1.54×1000=1540，所以1.54km=1540m;3200÷1000=3.2，所以3200g=3.2kg;26.54m=26m+0.54m,0.54m=54cm,所以26.54m=26m54cm。",
+        "knowledge_points":  [
+                                 "厘米和米之间的进率与换算",
+                                 "千米和米之间的进率与换算",
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "克，千克之间的换算与比较"
+                             ],
+        "source":  "24-25四年级下·河北秦皇岛·期末"
+    },
+    {
+        "id":  "399",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用小数表示下面图形的涂色部分.\n![](images/b8ab9f9d49476db057b92ec50bd0bcb6558f0c3cdccd9db054339fbd576dc0fa.jpg)()\n![](images/2026-03-08%20221739.jpg)()",
+        "options":  [
+
+                    ],
+        "answer":  "0.4;2.09",
+        "solution":  "【分析】按小数意义：十分之一对应0.1，百分之一对应0.01，分别统计两图中涂色份数.\n【详解】第一组图：左图涂2个0.1为0.2，右图涂20个0.01为0.2，合计0.4;第二组图：整涂2个正方形为2，另涂9个0.01为0.09，合计2.09。",
+        "knowledge_points":  [
+                                 "小数的意义"
+                             ],
+        "source":  "24-25四年级下·河北秦皇岛·期末"
+    },
+    {
+        "id":  "400",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "去掉0.329的小数点，这个数就扩大到原数的()倍;把1.2缩小到它的()是0.12。",
+        "options":  [
+
+                    ],
+        "answer":  "1000;1÷10",
+        "solution":  "【分析】去掉0.329小数点相当于小数点右移3位，数扩大1000倍;1.2变0.12相当于小数点左移1位，数缩小到1÷10.\n【详解】0.329→329，扩大1000倍;1.2→0.12，缩小到1÷10。",
+        "knowledge_points":  [
+                                 "小数点向左移动引起小数大小变化的规律",
+                                 "小数点向右移动引起小数大小变化的规律"
+                             ],
+        "source":  "24-25四年级下·陕西安康·期末"
+    },
+    {
+        "id":  "401",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "西安市是联合国教科文组织于1981年确定的\"世界历史名城\"，是中华文明和中华民族重要发祥地之一，古丝绸之路的起点，西安市城区面积约为(0.083)万平方千米，括号内的数是由( )个0.01和( )个0.001组成的。",
+        "options":  [
+
+                    ],
+        "answer":  "8;3",
+        "solution":  "【分析】根据小数的数位意义，在数位上的数字是几，就表示有几个对应的计数单位，据此作答.\n【详解】0.083的百分位是8，表示8个0.01;千分位是3，表示3个0.001。因此,0.083由8个0.01和3个0.001组成。",
+        "knowledge_points":  [
+                                 "小数的意义",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·陕西安康·期末"
+    },
+    {
+        "id":  "402",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "11.0152精确到百分位约是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "11.02",
+        "solution":  "【分析】根据题意，百分位是小数点后第二位数字。要精确到百分位，需要看第三位小数(千分位)进行四舍五入。11.0152的千分位上的数是5，大于等于5，就向百分位进一，以此答题即可.\n【详解】根据分析可知:11.0152≈11.02。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25四年级下·河南驻马店·期末"
+    },
+    {
+        "id":  "403",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "先把0.618扩大为原来的100000倍，再缩小到1÷100是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "618",
+        "solution":  "【分析】根据小数点移动引起小数大小变化的规律，把0.618扩大为原来的100000倍，小数点向右移动五位，得到61800，再缩小到1÷100，小数点向左移动两位，得到618。据此解答.\n【详解】由分析可知：先把0.618扩大为原来的100000倍，再缩小到1÷100是618。",
+        "knowledge_points":  [
+                                 "小数点向右移动引起小数大小变化的规律",
+                                 "小数点向左移动引起小数大小变化的规律"
+                             ],
+        "source":  "24-25四年级下·河南驻马店·期末"
+    },
+    {
+        "id":  "404",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "二十亿九千八百二十万写作( ),改写成以亿为单位的数是( ),保留一位小数约是( )亿。",
+        "options":  [
+
+                    ],
+        "answer":  "2098200000;20.982亿;21.0",
+        "solution":  "【分析】根据大数的写法，从高位写起：按照数位顺序，从高位到低位，一级一级地写。哪个数位上一个单位也没有，就在那个数位上写0占位。改写成用\"亿\"作单位的数，就是在亿位数的右下角点上小数点，然后把小数末尾的0去掉，在数的后面带上\"亿\"字;改成用\"亿\"作单位的数后，保留一位小数就是精确到十分位，要把百分位上的数进行四舍五入;据此解答.\n【详解】根据分析可知：二十亿九千八百二十万写作2098200000;2098200000=20.982亿;20.982亿的百分位上是8,8\u003e5，向十分位进1，十分位9+1=10，向个位进一，十分位为0，所以20.982亿≈21.0亿。",
+        "knowledge_points":  [
+                                 "亿以上数的读，写法",
+                                 "小数的近似数",
+                                 "亿以上数的改写"
+                             ],
+        "source":  "24-25四年级下·河南驻马店·期末"
+    },
+    {
+        "id":  "405",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "13厘米=( )分米;5350米=( )千米;2.3千克=( )克;2.95平方米=( )平方米( )平方分米",
+        "options":  [
+
+                    ],
+        "answer":  "1.3;5.35;2300;2;95",
+        "solution":  "【分析】根据1分米=10厘米,1千米=1000米,1千克=1000克,1平方米=100平方分米，高级单位转换成低级单位乘进率，低级单位转换成高级单位除以进率，据此换算即可.\n【详解】13÷10=1.3(分米),即13厘米=1.3分米;5350÷1000=5.35(千米),即5350米=5.35千米;2.3×1000=2300(克),即2.3千克=2300克;将2.95平方米看作2平方米和0.95平方米,0.95×100=95(平方分米),即0.95平方米=95平方分米，所以2.95平方米=2平方米95平方分米。",
+        "knowledge_points":  [
+                                 "克，千克之间的换算与比较",
+                                 "千米和米之间的进率与换算",
+                                 "平方厘米，平方分米，平方米之间的进率与换算",
+                                 "运用小数点移动解决小数的单位换算问题"
+                             ],
+        "source":  "24-25四年级下·河南驻马店·期末"
+    },
+    {
+        "id":  "406",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "1000千克小麦可出面粉850千克,1千克小麦可以出面粉( )千克。",
+        "options":  [
+
+                    ],
+        "answer":  "0.85",
+        "solution":  "【分析】根据题意，已知1000千克小麦可出面粉850千克，求1千克小麦可以出面粉多少千克直接用850除以1000即可解答.\n【详解】850÷1000=0.85(千克),所以1千克小麦可以出面粉0.85千克。",
+        "knowledge_points":  [
+                                 "小数点向左移动引起小数大小变化的规律"
+                             ],
+        "source":  "24-25四年级下·河南驻马店·期末"
+    },
+    {
+        "id":  "407",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "0.06扩大到原数的( )倍是6,85缩小到原数的( )是0.085。",
+        "options":  [
+
+                    ],
+        "answer":  "100;1÷1000",
+        "solution":  "【分析】小数点的移动规律是：小数点向右移动一位相当于把原数乘10，小数就扩大到原数的10倍;移动两位相当于把原数乘100，小数就扩大到原数的100倍;移动三位相当于把原数乘1000，小数就扩大到原数的1000倍。小数点向左移动一位相当于把原数除以10，小数就缩小到原数的1÷10;移动两位相当于把原数除以100，小数就缩小到原数的1÷100;移动三位相当于把原数除以1000，小数就缩小到原数的1÷1000，据此解答.\n【详解】0.06×100=6;85÷1000=0.085，所以0.06扩大到原数的100倍是6,85缩小到原数的1÷1000是0.085。",
+        "knowledge_points":  [
+                                 "小数点向右移动引起小数大小变化的规律",
+                                 "小数点向左移动引起小数大小变化的规律"
+                             ],
+        "source":  "24-25四年级下·河南驻马店·期末"
+    },
+    {
+        "id":  "408",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在5.50,5.050,5.005,5.009中，按从小到大的顺序排列是:( )\u003c( )\u003c( )\u003c( )。",
+        "options":  [
+
+                    ],
+        "answer":  "5.005;5.009;5.050;5.50",
+        "solution":  "【分析】根据小数的大小比较的方法：先比较整数部分，整数部分大的那个小数就大;如果整数部分相同，再比较小数部分，小数部分先比较十分位，十分位大的那个小数就大;如果十分位相同就比较百分位，以此类推，直到比较出小数的大小为止。以此答题即可.\n【详解】根据分析可知：在5.50,5.050,5.005,5.009中，按从小到大的顺序排列是:5.005\u003c5.009\u003c5.050\u003c5.50。",
+        "knowledge_points":  [
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "24-25四年级下·河南驻马店·期末"
+    },
+    {
+        "id":  "409",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "2014年我国在校小学生116897000人，改写成用\"亿\"作单位的数是( )亿人。(保留一位小数)",
+        "options":  [
+
+                    ],
+        "answer":  "1.2",
+        "solution":  "【分析】改写成用\"亿\"作单位的数，就是在亿位数的右下角点上小数点，然后把小数末尾的0去掉，再在数的后面写上\"亿\"字;保留一位小数，就是精确到十分位，就看百分位上的数字，百分位上比5小就舍去，如果大于或者等于5就向十分位进1，据此解答即可.\n【详解】116897000=1.16897亿;1.16897亿≈1.2亿。",
+        "knowledge_points":  [
+                                 "小数的改写",
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25四年级下·河南驻马店·期末"
+    },
+    {
+        "id":  "410",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "用0,2,5,8和小数点组成的最大两位小数是( ),最小三位小数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "85.20;0.258",
+        "solution":  "【分析】小数大小的比较：先看它们的整数部分，整数部分大的那个数就大;如果整数部分相同，十分位上的数大的那个数就大;如果十分位上相同，百分位上的数大的那个数就大，以此类推。把0,2,5,8四个数字按从大到小排列为:8\u003e5\u003e2\u003e0，小数点点在第二大数的右下角，这样组成的两位小数最大。把0,2,5,8四个数字按从小到大排列为:0\u003c2\u003c5\u003c8,0放在个位,2放在十分位,5放在百分位,8放在千分位，然后点上小数点，这样组成的三位小数最小;据此即可解答.\n【详解】用0,2,5,8和小数点组成的最大两位小数是85.20，最小三位小数是0.258。",
+        "knowledge_points":  [
+                                 "小数的数位和计数单位的认识",
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "24-25四年级下·河北唐山·期末"
+    },
+    {
+        "id":  "411",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把3.8扩大到它的100倍是( ),把3.8缩小到它的( )是0.0038。",
+        "options":  [
+
+                    ],
+        "answer":  "380;1÷1000",
+        "solution":  "【分析】根据小数点移动引起小数大小变化的规律，如果一个数扩大(或缩小)到原数的10倍(或1÷10),100倍(或1÷100),1000倍(或1÷1000)。只要把小数点向右(或向左)移动一位，两位，三位。位数不够时，要用\"0\"补足.\n【详解】3.8扩大到它的100倍，也就是把3.8的小数点向右移动两位，得380;3.8变为0.0038，是小数点向左移动三位，也就是缩小到它的1÷1000。",
+        "knowledge_points":  [
+                                 "小数点向右移动引起小数大小变化的规律",
+                                 "小数点向左移动引起小数大小变化的规律"
+                             ],
+        "source":  "24-25四年级下·河北唐山·期末"
+    },
+    {
+        "id":  "412",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "下表是小明，小强，晓峰，乐乐50米跑的成绩单，但成绩单上一些数字被墨迹弄脏了。根据表格中提供的数据，你能知道他们的成绩分别是多少吗。\n\n| **名次** | **姓名** | **成绩** |\n| -------- | -------- | -------- |\n| 第一名 | 小明 | 1★.49 秒 |\n| 第二名 | 小强 | 11.48 秒 |\n| 第三名 | 晓峰 | 11.★8 秒 |\n| 第四名 | 乐乐 | 11.5★ 秒 |\n\n小明:( )秒，晓峰:( )秒，乐乐:( )秒。",
+        "options":  [
+
+                    ],
+        "answer":  "10.49;11.58;11.59",
+        "solution":  "【分析】比较小数的大小，先比较整数部分，整数部分大的那个数就大，如果整数部分相同就比较十分位，十分位上的数大的那个数就大，如果十分位上的数也相同就比较百分位，百分位上的数大的那个数就大。\n【详解】在跑步比赛中，用时短的人跑得快，名次靠前。小明是第一名，小强是第二名，第一名用时要小于第二名，即1★.49\u003c11.48，所以小明成绩是10.49秒;晓峰是第三名，小强是第二名，乐乐是第四名，即11.48\u003c11.★8\u003c11.5★,所以晓峰的成绩是11.58秒，乐乐成绩大于晓峰成绩11.58秒，乐乐成绩十分位也是5，百分位要大于8，即乐乐的成绩是11.59秒。",
+        "knowledge_points":  [
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "24-25四年级下·山西吕梁·期末"
+    },
+    {
+        "id":  "413",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "近年来，电子支付以其方便快捷的优势深受人们喜爱，小明妈妈的微信余额是由5个千,4个一,3个十分之一,7个0.01组成的，这个数是( ),读作( )。",
+        "options":  [
+
+                    ],
+        "answer":  "5004.37;五千零四点三七",
+        "solution":  "【分析】根据小数的组成,5个千对应千位的5,4个一对应个位的4，整数部分中间空缺的百位，十位用0补全;3个十分之一对应十分位的3,7个0.01对应百分位的7，小数部分直接组合。小数的读法是小数的整数部分按照整数的读法读，小数点读作\"点\"，小数部分依次读出每个数字.\n【详解】这个数是5004.37，读作五千零四点三七。",
+        "knowledge_points":  [
+                                 "小数的读法和写法",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·山西吕梁·期末"
+    },
+    {
+        "id":  "414",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在()里填上\"\u003e\"、\"\u003c\"或\"=\"。\n720÷6+3×4( )720÷(6+3)×4;90g( )0.8kg;100×47( )47×55+45×47",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003c;=",
+        "solution":  "【分析】根据四则混合运算顺序先算乘除后算加减;单位不同先统一单位;再结合乘法分配律比较大小.\n【详解】720÷6+3×4=132,720÷(6+3)×4=320，所以132\u003c320;0.8kg=800g,所以90g\u003c0.8kg;47×55+45×47=(45+55)×47=100×47，所以相等。",
+        "knowledge_points":  [
+                                 "整数乘法分配律",
+                                 "克，千克之间的换算与比较",
+                                 "运用小数点移动解决小数的单位换算问题"
+                             ],
+        "source":  "24-25四年级下·广东汕头·期末"
+    },
+    {
+        "id":  "415",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "把0.6的小数点向左移动两位，再向右移动两位，结果是( );( )扩大到它的1000倍是695。",
+        "options":  [
+
+                    ],
+        "answer":  "0.6;0.695",
+        "solution":  "【分析】小数点位置移动引起数的大小变化规律：向右移扩大，向左移缩小。把0.6先左移两位再右移两位，净变化为0;另一个空可由695向左移动三位得到原数.\n【详解】把0.6的小数点向左移动两位，再向右移动两位，结果是0.6;695小数点向左移动三位为0.695。",
+        "knowledge_points":  [
+                                 "小数点向右移动引起小数大小变化的规律",
+                                 "小数点向左移动引起小数大小变化的规律"
+                             ],
+        "source":  "24-25四年级下·广东汕头·期末"
+    },
+    {
+        "id":  "416",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个小数，百位和百分位都是8，其他各位都是0，这个小数写作( ),读作( )。",
+        "options":  [
+
+                    ],
+        "answer":  "800.08;八百点零八",
+        "solution":  "【分析】百位是8，十位个位为0;小数部分十分位为0，百分位为8，据此写数并读数.\n【详解】这个小数写作800.08，读作八百点零八。",
+        "knowledge_points":  [
+                                 "小数的读法和写法",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·广东汕头·期末"
+    },
+    {
+        "id":  "417",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "7千米50米=( )千米;6角8分=( )元;3450千克=( )吨;3.5平方米=( )平方分米",
+        "options":  [
+
+                    ],
+        "answer":  "7.05;0.68;3.45;350",
+        "solution":  "【分析】1千米=1000米,1元=10角=100分,1吨=1000千克,1平方米=100平方分米;高低单位按进率换算.\n【详解】50÷1000=0.05，所以7千米50米=7.05千米;6÷10=0.6,8÷100=0.08，所以6角8分=0.68元;3450÷1000=3.45吨;3.5×100=350平方分米。",
+        "knowledge_points":  [
+                                 "元，角，分的认识及换算",
+                                 "吨，千克之间的换算与比较",
+                                 "千米和米之间的进率与换算",
+                                 "运用小数点移动解决小数的单位换算问题"
+                             ],
+        "source":  "24-25四年级下·河北邢台·期末"
+    },
+    {
+        "id":  "418",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把5.248扩大到它的100倍是( ),再缩小到它的1÷1000是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "524.8;0.5248",
+        "solution":  "【分析】小数点右移两位相当于乘100，左移三位相当于除以1000.\n【详解】5.248×100=524.8;524.8÷1000=0.5248。",
+        "knowledge_points":  [
+                                 "小数点向右移动引起小数大小变化的规律",
+                                 "小数点向左移动引起小数大小变化的规律"
+                             ],
+        "source":  "24-25四年级下·河北邢台·期末"
+    },
+    {
+        "id":  "419",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个小数由8个十,5个十分之一和4个百分之一组成，这个小数是( ),读作( )。",
+        "options":  [
+
+                    ],
+        "answer":  "80.54;八十点五四",
+        "solution":  "【分析】8个十是80,5个十分之一是0.5,4个百分之一是0.04，合起来是80.54。小数读法为整数部分按整数读，小数点读\"点\"，小数部分逐位读.\n【详解】这个小数是80.54，读作八十点五四。",
+        "knowledge_points":  [
+                                 "小数的读法和写法",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·河北邢台·期末"
+    },
+    {
+        "id":  "420",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "世界上最小的鸟，身长约0.05m,合( )cm;体重仅2g左右，合( )kg。",
+        "options":  [
+
+                    ],
+        "answer":  "5;0.002",
+        "solution":  "【分析】1m=100cm,1kg=1000g;大单位化小单位乘进率，小单位化大单位除进率.\n【详解】0.05×100=5cm;2÷1000=0.002kg。",
+        "knowledge_points":  [
+                                 "克，千克之间的换算与比较",
+                                 "毫米，厘米，分米，米之间的进率与换算",
+                                 "运用小数点移动解决小数的单位换算问题"
+                             ],
+        "source":  "24-25四年级下·山西晋中·期末"
+    },
+    {
+        "id":  "421",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "\"玉兔号\"月球车在某次任务中，将探测数据3.08GB扩大到原来的100倍后传回地球，地球得到的数据是( )GB;把一张45MB的月面高清照片压缩到原来大小的1÷1000，地球得到的照片是( )MB。",
+        "options":  [
+
+                    ],
+        "answer":  "308;0.045",
+        "solution":  "【分析】扩大100倍相当于乘100;缩小到1÷1000相当于除以1000.\n【详解】3.08×100=308(GB);45÷1000=0.045(MB)。",
+        "knowledge_points":  [
+                                 "小数点向右移动引起小数大小变化的规律",
+                                 "小数点向左移动引起小数大小变化的规律"
+                             ],
+        "source":  "24-25四年级下·山西晋中·期末"
+    },
+    {
+        "id":  "422",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\"\u003e\"\"\u003c\"或\"=\".\n3.07( )3.7;1÷100( )0.09;0.06×10( )6÷10;26×99( )26×100-1",
+        "options":  [
+
+                    ],
+        "answer":  "\u003c;\u003c;=;\u003c",
+        "solution":  "【分析】比较小数时先比整数部分，再比小数部分;把分数转成小数再比较;计算两边结果后比较;利用乘法分配律化简后比较.\n【详解】3.07\u003c3.7;1÷100=0.01\u003c0.09;0.06×10=0.6,6÷10=0.6，故相等;26×99=2574,26×100-1=2599，所以2574\u003c2599。",
+        "knowledge_points":  [
+                                 "多位小数的大小比较",
+                                 "整数乘法分配律",
+                                 "小数点向右移动引起小数大小变化的规律",
+                                 "小数点向左移动引起小数大小变化的规律"
+                             ],
+        "source":  "24-25四年级下·河北唐山·期末"
+    },
+    {
+        "id":  "423",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "0.6t=( )kg;5dm²=( )m²;8元5分=( )元;75400=( )万",
+        "options":  [
+
+                    ],
+        "answer":  "600;0.05;8.05;7.54",
+        "solution":  "【分析】1t=1000kg,1m²=100dm²,1元=100分,1万=10000，高级单位化低级单位乘进率，低级单位化高级单位除进率.\n【详解】0.6t=600kg;5dm²=0.05m²;8元5分=8.05元;75400=7.54万。",
+        "knowledge_points":  [
+                                 "元，角，分的认识及换算",
+                                 "吨，千克之间的换算与比较",
+                                 "平方厘米，平方分米，平方米之间的进率与换算",
+                                 "运用小数点移动解决小数的单位换算问题"
+                             ],
+        "source":  "24-25四年级下·河北唐山·期末"
+    },
+    {
+        "id":  "424",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "7千米50米=( )千米;6角8分=( )元;3450千克=( )吨;3.5平方米=( )平方分米",
+        "options":  [
+
+                    ],
+        "answer":  "7.05;0.68;3.45;350",
+        "solution":  "【分析】明确进率:1千米=1000米,1元=10角=100分,1吨=1000千克,1平方米=100平方分米;按高低单位换算规则计算.\n【详解】50÷1000=0.05,7千米50米=7.05千米;6÷10=0.6,8÷100=0.08,6角8分=0.68元;3450÷1000=3.45吨;3.5×100=350平方分米。",
+        "knowledge_points":  [
+                                 "元，角，分的认识及换算",
+                                 "吨，千克之间的换算与比较",
+                                 "千米和米之间的进率与换算",
+                                 "运用小数点移动解决小数的单位换算问题"
+                             ],
+        "source":  "24-25四年级下·河北邢台·期末"
+    },
+    {
+        "id":  "425",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个小数由8个十,5个十分之一和4个百分之一组成，这个小数是( ),读作( )。",
+        "options":  [
+
+                    ],
+        "answer":  "80.54;八十点五四",
+        "solution":  "【分析】按数位组成写出小数：十位是8，十分位是5，百分位是4，个位补0。读数时按小数读法.\n【详解】这个小数是80.54，读作八十点五四。",
+        "knowledge_points":  [
+                                 "小数的读法和写法",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·河北邢台·期末"
+    },
+    {
+        "id":  "426",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个小数，整数部分最小的计数单位是( ),小数部分最大的计数单位是( )。一个数的百位和百分位上都是3，其余数位都是0，这个数写作( )。",
+        "options":  [
+
+                    ],
+        "answer":  "1;0.1;300.03",
+        "solution":  "【分析】整数部分最小计数单位是个位\"1\";小数部分最大计数单位是十分之一\"0.1\"。百位与百分位都是3，其余位补0即可.\n【详解】填空为1,0.1,300.03。",
+        "knowledge_points":  [
+                                 "整数的数级，数位和计数单位的认识",
+                                 "小数的读法和写法",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·湖南长沙·期末"
+    },
+    {
+        "id":  "427",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上\"\u003e\"\"\u003c\"或\"=\".\n7÷100( )0.069;1.25千米( )1250米;(12+8)×5( )12+8×5",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;=;\u003e",
+        "solution":  "【分析】先把分数转为小数比较;单位统一后比较;按运算顺序分别算出两边再比较.\n【详解】7÷100=0.07,0.07\u003e0.069;1.25千米=1250米;(12+8)×5=100,12+8×5=52，故100\u003e52。",
+        "knowledge_points":  [
+                                 "多位小数的大小比较",
+                                 "带有小括号的混合运算",
+                                 "小数点向右移动引起小数大小变化的规律",
+                                 "运用小数点移动解决小数的单位换算问题"
+                             ],
+        "source":  "24-25四年级下·湖南长沙·期末"
+    },
+    {
+        "id":  "428",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "把0.05的小数点先向右移动三位，再向左移动两位，得到的数是原数的( )。",
+        "options":  [
+
+                    ],
+        "answer":  "10倍",
+        "solution":  "【分析】先右移三位再左移两位，净效果是小数点右移一位，即扩大10倍.\n【详解】得到的数是原数的10倍。",
+        "knowledge_points":  [
+                                 "小数点向右移动引起小数大小变化的规律",
+                                 "小数点向左移动引起小数大小变化的规律"
+                             ],
+        "source":  "24-25四年级下·湖南长沙·期末"
+    },
+    {
+        "id":  "429",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "山西省右玉县出土的胡傅温酒樽是古代用来盛酒的器物，高约2.5分米。实际测量高度是一个两位小数，它的实际高度最小可能是( )分米，最大可能是( )分米。",
+        "options":  [
+
+                    ],
+        "answer":  "2.45;2.54",
+        "solution":  "【分析】两位小数四舍五入到十分位是2.5时，最小边界为2.45，最大边界为2.54.\n【详解】实际高度最小可能是2.45分米，最大可能是2.54分米。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25四年级下·山西晋中·期末"
+    },
+    {
+        "id":  "430",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "榆次是晋中市政治，经济，文化，科教中心，古称\"魏榆\"，全区面积1318平方千米，约为( )万平方千米(保留两位小数)。榆次老城占地面积1.074平方千米，合( )公顷;后沟古村位于榆次区东赵乡，总面积1.33平方千米，比榆次老城占地面积( )(填\"大\"或\"小\")。",
+        "options":  [
+
+                    ],
+        "answer":  "0.13;107.4;大",
+        "solution":  "【分析】1318平方千米改写成\"万平方千米\"需除以10000，再保留两位小数;1平方千米=100公顷;比较1.33与1.074大小即可.\n【详解】1318÷10000=0.1318≈0.13(万平方千米);1.074×100=107.4(公顷);1.33\u003e1.074，所以填\"大\"。",
+        "knowledge_points":  [
+                                 "多位小数的大小比较",
+                                 "公顷，平方千米的进率与换算",
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25四年级下·山西晋中·期末"
+    },
+    {
+        "id":  "431",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "8.719精确到十分位是( ),把171000000改写成用\"亿\"作单位的数是( )亿。",
+        "options":  [
+
+                    ],
+        "answer":  "8.7;1.71",
+        "solution":  "【分析】精确到十分位看百分位,8.719的百分位是1，舍去;171000000改写成\"亿\"为小数点左移8位.\n【详解】8.719≈8.7;171000000=1.71亿。",
+        "knowledge_points":  [
+                                 "小数的改写",
+                                 "小数的近似数",
+                                 "亿以上数的改写"
+                             ],
+        "source":  "24-25四年级下·陕西安康·期末"
+    },
+    {
+        "id":  "432",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "4个十和5个千分之一组成的数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "40.005",
+        "solution":  "【分析】4个十是40,5个千分之一是0.005，合并即40.005.\n【详解】组成的数是40.005。",
+        "knowledge_points":  [
+                                 "小数的读法和写法",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·陕西安康·期末"
+    },
+    {
+        "id":  "433",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "0.80里面有( )个0.01。",
+        "options":  [
+
+                    ],
+        "answer":  "80",
+        "solution":  "【分析】0.80是两位小数，百分位计数单位是0.01,0.80表示80个0.01.\n【详解】0.80里面有80个0.01。",
+        "knowledge_points":  [
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·陕西安康·期末"
+    },
+    {
+        "id":  "434",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "由6个一,4个十分之一和3个千分之一组成的数写作( ),读作( ),它的计数单位是( ),它里面有( )个这样的计数单位。",
+        "options":  [
+
+                    ],
+        "answer":  "6.403;六点四零三;0.001;6403",
+        "solution":  "【分析】根据小数的组成和读写规则，整数部分为6，十分位是4，千分位是3，百分位补0，写作6.403。读法需依次读出各位数字，计数单位为最小数位对应的千分之一(0.001),总计数单位数为各部分换算后的总和.\n【详解】根据分析可知：由6个一,4个十分之一和3个千分之一组成的数写作6.403，读作六点四零三，它的计数单位是0.001，它里面有6403个这样的计数单位。",
+        "knowledge_points":  [
+                                 "小数的读法和写法",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·河北邢台·期末"
+    },
+    {
+        "id":  "435",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "![](images/da13a55563b2353f9bbe5f7398b5887e6e68f774dfa8457e8bd9e7d5aff76b28.jpg)\n(1)直线上所标的数a=( )。它是由16个一,( )个十分之一,( )个百分之一组成的。\n(2)直线上与a最接近的一位小数是( ),这就是将a用\"四舍五入\"法精确到( )位所得的近似数。",
+        "options":  [
+
+                    ],
+        "answer":  "16.28;2;8;16.3;十分",
+        "solution":  "【分析】(1)由题意可知直线上所标刻度一小格为百分之一，用小数表示是0.01;a所标的位置是16.2向后数8个小格，据此解答。(2)a所标的位置在16.2和16.3之间，从图上可以看出a离16.3更近，也就是把a\"四舍五入\"精确到十分位，保留一位小数的近似数.\n【详解】(1)直线上所标的数a=16.28。它是由16个一,2个十分之一,8个百分之一组成的。(2)直线上与a最接近的一位小数是16.3，这就是将a用\"四舍五入\"法精确到十分位所得的近似数。",
+        "knowledge_points":  [
+                                 "小数的近似数",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·河南濮阳·期末"
+    },
+    {
+        "id":  "436",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个两位小数，用四舍五入法取近似值为5.3，则这个小数原来最大是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "5.34",
+        "solution":  "【分析】小数的近似数：根据\"四舍五入\"求近似数的方法，找到要求保留的数位，看下一位;如果下一位的数字大于或等于5，要往前进一;如果下一位的数字小于5，要舍去。已知一个数的近似数，用\"四舍\"法得到最大的原数，用\"五入\"法得到最小的原数.\n【详解】要考虑5.3是一个两位小数的近似数，有两种情况：\"四舍\"得到的5.3，有5.31,5.32,5.33,5.34，其中最大是5.34;\"五入\"得到的5.3，有5.25,5.26,5.27,5.28,5.29，其中最小是5.25。填空如下：一个两位小数，用四舍五入法取近似值为5.3，则这个小数原来最大是5.34。",
+        "knowledge_points":  [
+                                 "小数的近似数",
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "2025·重庆渝北·小升初真题"
+    },
+    {
+        "id":  "437",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "8.05是由8个( ),5个( )组成的。",
+        "options":  [
+
+                    ],
+        "answer":  "1;0.01",
+        "solution":  "【分析】由数位表可知:8.05的整数部分8在个位上，表示8个1;小数部分的5在百分位上，表示5个0.01.\n【详解】8.05是由8个1,5个0.01组成的。",
+        "knowledge_points":  [
+                                 "小数的意义",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·黑龙江·期末"
+    },
+    {
+        "id":  "438",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "已知10千克芝麻可以榨出香油4.5千克,100千克芝麻可榨油( )千克。",
+        "options":  [
+
+                    ],
+        "answer":  "45",
+        "solution":  "【分析】先求1千克芝麻能榨出多少油:4.5÷10;再求100千克芝麻的榨油量：结果*100.\n【详解】4.5÷10=0.45(千克);0.45×100=45(千克);所以100千克芝麻可榨油45千克。",
+        "knowledge_points":  [
+                                 "小数点向右移动引起小数大小变化的规律",
+                                 "小数点向左移动引起小数大小变化的规律",
+                                 "单归一问题"
+                             ],
+        "source":  "24-25四年级下·河北唐山·期末"
+    },
+    {
+        "id":  "439",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "珠穆朗玛峰海拔是八千八百四十八点八六米，写作( )米，精确到十分位约是( )米。",
+        "options":  [
+
+                    ],
+        "answer":  "8848.86;8848.9",
+        "solution":  "【分析】写小数时整数部分按整数写法写，小数点写在个位右下角，小数部分顺次写出各数位数字。精确到十分位即保留一位小数，看百分位上的数四舍五入.\n【详解】写作8848.86;8848.86百分位上是6,6\u003e5，向十分位进1，所以精确到十分位约是8848.9。",
+        "knowledge_points":  [
+                                 "小数的读法和写法",
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25四年级下·河北唐山·期末"
+    },
+    {
+        "id":  "440",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把0.945的小数点先向右移动三位，再向左移动两位，结果是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "9.45",
+        "solution":  "【分析】先右移三位再左移两位，净效果是小数点向右移动一位，即扩大到原来的10倍.\n【详解】0.945的小数点最终向右移动一位，结果是9.45。",
+        "knowledge_points":  [
+                                 "小数点向右移动引起小数大小变化的规律",
+                                 "小数点向左移动引起小数大小变化的规律"
+                             ],
+        "source":  "24-25四年级下·安徽淮北·期末"
+    },
+    {
+        "id":  "441",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "2025年\"五一\"假期，全国国内出游共计(31400000)人次，把括号内的数改写成用\"亿\"作单位的数是( )亿人次，约是( )亿人次。(保留整数)",
+        "options":  [
+
+                    ],
+        "answer":  "0.314;0",
+        "solution":  "【分析】改写成以\"亿\"为单位，小数点向左移动8位;保留整数看十分位四舍五入.\n【详解】31400000=0.314亿;十分位是3，小于5舍去，约是0亿。",
+        "knowledge_points":  [
+                                 "小数的改写",
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25四年级下·安徽淮北·期末"
+    },
+    {
+        "id":  "442",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个数由5个十,4个十分之一和3个千分之一组成，这个数写作:( ),读作:( )。",
+        "options":  [
+
+                    ],
+        "answer":  "50.403;五十点四零三",
+        "solution":  "【分析】5个十即50,4个十分之一即0.4,3个千分之一即0.003，百分位为0占位，合成50.403;按小数读法读数.\n【详解】写作50.403，读作五十点四零三。",
+        "knowledge_points":  [
+                                 "小数的读法和写法",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·安徽淮北·期末"
+    },
+    {
+        "id":  "443",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\"\u003e\"、\"\u003c\"或\"=\"。\n10元( )9.99元;1日( )24小时;5分米( )4.7米;0.3( )0.03",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;=;\u003c;\u003e",
+        "solution":  "【分析】小数比较先比整数部分再比小数部分;单位换算先统一单位再比较.\n【详解】10元\u003e9.99元;1日=24小时;5分米=0.5米,0.5\u003c4.7，所以5分米\u003c4.7米;0.3\u003e0.03。",
+        "knowledge_points":  [
+                                 "年，月，日的认识及换算",
+                                 "多位小数的大小比较",
+                                 "毫米，厘米，分米，米之间的进率与换算"
+                             ],
+        "source":  "24-25三年级下·山西吕梁·期末"
+    },
+    {
+        "id":  "444",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "10kg黄豆可以磨24.3kg豆腐,1000kg黄豆可以磨( )kg豆腐。",
+        "options":  [
+
+                    ],
+        "answer":  "2430",
+        "solution":  "【分析】先求1千克黄豆能磨多少千克豆腐:24.3÷10=2.43;再求1000千克能磨:2.43×1000.\n【详解】24.3÷10=2.43(千克);2.43×1000=2430(千克)。所以1000kg黄豆可以磨2430kg豆腐。",
+        "knowledge_points":  [
+                                 "单归一问题",
+                                 "小数点向左移动引起小数大小变化的规律",
+                                 "小数点向右移动引起小数大小变化的规律"
+                             ],
+        "source":  "24-25四年级下·辽宁盘锦·期末"
+    },
+    {
+        "id":  "445",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数，精确到百分位后约是3.36，这个三位小数最小是( ),最大是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "3.355;3.364",
+        "solution":  "【分析】精确到百分位即保留两位小数，看千分位四舍五入;由此确定能得到3.36的三位小数范围.\n【详解】能得到3.36的三位小数最小是3.355，最大是3.364。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25四年级下·辽宁盘锦·期末"
+    },
+    {
+        "id":  "446",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把687530000改成用\"亿\"做单位的数是( ),精确到百分位是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "6.8753亿;6.88亿",
+        "solution":  "【分析】改写成\"亿\"作单位小数点左移8位;精确到百分位保留两位小数，看千分位四舍五入.\n【详解】687530000=6.8753亿;6.8753亿≈6.88亿。",
+        "knowledge_points":  [
+                                 "亿以上数的改写",
+                                 "小数的近似数",
+                                 "小数的改写"
+                             ],
+        "source":  "24-25四年级下·辽宁盘锦·期末"
+    },
+    {
+        "id":  "447",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "8平方米50平方分米=( )平方米;5kg60g=( )kg;3.5km=( )km( )m;4.67km²=( )公顷",
+        "options":  [
+
+                    ],
+        "answer":  "8.5;5.06;3;500;467",
+        "solution":  "【分析】1平方米=100平方分米,1千克=1000克,1千米=1000米,1平方千米=100公顷;按进率进行换算.\n【详解】50÷100=0.5,8平方米50平方分米=8.5平方米;60÷1000=0.06,5kg60g=5.06kg;0.5×1000=500,3.5km=3km500m;4.67×100=467,4.67km²=467公顷。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "公顷，平方千米的进率与换算",
+                                 "平方厘米，平方分米，平方米之间的进率与换算",
+                                 "千米和米之间的进率与换算"
+                             ],
+        "source":  "24-25四年级下·辽宁盘锦·期末"
+    },
+    {
+        "id":  "448",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "把0.03的小数点向左移动两位，得到的数是它的( )。",
+        "options":  [
+
+                    ],
+        "answer":  "1÷100",
+        "solution":  "【分析】小数点向左移动两位相当于除以100，即缩小到原来的1÷100.\n【详解】0.03小数点向左移动两位相当于0.03÷100，所以得到的数是它的1÷100。",
+        "knowledge_points":  [
+                                 "小数点向左移动引起小数大小变化的规律"
+                             ],
+        "source":  "24-25四年级下·安徽阜阳·期末"
+    },
+    {
+        "id":  "449",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "2吨70千克=( )千克;3.05千米=( )米;3米20厘米=( )米",
+        "options":  [
+
+                    ],
+        "answer":  "2070;3050;3.2",
+        "solution":  "【分析】1吨=1000千克,1千米=1000米,1米=100厘米;按进率换算即可.\n【详解】2吨=2000千克,2吨70千克=2070千克;3.05×1000=3050米;20÷100=0.2米，所以3米20厘米=3.2米。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "千米和米之间的进率与换算",
+                                 "厘米和米之间的进率与换算",
+                                 "吨，千克之间的换算与比较"
+                             ],
+        "source":  "24-25四年级下·安徽芜湖·期末"
+    },
+    {
+        "id":  "450",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\"\u003e\"、\"\u003c\"或\"=\"。\n5.901( )5.091;1.5小时( )150分钟;25×100-1( )25×99;4吨50千克( )4.5吨",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;\u003c;\u003e;\u003c",
+        "solution":  "【分析】小数大小比较先比整数部分再比小数部分;时间，质量先统一单位;乘法算式可分别计算或用分配律变形比较.\n【详解】5.901\u003e5.091;1.5小时=90分钟,90\u003c150，所以1.5小时\u003c150分钟;25×100-1=2499,25×99=2475，所以2499\u003e2475;4吨50千克=4.05吨,4.05\u003c4.5，所以4吨50千克\u003c4.5吨。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "吨，千克之间的换算与比较",
+                                 "整数乘法分配律",
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "24-25四年级下·安徽宣城·期末"
+    },
+    {
+        "id":  "451",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把4.8缩小到它原来的1÷100是( ),把0.0207扩大到它原来的( )倍是20.7。",
+        "options":  [
+
+                    ],
+        "answer":  "0.048;1000",
+        "solution":  "【分析】缩小到1÷100相当于除以100(小数点左移2位);0.0207扩大到20.7相当于小数点右移3位，即乘1000.\n【详解】4.8÷100=0.048;0.0207×1000=20.7，所以倍数是1000。",
+        "knowledge_points":  [
+                                 "小数点向左移动引起小数大小变化的规律",
+                                 "小数点向右移动引起小数大小变化的规律"
+                             ],
+        "source":  "24-25四年级下·安徽宣城·期末"
+    },
+    {
+        "id":  "452",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "6060克=( )千克( )克;8角5分=( )元;3.07米=( )厘米;709平方分米=( )平方米",
+        "options":  [
+
+                    ],
+        "answer":  "6;60;0.85;307;7.09",
+        "solution":  "【分析】1千克=1000克,1元=10角=100分,1米=100厘米,1平方米=100平方分米;按进率换算.\n【详解】6060克=6千克60克;8角5分=0.85元;3.07米=307厘米;709平方分米=7.09平方米。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "平方厘米，平方分米，平方米之间的进率与换算",
+                                 "毫米，厘米，分米，米之间的进率与换算",
+                                 "元，角，分的认识及换算"
+                             ],
+        "source":  "24-25四年级下·安徽宣城·期末"
+    },
+    {
+        "id":  "453",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把795500000改写成以\"亿\"作单位的数是( ),保留一位小数约是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "7.955亿;8.0亿",
+        "solution":  "【分析】改写成\"亿\"作单位小数点左移8位;保留一位小数看百分位四舍五入.\n【详解】795500000=7.955亿;7.955亿≈8.0亿。",
+        "knowledge_points":  [
+                                 "小数的近似数",
+                                 "小数的改写"
+                             ],
+        "source":  "24-25四年级下·安徽宣城·期末"
+    },
+    {
+        "id":  "454",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "3.0500化简后是( );3.8改写成大小不变的两位小数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "3.05;3.80",
+        "solution":  "【分析】小数的性质：小数末尾的0可以去掉或添加，不改变数的大小。化简去末尾0;改写成两位小数在末尾补0.\n【详解】3.0500=3.05;3.8=3.80。",
+        "knowledge_points":  [
+                                 "小数的性质"
+                             ],
+        "source":  "24-25四年级下·安徽淮北·期末"
+    },
+    {
+        "id":  "455",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "1.7平方千米=( )公顷;10元6分=( )元;3米46厘米=( )米;980千克=( )吨",
+        "options":  [
+
+                    ],
+        "answer":  "170;10.06;3.46;0.98",
+        "solution":  "【分析】1平方千米=100公顷,1元=100分,1米=100厘米,1吨=1000千克;按进率换算.\n【详解】1.7×100=170公顷;6÷100=0.06元,10元6分=10.06元;46÷100=0.46米,3米46厘米=3.46米;980÷1000=0.98吨。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题"
+                             ],
+        "source":  "24-25四年级下·安徽淮北·期末"
+    },
+    {
+        "id":  "456",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "由8个一,5个十分之一和7个千分之一组成的数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "8.507",
+        "solution":  "【分析】个位是8，十分位是5，千分位是7，百分位补0.\n【详解】组成的数是8.507。",
+        "knowledge_points":  [
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·安徽淮北·期末"
+    },
+    {
+        "id":  "457",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "把6.8的小数点向右移动三位，再向左移动一位后是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "680",
+        "solution":  "【分析】先右移三位相当于乘1000，再左移一位相当于除以10，净效果为乘100.\n【详解】6.8×1000=6800;6800÷10=680。",
+        "knowledge_points":  [
+                                 "小数点向右移动引起小数大小变化的规律",
+                                 "小数点向左移动引起小数大小变化的规律"
+                             ],
+        "source":  "24-25四年级下·四川南充·期末"
+    },
+    {
+        "id":  "458",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上合适的数.\n4200平方米=( )公顷;10元8角=( )元;6.04米=( )米( )厘米;7千克40克=( )千克",
+        "options":  [
+
+                    ],
+        "answer":  "0.42;10.8;6;4;7.04",
+        "solution":  "【分析】1公顷=10000平方米,1元=10角,1米=100厘米,1千克=1000克;按进率换算.\n【详解】4200÷10000=0.42公顷;10元8角=10.8元;6.04米=6米+0.04米=6米4厘米;7千克40克=7.04千克。",
+        "knowledge_points":  [
+                                 "元，角，分的认识及换算",
+                                 "克，千克之间的换算与比较",
+                                 "毫米，厘米，分米，米之间的进率与换算",
+                                 "运用小数点移动解决小数的单位换算问题"
+                             ],
+        "source":  "24-25四年级下·四川南充·期末"
+    },
+    {
+        "id":  "459",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个两位小数\"四舍五入\"后的近似数是6.0，这个两位小数最小是( ),最大是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "5.95;6.04",
+        "solution":  "【分析】近似数为6.0(保留一位小数),最小值来自\"五入\"边界5.95，最大值来自\"四舍\"边界6.04.\n【详解】最小是5.95，最大是6.04。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25四年级下·四川南充·期末"
+    },
+    {
+        "id":  "460",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "小数0.85的十分位上的数是( ),它表示( )个( )。",
+        "options":  [
+
+                    ],
+        "answer":  "8;8;十分之一或0.1",
+        "solution":  "【分析】小数点后的第一位是十分位,0.85的十分位数字为8，表示8个十分之一(0.1).\n【详解】十分位上的数是8，表示8个十分之一(或0.1)。",
+        "knowledge_points":  [
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·四川南充·期末"
+    },
+    {
+        "id":  "461",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "将15895000000改写成用\"亿\"作单位的数是( )亿，保留一位小数约是( )亿。",
+        "options":  [
+
+                    ],
+        "answer":  "158.95;159.0",
+        "solution":  "【分析】改写成\"亿\"作单位小数点左移8位;保留一位小数看百分位四舍五入.\n【详解】15895000000=158.95亿;158.95亿≈159.0亿。",
+        "knowledge_points":  [
+                                 "小数的改写",
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25四年级下·四川南充·期末"
+    },
+    {
+        "id":  "462",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "如图，铅笔的长度是( )cm,也就是( )cm( )mm.\n![](images/32fa12683cc2e67ce4b463de3a3c1da57b3f0752d3388337f59943f09d9f6c53.jpg)",
+        "options":  [
+
+                    ],
+        "answer":  "6.8;6;8",
+        "solution":  "【分析】读出直尺刻度得到长度;1厘米=10毫米，将小数厘米分解成厘米和毫米.\n【详解】铅笔的长度是6.8cm,也就是6cm8mm。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "毫米，厘米，分米，米之间的进率与换算",
+                                 "长度的测量方法"
+                             ],
+        "source":  "24-25四年级下·内蒙古呼和浩特·期末"
+    },
+    {
+        "id":  "463",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个两位小数，保留一位小数后是5.0，这个两位小数最大是( ),最小是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "5.04;4.95",
+        "solution":  "【分析】保留一位小数为5.0：最大来自\"四舍\"边界5.04;最小来自\"五入\"边界4.95.\n【详解】最大是5.04，最小是4.95。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25五年级上·海南三亚·期中"
+    },
+    {
+        "id":  "464",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "绿化250m²草坪需要8kg草籽，那么平均绿化1m²草坪需要( )kg草籽，平均1kg草籽可以绿化( )m²草坪。",
+        "options":  [
+
+                    ],
+        "answer":  "0.032;31.25",
+        "solution":  "【分析】平均每1m²需要草籽:8÷250;平均1kg草籽可绿化面积:250÷8.\n【详解】8÷250=0.032(kg);250÷8=31.25(m²)。",
+        "knowledge_points":  [
+                                 "小数的初步认识",
+                                 "三位数除以一位数，首位不够除"
+                             ],
+        "source":  "24-25五年级上·广西百色·期中"
+    },
+    {
+        "id":  "465",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在()里填上适当的数.\n5.05公顷=( )平方米;8吨100千克=( )吨;10.73千米=( )千米( )米;10元5分=( )元",
+        "options":  [
+
+                    ],
+        "answer":  "50500;8.1;10;730;10.05",
+        "solution":  "【分析】1公顷=10000平方米,1吨=1000千克,1千米=1000米,1元=100分;按进率换算并拆分整数部分与小数部分.\n【详解】5.05公顷=50500平方米;8吨100千克=8.1吨;10.73千米=10千米730米;10元5分=10.05元。",
+        "knowledge_points":  [
+                                 "元，角，分的认识及换算",
+                                 "吨，千克之间的换算与比较",
+                                 "千米和米之间的进率与换算",
+                                 "运用小数点移动解决小数的单位换算问题"
+                             ],
+        "source":  "24-25四年级下·湖南衡阳·期末"
+    },
+    {
+        "id":  "466",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "一个数由7个十,4个一,9个百分之一和5个千分之一组成，这个数是( ),读作( ),保留两位小数约是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "74.095;七十四点零九五;74.10",
+        "solution":  "【分析】十位是7，个位是4，十分位补0，百分位是9，千分位是5，写作74.095;读法按小数读法;保留两位小数看千分位,5需进位.\n【详解】74.095保留两位小数约是74.10;读作七十四点零九五。",
+        "knowledge_points":  [
+                                 "小数的读法和写法",
+                                 "小数的数位和计数单位的认识",
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25四年级下·湖南衡阳·期末"
+    },
+    {
+        "id":  "467",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把1265920000改写成以\"亿\"作单位的数是( ),省略这个数\"亿\"后面的尾数大约是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "12.6592亿;13亿",
+        "solution":  "【分析】改写成\"亿\"作单位小数点左移8位;省略\"亿\"后面的尾数相当于保留整数(到亿位)四舍五入.\n【详解】1265920000=12.6592亿;12.6592≈13，所以约是13亿。",
+        "knowledge_points":  [
+                                 "小数的近似数",
+                                 "小数的改写"
+                             ],
+        "source":  "24-25六年级下·广东阳江·期末"
+    },
+    {
+        "id":  "468",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "\"山为翠浪涌，水作玉虹流\"，一代文豪苏东坡的咏叹，让赣州这座山水之城的风韵一跃千年。其中江西峰山国家森林公园位于赣州市章贡区沙石镇，占地面积207352000m²,其中宝盖峰是峰山的主峰，海拔1016.4m。这里的丰富植被营造出平均每立方厘米(14.8)万个负离子含量的天然\"氧吧\"，是赣州中心城区的\"绿肺\"和广大市民休闲，度假的理想去处。(1)207352000改写成用\"亿\"作单位的数(保留一位小数)是( )亿。(2)207352000平方米=( )公顷;1016.4米=( )千米;(3)括号内的数读作( ),它的计数单位是( ),再添上( )个这样的计数单位正好是15。",
+        "options":  [
+
+                    ],
+        "answer":  "2.1;20735.2;1.0164;十四点八;0.1;2",
+        "solution":  "【分析】(1)将整数值改写成用\"亿\"作单位的数，在亿位右下角点小数点并保留一位小数(看百分位四舍五入)。(2)1公顷=10000平方米,1千米=1000米，按进率换算。(3)按小数读法读出14.8;一位小数计数单位是0.1;15里面有150个0.1,14.8里有148个0.1，相差2个0.1.\n【详解】(1)207352000改写成用\"亿\"作单位并保留一位小数是2.1亿。(2)207352000平方米=20735.2公顷;1016.4米=1.0164千米。(3)14.8读作十四点八，计数单位是0.1，再添上2个这样的计数单位正好是15。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "公顷，平方千米的进率与换算",
+                                 "小数的数位和计数单位的认识",
+                                 "小数的读法和写法"
+                             ],
+        "source":  "24-25四年级下·河北衡水·期末"
+    },
+    {
+        "id":  "469",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在括号里填上\"\u003e\"、\"\u003c\"或\"=\"。\n34.24( )34.209;7角5分( )0.8元;7.2米( )7米2厘米;65×100-1( )65×99;1000-156-244( )1000-(156+244)",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;\u003c;\u003e;\u003e;=",
+        "solution":  "【分析】小数比较先比整数部分再比小数部分;金额，长度先统一单位;乘法用分配律变形比较;减法用性质变形比较.\n【详解】34.24\u003e34.209;7角5分=0.75元,0.75\u003c0.8，所以7角5分\u003c0.8元;7.2米=7米2分米\u003e7米2厘米;65×99=65×(100-1)=65×100-65，故65×100-1\u003e65×99;1000-156-244=1000-(156+244)。",
+        "knowledge_points":  [
+                                 "多位小数的大小比较",
+                                 "整数减法的性质",
+                                 "整数乘法分配律",
+                                 "运用小数点移动解决小数的单位换算问题"
+                             ],
+        "source":  "24-25四年级下·湖北十堰·期末"
+    },
+    {
+        "id":  "470",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "47m²=( )dm²;6t20kg=( )t",
+        "options":  [
+
+                    ],
+        "answer":  "4700;6.02",
+        "solution":  "【分析】1m²=100dm²,1t=1000kg,高级单位化成低级单位就乘它们之间的进率，低级单位化成高级单位就除以它们之间的进率，依此换算.\n【详解】47×100=4700，即47m²=4700dm²;20÷1000=0.02，即20kg=0.02t,因此6t20kg=6.02t。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "平方厘米，平方分米，平方米之间的进率与换算",
+                                 "吨，千克之间的换算与比较"
+                             ],
+        "source":  "24-25四年级下·山东日照·期末"
+    },
+    {
+        "id":  "471",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "城市建设中采用了新型空气净化膜，经科学检测,1平方米的净化膜每天大约能吸附空气中的灰尘颗粒0.07千克，分解有害气体0.04千克。市政部门计划在广场，车站等场所安装这种净化膜，那么100平方米的净化膜每天大约能吸附( )千克灰尘颗粒。10000平方米的净化膜每天大约能分解( )千克有害气体。",
+        "options":  [
+
+                    ],
+        "answer":  "7;400",
+        "solution":  "【分析】根据题意,1平方米的净化膜每天吸附0.07千克灰尘颗粒，分解0.04千克有害气体。要求100平方米吸附的灰尘颗粒总量，用乘法计算;同理,10000平方米分解的有害气体总量也用乘法计算.\n【详解】0.07×100=7(千克);0.04×10000=400(千克)。所以100平方米的净化膜每天大约能吸附7千克灰尘颗粒,10000平方米的净化膜每天大约能分解400千克有害气体。",
+        "knowledge_points":  [
+                                 "小数点向右移动引起小数大小变化的规律"
+                             ],
+        "source":  "24-25四年级下·山东日照·期末"
+    },
+    {
+        "id":  "472",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "已知4.□6≈5,□里最小填( );若4.□6≈4,□里可以填( )。(写出所有可能的数)",
+        "options":  [
+
+                    ],
+        "answer":  "5;0、1、2、3、4",
+        "solution":  "【分析】一个小数，保留整数时，需要看十分位上数的大小，用\"四舍五入\"法求它的近似数。(1)已知4.□6≈5，近似数比原数大，说明原数\"五入\"后得到5，原数十分位上的数必须大于或等于5,□里最小填5。(2)若4.□6≈4，近似数比原数小，说明原数\"四舍\"后得到4，原数十分位上的数必须小于5,□里可以填0,1,2,3,4.\n【详解】已知4.□6≈5,□里最小填5;若4.□6≈4,□里可以填0,1,2,3,4。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25四年级下·内蒙古呼和浩特·期末"
+    },
+    {
+        "id":  "473",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "0.3里有3个( ),15个千分之一是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "十分之一;0.015",
+        "solution":  "【分析】根据小数的意义：一位小数就是十分之几;两位小数就是百分之几;三位小数就是千分之几。0.3是一位小数，也就是3÷10;15个千分之一是15÷1000，对应三位小数0.015.\n【详解】0.3是3÷10，里面有3个十分之一;15个千分之一是15÷1000，是一个三位小数，即0.015。",
+        "knowledge_points":  [
+                                 "小数的意义"
+                             ],
+        "source":  "24-25四年级下·内蒙古呼和浩特·期末"
+    },
+    {
+        "id":  "474",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "2025年5月呼和浩特市被授予\"全国文明城市\"的荣誉称号。呼和浩特市的总面积约1.72万平方千米，其中所辖9个旗，县，区中的赛罕区总面积约1015平方千米，把它改写成以万为单位的数，并保留两位小数约是( )万平方千米。",
+        "options":  [
+
+                    ],
+        "answer":  "0.10",
+        "solution":  "【分析】改写成以万作单位的数：小数点向左移动4位;保留两位小数看千分位四舍五入.\n【详解】1015平方千米=0.1015万平方千米;0.1015万平方千米≈0.10万平方千米。",
+        "knowledge_points":  [
+                                 "小数的近似数",
+                                 "小数的改写"
+                             ],
+        "source":  "24-25四年级下·内蒙古呼和浩特·期末"
+    },
+    {
+        "id":  "475",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在括号里填上\"\u003e\"、\"\u003c\"或\"=\"。10个0.1( )100个0.01;120÷(2+10)( )120÷2+10;6.3千米( )630米;5.02米( )5米20厘米",
+        "options":  [
+
+                    ],
+        "answer":  "=;\u003c;\u003e;\u003c",
+        "solution":  "【分析】(1)10个0.1和100个0.01都表示1。(2)分别按运算顺序计算两式。(3)(4)先统一单位再比较大小.\n【详解】10个0.1是1,100个0.01也是1，故相等;120÷(2+10)=10,120÷2+10=70,10\u003c70;630米=0.63千米,6.3\u003e0.63;5米20厘米=5.2米,5.02\u003c5.2。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "毫米，厘米，分米，米之间的进率与换算",
+                                 "除加，除减混合运算",
+                                 "多位小数的大小比较"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "476",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "90千克=( )吨;0.8元=( )分",
+        "options":  [
+
+                    ],
+        "answer":  "0.09;80",
+        "solution":  "【分析】1吨=1000千克,1元=100分;单位换算按进率移动小数点.\n【详解】90÷1000=0.09(吨);0.8×100=80(分)。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "477",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "小数5.04的计数单位是( ),它有( )个这样的计数单位，至少去掉( )个这样的计数单位就是一个整数。",
+        "options":  [
+
+                    ],
+        "answer":  "0.01;504;4",
+        "solution":  "【分析】两位小数的计数单位是0.01。5.04表示504个0.01;整数5表示500个0.01，二者相差4个0.01.\n【详解】5.04的计数单位是0.01,5.04里面有504个0.01;504-500=4，至少去掉4个这样的计数单位就是一个整数。",
+        "knowledge_points":  [
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "478",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "我国的《四库全书》是世界上页数最多的书，如果将书摊开来逐页相接，可绕地球一周多。全书约有(997000000)字，把括号内的数改写成用\"亿\"作单位的数是( )亿，精确到十分位是( )亿。",
+        "options":  [
+
+                    ],
+        "answer":  "9.97;10.0",
+        "solution":  "【分析】改写成\"亿\"作单位在亿位右下角点小数点;精确到十分位保留一位小数，看百分位四舍五入.\n【详解】997000000=9.97亿;9.97亿≈10.0亿。",
+        "knowledge_points":  [
+                                 "亿以上数的改写",
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "479",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在()里填上\"\u003e\"、\"\u003c\"或\"=\"。80÷(40×2)( )80÷40/2;0.46×100( )4600;7.079吨( )7.709吨;99×56( )100×56-99;7.90( )7.900;3米8厘米( )3.8米",
+        "options":  [
+
+                    ],
+        "answer":  "=;\u003c;\u003c;\u003e;=;\u003c",
+        "solution":  "【分析】用除法性质比较第一组;第二组先算0.46×100;第三组按小数大小比较;第四组用分配律变形;第五组用小数性质;第六组先把3米8厘米换算成米.\n【详解】80÷(40×2)=80÷40/2;0.46×100=46,46\u003c4600;7.079\u003c7.709;99×56=100×56-56\u003e100×56-99;7.90=7.900;3米8厘米=3.08米,3.08\u003c3.8。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "整数乘法分配律",
+                                 "多位小数的大小比较",
+                                 "小数的性质"
+                             ],
+        "source":  "24-25四年级下·四川广元·期末"
+    },
+    {
+        "id":  "480",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "截至2022年底，全国现有海水淡化工程150个，工程规模日均(235.7048)万吨。括号内的数读作( ),用\"四舍五入\"法保留两位小数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "二百三十五点七零四八;235.70",
+        "solution":  "【分析】小数的读法：整数部分按整数读法读，小数点读作\"点\"，小数部分依次读出每一位上的数字;保留两位小数看小数点后第三位数字，大于等于5进一，小于5舍去.\n【详解】235.7048读作二百三十五点七零四八;4\u003c5，用\"四舍五入\"法保留两位小数是235.70。",
+        "knowledge_points":  [
+                                 "小数的近似数",
+                                 "小数的读法和写法"
+                             ],
+        "source":  "24-25四年级下·四川广元·期末"
+    },
+    {
+        "id":  "481",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "由7个十,5个一,9个十分之一和3个百分之一组成的数是( ),它的计数单位是( ),再加上( )个这样的计数单位就是76。",
+        "options":  [
+
+                    ],
+        "answer":  "75.93;0.01;7",
+        "solution":  "[analysis]7个十,5个一,9个十分之一和3个百分之一说明十位上是7，个位上是5，十分位上是9，百分位上是3，据此写出数是75.93;这个数是两位小数，所以计数单位是0.01;再加上若干个0.01使其变为76.\n【详解】由7个十,5个一,9个十分之一和3个百分之一组成的数是75.93;它的计数单位是0.01;76-75.93=0.07，即再加上7个0.01就是76。",
+        "knowledge_points":  [
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·四川广元·期末"
+    },
+    {
+        "id":  "482",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "把2.8扩大到它的100倍是( ),把23缩小到它的( )是0.023。",
+        "options":  [
+
+                    ],
+        "answer":  "280;1÷1000",
+        "solution":  "【分析】把2.8扩大到100倍相当于小数点向右移动两位;23变为0.023相当于小数点向左移动三位，即缩小到原数的1÷1000.\n【详解】把2.8扩大到它的100倍是280，把23缩小到它的1÷1000是0.023。",
+        "knowledge_points":  [
+                                 "小数点向左移动引起小数大小变化的规律",
+                                 "小数点向右移动引起小数大小变化的规律"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "483",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "某超市举办优惠活动，一种薯片买5包送1包。这种薯片每包3.5元，李叔叔想买12包，他实际需要支付( )元。",
+        "options":  [
+
+                    ],
+        "answer":  "35",
+        "solution":  "【分析】买5送1即每6包付5包钱。12包相当于2组6包，共付10包钱;总价=3.5×10.\n【详解】5+1=6(包);12÷6=2;5×2=10(包);3.5×10=35(元)。所以实际需要支付35元。",
+        "knowledge_points":  [
+                                 "小数点向右移动引起小数大小变化的规律",
+                                 "经济问题"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "484",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "2024年全球智能手机出货量为1289000000部，把它改写成用\"亿\"作单位的数是( )亿部，保留整数约为( )亿部。",
+        "options":  [
+
+                    ],
+        "answer":  "12.89;13",
+        "solution":  "【分析】改写成以亿作单位的数：小数点向左移动8位;保留整数用\"四舍五入\"法.\n【详解】1289000000=12.89亿部，保留整数约为13亿部。",
+        "knowledge_points":  [
+                                 "小数的近似数",
+                                 "小数的改写"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "485",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "在下面的括号里填上\"\u003e\"、\"\u003c\"或\"=\"。7.526( )7.522;1460m( )1.5km;30万( )299792",
+        "options":  [
+
+                    ],
+        "answer":  "\u003e;\u003c;\u003e",
+        "solution":  "【分析】小数先比整数部分再比小数位;长度单位先统一;30万换算为300000后比较.\n【详解】7.526\u003e7.522;1.5km=1500m,1460m\u003c1500m;30万=300000,300000\u003e299792。",
+        "knowledge_points":  [
+                                 "多位小数的大小比较",
+                                 "大数的比较"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "486",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "0.76m²=( )dm²;5吨80千克=( )吨",
+        "options":  [
+
+                    ],
+        "answer":  "76;5.08",
+        "solution":  "【分析】1m²=100dm²,小数点右移两位;1吨=1000千克,80千克=0.08吨.\n【详解】0.76m²=76dm²;5吨80千克=5.08吨。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "平方厘米，平方分米，平方米之间的进率与换算",
+                                 "吨，千克之间的换算与比较"
+                             ],
+        "source":  "24-25四年级下·山东济宁·期末"
+    },
+    {
+        "id":  "487",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "2025年3月14日，自然资源部发布《2024年中国自然资源公报》。数据显示，全国共有湿地面积23519800公顷，改写成用\"万\"作单位的数是( )万公顷;耕地面积128608800公顷，用\"亿\"作单位并保留两位小数约是( )亿公顷。",
+        "options":  [
+
+                    ],
+        "answer":  "2351.98;1.29",
+        "solution":  "【分析】改写成\"万\"作单位小数点左移4位;改写成\"亿\"作单位小数点左移8位;保留两位小数看第三位四舍五入.\n【详解】23519800=2351.98万;128608800=1.286088亿≈1.29亿。",
+        "knowledge_points":  [
+                                 "小数的近似数",
+                                 "小数的改写"
+                             ],
+        "source":  "24-25四年级下·山西长治·期末"
+    },
+    {
+        "id":  "488",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "木星的轨道半径是778330000千米。把778330000改写成用\"亿\"作单位的数是( )亿，保留整数约是( )亿。",
+        "options":  [
+
+                    ],
+        "answer":  "7.7833;8",
+        "solution":  "【分析】改写成\"亿\"作单位小数点左移8位;保留整数看十分位四舍五入.\n【详解】778330000=7.7833亿;7.7833亿≈8亿。",
+        "knowledge_points":  [
+                                 "小数的改写",
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25四年级下·四川南充·期末"
+    },
+    {
+        "id":  "489",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "2025年端午假期(5月31日至6月2日),晋祠博物馆，天龙山石窟博物馆等22处山西省太原市文物局直属文博场馆全部对外开放，旅游人次与门票收入再创新高。太原市文博场馆(含蒙山景区及市文物局直属场馆)接待游客257600万人次，门票收入3611100元。蒙山景区接待2.7万人次，门票收入65.22万元;市文物局直属文博场馆接待游客23.05万人次，门票收入295.89万元.\n(1)23.05读作( ),把这个小数改写成以千分之一为计数单位的数是( ).\n(2)295.89小数点左边的\u00279\u0027表示( ),小数点右边的\u00279\u0027表示( ).\n(3)3611100改写为用\"万\"作单位的数是( ),257600省略\"万\"后面的尾数约是( ).\n(4)一个两位小数，用四舍五入法取近似数是2.7，这个数最大是( ),最小是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "二十三点零五;23.050;9个十;9个百分之一;361.11万;26万;2.74;2.65",
+        "solution":  "【分析】(1)按小数读法读数;改写成千分之一为计数单位需写成三位小数。(2)根据数位确定各数字的意义。(3)改写成\"万\"作单位需除以10000;省略万位后尾数相当于四舍五入到万位。(4)取近似数2.7(保留一位小数),最大来自\"四舍\"为2.74，最小来自\"五入\"为2.65.\n【详解】(1)23.05读作二十三点零五，把这个小数改写成以千分之一为计数单位的数是23.050。(2)小数点左边的\u00279\u0027在十位上，表示9个十;小数点右边的\u00279\u0027在百分位上，表示9个百分之一。(3)3611100改写为用\"万\"作单位的数是361.11万,257600省略\"万\"后面的尾数约是26万。(4)最大是2.74，最小是2.65。",
+        "knowledge_points":  [
+                                 "小数的读法和写法",
+                                 "小数的数位和计数单位的认识",
+                                 "小数的近似数",
+                                 "亿以内数的改写"
+                             ],
+        "source":  "24-25四年级下·山西晋中·期末"
+    },
+    {
+        "id":  "490",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.85,
+        "question":  "2024年天峨县实现地区生产总值(GDP)(七十五亿八千五百万)元，括号内的数写作( )元，改写成用\"亿\"作单位的数是( )亿元。",
+        "options":  [
+
+                    ],
+        "answer":  "7585000000;75.85",
+        "solution":  "【分析】先按整数写法写出\u0027七十五亿八千五百万\u0027;再改写成\"亿\"作单位，在亿位右下角点小数点并去掉末尾0.\n【详解】七十五亿八千五百万写作7585000000;7585000000=75.85亿。",
+        "knowledge_points":  [
+                                 "亿以上数的改写",
+                                 "小数的改写",
+                                 "亿以上数的读，写法"
+                             ],
+        "source":  "24-25六年级下·广西河池·期末"
+    },
+    {
+        "id":  "491",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "一个三位小数，用\"四舍五入\"法保留整数是5，保留一位小数是5.0，保留两位小数是5.00。已知这个小数各个数位上的数字相加的和是30，那么这个三位小数是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "4.998",
+        "solution":  "【分析】能四舍五入后得到5.00的三位小数范围是4.995到5.004;又因各数位数字和为30，满足条件的是4.998.\n【详解】近似数是5.00的三位小数最大是5.004，最小是4.995，并且数字和为30,4+9+9+8=30，所以这个三位小数是4.998。",
+        "knowledge_points":  [
+                                 "小数的近似数"
+                             ],
+        "source":  "24-25四年级下·浙江杭州·期末"
+    },
+    {
+        "id":  "492",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "100张纸叠在一起厚0.9厘米，平均每张纸厚( )厘米，一百万张这样的纸叠起来会有( )米。",
+        "options":  [
+
+                    ],
+        "answer":  "0.009;90",
+        "solution":  "【分析】平均每张厚度=0.9÷100;一百万张厚度=平均厚度*1000000，再把厘米换算成米(÷100).\n【详解】0.9÷100=0.009(厘米);0.009×1000000÷100=90(米)。",
+        "knowledge_points":  [
+                                 "毫米，厘米，分米，米之间的进率与换算",
+                                 "小数点向右移动引起小数大小变化的规律",
+                                 "小数点向左移动引起小数大小变化的规律"
+                             ],
+        "source":  "24-25四年级下·浙江杭州·期末"
+    },
+    {
+        "id":  "493",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "红星小学是一所秉承优良传统，充满青春活力的小学，校园优美而宽敞，总建筑面积为(二万九千八百八十四点九四)平方米。括号内的数写作( ),省略万位后的尾数约是( )万。",
+        "options":  [
+
+                    ],
+        "answer":  "29884.94;3",
+        "solution":  "【分析】先按小数写法写出29884.94;省略万位后的尾数即四舍五入到万位，看千位数字.\n【详解】写作29884.94;千位是9，进位后约为3万。",
+        "knowledge_points":  [
+                                 "小数的读法和写法",
+                                 "小数的改写"
+                             ],
+        "source":  "2025·重庆綦江·小升初真题"
+    },
+    {
+        "id":  "494",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "博物馆内有文物( )件，改写成以\"万\"为单位的数是18.45万件，省略万位后面的尾数得到的近似数是( )万件。",
+        "options":  [
+
+                    ],
+        "answer":  "184500;18",
+        "solution":  "【分析】18.45万还原为184500;省略万位后面的尾数相当于四舍五入到万位.\n【详解】18.45万=184500;184500≈18万。",
+        "knowledge_points":  [
+                                 "亿以内数的近似数",
+                                 "小数的改写"
+                             ],
+        "source":  "24-25六年级下·河南许昌·期末"
+    },
+    {
+        "id":  "495",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "2025年五一小长假期间，某市共接待游客(466700)人次，改写成以\"万人次\"作单位的数是( )万人次，实现旅游收入一亿七千四百万元，省略亿后面的尾数是( )亿元。",
+        "options":  [
+
+                    ],
+        "answer":  "46.67;2",
+        "solution":  "【分析】改写成\"万\"作单位小数点左移4位;一亿七千四百万元写作174000000元，省略亿后尾数即四舍五入到亿.\n【详解】466700人次=46.67万人次;174000000元省略亿后面的尾数是2亿元。",
+        "knowledge_points":  [
+                                 "小数的改写",
+                                 "亿以内数的改写",
+                                 "亿以上数的近似数"
+                             ],
+        "source":  "2025·甘肃武威·小升初真题"
+    },
+    {
+        "id":  "496",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "在()里填上适当的数.\n(1)把6缩小到它的1÷100是( ).\n(2)0.021里面有( )个千分之一.\n(3)3个百分之一写成小数是( ).\n(4)把0.004扩大到它原来的( )倍是4.\n(5)把58缩小到它原来的( )是0.058.\n(6)把2.6的小数点向左移动一位是( )。",
+        "options":  [
+
+                    ],
+        "answer":  "0.06;21;0.03;1000;1÷1000;0.26",
+        "solution":  "【分析】(1)缩小到1÷100即除以100，小数点左移两位。(2)0.021=21个0.001。(3)3个0.01为0.03。(4)扩大1000倍小数点右移三位。(5)58变0.058为除以1000，即缩小到1÷1000。(6)小数点左移一位即除以10.\n【详解】(1)6÷100=0.06。(2)0.021里面有21个千分之一。(3)3个百分之一写成小数是0.03。(4)0.004×1000=4。(5)58÷1000=0.058，所以缩小到原来的1÷1000。(6)2.6÷10=0.26。",
+        "knowledge_points":  [
+                                 "小数点向左移动引起小数大小变化的规律",
+                                 "小数点向右移动引起小数大小变化的规律",
+                                 "小数的数位和计数单位的认识"
+                             ],
+        "source":  "24-25四年级下·海南省直辖县级单位·期末"
+    },
+    {
+        "id":  "497",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "世界上最小的鸟是蜂鸟，它的蛋只有绿豆那么大，生物学家测得一枚蜂鸟重零点(五五克)。括号内的数写作( ),保留一位小数约( )。",
+        "options":  [
+
+                    ],
+        "answer":  "0.55;0.6",
+        "solution":  "【分析】\"零点五五\"写作0.55;保留一位小数看百分位,0.55的百分位是5，进一得0.6.\n【详解】写作0.55;0.55≈0.6。",
+        "knowledge_points":  [
+                                 "小数的近似数",
+                                 "小数的读法和写法"
+                             ],
+        "source":  "24-25四年级下·海南省直辖县级单位·期末"
+    },
+    {
+        "id":  "498",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "单位换算。5kg60g=( )kg;6.007m²=( )cm²;5元4角+6分=( )元;40008mL=( )L+( )mL",
+        "options":  [
+
+                    ],
+        "answer":  "5.06;60070;5.46;40;8",
+        "solution":  "【分析】1千克=1000克,1平方米=10000平方厘米,1元=10角=100分,1升=1000毫升;按进率换算并相加.\n【详解】60÷1000=0.06,5kg60g=5.06kg;6.007×10000=60070,6.007m²=60070cm²;4÷10=0.4,6÷100=0.06,5元4角+6分=5.46元;40008÷1000=40.008，所以40008mL=40L+8mL。",
+        "knowledge_points":  [
+                                 "运用小数点移动解决小数的单位换算问题",
+                                 "平方厘米，平方分米，平方米之间的进率与换算",
+                                 "吨，千克之间的换算与比较",
+                                 "元，角，分的认识及换算"
+                             ],
+        "source":  "24-25四年级下·上海松江·期末"
+    },
+    {
+        "id":  "499",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "某商场举办促销活动，一种牛奶买5瓶送1瓶，这种牛奶每瓶3.58元，李阿姨想买12瓶，需要( )元。",
+        "options":  [
+
+                    ],
+        "answer":  "35.8",
+        "solution":  "【分析】买5送1即每6瓶付5瓶钱;12瓶是2组6瓶，共付10瓶钱;总价=3.58×10.\n【详解】12÷(5+1)=2(组);2×5=10(瓶);3.58×10=35.8(元)。",
+        "knowledge_points":  [
+                                 "小数点向右移动引起小数大小变化的规律",
+                                 "经济问题"
+                             ],
+        "source":  "24-25四年级下·陕西商洛·期末"
+    },
+    {
+        "id":  "500",
+        "unit":  "第四单元小数意义",
+        "question_type":  "填空题",
+        "difficulty":  0.65,
+        "question":  "某商场举办促销活动，一种牛奶买5瓶送1瓶。这种牛奶每瓶3.58元，李阿姨想买12瓶，需要( )元钱。",
+        "options":  [
+
+                    ],
+        "answer":  "35.8",
+        "solution":  "【分析】5瓶送1瓶表示买5瓶的钱得到6瓶。12瓶=2组6瓶，需要付钱买10瓶;再用单价乘数量.\n【详解】12÷(5+1)=2(组);2×5=10(瓶);3.58×10=35.8(元)。所以需要35.8元钱。",
+        "knowledge_points":  [
+                                 "小数点向右移动引起小数大小变化的规律"
+                             ],
+        "source":  "24-25四年级下·陕西商洛·期末"
+    }
+]